Solution Manual for Gravitation (MTW) 0716703440, 9780716703440, 9780691177793, 9780716703341, 9789570911336, 9784621083277

Citation preview

Solution Manual for Gravitation (MTW) Yu-Yang Songsheng [email protected] Institute of High Energy Physics, Chinese Academy of Sciences

January 10, 2019

2

Contents

1 GEOMETRODYNAMICS IN BRIEF

5

1.1 CURVATURE OF A CYLINDER . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5

1.2 SPRING TIDE VS. NEAP TIDE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

5

1.3 KEPLER ENCAPSULATED . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6

2 FOUNDATIONS OF SPECIAL RELATIVITY

7

2.1 EXERCISE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7

2.2 LOWERING INDEX TO GET THE 1-FORM CORRESPONDING TO A VECTOR . . . . . . . .

7

2.3 RAISING INDEX TO RECOVER THE VECTOR . . . . . . . . . . . . . . . . . . . . . . . . .

8

2.4 VARIED ROUTES TO THE SCALAR PRODUCT . . . . . . . . . . . . . . . . . . . . . . . . .

8

2.5 ENERGY AND VELOCITY FROM 4-MOMENTUM . . . . . . . . . . . . . . . . . . . . . . . .

8

2.6 TEMPERATURE GRADIENT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

9

2.7 BOOST IN AN ARBITRARY DIRECTION

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

3 THE ELECTROMAGNETIC FIELD

10

13

3.1 EXERCISE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

13

3.2 TRANSFORMATION LAW FOR COMPONENTS OF A TENSOR

. . . . . . . . . . . . . . . .

13

3.3 RAISING AND LOWERING INDICES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14

3.4 TENSOR PRODUCT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

14

3.5 BASIS TENSORS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

15

3.6 Faraday MACHINERY AT WORK . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

15

3.7 MAXWELL’S EQUATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

16

3.8 CONTRACTION IS FRAME-INDEPENDENT . . . . . . . . . . . . . . . . . . . . . . . . . . .

16

3

3.9 DIFFERENTIATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

17

3.10 MORE DIFFERENTIATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

17

3.11 SYMMETRIES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

18

3.12 SYMMETRIZATION AND ANTISYMMETRIZATION . . . . . . . . . . . . . . . . . . . . . .

18

3.13 LEVI-CIVITA TENSOR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

20

3.14 DUALS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

22

3.15 GEOMETRIC VERSIONS OF MAXWELL EQUATIONS . . . . . . . . . . . . . . . . . . . . .

23

3.16 CHARGE CONSERVATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

24

3.17 VECTOR POTENTIAL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

24

3.18 DIVERGENCE OF ELECTROMAGNETIC STRESS-ENERGY TENSOR . . . . . . . . . . . . . .

25

4 ELECTROMAGNETISM AND DIFFERENTIAL FORMS

27

4.1 GENERIC LOCAL ELECTROMAGNETIC FIELD EXPRESSED IN SIMPLEST FORM . . . . . .

27

4.2 FREEDOM OF CHOICE OF 1-FORMS IN CANONICAL REPRESENTATION OF GENERIC LOCAL FIELD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

28

4.3 A CLOSED OR CURL-FREE 1-FORM IS A GRADIENT . . . . . . . . . . . . . . . . . . . . .

29

4.4 CANONICAL EXPRESSION FOR A ROTATION-FREE 1-FORM . . . . . . . . . . . . . . . . .

30

4.5 FORMS ENDOWED WITH POLAR SINGULARITIES . . . . . . . . . . . . . . . . . . . . . .

30

4.6 THE FIELD OF THE OSCILLATING DIPOLE . . . . . . . . . . . . . . . . . . . . . . . . . . .

31

4.7 THE 2-FORM MACHINERY TRANSLATED INTO TENSOR MACHINERY . . . . . . . . . . .

32

4.8 PANCAKING THE COULOMB FIELD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

32

4.9 COMPUTATION OF SURFACE INTEGRALS . . . . . . . . . . . . . . . . . . . . . . . . . . .

32

4.10 WHITAKER’S CALUMOID, OR, THE LIFE OF A LOOP . . . . . . . . . . . . . . . . . . . . .

33

4.11 DIFFERENTIAL FORMS AND HAMilTONIAN MECHANICS . . . . . . . . . . . . . . . . . .

35

4.12 SYMMETRY OPERATIONS AS TENSORS . . . . . . . . . . . . . . . . . . . . . . . . . . . .

36

5 STRESS-ENERGY TENSOR AND CONSERVATION LAWS

39

5.1 EXERCISE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

39

5.2 CHARGE CONSERVATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

39

5.3 PARTICLE PRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

40

4

5.4 INERTIAL MASS PER UNIT VOLUME . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

40

5.5 DETERMINANTS AND JACOBIANS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

42

5.6 CENTROIDS AND SIZES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

44

6 ACCELERATED OBSERVERS 6.1 A TRIP TO THE GALACTIC NUCLEUS

47 . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

47

6.2 ROCKET PAYLOAD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

48

6.3 TWIN PARADOX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

48

6.4 RADAR SPEED INDICATOR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

49

6.5 RADAR DISTANCE INDICATOR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

51

6.6 CLOCK RATES VERSUS COORDINATE TIME IN ACCELERATED COORDINATES . . . . . .

51

6.7 ACCELERATION OF LATTICE POINTS IN ACCELERATED COORDINATES . . . . . . . . . .

52

6.8 OBSERVER WITH ROTATING TETRAD . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

52

6.9 THOMAS PRECESSION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

55

7 INCOMPATIBILITY OF GRAVITY AND SPECIAL RELATIVITY

57

7.1 SCALAR GRAVITATIONAL FIELD, Φ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

57

7.2 VECTOR GRAVITATIONAL FIELD, Φµ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

60

7.3 SYMMETRIC TENSOR GRAVITATIONAL FIELD, hµν = hνµ . . . . . . . . . . . . . . . . . .

63

8 GRAVITATIONAL COLLAPSE

65

8.1 PRACTICE WITH TENSOR ALGEBRA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

65

8.2 COMMUTATORS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

65

8.3 COMPONENTS OF LEVI-CIVITA TENSOR IN NONORTHONORMAL FRAME . . . . . . . . .

66

8.4 PRACTICE IN WRITING COMPONENTS OF GRADIENT . . . . . . . . . . . . . . . . . . . .

67

8.5 A SHEET OF PAPER IN POLAR COORDINATES . . . . . . . . . . . . . . . . . . . . . . . . .

67

8.6 SPHERICAL COORDINATES IN FLAT SPACETIME . . . . . . . . . . . . . . . . . . . . . . .

68

8.7 SYMMETRIES OF CONNECTION COEFFICIENTS . . . . . . . . . . . . . . . . . . . . . . .

69

8.8 NEW DEFINITION OF ∇T COMPARED WITH OLD DEFINITION . . . . . . . . . . . . . . .

69

8.9 CHAIN RULE FOR ∇u T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

69

8.10 COVARIANT DERIVATIVE COMMUTES WITH CONTRACTION . . . . . . . . . . . . . . . .

70

5

8.11 ALGEBRAIC PROPERTIES OF ∇ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

70

8.12 CONNECTION COEFFICIENTS FOR 1-FORM BASIS . . . . . . . . . . . . . . . . . . . . . .

71

8.13 “ΓT ” CORRECTION TERMS FOR T βα;γ . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

71

8.14 METRIC IS COVARIANTLY CONSTANT . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

71

8.15 CONNECTION COEFFICIENTS IN TERMS OF METRIC . . . . . . . . . . . . . . . . . . . .

71

8.16 SOME USEFUL FORMULAS IN COORDINATE FRAMES . . . . . . . . . . . . . . . . . . . .

71

9 DIFFERENTIAL TOPOLOGY

73

9.1 COMPONENT MANIPULATIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

73

9.2 COMPONENTS OF GRADIENT, AND DUALITY OF COORDINATE BASES . . . . . . . . . . .

73

9.3 PRACTICE MANIPULATING TANGENT VECTORS . . . . . . . . . . . . . . . . . . . . . . .

74

9.4 MORE PRACTICE WITH TANGENT VECTORS . . . . . . . . . . . . . . . . . . . . . . . . .

74

9.5 PICTURE OF BASIS 1-FORMS INDUCED BY COORDINATES . . . . . . . . . . . . . . . . .

75

9.6 PRACTICE WITH DUAL BASES

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

75

9.7 PRACTICE WITH COMMUTATORS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

75

9.8 ANGULAR MOMENTUM OPERATORS . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

76

9.9 COMMUTATORS AND COORDINATE-INDUCED BASES . . . . . . . . . . . . . . . . . . . .

76

9.10 COMPONENTS OF COMMUTATOR IN NON-COORDINATE BASIS . . . . . . . . . . . . . .

78

9.11 LIE DERIVATIVE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

78

9.12 A CHIP OFF THE OLD BLOCK . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

79

9.13 ROTATION GROUP: GENERATORS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

79

9.14 ROTATION GROUP: STRUCTURE CONSTANTS . . . . . . . . . . . . . . . . . . . . . . . . .

82

10 AFFINE GEOMETRY: GEODESICS, PARALLEL TRANSPORT, AND COVARIANT DERIVATIVE

85

10.1 ADDITIVITY OF COVARIANT DIFFERENTIATION . . . . . . . . . . . . . . . . . . . . . . .

85

10.2 CHAIN RULE FOR COVARIANT DIFFERENTIATION . . . . . . . . . . . . . . . . . . . . . .

85

10.3 ANOTHER CHAIN RULE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

86

10.4 STILL ANOTHER CHAIN RULE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

86

10.5 ONE MORE CHAIN RULE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

87

10.6 GEODESIC EQUATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

87

6

10.7 COMPUTATION OF CONNECTION COEFFICIENTS . . . . . . . . . . . . . . . . . . . . . .

87

10.8 CONNECTION FOR 1-FORM BASIS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

88

10.9 SYMMETRY OF CONNECTION COEFFICIENTS

. . . . . . . . . . . . . . . . . . . . . . . .

88

10.10COMPONENTS OF GRADIENT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

88

10.11DIVERGENCE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

89

10.12VERIFICATION OF CHAIN RULE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

89

10.13TRANSFORMATION LAW FOR CONNECTION COEFFICIENTS . . . . . . . . . . . . . . . .

90

10.14POLAR COORDINATES IN FLAT 2-DIMENSIONAL SPACE . . . . . . . . . . . . . . . . . . .

90

10.15COMPONENTS OF PARALLEL-TRANSPORT LAW . . . . . . . . . . . . . . . . . . . . . . .

91

10.16GEODESICS IN POLAR COORDINATES . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

91

10.17ROTATION GROUP: GEODESICS AND CONNECTION COEFFICIENTS . . . . . . . . . . . .

92

11 GEODESIC DEVIATION AND SPACETIME CURVATURE

95

11.1 [∇A , ∇B ]C DEPENDS ON DERIVATIVES OF C . . . . . . . . . . . . . . . . . . . . . . . . .

95

11.2 PROOF THAT RIEMANN IS A TENSOR . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

95

11.3 COMPONENTS OF RIEMANN IN COORDINATE BASIS . . . . . . . . . . . . . . . . . . . .

97

11.4 COMPONENTS OF RIEMANN IN NONCOORDINATE BASIS . . . . . . . . . . . . . . . . . .

97

11.5 COPLANARITY OF CLOSED CURVES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

97

11.6 SYMMETRIES OF Riemann . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

97

11.7 GEODESIC DEVIATION MEASURES ALL CURVATURE COMPONENTS . . . . . . . . . . . .

98

11.8 GEODESIC DEVIATION IN GORY DETAIL . . . . . . . . . . . . . . . . . . . . . . . . . . . .

99

11.9 RIEMANN NORMAL COORDINATES IN GENERAL . . . . . . . . . . . . . . . . . . . . . . . 100 11.10BIANCHI IDENTITIES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 11.11CURVATURE OPERATOR ACTS ON 1-FORMS . . . . . . . . . . . . . . . . . . . . . . . . . . 102 11.12ROTATION GROUP: RIEMANN CURVATURE . . . . . . . . . . . . . . . . . . . . . . . . . . 102

12 NEWTONIAN GRAVITY IN THE LANGUAGE OF CURVED SPACETIME

103

12.1 RIEMANN CURVATURE OF NEWTONIAN SPACETIME . . . . . . . . . . . . . . . . . . . . 103 12.2 NEWTONIAN FIELD EQUATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 12.3 GEODESIC DEVIATION DERIVED . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104

7

12.4 CONNECTION COEFFICIENTS FOR ROTATING, ACCELERATING COORDINATES . . . . . . 104 12.5 EINSTEIN’S ELEVATOR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 12.6 GEODESIC DEVIATION ABOVE THE EARTH . . . . . . . . . . . . . . . . . . . . . . . . . . 105 12.7 FROM NEWTON TO CARTAN . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107 12.8 FROM CARTAN TO NEWTON . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108 12.9 SPATIAL METRIC ALLOWED BY OTHER AXIOMS . . . . . . . . . . . . . . . . . . . . . . . 109 12.10SPACETIME METRIC FORBIDDEN BY OTHER AXIOMS . . . . . . . . . . . . . . . . . . . . 110

13 RIEMANNIAN GEOMETRY: METRIC AS FOUNDATION OF ALL

111

13.1 TEST WHETHER SPACETIME IS LOCAL LORENTZ . . . . . . . . . . . . . . . . . . . . . . 111 13.2 PRACTICE WITH METRIC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 13.3 MATHEMATICAL REPRESENTATION OF LOCAL LORENTZ FRAME . . . . . . . . . . . . . . 112 13.4 CONSEQUENCES OF COMPATIBILITY BETWEEN g AND ∇ . . . . . . . . . . . . . . . . . . 112 13.5 ONCE TIMELIKE, ALWAYS TIMELIKE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112 13.6 SPACELIKE GEODESICS HAVE EXTREMAL LENGTH . . . . . . . . . . . . . . . . . . . . . . 113 13.7 METRIC TENSOR MEASURED BY LIGHT SIGNALS AND FREE PARTICLES . . . . . . . . . 113 13.8 RIEMANN ANTISYMMETRIC IN FIRST TWO INDICES . . . . . . . . . . . . . . . . . . . . 114 13.9 NUMBER OF INDEPENDENT COMPONENTS OF RIEMANN . . . . . . . . . . . . . . . . . . 115 13.10RIEMANN SYMMETRIC IN EXCHANGE OF PAIRS; COMPLETELY ANTISYMMETRIC PART VANISHES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115 13.11DOUBLE DUAL OF RIEMANN: EINSTEIN . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116 13.12RICCI AND EINSTEIN RELATED . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 13.13THE WEYL CONFORMAL TENSOR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117 13.14INERTIAL AND CORIOLIS FORCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 13.15ROTATION GROUP: METRIC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

14 CALCULATION OF CURVATURE

121

14.1 CURVATURE OF A TWO-DIMENSIONAL HYPERBOLOID . . . . . . . . . . . . . . . . . . . 121 14.2 RIEMANNIAN CURVATURE EXPRESSIBLE IN TERMS OF RICCI CURVATURE IN TWO AND THREE DIMENSIONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121 14.3 CURVATURE OF 3-SPHERE IN ORTHONORMAL FRAME . . . . . . . . . . . . . . . . . . . 122

8

14.4 EINSTEIN EQUATIONS FOR THE CLOSED FRIEDMANN UNIVERSE CALCULATED BY USING THE GEODESIC LAGRANGIAN METHOD . . . . . . . . . . . . . . . . . . . . . . . . . 123 14.5 EXTERIOR DERIVATIVE OF A PRODUCT OF FORMS . . . . . . . . . . . . . . . . . . . . . . 124 14.6 RELATIONSHIP BETWEEN EXTERIOR DERIVATIVE AND COMMUTATOR . . . . . . . . . . 124 14.7 CHRISTOFFEL FORMULA DERIVED FROM CONNECTION FORMS . . . . . . . . . . . . . . 125 14.8 RIEMANN-CHRISTOFFEL CURVATURE FORMULA RELATED TO CURVATURE FORMS . . . 126 14.9 MATRIX NOTATION FOR REVIEW OF CARTAN STRUCTURE EQUATIONS . . . . . . . . . . 126 14.10TRANSFORMATION RULES FOR CONNECTION FORMS IN COMPACT NOTATION . . . . . 128 14.11SPACE IS FLAT IF THE CURVATURE VANISHES . . . . . . . . . . . . . . . . . . . . . . . . 128 14.12SYSTEMATIC COMPUTATION OF CONNECTION FORMS IN ORTHONORMAL FRAMES . . 129 14.13SCHWARZSCHILD CURVATURE FORMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129 14.14MATRIX DISPLAY OF THE RIEMANN-TENSOR COMPONENTS . . . . . . . . . . . . . . . . 130 14.15RIEMANN MATRIX WITH VANISHING EINSTEIN TENSOR . . . . . . . . . . . . . . . . . . 131 14.16COMPUTATION OF CURVATURE FOR A PULSATING OR COLLAPSING STAR . . . . . . . . 131 14.17BIANCHI IDENTITY IN dR = 0 FORM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132 14.18LOCAL CONSERVATION OF ENERGY AND MOMENTUM: d ∗T = 0 MEANS ∇ · T = 0 . . . 133 15 BIANCHI IDENTITIES AND THE BOUNDARY OF A BOUNDARY

135

15.1 THE BOUNDARY OF THE BOUNDARY OF A 4-SIMPLEX . . . . . . . . . . . . . . . . . . . 135 15.2 THE BEL-ROBINSON TENSOR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 136

16 EQUIVALENCE PRINCIPLE AND MEASUREMENT OF GRAVITATIONAL FIELD

139

16.1 HYDRODYNAMICS IN A WEAK GRAVITATIONAL FIELD . . . . . . . . . . . . . . . . . . . . 139 16.2 WORLD LINES OF PHOTONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 140 16.3 NONCOM MUTATION OF COVARIANT DERIVATIVES . . . . . . . . . . . . . . . . . . . . . 141 16.4 PRECESSION OF THE EQUINOXES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 142 16.5 GRAVITY GRADIOMETER . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

17 HOW MASS-ENERGY GENERATES CURVATURE

145

17.1 UNIQUENESS OF THE EINSTEIN TENSOR . . . . . . . . . . . . . . . . . . . . . . . . . . . 145 17.2 NO TENSOR CONSTRUCT ABLE FROM FIRST DERIVATIVES OF METRIC . . . . . . . . . . 146

9

17.3 RIEMANN AS THE ONLY TENSOR CONSTRUCTABLE FROM, AND LINEAR IN SECOND DERIVATIVES OF METRIC . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 17.4 UNIQUENESS OF THE EINSTEIN TENSOR . . . . . . . . . . . . . . . . . . . . . . . . . . . 146 17.5 MAGNITUDE OF COSMOLOGICAL CONSTANT . . . . . . . . . . . . . . . . . . . . . . . . . 147 17.6 RAMIFICATIONS OF CORRESPONDENCE FOR GRAVITY IN A PASSIVE ROLE . . . . . . . . 147 17.7 CORRESPONDENCE IN THE LANGUAGE OF CURVED SPACETIME . . . . . . . . . . . . . . 148 17.8 EINSTEIN-FOKKER REDUCES TO NORDSTROM . . . . . . . . . . . . . . . . . . . . . . . . 148

18 WEAK GRAVITATIONAL FIELDS

149

18.1 GAUGE INVARIANCE OF THE RIEMANN CURVATURE . . . . . . . . . . . . . . . . . . . . 149 18.2 JUSTIFICATION OF LORENTZ GAUGE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149 18.3 EXTERNAL FIELD OF A STATIC, SPHERICAL BODY . . . . . . . . . . . . . . . . . . . . . . 150 18.4 SPACETIME CURVATURE FOR A PLANE GRAVITATIONAL WAVE . . . . . . . . . . . . . . . 152 18.5 A PRIMITIVE GRAVITATIONAL-WAVE DETECTOR . . . . . . . . . . . . . . . . . . . . . . . 153 18.6 BENDING OF LIGHT BY THE SUN . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 153 18.7 GRAVITATIONAL REDSHIFT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 155

19 MASS AND ANGULAR MOMENTUM OF A GRAVITATING SYSTEM

157

19.1 DERIVATION OF METRIC FAR OUTSIDE A WEAKLY GRAVITATING BODY . . . . . . . . . . 157 19.2 GYROSCOPE PRECESSION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 19.3 GRAVITATIONAL FIELD FAR FROM A STATIONARY, FUllY RELATIVISTIC SOURCE . . . . . 158

20 CONSERVATION LAWS FOR 4-MOMENTUM AND ANGULAR MOMENTUM

159

20.1 FLUX INTEGRAL FOR TOTAL MASS-ENERGY IN LINEARIZED THEORY . . . . . . . . . . . 159 20.2 FLUX INTEGRAL FOR ANGULAR MOMENTUM IN LINEARIZED THEORY . . . . . . . . . . 160 20.3 FLUX INTEGRALS FOR AN ARBITRARY STATIONARY SOURCE . . . . . . . . . . . . . . . . 160 µανβ 20.4 FORM OF HL−L FAR FROM SOURCE

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162

20.5 TOTAL MASS-ENERGY IN NEWTONIAN LIMIT . . . . . . . . . . . . . . . . . . . . . . . . 162 20.6 SIMPLE FEATURES OF THE ELECTROMAGNETIC FIELD AND ITS STRESS-ENERGY TENSOR163 20.7 THE STRESS-ENERGY TENSOR DETERMINES THE ELECTROMAGNETIC FIELD EXCEPT FOR ITS COMPLEXION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167

10

20.8 THE MAXWELL EQUATIONS CANNOT BE DERIVED FROM THE LAW OF CONSERVATION OF STRESS-ENERGY WHEN (E · B) = 0 OVER AN EXTENDED REGION . . . . . . . . . . . 169 20.9 EQUATION OF MOTION OF A SCALAR FIELD AS CONSEQUENCE OF THE EINSTEIN FIELD EQUATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

21 VARIATIONAL PRINCIPLE AND INITIAL-VALUE DATA

173

21.1 VARIATION OF THE DETERMINANT OF THE METRIC TENSOR . . . . . . . . . . . . . . . 173 21.2 STRESS-ENERGY TENSOR FOR A SCALAR FIELD . . . . . . . . . . . . . . . . . . . . . . . 174 21.3 FARADAY-MAXWELL STRESS-ENERGY TENSOR . . . . . . . . . . . . . . . . . . . . . . . . 174 21.4 SCALAR CURVATURE INVARIANT IN TERMS OF AREA DEFICIT . . . . . . . . . . . . . . . 175 21.5 EXTRINSIC CURVATURE TENSOR FOR SLICE OF FRIEDMANN GEOMETRY

. . . . . . . . 177

21.6 EVALUATION OF R(ei , ej )n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 21.7 EVALUATION OF THE COMMUTATOR [ej , n] . . . . . . . . . . . . . . . . . . . . . . . . . . 178 21.8 LIE DERIVATIVE OF A TENSOR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 21.9 EXPRESSION FOR DYNAMIC COMPONENTS OF THE CURVATURE TENSOR . . . . . . . . 180 (4)

i

21.10EXPRESSION OF R nin IN TERMS OF EXTRINSIC CURVATURE, PLUS A COVARIANT DIVERGENCE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 21.11FIRST EXPLOITATION OF THE ADM VARIATIONAL PRINCIPLE FOR THE ELECTROMAGNETIC FIELD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 21.12SECOND EXPLOITATION OF THE ADM VARIATIONAL PRINCIPLE FOR THE ELECTROMAGNETIC FIELD . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 21.13FARADAY-MAXWELL SOURCE TERM IN THE DYNAMIC EQUATIONS OF GENERAL RELATIVITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 21.14THE CHOICE OF Φ DOESN’T MATTER . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 21.15THE CHOICE OF SLICING OF SPACETIME DOESN’T MATTER . . . . . . . . . . . . . . . . 186 21.16POOR MAN’S WAY TO DO COSMOLOGY . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 21.17THIN-SANDWICH VARIATIONAL PRINCIPLE FOR THE SCALAR POTENTIAL IN ELECTRODYNAMICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 21.18THIN-SANDWICH VARIATIONAL PRINCIPLE FOR THE LAPSE AND SHIFT FUNCTIONS IN GEOMETRODYNAMICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 21.19CONDENSED THIN-SANDWICH VARIATIONAL PRINCIPLE . . . . . . . . . . . . . . . . . . 191 21.20POOR MAN’S WAY TO SCHWARZSCHILD GEOMETRY . . . . . . . . . . . . . . . . . . . . 192 21.21WHY THE WEYL CONFORMAL CURVATURE TENSOR VANISHES . . . . . . . . . . . . . . 193

11

21.22YORK’S CURVATURE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 21.23PULLING THE POYNTING FLUX VECTOR “OUT OF THE AIR” . . . . . . . . . . . . . . . . 197 21.24THE EXTREMAL ACTION ASSOCIATED WITH THE HILBERT ACTION PRINCIPLE DEPENDS ON CONFORMAL 3-GEOMETRY AND EXTRINSIC TIME . . . . . . . . . . . . . . . . . . . 198 21.25EQUATION OF MOTION FOR A SURFACE LAYER . . . . . . . . . . . . . . . . . . . . . . . 199 21.26THIN SHELLS OF DUST . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 21.27SPHERICAL SHELL OF DUST . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201

22 THERMODYNAMICS, HYDRODYNAMICS, ELECTRODYNAMICS, GEOMETRIC OPTICS, AND KI203 NETIC THEORY 22.1 DIVERGENCE OF FLOW LINES PRODUCES VOLUME CHANGES . . . . . . . . . . . . . . . 203 22.2 EQUATION OF CONTINUITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 203 22.3 CHEMICAL POTENTIAL FOR IDEAL FERMI GAS . . . . . . . . . . . . . . . . . . . . . . . . 204 22.4 PROJECTION TENSORS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 22.5 PRESSURE GRADIENT IN STATIONARY GRAVITATIONAL FIELD . . . . . . . . . . . . . . . 204 22.6 EXPANSION, ROTATION, AND SHEAR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 205 22.7 HYDRODYNAMICS WITH VISCOSITY AND HEAT FLOW . . . . . . . . . . . . . . . . . . . 206 22.8 THE VECTOR POTENTIAL FOR ELECTRODYNAMICS . . . . . . . . . . . . . . . . . . . . . 209 22.9 CHARGE CONSERVATION IN THE PRESENCE OF GRAVITY . . . . . . . . . . . . . . . . . 210 22.10INTERACTING ELECTROMAGNETIC FIELD AND CHARGED MATTER . . . . . . . . . . . . 210 22.11ELECTROMAGNETIC FIELD AND STRESS ENERGY . . . . . . . . . . . . . . . . . . . . . . 211 22.12POLARIZATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 22.13THE AREA OF A BUNDLE OF RAYS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 212 22.14FOCUSING THEOREM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 214 22.15INVERSE SQUARE LAW FOR FLUX . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 22.16BRIGHTNESS OF THE SUN . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215 22.17BLACK BODY RADIATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216 22.18STRESS-ENERGY TENSOR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 22.19KINETIC THEORY FOR NONIDENTICAL PARTICLES . . . . . . . . . . . . . . . . . . . . . . 217

23 SPHERICAL STARS

219

12

23.1 ISOTROPIC COORDINATES AND NEWTONIAN LIMIT . . . . . . . . . . . . . . . . . . . . . 219 23.2 PROPER REFERENCE FRAMES OF FLUID ELEMENTS . . . . . . . . . . . . . . . . . . . . . 220 23.3 LAW OF LOCAL ENERGY-MOMENTUM CONSERVATION . . . . . . . . . . . . . . . . . . . 220 23.4 EINSTEIN CURVATURE TENSOR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221 23.5 TOTAL NUMBER OF BARYONS IN A STAR . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 23.6 BUOYANT FORCE IN A STAR . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 222 23.7 GRAVITATIONAL ENERGY OF A NEWTONIAN STAR . . . . . . . . . . . . . . . . . . . . . . 223 23.8 NEWTONIAN STARS OF UNIFORM DENSITY . . . . . . . . . . . . . . . . . . . . . . . . . 224 23.9 GOOD BEHAVIOR OF r EXERCISES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 225 23.10CENTER OF STAR OCCUPIED BY IDEAL FERMI GAS AT EXTREME RELATIVISTIC LIMIT . 225

25 PARTICLE MOTION IN SCHWARZSCHILD GEOMETRY

227

25.1 CONSTANT OF MOTION OBTAINED FROM HAMIlTON’S PRINCIPLE . . . . . . . . . . . . 227 25.2 SUPER-HAMilTONIAN FORMALISM FOR GEODESIC MOTION . . . . . . . . . . . . . . . . 228 25.3 KILLING VECTORS IN FLAT SPACETIME . . . . . . . . . . . . . . . . . . . . . . . . . . . . 228 25.4 POISSON BRACKET AS KEY TO CONSTANTS OF MOTION . . . . . . . . . . . . . . . . . . 229 25.5 COMMUTATOR OF Killing VECTORS IS A Killing VECTOR . . . . . . . . . . . . . . . . . . 229 25.6 EIGENVALUE PROBLEM FOR KIllING VECTORS . . . . . . . . . . . . . . . . . . . . . . . . 230 25.7 RADIAL VELOCITY OF A TEST PARTICLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . 231 25.8 ROTATIONAL KILLING VECTORS FOR SCHWARZSCHILD GEOMETRY . . . . . . . . . . . 231 25.9 CONSERVATION OF TOTAL ANGULAR MOMENTUM OF A TEST PARTICLE . . . . . . . . . 233 25.10SELECTING EQUATION BY SELECTING WHAT IS VARIED . . . . . . . . . . . . . . . . . . 233 25.11MOTION DERIVED FROM SUPER-HAMILTONIAN . . . . . . . . . . . . . . . . . . . . . . . 234 25.12REDSHIFT BY TIMED PULSES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234 25.13QUALITATIVE FORMS OF PARTICLE ORBITS . . . . . . . . . . . . . . . . . . . . . . . . . . 235 25.14IMPACT PARAMETER

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235

25.15TIME TO FALL TO r = 2M . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236 25.16PERIASTRON SHIFT FOR NEARLY CIRCULAR ORBITS . . . . . . . . . . . . . . . . . . . . 236 25.17ANGULAR MOTION DURING INFALL . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 237

13

25.18MAXIMUM AND MINIMUM OF EFFECTIVE POTENTIAL . . . . . . . . . . . . . . . . . . . 238 25.19KEPLER LAW VALID FOR CIRCULAR ORBITS . . . . . . . . . . . . . . . . . . . . . . . . . 239 25.20HAMILTON-JACOBI FUNCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 25.21DEFLECTION BY GRAVITY CONTRASTED WITH DEFLECTION BY ELECTRIC FORCE . . . 240 25.22CAPTURE BY A BLACK HOLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241 25.23QUALITATIVE FORMS OF PHOTON ORBITS . . . . . . . . . . . . . . . . . . . . . . . . . . 242 25.24LIGHT DEFLECTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 25.25CAPTURE OF LIGHT BY A BLACK HOLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243 25.26RETURN OF LIGHT FROM A BLACK HOLE . . . . . . . . . . . . . . . . . . . . . . . . . . . 244 25.27ISOTROPIC STAR CLUSTER . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 25.28SELF-SIMILAR CLUSTER . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 245 25.29CLUSTER WITH CIRCULAR ORBITS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 246

26 STELLAR PULSATIONS

247

26.1 DRAGGING OF INERTIAL FRAMES BY A SLOWLY ROTATING STAR . . . . . . . . . . . . . 247

27 IDEALIZED COSMOLOGIES

251

27.1 ISOTROPY IMPLIES HOMOGENEITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251 27.2 SYNCHRONOUS COORDINATES IN GENERAL . . . . . . . . . . . . . . . . . . . . . . . . . 251 27.3 ARBITRARINESS IN THE EXPANSION FACTOR . . . . . . . . . . . . . . . . . . . . . . . . 252 27.4 UNIQUENESS OF METRIC FOR 3-SURFACE OF CONSTANT CURVATURE . . . . . . . . . . 253 27.5 METRIC FOR 3-SURFACE OF CONSTANT CURVATURE . . . . . . . . . . . . . . . . . . . . 254 27.6 PROPERTIES OF THE 3-SURFACES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 27.7 ISOTROPY IMPLIES HOMOGENEITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 27.8 MATTER-DOMINATED AND RADIATION-DOMINATED REGIMES Of FRIEDMANN COSMOLOGY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256 27.9 TRANSITION FROM RADIATION-DOMINATED REGIME TO MATTER-DOMINATED REGIME 256 27.10THE EXPANDING AND RECONTRACTING SPHERICAL WAVE FRONT . . . . . . . . . . . . 258 27.11ON SEEING THE BACK OF ONE’S HEAD . . . . . . . . . . . . . . . . . . . . . . . . . . . . 259 27.12DO THE CONSERVATION LAWS FORBID THE PRODUCTION OF PARTICLE-ANTIPARTICLE PAIRS OUT OF EMPTY SPACE BY TIDAL GRAVITATION FORCES? . . . . . . . . . . . . . . 259

14

27.13TURN-AROUND UNIVERSE MODEL NEGLECTING MATTER DENSITY . . . . . . . . . . . . 259 27.14“HESITATION” UNIVERSE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260 27.15UNIVERSE OPAQUE TO BLACK-BODY RADIATION AT A NONSINGULAR PAST TURN-AROUND REQUIRES IMPOSSIBLE PARAMETERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261 28 EVOLUTION OF THE UNIVERSE INTO ITS PRESENT STATE

263

28.1 UNCERTAINTY IN EVOLUTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263 29 PRESENT STATE AND FUTURE EVOLUTION OF THE UNIVERSE

265

29.1 IMPLICATIONS OF PARAMETER VALUES . . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 29.2 ALTERNATIVE DERIVATION OF REDSHIFT . . . . . . . . . . . . . . . . . . . . . . . . . . . 265 29.3 REDSHIFT OF PARTICLE DE BROGLIE WAVELENGTHS . . . . . . . . . . . . . . . . . . . . 266 29.4 m(z) DERIVED USING STATISTICAL PHYSICS . . . . . . . . . . . . . . . . . . . . . . . . . 266 29.5 DOPPLER SHIFT VERSUS COSMOLOGICAL REDSHIFT . . . . . . . . . . . . . . . . . . . . 267 29.6 SOURCE COUNTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 268 29.7 COSMIC-RAY DENSITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269 29.8 FRACTION OF SKY COVERED BY GALAXIES . . . . . . . . . . . . . . . . . . . . . . . . . . 269 31 SCHWARZSCHILD GEOMETRY

271

31.1 TIDAL FORCES ON IN FALLING EXPLORER . . . . . . . . . . . . . . . . . . . . . . . . . . 271 31.2 NONRADIAL LIGHT CONES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 31.3 THE CRACK OF DOOM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 31.4 HOW LONG TO LIVE? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272 31.5 EDDINGTON-FINKELSTEIN AND KRUSKAL-SZEKERES COMPARED . . . . . . . . . . . . . 273 31.6 ANOTHER COORDINATE SYSTEM . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 274 31.7 SCHWARZSCHILD METRIC IN ISOTROPIC COORDINATES . . . . . . . . . . . . . . . . . . 274 31.8 REISSNER-NORDSTROM GEOMETRY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 276 32 GRAVITATIONAL COLLAPSE

279

32.1 UNIQUENESS OF REISSNER-NORDSTROM GEOMETRY . . . . . . . . . . . . . . . . . . . 279 32.2 REDSHIFTS DURING COLLAPSE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282 32.3 EMBEDDING DIAGRAMS AND PHOTON PROPAGATION FOR COLLAPSING STAR . . . . . 285

15

32.4 MATCH OF FRIEDMANN INTERIOR TO SCHWARZSCHILD EXTERIOR . . . . . . . . . . . 285 32.5 STARS THAT COLLAPSE FROM INFINITY . . . . . . . . . . . . . . . . . . . . . . . . . . . 287 32.6 GENERAL SPHERICAL COLLAPSE: METRIC IN COMOVING COORDINATES . . . . . . . . 288 32.7 ADIABATIC SPHERICAL COLLAPSE: EQUATIONS OF EVOLUTION . . . . . . . . . . . . . . 289 32.8 ANALYTIC SOLUTIONS FOR PRESSURE-FREE COLLAPSE . . . . . . . . . . . . . . . . . . . 292 32.9 COLLAPSE WITH UNIFORM DENSITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 292 32.10PRICE’S THEOREM FOR A SCALAR FIELD . . . . . . . . . . . . . . . . . . . . . . . . . . . 293 32.11NEWMAN-PENROSE “CONSTANTS” . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298

33 BLACK HOLES

301

33.1 KERR DESCRIPTION OF KILLING VECTORS . . . . . . . . . . . . . . . . . . . . . . . . . . 301 33.2 OBSERVATIONS OF ANGULAR VELOCITY . . . . . . . . . . . . . . . . . . . . . . . . . . . 302 33.3 LOCALLY NONROTATING OBSERVERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 303 33.4 ORTHONORMAL FRAMES OF LOCALLY NONROTATING OBSERVERS . . . . . . . . . . . . 304 33.5 LOCAL LIGHT CONES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 306 33.6 SUPERHAMllTONIAN FOR CHARGED-PARTICLE MOTION . . . . . . . . . . . . . . . . . . 307 33.7 HAMILTON-JACOBI DERIVATION OF EQUATIONS OF MOTION . . . . . . . . . . . . . . . 308 33.8 KERR-SCHILD COORDINATES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311 33.9 NULL GENERATORS OF HORIZON . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 312 33.10ANGULAR MOMENTUM VECTOR FOR IN FALLING PARTICLE . . . . . . . . . . . . . . . . 313 33.11IRREDUCIBLE MASS IS IRREDUCIBLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314 33.12SURFACE AREA OF A BLACK HOLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 314 33.13ANGULAR VELOCITY OF A BLACK HOLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . 315 33.14SEPARATION OF VARIABLES FOR WAVE EQUATIONS . . . . . . . . . . . . . . . . . . . . . 316

34 GLOBAL TECHNIQUES, HORIZONS, AND SINGULARITY THEOREMS

319

34.1 FLAT SPACETIME IN ψ, ξ, θ, ϕ COORDINATES . . . . . . . . . . . . . . . . . . . . . . . . . 319 34.2 SCHWARZSCHILD SPACETIME IN ψ, ξ, θ, ϕ COORDINATES . . . . . . . . . . . . . . . . . 320 34.3 REISSNER-NORDSTROM SPACETIME

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321

34.4 A BLACK HOLE CAN NEVER BIFURCATE . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323

16

35 PROPAGATION OF GRAVITATIONAL WAVES

325

35.1 TRANSFORMATION OF PLANE WAVE TO TT GAUGE . . . . . . . . . . . . . . . . . . . . . 325 35.2 LIMITATION ON EXISTENCE OF TT GAUGE . . . . . . . . . . . . . . . . . . . . . . . . . . 326 35.3 A CYLINDRICAL GRAVITATIONAL WAVE . . . . . . . . . . . . . . . . . . . . . . . . . . . . 326 35.4 NON-TT PARTS OF METRIC PERTURBATION . . . . . . . . . . . . . . . . . . . . . . . . . 327 35.5 ALTERNATIVE CALCULATION OF RELATIVE OSCILLATIONS . . . . . . . . . . . . . . . . . 328 35.6 ROTATIONAL TRANSFORMATIONS FOR POLARIZATION STATES . . . . . . . . . . . . . . 329 35.7 ELLIPTICAL POLARIZATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 330 35.8 GLOBALLY WELL-BEHAVED COORDINATES FOR PLANE WAVE . . . . . . . . . . . . . . . 332 35.9 GEODESIC COMPLETENESS FOR PLANE-WAVE MANIFOLD . . . . . . . . . . . . . . . . . 333 35.10PLANE WAVE WITH TWO POLARIZATIONS PRESENT . . . . . . . . . . . . . . . . . . . . 333 35.11CONNECTION COEFFICIENTS AND CURVATURE TENSORS FOR A PERTURBED METRIC . 335 35.12GAUGE TRANSFORMATIONS IN A CURVED BACKGROUND . . . . . . . . . . . . . . . . . 337 35.13TRANSVERSE-TRACELESS GAUGE FOR GRAVITATIONAL WAVES PROPAGATING IN A CURVED BACKGROUND . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 338 35.14BRILL-HARTLE AVERAGE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339 35.15GEOMETRIC OPTICS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 340 35.16GRAVITONS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 343 35.17GRAVITATIONAL DEFLECTION OF GRAVITATIONAL WAVES . . . . . . . . . . . . . . . . . 343 (GW)

35.18GAUGE INVARIANCE OF Tµν (GW)

35.19Tµν

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344

EXPRESSED AS THE AVERAGE OF A STRESS-ENERGY PSEUDOTENSOR . . . . . . 345

35.20SHORTWAVE APPROXIMATION FROM A VARIATIONAL VIEWPOINT . . . . . . . . . . . . 346

36 GENERATION OF GRAVITATIONAL WAVES

349

36.1 GRAVITATIONAL WAVES FROM ROTATING BEAM . . . . . . . . . . . . . . . . . . . . . . . 349 36.2 GRAVITATIONAL WAVES FROM MATTER FALLING INTO A BLACK HOLE . . . . . . . . . . 350 36.3 GRAVITATIONAL WAVES FROM A VIBRATING NEUTRON STAR . . . . . . . . . . . . . . . 350 36.4 PULSAR SLOWDOWN . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 353 36.5 ENERGY AND ANGULAR MOMENTUM LOSSES DUE TO RADIATION REACTION

. . . . . 354

36.6 GRAVITATIONAL WAVES FROM BINARY STAR SYSTEMS . . . . . . . . . . . . . . . . . . . 355

17

36.7 MAGNITUDE OF tµν . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 356 36.8 PROOF THAT THE TRANVERSE TRACELESS PARTS OF . . . . . . . . . . . . . . . . . . . . 356 36.9 ENERGY AND ANGULAR MOMENTUM RADIATED . . . . . . . . . . . . . . . . . . . . . . 357

37 DETECTION OF GRAVITATIONAL WAVES

359

37.1 GENERAL PLANE WAVE IN TT GAUGE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 359 37.2 TEST-PARTICLE MOTION IN PROPER REFERENCE FRAME . . . . . . . . . . . . . . . . . . 360 37.3 CONNECTION COEFFICIENTS IN PROPER REFERENCE FRAME . . . . . . . . . . . . . . . 361 37.4 WHY THE a · x ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362 37.5 ORIENTATION OF POLARIZATION DIAGRAM . . . . . . . . . . . . . . . . . . . . . . . . . 362 37.6 RELATIVE MOTION OF FREELY FALLING BODIES AS A DETECTOR OF GRAVITATIONAL WAVES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 363 37.7 EARTH-MOON SEPARATION AS A GRAVITATIONAL-WAVE DETECTOR . . . . . . . . . . . 364 37.8 POWER RERADIATED . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365 37.9 CROSS SECTIONS CALCULATED BY DETAILED BALANCE . . . . . . . . . . . . . . . . . . 365 37.10NORMAL-MODE ANALYSIS OF VIBRATING, RESONANT DETECTORS . . . . . . . . . . . . 366 37.11SPECTRUM OF ENERGY RADIATED BY A SOURCE . . . . . . . . . . . . . . . . . . . . . . 370 37.12PATTERNS OF EMISSION AND ABSORPTION . . . . . . . . . . . . . . . . . . . . . . . . . 371 37.13MULTIMODE DETECTOR

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 372

37.14CROSS SECTION OF IDEALIZED MODEL OF EARTH FOR ABSORPTION OF GRAVITATIONAL RADIATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 373

39 OTHER THEORIES OF GRAVITY AND POST-NEWTONIAN APPROXIMATION

377

39.1 ORDERS OF MAGNITUDE IN GRAVITATIONALLY BOUND SYSTEMS . . . . . . . . . . . . . 377 39.2 PATTERN OF TERMS IN POST-NEWTONIAN EXPANSION . . . . . . . . . . . . . . . . . . . 377 39.3 NEWTONIAN APPROXIMATION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 378 39.4 A USEFUL FORMULA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379 39.5 STRESS TENSOR FOR NEWTONIAN GRAVITATIONAL FIELD . . . . . . . . . . . . . . . . . 380 39.6 NEWTONIAN VIRIAL THEOREMS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380 39.7 PULSATION FREQUENCY FOR NEWTONIAN STAR . . . . . . . . . . . . . . . . . . . . . . 382 39.8 ABSENCE OF “METRIC-GENERATES-METRIC” TERMS IN POST-NEWTONIAN LIMIT . . . 383

18

39.9 REMOVAL OF Σ TERM FROM g00 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 385 39.10VERIFICATION OF FORMS OF POST-NEWTONIAN CORRECTIONS . . . . . . . . . . . . . 386 39.11TRANSFORMATION TO MOVING FRAME . . . . . . . . . . . . . . . . . . . . . . . . . . . 387 39.12THE TRANSFORMATION BETWEEN COMOVING FRAME AND PPN FRAME . . . . . . . . 389 39.13EQUATIONS OF MOTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 390 39.14POST-NEWTONIAN APPROXIMATION TO GENERAL RELATIVITY . . . . . . . . . . . . . . 395 39.15MANY-BODY SYSTEM IN POST-NEWTONIAN LIMIT OF GENERAL RELATIVITY . . . . . . 399 40 SOLAR-SYSTEM EXPERIMENTS

405

40.1 PPN METRIC FOR IDEALIZED SUN . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405 40.2 TRAJECTORY OF LIGHT RAY IN SUN’S GRAVITATIONAL FIELD . . . . . . . . . . . . . . . 406 40.3 FERMAT’S PRINCIPLE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 407 40.4 DERIVATION OF PERIHELION SHIFT IN PPN FORMALISM . . . . . . . . . . . . . . . . . . 408 40.5 PERIHELION SHIFT FOR OBLATE SUN . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 409 40.6 PRECESSIONAL ANGULAR VELOCITY . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 411 40.7 OFF-DIAGONAL TERMS IN METRIC ABOUT THE EARTH . . . . . . . . . . . . . . . . . . . 412 40.8 SPIN-CURVATURE COUPLING . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 413 40.9 CAVENDISH CONSTANT FOR IDEALIZED SUN . . . . . . . . . . . . . . . . . . . . . . . . . 413 40.10CAVENDISH CONSTANT FOR ANY BODY . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415 41 SPINORS

417

41.1 ELEMENTARY FEATURES OF THE ROTATION MATRIX . . . . . . . . . . . . . . . . . . . . 417 41.2 ROTATION MATRIX HAS UNIT DETERMINANT . . . . . . . . . . . . . . . . . . . . . . . . 418 41.3 MORE PROPERTIES OF THE ROTATION MATRIX . . . . . . . . . . . . . . . . . . . . . . . 418 43 SUPERSPACE ARENA FOR THE DYNAMICS OF GEOMETRY

421

43.1 THE ACTION PRINCIPLE FOR A FREE PARTICLE IN NON RELATIVISTIC MECHANICS . . 421 43.2 THE ACTION FOR THE HARMONIC OSCILLATOR . . . . . . . . . . . . . . . . . . . . . . . 422 43.3 QUANTUM PROPAGATOR FOR HARMONIC OSCILLATOR . . . . . . . . . . . . . . . . . . . 423 43.4 QUANTUM PROPAGATOR FOR FREE ELECTROMAGNETIC FIELD . . . . . . . . . . . . . . 424 43.5 HAMILTON-JACOBI FORMULATION OF MAXWELL ELECTRODYNAMICS . . . . . . . . . . 425

19

20

Chapter 1

GEOMETRODYNAMICS IN BRIEF 1.1 CURVATURE OF A CYLINDER Show that the Gaussian curvature R of the surface of a cylinder is zero by showing that geodesics on that surface (unroll!) suffer no geodesic deviation. Give an independent argument for the same conclusion by employing the formula R = 1/ρ1 ρ2 where ρ1 and ρ2 are the principal radii of curvature at the point in question with respect to the enveloping Euclidean three-dimensional space.

Solution: The metric of the surface of a cylinder in cylindrical coordinates (ϕ, z) is ( g=

r2 0

) 0 . 1

The geodesic equation can be gotten by minimizing ∫ √ s= r2 d2 ϕ + d2 z.

(1.1.1)

(1.1.2)

The variation method gives dz(ϕ) = constant dϕ

(1.1.3)

for geodesics. Since the “slope” of the geodesics on the surface is a constant, two initially parallel geodesics will be parallel forever. So the Gaussian curvation of the surface of a cylinder is 0. The curvature of the curve ϕ = costant on the surface is 0. Thus, one of the princpal radii of curvature is infinite. We then have R = 1/ρ1 ρ2 = 0.

1.2 SPRING TIDE VS. NEAP TIDE Evaluate (1) in conventional units and (2) in geometrized units the magnitude of the Newtonian tideproducing acceleration Rm0n0 generated at the Earth by (1) the moon (mconv = 7.35 × 1025 g, r = 3.84 × 1010 cm) and (2) the sun (mconv = 1.989 × 1033 g, r = 1.496 × 1013 cm). By what factor do you expect spring tides to exceed neap tides?

21

Solution: In geometrized units, we have Rm0n0 = ∂ m ∂n ϕ,

(1.2.1)

where ϕ = −M/r. We can figure out that Rm0n0 =

M (δnm r2 − xm xn ) . r5

(1.2.2)

Suppose the coordinates of the earth are (r, 0, 0). We can get R2020 = R3030 =

M . r3

(1.2.3)

As for the moon, we have Gmconv M = = 9.64 × 10−35 cm−2 . r3 c2 r3

(1.2.4)

As for the sun, we have M = 4.41 × 10−35 cm−2 . (1.2.5) r3 Spring tide happens when moon and sun contribute together to the curvature of space-time at earth, while neap tide takes place when they cancels. So, we would expect spring tides to be three times stronger than neaps tides. The discussion in conventional units is straightforward and will be omitted here.

1.3 KEPLER ENCAPSULATED A small satellite has a circular frequency ω(cm−1 ) in an orbit of radius r about a central object of mass m(cm). From the known value of ω, show that it is possible to determine neither r nor m individually, but only the effective “Kepler density” of the object as averaged over a sphere of the same radius as the orbit. Give the formula for ω 2 in terms of this Kepler density.

Solution: Since ω2 r =

m , r2

m=

ω2 =

4 πρKepler . 3

4 πρKepler r3 , 3

(1.3.1)

we can get

22

(1.3.2)

Chapter 2

FOUNDATIONS OF SPECIAL RELATIVITY 2.1 EXERCISE Show that equation (2.14) is in accord with the quantum-mechanical properties of a de Broglie wave, [ ] ψ = eiϕ = exp i(⃗k · ⃗x − ωt) .

(2.1.1)

Solution: ˜ is The number of surfaces x pierces in the 4-momentum 1-form k ϕ(⃗x, t) − ϕ(0, 0) = ⃗k · ⃗x − ωt = k · x.

(2.1.2)

˜ x⟩. k · x = ⟨k,

(2.1.3)

So we have

2.2 LOWERING INDEX TO GET THE 1-FORM CORRESPONDING TO A VECTOR ˜ that corresponds to a vector u can be obtained by “lowering an index” The components uα of the 1-form u with the metric coefficients ηαβ : uα = ηαβ uβ ; i.e., u0 = −u0 , uk = uk .

Solution: ˜ eα ⟩ = u · eα = ηµν uµ eµα = ηµν uµ δαν = ηαβ uβ . uα = ⟨u,

23

(2.2.1)

2.3 RAISING INDEX TO RECOVER THE VECTOR One can return to the components of u by raising indices, uα = η αβ uβ ;

(2.3.1)

the matrix ||η αβ || is defined as the inverse of ||ηαβ ||, and happens to equal ||ηαβ ||: η αβ ηβγ ≡ δ αγ ;

η αβ = ηαβ for all α, β.

(2.3.2)

Solution: η αβ uβ = η αβ ηβγ uγ = δ αγ uγ = uα .

(2.3.3)

2.4 VARIED ROUTES TO THE SCALAR PRODUCT The scalar product of u with v can be calculated in any of the following ways: u · v = g(u, v) = uα v β ηαβ = uα vα = uα vβ η αβ .

(2.4.1)

The proposition is straightforward to prove by applying conclusions of the problems above.

2.5 ENERGY AND VELOCITY FROM 4-MOMENTUM A particle of rest mass m and 4-momentum p is examined by an observer with 4-velocity u. Show that just as (a) the energy he measures is E = −p · u; (2.5.1) so (b) the rest mass he attributes to the particle is m2 = −p2 ;

(2.5.2)

(c) the momentum he measures has magnitude |⃗ p| = [(p · u)2 + (p · p)]1/2 ;

(2.5.3)

(d) the ordinary velocity ⃗v he measures has magnitude |⃗v | =

|⃗ p| , E

(2.5.4)

where |⃗ p| and E are as given above; and (e) the 4-vector v, whose components in the observer’s Lorentz frame are v 0 = 0, v j = (dxj /dt)for particle = ordinary velocity, (2.5.5) is given by v=

p + (p · u)u . −p · u

24

(2.5.6)

Solution: (a) In the frame of observer, we have u = (1, 0, 0, 0) and p0 = E. Thus E = −p · u

(2.5.7)

holds in observer’s frame. Since the two sides of the equation are all geometric objects which is invariant in different reference frames, we have proved that E = −p · u.

(2.5.8)

(b) In observer’s frame, we have p = (E, p⃗). So the rest mass he attributes to the particle is m2 = E 2 − |⃗ p|2 = −p2 .

(2.5.9)

|⃗ p| = (E 2 − m2 )1/2 = [(p · u)2 + (p · p)]1/2 .

(2.5.10)

(c) (d) In observer’s frame, we have E = γm and p⃗ = γm⃗v , where γ = (1 − |⃗v |2 )−1/2 . Thus ⃗v =

p⃗ , E

|⃗v | =

|⃗ p| . E

(2.5.11)

(e) In observer’s frame, we have and

p0 + (p · u)u0 = E − E = 0,

(2.5.12)

p⃗ p⃗ + (p · u)⃗u = = ⃗v . −p · u E

(2.5.13)

Thus v=

p + (p · u)u . −p · u

(2.5.14)

holds in observer’s frame. Since the two sides of the equation are all geometric objects which is invariant in different reference frames, it holds in all frames.

2.6 TEMPERATURE GRADIENT To each event L inside the sun one attributes a temperature T (L), the temperature measured by a thermometer at rest in the hot gas there. Then T (L) is a function; no coordinates are required for its definition and discussion. A cosmic ray from outer space flies through the sun with 4-velocity u. Show that, as measured by the cosmic ray’s clock, the time derivative of temperature in its vicinity is dT /dτ = ∂u T = ⟨dT, u⟩.

(2.6.1)

In a local Lorentz frame inside the sun, this equation can be written 1 dT ∂T ∂T vj ∂T . = uα α = √ +√ j 2 2 dτ ∂x ∂t ∂x 1−v 1−v

(2.6.2)

Why is this result reasonable?

Solution: The equation (2.15) of the MTW tells T (L) = T (L0 ) + ⟨dT, L − L0 ⟩ + (nonlinear terms).

25

(2.6.3)

Suppose the world line of the observer is P (τ ), the we have L(τ ) − L(τ0 ) = (τ − τ0 )u + (nonlinear terms).

(2.6.4)

dT = ⟨dT, u⟩. dτ

(2.6.5)

In the limit τ → τ0 , we have

In a local Lorentz frame inside the sun, we have dT ∂T 1 ∂T vj ∂T = (dT )α uα = uα α = √ +√ . 2 2 dτ ∂x 1 − v ∂t 1 − v ∂xj

(2.6.6)

2.7 BOOST IN AN ARBITRARY DIRECTION An especially useful Lorentz transformation has the matrix components ′ 1 Λ0 0 = γ ≡ √ , 1 − β2 ′

Λ0 j = Λ j Λj

′

′

0

= −βγnj ,

′

k

= Λkj = (γ − 1)nj nk + δ jk ,

Λµν ′ = (same as Λν

′

µ

but withβ repalced by − β),

(2.7.1)

where β, n1 , n2 , and n3 are parameters, and n2 = n21 + n22 + n33 = 1. Show (a) that this does satisfy the condition ΛT ηΛ = η equired of a Lorentz transformation (see Box 2.4); (b) that the primed frame moves with ordinary velocity βn as seen in the unprimed frame; (c) that the unprimed frame moves with ordinary velocity −βn as seen in the primed frame; and (d) that for motion in the z direction, the transformation matrices reduce to the familiar form     γ 0 0 βγ γ 0 0 −βγ    0 ′ 1 0 0   , Λµ ′ =  0 1 0 0  . (2.7.2) Λν µ =  ν 0 0 1 0  0 0 1 0  −βγ 0 0 γ βγ 0 0 γ

Solution: (a) ′

′

′

′

′

′

′

β′

′

′

j′

′

Λα 0 ηα′ β ′ Λβ 0 = Λ0 0 η0′ 0′ Λ0 0 + Λj 0 ηj ′ k′ Λk 0 = −γ 2 + β 2 γ 2 nj nj = −1 Λα j ηα′ β ′ Λ

k

= Λ0 j η0′ 0′ Λ0

k

+Λ

η ′ ′Λ j j k

k

(2.7.3)

k

j i′

′

′

′

= −β γ n n + [(γ − 1)n n + δ ji ][(γ − 1)nk ni + δ ki ] 2 2 j k

= [−β 2 γ 2 + (γ − 1)2 + 2(γ − 1)]nj nk + δ kj = δjk ′

′

′

′

′

(2.7.4)

′

Λα 0 ηα′ β ′ Λβ j = Λ0 0 η0′ 0′ Λ0 j + Λj 0 ηj ′ k′ Λk j ′

′

′

= βγ 2 nj − βγni [(γ − 1)ni nj + δ ji ] (2.7.5)

=0 ′

′

So, we have prove that Λα 0 ηα′ β ′ Λβ 0 = ηαβ , i.e. ΛT ηΛ = η.

26

′

(b) xµ = (t′ , 0, 0, 0) is the world line of one observer who is static at the origin of primed frame. The coordinates of his world line in unprimed frame is ′

xµ = Λµν ′ xν = γt′ (1, βnj ).

(2.7.6)

So the velocity of the observer as seen by unprimed frame is vj =

xj γt′ βnj = = βnj . t γt′

(2.7.7)

(c) xµ = (t, 0, 0, 0) is the world line of one observer who is static at the origin of uprimed frame. The coordinates of his world line in primed frame is ′

′

xµ = Λµ ν xν = γt(1, −βnj ).

(2.7.8)

So the velocity of the observer as seen by primed frame is v ′j =

x′j −γtβnj = −βnj . = t′ γt

(2.7.9)

(d) For motion in the z direction, we have nj = (0, 0, 1). Substitute it into equation (29), it is easy to figure out     γ 0 0 βγ γ 0 0 −βγ    0 ′ 1 0 0   , Λµ ′ =  0 1 0 0  . (2.7.10) Λν µ =  ν    0 0 0 1 0 0 1 0 βγ 0 0 γ −βγ 0 0 γ

27

28

Chapter 3

THE ELECTROMAGNETIC FIELD 3.1 EXERCISE Derive equations (3.5) and (3.7) for the components of Faraday by comparing (3.4) with (3.2a,b), and by using definition (3.6).

Solution: By comparing

and

dp0 = eE · u dτ

(3.1.1)

dp0 = eF 0i ui , dτ

(3.1.2)

F 0i = E i .

(3.1.3)

dp = e(u0 E + u × B) dτ

(3.1.4)

dpi = eF i0 u0 + eF ij uj , dτ

(3.1.5)

F ij = ϵijk B k .

(3.1.6)

we have By comparing

and

we have Now it is straightforward to get equation (3.5) and (3.7).

3.2 TRANSFORMATION LAW FOR COMPONENTS OF A TENSOR From the transformation laws for components of vectors and 1-forms, derive the transformation law (3.14).

Solution: S(σ, ρ, v) is invariant under coordinate transformation. In the original frame, we have S(σ, ρ, v) = S αβγ σα ρβ v γ .

29

(3.2.1)

In the primed frame, we have

′ ′

′

S(σ, ρ, v) = S µ ν λ′ σµ′ ρν ′ v λ .

(3.2.2)

From the transformation laws for components of vectors and 1-forms, we have ′

′

σα = (Λ−1 )αµ σµ′ = Λµ α σµ′ ,

′

′

ρβ = (Λ−1 )β ν ρν ′ = Λν β σν ′ ,

Now, we can get

′ ′

′

′

v γ = (Λ−1 )γ λ′ v λ = Λλ′ γ σλ′ .

′

S µ ν λ′ = Λµ α Λν β Λλ′ γ S αβγ

(3.2.3) (3.2.4)

3.3 RAISING AND LOWERING INDICES Derive equations (3.16) from equation (3.15’) plus the law nα = ηαβ nβ for getting the components of the ˜ from the components of its corresponding vector n. 1-form n

Solution: S αβγ σα nβ v γ = S αβγ σα nβ v γ = S αµγ σα ηµβ nβ v γ .

(3.3.1)

S αβγ = ηµβ S αµγ .

(3.3.2)

S αµγ = η µν ηνβ S αβγ = η µν S ανγ = η µβ S αβγ .

(3.3.3)

Thus, we have And we can also get

3.4 TENSOR PRODUCT Given any two vectors u and v, one defines the second-rank tensor u ⊗ v (”tensor product of u with v”) to be a machine, with two input slots, whose output is the number (u ⊗ v)(σ, λ) = ⟨u, σ⟩⟨v, λ⟩

(3.4.1)

when 1-forms σ and λ are inserted. Show that the components of T = u ⊗ v are the products of the components of u and v: T αβ = uα v β , Tα β = uα v β , Tαβ = uα vβ . (3.4.2) Extend the definition to several vectors and forms, (u ⊗ v ⊗ β ⊗ w)(σ, λ, n, ξ) = ⟨u, σ⟩⟨v, λ⟩⟨β, n⟩⟨ξ, w⟩,

(3.4.3)

and show that the product rule for components still holds: S = u ⊗ v ⊗ β ⊗ w has components S µνλ ξ = uµ v ν βλ wξ .

(3.4.4)

Solution:

T αβ = T (ω α , ω β ) = ⟨u, ω α ⟩⟨v, ω β ⟩ = T αβ ; Tα β = ηαµ T µβ = ηαµ uµ v β = uα v β ; Tαβ = ηβµ Tα µ = ηβµ uα v µ = uα vβ .

30

(3.4.5)

S µνλ ξ = S(ω µ , ω ν , eλ , ω ξ ) = ⟨u, ω µ ⟩⟨v, ω ν ⟩⟨β, eλ ⟩⟨ω ξ , w⟩ = uµ v ν βλ wξ .

(3.4.6)

3.5 BASIS TENSORS Show that a tensor M with components M αβγ δ in a given Lorentz frame can be reconstructed from its components and from the basis 1-forms and vectors of that frame as follows: M = M αβγ δ eα ⊗ eβ ⊗ ω γ ⊗ eδ .

(3.5.1)

Solution: Because M µνρ σ eµ ⊗ eν ⊗ ω ρ ⊗ eσ (ω α , ω β , eγ , ω δ ) = M µνρ σ ⟨ω α , eµ ⟩⟨ω β , eν ⟩⟨ω ρ , eγ ⟩⟨ω δ , eσ ⟩ = M µνρ σ δ αµ δ βν δ ργ δ δσ = M αβγ δ ,

(3.5.2)

M (ω α , ω β , ω γ , ω δ ) = M αβγ δ ,

(3.5.3)

M = M αβγ δ eα ⊗ eβ ⊗ ω γ ⊗ eδ .

(3.5.4)

we have

3.6 Faraday MACHINERY AT WORK An observer with 4-velocity u picks out three directions in spacetime that are orthogonal and purely spatial (no time part) as seen in his frame. Let eˆ1 , eˆ2 , eˆ3 be unit vectors in those directions and let them be oriented in a righthanded way (eˆ1 ·eˆ2 ×eˆ3 = +1 in three-dimensional language). Why do the following relations hold? eˆj · u = 0, eˆj · ekˆ = δjk . (3.6.1) What vectors are to be inserted in the two slots of the electromagnetic field tensor Faraday if one wants to get out the electric field along eˆj ; as measured by this observer? What vectors must be inserted to get the magnetic field he measures along eˆj ?

Solution: As eˆj are orthogonal and purely spatial as seen by the observer with four velocity u, we have the following relations: (3.6.2) eˆj · u = 0, eˆj · ekˆ = δjk . The electric field along eˆj as measured by this observer is F (eˆj , u).

(3.6.3)

F (eˆi , ekˆ )

(3.6.4)

The magnetic field he measures along eˆj is where eˆj · eˆi × ekˆ = +1.

31

3.7 MAXWELL’S EQUATIONS Show, by explicit examination of components, that the geometric laws Fαβ,γ + Fβγ,α + Fγα,β = 0,

F αβ,β = 4πJ α ,

(3.7.1)

do reduce to Maxwell’s equations (3.26), (3.31), (3.34), (3.35), as claimed above.

Solution: If two among the indexes α, β and γ are the same, it is easy to find Fαβ,γ + Fβγ,α + Fγα,β = 0

(3.7.2)

would hold automatically due to the antisymmetry of the Faraday tensor. Now we assume all indexes are different. We have [ ] ∂B F0i,j + Fij,0 + Fj0,i = −Ei,j + ϵijk B˙ k + Ej,i = ϵijk ∇ × E + = 0, (3.7.3) ∂t k and Fij,k + Fjk,i + Fki,j = ϵijl Bl,k + ϵjkl Bl,i + ϵkil Bl,j = Bk,k + Bi,i + Bj,j = 3∇ · B = 0.

(3.7.4)

They reduce to the (3.26) and (3.31). As for the second equation, we have

and

F 0i,i = Ei,i = ∇ · E = 4πJ 0

(3.7.5)

] [ ∂E = 4πJi . F i0,0 + F ij,j = −E˙ i + ϵijk Bk,j = ∇ × B − ∂t i

(3.7.6)

They reduce to the (3.34) and (3.35).

3.8 CONTRACTION IS FRAME-INDEPENDENT Show that contraction, as defined in equation (3.40), does not depend on which Lorentz frame ea and ω a are taken from. Also show that equation (3.40) implies M (u, v) = Rαµαν uµ v ν .

(3.8.1)

Solution: Suppose in another frame, the base vector and 1-form are f and ξ. Then we have fα = Λαβ eβ ,

ξ α = Λαβ ω β .

(3.8.2)

Thus, R(fα , u, ξ α , v) = Λαβ Λαγ R(eβ , u, ω γ , v) = δγβ R(eβ , u, ω γ , v) = R(eβ , u, ω β , v).

(3.8.3)

So contraction is frame independent. And M (u, v) = R(eα , u, ξ α , v) = Rβµγ ν (eα )β uµ (ω α )γ v ν = Rαµαν uµ v ν .

32

(3.8.4)

3.9 DIFFERENTIATION (a) Justify the formula d(uµ v ν )/dτ = (d(uµ )/dτ )v ν + uµ (d(v ν )/dτ ).

(3.9.1)

by considering the special case µ = 0, ν = 1. (b) Explain why (T αβ vβ ),µ = T αβ,µ vβ + T αβ vβ,µ .

(3.9.2)

d(u0 v 1 )/dτ = d(u0 )/dτ v 1 + u0 d(v 1 )/dτ.

(3.9.3)

Solution: (a)

(b) The contraction of a tensor is a kind of summation as for the components. Since the order of differention and summation can be exchaged. So we have the equation 3.9.2.

3.10

MORE DIFFERENTIATION

(a) Justify the formula, d(uµ uν )/dτ = 2uµ (duµ /dτ )

(3.10.1)

by writing out the summation uµ uµ ≡ ηµν uµ uν explicitly. (b) Let δ indicate a variation or small change, and justify the formula δ(F αβ Fαβ ) = 2Fαβ δ(F αβ ). (3.10.2) (c) Compute (F αβ Fαβ ),µ = ?

Solution: (a) d(uµ uν )/dτ = d(ηµν uµ uν )/dτ = ηµν uµ (duν /dτ ) + ηµν uν (duµ /dτ ) = uν (duν /dτ ) + uµ (duµ /dτ ) = 2uµ (duµ /dτ ).

(3.10.3)

(b) δ(F αβ Fαβ ) = δ(ηµα ηνβ F αβ F µν ) = ηµα ηνβ F αβ δ(F µν ) + ηµα ηνβ F µν δ(F αβ ) = Fµν δ(F µν ) + Fαβ δ(F αβ ) = 2Fαβ δ(F αβ ).

(3.10.4)

(c) (F αβ Fαβ ),µ = 2Fαβ F αβ,µ .

33

(3.10.5)

3.11

SYMMETRIES

Let Aµν be an anbsymmetric tensor so that Aµν = −Aνµ ; and let S µν be a symmetrIc tensor so that S µν = S νµ . (a) Justify the equations Aµν S µν = 0 in two ways: first, by writing out the sum explIcitly (all sixteen terms) and showing how the terms in the sum cancel in pairs; second, by giving an argument to justify each equals sign in the following string: Aµν S µν = −Aνµ S µν = −Aνµ S νµ = −Aαβ S αβ = −Aµν S µν = 0.

(3.11.1)

(b) Establish the following two identities for any arbitrary tensor Vµν : V µν Aµν =

1 µν (V − V νµ )Aµν ; 2

V µν Sµν =

1 µν (V + V νµ )Sµν . 2

(3.11.2)

Solution: (a) i. Aµν S µν = A00 S 00 + A11 S 11 + A22 S 22 + A33 S 33 + (A01 S 01 + A10 S 10 ) + (A02 S 02 + A20 S 20 ) + (A03 S 03 + A30 S 30 ) + (A12 S 12 + A21 S 21 ) + (A13 S 13 + A31 S 31 ) + (A23 S 23 + A32 S 32 ) (3.11.3)

= 0.

ii. The first equals sign is due to the antisymmetry of Aµν . The second equals sign is due to the symmetry of Sµν . The third and fourth equals sign is a transformation of dumb indexes. The last equals sign is due to the fact that Aµν S µν = −Aµν S µν . (b) Since V µν Aµν = −V µν Aνµ = −V νµ Aµν

(3.11.4)

V µν Sµν = V µν Sνµ = V νµ Sµν ,

(3.11.5)

and we have V µν Aµν =

3.12

1 µν (V − V νµ )Aµν ; 2

V µν Sµν =

1 µν (V + V νµ )Sµν . 2

(3.11.6)

SYMMETRIZATION AND ANTISYMMETRIZATION

To “symmetrize” a tensor, one averages it with all of its transposes. The components of the new, symmetrized tensor are distinguished by round brackets: 1 (Vµν + Vνµ ); 2 1 ≡ (Vµνλ + Vνλµ + Vλµν + Vνµλ + Vλνµ + Vµλν ). 3!

V(µν) ≡ V(µνλ)

(3.12.1)

One “antisymmetrizes” a tensor (square brackets) similarly: 1 (Vµν − Vνµ ); 2 1 ≡ (Vµνλ + Vνλµ + Vλµν − Vνµλ − Vλνµ − Vµλν ). 3!

V[µν] ≡ V[µνλ]

34

(3.12.2)

(a) Show that such symmetrized and antisymmetrized tensors are, indeed, symmetric and antisymmetric under interchange of the vectors inserted into their slots: V(αβγ) uα v β wγ = +V(αβγ) v α uβ wγ = · · · V[αβγ] uα v β wγ = −V(αβγ) v α uβ wγ = · · ·

(3.12.3)

(b) Show that a second-rank tensor can be reconstructed from its symmetric and antisymmetric parts, Vµν = V(µν) + V[µν] ,

(3.12.4)

but that a third-rank tensor cannot; V(αβγ) and V[αβγ] contain together “less information” than Vαβγ . “Young diagrams” (see, e.g., Messiah [1961], appendix D) describe other symmetries, more subtle than these two, which contain the missing information. (c) Show that the electromagnetic field tensor satisfies F(αβ) = 0;

F[αβ] = 0.

(3.12.5)

(d) Show that Maxwell’s “magnetic” equations Fαβ,γ + Fβγ,α + Fγα,β = 0

(3.12.6)

F[αβ,γ] = 0.

(3.12.7)

can be rewritten in the form

Solution: (a) V(αβγ) uα v β wγ = V(βαγ) v β uα wγ = V(αβγ) uα v β wγ , V[αβγ] uα v β wγ = −V[βαγ] v β uα wγ = −V(αβγ) v α uβ wγ .

(3.12.8)

(b)

1 1 (Vµν + Vνµ ) + (Vµν − Vνµ ) = Vµν . (3.12.9) 2 2 The degree of freedom of Vαβγ , V(αβγ) , V[αβγ] is d3 , Cd3 + Cd2 + Cd1 , and Cd3 respectively. d is the dimension of the space. We have (4d + 7)(d − 1)d d3 − (2Cd3 + Cd2 + Cd1 ) = >0 (3.12.10) 6 if d > 1. So a third-rank tensor cannot be reconstructed from its symmetric and antisymmetric parts. V(µν) + V[µν] =

(c) Since Fαβ + Fβα = 0, it is easy to get F(αβ) = 0;

F[αβ] = 0.

(3.12.11)

(d) Since Fαβ + Fβα = 0, we have Fαβ,γ + Fβγ,α + Fγα,β = −(Fβα,γ + Fγβ,α + Fαγ,β ).

(3.12.12)

Fαβ,γ + Fβγ,α + Fγα,β = 0

(3.12.13)

Fαβ,γ + Fβγ,α + Fγα,β − (Fβα,γ + Fγβ,α + Fαγ,β ) = 6F[αβ,γ] = 0.

(3.12.14)

Thus, is equal to

35

3.13

LEVI-CIVITA TENSOR

The “Levi-Civita tensor” ϵ in spacetime is a fourth-rank, completely antisymmetric tensor: ϵ(u, u, v, w) changes sign when any two of the vectors are interchanged.

(3.13.1)

Choose an arbitrary but specIfic Lorentz frame, with e0 pointing toward the future and with e1 , e2 , e3 a righthanded set of spatial basis vectors. The covariant components of ϵ in this frame are ϵ0123 = ϵ(e0 , e1 , e2 , e3 ) = +1.

(3.13.2)

[Note: In an n-dimensional space, ϵ is the analogous completely antisymmetric rank-n tensor. Its components are ϵ1,··· ,n = ϵ(e1 , · · · , en ) = +1, (3.13.3) when computed on a “positively oriented,” orthonormal basis e1 , · · · , en .] (a) Use the antisymmetry to show that ϵαβγδ = 0 unless α, β, γ, δ are all different, { +1 for even permutations of 0, 1, 2, 3 and ϵπ0π1π2π3 = −1 for odd permutations.

(3.13.4)

(b) Show that ϵπ0π1π2π3 = −ϵπ0π1π2π3 .

(3.13.5)

¯ ¯

(c) By means of a Lorentz transformation show that ϵα¯ β¯γ δ and ϵα¯ β¯ ¯γ δ¯ have these same values in any other Lorentz frame with e0 pointing toward the future and with e1 , e2 , e3 a righthanded set. (d) What are the components of ϵ in a Lorentz frame with past-pointing e0 ? with lefthanded e1 , e2 , e3 ? (e) From the Levi-Civita tensor, one can construct several “permutation tensors.” In index notation: δ αβγµνλ ≡ −ϵαβγρ ϵµνλρ ; 1 1 δ αβµν ≡ δ αβλµνλ = − ϵαβλρ ϵµνλρ ; 2 2 1 αβλ 1 1 αβ α δ µ ≡ δ µβ = δ µβλ = − ϵαβλρ ϵµβλρ . 3 6 6 Show that: δ αβγµνλ

(3.13.6)

  +1 if αβγ is an even permutation of µνλ, = −1 if αβγ is an odd permutation of µνλ,   0 otherwise;

δ αβµν = δ αµ δ βν − δ αν δ βµ   +1 if αβ is an even permutation of µν, = −1 if αβ is an odd permutation of µν,   0 otherwise; { +1 if α = µ, δ αµ = 0 otherwise.

(3.13.7)

Solution: (a) If α, β, γ, δ are not all different, say α = β, we then have ϵαβγδ = −ϵβαγδ = ϵαβγδ .

36

(3.13.8)

Thus ϵαβγδ = 0 unless α, β, γ, δ are all different. If π is an even permutation, we have ϵπ0π1π2π3 = (−1)π ϵ0123 = 1.

(3.13.9)

similarly, for odd permutations of 0, 1, 2, 3, we have ϵπ0π1π2π3 = (−1)π ϵ0123 = −1.

(3.13.10)

(b) { ϵπ0π1π2π3 = g π0α g π1β g π2γ g π3δ ϵαβγδ =

det g = −1 if π is an even permutation − det g = 1 if π is an odd permutation

(3.13.11)

Therefore, ϵπ0π1π2π3 = −ϵπ0π1π2π3 .

(3.13.12)

(c) Firstly, we have ¯

¯ ¯

¯

ϵα¯ β¯γ δ = Λα¯ α Λβ β Λγ¯ γ Λδδ ϵαβγδ .

(3.13.13)

¯ γ¯ , δ¯ are not all different, say α If α ¯ , β, ¯ = β¯ = ¯0, we have ¯¯ ¯

¯

¯

¯

¯

¯

¯

¯

¯

¯

¯¯ ¯

ϵ00¯γ δ = Λ0α Λ0β Λγ¯ γ Λδδ ϵαβγδ = −Λ0β Λ0α Λγ¯ γ Λδδ ϵβαγδ = −Λ0α Λ0β Λγ¯ γ Λδδ ϵαβγδ = −ϵ00¯γ δ .

(3.13.14)

¯ ¯

¯ γ¯ , δ¯ are all different. If α ¯γ δ¯ is a permutation of ¯0¯1¯2¯3, we have Thus ϵα¯ β¯γ δ is 0 unless α ¯ , β, ¯ β¯ ¯ ¯

¯

¯

¯

¯

¯

¯

¯

¯

ϵα¯ β¯γ δ = Λπ0α Λπ1β Λπ2γ Λπ3δ ϵαβγδ = −(−1)π Λ0α Λ1β Λ2γ Λ3δ ϵαβγδ = −(−1)π det Λ.

(3.13.15)

Secondly, from the defination of Lorentz transformation, we have ΛT ηΛ = η. Take the determinant of the both side of the equation, we have det Λ det η det Λ = det η. (3.13.16) Thus det Λ = ±1. If e0 is pointing toward the future with e1 , e2 , e3 are a righthanded set. The transformation can be made up of successive infinitesimal boost or space rotation I + δω. Because the determinant of I + δω cannot be −1. We can conclude that det Λ = +1. Now, we have ¯ ¯ ¯ γ¯ , δ¯ are all different, ϵα¯ β¯γ δ = 0 unless α ¯ , β, { −1 for even permutations of ¯0, ¯1, ¯2, ¯3 and ¯ ¯ ¯ ¯ ϵπ0π1π2π3 = +1 for odd permutations.

(3.13.17)

¯ ¯

So ϵα¯ β¯γ δ have these same values in any other Lorentz frame with e0 pointing toward the future and with e1 , e2 , e3 a righthanded set. (d) In a Lorentz frame with past-pointing e0 or lefthanded e1 , e2 , e3 , we have det Λ = −1. So the components of ϵ are the opposite number of those in the original frame. (e) Firstly we focus on δ αβγµνλ ≡ −ϵαβγρ ϵµνλρ .

(3.13.18)

If α, β, γ or µ, ν, λ are not all different, ϵαβγρ or ϵµνλρ will be zero, and so δ αβγµνλ will be zero. If αβγ is not a permutation of µνλ, α, β, γ and µ, ν, λ are all different, then ρ cannot be different from α, β, γ and µ, ν, λ simutaneously. So δ αβγµνλ will be zero. If αβγ is an even permutation of µνλ, then αβγρ is also an even permutation of µνλρ if αβγρ (µνλρ) are all different (in this case, ρ has only one possible value). So − ϵαβγρ ϵµνλρ = −(±1)(∓1) = 1.

(3.13.19)

If αβγ is an odd permutation of µνλ, then αβγρ is also an odd permutation of µνλρ if αβγρ (µνλρ) are all different. So − ϵαβγρ ϵµνλρ = −(±1)(±1) = −1. (3.13.20)

37

As a result, we can get

δ αβγµνλ

  +1 if αβγ is an even permutation of µνλ, = −1 if αβγ is an odd permutation of µνλ,   0 otherwise.

(3.13.21)

Secondly, let us foucs on δ αβµν ≡

1 αβλ δ µνλ . 2

(3.13.22)

If αβ is not a permutation of µν, αβλ will not be a permutation of µνλ. So δ αβµν will be zero. If αβ is an even permutation of µν, then αβλ will be an even permutation of µνλ if α, β, λ and µ, ν, λ are all different. In this case, there are two possible values for λ and δ αβλµνλ = 2.

(3.13.23)

δ αβλµνλ = −2.

(3.13.24)

If αβ is an odd permutation of µν, we have

As a result, we can get

δ αβµν

  +1 if αβ is an even permutation of µν, = −1 if αβ is an odd permutation of µν,   0 otherwise.

(3.13.25)

δ αβµν = δ αµ δ βν − δ αν δ βµ is easy to verify according to the equation above. Finally, let us move to δ αµ ≡

1 αβ δ µβ . 3

(3.13.26)

If α ̸= µ, αβ can not be a permutation of µβ, and so δ αµ will be 0. If α = µ, αβ will be an even permutation of µβ if α, β and µ, β are all different. In this case, there are three possible values for β and δ αβµβ = 3. As a result, we can get

{ δ αµ =

3.14

(3.13.27)

+1 if α = µ, 0 otherwise.

(3.13.28)

DUALS

From any vector J , any second-rank antisymmetric tensor F and any third-rank antisymmetric tensor B, one can construct new tensors defined by ∗

Jαβγ = J µ ϵµαβγ ,

∗

Fαβ =

1 µν F ϵµναβ , 2

∗

Bα =

1 λµν B ϵλµνα 3!

(3.14.1)

One calls ∗J the “dual” of J , ∗F the dual of F , and ∗B the dual of B. [A previous and entirely distinct use of the word “dual” (section 2.7) called a set of basis one-forms ω α dual to a set of basIs vectors eα if ⟨ω α , eβ ⟩ = δ αβ . Fortunately there are no grounds for confusion between the two types of duality. One relates sets of vectors to sets of 1-forms. The other relates antisymmetric tensors of rank p to antisymmetric tensors of rank 4 − p.]

38

(a) Show that

∗∗

J = J,

∗∗

F = −F ,

∗∗

(3.14.2)

B = B.

so (aside from sign) one can recover any completely antisymmetric tensor H from its dual ∗H by taking the dual once again, ∗∗J . This shows that H and ∗H contain precisely the same information. (b) Make explicit this fact of same-information-content by writing out the components ∗Aαβγ in terms of Aα , also ∗F αβ in terms of F αβ , also ∗B α in terms of B αβγ .

Solution: (a) ∗∗ α

1∗ 1 1 J ϵλµνα = J β ϵβλµν ϵλµνα = − J β ϵβλµν ϵαλµν = J β δ αβ = J α ; 3! λµν 3! 6 1∗ 1 1 1 = Fµν ϵµναβ = F ρσ ϵρσµν ϵµναβ = F ρσ ϵρσµν ϵαβµν = − F ρσ δ αβρσ = −F αβ ; 2 4 4 2 1 1 1 = ∗Bµ ϵµαβγ = B ρσλ ϵρσλµ ϵµαβγ = − B ρσλ ϵρσλµ ϵαβγµ = B ρσλ δ αβγρσλ = B αβγ . 3 3! 3!

J =

∗∗

F αβ

∗∗

B αβγ

(b)

∗

A012 = A3 ,

∗

F 01 = −F 23 ,

∗

F 02 = F 13 , ∗

∗

A013 = −A2 ,

∗

F 03 = −F 12 ,

B 0 = B 123 ,

3.15

∗

∗

B 1 = B 023 ,

A023 = A1 ,

∗

F 12 = F 03 ,

∗

A123 = A0 .

∗

∗

F 13 = −F 02 ,

B 2 = −B 013 ,

(3.14.3)

(3.14.4) ∗

F 23 = F 01 . (3.14.5)

∗

B 3 = B 012 .

(3.14.6)

GEOMETRIC VERSIONS OF MAXWELL EQUATIONS

Show that, if F is the electromagnetic field tensor, then ∇ · ∗F = 0 is a geometric frame-independent version of the Maxwell equations Fαβ,γ + Fβγ,α + Fγα,β = 0. (3.15.1) Similarly show that ∇ · F = 4πJ (divergence on second slot of F ) is a geometric version of F αβ,β = 4πJ α .

Solution: From the defination of star operator, we have [∇ · ∗F ]α = ∗F αβ,β =

1 1 Fµν,β ϵµναβ Fµν,β ϵαµνβ . 2 2

(3.15.2)

So ∇ · ∗F = 0 is equal to the equation F[αβ,γ] = 0.

(3.15.3)

According to the conclusion of exercise 13(d), we can get the first pair of Maxwell equations, Fαβ,γ + Fβγ,α + Fγα,β = 0.

(3.15.4)

[∇ · F ]α = F αβ,β .

(3.15.5)

Similarly, we have Thus it is easy to see that ∇ · F = 4πJ is a geometric version of F αβ,β = 4πJ α .

39

3.16

CHARGE CONSERVATION

From Maxwell’s equations F αβ,β = 4πJ α , derive the “equation of charge conservation” J α,α = 0.

(3.16.1)

Show that this equation does, indeed, correspond to conservation of charge. It will be studied further in Chapter 5.

Solution: J α,α =

3.17

1 αβ 1 αβ 1 αβ F ,βα = [F ,βα + F βα,αβ ] = [F ,βα F αβ,αβ ] = 0. 4π 4π 4π

(3.16.2)

VECTOR POTENTIAL

The vector potential A of electromagnetic theory generates the electromagnetic field tensor via the geometric equation F = −(antisymmetric part of ∇A),

(3.17.1)

Fµν = Aν,µ − Aµ,ν .

(3.17.2)

i.e.,

(a) Show that the electric and magnetic fields in a specific Lorentz frame are given by B = ∇ × A,

E = −∂A/∂t − ∇A0 .

(3.17.3)

(b) Show that F will satisfy Maxwell’s equations if and only if A satisfies Aα,µ,µ − Aµ,µ,α = −4πJ α .

(3.17.4)

(c) Show that “gauge transformations” ϕ = arbitrary function,

ANEW = AOLD + dϕ,

(3.17.5)

leave F unaffected. (d) Show that one can adjust the gauge so that ∇·A=0

(“Lorentz gauge”),

□A = −4πJ .

(3.17.6)

Here □ is the wave operator (“d’ Alembertian ”): □A = Aα,µ,µ eα .

Solution:

40

(3.17.7)

(a) 1 ijk 1 ϵ Fjk = ϵijk [∂j Ak − ∂k Aj ] = ϵijk ∂j Ak , 2 2 E i = F 0i = −∂t Ai − ∂i A0 .

(3.17.8)

− F αµ,µ = −(Aµ,α − Aα,µ )µ = (Aα,µ,µ − Aµ,µ,α ) = −4πJ α .

(3.17.9)

Bi =

(b)

(c) µν µ,ν ν,µ µ,ν µν ,νµ ,µν FNEW = Aν,µ ) − (Aµ,ν ) = Aν,µ NEW − ANEW = (AOLD + ϕ OLD + ϕ OLD − AOLD = FOLD .

(3.17.10)

(d) If we want ∂µ AµNEW = ∂µ AµOLD + ∂µ ∂ µ ϕ = 0,

(3.17.11)

we should choose ϕ to satisfy the equation ∂µ ∂ µ ϕ = −∂µ AµOLD .

(3.17.12)

Aα,µ,µ − Aµ,µ,α = Aα,µ,µ = −4πJ α ,

(3.17.13)

□A = −4πJ .

(3.17.14)

In this case, we have i.e.

3.18

DIVERGENCE OF ELECTROMAGNETIC STRESS-ENERGY TENSOR

From an electromagnetic field tensor F , one constructs a second-rank, symmetric tensor T as follows: ( ) 1 1 µν µν µα ν αβ T = F F α − η Fαβ F . (3.18.1) 4π 4 As an exercise in index gymnastics: (a) Show that ∇ · T has components T µν,ν

[ ] 1 1 ,µ αβ µα ν µα ν F ,ν F α + F F α,ν − Fαβ F = . 4π 2

(b) Manipulate this expression into the form [ ] 1 1 αβ ν αν Tµ ,ν = −Fµα F ,ν − F (Fαβ,µ + Fµα,β + Fβµ,α ) ; 4π 2

(3.18.2)

(3.18.3)

note that the first term of (123) arises directly from the second term of (122). (c) Use Maxwell’s equations to conclude that T µν,ν = −F µα Jα .

41

(3.18.4)

Solution: (a) ( 1 F µα,ν F να 4π ( 1 = F µα,ν F να 4π ( 1 = F µα,ν F να 4π ( 1 = F µα,ν F να 4π

T µν,ν =

1 1 + F µα F να,ν − η µν Fαβ,ν F αβ − η µν Fαβ F αβ,ν 4 4 ) 1 1 + F µα F να,ν − Fαβ ,µ F αβ − Fαβ F αβ,µ 4 4 ) 1 1 + F µα F να,ν − Fαβ ,µ F αβ − F αβ Fαβ ,µ 4 4 ) 1 + F µα F να,ν − Fαβ ,µ F αβ . 2

)

(3.18.5)

(b) Tµ ν,ν

= = = = = =

1 4π 1 4π 1 4π 1 4π 1 4π 1 4π

(

) 1 αβ + Fµ − Fαβ,µ F 2 ( ) 1 να α ν αβ Fµα,ν F + Fµ F α,ν − Fαβ,µ F 2 ( ) 1 αβ α ν αβ −Fµα,β F + Fµ F α,ν − Fαβ,µ F 2 ( ) 1 1 1 αβ βα α ν αβ − Fµα,β F − Fαµ,β F + Fµ F α,ν − Fαβ,µ F 2 2 2 ( ) 1 1 1 αβ αβ α ν αβ − Fµα,β F − Fβµ,α F + Fµ F α,ν − Fαβ,µ F 2 2 2 [ ] 1 −Fµα F αν,ν − F αβ (Fαβ,µ + Fµα,β + Fβµ,α ) . 2 Fµ α,ν

F να

α

F να,ν

(3.18.6)

(c) We have the Maxwell equations: F αν,ν = 4πJ α ,

Fαβ,µ + Fµα,β + Fβµ,α = 0.

(3.18.7)

Thus Tµ ν,ν = −Fµα J α ,

(3.18.8)

T µν,ν = −F µα Jα .

(3.18.9)

i.e.

42

Chapter 4

ELECTROMAGNETISM AND DIFFERENTIAL FORMS 4.1 GENERIC LOCAL ELECTROMAGNETIC FIELD EXPRESSED IN SIMPLEST FORM In the laboratory Lorentz frame, the electric field is E, the magnetic field B. Special cases are: (1) pure electrIc field (B = 0); (2) pure magnetic field (E = 0); and (3) “radiation field” or “null field” (E and B equal in magnitude and perpendicular in direction). All cases other than (1), (2), and (3) are “generic”. In the generic case, calculate the Poynting density of flow of energy E × B/4π and the density of energy (E 2 + B 2 )/8π. Define the direction of a unit vector n and the magnitude of a velocity parameter α by the ratio of energy flow to energy density: n tanh 2α =

2E × B . E2 + B2

(4.1.1)

View the same electromagnetic field in a rocket frame moving in the direction of n with the velocity parameter α (not 2α; factor 2 comes in because energy flow and energy density are components, not of a vector, but of a tensor). By employing the formulas for a Lorentz transformation (equation 3.23), or ¯ and B ¯ otherwise, show that the energy flux vanishes in the rocket frame, with the consequence that E ¯ and B. ¯ Show are parallel. No one can prevent the z-axis from being put in the direction common to E that with thiS choice of direction, F araday becomes ¯z d¯ ¯z d¯ F =E z ∧ dt¯ + B x ∧ d¯ y

(4.1.2)

(only two wedge products needed to represent the generic local field; “canonical representation”; valid in one frame, valid in any frame).

Solution: By employing the formulas for a Lorentz transformation, we have ¯ = γ(E⊥ + β × B) + E∥ E = cosh α[E − (E · n)n + tanh α n × B] + (E · n)n = cosh α[E + tanh α n × B] = cosh α E + sinh α n × B.

43

(4.1.3)

¯ = γ(B⊥ − β × E) + B∥ B = cosh α[B − (B · n)n − tanh α n × E] + (B · n)n = cosh α[B − tanh α n × E] = cosh α B − sinh α n × E.

(4.1.4)

The cross product of electric field and magnetic field in the rocket frame is ¯ ×B ¯ = (cosh α E + sinh α n × B) × (cosh α B − sinh α n × E) E = cosh2 α E × B + sinh α cosh α[(n × B) × B + (n × E) × E] − sinh2 α(n × B) × (n × E). (4.1.5) Note thispagestyle (n × B) × B = (B · n)B − (B · B)n = −B 2 n; (n × E) × E = (E · n)E − (E · E)n = −E 2 n; (n × B) × (n × E) = [(n × B) · E]n − [(n × B) · n]E = [(B × E) · n]n =−

E2 + B2 tanh 2α n. 2

(4.1.6)

Finally we can get 2 2 2 2 ¯ ×B ¯ = E + B cosh2 α tanh 2α n − (E 2 + B 2 ) sinh α cosh α n + E + B sinh2 α tanh 2α n E 2 2 2 2 E +B = (cosh2 α tanh 2α + sinh2 α tanh 2α − 2 sinh α cosh α)n 2 E2 + B2 (cosh 2α tanh 2α − sinh 2α)n = 2 = 0. (4.1.7)

¯ and B ¯ are parallel in the rocket frame. If we put z-axis in the direction common to E ¯ and B, ¯ we Thus E have ¯z ; F12 = −F21 = B ¯z , F30 = F¯03 = E (4.1.8) and all other components vanish. So ¯z d¯ ¯z d¯ F =E z ∧ dt¯ + B x ∧ d¯ y.

(4.1.9)

4.2 FREEDOM OF CHOICE OF 1-FORMS IN CANONICAL REPRESENTATION OF GENERIC LOCAL FIELD Deal with a region so small that the variation of the field from place to place can be neglected. Write Faraday in canonical representation in the form F = dpI ∧ dq I + dpII ∧ dq II ,

(4.2.1)

where pA (A = I or II) and q A are scalar functions of position in spacetime. Define a “canonical transformation” to new scalar functions of position pA and qA by way of the “equation of transformation” ¯

pA dq A = dS + pA¯ dq A ,

(4.2.2) ¯

where the “generating function” S of the transformation is an arbitrary function of the q A and the q A : ¯

¯

dS = (∂S/∂q A )dq A + (∂S/∂q A )dq A .

44

(4.2.3)

(a) Derive expressions for the two pA ’s and the two pA¯ ’s in terms of S by equating coefficients of dq I , dq II , ¯ ¯¯ dq I , dq I I individually on the two sides of the equation of transformation. ¯ (b) Use these expressions for the pA ’s and pA¯ ’s to show that F = dpA ∧ dq A and F¯ = dpA¯ ∧ dq A , ostensibly different, are actually expressions for one and the same 2-form in terms of alternative sets of 1-forms.

Solution: (a) From the definition of S, we have ¯

¯

¯

¯

pA dq A = dS + pA¯ dq A = (∂S/∂q A )dq A + (∂S/∂q A )dq A + pA¯ dq A . Thus,

¯

(4.2.5)

¯¯

(4.2.6)

pA¯ = −∂S/∂q A .

pA = ∂S/∂q A ;

(4.2.4)

i.e. pI = ∂S/∂q I ;

¯

pI¯ = −∂S/∂q I

pI¯I¯ = −∂S/∂q I I .

pII = ∂S/∂q II ; (b)

( F = dpA ∧ dq A = d ¯ F¯ = dpA¯ ∧ dq A = −d

(

∂S ∂q A ∂S ∂q A¯

) ∧ dq A = )

∂2S ¯ dq A ∧ dq A . ∂q A ∂q A¯

¯

∧ dq A = −

∂2S ¯ dq A ∧ dq A . ∂q A ∂q A¯

(4.2.7) (4.2.8)

So we have F = F¯ .

4.3 A CLOSED OR CURL-FREE 1-FORM IS A GRADIENT Given a 1-form a such that dσ = 0, show that σ can be expressed in the form σ = df , where f is some scalar. The 1-form σ is said to be “curl-free,” a narrower category of 1-form than the “rotation-free” 1-form of the next exercise (expressible as σ = hdf ), and it in turn is narrower (see Figure 4.7) than the category of “1-forms with rotation” (not expressible in the form σ = hdf ). When the 1-form σ is expressed in terms of basis 1-forms dxα , multiplied by corresponding components σα , show that “curl-free” implies σ[α,β] = 0.

Solution: Suppose σ = σα dxα . Then dσ = 0 implies that dσ = σα,β dxβ ∧ dxα = 0,

(4.3.1)

σ[α,β] = 0.

(4.3.2)

i.e. From the theory of PDE, we can find a solution to the equations σα =

∂f ∂xα

(4.3.3)

if and only if σα,β = σβ,α .

(4.3.4)

Thus given a 1-form a such that dσ = 0, σ can be expressed in the form σ = df , where f is some scalar.

45

4.4 CANONICAL EXPRESSION FOR A ROTATION-FREE 1-FORM In three dimensions a rigid body turning with angular velocity ω about the z-axis has components of velocity νy = ωx, and νx = −ωy. The quantity curl ν = ∇ × ν has z-component equal to 2ω, and all other components equal to zero. Thus the scalar product of ν and curl ν vanishes: ν[i,j νk] = 0.

(4.4.1)

The same concept generalizes to four dimensions, ν[α,β νγ] = 0,

(4.4.2)

and lends itself to expression in coordinate-free language, as the requirement that a certain 3-form must vanish: dν ∧ ν = 0. (4.4.3) Any 1-form ν satisfying this condition is said to be “rotation-free.” Show that a 1-form is rotation-free if and only if it can be written in the form ν = hdf, (4.4.4) where h and f are scalar functions of position (the “Frobenius theorem”).

Solution: If ν = hdf , we have dν ∧ ν = hdh ∧ df ∧ df + h2 d2 f ∧ df = 0.

(4.4.5)

The condition ν ∧ dν = 0 on ν implies that its kernel V = {u ∈ T M : ν(u) = 0}

(4.4.6)

is an involutive distribution, meaning it is closed under the Lie bracket. To see this, let X, Y be smooth vector fields in V (i.e. such that ν(X) = ν(Y ) = 0) and compute 0 = ν ∧ dν(X, Y, Z) = ν(Z) ∧ dν(X, Y ) = ν(Z) ∧ ν([X, Y ]);

(4.4.7)

so choosing Z such that ν(Z) vanishes only where ν does we conclude that ν([X, Y ]) = 0. The Frobenius theorem tells us that an involutive distribution is tangent to a foliation by submanifolds. The idea of the proof is to construct a basis of the distribution consisting of pairwise commuting vector fields, so that the flows of these vector fields will provide natural adapted coordinates. Once we know V is tangent to a foliation by submanifolds, the local version of the claim follows: in adapted coordinates xi such that the surfaces x0 = constants are tangent to V , we know that ν(∂/∂xi ) for i ̸= 0; so ν must have component expansion ν = ν0 dx0 . Thus h = ν0 , f = x0 does the job. This proof is provided by Anthony Carapetis. You can find the original answer on the StackExchange.

4.5 FORMS ENDOWED WITH POLAR SINGULARITIES List the principal results on how such forms are representable, such as Φ1 =

dS ∧ Ψ 1 + θ1 , S

(4.5.1)

and the conditions under which each applies [for the meaning and answer to this exercise, see Lascoux (1968)].

46

Solution: This exercise is a proposition in the paper Perturbation Theory in Quantum Field Theory and Homology by Jean Lascoux (1968). The paper is included in the book Battelle Rencontres, 1967 Lectures in Mathematics and Physics, edited by Cecile M. DeWitt and John A. Wheeler. The complete statement of the proposition is as follows. Let X be a nonsingular complex analytic manifold dim X = l. Introduce local coordinates x = (x1 , x2 , · · · , xl ) for the complex structure, but for the real C ∞ -differentiable structure, we shall use x = (x1 , x ¯ 1 , · · · , xl , x ¯l ). We shall note, for all complex manifolds of codimension 1, their local irreducible equation as follows: S, in the neighborhood of y ∈ S, is given by s(x, y) = 0. Let (Si )i = 1, · · · ,∪m be such that dsi ̸= 0 for x = y and we shall say that (S1 , · · · , Sm ) are in general position at y ∈ SI = i∈I Si if the dsi are linearly independent. By a regular function, one means a function which is C ∞ -differentiable over X. A differential form ϕ(x) is regular if it has regular coefficients. If dϕ = 0, ϕ is closed. ϕ has a polar singularity of order p along S if ϕ(x) is regular on X − S and if there exists a number p such that s(u, x)p ϕ(x) is regular on X. Now, let us define the residue form. Let dϕ = 0 on X − S, let ϕ have a polar singularity of order p along S. Then ds ∧ψ+θ S where the restriction of the differential form ψ on S, ψ|S is closed. ϕ=

(4.5.2)

The proof of the proposition is too technical and out of my scope. So please refer to the original paper of Lascoux for more information.

4.6 THE FIELD OF THE OSCILLATING DIPOLE Verify that the expressions given for the electromagnetic field of an oscillating dipole in equations (4.23) and (4.24) satisfy dF = 0 everywhere and d ∗F everywhere except at the origin.

Solution: [ ( ) 1 iω dF = p1 e −2 sin θ − 2 dθ ∧ dr ∧ dt r3 r ( ) 2 2 2iω ω 3 + sin θ − 3 + 2 + − iω dr ∧ dθ ∧ dt r r r ( ) ] ω2 3 + sin θ − + iω dt ∧ dr ∧ dθ r iωr−iωt

(4.6.1)

= 0.

[ ( ) ( ) iω ω 2 −iω 2 3 2 + d F = p1 e sin θ − iω dr ∧ dt ∧ dϕ + 2 sin θ cos θ − ω dθ ∧ dt ∧ dϕ r2 r r ( ) ( ) 1 iω iω + 2 sin θ cos θ − 2 + + ω 2 dr ∧ dθ ∧ dϕ + 2 sin θ cos θ − − ω 2 dt ∧ dθ ∧ dϕ r r r ) ( ) ] ( 2 ω iω 1 iω 2 3 2 − + iω dt ∧ dϕ ∧ dr + 2 sin θ cos θ − ω dθ ∧ dϕ ∧ dr + sin θ − 2 − r r r2 r ∗

iωr−iωt

(4.6.2)

= 0.

47

4.7 THE 2-FORM MACHINERY TRANSLATED INTO TENSOR MACHINERY This exercise is stated at the end of the legend caption of Figure 4.1.

Solution: For general F , we have

∑

F =

Fij dxi ∧ dxj .

(4.7.1)

i 0.

(25.15.1)

The proper time to fall to r = 2M is ∫

( )−1/2 dτ 2 ˜ 2 − V˜max dr ≤ E (R − 2M ), dr

R

∆τ = 2M

(25.15.2)

which is finite. On the other hand, we also have ( ) ∗ 2 ˜ dr ˜ 2 − V˜ 2 (r∗ ) < E ˜2. E =E dt

(25.15.3)

The coordinate time to fall to r = 2M (r∗ = −∞) is ∫

R∗

dt ∗ dr ≥ dr∗

∆t = ∞

25.16

∫

R∗

∞

dr∗ → ∞.

(25.15.4)

PERIASTRON SHIFT FOR NEARLY CIRCULAR ORBITS

Rewrite equation (25.42) in the form ˜2 − E ˜02 )/L†2 (du/dϕ)2 + (1 − 6u0 )(u − u0 )2 − 2(u − u0 )3 = (E

(25.16.1)

˜ ˜0 in terms of u0 . Show for a nearly Express the constant u0 ≡ M/r0 in terms of L† ≡ L/M , and express E circular orbit of radius r0 that the angle swept out between two successive periastra (points of closest approach to the star) is ∆ϕ = 2π(1 − 6M/r0 )−1/2 . (25.16.2) Sketch the shape of the orbit for r0 = 8M .

Solution: First, we have equation (25.42) (

du dϕ

)2 =

˜ 2 − (1 − 2u)(1 + L†2 u2 ) E . L†2

(25.16.3)

Take the derivative of both sides and we have L†2

d2 u = (1 + L†2 u2 ) + (2u − 1)L†2 u. dϕ2

252

(25.16.4)

So for circular orbits, we know and

˜02 − (1 − 2u0 )(1 + L†2 u20 ) = 0 E

(25.16.5)

3L†2 u20 − L†2 u0 + 1 = 0.

(25.16.6)

Thus, we find u0 = and

L† −

√

L†2 − 12 6L†

(25.16.7)

˜ = √1 − 2u0 E 1 − 3u0

(25.16.8)

We can verify that ˜2 − E ˜2 ˜ 2 − (1 − 2u)(1 + L†2 u2 ) E E 0 = †2 †2 L L ˜2 ˜2 − E E 0 = †2 L ˜2 − E ˜2 E 0 = L†2 ˜2 − E ˜2 E 0 = L†2 ˜2 − E ˜2 E 0 = †2 L i.e.

(1 − 2u0 )(1 + L†2 u20 ) − (1 − 2u)(1 + L†2 u2 ) L†2 2(u − u0 ) + + u20 − u2 + 2(u3 − u30 ) L†2 +

+ 2(u0 − 3u20 )(u − u0 ) + u20 − u2 + 2(u3 − u30 ) + 2u3 − u2 + 2u0 (1 − 3u0 )u + 4u30 − u20 + 2(u − u0 )3 − (1 − 6u0 )(u − u0 )2 ,

˜2 − E ˜ 2 )/L†2 . (du/dϕ)2 + (1 − 6u0 )(u − u0 )2 − 2(u − u0 )3 = (E 0

(25.16.9) (25.16.10)

Take the derivative of both sides, we would arrive at

If |u − u0 | ≪ 1, we have

d2 u + (1 − 6u0 )(u − u0 ) − 3(u − u0 )2 = 0. dϕ2

(25.16.11)

d2 (u − u0 ) + (1 − 6u0 )(u − u0 ) = 0. dϕ2

(25.16.12)

u = u0 + ϵ sin(ωt + ϕ0 ),

(25.16.13)

The solution is where ω = 1 − 6u0 . Thus the angle swept out between two successive periastra is 2

∆ϕ = 2π(1 − 6M/r0 )−1/2 .

25.17

(25.16.14)

ANGULAR MOTION DURING INFALL

From equation (25.42), deduce that the total angle ∆ϕ swept out on a trajectory falling into r = 0 is finite. The computation is straightforward; but the interpretation, in view of the behavior of t(λ) on the same trajectory (equation 25.32 and exercise 25.15), is not. The interpretation will be elucidated in Chapter 31.

Solution: ˜ 2 > V˜ 2 , i.e. If the particle can fall into r = 0, then we have E max ( )2 2 ˜ du E − V˜max (u) ≥ . dϕ L†2

253

(25.17.1)

And when u > 1, we have

− (1 − 2u)(1 + L†2 u2 ) > L†2 u3

i.e.

So

(

∫

∞

∆ϕ = µ0

dϕ du < du

∫

∞

−3/2

u

∫

1

du +

1

du dϕ

(25.17.2)

)2 > u3

(25.17.3)

˜ 2 − V˜max (u)]−1/2 du = 2 + L† [E ˜ 2 − V˜max (u)]−1/2 (1 − µ0 ), L† [E

µ0

(25.17.4) i.e. the total angle ∆ϕ swept out on a trajectory falling into r = 0 is finite.

25.18

MAXIMUM AND MINIMUM OF EFFECTIVE POTENTIAL

Derive the expressions given in the caption of Figure 25.2 for the locations of the maximum and the minimum of the effective potential as a function of angular momentum. Determine also the limiting form ˜ is very large compared of the dependence of barrier height on angular momentum in the limit in which L to M .

Solution: From the expression of V˜ (r), we have ( ) ( ) ˜2 ˜2 ˜ 2 r + 6M L ˜2 dV˜ 2 2M L 2M L 2M r2 − 2L = 2 1+ 2 −2 1− = . 3 4 dr r r r r r So the rextrem s satisfy that The solution is rextrem =

˜2 ± L

√

2 ˜ 2 rextrem + 3M L ˜ 2 = 0. M rextrem −L

˜ 4 − 12M 2 L ˜2 L 1± = 2M

√

6M 1 − 12L†−2 √ M L†2 = . 2 1 ∓ 1 − 12L†−2

Further more, the second derivative of V˜ 2 (r) at rextrem is √ ˜ 4 − 12M 2 L ˜2 4M rextrem − 2L†2 ±2 L d2 V˜ 2 = = . 4 4 2 dr rextrem rextrem

(25.18.1)

(25.18.2)

(25.18.3)

(25.18.4)

rextrem

So the maximum of the V˜ (r) is at rmax =

˜2 − L

√

˜ 4 − 12M 2 L ˜2 L 6M √ = , 2M 1 + 1 − 12L†−2

(25.18.5)

while the minimum of the V˜ (r) is at rmin = The barrier height is 2 V˜max

1−

√

6M . 1 + 12L†−2

) ( )( ˜2 2M L = 1− 1+ 2 rmax rmax ( )( ) √ √ 1 + 1 − 12L†−2 1 − 6L†−2 + 1 − 12L†−2 †2 = 1− 1+ L 3 18 √ L†2 + 36 + (L†2 − 12) 1 − 12L†−2 . = 54

254

(25.18.6)

(25.18.7)

In the limit L† → ∞, we have ( ) L†2 + 36 + (L†2 − 12) 1 − 6L†−2 − 18L†−4 L†2 + 9 + 27L†−2 2 V˜max ≈ ≈ . 54 27

25.19

(25.18.8)

KEPLER LAW VALID FOR CIRCULAR ORBITS

From dϕ/dτ of (25.17) and dt/dτ of (25.18), deduce an expression for the circular frequency of revolution as seen by a faraway observer; and from the results of exercise 25.18 (or otherwise) show that it fulfills exactly the Kepler relation ω2 r3 = M

(25.19.1)

for any circular orbit of Schwarzschild r-value equal to r, whether stable (potential minimum) or unstable (potential maximum).

Solution: From equation (25.17) and (25.18), we have ω=

˜ L dϕ = ˜ dt r2 E

( 1−

2M r

) =

1 L† ˜ M x2 E

( 1−

2 x

) ,

(25.19.2)

where x ≡ r/M . For circular orbits, we have ( ˜2 = E

2 1− x

)(

L†2 1+ 2 x

) (25.19.3)

and x2 − L†2 x + 3L†2 = 0,

(25.19.4)

x2 + L†2 = L†2 x − 2L†2 .

(25.19.5)

i.e.

Thus we find ω 2 r 3 = ω 2 x3 M 3 =

25.20

L†2 (x − 2) M = M. x2 + L†2

(25.19.6)

HAMILTON-JACOBI FUNCTION

˜ r, θ) = 0 for t = 0 and for Construct the locus in the r, θ diagram dynamic phase S(t, √of points of constant 1/2 ˜ ˜ ˜ ˜ values L = 4M , E = 1 (or for L = 2 3M , E = (8/9) , or for some other equally simple set of values for these two parameters). Show that the whole set of surfaces of constant S˜ can be obtained by rotating the foregoing locus through one angle, then another and another, and recopying or retracing. Interpret physically the principal features of the resulting pattern of curves. The meaning of the exercise is not very clear. Any help is appreciated.

255

25.21

DEFLECTION BY GRAVITY CONTRASTED WITH DEFLECTION BY ELECTRIC FORCE

A test particle of arbitrary velocity β flies past a mass M at an impact parameter b so great that the deflection is small. Show that the deflection is θ=

2M (1 + β 2 ) bβ 2

(25.21.1)

Derive the deflection according to Newtonian mechanics for a particle moving with the speed of light. Show that (25.49) in the limit β → 1 is twice the Newtonian deflection. Derive also (flat-space analysis) the contrasting formula for the deflection of a fast particle of rest mass µ and charge e by a nucleus of charge Ze, 2Ze2 θ= (1 − β 2 )1/2 . (25.21.2) µbβ 2 How feasible is it to rule out a “vector” theory of gravitation [see, for example, Brillouin (1970)], patterned after electromagnetism, by observations on the bending of light by the sun?

Solution: In Schwarzschild metric, the orbit of a test particle satisfies the equation (

du dϕ

)2 =

˜ 2 − (1 − 2u)(1 + L†2 u2 ) E , L†2

(25.21.3)

where u ≡ M/r. Differentiate once with respect to ϕ to convert into a second-order equation. Rearrange to put on the left all those terms that would be there in the absence of gravity, and on the right all those that originate from the −2u term (gravitation) in the factor (1 − 2u). Now we have d2 u 1 + u = 3u2 + †2 . 2 dϕ L

(25.21.4)

Neglecting the right-hand side of the equation, we have the straight line solution u = A sin(ϕ − ϕ0 ).

(25.21.5)

By redefine the starting point of ϕ, the solution can be simplified into u = A sin ϕ. When ϕ = 0 or π, we ˜ 2 − 1)/L†2 . find that u = 0 and du/dϕ = ±A. From equation √ 25.21.3, we also find that (du/dϕ)2 = (E ˜ 2 − 1b/M . So the straight line solution is According to equation 25.14.1, we also have L† = E u=

M sin ϕ. b

(25.21.6)

Evaluate the perturbing term ∆ ( of the order b−2 ) on the right as a function of ϕ by inserting in it the unperturbed expression for u(ϕ). We have 3M 2 M2 d2 ∆ 2 + ∆ = sin ϕ + . ˜ 2 − 1) dϕ2 b2 b2 ( E The solution is ∆=

M2 M2 2 (2 − sin ϕ) + . ˜ 2 − 1) b2 b2 ( E

(25.21.7)

(25.21.8)

At the infinity, we have u = 0 and so M M2 M2 = 0. sin ϕ + 2 (2 − sin2 ϕ) + ˜ 2 − 1) b b b2 (E

256

(25.21.9)

The solution of sin ϕ to the order of b−1 is sin ϕ = −

M (1 + β 2 ) 2M M − =− . ˜ 2 − 1) b bβ 2 b(E

(25.21.10)

So ϕ = −M (1 + β 2 )/bβ 2 and π + M (1 + β 2 )/bβ 2 . And the deflection angle is ∆ϕ =

2M (1 + β 2 ) . bβ 2

(25.21.11)

4M . b

(25.21.12)

In the limit of β → 1, we have ∆ϕ = In Newtonian mechanics, the orbit is hyperbolic, i.e.

p = 1 + e sin ϕ, r

(25.21.13)

where p = b2 β 2 /M and e = (1 + b2 β 4 /M 2 )1/2 . At the infinity, we have sin ϕ = −

1 M = − 2 + O(b−2 ). e bβ

(25.21.14)

And the deflection angle is ∆ϕ =

2M . bβ 2

(25.21.15)

∆ϕ =

2M . b

(25.21.16)

In the limit of β → 1, we have

To the zeroth order, the orbit of a fast particle of rest mass µ and charge e in the field of a nucleus of charge Ze is a straight line. When it passes the nucleus, it will gain momentum in the direction perpendicular to the line. And we have ∫ ∞ ∫ ∞ γZe2 b 2γZe2 µ dpy dτ µ dx = dx = . (25.21.17) ∆py = 2 2 3/2 p bpx x −∞ (b + x ) −∞ dτ dx So the deflection angle is ∆ϕ =

py 2γZe2 µ 2Ze2 = = (1 − β 2 )1/2 . 2 px bpx µbβ 2

(25.21.18)

In the limit of β → 1, we have ∆ϕ = 0, i.e. the orbit of the particle is not bent. So we can rule out a “vector” theory of gravitation patterned after electromagnetism, by observations on the bending of light by the sun.

25.22

CAPTURE BY A BLACK HOLE

Over and above any scattering of particles by a black hole, there is direct capture into the black hole. Show that the cross section for capture is πb2crit with the critical impact parameter bcrit given by Lcrit /(E 2 −µ2 )1/2 . From the formulas in the caption of Fig. 25.2 or otherwise, show that for high-energy particles this cross section varies with energy as ( ) 2 σcapt = 27πM 2 1 + + ··· (25.22.1) ˜2 3E ˜ → ∞) and for low energies as (photon limit for E σcapt = 16πM 2 /β 2 ,

257

(25.22.2)

where β is the velocity relative to the velocity of light [Bogorodsky (1962)].

Solution: From equation 25.18.7, the barrier height of the potential is √ L†2 + 36 + (L†2 − 12) 1 − 12L†−2 2 V˜max = . 54

(25.22.3)

˜ > 1. If the particle is captured by black hole, we have The energy of incident particles must satisfy that E √ L†2 + 36 + (L†2 − 12) 1 − 12L†−2 2 ˜2 V˜max = ≤E (25.22.4) 54 √ †2 † † †−2 † ˜2 ˜2 Define L†crit by L†2 crit + 36 + (Lcrit − 12) 1 − 12Lcrit = 54E . If L < Lcrit , i.e. b < Lcrit M/(E − 1) ≡ bcrit , the particle will be captured by black hole. The cross section is σcrit = πb2crit =

L†2 crit πM 2 . ˜2 − 1 E

And

√ ˜ 2 − 1)x + 36 + [(E ˜ 2 − 1)x − 12] (E

1−

12 ˜2, = 54E 2 ˜ (E − 1)x

(25.22.5)

(25.22.6)

˜ → ∞, the zeroth order approximation of the x is 27. To linear where x ≡ σcrit /πM 2 . In the limit of E order, we have √ 12 2 2 ˜ ˜ ˜2, E x − 27 + 36 + (E x − 27 − 12) 1 − = 54E (25.22.7) ˜2 27E The solution is 18 x = 27 + +O ˜2 E i.e.

(

1 ˜ E4

) ,

( ) 2 σcrit = 27πM 2 1 + + ··· . ˜2 3E

In the low energy limit, we have

√ 12 y + (y − 12) 1 − = 18, y

(25.22.8)

(25.22.9)

(25.22.10)

˜ 2 − 1)x. The solution is y = 16. As E ˜ 2 − 1 = γ 2 − 1 = 1/(1 − β 2 ) − 1 ≈ β 2 , we have where y ≡ (E σcapt = 16πM 2 /β 2 .

25.23

(25.22.11)

QUALITATIVE FORMS OF PHOTON ORBITS

Verify all the statements about orbits for particles of zero rest mass made in Box 25.7. The proof for each statements about photon orbits made in part Box 25.7 is trivial. I will not write down detailed demonstration for all of them here.

258

25.24

LIGHT DEFLECTION

Using the dimensionless variable u = M/r in place of r itself, and ub = M/b in place of the impact parameter, transform (25.55) into the first-order equation ( )2 du + (1 − 2u)u2 = u2b (25.24.1) dϕ and thence, by differentiation, into d2 u + u = 3u2 . (25.24.2) dϕ2 (a) In the large-impact-parameter or small-u approximation, in which the term on the right is neglected, show that the solution of 25.24.2 yields elementary rectilinear motion (zero deflection). (b) Insert this zero-order solution into the perturbation term 3u2 on the righthand side of 25.24.2, and solve anew for u (“rectilinear motion plus first-order correction”). In this way, verify the formula for the bending of light by the sun given by putting β = 1 in equation 25.21.1.

Solution: (a) Neglecting the right-hand side of the equation 25.24.2, we have the straight line solution u = A sin(ϕ − ϕ0 ).

(25.24.3)

By redefine the starting point of ϕ, the solution can be simplified into u = A sin ϕ. When ϕ = 0 or π, we find that u = 0 and du/dϕ = ±A. According to equation 25.24.1, we also have du/dϕ = ub . So the straight line solution is u = ub sin ϕ. (25.24.4) (b) Inserting the zero-order solution into the perturbation term 3u2 on the righthand side of 25.24.2, we have d2 ∆ + ∆ = 3u2b sin2 ϕ, (25.24.5) dϕ2 where ∆ = u − ub sin ϕ. The solution is ∆ = u2b (2 − sin2 ϕ),

(25.24.6)

i.e. u = ub sin ϕ + − sin ϕ). When u = 0, we have sin ϕ ≈ −2ub , and so ϕ ≈ −2ub or π + 2ub . The deflection angle is 4ub = 4M/b. u2b (2

25.25

2

CAPTURE OF LIGHT BY A BLACK HOLE

Show that a Schwarzschild black hole presents a cross section σcapt = 27πM 2 for capture of light.

Solution: From equation (25.66),

(

dr dλ

)2 +

1 1 − 2M/r = 2, r2 b

the effective potential is

(25.25.1)

1 − 2M/r . (25.25.2) V˜ 2 (r) = r2 The maximum of V˜ 2 (r) is at r = 3M and equals 1/27M 2 . If the photon is captured by the black hole, we must have 1/b2 > 1/27M 2 , i.e. b < 33/2 M . So the cross section is 27πM 2 .

259

25.26

RETURN OF LIGHT FROM A BLACK HOLE

Show that flashing a powerful pulse of light onto a black hole leads in principle to a return from rings of brightness located at b/M − 33/2 = 0.151, 0.00028, · · · How can one evaluate the difference in time delays of these distinct returns? Show that the intensity I of the return (erg/cm2 ) as a function of the energy E0 (erg/steradian) of the original pulse, the mass M (cm) of the black hole, the distance R to it, and the lateral distance r from the “flashlight” to the receptor of returned radiation is ∑ 2bdb E0 M 2 E0 I= 3 (25.26.1) dΘ = R3 r (1.75 + 0.0029 + 0.0000055 + · · · ) R r Θ=(2N +1)π

under conditions where diffraction can be neglected.

Solution: From equation 25.24.1, the deflection angle of the photon is ∫ u0 ∫ u0 dϕ 1 √ Θ=2 du − π = 2 − π, 2 du ub − (1 − 2u)u2 0 0

(25.26.2)

where (1 − 2u0 )u20 = u2b . With the help of WolframAlpha, we have √ √ ( ) u0 ∫ u0 1 2v0 2 2u0 + 4u + v0 − 1 −1 (25.26.3) √ F sin =2 2 2 v + 6u − 1 2v 6u + v − 1 ub − (1 − 2u)u 0 0 0 0 0 0 0 √ where v0 ≡ −12u20 + 4u0 + 1 and F (ϕ|m) is the incomplete elliptic integral of the first kind. When b/M − 33/2 √ is small, u2b − 1/27 is also small, as well as 1/3 − u0 , v0 − 1, 1 − (6u0 + v0 − 1)/2v0 and −1 π/2 − sin (6u0 + v0 − 1)/(2v0 ). Note that F (π/2, 1) is infinite. So we have the approximation ( √ ) ( ) Θ+π 1 1 −1 −1 ≈ 2F sin (1 − δ) (25.26.4) − 2F sin 1 , 2 (1 − δ)2 3 where δ = 1 −

we have

√

(6u0 + v0 − 1)/(2v0 ). Using the expansion ) ( ) ( 1 δ 1 −1 = − ln + O(δ), F sin (1 − δ) (1 − δ)2 2 8 (

Θ+π δ ≈ 8 exp − − 2F 2

( −1

sin

(25.26.5)

√ )) ( ) 1 Θ 1 ≈ 0.44561 exp − 3 2

On the other hand, we have the expansion [( √ ( ) )2 ] 1 6u0 + v0 − 1 1 +O u0 − 1−δ = = 1 + 2 u0 − 2v0 3 3 and 39/2 b 1 1 3/2 = 3 + = =√ M ub 2 (1 − 2u0 )u20

(

1 u0 − 3

[(

)2 +O

1 u0 − 3

(25.26.6)

(25.26.7)

)3 ] .

(25.26.8)

So we have

b 39/2 2 ≈ 33/2 + δ ≈ 33/2 + 3.482e−Θ M 8 If the photon returns, we have Θ = (2N + 1)π. And so b − 33/2 = 0.150, 2.81 × 10−4 , 5.25 × 10−7 , · · · M I do not know how to answer the second and third question. Any help is appreciated.

260

(25.26.9)

(25.26.10)

25.27

ISOTROPIC STAR CLUSTER

For a cluster with distribution function independent of angular momentum, derive properties B.1 to B.6 of Box 25.8.

Solution: The statement 1 and 2 is trivial. If F = F (E, µ), we have ∫

ˆ

ˆ

(p0 )2 pT Tˆ ˆ0 dp dp dµ prˆ ∫ ∫ ˆ pT ˆ ˆ ˆ ϕ ˆ 0 0 2 0 = 4π F (e p , µ)(p ) dp dµ √ dpT (pˆ0 )2 − µ2 − (pTˆ )2 ∫ √ ˆ ˆ ˆ = 4π F (eϕ p0 , µ)(p0 )2 (pˆ0 )2 − µ2 dp0 dµ;

ρ = 4π

ˆ

F (eϕ p0 , µ)

∫

ˆ

ˆ

ˆ

ˆ

F (eϕ p0 , µ)(prˆpT )dpT dp0 dµ ∫ ∫ √ ˆ ˆ ˆ ϕ ˆ 0 0 = 4π F (e p , µ)dp dµ (pˆ0 )2 − µ2 − (pTˆ )2 pT dpT ∫ 4π ˆ ˆ ˆ = F (eϕ p0 , µ)((p0 )2 − µ2 )3/2 dp0 dµ; 3

Pr = 4π

(25.27.1)

(25.27.2)

and ∫

ˆ

(pT )3 Tˆ ˆ0 dp dp dµ prˆ ∫ ∫ ˆ (pT )3 ˆ ˆ ˆ dpT = 2π F (eϕ p0 , µ)dp0 dµ √ (pˆ0 )2 − µ2 − (pTˆ )2 ∫ 4π ˆ ˆ ˆ = F (eϕ p0 , µ)((p0 )2 − µ2 )3/2 dp0 dµ. 3

PT = 2π

ˆ

F (eϕ p0 , µ)

(25.27.3)

So statement 3 and 4 is true. Finally, note that ˆ

T αˆ β = diag(ρ, P, P, P )

(25.27.4)

ds2 = −e2Φ dt2 + e2Λ dr2 + r2 dθ2 + r2 sin2 θdϕ2 .

(25.27.5)

and The form of metric and energy momentum tensor of cluster is the same as that of gas sphere. So the statement 5 and 6 must be true.

25.28

SELF-SIMILAR CLUSTER

(a) Find a solution to the equations of structure for a spherical star of infinite central density, with the equation of state P = γρ, where γ is a constant (0 < γ < 1/3). (b) Find an isotropic distribution function F (E, L) that leads to a star cluster with the same distributions of ρ, P , m, and Φ as in the gas sphere of part (a).

261

Solution: Substitute the equation of state into OV equation. Then we have d ln ρ (γ + 1)(m + 4πγr3 ρ) =− . dr γr(r − 2m)

(25.28.1)

Try power-law solution ρ = Crα . We have ∫ r 4πC 3+α m= 4πCr2+α dr = r . 3 +α 0 And so

(25.28.2)

4πC 3+α r + 4πCγr3+α ) (γ + 1)( 3+α α =− . 8πC 3+α r γr(r − 3+α r )

(25.28.3)

The solution is α = −2 and C = γ/2π(γ 2 + 6γ + 1), i.e. ρ= And so e2Λ = From

γ2 1 . γ 2 + 6γ + 1 2πr2

(25.28.4)

γ 2 + 6γ + 1 1 1 . 2m = 1 − 8πC = (1 + γ)2 1− r

(25.28.5)

dP dΦ = −(ρ + P ) , dr dr

(25.28.6)

we can get 4γ

e2Φ = Br γ+1 ,

(25.28.7)

B = const.

(b) From equation 25.27.1, we have ∫ 2γ −2 2 2 Cr = 4π F (B 1/2 r γ+1 Elocal , µ)[Elocal − µ2 ]1/2 Elocal dElocal dµ.

(25.28.8)

From equation 25.27.2, we have γCr−2 = So we should let

4π 3

∫

2γ

2 F (B 1/2 r γ+1 Elocal , µ)[Elocal − µ2 ]3/2 dElocal dµ.

2γ

2γ

γ

F (B 1/2 r γ+1 Elocal , µ) = Aδ(µ − µ0 )[B 1/2 r γ+1 Elocal ]− γ+1 .

(25.28.9)

(25.28.10)

A and µ0 are constants determined by normalization condition from two equations above.

25.29

CLUSTER WITH CIRCULAR ORBITS

What must be the form of the distribution function to guarantee that all stars move in circular orbits? Specialize the equations of structure to this case. Analyze the stability of the orbits of individual stars in the cluster, using an effective-potential diagram. What conditions must the distribution function satisfy if all orbits are to be stable? The description of the question is vague and I do not know how to solve it. Any help is appreciated. Please read the paper Can the Redshifts of Quasi-Stellar Objects Be Gravitational? (Zapolsky, 1968) for more information on a spherical cluster composed of circular moving stars and its stability.

262

Chapter 26

STELLAR PULSATIONS 26.1

DRAGGING OF INERTIAL FRAMES BY A SLOWLY ROTATING STAR

A fluid sphere rotates very slowly. Analyze its rotation using perturbation theory; keep only effects and terms linear in the angular velocity of rotation.

Solution: The answer is taken from papers Variational Principle for the Equilibrium of a Relativistic, Rotating Star (Hartle & Sharp, 1967) and Slowly Rotating Relativistic Stars. I. Equations of Structure (Hartle, 1967). For the description of an axially symmetric equilibrium configuration of a rotating fluid, two coordinates are conveniently taken to be the time t = x0 and an angle ϕ = x3 . The stationary nature of the configuration and its axial symmetry are then expressed by requiring that the metric coefficients be independent of t and ϕ, gαβ = gαβ (x1 , x2 ). (26.1.1) The metric remains a function of x1 and x2 alone under coordinate transformations of the form t = t′ + f1 (x1 , x2 ),

ϕ = ϕ′ + f2 (x1 , x2 ).

(26.1.2)

By an appropriate choice of f1 and f2 , one can always find a coordinate system in which g13 = g23 = 0.

(26.1.3)

In addition to the independence of the metric on the angle, the assumption of axial symmetry also entails that the velocity distribution have only an angular component, so that the four-velocity may be written as uµ = u0 (1, 0, 0, Ω).

(26.1.4)

This means that the source Tµν of the gravitational field is left unchanged by the simultaneous inversion of the time and the azimuthal angle. This symmetry must be reflected in the solutions of the field equations, and consequently we must require the line element to be left unchanged under this inversion. This requires us to set g01 = g02 = 0. (26.1.5) Now the line element must have the general form: ds2 = g00 (dt)2 + 2g03 dtdϕ + g33 (dϕ)2 + g11 (dx1 )2 + 2g12 dx1 dx2 + g22 (dx2 )2 .

263

(26.1.6)

This form of the line element remains unchanged under coordinate transformations of the form ′

′

x1 = x2 (x1 , x2 ),

′

′

x2 = x2 (x1 , x2 ).

(26.1.7)

By an appropriate coordinate transformations, the line element can be written as ds2 = g00 (dt)2 + 2g03 dtdϕ + g33 (dϕ)2 + g11 (dx1 )2 + g22 (dx2 )2 .

(26.1.8)

g00 , g11 , g22 and g33 are invariant under ϕ → −ϕ, while g03 changes sign. So g00 , g11 , g22 and g33 are even functions of Ω while g03 is odd functions of Ω. In the case of a slowly rotating star, Ω is small and metric coefficients can be expanded as series of Ω. And so g00 , g11 , g22 and g33 contain only even powers of Ω while g03 contains odd powers. If we keep only the terms linear in the angular velocity of rotation, we have (26.1.9) ds2 = −e2Φ0 dt2 + e2Λ0 dr2 + r2 (dθ2 + sin2 θdϕ2 ) − 2(r2 sin2 θ)ω(r, θ)dϕdt, where ω is linear in Ω, and Φ0 and Λ0 are unperturbed potential in non rotating star. Then we can get the Einstein tensor from the metric. To the first order, all components of the Einstein tensor are the same as unperturbed ones, except that 1 Gtϕ = − [r2 ω ′′ − (r2 Λ′0 + r2 Φ′0 − 4r)ω ′ ] sin2 θe−2Λ0 −2Φ0 2 [ ] 2 1 ∂ω 2 ∂ ω − 3 sin θ cos θ + sin θ 2 e−2Φ0 2 ∂θ ∂θ

(26.1.10)

The four velocity of the fluid satisfies that (gtt + 2gtϕ Ω + gϕϕ Ω2 )(ut )2 = −1.

(26.1.11)

ut = e−Φ0 , uϕ = Ωe−Φ0

(26.1.12)

ut = −eΦ0 , uϕ = r2 sin2 θ(Ω − ω)e−Φ0 .

(26.1.13)

To the first order, we have and so

The tϕ component of stress energy tensor is Tϕt = (ρ + p)ut uϕ ≈ (ρ0 + p0 )e−2Φ0 r2 sin2 θ(Ω − ω).

(26.1.14)

Define ω ¯ ≡ Ω − ω. The Einstein equation Gtϕ = 8πTϕt gives [ ] [ ] 1 ∂ ¯ eΛ0 −Φ0 ∂ ¯ 3 ∂ω 4 −Φ0 −Λ0 ∂ ω r e + 2 3 sin θ = 16π(ρ0 + p0 )eΛ0 −Φ0 ω ¯. r4 ∂r ∂r ∂θ r sin θ ∂θ

(26.1.15)

Defining j(r) ≡ exp(−Λ0 − Φ0 ) and using equations of structure of non-rotating spherical stars, we have the equation ( ) [ ] 1 ∂ 4 dj eΛ0 −Φ0 1 ∂ ¯ ¯ 3 ∂ω 4 ∂ω r + ω ¯ + sin θ = 0. (26.1.16) j r4 ∂r ∂r r dr r2 sin3 θ ∂θ ∂θ Using the expansion ω ¯ (r, θ) =

∞ ∑ l=1

we have 1 d r4 dr

( r4 j

d¯ ωl dr

)

[ +

) ( 1 dPl , ω ¯ l (r) − sin θ dθ

] l(l + 1) − 2 4 dj ω ¯ l = 0. − eΛ0 −Φ0 r dr r2

(26.1.17)

(26.1.18)

At small r, j(r) is a regular function so that the differential equation admits a small r behavior of the form ω ¯ l (r) → const. rS+ + const. rS− , r → 0, ] [ 3 9 S± = − ± + e2Λ0 (l2 + l − 1) . 2 4

264

(26.1.19)

If the geometry is to be regular at the origin we must demand that the coefficient of rS− vanish. At large r, j(r) becomes unity and ω ¯ has the form ω ¯ l (r) → const. r−l−2 + const. rl−1 ,

r → 0,

(26.1.20)

The behavior has already been fixed at the origin. Consequently the ratio of the two constants in equation above is determined and neither will vanish unless they both do. If space is to be flat at large r, ω ¯ l (r) must decrease faster than r−3 , so that ω ¯ (r) = Ω − ω approaches Ω. Thus all coefficients in the Legendre expansion of ω ¯ vanish except l = 1. Consequently ω ¯ is a function of r alone. It obeys the differential equation ( ) 1 d d¯ ω 4 dj r4 j + ω ¯ = 0. (26.1.21) 4 r dr dr r dr This equation is to be integrated outward starting with ω(0) = const. Outside the star j(r) = 1 and the solution has the form 2J ω ¯ = Ω − 3 , r > a, (26.1.22) r i.e. ω(r, θ) = 2J/r3 . The constant J can be identified with the total angular momentum of the star. And the metric far from the star is ( ) 2M dr2 4J sin2 θ 2 dt2 + + r2 (dθ2 + sin2 θdϕ2 ) − dϕdt. (26.1.23) ds = − 1 − r 1 − 2M/r r In Cartesian coordinates, we have −

4J sin2 θ 4J dϕdt = − 3 (−ydxdt + xdydt), r r

(26.1.24)

4J sin2 θ xl dϕdt = −4ϵjkl S k 3 dtdxj . r r

(26.1.25)

or generally, −

265

266

Chapter 27

IDEALIZED COSMOLOGIES 27.1

ISOTROPY IMPLIES HOMOGENEITY

Use elementary thought experiments to show that isotropy of the universe implies homogeneity.

Solution: For any two points in space, we can find a third point whose distance to these two points are equal. If the universe is isotropy from the point of view of any points, then these two points must be equivalent. Now we can say the universe is homogeneous.

27.2

SYNCHRONOUS COORDINATES IN GENERAL

In an arbitrary spacetime manifold (not necessarily homogeneous or isotropic), pick an initial spacelike hypersurface SI , place an arbitrary coordinate grid on it, eject geodesic world lines orthogonal to it, and give these world lines the coordinates (x1 , x2 , x3 ) = constant,

x0 ≡ t = tI + τ,

(27.2.1)

where τ is proper time along the world line, beginning with τ on SI . Show that in this coordinate system the metric takes on the synchronous (Gaussian normal) form (27.14).

Solution: For any point in surface t = tI , we have g0i = 0 as geodesics 27.2.1 are orthogonal to the surface. For geodesics 27.2.1, we have uµ = dxµ /dτ = (1, 0, 0, 0). We can get g00 = −1 from gαβ uα uβ = −1. If we substitute geodesics 27.2.1 into geodesic equation dxβ dxγ d2 xα α + Γ = 0, βγ dτ 2 dτ dτ

(27.2.2)

we have Γα00 = 0 and so Γα00 = 0. And so g00,0 = 2Γ000 = 0.

(27.2.3)

Now we know g00 = −1 throughout the whole spacetime. Note that 1 Γi00 = gi0,0 − g00,i . 2

267

(27.2.4)

As g00,i = 0, we have gi0,0 = 0. And we know g0i = 0 throughout the whole spacetime. In this coordinate system the metric takes on the synchronous (Gaussian normal) form ds2 = −dt2 + gij dxi dxj .

27.3

(27.2.5)

ARBITRARINESS IN THE EXPANSION FACTOR

How much arbitrariness is there in the definition of the expansion factor a(t)? Civilization A started long ago at time tA . For it, the expansion factor is proper distance betweee two particles of the “cosmological fluid” at time t = aA (t). proper distance between the same two particles at time tA

(27.3.1)

Subsequently men planted civilization B at time tB on a planet in a nearby galaxy. [At this time, the expansion factor aA had the value aA (tB )]. Civilization B defines the expansion factor relative to the time of its own beginning: proper distance betweee two particles of the “cosmological fluid” at time t = aB (t). proper distance between the same two particles at time tB

(27.3.2)

At two subsequent events, C and D, of which both civilizations are aware, they assign to the universe in their bookkeeping by no means identical expansion factors, aA (tC ) ̸= aB (tC ),

aA (tD ) ̸= aB (tD ).

(27.3.3)

Show that the relative expansion of the model universe in passing from stage C to stage D in its evolution is nevertheless the same in the two systems of bookkeeping: aA (tD ) aB (tD ) = . aA (tC ) aB (tC )

(27.3.4)

Solution:

aA (tD ) proper distance betweee two particles of the “cosmological fluid” at time tD = aA (tC ) proper distance between the same two particles at time tA proper distance between the same two particles at time tA × proper distance betweee two particles of the “cosmological fluid” at time tC proper distance betweee two particles of the “cosmological fluid” at time tD = . proper distance betweee two particles of the “cosmological fluid” at time tC (27.3.5) Similarly, we have aB (tD ) proper distance betweee two particles of the “cosmological fluid” at time tD = . aB (tC ) proper distance betweee two particles of the “cosmological fluid” at time tC

(27.3.6)

So the relative expansion of the model universe in passing from stage C to stage D in its evolution is the same in the two systems of bookkeeping.

268

27.4

UNIQUENESS OF METRIC FOR 3-SURFACE OF CONSTANT CURVATURE ′

Let γij and γi′ j ′ be two sets of metric coefficients, in coordinate systems {xi } and {xi }, that have Riemann curvature tensors of the constant-curvature type (27.21). Let it be given in addition that the curvature parameters K and K ′ are equal. Show that γij and γi′ j ′ are related by a coordinate transformation.

Solution: ′

γij and γi′ j ′ are equivalent if we have a set of independent solutions xi = xi (xi ) for i = 1, · · · , n that γi′ j ′ = γij

∂xi ∂xj . ∂xi′ ∂xj ′

(27.4.1)

If we put ′ ′ ∂xi = pii′ (x1 , · · · , xn ), ∂xi′ the transformation law for affine connections gives

∂pij ′ ′ = Γi j ′ k′ pii′ − Γijk pj j ′ pkk′ . ∂xk′

(27.4.2)

(27.4.3)

Hence the problem reduces to the determination of n(n + 1) functions xi , pii′ , satisfying differential equations 27.4.2 and 27.4.3 and also the n(n + 1)/2 finite equations γi′ j ′ − γij pii′ pj j ′ = 0.

(27.4.4)

The conditions of integrability of 27.4.2 are satisfied identically in consequence of 27.4.3. If 27.4.3 holds, it is easily to induce that ∂pij ′ ∂pik′ = (27.4.5) ′ ∂xk ∂xj ′ Notice that ∂pij ′ j j j i′ i i k i′ i i k i k ′ = Γ j ′ k ′ ,l′ p i′ − Γ jk,l′ p j ′ p k ′ + Γ j ′ k ′ p i′ ,l′ − Γ jk p j ′ ,l′ p k ′ − Γ jk p j ′ ,l′ p k ′ ,l′ ′ l k ∂x ∂x ′ = Γi j ′ k′ ,l′ pii′ − Γijk,l pj j ′ pkk′ pll′ − Γijk pj j ′ ,l′ pkk′ ,l′ ′

′

′

+ Γi j ′ k′ (Γm i′ l′ pim′ − Γijk pj i′ pkl′ ) − Γijk (Γm j ′ l′ pj m′ − Γimn pmj ′ pnl′ )pkk′ ′

′

′

= (Γi j ′ k′ ,l′ + Γi m′ l′ Γm j ′ k′ )pii′ − (Γijk,l − Γimk Γmjl )pj j ′ pkk′ pll′ ′

′

− Γi j ′ k′ Γijk pj i′ pkl′ − Γi j ′ l′ Γijk pj i′ pkk′ − Γijk pj j ′ ,l′ pkk′ ,l′ .

(27.4.6)

So we have

∂pij ′ ∂pij ′ ′ − = Ri j ′ k′ l′ pii′ − Rijkl pj j ′ pkk′ pll′ . ∂xk′ ∂xl′ ∂xl′ ∂xk′ The conditions of integrability of 27.4.3 are Ri′ j ′ k′ l′ = Rijkl pii′ pj j ′ pkk′ pll′ .

(27.4.7)

(27.4.8)

If Riemann curvature tensors in two coordinates have the form Rijkl = K(γik γjl − γil γjk ),

Ri′ j ′ k′ l′ = K(γi′ k′ γj ′ l′ − γi′ l′ γj ′ k′ ),

(27.4.9)

and equation 27.4.4 holds, equation 27.4.8 will be satisfied identically. Moreover, the left hand of 27.4.4 is constant if 27.4.2 and 27.4.3 hold. So we can adjust the value of pii′ at one point to make 27.4.4 holds at that point. Then we can solve the differential equations 27.4.2 and 27.4.3 whose conditions of integrability are already satisfied consistently. And condition 27.4.4 will be satisfied automatically. So we prove that γij and γi′ j ′ are related by a coordinate transformation.

269

27.5

METRIC FOR 3-SURFACE OF CONSTANT CURVATURE

(a) Show that the following metric has expression (27.21) as its curvature tensor ( γij =

1 1 + Kδkl xk xl 4

)−2 δij .

(27.5.1)

(b) By transforming to spherical coordinates (R, θ, ϕ) and then changing to a Schwarzschild radial coordinate (2πr = “proper circumference”), transform this metric into the form dσ 2 =

dr2 + r2 (dθ2 + sin2 θdϕ2 ). 1 − Kr2

(27.5.2)

(c) Find a further change of radial coordinate that brings the metric into the form (27.22).

Solution: (a) The derivative of the metric is )−3 ( 1 δij xk , γij,k = −K 1 + KR2 4

(27.5.3)

where R2 = δkl xk xl . So the connection is Γijk = −

K 2

(

1 1 + KR2 4

)−1 (δij xk + δik xj − δjk xi ).

(27.5.4)

The curvature is Rijkl = Γijl,k − Γijk,l + Γikm Γmjl − Γilm Γmjk )−2 )−1 ( ( K K2 1 1 xk (δij xl + δil xj − δjl xi ) − (δij δkl + δil δjk − δjl δik ) = 1 + KR2 1 + KR2 4 4 2 4 )−2 )−1 ( ( K2 1 K 1 2 2 − 1 + KR xl (δij xk + δik xj − δjk xi ) + 1 + KR (δij δkl + δik δjl − δjk δil ) 4 4 2 4 )−2 ( K2 1 + (δik xm + δim xk − δmk xi )(δmj xl + δml xj − δjl xm ) 1 + KR2 4 4 ( )−2 K2 1 2 − 1 + KR (δil xm + δim xl − δml xi )(δmj xk + δmk xj − δjk xm ) 4 4 ( ( )−2 )−1 1 K2 1 (δil xj xk − δjl xi xk − δik xj xl + δjk xi xl ) − K 1 + KR2 (δil δjk − δjl δik ) = 1 + KR2 4 4 4 ( )−2 K2 1 2 + 1 + KR (δik xl xj − δik δjl R2 + δil δjk R2 − δil xk xj − δkj xi xl + δlj xi xk ) 4 4 ( )−2 ( )−1 1 1 K2 1 + KR2 (−δik δjl + δil δjk )R2 − K 1 + KR2 (δil δjk − δjl δik ) = 4 4 4 )−2 ( 1 2 (δik δjl − δil δjk ), = K 1 + KR (27.5.5) 4 or equivalently, Rijkl

( )−4 1 2 = K 1 + KR (δik δjl − δil δjk ) = K(γik γjl − γil γjk ). 4

270

(27.5.6)

(b) In spherical coordinates, we have ( )−2 1 2 dσ = 1 + KR [dR2 + R2 (dθ2 + sin2 θdϕ2 )]. 4 2

Define

(

1 1 + KR2 4

)−2 dR2 =

1 1 + KR2 4

R.

(27.5.8)

dr =

1 − KR2 /4 dR. (1 + KR2 /4)2

(27.5.9)

(1 + KR2 /4)2 2 (1 + KR2 /4)2 1 dr = dr2 = dr2 . (1 − KR2 /4)2 (1 + KR2 /4)2 − KR2 1 − Kr2

The metric becomes dσ 2 =

dr2 + r2 (dθ2 + sin2 θdϕ2 ). 1 − Kr2

(c) If K > 0, define

√ dχ ≡

So we have χ = arcsin

√

)−1

r≡ We have

Thus (

K dr. 1 − Kr2

(27.5.10)

(27.5.11)

(27.5.12)

Kr. The metric becomes dσ 2 = K −1 [dχ2 + sin2 χ(dθ2 + sin2 θdϕ2 )].

If K < 0, define

√ dχ ≡

So we have χ = sinh−1

(27.5.7)

√

−K dr. 1 + (−K)r2

(27.5.13)

(27.5.14)

−Kr. The metric becomes dσ 2 = (−K)−1 [dχ2 + sinh2 χ(dθ2 + sin2 θdϕ2 )].

(27.5.15)

If K = 0, define χ = r, and we have dσ 2 = dχ2 + χ2 (dθ2 + sin2 θdϕ2 ).

27.6

(27.5.16)

PROPERTIES OF THE 3-SURFACES

Verify all statements made in Box 27.2. Each statements made in section Box 27.2 are intuitive and easy to verify. I will not write down detailed demonstration for all of them here.

27.7

ISOTROPY IMPLIES HOMOGENEITY

Use the contracted Bianchi identity

(3)

G

ik |k

on the 3-geometry alone) to show (1) that

= 0 (where the stroke indicates a covariant derivative based ∇ K = 0 in equation (27.21), and therefore to show (2)

(3)

271

that direction-independence of the curvature [isotropy; curvature of form (27.21)] implies and demands homogeneity (K constant in space).

Solution: Firstly, we have Rbd = γ ac Rabcd = Kγ ac (γac γbd − γad γbc ) = 2Kγbd .

(27.7.1)

So the curvature scalar is R = γ bd Rbd = 6K. The Einstein tensor is therefore 1 Gbd = Rbd − γbd R = −Kγbd . 2 And so we have 0= i.e.

(3)

G

bd |d

= −(Kγ bd )|d = −K |b ,

(27.7.2)

(27.7.3)

∇ K = 0.

(3)

27.8

MATTER-DOMINATED AND RADIATION-DOMINATED REGIMES Of FRIEDMANN COSMOLOGY

Derive the results listed in the last two columns of Box 27.3, except for the focusing properties of the curved space. The results listed in the last two columns of Box 27.3 are straightforward and easy to verify. I will not write down detailed demonstration for all of them here.

27.9

TRANSITION FROM RADIATION-DOMINATED REGIME TO MATTERDOMINATED REGIME

Including both the radiation and the matter terms in equation (27.51), restate the equation in terms of the arc parameter η (with dη = dt/a) as independent variable, and integrate to find a = amax /2 − [(amax /2)2 + a∗2 ]1/2 cos(η + δ)

where

t = (amax /2)η − [(amax /2)2 + a∗2 ]1/2 [sin(η + δ) − sin δ],

(27.9.1)

δ = arctan[a∗ /(amax /2)].

(27.9.2)

(a) Verify that under appropriate conditions these expressions reduce at early times to a “circle” relation between radius and time and to a “cycloid” relation later. (b) Assign to a∗2 the value a0 amax /10000 (why?) and construct curves for the dimensionless measures of density,    ρm   2 8πamax , ρr log10  (27.9.3)   3 ρm + ρr , as a function of the dimensionless measure of time, log10 (t/amax ). What conclusions emerge from inspecting the logarithmic slope of these curves?

272

(27.9.4)

Solution: (a) When η ≪ 1, we have a ≈ amax /2 − [(amax /2)2 + a∗2 ]1/2 [cos δ − η sin δ] = a∗ η ≈ a∗ sin η, and

t ≈ (amax /2)η − [(amax /2)2 + a∗2 ]1/2 [sin δ + η cos δ − sin δ] = 0 ≈ a∗ (1 − cos η)

(27.9.5) (27.9.6)

to the first order of η. When η + δ − π ≪ 1, we have amax /2 − [(amax /2)2 + a∗2 ]1/2 (−1)

a≈ = amax /2 + [(amax /2)2 + a∗2 ]1/2 =

amax /2 + [(amax /2)2 + a∗2 ]1/2 [1 − cos(η + δ)] 2

(27.9.7)

and t≈

(amax /2)(π − δ) + (amax /2)(η + δ − π) − [(amax / ∗

∗2 1/2

= (amax /2)(π − δ) + a + {amax /2 + [(amax /2) + a ] 2

}(η + δ − π)

amax /2 + [(am

≈ + (amax /2)(π − δ) + a∗ −

amax /2 + [(amax /2)2 + a∗2 ]1/2 π 2

(27.9.8)

to the first order of η + δ − π. If we shift the t with a constant, we will have a “cycloid” relation between radius and time. (b) The ratio of matter density and radiation density now is ρm0 a0 amax amax /a3 . = ∗2 40 = ρr0 a /a0 a∗2

(27.9.9)

Assigning to a∗2 the value a0 amax /10000 means the ratio of matter density and radiation density now is about 10000, which may be compatible with the observation in 1970s. (The observation todya, i.e. 2010s, gives H0 = 69.6km/s/Mpc, Ωm = 0.286, Tr = 2.72528K, the ratio of matter density and radiation density now is about 5600.) Firstly, we have ρm =

3 amax , 8π a3

ρr =

3 a∗2 . 8π a4

(27.9.10)

Given a∗2 = a0 amax /10000, we have 8πa2max a3 ρm = max ={ 3 a3 1 2

8πa2max a2 a∗2 ={ ρr = max4 3 a 1 2

t amax

1 = η− 2

−

[( ) 1 2 2

−

[( ) 1 2 2

1 +

1 +

a0 10000amax

a0 10000amax

]1/2

}3 cos(η + δ) }4

]1/2 cos(η + δ)

a0 10000amax

[( ) ]1/2 2 a0 1 + [sin(η + δ) − sin δ], 2 10000amax √

and δ = arctan 2

a0 . 10000amax

(27.9.11)

(27.9.12)

(27.9.13)

(27.9.14)

Take the value from Box 27.4, we have a0 /10000amax = 2.55 × 10−21 . Curves are shown in Fig 1. When ρr ≫ ρm , the logarithmic slope of ρm and ρr are −1.5 and −2 respectively, meaning t ∝ a2 . When ρr ≪ ρm , the logarithmic slope of ρm and ρr are −2 and −2.67 respectively, meaning t ∝ a3/2 . The logarithmic slope of ρm + ρr is −2 before the universe contracts.

273

matter radiation

1093 1078

Density

1063 1048 1033 1018 103 10−12

10−47

10−40

10−33

10−26 Time

10−19

10−12

10−5

102

Figure 27.1: Curves for the dimensionless measures of density as a function of the dimensionless measure of time.

27.10

THE EXPANDING AND RECONTRACTING SPHERICAL WAVE FRONT

An explosion takes place at the “N-pole” of the matter-dominated Friedmann model universe at the value of the “arc parameter time” η = π/3, when the radius of the universe has reached half its peak value. The photons from the explosion race out on a spherical wave front. Through what fraction of the “cosmological fluid” has this wave front penetrated at that instant when the wave front has its largest proper surface area?

Solution: In a matter dominated universe, we have a(η) =

amax (1 − cos η). 2

(27.10.1)

Note when η = π/3, the radius of the universe reaches one quarter of its peak value. The statements in exercise is wrong. The world line of the light is χ = η − π/3.

(27.10.2)

A = 4πa2 sin2 χ = πa2max (1 − cos η)2 sin2 (η − π/3).

(27.10.3)

The proper surface area of the wave front is

274

When η = 8π/9, A reaches its maximum value. At that instant, this wave front has penetrated − sin(16π/9)/4 + 4π/9 ≈ 99.1% π/2

27.11

(27.10.4)

ON SEEING THE BACK OF ONE’S HEAD

Can a being at rest relative to the “cosmological fluid” ever see the back of his head by means of photons that travel all the way around a closed model universe that obeys the Friedmann cosmology and has a non-zero cosmological constant?

Solution: In a matter dominated and close universe, if the value of cosmological constant is much smaller than the critical value Λcirc , the universe will contract before cosmological constant takes a significant role. So the evolution of scale factor would be essentially identical to that in a matter dominated and close universe without cosmological constant. From Penrose diagram for a matter dominated and close universe without cosmological constant, seeing the back of one’s head is possible So we will state that it is also possible to see the back of one’s head in a matter dominated and close universe with a non-zero cosmological constant.

27.12

DO THE CONSERVATION LAWS FORBID THE PRODUCTION OF PARTICLE-ANTIPARTICLE PAIRS OUT OF EMPTY SPACE BY TIDAL GRAVITATION FORCES?

Find out what is wrong with the following argument: “The classical equations Gαβ = 8πTαβ

(27.12.1)

are not compatible with the production of pairs, since they lead to the identity Tα β;β = 0. Let the initial state be vacuum, and let Tαβ and its derivative be equal to zero on the hypersurface t = const or t = −∞. It then follows from , that the vacuum is always Tα β;β = 0, that the vacuum is always conserved.

Solution: The argument is wrong because there may exist singularities in the solution of Einstein equation.

27.13

TURN-AROUND UNIVERSE MODEL NEGLECTING MATTER DENSITY

If turn-around (minimum radius) occurs far to the right (large a) of the maximum of the potential V (a) in equation (27.75), the matter terms will be negligible. Let ρm0 = ρr0 = 0. Then, solve to show that Λ = 3(amin )−2 , H = (amin )−1 tanh(t/amin ) near turnaround (t = 0) and that the deceleration parameter q ≡ −(1/H 2 a)(d2 a/dt2 ) has the value −1 q = −a2 (a2 − a−2 < −1. min )

275

(27.13.1)

Solution: The Friedmann equation in this case is 1 1 = Λ. a2 3 −2 We have H = 0 when a = amin . And so Λ = 3(amin ) . The equation can now be written as ( )2 da a2 +1= 2 dt amin H2 +

(27.13.2)

(27.13.3)

The solution of the equation is a t = cosh . amin amin

(27.13.4)

1 t 1 da = tanh a dt amin amin

(27.13.5)

It is easy to verify that H= and q=−

amin cosh(t/amin ) 1 t cosh2 (t/amin ) a2 × cosh = − = − < −1. amin amin a2 − a2min sinh2 (t/amin ) cosh2 (t/amin ) − 1

27.14

(27.13.6)

“HESITATION” UNIVERSE

Neglect radiation in equation (27.75) but assume K0 and Λ to be chosen so that the universe spent a very long time with a(t) near ah (ah measures location of highest point of the barrier, or the size of the universe at which the universe is most sluggish). Choose ah = a0 /3 to produce an abnormally great number of quasar redshifts near z = 2 [as Burbidge and Burbidge (1969a,b) believe to be the case, though their opinion is not shared by all observers]. Show (a) that the density of matter now would account for only 2 10 per cent of the value of H02 = (a/a) ˙ now in equation (27.75) [“missing matter”, i.e., K0 and Λ terms, account for 90 per cent], (b) that ah ≈ 201/2 H0−1 , and (c) that the deceleration parameter defined in the previous exercise, as evaluated “today”, has the value q0 = −13/10.

Solution: The generalized Friedmann equation now becomes [ ( )]2 ( ) d a(t) a +V = −K0 . dt a0 a0 and

( V

a a0

)

8π a0 1 = − ρm0 − Λ 3 a 3

(

a a0

(27.14.1)

)2 (27.14.2)

The location of highest point of the barrier satisfies that 2 ah 8π a0 ρm0 2 − Λ 2 = 0. 3 ah 3 a0 The solution is

( ah =

4πρm0 Λ

)1/3 a0

and we demand that ah = a0 /3. Then we have 4πρm0 /Λ = 1/27 and [ ( )2 ] 1 a 2 a0 . V =− Λ + 3 27 a a0

276

(27.14.3)

(27.14.4)

(27.14.5)

When a = a0 /3, we have Vmax = −Λ/9. In the case of hesitation, we have K0 = Λ/9. So H02 = −K0 − V (1) =

20 Λ 81

(27.14.6)

while 8πρm0 /3 = 2Λ/81. So the density of matter now would account for only 10 per cent of the value of H02 . As K0 = 1/a20 = Λ/9, we have ah = a0 /3 = Λ−1/2 = (81H02 /20)−1/2 = 201/2 /9H0 . The acceleration of a satisfies that

So we have

da 2 d2 a da +V′ = 0. a20 dt2 dt dt

(27.14.7)

( ) d2 a 1 1 a30 = Λ a− = dt2 3 27 a2

(27.14.8)

The deceleration parameter is q0 = −

27.15

( ) 1 1 1 13 Λ a − a =− . 0 0 a0 H02 3 27 10

(27.14.9)

UNIVERSE OPAQUE TO BLACK-BODY RADIATION AT A NONSINGULAR PAST TURN-AROUND REQUIRES IMPOSSIBLE PARAMETERS

From a plot like that in Box 27.5, construct a model of the universe that contains 2.7K black-body radiation at the present, but, with k = +1 and Λ > 0, had a past turn-around (minimum radius) at which the blackbody temperature reached 3000K where hydrogen would be ionized. Try to use values of H0−1 and ρm0 that are as little as possible smaller than presently accepted values.

Solution: The conditions above implies that H02 +

and

1 8π Λ 8π ρm0 + ρr0 + , = a20 3 3 3

(27.15.1)

1 a0 8π a2 Λ a2t 8π = , ρm0 + ρr0 02 + 2 a0 3 at 3 at 3 a20

(27.15.2)

a0 3000 10000 > = . at 2.7 9

(27.15.3)

In order to make H0−1 and ρm0 as little as possible, we can take ρm0 = 0, then we have ) ( 8π Λ 1 2 2 H0 = ρr0 (1 − x ) + 1− 2 , 3 3 x where x > 10000/9. If we make Λ as large as possible, we can get H0 as little as possible.

277

(27.15.4)

278

Chapter 28

EVOLUTION OF THE UNIVERSE INTO ITS PRESENT STATE 28.1

UNCERTAINTY IN EVOLUTION

Current observations, plus the assumption of complete homogeneity and isotropy at the beginning of expansion, plus the assumption that the excess of leptons over antileptons is less than or of the order of the excess of baryons over antibaryons, place the following limits on the cosmological parameters today: • Matter density today = ρm0 between 10−28 and 2 × 10−31 g/cm3 ; • k = 0 or +1 or −1; • temperature of electromagnetic radiation today = 2.7 ± 0.1K; • Total radiation density [observed photons, plus neutrinos and gravitons that presumably originated in big bang in thermal equilibrium with photons] = ρr0 between 0.7 × 10−33 and 1.2 × 10−33 g/cm3 . Use the equations in section 27.10 to calculate the uncertainties in the evolutionary history (Figure 28.1) caused by these uncertainties in the present state of the universe.

Solution: Define ρc0 ≡

3H02 8π

and Ωr0 ≡

ρr0 , ρc0

Ωm0 ≡

ρr0 , ρm0

ΩK0 ≡ −

(28.1.1) 3k/8πa20 = 1 − Ωr0 − Ωm0 . ρc0

The evolution of scale factor is given by ∫ ∞ ′ dz 1 √ η(z) = ′ 4 H Ωr0 (1 + z ) + Ωm0 (1 + z ′ )3 + ΩK0 (1 + z ′ )2 0 z

(28.1.2)

(28.1.3)

where dt = adη and z + 1 = a0 /a. a(t) can then be obtained by one more integration. The evolution of ρr , ρm and T is given by ( )−4 ( )−3 a a a0 ρr = ρr0 , ρm = ρm0 , T = T0 . (28.1.4) a0 a0 a

279

The numeric calculation of the uncertainties in the evolutionary history can be performed by Monte Carlo method and is eliminated here. A useful website for calculation is Cosmology Calculator.

280

Chapter 29

PRESENT STATE AND FUTURE EVOLUTION OF THE UNIVERSE 29.1

IMPLICATIONS OF PARAMETER VALUES

Derive the results quoted in Box 29.1. Each statements in Box 29.1 are straightforward and trivial to prove. I will not write down detailed demonstration for all of them here.

29.2

ALTERNATIVE DERIVATION OF REDSHIFT

Notice that the only part of the line element that is relevant for the light ray is ds2 = −dt2 + a2 (t)dχ2 ,

(29.2.1)

since dθ = dϕ = 0 along its world line (spherical symmetry!). Regard the light ray as made of photons with 4-momenta p. From the geodesic equation (or, for the reader who has studied chapter 25, from arguments about Killing vectors), show that pχ ≡ p · (∂/∂χ) (29.2.2) is conserved along the photon’s world line. Use this fact, the fact that a photon’s 4-momentum is null, p · p = 0, and the equation E = −p · u for the energy measured by an observer with 4-velocity u, to derive the redshift equation (29.11).

Solution: In this coordinates, Γχµν is not zero for Γχχt = Γχtχ =

1 gχχ,t . 2

(29.2.3)

For vector χµ = (0, 1), χµ = (0, a2 ), we have

So

χµ;ν = χµ,ν − Γαµν χα = χµ,ν − Γχµν .

(29.2.4)

1 1 1 χχ;t = χχ,t − Γχχt = (a2 ),t − gχχ,t = gtt , χt;χ = −Γχtχ = − gtt . 2 2 2

(29.2.5)

281

All other components vanish. We now have χµ;ν +χν;µ = 0 and so χ is a killing vector. Thus pχ is conserved along the photon’s world line. From p · p = 0, we have p2t − a−2 p2χ = 0. So apt = pχ is a constant. Recall that E = −pt . We have λ1 E2 a1 = = . (29.2.6) λ2 E1 a2

29.3

REDSHIFT OF PARTICLE DE BROGLIE WAVELENGTHS

A particle of finite rest mass µ moves along a geodesic world line through the expanding cosmological fluid. Let µv p≡ (29.3.1) (1 − v 2 )1/2 be the spatial 4-momentum of the particle as measured by observers at rest in the fluid. (The ordinary velocity they measure in their proper reference frames is v.) The associated “de Broglie wavelength” of the particle is λ ≡ h/p. (a) Show that this de Broglie wavelength is redshifted in precisely the same manner as a photon wavelength: λ/a = constant. (29.3.2) (b) Employing this result, show that, for the molecules of an ideal gas that fills the universe, their mean kinetic energy decreases in inverse proportion to a2 when the gas is nonrelativistic and (like photon energies) in inverse proportion to a when the gas is highly relativistic.

Solution: (a) Similar to the case in exercise 29.2, we neglect coordinate θ and ϕ. So pχ is still a constant along geodesic. And we also have the normalization condition p2t − a−2 p2χ = µ2 , leading to p = a−1 pχ . Note that λ ∝ 1/p. We have λ/a = constant. (29.3.3) (b) The energy distribution of particles is characterized by exp(−(E − µ)/T ), where T is the temperature of particles. When the gas is nonrelativistic, we have E ≈ µ + p2 /2µ, and so T ∝ p2 ∝ a−2 . When the gas is relativistic, we have E ≈ p ≫ µ, and so T ∝ p ∝ a−1 .

29.4

m(z) DERIVED USING STATISTICAL PHYSICS

Derive the magnitude-redshift relation using a statistical description of the photon distribution [cf. eq. (22.49) and associated discussion].

Solution: Equation 22.49 in the textbook states that the phase space density of photon gas is N = h−4 (Iν /ν 3 ),

(29.4.1)

where Iν ≡

d(energy) . d(time)d(area)d(frequency)d(solid angle)

282

(29.4.2)

And N is constant along geodesic. The flux can be calculated as ∫ F = Iν dν∆Ω,

(29.4.3)

where ∆Ω is the angle extended by the object, ∆Ω = So we have ∫ F = ∆Ω Iνr dνr ∝

1 a(te )2 Σ2 (χr − χe )

∫

A . [a(te )Σ(χr − χe )]2

1 ν 3 νr Iνe r3 dνe = νe νe (1 + z)4 a(te )2 Σ2 (χr − χe )

(29.4.4)

∫ Ie dνe . (29.4.5)

Note that (1 + z)ate = a(tr ) and R ≡ a(tr )Σ(χr − χe ). We obtain F ∝

29.5

1 . (1 + z)2 R2

(29.4.6)

DOPPLER SHIFT VERSUS COSMOLOGICAL REDSHIFT

(a) Consider, in flat spacetime, a galaxy moving away from the Earth with velocity v, and emitting light that is received at Earth. Let the distance between Earth and galaxy, as measured in the Earth’s Lorentz frame at some specific moment of emission, be r; and let the Doppler shift of the radiation when it is eventually received be z = ∆λ/λ. Show that the flux of energy S received at the Earth is related to the galaxy’s intrinsic luminosity L by L S= . (29.5.1) 2 4πr (1 + z)4 (b) Compare this formula for the flux with formula (29.27), where the redshift is of cosmological origin. Why is the number of factors of 1 + z different for the two formulas?

Solution: (a) Suppose there is an observer located at the same place with earth and stationary relative to the galaxy. The flux of energy S received by the observer is S′ =

L . 4πr′2

(29.5.2)

The projected area of the galaxy in two frames are equivalent. Thus the solid angle extended by the galaxy in two frames are related by ′ r2 ∆Ω = r 2 ∆Ω′ . (29.5.3) From equation (22.49), the intensity of light in two frames are related by Iν ′ Iν = ′3 . 3 ν ν

(29.5.4)

Now, we have ∫ S = ∆Ω

Iν dν =

r′2 ∆Ω′ r2

∫

ν3 ν ′ r′2 L ′ I dν = S′ = . ν ′3 ′ 4 2 2 ν ν (1 + z) r 4πr (1 + z)4

(b) Because there is a beaming effect in the case of Doppler redshift.

283

(29.5.5)

29.6

SOURCE COUNTS

Suppose that one could find (which one cannot) a family of light or radio sources that (1) are all identical with intrinsic luminosities L, (2) are distributed uniformly throughout the universe, and (3) are born at the same rate as they die so that the number in a unit comoving coordinate volume is forever fixed. (a) Show that the number of such sources N (z) with redshifts less than z, as observed from Earth today, would be ] [ 3 3 2 N (z) = (constant) · z 1 − (1 + q0 )z + O(z ) . (29.6.1) 2 (b) Show that the number of sources N (S) with fluxes greater than S as observed at Earth today would be ( )3/2 [ ( )1/2 ( )] LH02 LH02 LH02 N (S) = (constant) · 1−3 +O . (29.6.2) 4πS 4πS 4πS

Solution: (a) The number of sources is proportional to the comoving volume  1 1  ∫ χ  2 χ − 4 sin(2χ) 1 ′ ′ = χ3 + O(χ5 ). V = 4πΣ(χ )dχ = 13 χ3  3 0  1 − 2 χ + 41 sinh(2χ)

(29.6.3)

Note the equation (29.12) and (29.16) in the textbook: ℓ = a0 χ, We can get

Now we have

1 z = H0 ℓ + (H0 ℓ)2 + O([H0 ℓ]3 ). 2

(29.6.4)

] [ z 1 + q0 2 z + O(z ) . χ= 1− a0 H0 2

(29.6.5)

[ ] 3 N ∝ χ3 + O(χ5 ) ∝ z 3 1 − (1 + q0 )z + O(z 2 ) . 2

(29.6.6)

(b) The flux of the source received by us is L . 4π(1 + z)2 Σ2 a20

(29.6.7)

If the flux must be larger that S, the maximum redshift of the object would satisfy that ( (1 + z)Σ =

L 4πa20 S

)1/2 .

Note that Σ = χ + O(χ3 ) and equation 29.6.5. We have [ ] z 1 − q0 2 (1 + z)Σ = 1+ z + O(z ) . a0 H0 2

(29.6.8)

(29.6.9)

Now we arrive at the equation [ )1/2 ] ( 1 − q0 LH02 2 z 1+ . z + O(z ) = 2 4πS

284

(29.6.10)

The solution for z is ( z=

LH02 4πS

)1/2 [

q0 − 1 1+ 2

(

LH02 4πS

)1/2

( +O

LH02 4πS

)] .

(29.6.11)

Substitute 29.6.11 into 29.6.1. We can obtain ( N (S) = (constant) ·

29.7

LH02 4πS

)3/2 [ ( )1/2 ( )] LH02 LH02 1−3 +O . 4πS 4πS

(29.6.12)

COSMIC-RAY DENSITY

Suppose the universe has contained the same number of galaxies indefinitely into the past. Suppose further that the cosmic rays in the universe were created in galaxies and that a negligible fraction of them have been degraded or lost since formation. Derive an expression for the average density of energy in cosmic rays in the universe today in terms of: (1) the number density of galaxies, N0 , today; and (2) the nonconstant rate, dE/dz, at which the average galaxy created cosmic-ray energy during the past history of the universe.

Solution: At redshift z in range dz, the average galaxy liberates energy (dE/dz)dz into cosmic rays. Taking the redshift of cosmic ray into account, the average density of energy in cosmic rays in the universe today is ∫ zmax 1 dE ϵ = N0 dz. (29.7.1) 1 + z dz 0

29.8

FRACTION OF SKY COVERED BY GALAXIES

Assume that the redshifts of quasars are cosmological. Let the number of galaxies per unit physical volume in the universe today be N0 , and assume that no galaxies have been created or destroyed since a redshift of > 7. Let D be the average angular diameter of a galaxy. Calculate the probability that the light from a quasar at redshift z, has passed through at least one intervening galaxy during its travel to Earth.

Solution: I think it is the physical diameter of a galaxy that matters and can be averaged. So I will assume the D to be the average physical diameter of a galaxy. Suppose we sit at χ = 0, then the light at redshift z will be at χ(z) =

0 dz ′ 1 √ . z H0 Ωr0 (1 + z ′ )4 + Ωm0 (1 + z ′ )3 + ΩK0 (1 + z ′ )2 + ΩΛ

(29.8.1)

The physical distance from z + dz to z is dl(z) = a

285

dχ dz dz

(29.8.2)

The probability that the light from a quasar at redshift z, has passed through at least one intervening galaxy during its travel to earth is therefore ∫ z P = N0 Ddl(z). (29.8.3) 0

286

Chapter 31

SCHWARZSCHILD GEOMETRY 31.1

TIDAL FORCES ON IN FALLING EXPLORER

(a) Carry out the details of the derivation of the Riemann tensor components (31.6). (b) Calculate, roughly, the critical mass Mcric such that, if M > Mcric the explorer’s body (a human body made of normal flesh and bones) can withstand the tidal forces at r = 2M , but if M < Mcric his body is mutilated by them.

Solution: (a) Focusing on the tˆ and rˆ, the transformation matrix is ( α ˆ

Λαˆ ′ =

γ γβ

γβ γ

) (31.1.1)

where β = v rˆ and γ = (1 − β 2 )−1/2 . So ˆ

ˆ

β γ ˆ α ˆ δ Rτˆρˆ ˆγ δˆ ˆτ ρˆ = Λτˆ Λρˆ Λτˆ Λρˆ Rα ˆ βˆ ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

ˆ

= Λτˆt ΛρˆrˆΛτˆt ΛρˆrˆRtˆrˆtˆrˆ + ΛτˆrˆΛρˆt Λτˆt ΛρˆrˆRrˆtˆtˆrˆ + Λτˆt ΛρˆrˆΛτˆrˆΛρˆt Rtˆrˆrˆtˆ + ΛτˆrˆΛρˆt ΛτˆrˆΛρˆt Rrˆtˆrˆtˆ = γ 4 (1 − 2β 2 + β 4 )Rtˆrˆtˆrˆ (31.1.2)

= Rtˆrˆtˆrˆ. Other equations in (31.6) can be verified by the same procedure. (b) The tidal force per unit mass near the event horizon is of the order g∼

l M l ∼ 2, 3 r M

(31.1.3)

where l is the length scale of the human. If we take gcric = 102 m/s2 and l = 1m, we can get

Mcric

c3 ∼ G

√ l ≈ 4 × 1034 kg ≈ 2 × 104 Msun . g

287

(31.1.4)

31.2

NONRADIAL LIGHT CONES

Show that the world line of a photon traveling nonradially makes an angle less than 45 degrees with the vertical v-axis of a Kruskal-Szekeres coordinate diagram. From this, infer that particles with finite rest mass, traveling nonradially or radially, must always move “generally upward” (angle less than 45 degrees with vertical v-axis).

Solution: The metric in Kruskal-Szekeres coordinate is ds2 =

32M 3 −r/2M e (−dv 2 + du2 ) + r2 (dθ2 + sin2 θdϕ2 ). r

(31.2.1)

For photons, we have 32M 3 −r/2M e [−(pv )2 + (pu )2 ] + r2 [(pθ )2 + (pϕ )2 ] = 0. r

(31.2.2)

Thus, we must have pv > |pu |, i.e. the world line of a photon traveling nonradially makes an angle less than 45 degrees with the vertical v-axis of a Kruskal-Szekeres coordinate diagram. And also, articles with finite rest mass, traveling nonradially or radially, must always move “generally upward”.

31.3

THE CRACK OF DOOM

Use a Kruskal diagram to show the following. (a) If a man allows himself to fall through the gravitational radius r = 2M , there is no way whatsoever for him to avoid hitting (and being killed in) the singularity at r = O. (b) Once a man has fallen inward through r = 2M , there is no way whatsoever that he can send messages out to his friends at r > 2M , but he can still receive messages from them.

Solution: In Kruskal diagram, all particles, traveling nonradially or radially, must always move “generally upward”, i.e. angle less than 45 degrees with vertical v-axis. The horizon r = 2M is v = ±u in the diagram. By simple geometrical analysis, we can verify two statements above.

31.4

HOW LONG TO LIVE?

Show that once a man falling inward reaches the gravitational radius, no matter what he does subsequently (no matter in what directions, how long, and how hard he blasts his rocket engines), he will be pulled into the singularity and killed In a proper time of τ < τmax = πM = 1.54 × 10−5

Solution:

288

M seconds. M⊙

(31.4.1)

The trajectory of longest proper time lapse must be a geodesic. The radial motion of a free particle is described by ) ( )2 ( )( ˜2 dr 2M L ˜2 − 1 − =E 1+ 2 (31.4.2) dτ r r ˜ = 0 and L ˜ = 0, i.e. In order to make τ as large as possible, we must have E (

So

∫ τmax = 0

31.5

2M

√

dr 2M r

−1

dr dτ

= πM = π

)2 =

2M −1 r

(31.4.3)

GM⊙ M M = 1.54 × 10−5 seconds. c3 M ⊙ M⊙

(31.4.4)

EDDINGTON-FINKELSTEIN AND KRUSKAL-SZEKERES COMPARED

Use coordinate diagrams to compare the ingoing and outgoing Eddington-Finkelstein coordinates of Box 31.2 with the Kruskal-Szekeres coordinates. Pattern the comparison after that between Schwarzschild and Kruskal-Szekeres in Figures 31.3 and 31.4.

Solution:

Outgoing E-F coordinates in K-S coordinates

2

2

1

1

v

v

Ingoing E-F coordinates in K-S coordinates

0

0

−1

−1

−2

−2

−2

−1

0

1

2

−2

u

−1

0

1

2

u

˜ in the right. The red lines are constant r, while the blue lines are constant V˜ in the left and constant U

289

31.6

ANOTHER COORDINATE SYSTEM

˜ , V˜ , θ, ϕ coordinate system of Box 31.2. Show such features as Construct a coordinate diagram for the U (1) the relationship to Schwarzschild and to Kruskal-Szekeres coordinates; (2) the location of r = 2M ; and (3) radial geodesics.

Solution: (1) The relationship to Schwarzschild coordinates is ˜ = t − r − 2M ln |r/2M − 1|, U

V˜ = t + r + 2M ln |r/2M − 1|.

(31.6.1)

The relationship to Kruskal-Szekeres coordinates is ˜ = −4M ln(u − v), U

V˜ = 4M ln(u + v).

(31.6.2)

˜ = ∞. (2) The location of r = 2M is V˜ = −∞ and U (3) The line elements for constant θ and ϕ is ) ( 2M ˜ dV˜ . dU ds = − 1 − r 2

(31.6.3)

Connection in this coordinates is ΓV˜ U˜ U˜ = gV˜ U˜ ,U˜ ,

ΓU˜ V˜ V˜ = gV˜ U˜ ,V˜ .

(31.6.4)

All other components vanish. And so ˜

ΓU U˜ U˜ =

gV˜ U˜ ,U˜ gV˜ U˜

,

˜

ΓV V˜ V˜ =

gV˜ U˜ ,V˜ gV˜ U˜

.

The radial geodesics for particles with finite mass are ( )2 ˜ ˜ d2 U dU ˜ U + Γ U˜ U˜ =0 dτ 2 dτ and d2 V˜ ˜ + ΓV V˜ V˜ dτ 2

(

dV˜ dτ

(31.6.5)

(31.6.6)

)2 = 0.

(31.6.7)

And we also have normalization

˜ dV˜ dU = 1. dτ dτ ˜ = constant or V˜ = constant. While the radial geodesics for photons are U gU˜ V˜

31.7

(31.6.8)

SCHWARZSCHILD METRIC IN ISOTROPIC COORDINATES

(a) Show that, rewritten in the isotropic coordinates of Exercise 23.1, the Schwarzschild metric reads ( )2 )4 ( M 1 − M/2¯ r dt2 + 1 + [d¯ r2 + r¯2 (dθ2 + sin2 θdϕ2 )]; (31.7.1) ds2 = − 1 + M/2¯ r 2¯ r and derive the transformation

( )2 M r = r¯ 1 + 2¯ r

290

(31.7.2)

between the two radial coordinates. (b) regions of spacetime (I, II, III, IV; see Figure 31.3) are covered by the isotropic coordinate patch, and which are not? (c) Calculate and construct an embedding diagram for the spacelike hypersurface t = 0, 0 < r¯ < ∞. (d) Find a coordinate transformation that interchanges the region near r = 0 with the region near r¯ = ∞, while leaving the metric coefficients in their original form.

Solution: (a) The metric in Schwarzschild coordinates is ( ) ( )−1 2M 2M ds2 = − 1 − dt2 + 1 − dr2 + r2 dΩ2 . r r

(31.7.3)

The metric in isotropic coordinates must be the form of ds2 = −f 2 (¯ r)dt2 + g 2 (¯ r)(d¯ r2 + r¯2 dΩ2 ). So we have

( )−1/2 2M g(¯ r)d¯ r = 1− dr, r

g(¯ r)¯ r = r, i.e.

(31.7.4)

(31.7.5)

d¯ r dr = √ . r¯ r 1 − 2M/r

(31.7.6)

By choosing r¯ = M/2 when r = 2M , we have √ 1 + 1 − 2M/r 2¯ r √ = . M 1 − 1 − 2M/r Now it is easy to derive

(31.7.7)

( )2 M r = r¯ 1 + . 2¯ r

√ 1 − M/2¯ r 1 − 2M/r = , 1 + M/2¯ r

(31.7.8)

And the metric in isotropic coordinates is ( ds = − 2

1 − M/2¯ r 1 + M/2¯ r

)2

(

M dt + 1 + 2¯ r

)4

2

(d¯ r2 + r¯2 dΩ2 ).

(31.7.9)

(b) When r¯ = M/2,r(¯ r) has minimum 2M . When r¯ → 0 or r¯ → ∞, r → ∞. So region I and III are covered by isotropic coordinate patch. (c) The embedding function z(¯ r) is given by ( dr2 + dz 2 = The solution is

(d) The transformation r¯ →

1+

M 2¯ r

(

)4 d¯ r2 ,

√ z = 4M r/2M − 1 = 4M M2 4¯ r

r= √

1+

M 2¯ r

)2

M 1 r¯ + − . 2M 8¯ r 2

r¯.

(31.7.10)

(31.7.11)

leaves z(¯ r) and r(¯ r) unchanged. So it will also leave metric unchanged.

(e) Similarly explore the spacetime geometry for Q = M .

291

31.8

REISSNER-NORDSTROM GEOMETRY

(a) Solve the Einstein field equations for a spherically symmetric, static gravitational field ds2 = −e2Φ(r) dt2 + e2Λ(r) dr2 + r2 (dθ2 + sin2 θdϕ2 ),

(31.8.1)

with no matter present, but with a radial electric field B = 0, E = f (r)erˆ in the static orthonormal frame ˆ

ω t = eΦ dt,

ˆ

ω rˆ = eΛ dr,

ω θ = rdθ,

ˆ

ω ϕ = r sin θdϕ.

(31.8.2)

Use as a source in the Einstein field equations the stress-energy of the electric field. (b) Show that the constant Q is the total charge as measured by a distant observer (r ≫ 2M and r ≫ Q), who uses a Gaussian flux integral, or who studies the coulomb-force-dominated orbits of test charges with charge-to-mass ratio e/µ ≫ M/Q. What is the charge-to-mass ratio, in dimensionless units, for an electron? Show that the constant M is the total mass as measured by a distant observer using the Keplerian orbits of electrically neutral particles. (c) Show that for Q > M , the Reissner-Nordstrom coordinate system is well behaved from r = ∞ down to r = 0, where there is a physical singularity and infinite tidal forces. (d) Explore the nature of the spacetime geometry for Q < M , using all the techniques of this chapter (coordinate transformations, Kruskal-like coordinates, studies of particle orbits, embedding diagrams, etc.). (e) Similarly explore the spacetime geometry for Q = M . (f) For the case of a large ratio of charge to mass, show that the region near r = 0 is unphysical. More precisely, show that any spherically symmetric distribution of charged stressed matter that gives rise to the fields E = Q/r2 er outside its boundary must modify these fields for r < r0 = Q2 /2M .

Solution: (a) In the static orthonormal frame, the stress energy tensor of the EM field is Ttˆtˆ =

f2 , 8π

Trˆrˆ = −

f2 , 8π

Tθˆθˆ =

f2 , 8π

Tϕˆϕˆ =

f2 . 8π

(31.8.3)

All other terms vanish. On the other hand, we have Gtˆtˆ = r−2 (1 − e−2Λ ) + 2r−1 e−2Λ Λ′ , Grˆrˆ = −r−2 (1 − e−2Λ ) + 2r−1 e−2Λ Φ′ [ ] Gθˆθˆ = Gϕˆϕˆ = e−2Λ r−1 (Φ′ − Λ′ ) + Φ′′ + Φ′2 − Φ′ Λ′ .

(31.8.4)

Using Einstein equation, we have Gtˆtˆ + Grˆrˆ = 0, from which we can obtain Λ′ + Φ′ = 0. By Gtˆtˆ = Gθˆθˆ, we can get (r2 e−2Λ )′′ = 2. The general solution is r2 e−2Λ = r2 −2M r+Q2 . And so e2Φ = C(1−2M/r+Q2 /r2 ). By rescaling t coordinates, we can get )−1 ( Q 2M Q2 2M Q2 2Φ 2Λ . (31.8.5) f (r) = 2 , e = 1 − + 2 , e = 1− + 2 r r r r r The metric is therefore ) ( )−1 ( 2M Q2 Q2 2M 2 2 dr2 + r2 (dθ2 + sin2 θdϕ2 ). + 2 dt + 1 − + 2 ds = − 1 − r r r r (b) Using Gaussian theorem, we have

I

∫ E · dS =

4πQtotal = S

Q 2 · r sin θdθdϕ = 4πQ, r2

292

(31.8.6)

(31.8.7)

i.e. Q is the total charge as measured by a distant observer. For electron, the charge-to-mass ratio in dimensionless units is √ √ e ke Ge/c2 ke e = = = 2.04 × 1021 . me Gme /c2 G me

(31.8.8)

For electrically neutral particles, there orbits will follow geodesics of the metric. For a distant observer, the r−2 term in the metric can be neglected. So the orbit of a electrically neutral particle will be essentially the same as that in Schwarzschild metric. Thus M can be viewed as the total mass of the system. (c) If Q > M , then we have

1−

2M Q2 (r − M )2 + Q2 − M 2 + 2 = > 0. r r r2

(31.8.9)

So the Reissner-Nordstrom coordinate system is well behaved from r = ∞ down to r = 0, where there is a physical singularity and infinite tidal forces. √ (d) If Q < M , there will be two horizons r± = M ± M 2 − Q2 . In the region r > r+ and r < r− , the t direction (constant r) is time like. While In the region r− < r < r+ , the t direction is time like. The conformal diagram for the whole spacetime is given by the figure below.

(e) There is only one horizon r = M in this case. And the t direction is spacelike when r < M and r > M (r = M is null space). The conformal diagram for the whole spacetime is given by the figure below.

293

(f) For the case of a large ratio of charge to mass, r = 0 is a naked singularity which is observable from the outside. I known little about the naked singularities and their properties. Any help is appreciated. Please refer to Gravitational collapse (Penrose, R.) and Generalized strong curvature singularities and weak cosmic censorship in cosmological space-times (W. Rudnicki et al.) for more information.

294

Chapter 32

GRAVITATIONAL COLLAPSE 32.1

UNIQUENESS OF REISSNER-NORDSTROM GEOMETRY

Prove the following generalization of Birkhoff’s theorem. Let the geometry of a given region of spacetime (1) be spherically symmetric, and (2) be a solution to the Einstein field equations with an electromagnetic field as source. Then that geometry is necessarily a piece of the Reissner-Nordstrom geometry [equation (31.24b)] with electric and magnetic fields, as measured in the standard static orthonormal frames E=

Qe erˆ, r2

B=

Qm erˆ, r2

Q = (Q2m + Q2e )1/2 .

(32.1.1)

Solution: (1) First consider regions of spacetime in which (∇r)2 ̸= 0. In this case, one can introduce Schwarzschild coordinates ds2 = −e2Φ dt2 + e2Λ dr2 + r2 (dθ2 + sin2 θdϕ2 ),

Φ = Φ(t, r),

Λ = Λ(t, r),

(32.1.2)

according to Box 23.3 of the textbook. Using the orthonormal components of the Einstein tensor as derived in exercise 14.16, we have Gtˆtˆ = r−2 (1 − e−2Λ ) + 2(Λ,r /r)e−2Λ , Gtˆrˆ = Grˆtˆ = 2(Λ,t /r)e−(Λ+Φ) , Grˆrˆ = 2(Φ,r /r)e−2Λ − r−2 (1 − e−2Λ ), Gθˆθˆ = Gϕˆϕˆ = (Φ,rr + Φ2,r − Φ,r Λ,r + Φ,r /r − Λ,r /r)e−2Λ − (Λ,tt + Λ2,t − Λ,t Φ,t )e−2Φ , Gtˆθˆ = Gθˆtˆ = Gtˆϕˆ = Gϕˆtˆ = Grˆθˆ = Gθˆ ˆr = Grˆϕ ˆ = Gϕˆ ˆr = Gθˆϕ ˆ = Gϕ ˆθˆ = 0.

(32.1.3)

The Einstein equation states that Gµˆνˆ = 8πTµˆνˆ , where Tµˆνˆ is the stress energy of an an electromagnetic field. According to exercise 20.6 of the textbook, the Maxwell stress-energy tensor has zero trace, and that its square is a multiple of the unit tensor. Thus we have Gµˆνˆ = 0,

Gµˆαˆ Gαˆ νˆ ∝ δνµˆˆ .

(32.1.4)

Or explicitly, − Gtˆtˆ + Grˆrˆ + 2Gθˆθˆ = 0,

Gtˆrˆ(Gtˆtˆ − Grˆrˆ) = 0,

G2tˆtˆ − G2tˆrˆ = G2rˆrˆ − G2tˆrˆ = G2θˆθˆ.

If Gtˆrˆ = 0, then it is easy to derive that Gtˆtˆ = −Grˆrˆ = Gθˆθˆ.

295

(32.1.5)

Firstly, we have Λ,t = 0, according to Gtˆrˆ = 0. Next, Gtˆtˆ + Grˆrˆ = 0 leads to Λ,r + Φ,r = 0. Furthermore, using Gtˆtˆ = Gθˆθˆ, we find that (r2 e−2Λ ),rr = 2. The general solution is e−2Λ = 1 −

2M Q2 + 2. r r

(32.1.6)

Then applying Λ,r + Φ,r = 0, we have ) ( Q2 2M e2Φ = C 2 (t) 1 − + 2 r r ∫t By redefining the time coordinate tnew = C(t)dt, we will obtain the R-N metric (

2M Q2 ds = − 1 − + 2 r r 2

)

(

Q2 2M + 2 dt + 1 − r r

(32.1.7)

)−1 dr2 + r2 (dθ2 + sin2 θdϕ2 ).

2

The stress energy tensor of EM field in the local Lorentz coordinate frame is [ ] (E × B)ˆi E2 + B2 1 1 2 2 Ttˆtˆ = , Ttˆˆi = Tˆitˆ = − , Tˆj kˆ = −(Eˆj Ekˆ + Bˆj Bkˆ ) + (E + B )δˆj kˆ . 8π 4π 4π 2

(32.1.8)

(32.1.9)

It is easy to verify that E and B must be radial to produce the required from of stress energy tensor in this case. Suppose the electromagnetic field is E = f (t, r)erˆ,

(32.1.10)

B = g(t, r)erˆ.

Then we have

f 2 + g2 . 8π The field equation gives that f 2 + g 2 = Q2 /r4 . So the EM field is Ttˆtˆ = −Trˆrˆ = Tθˆθˆ = Tϕˆϕˆ =

E=

Qe erˆ, r2

B=

Qm erˆ, r2

(32.1.11)

Q = (Q2m + Q2e )1/2 .

(32.1.12)

If Grˆrˆ ̸= 0, then it is easy to derive that Gtˆtˆ = Grˆrˆ = |Gtˆrˆ| and Gθˆθˆ = Gϕˆϕˆ = 0. The spacetime will be described by Vaidya metric, [ ] 2M (u) ds2 = − 1 − du2 − 2dudr + r2 dΩ2 (outgoing), r [ ] 2M (v) (32.1.13) ds2 = − 1 − dv 2 + 2dvdr + r2 dΩ2 (ingoing) r in null-radial coordinates. For outgoing Vaidya metric, the Einstein tensor is Guu = −2

M,u . r2

(32.1.14)

All other components vanish. Suppose the coordinate transformation t(u, r) can bring the metric into diagonal coordinates. Then we have ( )−1 ( )2 ( )−1 ∂u 2M ∂u ∂u ∂u =− 1− , gtt = , grr = − . (32.1.15) ∂r r ∂t ∂r ∂r The Einstein tensor in t − r coordinate is ( )2 ∂u Gtt = Ruu , ∂t

Gtr

∂u ∂u = Ruu , ∂t ∂r

( Grr =

∂u ∂r

)2 Ruu .

(32.1.16)

Written in orthonormal frame, we have Gtˆtˆ = −Gtˆrˆ = Grˆrˆ = −

∂u −2M,u Guu = 2 . ∂r r (1 − 2M/r)

296

(32.1.17)

In the case of ingoing Vaidya metric, the analysis can be carried in the same way. Please refer to the paper Vaidya spacetime in the diagonal coordinates (Berezin et al. 2017) for more discussion on coordinate transformation. If the source of stress energy tensor is EM field, we can easily deduce that Erˆ = Brˆ = 0, E · B = 0 and |E| = |B|. It seems that a EM wave propagating radially can generate the Vaidya spacetime. However, according to Vaidya metric, the Vaidya field is a pure radiation field rather than electromagnetic fields. The emitted particles or energy-matter flows have zero rest mass and thus are generally called “null dusts", typically such as photons and neutrinos, but cannot be electromagnetic waves because the MaxwellNewman-Penrose equations are not satisfied. I know little about Maxwell-Newman-Penrose equations and the detailed proof is beyond my ability. From know on, I will take it for granted that EM field is non-null. The following discussion is based on the “Hint” of the exercise in the textbook. (2) Any region of dimensionality less than four, In which (∇r)2 = 0 (e.g., the Schwarzschild radius), can be treated as the join between four-dimensional regions with (∇r)2 ̸= 0. Moreover, the geometry of such a region is determined uniquely by the geometry of the adjoining four-dimensional regions (“junction conditions”; section 21.13). Since the adjoining regions are necessarily Reissner-Nordstrom, then so are such “sandwiched” regions. (3) Next consider four-dimensional regions in which (∇r)2 = 0 is null and nonzero. According to Box 23.3, the metric of spherically symmetric spacetime has the form ds2 = gtt dt2 + grr dr2 + R2 (r, t)dΩ2 .

(32.1.18)

Now we build a new coordinate system (T, R). If (∇R)2 = 0, then g RR = g tt (

∂R 2 ∂R 2 ) + g rr ( ) = (∇R)2 = 0. ∂t ∂r

(32.1.19)

We can choose T (r, t) satisfying g tt (

∂T 2 ∂T 2 ) + g rr ( ) = 0, ∂t ∂r

(32.1.20)

i.e. choose another null surface of the spacetime to be the T coordinate plane. We now have g T T = g RR = 0, and so gT T = gRR = 0. By a change of notation, the metric has the form ds2 = −2Ψdrdt + r2 (dθ2 + sin2 θdϕ2 ),

(32.1.21)

where Ψ = Ψ(r, t). It is straight to get Rθθ = 1 and Rϕϕ = sin2 θ from the metric above. In orthonormal frame, we have Rθˆθˆ = Rϕˆϕˆ = 1/r2 . Whereas the stress-energy tensor for a spherically symmetric electromagnetic field has ( ) 1 ˜ 8π Tθˆθˆ ≡ 8π Tθˆθˆ − gθˆθˆT = Q2 /r4 , Q = constant. (32.1.22) 2 These quantities, Rθˆθˆ and 8π T˜θˆθˆ must be equal (Einstein’s field equation) but cannot be because of their different r-dependence. Thus, an electromagnetic field cannot generate regions with dr = 0, dr · dr = 0. (4) Finally, consider four-dimensional regions in which dr = 0. Denote the constant value of r by a. Starting from equation 32.1.18, we have ds2 = gt′ t′ dt′2 + gr′ r′ dr′2 + a2 dΩ2 .

(32.1.23)

Clearly, (t′ , r′ ) and (θ, ϕ) are decoupled with each other. So we can neglect (θ, ϕ) and only focus on the (t′ , r′ ) coordinates. In two dimensional manifold, there are only 3 independent components in metric. The coordinate transformation can further eliminate two degrees of freedom. So the metric can be brought into the form ds2 = a2 (−d˜ τ 2 + e2λ dz 2 + dΩ2 ). (32.1.24)

297

˙ We choose the initial point of τ˜ coordinate for every z = constant in order to make λ(z) = 0 at that point, and rescale the line τ˜ = 0 to make λ(z) = 0 at that point. Then we have λ(˜ τ , z) = 0,

˙ τ , z) = 0. λ(˜

(32.1.25)

In the associated orthonormal frame, spherical symmetry demands E=

Qe ezˆ, a2

B=

Qm ezˆ, a2

Q = (Q2m + Q2e )1/2 .

(32.1.26)

The Ricci tensor in orthonormal coordinates are ¨ Rzˆzˆ = a−2 (λ˙ 2 + λ), ¨ R ˆˆ = R ˆ ˆ = a−2 . Rτˆτˆ = −a−2 (λ˙ 2 + λ), θθ ϕϕ

(32.1.27)

¨ = −1. For gravitation field with E-M field as source, the trace of Einstein tensor vanish, leading to λ˙ 2 + λ Under the constraint 32.1.25, we have the unique solution λ = ln cos τ˜. Also note that 8πTθˆθˆ = Q2 /a4 . We must have Q = a. Now the metric becomes ds2 = Q2 (−d˜ τ 2 + cos2 τ˜dz 2 + dΩ2 ).

(32.1.28)

(5) The Reissner-Nordstrom solution for the special case Q = M is ( )2 ( )−2 Q Q ds2 = − 1 − dt2 + 1 − dr2 + r2 dΩ2 . r r

(32.1.29)

Perform the following coordinate transformation on the Reissner-Nordstrom throat region (|r − Q| ≪ Q): r − Q = Qe−z cos τ˜,

t = Qez tan τ˜.

(32.1.30)

Firstly, we have ( gτ˜τ˜ =

∂t ∂ τ˜

)2

( gtt +

∂r ∂ τ˜

)2

[ grr = −Q2

1 sin2 τ˜(1 + e−z cos τ˜)2 − 2 −z 2 cos τ˜(1 + e cos τ˜) cos2 τ˜

] (32.1.31)

Note that z ≫ 0 in the throat region. If we keep the zero order, we have gτ˜τ˜ = −Q2 . Similarly, we have [ ] ∂r ∂r 1 ∂t ∂t −z 2 gτ˜z = gtt + grr = Q2 tan τ˜ − + (1 + e cos τ ˜ ) ≈0 (32.1.32) ∂ τ˜ ∂ z˜ ∂ τ˜ ∂ z˜ (1 + e−z cos τ˜)2 and ( gzz =

∂t ∂z

)2

( gtt +

∂r ∂z

)2

[ grr = Q2 (1 + e−z cos τ˜)2 −

] sin2 τ˜ ≈ Q2 cos2 τ˜. (1 + e−z cos τ˜)2

(32.1.33)

So the solution of the field equations in (4) is actually the throat of the Reissner-Nordstrom solution for the special case Q = M . (6) Thus, each possible case leads either to no solution at all, or to a segment of the Reissner-Nordstrom geometry.

32.2

REDSHIFTS DURING COLLAPSE

(a) Let a radio transmitter on the surface of a collapsing spherical star emit monochromatic waves of wavelength λe ; and let a distant observer, at the same θ, ϕ, as the transmitter, receive the waves. Show that at late times the wavelength received varies as λrec /λem ∝ et/4M ,

298

(32.2.1)

where t is proper time as measured by the distant observer. (b) Use kinetic theory for the outgoing photons (conservation of density in phase space: Liouville’s theorem; 22.6) to show that the energy flux of the radiation received (ergs/cm2 sec) varies as F ∝ e−t/M .

(32.2.2)

(c) Suppose that nuclear reactions at the center of the collapsing star generate neutrinos of energy Ee , and that these neutrinos flow freely outward (negligible absorption in star). Show that the energy of the neutrinos received by a distant observer decreases at late times as Erec /Eem ∝ e−t/4M .

(32.2.3)

(d) Show that the flux of neutrino energy dies out at late times as F ∝ e−t/2M .

(32.2.4)

(e) Explain in elementary terms why the decay laws 32.2.1 and 32.2.3 for energy are the same, but the decay laws 32.2.2 and 32.2.4 for energy flux are different. (f) Let a collapsing star emit photons from its surface at the black-body rate ( ) dN 11 photons = 1.5 × 10 × (surface area of star) × (temperature of surface)3 . dτ cm2 sec K3

(32.2.5)

Let a distant observer count the photons as they pass through his sphere of radius r ≫ M . Let him begin his count (time t = 0) when he sees (via photons traveling radially outward) the center of the star’s surface pass through the radius r = 3M . Show that, in order of magnitude, the time he and his associates must wait, until the last photon that will ever get out has reached them, is t = (M/M⊙ [8 × 10−4 + 5 × 10−5 log10 (T11 M/M⊙ )]seconds,

(32.2.6)

where T11 is the star’s surface temperature in units of 1011 K.

Solution: Solutions to exercise (a) and (b) are based on section 15 of the paper Relativistic astrophysics. I (Zel’dovich et al. 1964). (a) In the crudest approximation, we can neglect the effects of pressure and put P = 0. Let us consider the surface of a collapsing star. During the course of contraction the mass M dose not change, and therefore the particle on the surface simply falls under the influence of the gravitation of the mass M . Consequently, in order to explain the character of the collapse, it is sufficient to consider the free fall of a trial particle in the field of the M . For the rate of free fall in a Schwarzschild field, we have ( )( )1/2 dr 2M 1 − 2M/r , =− 1− 1− dt r 1 − 2M/r0

(32.2.7)

where r0 is the distance from which the fall begins and at which dr/dt = 0. The local stationary observer, situated side the falling body, will measure its velocity as follows ( )1/2 1 − 2M/r v ≡ 1/2 = − 1 − . 1 − 2M/r0 g00 dt 1/2

grr dr

299

(32.2.8)

The orbit for a radially propagating photon satisfy that dr 2M =1− . dt r The integration gives ∆t = 2M ln

(32.2.9)

R − 2M + R − r, r − 2M

(32.2.10)

where R is the position of the observer. With the aid of expression 32.2.7 we find r = r(t′ ), i.e., the position of the trial particle at the instant t′ as measured by the clock of the remote observer. But this is not the same place where this observer sees the particle at the instant t′ , since it takes the light some time ∆t to cover the path from the particle to the observer. This time can be readily calculated by formula 32.2.10. Denoting the time of arrival of the light at the observer by t: t = t′ + ∆t (32.2.11) With the aid of the expressions given above we can easily obtain the formula r = r(t) for the falling particle. At late time of collapse, r → 2M , the asymptotic form of the formula is: r = 2M + (r1 − 2M )e−

t−t1 4M

,

(32.2.12)

where r1 is the position of the particle at the instant t1 . During the propagation of photons in Schwarzschild metric, p0 is constant. The four momentum of photon is pα = [p0 , p0 (1 − 2M/r)−1 , 0, 0]. The four velocity of surface particle at r is uα = [(1 − 2M/r)−1/2 γ, (1 − 2M/r)1/2 γv, 0, 0], where γ ≡ (1 − v 2 )−1/2 . So the energy of photon according to the emitter is ϵem = −pα uα =

γ(1 − v) p0 . (1 − 2M/r)1/2

(32.2.13)

The four velocity of observer at R is uα = [(1 − 2M/R)−1/2 , 0, 0, 0]. So the energy of photon measured by the observer is ϵrec = (1 − 2M/R)−1/2 p0 . (32.2.14) So the redshift of the photon is 1+z =

γ(1 − v)(1 − 2M/R)1/2 (1 − 2M/r)1/2

(32.2.15)

From 32.2.8 it follows that v → −1 and γ → (1 − 2M/r)−1/2 as r → 2M . Now we have λrec ∝ (1 − 2M/r)−1 ∝ et/4M . λem

(32.2.16)

(b) The kinetic theory for the outgoing photons ensure the conservation of density in phase space. So Iν /v 3 is a constant during propagation. The flux from the central point of the visible disc of a contracting star received by observer is ∫ ∫ ∫ Iem 3 νrec −4 F = ∆Ω Irec dνrec = ∆Ω ν dν = z ∆Ω Iem dνem , (32.2.17) em rec 3 νem νem where ∆Ω is kept as a constant during observation. So we have F ∝ e−t/M .

(32.2.18)

For the entire disc, the deductions are much more complicated, since it is necessary to consider rays moving at a large angle to the radius, and the paths of such rays near the star are quite complicated. An analysis (Podurets, 1964) of this question shows that the formula for the luminosity of the entire star F is ( ) t F ∝ exp − √ (32.2.19) 3 3M

300

(c) We will assume that collapsing star is composed of dust with uniform density and negligible pressure. The interior of the star can be described by closed Friedmann spacetime ds2 = −dτ 2 + a2 (τ )[dχ2 + sin2 χ dΩ2 ],

(32.2.20)

where χ = χ0 is the surface star. And we have r = a sin χ0 , where r is the surface of star in Schwarzschild metric. Suppose the neutrino is produced at χ = 0, η = η0 with energy Eem . It will propagate to the surface of star at η = η0 + χ0 with energy Eem aη0 /a(η0 + χ0 ) measured by observer fixed at χ = χ0 , i.e. observer comoving with star surface. When the neutrino propagates to the observer far way observer at R, the energy measured measured by the observer will be Erec =

r(η0 )γ(1 − v)(1 − 2M/R)1/2 Eem , r(1 − 2M/r)1/2

(32.2.21)

caused by gravitational redshift and Doppler effect. When r → 2M , r(η0 )/r can be approximated as a constant. So similarly, we will have Erec =∝ e−t/4M . (32.2.22) Eem (d) The energy of every neutrino measured by observer will be ϵem (1 + z)−1 due to gravitational redshift. The number of neutrinos received by observer per unit time will be (1 + z)−1 of that produced by nuclear reaction per unit time due to gravitational time dilation. So the flux of neutrinos will scale as (1 + z)−2 , i.e. F ∝ (1 + z)−2 ∝ e−t/2M . (32.2.23) (e) When calculating the flux from the central point of the visible disc of a contracting star. Only the photon from a fixed solid angle is taken into account. The bending of the ray trajectories in the gravitational field and the beaming effect caused by Doppler effect must be considered. Neutrino sources are situated in the center of the contracting star. There are no light ray bending or beaming effect. Only redshift and time dilation should be taken into account. (f) I do not know how to tackle this problem yet.

32.3

EMBEDDING DIAGRAMS AND PHOTON PROPAGATION FOR COLLAPSING STAR

Verify in detail the features of homogeneous collapse described in Box 32.1. The features of homogeneous collapse described in Box 32.1 are intuitive from diagrams in the textbook. I think a detailed demonstration is unnecessarily since there is little mathematics involved in it.

32.4

MATCH OF FRIEDMANN INTERIOR TO SCHWARZSCHILD EXTERIOR

The Einstein field equations are satisfied on a star’s surface if and only if the intrinsic and extrinsic geometries of the surface’s three-dimensional world tube are the same, whether measured on its interior or on its exterior. Verify that for the collapsing star discussed above, the intrinsic and extrinsic geometries match at the join between the Friedmann interior and the Schwarzschild exterior.

Solution:

301

The original discussion can be found in Terminal Configurations of Stellar Evolution (Beckedorff et al. 1962). (a) The intrinsic geometries of the surface’s three-dimensional world tube measured on its exterior are ( ) ( )−1 2M 2M ds2 = − 1 − dt2 + 1 − dR2 + R2 (t)dΩ2 , R(t) R(t)

(32.4.1)

( )( )1/2 dR 2M 1 − 2M/R =− 1− 1− . dt R 1 − 2M/Ri

(32.4.2)

with

Eliminate t and we can get ds2 = −

dR2 + R2 dΩ2 . 2M (1/R − 1/Ri )

(32.4.3)

The intrinsic geometries measured on its interior are ds2 = a2 (η)(−dη 2 + sin2 χ0 dΩ2 ).

(32.4.4)

And we have R = a sin χ0 = Ri (1 + cos η)/2, As (

dR dη

)2

2M/Ri = sin2 χ0 .

[ ( )2 ] ) ( 2 Ri2 R 2R 1 1 2 i 2 = sin η = 1− − −1 = R Ri 4 4 Ri R Ri ) ( ) ( 1 1 1 1 − − = a2 sin2 χ0 Ri = 2M a2 , R Ri R Ri

(32.4.5)

(32.4.6)

the intrinsic geometries are the same, whether measured on its interior or on its exterior. (b) The inner metric is ds2 = a(η)2 (−dη 2 + dχ2 + sin2 χdΩ2 ).

(32.4.7)

The world tube of star surface is χ = χ0 . The normal vector of the surface is nα = (0, a, 0, 0). The extrinsic curvature is β µ χ α α β Kab = nα;β eα (32.4.8) a eb = nα,b ea − Γ αβ nµ ea eb = −aΓ ab . Given the Γχab , we can get Kηη = Kηθ = Kηϕ = Kθϕ = 0, The outer metric is

(

2M ds = − 1 − r

)

2

Kθθ = Kϕϕ / sin2 θ = a sin χ0 cos χ0 . (

2M dt + 1 − r

(32.4.9)

)−1

2

dr2 + r2 dΩ2 .

(32.4.10)

The world tube of star surface can be parameterize as [ ] cot χ0 + tan(η/2) 1 2 t = 2M ln + 2M cot χ0 η + csc χ0 (η + sin η) , cot χ0 − tan(η/2) 2 r = 2M csc2 χ0 cos2 (η/2).

(32.4.11)

The tangent vector of the surface is eα η = r,η (t,R , 1, 0, 0),

eα θ = (0, 0, 1, 0),

eα ϕ = (0, 0, 0, 1).

(32.4.12)

The normal vector of the surface is √

1 − 2M/R(1, −t,R , 0, 0) nα = √ . −1 + (1 − 2M/R)2 t2,R

302

(32.4.13)

Note that t,R =

1 −(1 − 2M/Ri )1/2 = . dR/dt (1 − 2M/R)(2M/R − 2M/Ri )1/2

(32.4.14)

We can deduce that [( nα =

2M 2M − R Ri

] )1/2 ( )1/2 ( )−1 2M 2M , 1− 1− , 0, 0 Ri R

(32.4.15)

The extrinsic curvature is µ α β Kab = nα,b eα a − Γ αβ nµ ea eb .

(32.4.16)

Kab = −Γtαβ nt − Γrαβ nr .

(32.4.17)

If one of a and b are θ or ϕ, we have

For example, Γrθθ = 2M − r, Γtθθ = 0. Thus, ( )1/2 2M Kθθ = R 1 − = a sin χ0 cos χ0 . Ri

(32.4.18)

Similar, we can derive that Kϕϕ = a sin χ0 cos χ0 sin2 θ, Kθϕ = Kηθ = Kηϕ = 0. Next, we calculate Kηη . We have 2 Kηη = nt,η t,η + nr,η r,η − Γrtt nr t2,η − 2Γttr nt t,η r,η − Γrrr nr r,η 2 2 = nt,R r,η t,η + nr,R r,η − Γrtt nr t2,η − 2Γttr nt t,η r,η − Γrrr nr r,η −1 r r t = (nt,R + nr,R t−1 ,R − Γ tt nr t,R − 2Γ tr nt − Γ rr nr t,R )r,η t,η

(32.4.19)

= 0.

So we demonstrate that extrinsic geometries are the same, whether measured on its interior or on its exterior.

32.5

STARS THAT COLLAPSE FROM INFINITY

(a) Patch together a truncated Schwarzschild geometry and the geometry of a truncated “flat” (k = 0), dust-filled Friedmann universe to obtain a model of a star that collapses from rest at an infinite initial radius. (b) Similarly patch together a truncated Schwarzschild geometry and the geometry of a truncated “open” (k = −1), dust-filled Friedmann universe to obtain a star which collapses from infinity with finite initial inward velocity.

Solution: (a) For a particle which free falls from infinity with zero initial zero velocity, we have √ dt 1 dr 2M = , =− . dτ 1 − 2M/r dτ r

(32.5.1)

The world tube of the star surface is r = R(t) given by equations above. If we parameters the 3-surface with τ , θ, ϕ, we can get ds2 = −dτ 2 + R2 (τ )dΩ2 . (32.5.2)

303

If R = Ri at τ = 0, we can deduce that [( )3/2 ( )3/2 ] 2 Ri R τ − = . 3 2M 2M 2M

(32.5.3)

The metric in the star is ds2 = −dτ 2 + a2 (τ )(dχ2 + χ2 dΩ2 ).

(32.5.4)

The star surface is given by χ = χ0 . The evolution of a is determined by friedmann equation a˙ 2 8πρ0 = 2 a 3

(

a a0

)−3 (32.5.5)

,

where ρ0 and a0 is the density of the dust and scale factor at τ = 0. The solution is [ ] τ 2 ( a0 χ0 )3/2 ( aχ0 )3/2 − = , 3 2m 2m 2m

(32.5.6)

where m = 4πρ0 χ30 a30 /3. Two surfaces can be connected if M = m and a0 χ0 = Ri . (b) For a particle which free falls from infinity with zero initial zero velocity, we have √ ˜ dt E dr ˜ 2 − 1 + 2M . = , =− E dτ 1 − 2M/r dτ r

(32.5.7)

The world tube of the star surface is r = R(t) given by equations above. If we parameters the 3-surface with τ , θ, ϕ, we can get ds2 = −dτ 2 + R2 (τ )dΩ2 . (32.5.8) The metric in the star is ds2 = −dτ 2 + a2 (τ )(dχ2 + sinh2 χdΩ2 ).

(32.5.9)

The star surface is given by χ = χ0 . The evolution of a is determined by friedmann equation a˙ 2 1 8πρ0 − 2 = 2 a a 3

(

a a0

)−3 ,

(32.5.10)

where ρ0 and a0 is the density of the dust and scale factor at τ = 0. The equation above can be simplified to √ da 2m sinh χ0 =− + sinh2 χ0 , (32.5.11) dτ a sinh χ0 ˜ = cosh χ0 and r = a sinh χ0 . where m = 4πρ0 sinh3 χ0 a30 /3. Two surfaces can be connected if M = m, E

32.6

GENERAL SPHERICAL COLLAPSE: METRIC IN COMOVING COORDINATES

Consider an inhomogeneous star with pressure, undergoing spherical collapse. Spherical symmetry alone is enough to guarantee the existence of a Schwarzschild coordinate system t, r, θ, ϕ throughout the interior and exterior of the star [see equation (32.2) and preceding discussion]. Label each spherical shell of the star by a parameter a, which tells how many baryons are contained interior to that shell. Then r(a, t) is the world line of the shell with label a. The expression for these world lines can be inverted to obtain a(r, t), the number of baryons interior to radius r at time t. Show that there exists a new time coordinate t˜(t, r), such that the line element (32.2), rewritten in the coordinates (t˜, a, θ, ϕ), has the form ˜ 2Φ

ds = −e 2

[ ]2 (∂r/∂a)t˜da 2 ˜ dt + + r2 dΩ2 , Γ

˜ = Φ( ˜ t˜, a), r = r(t˜, a), Γ = Γ(t˜, a). Φ

304

(32.6.1)

These are “comoving, synchronous coordinates” for the stellar interior.

Solution: Suppose

[ ] dt˜ = eψ e−2Λ a,r dt + e−2Φ a,t dr ,

(32.6.2)

eψ is the integrating factor to ensure the existence of t˜. The coordinate transformation (t, r) → (t˜, a) bring equation (32.2) into ds2 = −e−2ψ+2Φ+2Λ (e−2Λ a2,r − e−2Φ a2,t )−1 dt˜2 + (e−2Λ a2,r − e−2Φ a2,t )−1 da2

(32.6.3)

˜ and Γ. In local If e−2Λ a2,r − e−2Φ a2,t > 0, then we can bring 32.6.3 into 32.6.1 by choosing proper Φ normalized Lorentz frame, the condition can be rewritten as a2,ˆr > a2,tˆ.

(32.6.4)

Suppose the radial velocity of the baryon is v. After time ∆tˆ, the shell will move to rˆ − v∆tˆ. Then we have a,ˆr v∆tˆ = a,ˆr v∆tˆ.

(32.6.5)

Since v < 1, we must have a2,ˆr > a2,tˆ. Thus we prove the existence of comoving coordinates.

32.7

ADIABATIC SPHERICAL COLLAPSE: EQUATIONS OF EVOLUTION

Describe the interior of a collapsing star by the comoving, synchronous metric 32.6.1, by the number density of baryons n, by the total density of mass-energy ρ, and by the pressure p. The 4-velocity of the star’s fluid is ˜ u = e−Φ ∂/∂ t˜, (32.7.1) since the fluid is at rest in the coordinate system. Let a dot denote a proper time derivative as seen by the fluid-e.g., ˜ n˙ ≡ u[n] = e−Φ (∂n/∂ t˜)a ;

(32.7.2)

and let a prime denote a partial derivative with respect to baryon number,-e.g. n′ ≡ (∂n/∂a)t˜.

(32.7.3)

Denote by V the rate of change of (1/2π)× (circumference of shell), as measured by a man riding in a given shell: U ≡ r; ˙

(32.7.4)

and denote by m(t˜, a) the “total mass-energy interior to shell a at time t˜”: ∫ m(t˜, a) ≡

a

4πr2 ρ(t˜, a)r′ da.

(32.7.5)

0

(See Box 23.1 for discussion of this method of localizing mass-energy.) Assume that the collapse is adiabatic (no energy flow between adjacent shells; stress-energy tensor entirely that of a perfect fluid). (a) Show that the equations of collapse [baryon conservation, (22.3); local energy conservation, (22.11a); Euler equation, (22.13); and Einstein field equations (ex. 14.16)] can be reduced to the following eight

305

˜ Γ, r, n, ρ, p, V , m: equations for the eight functions Φ, (32.7.6a)

r˙ = U, 2 ·

′

(nr ) U =− ′, 2 nr r ρ˙ n˙ = , ρ+p n 2 Γ p′ m + 4πr3 p U˙ = − − , ρ + p r′ r2 p = p(n, ρ),

(32.7.6b) (32.7.6c) (32.7.6d) (32.7.6e)

′

˜′ = − p , Φ ˜ = 0 at star’s surface, Φ ρ+p m′ = 4πr2 ρr′ , ′

Γ = sign (r )(1 + U − 2m/r) 2

1/2

(32.7.6f) (32.7.6g) (32.7.6h)

.

(b) The preceding equations are in a form useful for numerical calculations. For analytic work it is often useful to replace 32.7.6b by Γ n= , (32.7.7) 4πr2 r′ and 32.7.6d by m ˙ = −4πr2 pU. (32.7.8) Derive these equations. (c) Explain why equations 32.7.6g and 32.7.8 justify the remarks made in Box 23.1 about localizability of energy.

Solution: (a) Equation 32.7.6a, 32.7.6e and 32.7.6g holds automatically by the definition in the exercise. The baryon conservation law gives 0 = ∇ · (nu) = n˙ + n∇ · u = n˙ + n(−g)−1/2 [(−g)1/2 uα ],α = n˙ +

Γn r2 r′

(

r′ r2 Γ

)· .

(32.7.9)

It can be rewritten as

(nr2 )· (r′ )· Γ˙ + ′ − = 0. (32.7.10) 2 nr r Γ For adiabatic flow, we have ds/dτ = 0 except in shock waves. The first law of thermodynamics gives ρ˙ = Note that

ρ+p n. ˙ n

˜ uα;β uβ = u˙α + Γα00 e−2Φ

(32.7.11)

(32.7.12)

and

Γ2 2Φ˜ ˜ ′ e Φ , Γ200 = Γ300 = 0. (32.7.13) r′2 ˜ ′ /r′2 . The Euler equation then gives The only none zero component of ∇u u is (∇u u)1 = Γ2 Φ ˜˙ Γ000 = eΦ Φ, ˜

Γ100 =

˜ ′ + p′ = 0. (ρ + p)Φ

(32.7.14)

˜ = 0 at infinity, we must have Φ ˜ = 0 at the star’s And so Φ′ = 0 outside the star. If we demand that Φ surface.

306

The stress energy of the fluid is Tαβ = (p + ρ)uα uβ + pgαβ . Explicitly, we have ˜

T00 = e2Φ ρ,

T11 =

pr′2 , Γ2

T22 = pr2 ,

T33 = pr2 sin2 θ.

(32.7.15)

All other components vanish. Einstein metric can be calculated from metric. Firstly, from G01 = 0 we have ˙ ′ − ΓΦ ˜ ′ U = 0. Γr

(32.7.16)

U′ (r′ )· Γ˙ = ′ − . ′ r r Γ

(32.7.17)

Note that U ′ = (r′ )· − U Φ′ . We have

Substituting 32.7.17 into 32.7.10, we can get 32.7.6b. From G00 = 8πT00 and 32.7.17, we can get − 2ΓΓ′ r + U 2 r′ + 2U U ′ r + (1 − Γ2 )r′ = 8πρr2 r′ ,

(32.7.18)

[r(1 − Γ2 ) + U 2 r)]′ = 2m′ .

(32.7.19)

i.e.

Noting that m(0) = 0, the integration of equation above will give 32.7.6h. From G11 = 8πT11 , we can get ˜′ 2Γ2 rΦ − U 2 − 2rU˙ + Γ2 − 1 = 8πpr2 . r′

(32.7.20)

Using equation 32.7.6f and 32.7.6h, we can obtain 32.7.6d. (b)Equation 32.7.10 can be rewritten as ( ) d nr2 r′ = 0. Γ dt˜

(32.7.21)

So we have n = f (a)

Γ , r2 r′

(32.7.22)

where f (a) is an arbitrary function of baryon number a. On the other hand, from the definition of baryon number and number density we have ∫ a r′ a= 4πnr2 da. (32.7.23) Γ 0 It is easy to get f = 1/4π and so we have equation 32.7.7. Equation 32.7.6h can be rewritten as 2m = r + rU 2 − rΓ2 .

(32.7.24)

˙ 2m ˙ = U + U 3 + 2rU U˙ − U Γ2 − 2rΓΓ.

(32.7.25)

Take the time derivatives and we can get

Using 32.7.16 and 32.7.20, we can obtain equation 32.7.8. (c) From equation 32.7.6g and 32.7.8, we find that the mass-energy m associated with a given ball of matter (fixed baryon number) can change in time only to the extent that locally measurable energy fluxes can be detected at the boundary of the ball. And energy fluxes in this case it the power expended by pressure forces against the moving boundary surface. Thus the energy m is localized by the circumstance that transfer of energy is detectable by local measurements.

307

32.8

ANALYTIC SOLUTIONS FOR PRESSURE-FREE COLLAPSE

Show how the general solution to the equations of collapse in the case of zero pressure can be generated.

Solution: In the case of zero pressure, the equations of collapse can be simplified to the following equations. (32.8.1a)

r˙ = U, Γ , 4πr2 r′ n˙ ρ˙ = , ρ n m ˙ = 0, ˜ = 0, Φ

(32.8.1b)

n=

′

(32.8.1c) (32.8.1d) (32.8.1e) ′

2

(32.8.1f)

m = 4πr ρr , ′

Γ = sign (r )(1 + U − 2m/r) 2

1/2

.

(32.8.1g)

(1) At first, we specify mass inside shell a, m(a) at the beginning. It will not change with time according to 32.8.1d. (2) From 32.8.1c, we have (ρ/n)· = 0. ρ and n can be related by (32.8.2)

ρ = µ(a)n. Here µ(a) can be seen as rest masses of the particles at shell a. (3) Using 32.8.1b, 32.8.1f and 32.8.2, we can obtain Γ=

m′ . µ

(32.8.3)

And so Γ is also independent of t˜. (4) From 32.8.1e, we know f˙ = ∂f /∂ t˜. Now using 32.8.1a, 32.8.1g, we can obtain (

∂r ∂ t˜

)2 −

2m(a) = Γ2 (a) − 1. r

(32.8.4)

If we specify the r(a) at t˜ = 0, r(t˜, a) can be solved by the equation 32.8.4. The metric in this case therefore ( ds2 = −dt˜2 +

r′ da Γ

)2 + r2 dΩ2 .

(32.8.5)

The remaining quantities of interest can be calculated easily.

32.9

COLLAPSE WITH UNIFORM DENSITY

Recover the Friedmann-Schwarzschild solution for collapse with uniform density and zero pressure by specifying appropriate forms for m(a) and r(a) in the prescription of exercise 32.8. In the interior of

308

the star, give the dust particles nonzero rest masses, µ = constant ̸= 0; in the exterior give them zero rest masses, µ = 0 (“imaginary dust particles” in vacuum). Reduce the resulting metric 32.8.5 to that of Friedmann inside the star, and to that of Novikov for the Schwarzschild geometry outside the star.

Solution: The surface of the star is a = as . When a < s, we assume r(t˜, a) = f (t˜)R(a), Γ2 = 1 − R2 and m/R3 = 2M/Rs3 , where M is a constant and Rs = R(as ). The form of R(a) is determined by 32.8.3, i.e. 6M R2 dR µ(1 − R2 ) = . (32.9.1) Rs3 da The solution is arctanh R − R = The form of f is determined by 32.8.4, i.e. (

∂f ∂ t˜

)2 =

µRs3 a. 6M

2M 1 − 1, R03 f

(32.9.2)

(32.9.3)

and we suppose ∂f /∂ t˜ = 0 when t˜ = 0. So the initial distribution of r is 2M r0 (a) ≡ r(t˜ = 0, a) = f (t˜ = 0)R(a) = 3 R(a). Rs

(32.9.4)

Especially, we have rs0 ≡ r0 (as ) = and so Γ2s = 1 −

2M Rs2

2M . rs0

(32.9.5)

(32.9.6)

The metric becomes

dR2 ds2 = −dt˜2 + f 2 + f 2 R2 dΩ2 . 1 − R2 It is clear that the resulting metric is Friedmann inside the star.

(32.9.7)

When a > a0 , m(a) is constant. And so we have m(a) = M by continuity. As m′ = 0 and µ = 0 outside the star, the value of Γ is arbitrary. And r(t˜, a) must satisfy that ( )2 ∂r 2M + Γ2 − 1. = (32.9.8) ˜ r ∂t And r(t, a ˜) satisfy the free fall equation for a particle in Schwartz metric with energy Γ(a). Suppose ∂r/∂ t˜ = 0 at t˜ = 0. We then have Γ2 = 1 − 2M/r0 . Note that Γ is also continuous across the star surface. The metric now becomes ( )2 1 ∂r 2 2 ˜ ds = −dt + da2 + r2 dΩ2 . (32.9.9) 1 − 2M/r0 ∂a Compare it with equation (31.11) and (31.12) in the textbook. It is clear that the resulting metric is Novikov outside the star.

32.10

PRICE’S THEOREM FOR A SCALAR FIELD

A collapsing spherical star, with an arbitrary nonspherical “scalar charge distribution,” generates an external scalar field Φ. The vacuum field equation for Φ is □Φ = Φ,αα = 0. Ignore the back-reaction of the field’s stress-energy on the geometry of spacetime.

309

(a) Resolve the external field into scalar spherical harmonics, using Schwarzschild coordinates for the external Schwarzschild geometry: ∑1 Φ= Ψl (t, r)Ylm (θ, ϕ). (32.10.1) r ℓ

Show that the vacuum field equation reduces to ( )( ) 2M 2M ℓ(ℓ + 1) − Ψℓ,tt + Ψℓ,r∗ r∗ = 1 − + Ψℓ . r r3 r2

(32.10.2)

where r∗ is the “tortoise coordinate” of section 25.5 and Figure 25.4: ( r ) r∗ = r + 2M ln −1 . 2M

(32.10.3)

Notice that 32.10.2 is a flat-space, one-dimensional wave equation with effective potential ( )( ) 2M 2M ℓ(ℓ + 1) Veff = 1 − + . r r3 r2

(32.10.4)

Part of this effective potential [ℓ(ℓ + 1)/r2 ] is the “centrifugal barrier”, and part [2M/r] is due to the curvature of spacetime. Notice the similarity of this effective potential for scalar waves, to the effective potentials for particles and photons moving in the Schwarzschild geometry, ) ( )( ˜2 2M L 2 e (V )particles = 1 − 1+ 2 , r r ) ( 2M r−2 (32.10.5) (B −2 )photons = 1 − r (Boxes 25.6 and 25.7). The scalar-wave potential, like the photon potential, is positive for all r > 2M . It rises, from 0 at r = 2M , to a barrier summit; then falls back to 0 at r = ∞. (b) Show that there exist no physically acceptable static scalar-wav perturbations of a Schwarzschild black hole. [More precisely, show that all static solutions to equation 32.10.2 become infinite at either the horizon (r = 2M , r∗ = −∞) or at radial infinity.] This suggests that somehow the black hole formed by collapse must divest itself of the star’s external scalar field before it can settle down into a quiescent state. (c) The general solution to the wave equation 32.10.2 can be written in terms of a Fourier transform. For waves that begin near the horizon, propagate outward, and are partially transmitted and partially reflected (“rightward-propagating waves”), show that the general solution is ∫ ∞ Ψℓ = A(k)Rkℓ (r∗ )e−ikt dk, (32.10.6) −∞

where d2 Rkℓ /dr∗2 = [−k 2 + Veff (r)]Rkℓ , ∗

(32.10.7)

∗

Rkℓ → eikr + Γk e−ikr as r∗ → −∞, (R)

∗

Rkℓ → Tk eikr as r∗ → ∞. (R)

(32.10.8) (R)

Show that the “reflection and transmission coefficients for rightward-propagating waves”, Γk have the following asymptotic forms for |k| ≪ 1/M (short wave number; long wavelength): (R)

Γk

= −1 + α2M ik,

(R)

Tk

=

β (2M ik)ℓ+1 , (2ℓ − 1)!!

(R)

and Tk ,

(32.10.9)

where α and β are constants of order unity. Give a similar analysis for waves that impinge on a Schwarzschild black hole from outside (“leftward-propagating waves”).

310

(d) Show that, as the star collapses into the horizon, the world line of its surface in (t, r∗ ) coordinates is r∗ = R∗ (t) = −t − R0∗ exp(−t/2M ) + const.,

(32.10.10)

where R0∗ is related to the magnitude a of the surface’s 4-acceleration (a > 0 for outward 4-acceleration) by { [ ( )1/2 ]} 1 8M ∗ 2 2 R0 = 1 + 16M a M a + M a + . (32.10.11) e 8 Thus, the world line of the surface appears to become null near the horizon (t + r∗ ≡ Ve = constant); of course, this is due to pathology of the coordinate system there. Show, further, that the scalar field on the star’s surface (Ve = constant) must vary as e

Ψℓ = Qℓ0 + Qℓ1 e−U /4M ,

e = t − r∗ , U

(32.10.12)

e → ∞), in order that the rate of change when the star is approaching the horizon (t → ∞, r∗ → −∞, U of Ψℓ be finite as measured on the star’s surface. Notice that Qℓ0 is the “final value” of the scalar field on the star’s surface. It can be regarded as an outgoing wave with zero wave number (infinite wavelength); and, consequently, it gets completely and destructively reflected by the effective potential. Conclusion: All multipoles of the scalar field die out at finite r∗ as t → ∞ (Price’s theorem for a scalar field.)

Solution: (a) Generally, we have Φ,αα = g αβ Φ,αβ − g αβ Γµαβ Φ,µ . In Schwarzschild metric, we have g

αβ

Γµαβ

(32.10.13)

] [ cos θ 2(M − r) ,− 2 ,0 . = 0, r2 r sin θ

(32.10.14)

So the field equation can be written as ( )−1 ( ) 2M 2M 2(M − r) 1 1 − 1− Φ,tt + 1 − Φ,rr − Φ,r + 2 (sin θΦ,θ ),θ + 2 2 Φ,ϕϕ = 0. r r r2 r sin θ r sin θ (32.10.15) Substitute 32.10.1 into the field equation, we can get ( )−1 ( ) 2M 2M 2M 2M ℓ(ℓ + 1) − 1− Ψℓ,tt + 1 − Ψℓ,rr + 2 Ψℓ,r − 3 Ψℓ − Ψℓ = 0. (32.10.16) r r r r r2 Using

dr∗ 1 = , dr 1 − 2M/r

(32.10.17)

we can get the field equation in tortoise coordinate, i.e. 32.10.2. (b) For static solutions, we have Ψℓ,r∗ r∗ =

(

2M 1− r

)(

ℓ(ℓ + 1) 2M + r3 r2

) Ψℓ .

(32.10.18)

When r∗ → −∞ (r → 2M ), we have Ψℓ,r∗ r∗ → 0, and the asymptotic behavior of the Ψ is Ψ ∼ r∗ or Ψ ∼ constant.

(32.10.19)

If the solution is regular when r∗ → −∞, the only possibility is Ψ ∼ constant. When r∗ → ∞ and ℓ > 0, we have Ψℓ,r∗ r∗ → ℓ(ℓ + 1)Ψl /r∗2 , and the asymptotic behavior of the Ψ is Ψ ∼ r∗ℓ+1 or Ψ ∼ r∗−ℓ .

311

(32.10.20)

If the solution is regular when r∗ → ∞, the only possibility is Ψ ∼ r∗−ℓ . Now we want to connect two pieces of solution at infinity. We suppose Ψ is positive when r∗ → −∞. Then we have Ψℓ,r∗ r∗ > 0. So Ψℓ must be increasing as r increases, which is contradict to its behavior at r∗ → ∞. The analysis for the opposite case is similar. If ℓ = 0, we have

( 1−

2M r

)

2M 2M Ψ0,r − 3 Ψ0 = 0. 2 r r

Ψ0,rr +

(32.10.21)

( ) 2M Ψ0 = c1 r + c2 r ln 1 − . r

We have exact solution

(32.10.22)

Clearly, it is not regular at r∗ → ±∞. Thus all static solutions to equation 32.10.2 become infinite at either the horizon (r = 2M , r∗ = −∞) or at radial infinity. (c) If we substitute Fourier decomposition 32.10.36 into field equation 32.10.2, it is easy to obtain 32.10.37. When r∗ → ±∞, i.e. r∗ ≫ k −1 ≫ M , we have Veff → 0 and so ∗

∗

Rkℓ = Cl eikr + Dl e−ikr .

(32.10.23)

If we assume the waves are rightward-propagating, Dl should be taken to 0 at r∗ → ∞. So we have 32.10.38 at infinity, up to a renormalization constant. At r∗ ≫ M (but r∗ ≪ k −1 , or r∗ ∼ k −1 , or r∗ ≫ k −1 ), we have [ ] d2 Rkℓ ℓ(ℓ + 1) 2 = −k + Rkℓ . dr∗2 r∗2 The solution is

(32.10.24)

Rkℓ = Cl r∗ hl (kr∗ ) + Dl r∗ hl (kr∗ ), (1)

(2)

(32.10.25)

where hl (kr∗ ) and hl (kr∗ ) are the first and second type spherical Hankel function. When kr ≫ 1, we have asymptotic approximation (1)

(2)

∗

∗

hl (kr∗ ) = i−(ℓ+1) (1)

eikr , kr∗

hl (kr∗ ) = iℓ+1 (2)

In order to match (R)

Rkℓ = Tk eikr at r ≫ k

−1

. We must have

∗

Rkℓ = Tk iℓ+1 kr∗ hl (kr∗ ) (R)

(1)

e−ikr . kr∗

(32.10.26)

(32.10.27) (32.10.28)

at r∗ ≫ M . At |r∗ | ≪ k −1 (but |r∗ | ∼ M or |r∗ | ≫ M ), Rℓk = 2M iky, where y = y(r∗ ) satisfies static field equation. d2 y = Veff y. dr∗2

(32.10.29)

And according to discussion in (b), we have y =α+ and

r∗ as r∗ → −∞ M

y = β(2M/r∗ )l as r∗ → +∞.

where α and β depend on multipole, but are both of order unity.

312

(32.10.30)

(32.10.31)

Now, by matching 32.10.28 and 32.10.31, noting that hℓ (kr∗ ) ∼ −i (1)

(2l − 1)!! at kr∗ ∼ 0, (kr∗ )l+1

(32.10.32)

we can get (R)

Tk

= (−1)l

β (2M ik)ℓ+1 . (2ℓ − 1)!!

(32.10.33) (R)

Thus waves have no transmission in limit k → 0. I have an extra (−1)l factor for Tk . Please check it. ∗

∗

By matching 32.10.30 and eikr + Γk e−ikr , noting that (R)

∗

∗

eikr + Γk e−ikr ∼ 1 + Γk (R)

(R)

at kr∗ ∼ 0,

(32.10.34)

we can get (R)

Γk

= −1 + α2M ik.

(32.10.35)

So waves have complete reflection and complete destructive interference in limit k → 0. Analysis for leftward-propagating waves can be carried similarly. The general solution is ∫ ∞ Ψℓ = A(k)Lℓk (r∗ )e−ikt dk,

(32.10.36)

−∞

where d2 Lℓk /dr∗2 = [−k 2 + Veff (r)]Lℓk , Lℓk

→e

(32.10.37)

−ikr ∗

Lℓk →

∗ (R) + Γk eikr as r∗ → ∞, ∗ (R) Tk e−ikr as r∗ → −∞.

(32.10.38) (R)

The “reflection and transmission coefficients for leftward-propagating waves”, Γk following asymptotic forms for |k| ≪ 1/M (short wave number; long wavelength): (R)

Γk

= (−1)ℓ+1 +

α (2M ik)2ℓ+1 , (2l − 1)!!

(R)

Tk

(R)

and Tk , have the

= β(2M ik)ℓ+1

(32.10.39)

where α and β are constants of order unity. Please see page 249 of the book Magic without Magic (Klauder et al .1972) for a more detailed discussion. (4) I do not know how to get the relation between R0∗ and four acceleration a. Any help is appreciated. Introduce Kruskal coordinates e

u ˜ = −4M e−U /4M ,

e

v˜ = 4M eV /4M .

(32.10.40)

Since u ˜ and v˜ are well-behaved coordinates at r = 2M , the partial derivatives ∂Ψ/∂ u ˜ and ∂Ψ/∂˜ v should e must fall off sharply at the event horizon because be finite. This implies that ∂Ψ/∂ U ∂Ψ ∂Ψ −Ue /4M = e , e ∂u ˜ ∂U

(32.10.41)

e → ∞ at r = 2M . The advanced time Ve is finite r = 2M so that ∂Ψ/∂ Ve is finite. and U e ) or v˜(˜ If we picture the path of the surface through space-time as Ve (U u), then near r = 2M we have

Thus we have

e d˜ v dVe dU dVe d˜ v −(Ue +Ve )/4M e = / = e ∼ e−U /4M e d˜ u d˜ v d˜ u d˜ u dU

(32.10.42)

∂Ψ dVe ∂Ψ dΨ e = + ∼ e−U /4M . e e e e dU ∂ V dU ∂U

(32.10.43)

313

e . Ψ varies as for large U

e

Ψ = Q0 + Q1 e−U /4M

(32.10.44)

on star’s surface as R∗ → −∞. e

The time-varying part of this surface field, Ψ = Q1 e−U /4M , is of medium wavelength, so much of it gets through the potential and propagates off to Ve → ∞. But the static part, Ψ = Q0 , is of infinite wavelength and is completely reflected by the potential. Put in more physical terms, the curvature of spacetime prevents any information whatsoever about the final surface field T = Q0 from propagating off to a distant external observer.

32.11

NEWMAN-PENROSE “CONSTANTS”

Wheeler (1955) showed that Maxwell’s equations for an ℓ-pole electromagnetic field residing in the Schwarzschild geometry can be reduced to the wave equation ( ) 2M ℓ(ℓ + 1) − Ψℓ,tt + Ψℓ,r∗ r∗ = 1 − Ψℓ . (32.11.1) r r2 After this equation has been solved, the components of the electromagnetic field can be obtained by applying certain differential operators to Ψℓ (t, r∗ )Yℓm (θ, ϕ). (a) Show that the general solution to the electromagnetic wave equation 32.11.1 for dipole (ℓ = 1) fields, with outgoing-wave boundary conditions at r∗ → ∞, has the form e e e ) + f1 (U ) + f2 (U ) + · · · Ψ1 = f0 (U 2 r r

(32.11.2)

e = t − r∗ is “retarded time”, and where U f1′ = f0 , f2′ = 0, · · · , fn′ = −

(n + 1)(n − 2) fn−1 + (n − 2)M fn−2 . 2n

(32.11.3)

When spacetime is flat (M = 0), this solution becomes e ) + f1 ( U e )/r. Ψ1 = f1′ (U

(32.11.4)

e )/r2 +· · · which are absent in flat spacetime, are attributable to backscatter of the outgoing The terms f2 (U waves by the curvature of spacetime. They are sometimes called the “tail” of the waves. (b) Show that the general static dipole field has the form 32.11.2 with (f0 )static = 0;

(f1 )static ≡ D = dipole moment;

(f2 )static =

3 M D; · · · 2

(32.11.5)

(c) Consider a star (not a black hole!) with a dipole field that is initially static. At time t = 0, let the star suddenly change its dipole moment to a new static value D′ . Equations 32.11.3 demand that f2 be conserved [“Newman-Penrose (1965) constant”]. Hence, f2 will always exhibit a value 3M D/2 corresponding to the old dipole moment; it can never change to 3M D′ /2. This is a manifestation of the tail of the waves that are generated by the sudden change in dipole moment. To understand this tail effect more clearly, and to discover an important flaw in the above result, evaluate the solution 32.11.2 for retarded time U > 0, using the assumptions e < 0, • field has static form 32.11.5 for U e > 0. • f1 = D′ for U

314

Put the answer in the form ∞ e n−2 D′ 3M D ∑ 2M (D′ − D)(−1)n+1 (n + 1)U Ψ1 = + + +O 2 n r 2r (2r) n=3

(

e M 2 M 2U , 4 3 r r

) .

(32.11.6)

e /r, so long as r ≫ M .) Evidently, so long (The terms neglected are small compared to those kept for all U as the series converges the Newman- Penrose “constant” (coefficient of 1/r2 ) remembers the old D value e > 2r -i.e., it diverges and is conserved, as claimed above. Show, however, that the series diverges for U inside a sphere that moves outward with asymptotically 1/3 the speed of light. Thus, the Newman-Penrose “constant” is well-defined and conserved only outside the “1/3-speed-of-light cone”. ˜ > 2r: (d) Sum the series in 32.11.6 to obtain a solution valid even for U ( 2) e + 3r) D′ M 3M D′ 2M (D′ − D) (U Ψ= +O + − 2 e r 2r2 r r3 (U + 2r) e < 2r(domain of convergence of that series) = the series 32.11.6 for U ( ) D′ 3M D′ M M2 e ≫ r ≫ M. = + , + O for U e r r3 r 2r2 U

(32.11.7)

From this result conclude that at fixed r and late times the electromagnetic field becomes asymptotically static and the Newman-Penrose “constant” assumes the new value 3M D′ /2 appropriate to the new dipole moment.

Solution: (a) The expansion for outgoing waves assumes that at sufficiently large r the radial dependence can be analyzed as a power series in r−1 . Thus we have the expansion 32.11.2. If we substitute 32.11.2 into 32.11.1, we then have ∞ ∑ e) 2kf ′ (U k=1

k rk+1

+

∞ e ) 2M k(k + 2)fk (U e) ∑ e) k(k + 1)fk (U 2fk (U − = . k+2 k+3 k+2 r r r

(32.11.8)

k=0

By equating the coefficients of r−k of both side of the equation above, we can get 2f1′ = 2f0 4f2′ + 2f1 = 2f1 2kfk′ + k(k − 1)fk−1 − 2M k(k − 2)fk−2 = 2fk−1 (k ≥ 3).

(32.11.9)

Some simple algebra manipulation will give 32.11.3. When space time is flat, we have f1′ = f0 , f2′ = 0, · · · , fn′ = −

(n + 1)(n − 2) fn−1 . 2n

(32.11.10)

˜ /3 + f3 (U ˜ = 0), which is unbound. Thus we must have f2 = 0. By similar If f2 ̸= 0, then f3 = −2f2 U e ) + f1 (U e )/r. analysis, we can show that all higher terms vanish. The solution is therefore Ψ1 = f1′ (U (b) For static dipole field, we have fn′ = 0, i.e. f0 = 0, fn−1 =

2nM fn−2 (n ≥ 3). n+1

(32.11.11)

If f1 = D, we can get fn =

3D (2M )n−1 (n ≥ 2). n+2

(32.11.12)

(c) When f1 changed form D to D′ , we have f3′ = M ∆f1 = M (D′ − D)

315

(32.11.13)

and so f3 =

12DM 2 e. + M (D′ − D)U 5

(32.11.14)

˜ ≫ M and r ≫ M , we can neglect the constant term in f3 . Generally, we can only keep In the case of U e . Suppose fn = Cn M (D′ − D)U e n−2 . We have terms with highest order of U fn′ = −

(n + 1)(n − 2) e n−3 Cn−1 M (D′ − D)U 2n

(32.11.15)

(n + 1) Cn−1 . 2n

(32.11.16)

n+1 (−1)n+1 . 2n−1

(32.11.17)

and so Cn = − As C3 = 1, we can obtain Cn = Now the field can be written as 32.11.6.

e /2r)n−2 is infinite as n → ∞ and so the summation e > 2r, the term 2M (D′ − D)(−1)n+1 (n + 1)r−2 (U If U dose not converge mathematically. Though the value of the field may be finite by cancellation between infinite terms (called analytic continuation mathematically or renormalization physically), the Newman˜ > 2r. Penrose “constant” become ill defined and cannot be evaluated uniquely from data in region U ˜ > 2r is equivalent to r < t/3, so the divergent sphere move outward with 1/3 of speed When r ≫ M , U of light. (4)Equation 32.11.6 can be rewritten as ( )n−2 ( ) ∞ e e U 3M D′ M (D′ − D) ∑ M 2 M 2U D′ (n + 1) − + − +O , 4 . Ψ1 = r 2r2 2r2 2r r3 r n=2 Note that

∞ ∑

(n + 3)an =

n=0

3 − 2a for |a| < 1. (1 − a)2

(32.11.18)

(32.11.19)

By analytic continuation, we can obtain ˜ + 3r D′ 3M D′ 2M (D′ − D) U Ψ1 = + − +O ˜ + 2r)2 r 2r2 r (U e ≫ r ≫ M , we have If U

D′ 3M D′ Ψ1 = + +O r 2r2

(

(

e M 2 M 2U , r3 r4

e M 2 M 2U , 4 3 r r

) .

(32.11.20)

) .

(32.11.21)

At fixed r and late times the electromagnetic field becomes asymptotically static and the Newman-Penrose “constant” assumes the new value 3M D′ /2 appropriate to the new dipole moment.

316

Chapter 33

BLACK HOLES 33.1

KERR DESCRIPTION OF KILLING VECTORS

(a) Use the transformation law from Boyer-Lindquist coordinates to Kerr coordinates [equation (4) of Box 33.2] to show that ( ( ) ) ∂ ∂ = ξ(t) ≡ (33.1.1a) ∂t r,θ,ϕ ∂ Ve r,θ,ϕe ( ) ( ) ∂ ∂ . ξ(ϕ) ≡ = (33.1.1b) ∂ϕ t,r,θ ∂ ϕe Ve ,r,θ Verify explicitly by examining metric components that gVe Ve = gtt ,

gVe ϕe = gtϕ ,

gϕeϕe = gϕϕ ,

(33.1.2)

In accordance with equations 33.1.1. (b) Show that for a stationary observer (world line of constant r, θ), the angular velocity expressed in terms of Kerr coordinates is e dϕ dϕe uϕ Ω≡ = = e. (33.1.3) dt uV dVe

Solution: (a) The transformation matrix is  1 (r2 + a2 )/∆ 0 α ˜  ∂x 0 1 0 = 0 0 1 ∂xβ a/∆ 0 0

 0 0  . 0 1 αβ ˜

(33.1.4)

It is easy to see that ∂ ∂ ∂xα˜ ∂ = = ∂t ∂t ∂xα˜ ∂xV˜

(33.1.5)

∂ ∂ ∂xα˜ ∂ = = . ∂ϕ ∂ϕ ∂xα˜ ∂xϕ˜

(33.1.6)

and

317

In Boyer-Lindquist coordinates, we have gtt = −

gtϕ =

∆ a2 sin2 θ r2 − 2M r + a2 + Q2 − a2 sin2 θ −2M r + Q2 + = − = −1 − , ρ2 ρ2 r2 + a2 cos2 θ r2 + a2 cos2 θ

a∆ sin2 θ a(a2 + r2 ) sin2 θ (a2 + r2 − ∆)a sin2 θ (2M r − Q2 )a sin2 θ − = − = − ρ2 ρ2 ρ2 ρ2

and gϕϕ = −

∆a2 sin4 θ (r2 + a2 )2 sin2 θ [(r2 + a2 )2 − ∆a2 sin2 θ] sin2 θ + = . ρ2 ρ2 ρ2

In Kerr coordinates, we have gVe Ve = −1 + gVe ϕe = − and gϕeϕe =

2M r − Q2 , ρ2

a(2M r − Q2 ) sin2 θ ρ2

[(r2 + a2 )2 − ∆a2 sin2 θ] sin2 θ . ρ2

(33.1.7)

(33.1.8)

(33.1.9)

(33.1.10) (33.1.11)

(33.1.12)

Now, we have verified equation 33.1.2. (b) The transformation matrix gives dVe = dt + (r2 + a2 )(dr/∆),

dϕe = dϕ + a(dr/∆).

(33.1.13)

For stationary object, we have dr = 0 and so dVe = dt,

dϕe = dϕ.

(33.1.14)

Now, we have verified equation 33.1.3.

33.2

OBSERVATIONS OF ANGULAR VELOCITY

An observer, far from a black hole and at rest in the hole’s asymptotic Lorentz frame, watches (with his eyes) as a particle moves along a stationary (nongeodesic) orbit near the black hole. Let Ω = dϕ/dt be the particle’s angular velocity, as defined and discussed above. The distant observer uses his stopwatch to measure the time required for the particle to make one complete circuit around the black hole (one complete circuit relative to the distant observer himself; i.e., relative to the hole’s asymptotic Lorentz frame). (a) Show that the circuit time measured is 2π/Ω. Thus, Ω can be regarded as the particle’s “angular velocity as measured from infinity.” (b) Let the observer moving with the particle measure its circuit time relative to the asymptotic Lorentz frame, using his eyes and a stopwatch he carries. Show that his answer for the circuit time must be ∆τ =

2π (−gtt − 2Ωgtϕ − Ω2 gϕϕ )1/2 . Ω

(33.2.1)

Solution: (a) Suppose the observer is at (R, θ0 , ϕ0 ) and R ≫ M . At time t, the particle pass ϕ0 and emit a photon to observer. The observer receive the photon at t + ∆t. At time t + 2π/Ω, the particle pass ϕ0 again and emit

318

a photon to observer. The observer then will receive the photon at t + 2π/Ω + ∆t. The time lag between two signals received by observer is therefore 2π/Ω, i.e. the circuit time measured is 2π/Ω. (b) The four velocity of the particle is ut (1, 0, 0, Ω). Normalization gives (gtt + 2gtϕ Ω + gϕϕ Ω2 )ut2 = −1.

(33.2.2)

dϕ = Ωut = Ω(−gtt − 2Ωgtϕ − Ω2 gϕϕ )−1/2 . dτ

(33.2.3)

We can get

The circuit time is ∆τ =

33.3

2π 2π = (−gtt − 2Ωgtϕ − Ω2 gϕϕ )1/2 . dϕ/dτ Ω

(33.2.4)

LOCALLY NONROTATING OBSERVERS

(a) Place a rigid, circular mirror (“ring mirror”) at fixed (r, θ) around a black hole. Let an observer at (r, θ) with angular velocity Ω emit a flash of light. Some of the photons will get caught by the mirror and will skim along its surface, circumnavigating the black hole in the positive-ϕ direction. Others will get caught and will skim along in the negative-ϕ direction. Show that the observer will receive back the photons from both directions simultaneously only if his angular velocity is Ω = ω(r, θ) ≡

gϕt 1 (2M r − Q2 )a (Ωmin + Ωmax ) = − . = 2 2 gϕϕ (r + a2 )2 − ∆a2 sin2 θ

(33.3.1)

Thus in this case, and only in this case, can the observer regard the +ϕ and −ϕ directions as equivalent in terms of local geometry. Put differently, in this case and only in this case is the observer “nonrotating relative to the local spacetime geometry”. Thus, it is appropriate to use the name “locally nonrotating observer” for an observer who moves with the angular velocity Ω = ω(r, θ). (b) Associated with the axial symmetry of a black hole is a conserved quantity, pϕ ≡ p · ξ(ϕ) for geodesic motion. This quantity for any particle-whether it is moving along a geodesic or not-is called the “component of angular momentum along the black hole’s spin axis”, or simply the particle’s “angular momentum”. Show that of all stationary observers at fixed (r, θ), only the “locally nonrotating observer” has zero angular momentum. [Note: Bardeen, Press, and Teukolsky (1972) have shown that the “locally nonrotating observer” can be a powerful tool in the analysis of physical processes near a black hole.]

Solution: (a) Suppose the angular velocity of the photon is Ωph . The four velocity of the photon is ut (1, 0, 0, Ωph ). Using normalization condition, we have gtt + 2gtϕ Ωph + gϕϕ Ω2ph = 0.

(33.3.2)

There are two solutions Ωmin and Ωmin , satisfying Ωmin + Ωmax = −2

gtϕ . gϕϕ

(33.3.3)

If the observer will receive back the photons from both directions simultaneously, we have 2π 2π = , Ωmax − Ω Ω − Ωmin i.e. Ω = ω(r, θ).

319

(33.3.4)

(b) The four momentum of the stationary observer is pt (1, 0, 0, Ω). So we have pϕ = gϕα pα = pt (gtϕ + gϕϕ Ω). If pϕ = 0, we then have Ω=−

33.4

(33.3.5)

gtϕ = ω(r, θ). gϕϕ

(33.3.6)

ORTHONORMAL FRAMES OF LOCALLY NONROTATING OBSERVERS

(a) Let spacetime be filled with world lines of locally nonrotating observers, and let each such observer carry an orthonormal frame with himself. Show that the spatial orientations of these frames can be so chosen that their basis 1-forms are ˆ

ˆ

ˆ

1/2

ω t = |gtt − ω 2 gϕϕ |1/2 dt, ω ϕ = gϕϕ (dϕ − ωdt), ω rˆ = (ρ/∆1/2 )dr, ω θ = ρdθ.

(33.4.1)

More specifically, show that these 1-forms are orthonormal and that the dual basis has ∂ = u ≡ 4 − velocity of locally nonrotating observer. ∂ tˆ

(33.4.2)

Show that −ω tˆ is a rotation-free field of 1-forms [dω tˆ ∧ ω tˆ = 0; exercise 4.4]. (b) One sometimes meets the mistaken notion that a “locally nonrotating observer” is in some sense locally inertial. To destroy this false impression, verify that: (i) such an observer has nonzero 4-acceleration, a = Γˆj tˆtˆeˆj =

1 ∇ ln |gtt − ω 2 gϕϕ |; 2

(33.4.3)

(ii) if such an observer carries gyroscopes with himself, applying the necessary accelerations at the gyroscope centers of mass, he sees the gyroscopes precess relative to his orthonormal frame with angular velocity [ ] 1/2 gϕϕ 1 ω,θ ∆1/2 ω,r precess Ω = Γθˆϕˆtˆerˆ + Γϕˆ erˆ − eθˆ . (33.4.4) ˆr tˆeθˆ = 2 |gtt − ω 2 gϕϕ |1/2 ρ ρ

Solution: (a) The metric of the spacetime is    gtt 0 0 −ωgϕϕ 1/(gtt − ω 2 gϕϕ ) 0 2 2  0  µν  ρ /∆ 0 0 0 ∆/ρ , g =  gµν =  2  0  0 0 ρ 0  0 −ωgϕϕ 0 0 gϕϕ ω/(gtt − ω 2 gϕϕ ) 0

0 0 1/ρ2 0

 ω/(gtt − ω 2 gϕϕ )  0 .  0 gtt /[gϕϕ (gtt − ω 2 gϕϕ )] (33.4.5)

Note that ˆ

(ω t )α = |gtt − ω 2 gϕϕ |1/2 (1, 0, 0, 0), ˆ

1/2

(ω ϕ )α = gϕϕ (−ω, 0, 0, 1), (ω rˆ)α = ρ/∆1/2 (0, 1, 0, 0), ˆ

(ω θ )α = ρ(0, 0, 1, 0).

(33.4.6) ˆ

ˆ

Direct calculation can show that these 1-forms are orthonormal. Also note that (∂/∂ tˆ)α (ω β )α = δtˆβ . We can get ( )αˆ ∂ = |gtt − ω 2 gϕϕ |−1/2 (1, 0, 0, ω). (33.4.7) ∂ tˆ

320

Using dϕ/dt = ω(r, θ) for nonrotating observer, we know ∂/∂ tˆ is his four velocity. Since we have ˆ

1/2

dω t = |f |1/2 ,r dr ∧ dt + |f |,θ dθ ∧ dt

(33.4.8)

where f = gtt − ω 2 gϕϕ , we can easily deduce that dω tˆ ∧ ω tˆ = 0, i.e. −ω tˆ is a rotation-free field of 1-forms. (b) Recall section 5 of the chapter 14. Note that 1 1 ˆ ˆ ˆ (ln |f |),r ∆1/2 ρ−1 ω rˆ ∧ ω t + (ln |f |),θ ρ−1 ω θ ∧ ω t 2 2 1 1 ˆ ˆ ˆ ˆ dω ϕ = (ln gϕϕ ),r ∆1/2 ρ−1 ω rˆ ∧ ω ϕ + (ln gϕϕ ),θ ρ−1 ω θ ∧ ω ϕ 2 2 ˆ ˆ ˆ 1/2 1/2 − gϕϕ ω,r |f |−1/2 ∆1/2 ρ−1 ω rˆ ∧ ω t − gϕϕ ω,θ |f |−1/2 ρ−1 ω θ ∧ ω t ( ) ˆ dω rˆ = ln ρ/∆1/2 ρ−1 ω θ ∧ ω rˆ ˆ

dω t =

,θ

θˆ

dω = ln(ρ),r ∆

1/2 −1

ρ

ˆ

ω rˆ ∧ ω θ .

(33.4.9)

We then find that ˆ

1 1 1 ˆ ˆ ϕ 1/2 −1 (ln |f |),r ∆1/2 ρ−1 , ctˆθˆt = (ln |f |),θ ρ−1 , cϕˆ ρ , ˆr = (ln gϕϕ ),r ∆ 2 2 2 1 ˆ ˆ 1/2 1/2 = (ln gϕϕ ),θ ρ−1 , crˆtˆϕ = gϕϕ ω,r |f |−1/2 ∆1/2 ρ−1 , cθˆtˆϕ = gϕϕ ω,θ |f |−1/2 ρ−1 , 2( )

ctˆrˆt = ˆ

cϕˆθˆϕ

crˆθˆrˆ = ln ρ/∆1/2

ˆ

,θ

θ 1/2 −1 ρ . ρ−1 , cθˆ ˆr = ln(ρ),r ∆

(33.4.10)

Since Γµˆνˆαˆ =

1 (cµˆνˆαˆ + cµˆαˆ ˆ ν − cν ˆαˆ ˆ µ ), 2

(33.4.11)

we have

Γtˆθˆϕˆ

1 1/2 1 1 Γtˆrˆtˆ = − (ln |f |),r ∆1/2 ρ−1 , Γtˆrˆϕˆ = − gϕϕ ω,r |f |−1/2 ∆1/2 ρ−1 , Γtˆθˆtˆ = − (ln |f |),θ ρ−1 , 2 2 2 1 1/2 1 1/2 1 1/2 −1/2 −1 −1/2 1/2 −1 = − gϕϕ ω,θ |f | ρ , Γtˆϕˆ ∆ ρ , Γtˆϕˆθˆ = − gϕϕ ω,θ |f |−1/2 ρ−1 , ˆr = − gϕϕ ω,r |f | 2 2 2 ( ) 1 1/2 1/2 −1 1/2 −1 Γrˆθˆ ρ . Γrˆθˆθˆ = − ln(ρ),r ∆ ρ , Γrˆϕˆtˆ = gϕϕ ω,r |f |−1/2 ∆1/2 ρ−1 ˆr = ln ρ/∆ 2 ,θ 1 1/2 1 1 1/2 −1 −1/2 −1 ρ , Γθˆϕˆϕˆ = − (ln gϕϕ ),θ ρ−1 . Γrˆϕˆϕˆ = − (ln gϕϕ ),r ∆ ρ , Γθˆϕˆtˆ = gϕϕ ω,θ |f | 2 2 2 (33.4.12)

All other terms either can be calculated by Γmuˆ ˆ να ˆ = −Γν ˆµ ˆα ˆ or vanish. The acceleration of the observer is a = u · ∇u = Γαtˆtˆeα = Γˆj tˆtˆej =

1 1 1 (ln |f |),r ∂r + (ln |f |),θ ∂θ = ∇|gtt − ω 2 gϕϕ |. 2 2 2

(33.4.13)

The angular velocity of the observer’s spatial frame is Ω=

1 1 eˆi ϵˆiˆj kˆ ek · ∇u eˆj = ϵˆiˆj kˆ Γkˆˆj tˆeˆi = Γϕˆθˆtˆerˆ + Γrˆϕˆtˆeθˆ. 2 2

(33.4.14)

The gyroscopes precess relative to his orthonormal frame with angular velocity precess

Ω

[ ] 1/2 gϕϕ ω,θ 1 ∆1/2 ω,r = −Ω = Γθˆϕˆtˆerˆ + Γϕˆ erˆ − eθˆ . ˆr tˆeθˆ = 2 |gtt − ω 2 gϕϕ |1/2 ρ ρ

321

(33.4.15)

33.5

LOCAL LIGHT CONES

Calculate the shapes of the light cones depicted in the Kerr diagram for an uncharged Kerr black hole (part II.F of Box 33.2). In particular, introduce a new time coordinate t˜ = V˜ − r

(33.5.1)

for which the slices of constant t˜ are horizontal surfaces in the Kerr diagram. Then the Kerr diagram plots t˜ vertically, r radially, and ϕ˜ azimuthally, while holding θ = π/2 (“equatorial slice through black hole”). (a) Show that the light cone emanating from given t˜, r and ϕ˜ has the form √ ( ) ˜ t˜)2 dr 1 r2 (dϕ/d dϕ˜ 2M/r ± − . =a − 1 + 2M/r (1 + 2M/r)2 1 + 2M/r dt˜ dt˜

(33.5.2)

(b) Show that the light cone slices through the surface of constant radius along the curves dr/dt˜ = 0,

˜ t˜ = Ωmin and Ωmax , dϕ/d

(33.5.3)

where Ωmin and Ωmax are given by expressions (33.15a,b) (minimum and maximum allowed angular velocities for stationary observers). ˜ (c) Show that at the static limit, r = r0 (π/2), the light cone is tangent to a curve of constant r, θ and ϕ. (d) Show that the light cone slices the surface of constant ϕ˜ along the curves dϕ˜ = 0, dt˜

dr 1 − 2M/r = −1 and . ˜ 1 + 2M/r dt

(33.5.4)

(e) Show that the light cone is tangent to the horizon. (f) Make pictures of the shapes of the light cone as a function of radius. (g) Describe qualitatively how the light cone must look near the horizon in Boyer-Lindquist coordinates. (Note: it will look “crazy” because the coordinates are singular at the horizon.)

Solution: (a)Take θ = π/2 and Q = 0, the metric in Kerr frame is ) )] ( [ ( 2M 2M 2 2 2 2 ˜ − 4M a dΦd ˜ ˜ V˜ . ˜ ds = − 1 − dV + 2drdV + r + a 1 + dϕ˜2 − 2adϕdr r r r

(33.5.5)

In new coordinates t˜ = V˜ − r, we have dV˜ = d˜ r + dt˜, and so ( ) ( ) [ ( )] ( ) 2M 2M 4M 2M 2M 4M a ˜ ˜ 2 2 2 2 2 2 ˜ ˜ ds = − 1 − dt˜ + 1 + dr + drdt˜+ r + a 1 + dϕ −2a 1 + dΦdr− dϕdt. r r r r r r (33.5.6) For null geodesics, we have ds2 = 0 and so   ] ( ) ( )2 [ ( ) [ ( )] ( ˜ )2 ) ˜ ( 2M 4M a dr 4M 2M dϕ˜ dr  2 2M d ϕ d ϕ 2M 1+ + − − 2a 1 + + r + a2 1 + − 1− = 0. r r r dt dt˜  r r dt˜ r  dt˜ dt˜ (33.5.7) Solving the quadratic equation for dr/dt˜ above, we can obtain 33.5.2.

322

(b) Using 33.5.6 and dr/dt˜ = 0, we can get [ ( )] ( ˜ )2 ( ) 2M dϕ 4M a dϕ˜ 2M 2 2 r +a 1+ − − 1− = 0. r r dt˜ r dt˜ Note that

( ) 2M gtt = − 1 − , r

gtϕ = −

2M a , r

( ) 2M gϕϕ = r2 + a2 1 + r

(33.5.8)

(33.5.9)

when Q = 0 and θ = π/2 in Boyer-Lindquist coordinates. We can conclude that Ωmin and Ωmax are the solution of 33.5.8. ˜ t˜ = 0 when dr/dt˜ = 0 for null geodesics. And (c) At static limit, we have Ωmin = 0, i.e. we can have dϕ/d ˜ so the light cone is tangent to a curve of constant r, θ and ϕ. (d) According to 33.5.2, we have 1 1 − 2M/r dr 2M/r ± = −1 and . =− ˜ 1 + 2M/r 1 + 2M/r 1 + 2M/r dt

(33.5.10)

a dϕ˜ = Ωmin = Ωmax = 2 2 a + r+ dt˜

(33.5.11)

˜ t˜ = 0. when dϕ/d (e) At horizon, we have

when dr/dt˜ = 0. So the light cone is tangent to the horizon.

33.6

SUPERHAMllTONIAN FOR CHARGED-PARTICLE MOTION

Show that Hamilton’s equations (33.28a) for the Hamiltonian (33.28b) reduce to equation (33.29a) for the value of the generalized momentum, and to the Lorentz force equation (33.29b).

Solution: The Hamiltonian is 1 µν g (πµ − eAµ )(πν − eAν ). 2

(33.6.1)

dxµ ∂H = = g µν (πν − eAν ) = π µ − eAµ dλ ∂πµ

(33.6.2)

H= So we have pµ ≡ and

dπµ 1 ∂H = − µ = − g αβ,µ (πα − eAα )(πβ − eAβ ) + eAα,µ (π α − eAα ). dλ ∂x 2

(33.6.3)

0 = (g αβ gβγ ),µ = gβγ g αβ,µ + g αβ gβγ,µ .

(33.6.4)

Note that

323

We have dg µν dpµ dpν = pν + g µν dλ dλ dλ d = g µν,α pα pν + g µν (πν − eAν ) dλ [ ] 1 αβ µρ α β µν α = −g gρβ,α p p + g − g ,ν pα pβ + eAα,ν p − eg µν Aν,α pα 2 ( ) 1 gαβ,ν − gνβ,α pα pβ + eg µν Fνα pα = g µν 2 1 = − g µν (gνα,β + gνβ,α − gαβ,ν )pα pβ + eF µα pα 2 = −Γµαβ pα pβ + eF µα pα .

33.7

(33.6.5)

HAMILTON-JACOBI DERIVATION OF EQUATIONS OF MOTION

Derive the first-order equations of motion (33.32) for a charged particle moving in the external fields of a Kerr-Newman black hole. Use the Hamilton-Jacobi method.

Solution: (a) We first replace the generalized momentum πa by the gradient ∂S/∂xα a of the Hamilton-Jacobi function hroughout the superhamiltonian H of equation (33.28b) (b) Then we write down the Hamilton-Jacobi equation in the form [ ] ( )( ) 1 αβ ∂S ∂S ∂S α ∂S =H x , α = g − eAα − eAβ . − ∂λ ∂x 2 ∂xα ∂xβ

(33.7.1)

(c) In Boyer-Lindquist coordinates, the metric restricted to (t, ϕ) for a Kerr-Newman black hole is ( ) −∆ + a2 sin2 θ a(∆ − r2 − a2 ) sin2 θ −2 gAB = ρ (33.7.2) a(∆ − r2 − a2 ) sin2 θ [(r2 + a2 )2 − ∆a2 sin2 θ] sin2 θ The inverse metric is g AB = −

1 ∆ρ2 sin2 θ

(

[(r2 + a2 )2 − ∆a2 sin2 θ] sin2 θ a(r2 + a2 − ∆) sin2 θ

a(r2 + a2 − ∆) sin2 θ −∆ + a2 sin2 θ

)

Put back r and θ coordinates and we can get [ [ ]2 ]2 ∂ ∂ ∂ 1 ∂ ∂ 1 ∂ g ≡ g αβ α β = − 2 (r2 + a2 ) + a + 2 2 + a sin2 θ ∂x ∂x ∆ρ ∂t ∂ϕ ∂t ρ sin θ ∂ϕ ( )2 ( )2 1 ∂ ∆ ∂ + 2 . + 2 ρ ∂r ρ ∂θ

(33.7.3)

(33.7.4)

(d) The vector potential of EM field is Aα = −

Qr (1, 0, 0, −a sin2 θ). ρ2

(33.7.5)

So we have (r2 + a2 )At + aAϕ = −Qr,

324

Aϕ + a sin2 θAt = 0

(33.7.6)

Now the Hamilton-Jacobi equation can be written as [ [ ]2 ]2 ∂S 1 1 ∂S 1 ∂S 1 2 ∂S 2 2 ∂S − =− (r + a ) +a + eQr + + a sin θ ∂λ 2 ∆ρ2 ∂t ∂ϕ 2 ρ2 sin2 θ ∂ϕ ∂t ( )2 ( )2 1 ∆ ∂S 1 1 ∂S + . + 2 ρ2 ∂r 2 ρ2 ∂θ

(33.7.7)

(e) Because the equation has no explicit dependence on λ, ϕ, or t, the solution must take the form S=

1 2 µ λ − Et + Lz ϕ + Sr (r) + Sθ (θ), 2

(33.7.8)

where the values of the “integration constants” follow from ∂S/∂λ = −H, ∂S/∂ϕ = πϕ and ∂S/∂t = πt . Insert this assumed form into (33.35), we have ]2 [ ]2 1 [ 1 −(r2 + a2 )E + aLz + eQr + 2 2 Lz − aE sin2 θ 2 ∆ρ ρ sin θ ( )2 ( )2 1 ∂Sθ ∆ ∂Sr + 2 . + 2 ρ ∂r ρ ∂θ

−µ2 = −

(33.7.9)

Multiply both sides of equation above by ρ2 ≡ r2 + a2 cos2 θ, ( ) ]2 1 [ cos2 θ −(r2 + a2 )E + aLz + eQr + L2z 1 + − 2aLz E + a2 E 2 (1 − cos2 θ) ∆ sin2 θ )2 ( )2 ( ∂Sθ ∂Sr + . (33.7.10) +∆ ∂r ∂θ

−(r2 + a2 cos2 θ)µ2 = −

Now we can separate the variable and get L2 cos2 θ − a cos θµ + L = z 2 − a2 E 2 cos2 θ + sin θ 2

and − r 2 µ2 − L = −

2

2

(

∂Sθ ∂θ

]2 1 [ −(r2 + a2 )E + aLz + eQr + (Lz − aE)2 + ∆ ∆

)2

(

(33.7.11)

∂Sr ∂r

)2 ,

(33.7.12)

where the constant L arises naturally as a “separation-of-variables constant” in this procedure. Using the definition [ ] L2z Θ ≡ L − cos2 θ a2 (µ2 − E 2 ) + , sin2 θ P ≡ E(r2 + a2 ) − Lz a − eQr, R ≡ P 2 − ∆[µ2 r2 + (Lz − aE)2 + L],

(33.7.13)

we can obtain the solution for Sr and Sθ : ∫ √ Sr =

R dr, ∆

Sθ =

∫ √

Θdθ.

(33.7.14)

(f) We will use L + (Lz − aE)2 as a new constant of motion instead of L. Therefore X = L + (Lz − aE)2 , µ2 , E or Lz are new canonical momentum brought by generating function S. New canonical coordinates Y = ∂S/∂X are also constant of motion. The equation of motion is given by Y = ∂S/∂X. We can set all Y to 0, as they can be absorbed into the integration of r and θ or eliminated by the redefinition of the origin of t and ϕ. By successively setting ∂S/∂X to zero, we can obtain the equations describing the test-particle orbits. Note that Θ = L + (Lz − aE)2 − a2 µ2 cos2 θ −

325

)2 1 ( Lz − aE sin2 θ . 2 sin θ

(33.7.15)

We have ∂R ∂Θ = −∆, = 1. 2 ∂[L + (Lz − aE) ] ∂[L + (Lz − aE)2 ] ∂R ∂Θ = −∆r2 , = −a2 cos2 θ. 2 ∂µ ∂µ2 ∂Θ ∂R = 2P (r2 + a2 ), = 2a(Lz − aE sin2 θ). ∂E ∂E ∂R ∂Θ = −2aP, = −2(Lz − aE sin2 θ)/ sin2 θ. ∂Lz ∂Lz

(33.7.16a) (33.7.16b) (33.7.16c) (33.7.16d)

Using

∫ ∫ ∂Sr dr ∂R ∂Sθ dr ∂Θ √ √ = , = , ∂X ∂X ∂X 2∆ R 2 Θ ∂X we have the equation of motion ∫ θ ∫ r dθ dr √ = √ , Θ R ∫ θ 2 ∫ r 2 a cos2 θ r √ √ λ= + Θ R ∫ θ ∫ r a(Lz − aE sin2 θ) P (r2 + a2 ) √ √ t= + Θ ∆ R ∫ θ ∫ r Lz − aE sin2 θ aP √ √ . ϕ= + sin2 θ Θ ∆ R

(g) By differentiating these equations, we have √ dθ Θ = , dr R ρ2 dλ a2 cos2 θ dθ r2 = √ +√ =√ , dr Θ dr R R a(Lz − aE sin2 θ) + P (r2 + a2 )∆−1 dt a(Lz − aE sin2 θ) dθ P (r2 + a2 ) √ √ √ = , = + dr dr Θ ∆ R R Lz sin−2 θ − aE + aP ∆−1 dϕ Lz − aE sin2 θ dθ aP √ √ √ = + = . dr sin2 θ Θ dr ∆ R R

(33.7.17)

(33.7.18a) (33.7.18b) (33.7.18c) (33.7.18d)

(33.7.19a) (33.7.19b) (33.7.19c) (33.7.19d)

After combining them, we have dθ dλ 2 dr ρ dλ 2 dt ρ dλ 2 dϕ ρ dλ ρ2

= =

√ √

Θ,

(33.7.20a)

R,

(33.7.20b)

= a(Lz − aE sin2 θ) + P (r2 + a2 )∆−1 ,

(33.7.20c)

= Lz sin−2 θ − aE + aP ∆−1 .

(33.7.20d) (33.7.20e)

(h) Using πα = ∂S/∂xα , we have 1 1 − µ2 = g αβ (πα − eAα )(πβ − eAβ ), 2 2 √

and πθ =

πt = −E,

] [ L2z . L − cos2 θ a2 (µ2 − E 2 ) + sin2 θ

326

π ϕ = Lz

(33.7.21)

(33.7.22)

Using πα − eAα = pα , we have g αβ pα pβ = −µ2 ,

(33.7.23)

Lz = pϕ + eAϕ

[ ] L2z L = p2θ + cos2 θ a2 (µ2 − E 2 ) + . sin2 θ

and

33.8

E = −(pt + eAt ),

(33.7.24)

KERR-SCHILD COORDINATES

(a) Show that In Kerr coordinates the ingoing null congruence (33.39) has the form (33.42a). Also show that the covariant components of the wave vector - after changing to a new affine parameter λnew = λold E - are (in) (in) (in) kr(in) = 0, kθ = 0, kϕ˜ = a sin2 θ, kV˜ = −1. (33.8.1) (b) Introduce new coordinates t˜, x, y, z, defined by ˜

x + iy = (r + ia)eiϕ sin θ,

z = r cos θ,

t˜ = V˜ − r;

(33.8.2)

and show that in tills “Kerr-Schild coordinate system” the metric takes the form (in)

ds2 = (ηαβ + 2Hkα(in) kβ )dxα dxβ , where H=

(33.8.3)

M r − 12 Q2 , r2 + a2 (z/r)2

(33.8.4)

r(xdy + ydx) − a(xdy − ydx) zdz − − dt˜. r2 + a2 r For the transformation to analogous coordinates in which kα(in) dxα = −

(out)

ds2 = (ηαβ + 2Hkα(out) kβ

(33.8.5)

)dxα dxβ ,

(33.8.6)

see, e.g., Boyer and Lindquist (1967).

Solution: (a) In Boyer-Lindquist coordinates, the four momentum of ingoing photons whose trajectories belong to principal null congruences are kt =

(r2 + a2 )E r aE , k = −E, k θ = 0, k ϕ = , ∆ ∆

(33.8.7)

Transformation between Boyer-Lindquist coordinates Kerr coordinates are described by dVe = dt + (r2 + a2 )(dr/∆),

dϕe = dϕ + a(dr/∆).

(33.8.8)

And so we can get r2 + a2 r a ˜ k = 0, k ϕ = k ϕ + k r = 0, ∆ ∆ i.e. k α˜ = (0, −E, 0, 0). The metric in Kerr coordinates is    −1 1 0 0 1 0 2  2M r − Q2  1 0 0 −a sin θ 0 0 +  gα˜ β˜ =   0  0 ρ2 0 0 0 ρ2 0 −a sin2 θ 0 (r2 + a2 ) sin2 θ −a sin2 θ 0 ˜

kV = kt +

327

(33.8.9)

0 0 0 0

 −a sin2 θ  0  . (33.8.10)  0 4 2 a sin θ

˜

So we have kα˜ = gα˜ β˜ k β = −gαr ˜ E = −E(1, 0, 0, −a sin θ). Changing to a new affine parameter λnew = λold E, we can get equation 33.8.1. (b)The transformation matrix between new coordinates α = t˜, x, y, z and Kerr coordinates α ˜ = V˜ , r, θ, ϕ˜ is   1 −1 0 0 α ˜ cos θ −(r sin ϕ˜ + a cos ϕ) ˜ sin θ 0 cos ϕ˜ sin θ (r cos ϕ˜ − a sin ϕ) ∂x . = (33.8.11)  α ˜ ˜ ˜ ˜ ˜ ˜ 0 sin ϕ sin θ (r sin ϕ + a cos ϕ) cos θ (r cos ϕ − a sin ϕ) sin θ  ∂x 0 cos θ −r sin θ 0 (in) (in)

Note that in Kerr coordinates, ηαβ + 2Hkα kβ

will be transformed to

∂xα ∂xβ (in) (in) η + 2Hkα˜ kβ˜ . αβ ∂xα˜ ∂xβ˜

(33.8.12)

It is easy to verify that  −1 1 ∂x ∂x  ηαβ ˜ =  0 ∂xα˜ ∂xβ 0 α

β

and

1 0 0 −a sin2 θ

0 0 ρ2 0



(in) (in)

2Hkα˜ kβ˜

1 2M r − Q2  0  =  0 ρ2 −a sin2 θ

 0 2 −a sin θ    0 2 2 2 (r + a ) sin θ 0 0 0 0 0 0 0 0

 −a sin2 θ  0 .  0 4 2 a sin θ

(33.8.13)

(33.8.14)

So in tills Kerr-Schild coordinate system the metric takes the form of 33.8.3.

33.9

NULL GENERATORS OF HORIZON

(a) Show that in Kerr coordinates the outgoing principle null congruence is described by the tangent vector dr dϕ˜ E dV˜ E dθ = 0, = E, = 2a , = 2(r2 + a2 ) . dλ dλ dλ ∆ dλ ∆

(33.9.1)

(b) These components of the wave vector become singular at the horizon (∆ = 0), not because of a singularity in the coordinate system-the coordinates are well-behaved! - but because of poor normalization of the affine parameter. For each outgoing geodesic, let ∆0 be a constant, defined as the value of ∆ at the event where the geodesic slices the hypersurface V˜ = 0. Then renormalize the affine parameter for each geodesic E λnew = λold . (33.9.2) ∆0 Show that the resulting wave vectors dθ dr dϕ˜ ∆0 dV˜ ∆0 = 0, = ∆0 , = 2a , = 2(r2 + a2 ) dλ dλ dλ ∆ dλ ∆

(33.9.3)

are well-behaved as one approaches the horizon; and show that the geodesics on the horizon have the form 2 θ = const., r = r+ = const., ϕ˜ = 2aλ, V˜ = 2(r+ + a2 )λ. (33.9.4) (c) Show that these are the only test-particle trajectories that remain forever on the horizon.

328

Solution: (a) In Boyer-Lindquist coordinates, the outgoing principle null congruence is described by the tangent vector dr dϕ aE dt (r2 + a2 )E dθ = 0, = E, = . = . (33.9.5) dλ dλ dλ ∆ dλ ∆ Using dVe = dt + (r2 + a2 )(dr/∆), dϕe = dϕ + a(dr/∆), (33.9.6) we can get E dV˜ E dϕ˜ = 2a , = 2(r2 + a2 ) . dλ ∆ dλ ∆

(33.9.7)

(b) When the photon approaches the horizon, we have ∆0 → 0 and so dr/dλ → 0. ∆(r) will be a constant equal to ∆0 . So we have dθ dr dϕ˜ dV˜ = 0, = 0, = 2a, = 2(r2 + a2 ). (33.9.8) dλ dλ dλ dλ The geodesic is well behaved and can be parameterized by 33.9.4. (c) The line element at horizon with dr = 0 is ˜ V˜ . ds2 = −[1 − ρ−2 (2M r+ − Q2 )]dV˜ 2 + ρ2 dθ2 + ρ−2 (r2 + a2 )2 sin2 θdϕ˜2 − 2aρ−2 (2M r+ − Q2 ) sin2 θdϕd (33.9.9) 2 2 + a2 cos2 θ, we can obtain + a2 + Q2 − 2M r+ = 0 and ρ2 = r+ Using ∆ = r+ 2 ˜ 2 + ρ2 dθ2 . ds2 = ρ−2 sin2 θ[adV˜ − (r+ + a2 )dϕ]

(33.9.10)

2 ˜ dθ = 0, adV˜ = (r+ + a2 )dϕ.

(33.9.11)

As ds2 ≤ 0, we must have

So 33.9.4 are the only test-particle trajectories that remain forever on the horizon.

33.10

ANGULAR MOMENTUM VECTOR FOR IN FALLING PARTICLE

Derive equations (33.49d,e,f) for the components Lx and Ly of the orbital angular momentum of a particle falling into a black hole. Assume negligible initial speed, E 2 − µ2 ≈ 0.

Solution: At infinity, we have x = r sin θ cos ϕ, y = r sin θ sin ϕ, z = r cos θ.

(33.10.1)

Assume negligible initial speed, we can get ˙ Lx /µ = y z˙ − z y˙ = −r2 sin ϕθ˙ − r2 cos θ sin θ cos ϕϕ, 2 2 ˙ Ly /µ = z x˙ − xz˙ = r cos ϕθ˙ − r cos θ sin θ sin ϕϕ,

(33.10.2b)

˙ Lz /µ = xy˙ − y x˙ = r2 sin2 θϕ.

(33.10.2c)

(33.10.2a)

Using equations 33.7.13 and 33.7.18 with conditions r ≫ M, a, Q and E 2 − µ2 ≈ 0, we have Θ = L − L2z cot2 θ, P = Er2 , ∆ = r2 , ρ2 = r2 and so r2

√ dθ = Θ, dλ

r2

dϕ Lz , = dλ sin2 θ

329

r2

dt = r2 E. dλ

(33.10.3)

(33.10.4)

Now we have Lx = − sin ϕ∞

√ Θ∞ − cot θ∞ cos ϕ∞ Lz ,

and

33.11

Ly = cos ϕ∞

√

Θ∞ − cot θ∞ sin ϕ∞ Lz

√ √ √ L2x + L2y = Θ∞ + cot2 θ∞ L2z = L.

(33.10.5)

(33.10.6)

IRREDUCIBLE MASS IS IRREDUCIBLE

Show that condition (33.56) is equivalent to δMir ≥ 0.

Solution: The irreducible mass is defined as

( Mir ≡

A 16π

)1/2 ,

(33.11.1)

2 A = 4π(r+ + a2 ).

(33.11.2)

where A is the surface area of horizon, √ So δMir ≥ 0 is equivalent to r+ δr+ + aδa ≥ 0. Using r+ = M + M 2 − Q2 − a2 , we have ) ( M δM − QδQ − aδa r+ δr+ + aδa = r+ δM + + aδa r+ − M r2 δM − r+ QδQ − ar+ δa a(r+ − M )δa = + + r+ − M r+ − M 2 δM − r+ QδQ − aM δ(S/M ) r+ = r+ − M 2 + a2 )δM − r+ QδQ − aδS (r+ = . r+ − M

(33.11.3)

So δMir ≥ 0 is equivalent to δM ≥

33.12

r+ QδQ + aδS . 2 + a2 r+

(33.11.4)

SURFACE AREA OF A BLACK HOLE

Show that the surface area of the horizon of the Kerr-Newman geometry [area of surface r = r+ and t = const (Boyer-Lindquist coordinates) or V˜ = const (Kerr coordinates)] is 16πMir2 .

Solution: In Boyer-Lindquist or Kerr coordinates, the line element on the horizon is 2 ds2 = ρ2 dθ2 + ρ−2 (r+ + a2 )2 sin2 θdϕ2 .

(33.12.1)

2 dS = (r+ + a2 ) sin θdθdϕ.

(33.12.2)

The surface element is therefore

330

The surface area is

∫ A=

2 (r+ + a2 ) sin θdθdϕ

2 = 4π(r+ + a2 ) √ = 4π[(M + M 2 − Q2 − a2 )2 + a2 ] √ = 4π(2M 2 − Q2 + 2 M 4 − Q2 M 2 − S 2 ) √ = 4π(2M 2 − Q2 + (2M 2 − Q2 )2 − Q4 − 4S 2 )

Note that

( M2 =

Mir +

Q2 4Mir

)2 +

S2 . 4Mir2

(33.12.3)

(33.12.4)

We can obtain A = 16πMir2 .

33.13

(33.12.5)

ANGULAR VELOCITY OF A BLACK HOLE

A general theorem [Hartle (1970) for relativistic case; Ostriker and Gunn (1969) for nonrelativistic case] says that, if one injects angular momentum into a rotating star while holding fixed all other contributions to its total mass-energy (contributions from entropy and from baryonic rest mass), then the injection produces a change in total mass-energy given by δ(mass-energy) = (angular velocity of star at point of injection) δ(angular momentum).

(33.13.1)

By analogy, if one injects an angular momentum δS into a rotating black hole while holding fixed all other contributions to its total mass-energy (contributions from irreducible mass and from charge), one identifies the coefficient Ωh in the equation δM = Ωh δS (33.13.2) as the angular velocity of the hole:

( Ωh =

∂M ∂S

) .

(33.13.3)

Q,Mir

(a) Show that the angular velocity of a black hole is equal to Ωh =

a 2 + a2 . r+

(33.13.4)

Notice that this is precisely the angular velocity of photons that live forever on the horizon [equation (33.42b); “barber-pole twist” of null generators of horizon]. (b) Show that any object falling into a black hole acquires an angular velocity (relative to Boyer-Lindquist coordinates) of Ω = dϕ dt in the late stages, as it approaches the horizon. (Recall that the horizon is a singularity of the Boyer-Lindquist coordinates. This is the reason that every object, regardless of its Lz , E, e, µ, L can approach and does approach Ω = Ωh .)

Solution: (a) Generally, we have equation 33.11.4. If we demand that δMir = 0 and δQ = 0, we will have δM =

a 2 + a2 δS. r+

And so we have 33.13.4.

331

(33.13.5)

(b) From equation 33.7.20, we have dϕ (Lz / sin2 θ − aE)∆ + aP = . dt (Lz − aE sin2 θ)a∆ + (r2 + a2 )P

(33.13.6)

As the particle approaches the horizon, ∆ → 0, and so we have dϕ a → 2 . dt r+ + a2

33.14

(33.13.7)

SEPARATION OF VARIABLES FOR WAVE EQUATIONS

This chapter has studied extensively the motion of small objects in the external fields of black holes. Of almost equal importance, but not so well-understood yet because of its complexity, is the evolution of weak electromagnetic and gravitational perturbations (“waves”) in the Kerr-Newman geometry. Just as one had no a priori reason to expect a “fourth constant” for test particle motion in the Kerr-Newman geometry, so one had no reason to expect separability for Maxwell’s equations, or for the wave equations describing gravitational perturbations-or even for the scalar wave equation □ψ ≡ −ψ,α α . Thus it came as a great surprise when Carter (1968c) proved separability for the scalar wave equation, and later when Teukolsky (1972, 1973) separated both Maxwell’s equations and the wave equations for gravitational perturbations. Show that separation of variables for the scalar-wave equation in the (uncharged) Kerr geometry yields solutions of the form ψ = (r2 + a2 )−1/2 uℓm (r)Smℓ (−iωa, cos θ)ei(mϕ−ωt) , (33.14.1) where m and ℓ are integers with 0 < |m| < ℓ; Smℓ is a spheroidal harmonic [see Meixner and Schärfke (1954)]; and uℓm satisfies the differential equation −

d2 u + V u = 0. dr∗2

(33.14.2)

In order to put the equation in this form, define a Regge-Wheeler (1957) “tortoise” -type radial coordinate r∗ by dr∗ = ∆−1 (r2 + a2 )dr, (33.14.3) and find an effective potential V (r∗ ) given by ( )2 ma V =− ω− 2 + [(m − ωa)2 + L](r2 + a2 )−2 ∆ r + a2 + 2(M r − a2 )(r2 + a2 )−3 ∆ + 3a2 (r2 + a2 )−4 ∆2 .

(33.14.4)

In this radial equation L is a constant (analog of Carter’s constant for particle motion), given in terms of m and ℓ by L ≡ λmℓ − m2 ; (33.14.5) λmℓ is eigenfunction of spheroidal harmonic.

Solution: The field equation □ψ = 0 is equivalent to g αβ ψ,αβ − g αβ Γµαβ = 0.

(33.14.6)

The components of metric and inverse metric in Boyer-Lindquist coordinates are given by 33.7.2, 33.7.3 and 33.7.4. Detailed calculations show that g αβ Γµαβ =

1 (0, 2M − 2r, cot θ, 0), ρ2

332

(33.14.7)

where ρ2 = r2 + a2 cos2 θ. The field equations can be written explicitly: 4M ar a2 1 (r2 + a2 )2 ψ,tt +a2 sin2 θψ,tt − ψ,tϕ − ψ,ϕϕ + 2 ψ,ϕϕ +∆ψ,rr +2(r−M )ψ,r +ψ,θθ +cot θψ,θ = 0, ∆ ∆ ∆ sin θ (33.14.8) where ∆ = r2 − 2M r + a2 . −

Suppose the field equation can be solved by separating variables and note that the field equation does not contain t and ϕ explicitly. We would like to assume ψ = R(r)Θ(θ)eimϕ e−iωt .

(33.14.9)

Substitute 33.14.9 into 33.14.8. We can obtain Θ,θθ + cot θΘ,θ (r2 + a2 )2 ω 2 − 4M armω + a2 m2 ∆R,rr + 2(r − M )R,r m2 + + − a2 ω 2 sin2 θ − . ∆ R Θ sin2 θ (33.14.10) The first two terms of the equation functions of r only while the last three terms are functions of θ only. So we can introduce a constant L such that m2 Θ,θθ + cot θΘ,θ − a2 ω 2 sin2 θ − = −a2 ω 2 − m2 − L. Θ sin2 θ The equation above can be rewritten as ( ) ( ) 1 ∂ ∂Θ m2 2 2 2 sin θ + a ω cos θ − Θ = −(L + m2 )Θ. sin θ ∂ ∂θ sin2 θ

(33.14.11)

(33.14.12)

The general solution is spheroidal harmonic Θ = Smℓ (−iωa, cos θ). And λmℓ = L + m2 is the eigenvalue of this Sturm-Liouville differential equation and fixed by the requirement that Smℓ (−iωa, cos θ) must be finite for | cos θ| = 1. The differential equation for R(r) is ∆[∆R,rr + 2(r − M )R,r ] + [(r2 + a2 )2 ω 2 − 4M armω + a2 m2 ]R = (L + a2 ω 2 + m2 )∆R

(33.14.13)

Now we introduce u = R(r2 + a2 )1/2 and note that d/dr = (r2 + a2 )∆−1 d/dr∗ . We have R,r = u,r∗

r (r2 + a2 )1/2 −u 2 ∆ (r + a2 )3/2

(33.14.14)

and (r2 + a2 )3/2 u,r∗ r 2(r − M )(r2 + a2 )1/2 − u + ,r∗ 2 2 2 ∆ ∆ ∆ (r + a2 )1/2 2 r u 3ur − 2 + 2 − u,r∗ 2 2 1/2 2 3/2 ∆(r + a ) (r + a ) (r + a2 )5/2

R,rr = u,r∗ r∗

(33.14.15)

It is easy to get ∆[∆R,rr + 2(r − M )R,r ] = (r2 + a2 )3/2 u,r∗ r∗ − And so we have −

3ur2 ∆2 u∆2 + 2u∆(r − M )r + . (r2 + a2 )3/2 (r2 + a2 )5/2

d2 u + V u = 0, dr∗2

(33.14.16)

(33.14.17)

where V = −ω 2 +

3r2 ∆2 4M armω − a2 m2 + (L + a2 ω 2 + m2 )∆ ∆2 + 2∆(r − M )r + − . (r2 + a2 )2 (r2 + a2 )3 (r2 + a2 )4

A little calculations can lead to 33.14.4.

333

(33.14.18)

334

Chapter 34

GLOBAL TECHNIQUES, HORIZONS, AND SINGULARITY THEOREMS 34.1

FLAT SPACETIME IN ψ, ξ, θ, ϕ COORDINATES

(a) Derive equation (34.2c) from (34.1) and (34.2a,b). (b) Show that the regions I + , I − , I 0 , I + , and I − of flat spacetime are located at I + : ψ = π, ξ = 0, I − : ψ = −π, ξ = 0, I 0 : ψ = 0, ξ = π, I + : ψ + ξ = π, −π < ψ − ξ < π, I − : ψ − ξ = −π, −π < ψ + ξ < π.

(34.1.1)

[see equatIons (34.2)]. These are the regions depicted in Figure 34.2. (c) Show that in flat spacetime, in a ψ, ξ coordinate diagram (Figure 34.2), radial null lines make angles of 45° with the vertical axis, and nonradial null lines make angles of less than 45°.

Solution: (a) From (34.2a,b), we have dt + dr =

dψ + dξ , 2 cos2 12 (ψ + ξ)

dt − dr =

dψ − dξ . 2 cos2 21 (ψ − ξ)

(34.1.2)

We can obtain − dt2 + dr2 = −(dt − dr)(dt + dr) =

−dψ 2 + dξ 2 . + ξ) cos2 12 (ψ − ξ)

4 cos2 21 (ψ

(34.1.3)

Now we can derive equation (34.2c). (b) From (34.2a,b), we have ψ = arctan(t + r) + arctan(t − r),

ξ = arctan(t + r) − arctan(t − r).

(34.1.4)

For future timelike infinity I + , we have t → +∞ while r is finite. So we can get ψ = π and ξ = 0. For past timelike infinity I − , we have t → −∞ while r is finite. So we can get ψ = −π and ξ = 0. For spacelike infinity I 0 , we have r → +∞ while t is finite. So we can get ψ = 0 and ξ = π.

335

Note that ψ + ξ = 2 arctan(t + r), ψ − ξ = 2 arctan(t − r).

(34.1.5)

For future null infinity I , we have t + r → +∞ while t − r is finite. So we can get ψ + ξ = π and ψ − ξ ∈ (−π, π). For past null infinity I − , we have t − r → +∞ while t + r is finite. So we can get ψ − ξ = π and ψ + ξ ∈ (−π, π). +

(c) For null geodesics, we have ds2 = 0. From equation (34.2c), we can get 1 1 dψ 2 = dξ 2 + 4 cos2 (ψ + ξ) cos2 (ψ − ξ)r2 (dθ2 + sin2 θdϕ2 ), 2 2

(34.1.6)

and so dψ 2 = dξ 2 for radial null lines while dψ 2 > dξ 2 for non-radial null lines. So radial null lines make angles of 45° with the vertical axis, and nonradial null lines make angles of less than 45°.

34.2

SCHWARZSCHILD SPACETIME IN ψ, ξ, θ, ϕ COORDINATES

(a) Derive equations (34.3c,d) from (34.3a,b) and the Kruskal-Szekeres equations (31.14). (b) Use equations (34.3) to justify the precise form of the coordinate diagram in Figure 34.3.

Solution: (a) The Schwarzschild geometry in Kruskal-Szekeres coordinates is ds2 =

32M 3 −r/2M e (−dv 2 + du2 ) + r2 (dθ2 + sin2 θdϕ2 ). r

Here r is to be regarded as a function of u and v defined implicitly by ) ( r − 1 er/2M = u2 − v 2 . 2M

(34.2.1)

(34.2.2)

From (34.3a,b), we have 1 v + u = tan (ψ + ξ), 2 So we have

(

1−

1 v − u = tan (ψ − ξ). 2

r ) r/2M 1 1 e = (v + u)(v − u) = tan (ψ + ξ) tan (ψ − ξ). 2M 2 2

Note that dv + du =

dψ + dξ , 2 cos2 12 (ψ + ξ)

dv − du =

dψ − dξ . 2 cos2 12 (ψ − ξ)

(34.2.3)

(34.2.4)

(34.2.5)

The metric becomes ds2 =

−dψ 2 + dξ 2 32M 3 −r/2M e + r2 (dθ2 + sin2 θdϕ2 ). r 4 cos2 12 (ψ + ξ) cos2 12 (ψ − ξ)

(34.2.6)

(b) Using 34.2.4, in ψ − ξ coordinates, the horizon r = 2M becomes 1 1 tan (ψ + ξ) tan (ψ − ξ) = 0, 2 2

(34.2.7)

ψ + ξ = 0 or ψ − ξ = 0.

(34.2.8)

1 1 tan (ψ + ξ) tan (ψ − ξ) = 1. 2 2

(34.2.9)

i.e. The singularity r = 0 becomes

336

Note that 1/ tan α = cot α, we have

i.e.

34.3

1 1 tan (ψ + ξ) = cot (ψ − ξ), 2 2

(34.2.10)

1 1 π π (ψ + ξ) + (ψ − ξ) = ψ = or − . 2 2 2 2

(34.2.11)

REISSNER-NORDSTROM SPACETIME

(a) Show that there exists a coordinate system in which the Reissner-Nordstrom geometry with 0 < |Q| < M (exercises 31.8 and 32.1) has the form ds2 = F 2 (−dψ 2 + dξ 2 ) + r2 dΩ2 ,

F = F (ψ, ξ), r = r(ψ, ξ),

(34.3.1)

and in which the honzons and infinities are as shown in Figure 34.4. (b) Use Figure 34.4 to deduce that the Reissner-Nordstrom geometry describes a “worm-hole” or bridge, connecting two asymptotically flat spacetimes, which: (i) expands to a state of maximum circumference; (ii) recontracts toward a state of minimum circumference, and in the process disconnects its outer regions from the two I 0 ’s (spatial infinity) and reconnects them to a pair of r = 0 singularities; (iii) bounces; (iv) reexpands, and in the process disconnects its “outer regions” from the two singularities and reconnects them to a pair of I 0 ’s in two new asymptotically flat universes; (v) slows its expansion to a halt; (vi) recontracts toward a state of minimum circumference, and in the process disconnects its outer regions from the two I 0 ’s and reconnects them to a new pair of r = 0 singularities; etc. ad infinitum.

Solution: (a) The metric of Reissner-Nordstrom geometry in t − r coordinates is ( ) ( )−1 2M Q2 2M Q2 ds2 = − 1 − + 2 dt2 + 1 − + 2 dr2 + r2 dΩ2 . r r r r

(34.3.2)

Define r∗ (r) by

dr∗ = f −1 , dr where f = (1 − 2M/r + Q2 /r2 ). The transformation matrix of coordinate transformation { V˜ = t + r∗ ˜ = t − r∗ U is

∂(t, r) = ˜) ∂(V˜ , U

(

1/2 f /2

The new components of the metric are therefore ( )( )( 1/2 f /2 −f 0 1/2 1/2 −f /2 0 f −1 f /2 i.e.

) 1/2 . −f /2 ) ( 1/2 0 .= −f /2 −f /2

) ( Q2 2M ˜ + r2 dΩ2 . + 2 dV˜ dU ds = − 1 − r r 2

Define

u ˜ = ∓sgn(f )e−U /4M , ˜

337

˜

v˜ = ±eV /4M .

(34.3.3)

(34.3.4)

(34.3.5)

) −f /2 , 0

(34.3.6)

(34.3.7)

(34.3.8)

We have

∗ ˜ dV˜ = sgn(f )16M 2 e(U˜ −V˜ )/4M d˜ dU ud˜ v = sgn(f )16M 2 e−r /2M d˜ ud˜ v.

(34.3.9)

Now introducing new coordinates 1 u ˜ = tan (ψ − ξ), 2

1 v˜ = tan (ψ + ξ). 2

(34.3.10)

We have d˜ ud˜ v=

dψ + dξ dψ − dξ −dψ 2 + dξ 2 =− . 1 1 1 2 2 2 2 cos 2 (ψ + ξ) −2 cos 2 (ψ − ξ) 4 cos 2 (ψ − ξ) cos2 21 (ψ + ξ)

(34.3.11)

ds2 = F 2 (−dψ 2 + dξ 2 ) + r2 dΩ2 ,

(34.3.12)

The metric is where F = 2M e

−r ∗ /M

1/2 1 1 Q2 2M sec (ψ + ξ) sec (ψ − ξ) 1 − + 2 . 2 2 r r

Note that f=

(34.3.13)

(r − r+ )(r − r− ) r2

(34.3.14)

The integration gives r∗ =

2 2 r+ log |r/r+ − 1| − r− log |r/r− − 1| + r. r+ − r−

(34.3.15)

So we have r∗ → ∞ when r → ∞; r∗ → −∞ when r → r+ ; r∗ → ∞ when r → r− ; r∗ → 0 when r → 0. Also note that u ˜v˜ = −sgn(f )er

∗

/2M

, v˜/˜ u = −sgn(f )et/2M , v˜ = e(r

∗

+t)/4M

, u ˜ = −sgn(f )e(r

∗

−t)/4M

.

(34.3.16)

The spacelike infinity is u ˜v˜ → −∞ while v˜/˜ u is finite. In ψ − ξ coordinates, we have ψ = 2nπ while ξ = ±π. We limit −π ≤ ξ ≤ π in order to separate disconnect spacetime. The future timelike infinity is t → ∞ for ξ > 0 and t → −∞ for ξ < 0, while r is finite. So we have v˜/˜ u → −∞ for ξ > 0 and v˜/˜ u → 0 for ξ < 0 while v˜u ˜ is finite. In ψ − ξ coordinates, we can get ψ = (2n + 1/2)π and ξ = ±π/2. Similarly, we have ψ = (2n − 1/2)π and ξ = ±π/2 for past time like infinity. The future null infinity is t + r → ∞ for ξ > 0 and r − t → ∞ for ξ < 0. So we have v˜ → ∞ for ξ > 0 and u ˜ → ∞ for ξ < 0. In ψ − ξ coordinates, we can get ψ + ξ = (2n + 1)π for ξ > 0 and ψ − ξ = (2n + 1)π for ξ < 0 Similarly, we have ψ − ξ = (2n + 1)π for ξ > 0 and ψ − ξ = (2n + 1)π for ξ < 0 in the case of past null infinity. Note we must keep the line representing the future (past) null infinity between spacelike infinity and future (past) timelike infinite. Horizon r = r− sits at t + r∗ → ∞ with t − r∗ finite or r∗ − t → ∞ with t + r∗ finite since horizon is null surface. So they are also lines ψ ± ξ = (2n + 1)π in ψ − ξ coordinates. But we must keep the line representing the inner horizon beyond spacelike infinity and timelike infinite. Horizon r = r+ sits at r∗ − t → −∞ with t + r∗ infinite or r∗ + t → ∞ with t − r∗ finite, corresponding to v˜ = 0 or u ˜ = 0. So they are lines ψ ± ξ = 2nπ. Singularity r = r∗ = 0 corresponds to u ˜v˜ = −1. It is easily to get ξ = ±π/2. Note we must keep it with the inner horizon. Put all the discussion together, we can get the following spacetime diagram copied from textbook.

338

(b) Consider the hypersurface ψ = constant. At ψ = 0, the worm whole expands to a state of maximum circumference 2πr+ . When 0 < ψ < π, the worm hole recontracts. At ψ = π/2, the worm hole disconnects its outer regions from the two I 0 s and reconnects them to a pair of r = 0 singularities. At ψ = π, the worm hole recontracts to a state of minimum circumference 2πr− , and then bounces. When π < ψ < 2π, the worm hole reexpands. At ψ = π/2, the worm hole disconnects its “outer regions” from the two singularities and reconnects them to a pair of I 0 s in two new asymptotically flat universes. At ψ = 2π, the worm hole slows its expansion to a halt and expands to a state of maximum circumference 2πr+ ...

34.4

A BLACK HOLE CAN NEVER BIFURCATE

Make plausible the theorem that no matter how hard one “zaps” a black hole, and no matter what one “zaps” it with, one can never make it bifurcate into two black holes.

Solution: The solution is based on chapter 9.2 of the book The large scale structure of spacetime (S.W.Hawking &

339

G.F.R.Ellis) Let B1 (τ1 ) be a black hole on S (τ1 ). Let B2 (τ2 ) and B3 (τ2 ) be black holes on a later surface S (τ2 ). If B1 (τ1 ) bifurcates into B2 (τ2 ) and B3 (τ2 ), then B2 (τ2 ) and B3 (τ2 ) both intersect J + (B1 (τ1 )), and does not intersect with each other. Other the other hand, every future-directed inextendible timelike curve from B1 (τ1 ) will intersect S (τ2 ). Thus J + (B1 (τ1 )) ∩ S (τ2 ) (34.4.1) is connected and will be contained in a connected component of B(τ2 ), or the signal from B1 (τ1 ) can escape to future null infinite I + . If B2 (τ2 ) and B3 (τ2 ) both intersect J + (B1 (τ1 )), then we have J + (B1 (τ1 )) ∩ S (τ2 ) ⊂ B2 (τ2 )

(34.4.2)

J + (B1 (τ1 )) ∩ S (τ2 ) ⊂ B3 (τ2 ),

(34.4.3)

and which is contradictory with the fact that B2 (τ2 ) and B3 (τ2 ) does not intersect with each other. So a black hole can never bifurcate.

340

Chapter 35

PROPAGATION OF GRAVITATIONAL WAVES 35.1

TRANSFORMATION OF PLANE WAVE TO TT GAUGE

Let a plane wave of the form (35.4) be given, in some arbitrary gauge of linearized theory. Exhibit explicitly the transformation that puts it into the TT gauge.

Solution: For plane wave ¯ µν = Aµν exp(ikα xα ), h α

(35.1.1)

α

where k kα = 0, Aµα k = 0, the gauge transformation generated by ξ µ = −iC µ exp(ikα xα )

(35.1.2)

will be Aµν,new = Aµν,old − Cµ kν − Cν kµ + ηµν kα C α . µ

0

(35.1.3) j

Work in a Lorentz frame where the 4-velocity u , of the TT gauge is u = 1, u = 0. The the new Aµν must satisfy Aµ0 = 0 and Akk = 0. We can obtain the equation A00,old − C0 k0 − Ci ki = 0,

(35.1.4a)

Ai0,old − Ci k0 − C0 ki = 0,

(35.1.4b)

Aii,old − 3C0 k0 + Ci ki = 0.

(35.1.4c)

From 35.1.4b, we have ki Ai0,old = ki Ci k0 + C0 ki2 = ki Ci k0 + C0 k02 = k0 (ki Ci + C0 k0 ) = k0 A00,old

(35.1.5)

using k02 = ki2 and A00 k0 = A0i ki . Noting that k0 ̸= 0, we conclude that 35.1.4a can be deduced from 35.1.4b. Equation 35.1.4b and 35.1.4c can be written as       A10,old C0 k1 k0  k2  C1  A20,old  k0   · = (35.1.6)  k3 k0  C2  A30,old  Aii,old C3 3k0 −k1 −k2 −k3 The determine of the matrix is −4k04 ̸= 0. So we must have a solution Cµ , i.e. we can put plane wave into the TT gauge.

341

35.2

LIMITATION ON EXISTENCE OF TT GAUGE

Although the metric perturbation hµν for any gravitational wave in linearized theory can be put into the TT form (35.8), nonradiative hµν ’s cannot. Consider, for example, the external field of a rotating, spherical star, which cannot be written as a superposition of plane waves: h00 =

2M , r

hjk =

2M δjk , r

h0k = −2ϵklm

S l xm , r3

(35.2.1)

√ where r = x2 + y 2 + z 2 , M is the star’s mass and S is its angular momentum. Show that this cannot be put into TT gauge.

Solution: Suppose the external field of a rotating can be put into TT gauge, then we have T hTjk,00 = −2Rj0k0 .

(35.2.2)

Rj0k0 is gauge invariant, so it can be calculated using original hµν . To linear order, we have 1 M (r2 δjk − 3xj xk ) Rj0k0 = Γj00,k − Γj0k,0 = − h00,jk = . 2 r5 Thus

2M (3xk xj − r2 δjk ) . r5 in both the original gauge and the new gauge. In TT gauge, we have T hTjk,00 =

Now we calculate R0xyz

1 T 1 T R0xyz = Γ0xz,y − Γ0xy,z = − hTxz,0y + hTxy,0z . 2 2

(35.2.3)

(35.2.4)

(35.2.5)

Clearly, R0xyz calculated in this way must be a function of time While R0xyz calculated using original hµν can not be a function of time. This is contradictory with the fact that Rµνρσ is gauge invariant. So the external field of a rotating can not be put into TT gauge.

35.3

A CYLINDRICAL GRAVITATIONAL WAVE

To restore one’s faith, which may have been shaken by exercise 35.2, one can consider the radiative solution whose only nonvanishing component hµν is √ ¯ zz = 4A cos(ωt)J0 (ω x2 + y 2 ), (35.3.1) h where J0 is the Bessel function. This solution represents a superposition of ingoing and outgoing cylindrical gravitational waves. For this gravitational field calculate Rj0k0 and from it infer hTjkT . Then calculate several other components of Rαβγδ (e.g., Rxyxy ) in the original gauge and in TT gauge, and verify that the answers are the same.

Solution: In the original gauge, we have

hµν

 1 0 = 0 0

0 −1 0 0

 0 0 0 0  2A cos(ωt)J0 (ωρ), −1 0 0 1

342

(35.3.2)

where ρ =

√

x2 + y 2 . So the curvature tensor 1 T 1 Rj0k0 = Γj00,k − Γj0k,0 = − (h00,jk + hjk,00 ) = − hTjk,00 . 2 2

(35.3.3)

hTjkT = −ω −2 h00,jk + hjk .

(35.3.4)

We can infer that

Then we calculate several other components of Rαβγδ in the original gauge and in TT gauge. For example, in the original gauge, we have 1 Rxyxy = Γxyy,x − Γxyx,y = − (hyy,xx + hxx,yy ). 2

(35.3.5)

In TT gauge, we have 1 T T T Rxyxy = − (hTyy,xx + hTxx,yy − 2hTxy,xy ) 2 1 1 = − (hyy,xx + hxx,yy ) + ω −2 (h00,yyxx + h00,xxyy − 2h00,xyxy ) 2 2 1 = − (hyy,xx + hxx,yy ). 2

(35.3.6)

By similar procedure, we can verify that other components of Rαβγδ in the original gauge and in TT gauge are the same.

35.4

NON-TT PARTS OF METRIC PERTURBATION

From Box 35.1 establish the formula hT = ∇−2 (hkk,ll − hkl,kl ); then verify the gauge invariance of hT directly, by showing that hkk,ll − hkl,kl is gauge-invariant. Use δhij = ξi,j + ξj,i . Show similarly that the ˜ 0k defined by quantities h ˜ 0k = h ¯ 0k − ∇−2 (h ¯µ + h ¯ kl,l0 ) h (35.4.1) 0,µk are gauge-invariant. Show from the gauge-invariant linearized field equations (18.5) that ∇2 hT = −16πT 00 ,

˜ 0k = −16πT0k , ∇2 h

(35.4.2)

˜ 0k must vamsh for waves in empty space. so hT and h

Solution: The projection operator to transverse direction is Pjk = δjk − ∇−2 ∂j ∂k .

(35.4.3)

hT ≡ Pkl hkl = hkk − ∇−2 hkl,kl = ∇−2 (hkk,ll − hkl,kl ).

(35.4.4)

So we have

Under gauge transformation, we have δ(hkk,ll − hkl,kl ) = 2ξk,kll − ξk,lkl − ξl,kkl = 0. So hT is gauge invariant.

343

(35.4.5)

˜ 0k , we have As for h

[ ] ˜ 0k = δh0k − ∇−2 δhµ − 1 δ µ (−δh00,µk + δhii,µk ) + δhkl,l0 − 1 δkl (−δh00,l0 + δhii,l0 ) δh 0,µk 2 0 2 ] [ 1 1 = δh0k − ∇−2 −δh00,0k + δhi0,ik − (−δh00,0k + δhii,0k ) + δhkl,l0 − (−δh00,k0 + δhii,k0 ) 2 2 = δh0k − ∇−2 (δhi0,ik − δhii,0k + δhkl,l0 ) = ξ0,k + ξk,0 − ∇−2 (ξi,0ik + ξ0,iik − 2ξi,i0k + ξk,ll0 + ξl,kl0 ) = ξ0,k + ξk,0 − ∇−2 (ξ0,iik + ξk,ii0 ) (35.4.6)

= 0. ˜ 0k is gauge invariant. So h The gauge-invariant linearized field equations is hµα,ν α + hνα,µα − hµν,αα − h,µν − ηµν (hαβ ,αβ − h,β β ) = 16πTµν .

(35.4.7)

The 00 component of the equation is 2h0α,0α − h00,αα − h,00 + (hαβ ,αβ − h,β β ) = − 2h00,00 + 2h0i,0i + h00,00 − h00,ii + h00,00 − hii,00 + (h00,00 − 2h0i,0i + hij,ij ) + (−h00 + hii ),00 − (−h00 + hii ),jj =hij,ij − hii,jj = 16πT00 .

(35.4.8)

Note that ∇2 hT = hkk,ll − hkl,kl .

(35.4.9)

∇2 hT = −16πT00 .

(35.4.10)

We have

The linearized field equations can also be written as α ¯ ¯ αβ + h ¯ α +h ¯ α = 16πTµν . −h µν,α − ηµν hαβ, µα, ν να, µ

(35.4.11)

The 0k component of the equation is α ¯ ¯ α ¯ α −h 0k,α + h0α, k + hkα, 0 ¯ ¯ ¯α − h ¯ =h −h +h 0k,00

0k,jj ¯ ¯α h0k,jj + h 0,αk

0,αk

k0,00

¯ +h kj,j0

¯ +h kj,j0 = 16πT0k .

(35.4.12)

˜ 0k = h ¯ 0k,jj − (h ¯µ + h ¯ kl,l0 ). ∇2 h 0,µk

(35.4.13)

˜ 0k = −16πT0k . ∇2 h

(35.4.14)

=− Note that We have

35.5

ALTERNATIVE CALCULATION OF RELATIVE OSCILLATIONS

Introduce a TT coordinate system in which, at time t = 0, the two particles are both at rest. Use the geodesic equation to show that subsequently they both always remain at rest in the TT coordinates, despite the action of the wave. This means that the contravariant components of the separation vector are always constant in the TT coordinate frame: nj = xjB − xjA = const.

344

(35.5.1)

Call this constant xjB(0) . Transform these components to the comoving orthonormal frame; the answer should be equation (35.15).

Solution: The geodesic equation particles is d2 xj dxα dxβ + Γj αβ = 0. 2 dτ dτ dτ

(35.5.2)

In TT gauge, Γj αβ = 0 unless if α = 0 or β = 0. However, dxα /dτ ̸= 0 unless α = 0. If the particle is at rest at t = 0, we have d2 xj /dτ 2 = 0 at t = 0. And so it will remain at rest. Assuming xA = 0, the coordinate transformation to local Lorentz frame of A is

So we have

1 ˆ xi = xi + hTijT xj + O(h2 , x2 ) 2

(35.5.3)

1 ˆ xiB = xiB + hTijT xjB . 2

(35.5.4)

Supposing hTijT vanish when τ = 0, we have 1 ˆ ˆ ˆ xiB (τ ) = xiB(0) + hTijT xiB(0) . 2

35.6

(35.5.5)

ROTATIONAL TRANSFORMATIONS FOR POLARIZATION STATES

Consider two Lorentz coordinate systems, one rotated by an angle θ about the z direction relative to the other: t′ = t, x′ = x cos θ + y sin θ, y ′ = y cos θ − y ′ sin θ, z ′ = z. (35.6.1) Let | ↑⟩ and | ↓⟩ be quantum-mechanical states of a neutrino with spin-up and spin-down relative to the x direction; and similarly for | ↑′ ⟩ and | ↓′ ⟩. Let ex , ey , ex′ , ey′ , be the unit polarization vectors in the two coordinate systems for an electromagnetic wave traveling in the z-direction; and similarly e+ , e× , e+′ , e×′ for a gravitational wave in linearized theory. Derive the following transformation laws: 1 1 1 1 | ↑′ ⟩ = | ↑⟩ cos θ + | ↓⟩ sin θ; | ↓′ ⟩ = −| ↑⟩ sin θ + | ↓⟩ cos θ; 2 2 2 2 ex′ = ex cos θ + ey sin θ; ey′ = −ex sin θ + ey cos θ; e+′ = e+ cos 2θ + e× sin 2θ; e×′ = −e+ sin 2θ + e× cos 2θ.

(35.6.2)

What is the generalization to the linear-polarization basis states for a radiation field of arbitrary spin S?

Solution: Using | ↑⟩ and | ↓⟩ as base vector, the operator for spin in x and y direction is ( ( ) ) 1 1 0 1 0 1 Sx = ; Sy = . 2 0 −1 2 1 0

(35.6.3)

So the spin operate in x′ direction is S

x′

1 = Sx cos θ + Sy sin θ = 2

345

( cos θ sin θ

sin θ − cos θ.

) (35.6.4)

The eigenvector is (

1 1 cos θ, sin θ 2 2

) for λ =

(

1 ; 2

1 1 − sin θ, cos θ 2 2

)

1 for λ = − , 2

(35.6.5)

i.e. 1 1 | ↑′ ⟩ = | ↑⟩ cos θ + | ↓⟩ sin θ; 2 2

1 1 | ↓′ ⟩ = −| ↑⟩ sin θ + | ↓⟩ cos θ. 2 2

(35.6.6)

The unit polarization vectors for an electromagnetic wave is a normal vector in space. The transformation laws are identical to that of coordinates. So we have ey′ = −ex sin θ + ey cos θ.

ex′ = ex cos θ + ey sin θ;

(35.6.7)

Note that ′ ′

′

hi j =

′

∂xi ij ∂xj h . ∂xi ∂xj

(35.6.8)

We have e

+′

( cos θ = sin θ

and

( e×′ =

cos θ sin θ

− sin θ cos θ

− sin θ cos θ

)(

)(

1 0

0 −1

0 1

1 0

)(

)(

cos θ − sin θ

cos θ − sin θ

sin θ cos θ

sin θ cos θ

)

( =

)

( =

)

cos 2θ sin 2θ

sin 2θ − cos 2θ

− sin 2θ cos 2θ

) cos 2θ , sin 2θ

(35.6.9)

(35.6.10)

i.e. e×′ = −e+ sin 2θ + e× cos 2θ.

e+′ = e+ cos 2θ + e× sin 2θ;

(35.6.11)

The generalization to the linear-polarization basis states for a radiation field of arbitrary spin S is eb = −ea sin Sθ + eb cos Sθ.

ea = ea cos Sθ + eb sin Sθ;

35.7

(35.6.12)

ELLIPTICAL POLARIZATION

Discuss elliptically polarized gravitational waves in a manner analogous to the discussion of linearly and circularly polarized waves in Figure 35.2.

Solution: √ For elliptical polarized waves e = (2e+ + ie× )/ 5, deformation of a ring of test particles are shown in the following figure.

346

1.0 0.5 0.0 −0.5 −1.0 1.0

−1

0

1

−1

0

1

−1

0

1

−1

0

1

0.5 0.0 −0.5 −1.0 1.0 0.5 0.0 −0.5 −1.0 1.0 0.5 0.0 −0.5 −1.0

347

35.8

GLOBALLY WELL-BEHAVED COORDINATES FOR PLANE WAVE

Find a coordinate transformation similar to (35.35), which puts the exact plane-wave solution (35.29a), (35.31), into the form ds2 = dX 2 + dY 2 − dU dV + (X 2 − Y 2 )F dU 2 ,

F = F (U ) completely arbitrary.

(35.8.1)

This coordinate system has the advantage of no coordinate singularities anywhere; while the original coordinate system has the advantages of an easy transition to linearized theory, and easy interpretation of the action of the wave on test particles.

Solution: The coordinate transformation can be given by X , Leβ y y= , Le−β u = U, ( ′ ) ( ′ ) L L ′ 2 ′ v=V − +β X − − β Y 2, L L x=

(35.8.2a) (35.8.2b) (35.8.2c) (35.8.2d)

where L′′ + (β ′ )2 L = 0. Then we can get ) ] [ ( ′ 1 L ′ dx = + β dU , dX − X Leβ L ) ] [ ( ′ 1 L ′ dy = − β dU , dY − Y Le−β L du = dU,

(

′

)

(

(35.8.3a) (35.8.3b) )

′

L L dv = dV − 2X + β ′ dX − 2Y − β ′ dY L L ) ( ′′ )] [ ( ′′ L′2 L L′2 L ′′ 2 ′′ 2 − 2 +β +Y − 2 −β dU. − X L L L L

(35.8.3c)

(35.8.3d)

The metric of the spacetime is therefore ds2 = L2 e2β dx2 + L2 e−2β dy 2 − dudv ) ]2 [ ( ′ ) ]2 [ ( ′ L L ′ ′ = dX − X + β dU + dY − Y − β dU L L ( ′ ) ( ′ ) L L ′ ′ − dU dV + 2X + β dXdU + 2Y − β dU dY L L ) ( )] [ ( ′′ L′′ L L′2 L′2 ′′ 2 ′′ 2 +Y dU 2 + X − 2 +β − 2 −β L L L L ( ′ ′ ) 2L β 2 2 2 2 ′′ = dX + dY − dU dV + (X − Y ) + β dU 2 . L

(35.8.4)

If we assume F (U ) = 2L′ β ′ /L + β ′′ , then we have ( ds = dX + dY − dU dV + F (U )(X − Y ) 2

2

2

2

348

2

) 2L′ β ′ ′′ + β dU 2 . L

(35.8.5)

35.9

GEODESIC COMPLETENESS FOR PLANE-WAVE MANIFOLD

Prove that the coordinate system (X, Y, U, V ) of exercise 35.8 completely covers its spacetime manifold. More specifically, show that every geodesic can be extended in both directions for an arbitrarily large affine-parameter length without leaving the X, Y , U , V coordinate system. This property is called geodesic completeness. [Hint: Choose an arbitrary event and an arbitrary tangent vector d/dλ there. They determine an arbitrary geodesic. Perform a coordinate transformation that leaves the form of the metric unchanged and puts ˜ , V˜ ) = constant 2-surface, or in the (X, ˜ Y˜ ) = constant 2-surface. Verify that the d/dλ either in the (U two coordinate systems cover the same region of spacetime. Then analyze completeness of geodesic in ˜ Y˜ , U ˜ , V˜ ) coordinates.] (X, I do not know how to find a coordinate transformation that leaves the form of the metric unchanged and ˜ , V˜ ) = constant 2-surface, or in the (X, ˜ Y˜ ) = constant 2-surface. puts an arbitrary vector either in the (U Any help is appreciated.

35.10

PLANE WAVE WITH TWO POLARIZATIONS PRESENT

The exact plane-wave solution (35.29) has polarization e+ . Construct a similar solution, containing two arbitrary amplitudes, β(u) and γ(u), for polarizations e+ and e× . Extend the discussions of section 35.935.12 to this solution.

Solution: Suppose the metric is ds2 = L2 (e2β dx2 + 4γdxdy + e−2β dy 2 ) − dudv.

(35.10.1)

The only component of the Ricci tensor that does not vanish identically is [ ] 2 (1 − 4γ 2 )Lβ ′2 − (1 + 4γ 2 )Lγ ′2 − 2γ(1 − 4γ 2 )Lγ ′′ − 4γ(1 − 4γ 2 )γ ′ L′ + (1 − 4γ 2 )2 L′′ Ruu = − . (1 − 4γ 2 )2 L (35.10.2) The field equation is [ ] (1 − 4γ 2 )2 L′′ − 4(1 − 4γ 2 )γγ ′ L′ + (1 − 4γ 2 )β ′2 − (1 + 4γ 2 )γ ′2 − 2(1 − 4γ 2 )γγ ′′ L = 0.

(35.10.3)

The linearized version of this equation is L′′ = 0. Therefore the solution corresponding to linearized theory is L = 1, β(u) and γ(u) arbitrary but small. (35.10.4) The corresponding metric is ds2 = (1 + 2β)dx2 + (1 − 2β)dy 2 + 4γdxdy + dz 2 − dt2 ,

β = β(t − z), γ = γ(t − z).

(35.10.5)

Notice that this is a plane wave containing two arbitrary amplitudes, β(u) and γ(u), for polarizations e+ and e× . Return attention to the exact plane wave, and focus on the case where the “wave factor” β(u) and γ(u) is a pulse of duration 2T , and |β ′ |, |γ ′ | ≪ 1/T throughout the pulse. Then the exact solution is: (1) for u < T (flat spacetime; pulse has not yet arrived), β = γ = 0, L = 1;

349

(35.10.6)

(2) for −T < u < T (interior of pulse), β = β(u) is arbitrary, except that |β ′ | ≪ 1/T, ∫ L(u) = 1 −

u

{∫

−T

u ¯

−T

γ = γ(u) is arbitrary, except that |γ ′ | ≪ 1/T, } ′2 ¯ ′2 ¯ ′′ ¯ ¯ ¯ [β (u) − γ (u) − 2γ(u)γ (u)]du d¯ u + O(h4 ),

(35.10.7)

where h ∼ β, γ, β ′ T , γ ′ T and γ ′′ T 2 . (3) for u > T (after the pulse has passed), β = 0, γ = 0 L = 1 −

u , a

a ≡ ∫T T

1 + O(h2 ) (β ′2 − γ ′2 − 2γγ ′′ )du

.

(35.10.8)

Perform the coordinate transformation as equation (35.35), we can show that spacetime is completely flat in regions where the “wave factor” vanishes - which is everywhere outside the pulse. Consider a family of particles that are all at rest in the original t, x, y, z coordinate system before the pulse arrives. Then even while the pulse is passing, and after it has gone, the particles remain at rest in the coordinate system from the fact that Γµ00 = 0. Two particles whose separation is in the direction of propagation of the pulse (z-direction) have constant proper separation since gzz = 1. Plane wave is completely transverse, like a plane wave of linearized theory. Neighboring particles transverse to the propagation direction have a proper separation that wiggles as the pulse passes ∆s2 ≈ L[(1 + 2β)(∆x)2 + (1 − 2β)(∆y)2 + 4γ∆x∆y]. (35.10.9) Superimposed on the usual linearized-theory type of wiggling is a very small net acceleration of the particles toward each other due to L. This is an acceleration of almost Newtonian type, caused by the gravitational attraction of the energy that the gravitational wave carries between the two particles. The total effect of all the energy that passes is to convert the particles from an initial state of relative rest, to a final state of relative motion with speed vfinal =

d∆s L∆si ∆si =d =− , dt dt a

(35.10.10)

where ∆si is the initial particle separation. Think of the exact gravitational plane-wave solution as ripples in the spacetime curvature, described by β(u) and γ(u), propagating on a very slightly curved background spacetime, characterized by L(u). By similar discussion in section 12, we can show that the wavelength of gravitational wave is much smaller than the radius of curvature of background. This difference in scales enables one to separate out the background from the ripples. The metric for the background of the gravitational plane wave is (B) ds2 = gµν dxµ dxν = L2 (dx2 + dy 2 ) − dudv.

(35.10.11)

The metric for the full spacetime is (B) ds2 = (gµν + hµν )dxµ dxν ,

hxx = −hyy = 2β, hxy = hyx = 2γ, all other hµν = 0.

(35.10.12)

One can think of the ripples as a transverse, traceless, Guu = −2L′′ /Lsymmetric tensor field hµν propagating in the background geometry. Suppose the background curvature is produced the gravitational wave ripples hµν . We have G(B) uu = −

2L′′ (EFF) = 8πTuu . L

350

(35.10.13)

Using equation 35.10.3, to the second order of h, we have (EFF) Tuu =

In averaging sense, we have

(35.10.14)

0 = ⟨(γ 2 )′′ ⟩ = 2⟨γ ′2 ⟩ + 2⟨γγ ′′ ⟩. ⟨

So (EFF) ⟨Tuu ⟩=

35.11

β ′2 − γ ′2 − 2γγ ′′ . 4π

β ′2 + γ ′2 4π

⟩

⟨ =

⟩ 1 hjk,u hjk,u . 32π

(35.10.15) (35.10.16)

CONNECTION COEFFICIENTS AND CURVATURE TENSORS FOR A PERTURBED METRIC

In a specific coordinate frame of an arbitrary spacetime, write the metric coefficients in covariant representation in the form (B) gµν = gµν + hµν . (35.11.1) (At the end of the calculation, one can split hµν into two parts, hµν → hµν + jµν and out of this split obtain (B) the formulas used in the text.) Assume that the typical components of hµν are much less than those of gµν ; so one can expand Christoffel symhols and curvature tensors in hµν . Raise and lower indices of hµν with (B) (B) and by a “;” covariant derivatives relative to ; and denote by a “|” covariant derivatives relative to gµν gµν gµν . (B) can be thought of as two different metrics coexisting in the spacetime manifold. Show (a) Here gµν and gµν that the difference between the corresponding covariant derivatives, ∇ − ∇(B) -indeed, the difference between any two covariant derivatives!- is a tensor with components

S αβγ = Γαβγ − Γ(B)αβγ .

(35.11.2)

µν − hµν + hµα hνα − hµα hβα hνβ + · · · g µν = g(B)

(35.11.3)

µν g µν = g(B) − hµν + hµα hνα − hµα hβα hβγ g γν .

(35.11.4)

(b) Show that and also that

(B) (c) By calculating in a local Lorentz frame of gµν and then transforming back to the original frame, show that 1 S µβγ = g µα (hαβ|γ + hαγ|β − hβγ|α ), (35.11.5a) 2 (35.11.5b) Rαβγδ − R(B)αβγδ = S αβδ|γ − S αβγ|δ + S αµγ S µβδ − S αµδ S µβγ ,

Rβδ − R(B)βδ = S αβδ|α − S αβα|δ + S αµα S µβδ − S αµδ S µβα .

(35.11.5c)

(d) Show that expression 35.11.5c reduces to Rβδ = R(B)βδ + R

(1) βδ

+R

(2) βδ

+ ···

(35.11.6)

where R(1) and R(2) are defined by equations (35.58).

Solution: (a) For any vector field u = uα eα , we have (∇ − ∇(B) )u = (uα;β − uα|β )eα ⊗ ω β = (Γαβγ − Γ(B)αβγ )uγ eα ⊗ ω β .

351

(35.11.7)

Suppose that S ≡ S αβγ eα ⊗ ω β ⊗ ω γ = (Γαβγ − Γ(B)αβγ )uγ eα ⊗ ω β ⊗ ω γ .

(35.11.8)

(∇ − ∇(B) )u = S · u.

(35.11.9)

We have Since (∇ − ∇

(B)

)u is a tensor, S · u is a tensor for any vector field u. And so S is a tensor with components S αβγ = Γαβγ − Γ(B)αβγ .

(35.11.10)

(B) gµν = gµν + ϵhµν .

(35.11.11)

µν g µν = g(B) + ϵx(1)µν + · · · + ϵn x(n)µν + · · · .

(35.11.12)

(b) Introduce a formal parameter ϵ in And we have the expansion

Using gµα g αν = δµν , we can obtain αν (B) (1)αν (B) (n)αν δµν = δµν + ϵ(hµα g(B) + gµα x ) + · · · + ϵn (hµα x(n−1)αν + gµα x ) + ···

(35.11.13)

Thus coefficients for all orders of ϵ must vanish, i.e. x(n)µν = −hµα x(n−1)αν ,

µν x(0)µν = g(B) .

(35.11.14)

Explicitly, we have x(1)µν = −hµν ;

x(2)µν = hµα hαν ;

βν x(3)µν = −hµα hα ··· βh

(35.11.15)

i.e. µν − hµν + hµα hνα − hµα hβα hνβ + · · · g µν = g(B)

(35.11.16)

(n−2)βν β (n−3)γν x(n)µν = −hµα x(n−1)αν = hµα hα = −hµα hα , βx β hγ x

(35.11.17)

If we note that we have µν β (0)γν + x(1)µν + x(2)µν − hµα hα + x(1)γν + · · · ) g µν = g(B) β hγ (x µν = g(B) − hµν + hµα hνα − hµα hβα hβγ g γν .

(35.11.18)

(B) (c) In local Lorentz frame of gαβ , at the origin we have

Γ(B)αβγ = 0,

Γαβγ = S αβγ ,

(B) = 0, gαβ,γ

gαβ,γ = hαβ,γ = hαβ|γ .

(35.11.19)

So we can get S µβγ = Γµβγ =

1 µα 1 g (gαβ,γ + gαγ,β − gβγ,α ) = g µα (hαβ|γ + hαγ|β − hβγ|α ), 2 2

(35.11.20a)

Rαβγδ − R(B)αβγδ = Γαβδ,γ − Γαβγ,δ + Γαµγ Γµβδ − Γαµδ Γµβγ + Γ(B)αβδ,γ − Γ(B)αβγ,δ = S αβδ,γ − S αβγ,δ + S αµγ S µβδ − S αµδ S µβγ = S αβδ|γ − S αβγ|δ + S αµγ S µβδ − S αµδ S µβγ , Rβδ − R(B)βδ = S αβδ|α − S αβα|δ + S αµα S µβδ − S αµδ S µβα .

(35.11.20b) (35.11.20c)

Because the leftmost and rightmost hand of equations 35.11.20 are tensors. So they will hold in any other frames.

352

(d) Using equations 35.11.20a and 35.11.16, we have S µβγ =

1 µα (g − hµα + · · · )(hαβ|γ + hαγ|β − hβγ|α ) 2 (B)

(35.11.21)

Thus, we can get ] 1 [ αρ (g(B) − hαρ + · · · )(hρβ|δ + hρδ|β − hβδ|ρ ) |α 2 1 = (hαβ|δα + hαδ|βα − hβδ|αα ) 2 1 αρ 1 − h |α (hρβ|δ + hρδ|β − hβδ|ρ ) − hαρ (hρβ|δα + hρδ|βα − hβδ|ρα ) + · · · 2 2 ] 1 [ αρ = (g(B) − hαρ + · · · )(hρβ|α + hρα|β − hβα|ρ ) |δ 2 ] 1 [ αρ = (g(B) − hαρ + · · · )hρα|β |δ 2 1 1 1 = h|βδ − hαρ|δ hρα|β − hαρ hαρ|βδ + · · · 2 2 2 1 α |µ µ µ = h α|µ (h β|δ + h δ|β − hβδ ) + · · · 4 1 |α |µ = (hαµ|δ + hαµ|δ − hµδ )(hµβ|α + hµα|β − hβα ) + · · · 4

S αβδ|α =

S αβα|δ

S αµα S µβδ S αµδ S µβα

(35.11.22a)

(35.11.22b) (35.11.22c) (35.11.22d)

It is easy to obtain that R

(1) βδ

=

1 α (h β|δα + hαδ|βα − hβδ|αα − h|βδ ) 2

(35.11.23)

and R

(2) βδ

1 1 1 1 = − hαρ|α (hρβ|δ + hρδ|β − hβδ|ρ ) − hαρ (hρβ|δα + hρδ|βα − hβδ|ρα ) + hαρ|δ hρα|β + hαρ hαρ|βδ 2 2 2 2 1 1 α |µ |α |µ µ µ µ µ α + h|µ (h β|δ + h δ|β − hβδ ) − (h µ|δ + h δ|µ − hµδ )(h β|α + h α|β − hβα ). (35.11.24) 4 4

If we note that |α

|µ

(hαµ|δ + hαδ|µ − hµδ )(hµβ|α + hµα|β − hβα ) = hαµ|β hαµ|δ + 2hβµ|α hδ

α|µ

− 2hβα|µ hδ

We can simplify equation 35.11.24 into [ 1 1 (2) hαρ|β hαρ|δ + hαρ (hβδ|αρ + hαρ|βδ − hρβ|δα − hρδ|βα ) R βδ = 2 2 ( ) ] 1 |ρ α|µ ρα +hδ (hβα|µ − hβµ|α ) − h |α − h (hρβ|δ + hρδ|β − hβδ|ρ ) 2

35.12

α|µ

. (35.11.25)

(35.11.26)

GAUGE TRANSFORMATIONS IN A CURVED BACKGROUND

(a) Show that the infinitesimal coordinate transformatIon (35.65a) induces the change (35.65b) in the functional form of the metric perturbation. (b) Discuss the relationship between this gauge transformation and the concept of a Killing vector.

Solution:

353

(a) The transformation law gives ( )( ) ∂ξ α ∂xβ α β g¯µν (¯ x) = δ µ − δν − gαβ (¯ x − ξ) ∂x ¯µ ∂x ¯ν ∂ξ α ∂ξ β g (¯ x ) − gµβ (¯ x) αν ∂xµ ∂xν α ∂ξ (B) ∂ξ α (B) (B) (B) ≈ gµν (¯ x) − gµν,ρ ξ ρ (¯ x) + hµν (¯ x) − g (¯ x ) − g (¯ x) ∂xµ αν ∂xν µα α ∂ξ (B) ∂ξ α (B) (B) (B) α(B) (B) α(B) g (¯ x ) − g (¯ x) = gµν (¯ x) − (gµα Γνρ + gαν Γµρ )ξ ρ (¯ x) + hµν (¯ x) − ∂xµ αν ∂xν µα (B) (35.12.1) = gµν (¯ x) + hµν (¯ x) − ξµ|ν (¯ x) − ξν|µ (¯ x). ≈ gµν (¯ x − ξ) −

So we have hnew µν = hµν − ξµ|ν − ξν|µ .

(35.12.2)

(b) If ξ µ is a Killing field, then the infinitesimal transformation xnewµ = xµ + ϵξ µ will leave hµν invariant.

35.13

TRANSVERSE-TRACELESS GAUGE FOR GRAVITATIONAL WAVES PROPAGATING IN A CURVED BACKGROUND

¯ α = 0 is preserved (a) Show that, in vacuum in a curved background spacetime, the gauge condition h µ |α by transformations whose generator satisfies the wave equation ξµ

|α α

= 0.

(b) Locally (over distances much smaller than R) linearized theory is applicable; so there exists such a transformation which makes ¯ = 0 + error, h ¯ µα uα = 0 + error. h (35.13.1) Here uα is a vector field that is as nearly covariantly constant as possible (uα|β = 0); i.e., it is a constant vector in the inertial coordinates of linearized theory; and the errors are small over distances much less ¯ = 0 can be imposed globally along with h ¯ α = 0; i.e., show that, if it is imposed on than R. Show that h µ |α an initial hypersurface, the propagation equation (35.68) preserves it. (c) Show that in general, the background curvature prevents any vector field from being covariantly con¯ µα uα = 0 cannot be imposed globally along with stant (uαˆ |βˆ ∼ uαˆ /R at best); and from this show that h ¯ α = 0. h µ |α

Solution: (a) After a gauge transformation, we have ¯ new |α = hold |α − ξ α − ξ α − 1 gµα (h ¯ old − 2ξ β )|α h µα µα µ|α α|µ |β 2 old |α α α α = h µα − ξµ|α − ξα|µ + ξ |αµ (B) ν = holdµα|α − ξµ|α α − Rµν ξ .

(35.13.2)

In vacuum, the only source of a nonvanishing Ricci tensor is the stress-energy carried by the gravitational (B) waves thenlselves; hence Rµν ∼ A2 /λ2 and can be neglected. So the gauge condition is preserved by transformations whose generator satisfies the wave equation ξµ

|α α

= 0.

¯ α = 0, we (b) Take the trace of the propagation equation (35.64) and impose the gauge condition h µ |α have ¯ α = 0. h (35.13.3) |α

354

¯ = 0 is t = t0 . Now, let us solve the wave equation Choose the coordinates in which the hypersurface with h (B) ν ξµ |νν + Rµν ξ =0

(35.13.4)

with initial conditions ξ µ|µ = 0,

ξ µ|µt = 0

(35.13.5)

on hypersurface t = t0 . Equation 35.13.5 can be rewritten as ξ t|t = −ξ i|i ,

ξ i|ti =

1¯ (B) α h,t + Rtα ξ − ξ t|tt . 2

(35.13.6)

Note that ξ t|tt will be fixed by ξ µ and ξ µ|t through 35.13.4. In principle, we can find appropriate initial value of ξ µ and ξ µ|t to make 35.13.5 holds. And then will can solve equation 35.13.4 throughout the ¯ = 0 on hypersurface whole spacetime. Using the gauge transformation generated by ξ µ , we can keep h α ¯ ¯ and hµ |α = 0 in whole spacetime, as well as make h,t = 0 on this surface. Now the propagation equation ¯ = 0 in whole spacetime. 35.13.3 will ensure that h (c) If there exist a vector field uα whose covariant derivatives vanish everywhere, then we have uα|βγ = uα|γβ , and so we can derive that Rα(B)ρβγ uρ = 0. (35.13.7) But this is impossible for a generic background curvature without special geometry.

35.14

BRILL-HARTLE AVERAGE

Isaacson (1968b) introduces the following averaging scheme, which he names “Brill-Hartle averaging”. (a) In the small region, of size several times it, where the averaging occurs, there will be a unique geodesic (B) connecting any two points P′ and P; so given a tensor E(P′ ) at P′ , one can parallel transport it of gµν along this geodesic to P′ , getting there a tensor E(P′ )→P . (b) Let f (P′ , P) be a weighting function that falls smoothly to zero when P′ and P are separated by many wavelengths, and such that ∫ √ f (P′ , P) −g (B) (P′ )d4 x′ = 1. (35.14.1) (c) Then the average of the tensor field E(P′ ) over several wavelengths about the point P is ∫ √ ⟨E⟩P ≡ E(P′ )→P f (P′ , P) −g (B) (P′ )d4 x′ .

(35.14.2)

′

(i) Show that there exists an entity gµ(B)α (P, P′ ), whose primed index transforms as a tensor at P and whose unprimed index transforms as a tensor at P′ , such that (for E second rank) ′

′

Eαβ (P′ )→P = gα(B)µ gβ(B)ν Eµ′ ν ′ (P′ ).

(35.14.3)

This entity is called the “bivector of geodesic parallel displacement”; see DeWitt and Brehme (1960) or Synge (1960a). (ii) Rewriting expression 35.14.2 in coordinate language as ∫ √ ′ ′ ⟨Eαβ (x)⟩ = gα(B)µ (x, x′ )gβ(B)ν (x, x′ )Eµ′ ν ′ (x′ )f (x, x′ ) −g (B) (x′ )d4 x′ , derive the three averaging rules cited at the beginning of section 35.15.

355

(35.14.4)

Solution: ′

(i) The matrix gµ(B)α (x, x′ ) is defined as the solution of the equation ′

σ ′ dx dgα(B)µ = Γρασ gρ(B)µ dτ dτ

(35.14.5)

′

with initial conditions that gµ(B)α (x, x′ ) is an identical matrix when x = x′ . Here xµ (τ ) is a geodesic from x′ . Now we have dEαβ d ( (B)µ′ (B)ν ′ ) = g gβ Eµ′ ν ′ dτ dτ α ( ) σ σ σ ′ ′ dx ′ dx ′ dx ρ (B)ν ρ (B)µ (B)ν (B)µ = Γ ασ gρ gβ + Γ βσ gρ gα Eµ′ ν ′ dτ dτ dτ ( ) dxσ = Γρασ Eρβ + Γρβσ Eαρ , (35.14.6) dτ i.e. E is parallel transport along the geodesic. Since Eµ′ ν ′ is a tensor at x′ while Eαβ is a tensor at x, it is ′ obvious that gµ(B)α (x, x′ )’s primed index transforms as a tensor at x and whose unprimed index transforms as a tensor at x′ . (ii) Firstly, we have hµν|αβ − hµν|βα = Rρµαβ hρν + Rρναβ hµρ

(35.14.7)

And the Riemann tensor is of the order ∂ 2 g (B) or (∂g (B) )2 . Suppose the background metric varies over the region with scale R, then the average value of Riemann tensor will be of the order (R)−2 while ∂ 2 h will be of the order hλ−2 , i.e. the fractional errors made by freely commuting are of the order (λ/R)2 . Secondly, for any tensor S, the covariant derivative ∇S is of the form (∂ + Γ). From equation 35.14.5, the ′ (B) matrix gµ(B)α (x, x′ ) is of the order 1 + λ/R and its derivative is of the order 1/R. The derivative of gαβ is also of the order 1/R. So we have ∫ √ ⟨∇S⟩ = g (B) (x, x′ )∇S(x′ )f (x′ ) −g (B) d4 x′ ( ) ∫ √ λ = ∂S(x′ )f (x′ ) −g (B) d4 x + O R ( ) ∫ ( √ ∫ ) √ λ 4 (B) (B) 4 = ∂ Sf −g d x − S∂f −g d x + O (35.14.8) R The first term may be converted to a surface integral taken in the region where f = 0. As for second term, since f (P′ , P) is a weighting function that falls smoothly to zero when P′ and P are separated by many wavelengths, S∂f can be neglected when compare with f ∂S, since ∂S varies a lot over λ, i.e. ( ) λ ⟨∇S⟩ ∼ O . (35.14.9) R

35.15

GEOMETRIC OPTICS

Develop geometric optics for gravitational waves of small amplitude propagating in a curved background. ¯ µν have an amPattern the analysis after geometric optics for electromagnetic waves. In particular, let h plitude that varies slowly (on a scale ℓ ≲ R) and a phase θ that varies rapidly (θαˆ ∼ 1/λ). Expand the amplitude in powers of λ/ℓ, so that ¯ µν = Re{Aµν + ϵBµν + ϵ2 Cµν + · · · }eiθ/ϵ . h

356

(35.15.1)

Here ϵ is a formal expansion parameter, actually equal to unity, which reminds one that the terms attached to ϵn are proportional to (λ/R)n . Define the following quantities (with A∗µν denoting the complex conjugate of Aµν ): “wave vector”: kα ≡ θ,α ( )1/2 1 ∗ µν “scalar amplitude”: A ≡ Aµν A 2 “polarization”: eµν ≡ Aµν /A.

(35.15.2a) (35.15.2b) (35.15.2c)

By inserting expression 35.15.1 into the gauge condition (35.66) and the propagation equation (35.68), derive the fundamental equations of geometrical optics as follows. (a) The rays (curves perpendicular to surfaces of constant phase) are null geodesics; i.e. kα k α = 0, β

kα|β k = 0.

(35.15.3a) (35.15.3b)

(b) The polarization is orthogonal to the rays and is parallel transported along them; eµα k α = 0,

(35.15.3c)

eµν|α k α = 0.

(35.15.3d)

(c) The scalar amplitude decreases as the rays diverge away from each other in accordance with 1 A,α k α = − k α|α A. 2

(35.15.3e)

(A2 k α )|α = 0(“conservation of gravitons”).

(35.15.3f)

i.e.

(d) The correction Bµν to the amplitude obeys Bµα k α = iAµα|α , 1 1 (B) Bµν|α k α = − k α|α Bµν + iAµν|αα + iRαµβν Aαβ . 2 2

(35.15.3g) (35.15.3h)

In accordance with exercise 35.13, specialize the gauge so that h = 0, i.e., eαα = 0.

(35.15.3i)

Then show that the stress-energy tensor (35.70’) for the waves is (GW) Tµν =

1 2 A kµ kν . 32π

(35.15.3j)

This has the same form as the stress-energy tensor for a beam of particles with zero rest mass. Show explicitly that T (GW)µν|ν = 0.

Solution: Firstly, we have i hµν|α = θ,α Aµν eiθ/ϵ + Aµν|α eiθ/ϵ , (35.15.4) ϵ where Aµν = Aµν + ϵBµν + ϵ2 Cµν + · · · . Using the definition 35.15.2a and contracting ν and α, the gauge condition becomes ( ) i µν µν µν h |ν = kν A + A |ν eiθ/ϵ = 0. (35.15.5) ϵ

357

Taking another derivative, we have ( ) i 1 i i hµν|αβ = θ|αβ Aµν − 2 θ,α θ,β Aµν + θ,α Aµν|β + θ,β Aµν|α + Aµν|αβ eiθ/ϵ . ϵ ϵ ϵ ϵ Using the definition 35.15.2a and contracting α and β, we can arrive at ( ) hµν|αα = iϵ−1 k α|α Aµν − ϵ−2 kα k α Aµν + 2iϵ−1 k α Aµν|α + Aµν|α α eiθ/ϵ .

(35.15.6)

(35.15.7)

The wave equation (35.68) now becomes (B) A αβ = 0. iϵ−1 k α|α Aµν − ϵ−2 kα k α Aµν + 2iϵ−1 k α Aµν|α + Aµν|α α + 2Rαµβν

(35.15.8)

(a) Take the term of the order ϵ−2 from 35.15.8, and we can get kα k α Aµν = 0,

(35.15.9)

i.e. kα k α = 0. As a result, we also have kα|β k β = θ|αβ k β = θ|βα k β = kβ|α k β =

) 1( kβ k β |α = 0. 2

(35.15.10)

(b) & (c) Take the term of the order ϵ−1 from 35.15.5, and we can get kν Aµν = 0,

(35.15.11)

i.e. eµα k α = 0. Take the term of the order ϵ−1 from 35.15.8, and we can get k α|α Aµν + 2k α Aµν|α = 0.

(35.15.12)

Substitute Aµν = Aeµν into the equation above, we have ( ) k α|α A + 2k α A,α eµν + 2k α Aeµν|α = 0.

(35.15.13)

Multiplying both sides of the equation above with e∗µν , we have ( ) ( ) 0 = k α|α A + 2k α A,α eµν e∗µν + k α A (e∗µν eµν )|α = 2 k α|α A + 2k α A,α .

(35.15.14)

i.e.

1 A,α k α = − k α|α A. 2

(35.15.15)

2k α Aeµν|α = 0,

(35.15.16)

Then we can get i.e. eµν|α = 0. (d) Take the term of the order O(1) from 35.15.5, and we can get

i.e.

ikν B µν + Aµν |ν = 0,

(35.15.17)

Bµα k α = iAµα|α .

(35.15.18)

Take the term of the order O(1) from 35.15.8, and we can get (B) Aαβ = 0, ik α|α Bµν + 2ik α Bµν|α + Aµν|αα + 2Rαµβν

358

(35.15.19)

i.e.

1 1 (B) Bµν|α k α = − k α|α Bµν + iAµν|αα + iRαµβν Aαβ . 2 2 Equation (35.70’) states that (GW) Tµν =

1 ¯ ¯ αβ ⟩ if h ¯ α =h ¯ = 0. ⟨hαβ|µ h µ |α |ν 32π

(35.15.20)

(35.15.21)

To the order of ϵ−2 , we have ¯ αβ|µ h ¯ ∗αβ = kµ kν Aαβ A∗αβ = 2A2 kµ kν h |ν Note that

(35.15.22)

1 XY ∗ (35.15.23) 2 and θ varies quickly when compared with the scale of the domain of averaging. So we have ⟨Re X Re Y ⟩ =

if X, Y ∝ eiθ

(GW) Tµν =

1 2 A kµ kν . 32π

(35.15.24)

Using equation 35.15.3b and 35.15.3e, we have T (GW)µν|ν =

35.16

1 A(2A,ν k ν k µ + Ak µ|ν k ν + Ak µ k ν|ν ) = 0. 32π

(35.15.25)

GRAVITONS

Show that geometric optics, as developed in the preceding exercise, is equivalent to the following: “A graviton is postulated to be a particle of zero rest mass and 4-momentum p, which moves along a null geodesic. It parallel transports with itself a transverse traceless polarization tensor e. Geometric optics is the theory of a stream of such gravitons moving through spacetime.” Exhibit the relationship between the quantities in this version of geometric optics and the quantities in the preceding version.

Solution: A graviton is postulated to be a particle of zero rest mass and 4-momentum p, which moves along a null geodesic. This is equivalent to p · p = 0, ∇p p = 0. (35.16.1) It parallel transports with itself a transverse traceless polarization tensor e. This is equivalent to e · p = 0,

eαα = 0,

∇p e = 0.

(35.16.2)

If we assume p is proportional to k, geometric optics developed in the preceding exercise is equivalent to the statements in this exercise.

35.17

GRAVITATIONAL DEFLECTION OF GRAVITATIONAL WAVES

Show that gravitational waves of short wavelength passing through the solar system experience the same redshift and gravitational deflection as does light.

Solution: Gravitons (in short wavelength approximation) and photons both follow null geodesic in spacetime. So gravitational waves of short wavelength passing through the solar system must experience the same redshift and gravitational deflection as does light.

359

35.18

(GW)

GAUGE INVARIANCE OF Tµν (GW)

Show that the stress-energy tensor Tµν the form (35.65).

of equation (35.70) is invariant under gauge transformations of

Solution: Under gauge transformation, we have ¯ new = h ¯ old − ξµ|ν − ξν|µ + g (B) ξ α , h µν µν µν |α

¯ new = h ¯ old + 2ξ α . h |α

The stress-energy tensor of gravitational waves is ⟨ ⟩ 1 (GW) ¯ αβ|µ h ¯ αβ − 1 h ¯ |µ h ¯ |ν − 2h ¯ αβ h ¯ Tµν = h |ν |β α(µ|ν) . 32π 2

(35.18.1)

(35.18.2)

Under gauge transformation, we have ¯ αβ|µ h ¯ αβ → h ¯ αβ|µ h ¯ αβ h |ν |ν (B) ρ ¯ αβ − (ξα|βµ + ξβ|αµ − gαβ ξ |ρµ )h |ν

αβ σ ¯ αβ|µ (ξ α|β + ξ β|α − g(B) ξ |σν ) −h ν ν αβ σ (B) ρ + (ξα|βµ + ξβ|αµ − gαβ ξ |ρµ )(ξ α|βν + ξ β|αν − g(B) ξ |σν )

1¯ ¯ 1¯ ¯ ρ ρ σ ¯ ¯ σ h|µ h|ν → h |µ h|ν + ξ |ρµ h|ν + h|µ ξ |σν + 2ξ |ρµ ξ |σν 2 2 ¯ αβ h ¯ ¯ αβ ¯ h |β αµ|ν → h |β hαµ|ν − (ξ

α|β β

+ξ

β|α β

(35.18.3) (35.18.4)

αβ ρ ¯ ξ |ρβ )h − g(B) αµ|ν

¯ αβ (ξα|µν + ξµ|αν − g (B) ξ σ ) −h αµ |σν |β + (ξ

α|β β

+ξ

β|α β

αβ ρ (B) σ ξ |ρβ )(ξα|µν + ξµ|αν − gαµ ξ |σν ) − g(B)

(35.18.5)

¯ αβ h ¯ ¯ αβ ¯ h |β αν|µ → h |β hαν|µ − (ξ

α|β β

+ξ

β|α β

αβ ρ ¯ ξ |ρβ )h − g(B) αν|µ

¯ αβ (ξα|νµ + ξν|αµ − g (B) ξ σ ) −h αν |σµ |β + (ξ

α|β β

+ξ

β|α β

αβ ρ (B) σ − g(B) ξ |ρβ )(ξα|νµ + ξν|αµ − gαν ξ |σµ )

(35.18.6)

So we have ¯ |ν − 2h ¯ αβ h ¯ αβ|µ h ¯ αβ − 1 h ¯ |µ h ¯ h |β α(µ|ν) |ν 2 ¯ αβ|µ h ¯ αβ − 1 h ¯ ¯ |µ h ¯ |ν − 2h ¯ αβ h →h |ν |β α(µ|ν) 2 − 2ξα|βµ hαβ |ν − 2ξα|βν hαβ |µ + 2ξα|βµ ξ α|βν + 2ξα|βµ ξ β|αν − 4ξ ρ|ρµ ξ σ|σν +ξ

α|β ¯ β hαµ|ν

ρ ρ ¯β ¯ αβ ¯ αβ ¯β ¯ + ξ α|βα h βµ|ν − ξ |ρβ h µ|ν + h |β ξα|µν + h |β ξµ|αν − ξ |ρν h µ|β

−ξ

α|β β ξα|µν

−ξ

+ξ

α|β ¯ β hαν|µ

ρ ρ ¯ ¯β ¯ αβ ¯ αβ ¯β + ξ α|βα h βν|µ − ξ |ρβ h ν|µ + h |β ξα|νµ + h |β ξν|αµ − ξ |ρµ h ν|β

−ξ

α|β β ξα|νµ

−ξ

α|β β ξµ|αν

α|β β ξν|αµ

− ξ α|βα ξβ|µν − ξ α|βα ξµ|βν + ξ ρ|ρβ ξ β|µν + ξ ρ|ρβ ξµ |βν + ξ ρ|ρν ξ β|µβ + ξ ρ|ρν ξµ

|β β

|β

− ξ α|βα ξβ|νµ − ξ α|βα ξν|βµ + ξ ρ|ρβ ξ β|νµ + ξ ρ|ρβ ξν |βµ + ξ ρ|ρµ ξ β|νβ + ξ ρ|ρµ ξν β (35.18.7)

As stated in the exercise 35.14, there are some useful rules for manipulating quantities inside the averaging brackets:

360

• Covariant derivatives commute; • Gradients average out to zero. ¯ αβ ξµ|αν because there is no other terms ever containing ξµ . Some I think it is impossible to eliminate the h |β other difficulties also exist. Any help is appreciated. I have posted the question on the Stackexchange. Please check it.

35.19

(GW)

Tµν EXPRESSED AS THE AVERAGE OF A STRESS-ENERGY PSEUDOTENSOR

Calculate the average over several wavelengths of the Landau-Lifshitz stress-energy pseudotensor [equa(GW) tion (20.22)] for gravitational waves with λ/R ≪ 1. The result should be equal to Tµν .

Solution: The Landau-Lifshitz pseudotensor is { 1 1 (−g)tαβ = gαβ ,λ gλµ,µ − gαλ,λ gβµ,µ + g αβ gλµ gλν ,ρ gρµ,ν L−L 16π 2 − (g αλ gµν gβν ,ρ gµρ,λ + g βλ gµν gαν ,ρ gµρ,λ ) + g λµ gνρ gαλ,ν gβµ,ρ } 1 αλ βµ αβ λµ ντ ρσ + (2g g − g g )(2gνρ gστ − gρσ gντ )g ,λ g ,µ , 8

(35.19.1)

where gαβ = (−g)1/2 g αβ . For simplicity, we will perform the calculation in flat background and TT gauge, i.e. gµν = ηµν + hµν , hµν ,ν = 0, hµµ = 0. (35.19.2) As a result, we have − g = 1 + O(h2 ),

gµν ,ρ = −hµν ,ρ + O(h2 ).

(35.19.3)

Up to the second order of h, we have { 1 1 αβ tL−L = + η αβ ηλµ hλν ,ρ hρµ,ν − (η αλ ηµν hβν ,ρ hµρ,λ + η βλ ηµν hαν ,ρ hµρ,λ ) + η λµ ηνρ hαλ,ν hβµ,ρ 16π 2 } 1 αλ βµ αβ λµ ντ ρσ + (2η η − η η )ηνρ ηστ h ,λ h ,µ + O(h3 ). (35.19.4) 4 When taking the average over the region of which the scale is several times that of the typical wavelength of the gravitational wave, we have ⟨a,m b⟩ = −⟨ab,m ⟩. (35.19.5) So we have ⟨hλν ,ρ hρµ,ν ⟩ = −⟨hλν hρµ,νρ ⟩ = 0,

(35.19.6)

⟨ηνρ hαλ,ν hβµ,ρ ⟩ = −⟨ηνρ hαλ,νρ hβµ ⟩ = 0

(35.19.7)

using Lorentz gauge condition

using field equation. Finally, we can get ⟨tαβ L−L ⟩ =

1 1 αλ βµ . η η ηνρ ηστ ⟨hντ ,λ hρσ,µ ⟩ = ⟨h hρσ,β ⟩ = T(αβ GW) 32π 32π ρσ,α

361

(35.19.8)

35.20

SHORTWAVE APPROXIMATION FROM A VARIATIONAL VIEWPOINT

Readers who have studied the variational approach to gravitation theory in Chapter 21 may find attractive the following derivation of the basic equations of the shortwave approximation. It was devised, independently, by Sandor Kovacs and Bernard Schutz, and by Bryce DeWitt (unpublished, 1971). MacCallum and Taub (1973) give a “non-Palatini” version. (a) Define ¯ µν ≡ hµν − 1 g (B) h h 2 µν ) 1 µα ( ≡ g(B) hαβ|γ + hαγ|β − hβγ|α . 2

(B) gµν ≡ gµν + hµν ,

(35.20.1a)

W µβγ

(35.20.1b)

Raise and lower indices on hµν and W µβγ with the background metric. Using the results of exercise 35.11, derive the following expression for the Lagrangian of the gravitational field: ( ) ( ) 1 ∂L A3 (B) L≡ (−g)1/2 R = L′ + perfect divergence of form , R A, and smaller , + corrections of order µν 16π ∂xα λ2 (35.20.1c) where [ ( ) )] 1 1/2 µν ( α α β α β ¯ µν W α L′ ≡ (−g (B) ) R(B) − h − W W W − W + g W . (35.20.1d) (B) βν µα βα µν µν|α µα|ν 16π Drop the corrections of order A3 /λ2 from L; and, knowing in advance that the field equations will demand (B) (B) ∼ A2 /λ2 , drop also the corrections of order Rµν A. Knowing that a perfect divergence contributes Rµν nothing in an extremization calculation, drop the divergence term from L. Then L′ is the only remaining part of L. ∫ (b) Extremize I ≡ L′ d4 x by the Palatini method (section 21.2); i.e., abandon (temporarily) definition µ ¯ µν = h ¯ νµ 35.20.1b of W βγ , and extremize I with respect to independent variations of W µβγ = W µγβ , h µ νµ µν and g(B) = g(B) . Show that extremization with respect to W βγ leads back to equation 35.20.1b for W µβγ ¯ µν , when combined with equations (35.20.1a, in terms of hµν . Show that extremization with respect to h 35.20.1b), leads to the propagation equation for gravitational waves (35.64). Show that extremization with respect to g (B)µν , when combined with equations (35.20.1a, 35.20.1b) and with the propagation equation (35.64), and when averaged over several wavelengths, leads to (GW) G(B) , µν = 8πTµν (GW)

where Tµν

(35.20.1e)

is given by equation (35.70).

Solution: (a) From equation 35.11.20, we have S µβγ =

1 µα g (hαβ|γ + hαγ|β − hβγ|α ), 2

α β α β α µν α R = g µν R(B) µν + g (S µν|α − S µα|ν + S βα S µν − S βν S µα ).

(35.20.2a) (35.20.2b)

(B) From gµν = gµν + hµν , we can get

µν g µν = g(B) − hµν + O(h2 ), g = g (B) (1 + h) + O(h2 ), 1 S µβγ = W µβγ − hµα (hαβ|γ + hαγ|β − hβγ|α ) + O(h3 ) 2

362

(35.20.3)

So

( ) h (−g)1/2 R = (−g (B) )1/2 1 + R + O(h3 , R(B) h2 ) 2 ] [ µν (B) µν = (−g (B) )1/2 g(B) R µν + (g(B) − hµν )(S αµν|α − S αµα|ν + W αβα W βµν − W αβν W βµα ) 1 µν ( α h W µν|α − W αµα|ν + O(h3 , R(B) h) + (−g (B) )1/2 g(B) 2[ ] µν (B) 1/2 = (−g ) R(B) − hµν (W αµν|α − W αµα|ν ) + g(B) (W αβα W βµν − W αβν W βµα ) + (−g (B) )1/2 (S αµµ|α − S αµα|µ ) ) 1 µν ( α + (−g (B) )1/2 g(B) h W µν|α − W αµα|ν + O(h3 , R(B) h) 2[ ] µν (B) 1/2 α α β α β ¯ µν (W α = (−g ) R(B) − h − W ) + g (W W − W W ) (B) βα µν βν µα µν|α µα|ν ] ∂ [ + (−g (B) )1/2 (S αµµ − S µαµ ) + O(h3 , R(B) h). (35.20.4) ∂xα

Now we can get the equation 35.20.1c and 35.20.1d. (b) The extremization with respect to W µβγ gives α (B) 1/2 µν ¯ µν (δW α 16πδL′ = −(−g (B) )1/2 h ) g(B) δ(W αβα W βµν − W αβν W βµα ) µν|α − δW µα|ν ) + (−g

¯ µν δW α − (−g (B) )1/2 h ¯ µν δW α = (−g (B) )1/2 h µν µα |α |ν µν + (−g (B) )1/2 g(B) (W αβα δW βµν + W βµν δW αβα − W αβν δW βµα − W βµα δW αβν ) [ ] µν ρσ µσ ρν ¯ µν − h ¯ µβ δ ν + g(B) = (−g (B) )1/2 h W βαβ + g(B) W µρσ δαν − g(B) W νασ − g(B) W µρα δW αµν |α |β α

(35.20.5) So we have the equation of motion ρν µσ ρσ µν ¯ µν − h ¯ µβ δ ν + g(B) W µρα = 0. W νασ − g(B) W µρσ δαν − g(B) W βαβ + g(B) h |α |β α

(35.20.6)

By contracting α and ν, we can get ρσ ¯ µν = g(B) W µρσ . h |ν

(35.20.7)

µν µσ ρν ¯ µν + g(B) h W βαβ − g(B) W νασ − g(B) W µρα = 0. |α

(35.20.8)

And so we have By contracting µ and ν, we can get ¯ + 2W β = 0. h |α αβ

(35.20.9)

And so we have µσ ρν W νασ − g(B) hµν |α − g(B) W µρα = 0,

(35.20.10)

hµν|α = Wναµ + Wµνα .

(35.20.11)

i.e. Now it is easily to derive that W αµν =

1 αβ + hβν|µ − hµν|β ). g (h 2 (B) βµ|ν

(35.20.12)

¯ µν gives The extremization with respect to h ¯ µν . 16πδL′ = −(−g (B) )1/2 (W αµν|α − W αµα|ν )δ h

(35.20.13)

So we have the equation of motion W αµν|α − W αµα|ν = 0.

363

(35.20.14)

Using 35.20.1b, we can get [ ] αβ g(B) hβµ|να + hβν|µα − hµν|βα − (hβµ|αν + hβα|µν − hµα|βν ) = 0,

(35.20.15)

i.e. hαµ|ν α + hαν|µα − hµν|αα − h|µν = 0.

(35.20.16)

µν The extremization with respect to g(B) gives [ ] [ ] µν α (B) 1/2 ¯ µν (W α 16πδL′ = δ (−g (B) )1/2 R(B) + (W αβα W βµν − W αβν W βµα )δ (−g (B) )1/2 g(B) −h ) µν|α − W µα|ν )δ(−g } { ] 1[ α ρσ α β α β β α β ρσ α α (B) ¯ (W W − W W )g − h (W − W ) g = G(B) + (W W − W W ) − (B) βα ρσ βσ ρα µν µν βα µν βν µα ρσ|α ρα|σ 2 µν × (−g (B) )1/2 δg(B) .

(35.20.17)

So we have the equation of motion ] 1[ α ρσ α (B) ¯ ρσ (W α (W βα W βρσ − W αβσ W βρα )g(B) −h − W ) gµν = 0. ρσ|α ρα|σ 2 (35.20.18) = 0 from 35.20.14. So we can get

Gµν +(W αβα W βµν −W αβν W βµα )− Note that W αρσ|α − W αρα|σ

(B) Rµν + (W αβα W βµν − W αβν W βµα ) = 0.

(35.20.19)

Using 35.20.1b, we have ] 1 |β 1 [ αβ h (hβµ|ν + hβν|µ − hµν|β ) − h |µ hαβ|ν + 2hµβ|α hν α|β − 2hµα|β hν α|β . 4 4 (35.20.20) If we use the TT gauge and take the average, it is easy to derive that W αβα W βµν − W αβν W βµα =

1 ⟨W αβα W βµν − W αβν W βµα ⟩ = − ⟨hαβ |µ hαβ|ν ⟩. 4

(35.20.21)

We can also find that ⟨hαβ|µ hαβ|µ = 0. So we have G(B) µν =

1 ⟨hαβ |µ hαβ|ν ⟩. 32π

364

(35.20.22)

Chapter 36

GENERATION OF GRAVITATIONAL WAVES 36.1

GRAVITATIONAL WAVES FROM ROTATING BEAM

A long steel beam of length l and mass M rotates end over end with angular velocity ω. Show that the power it radiates as gravitational waves is LGW =

2 24 6 M l ω . 45

(36.1.1)

Use this formula to verify that the rod described in the text radiates 2.2 × 10−22 ergs/sec.

Solution: Set the mass center of the beam as the origin and the line perpendicular to the rotation plain as z axis. The second moment of the mass distribution of the beam is ∫

l/2

M ′ 1 l cos θl′ cos θ dl′ = M l2 cos2 θ, l 12 −l/2 1 1 2 = M l sin θ cos θ, Iyy = M l2 sin2 θ, Ixz = Iyz = 0. 12 12 Ixx ≡

Ixy = Iyx

(36.1.2)

Izz is a constant can be neglected when calculating power of GW wave. So its trace free part is  2 cos θ − 13 1 2 = Ml sin θ cos θ 12 0

sin θ cos θ sin2 θ − 13 0

 sin 2θ ... 1 I jk = M l2 ω 3 cos 2θ 3 0

− cos 2θ − sin 2θ 0

I jk

Thus

 0 0  − 13  0 0 . 0

(36.1.3)

(36.1.4)

We then have LGW =

1 ⟨... ... ⟩ 2 24 6 2G 2 4 6 M l ω ≈ 2.26 × 10−22 erg s−1 . I jk I jk = M l ω = 5 45 45c5

365

(36.1.5)

36.2

GRAVITATIONAL WAVES FROM MATTER FALLING INTO A BLACK HOLE

A lump of matter with mass m falls into a black hole of mass M . Show that a burst of gravitational waves is emitted with duration M and power LGW ∼ (m/M )2 Lo , so that the total energy radiated is given in crude order of magnitude by equation (36.13).

Solution: The non spherical part of kinetic energy is Eint ∼ M m/R. The period of the system is T ∼ (M/R3 )−1/2 . So we have ( )2 Eint M 3 m2 LGM ∼ Lo ∼ Lo . (36.2.1) T R5 The infalling matter will radiate only weakly when it is far from the gravitational radius; but as it falls through the gravitational radius, i.e. R ∼ M ,it should emit a strong burst. And so we have LGM ∼

m2 Lo . M2

(36.2.2)

The total energy emitted is Eradiated ∼ m2 /M.

(36.2.3)

So the burst of gravitational waves is emitted with duration τ ∼ Eradiated /LGM ∼ M .

36.3

GRAVITATIONAL WAVES FROM A VIBRATING NEUTRON STAR

Idealize a neutron star as a sphere of incompressible fluid of mass M and radius R, with structure governed by Newton’s laws of gravity. Let the star pulsate in its fundamental quadrupole mode. Using Newtonian ⟨... ⟩ theory, calculate: the angular frequency of pulsation, ω; the energy of pulsation Epuls ; the quantity 2 I /5, which, according to equation (36.1), is the power radiated in gravitational waves, LGW ; and the e-folding time, τ = Epuls /LGW for radiating away the energy of the pulsations. Compare the answers with equations (36.15) - which are based on a much cruder approximation-and with the results in Box 36.1, which are based on much better approximations.

Solution: The fundamental equations in Newtonian fluid mechanics are dρ = −ρ∇ · v, dt

ρ

dv = −ρ∇ϕ − ∇p, dt

∇2 ϕ = 4πρ,

(36.3.1)

where d/dt is the time derivative comoving with the matter d ∂ = + v · ∇. dt ∂t

(36.3.2)

dρ = 0. dt

(36.3.3)

For incompressible fluid, we have

The spherically symmetric and static solution for an incompressible fluid is { { { 2π 2π 2 2 2 ρs r2 − 2πρs R2 if r < R ρs if r < R ρ (R − r ) if r < R ρ0 (r) = p0 (r) = 3 s ϕ0 (r) = 3 4πρs R3 0 if r > R 0 if r > R if r > R − 3r (36.3.4)

366

In the limit of small oscillations, we will apply first order perturbation theory and neglect all higher terms. Suppose that the perturbation of velocity, density, pressure and gravitational potential are δv, δρ, δp and δϕ respectively. The first order perturbations of the equation 36.3.1 and 36.3.3 are ∂δρ + δv · ∇ρ0 = −ρ0 ∇ · δv = 0, ∂t ∂δv ρ0 = −ρ0 ∇δϕ − δρ∇ϕ0 − ∇δp, ∂t 2 ∇ δϕ = 4πδρ.

(36.3.5) (36.3.6) (36.3.7)

Suppose the material particles are moved from their equilibrium position r0 by a displacement s(r0 , t), and ∥s∥ ≪ 1. To the first order, we have ∂s δv = . (36.3.8) ∂t For a point inside the star (r < R), the perturbation of density vanish because the fluid is imcompressible. For a point at the surface of the star (r = R), the mass of the bump at a surface element ∆S is ∆m = ρs ∆S · s = ρs sr ∆S, causing a surface density perturbation δσ = ρs sr . We can conclude that δρ = ρs sr δ(r − R).

(36.3.9)

It is compatible with ∂δρ + δv · ∇ρ0 = 0, ∂t noting that ∇ρ0 = ρs δ(r − R)ˆ r . As a result, we have

(36.3.10)

∇2 δϕ = 4πρs sr δ(r − R).

(36.3.11)

Taking the divergence of equation 36.3, we have ∇2 δp = 0

for

(36.3.12)

r < R.

Because the pressure vanish at the surface of the star, we have δp +

∂p0 sr = 0 ∂r

at r = R.

(36.3.13)

For r < R, equation can be written as ρs

∂2s = −ρs ∇δϕ − ∇δp. ∂t2

(36.3.14)

So we also have ∇ × s = 0 when r < R. s can be expanded using vector spherical harmonics, ∑ s= (Ulm (r)Ylm + Vlm (r)Ψlm + Wlm (r)Φlm ) e−iωt ,

(36.3.15)

l,m

where Ylm = Ylm rˆ,

Ψlm = r∇Ylm ,

Φlm = r × ∇Ylm .

(36.3.16)

Using ∇ · s = 0 and ∇ × s = 0 (r < R), we have dUlm 2 l(l + 1) + Ulm − Vlm = 0, dr r r

Wlm = 0

(r < R).

δp and δρ can be expanded using spherical harmonics, ∑ ∑ δp = Plm (r)Ylm e−iωt , δϕ = Φlm (r)Ylm e−iωt . lm

lm

367

(36.3.17)

(36.3.18)

Substituting 36.3.15 and 36.3.18 into 36.3.14, we can obtain ρs ω 2 Ulm = ρs

dΦlm dPlm + , dr dr

ρs ω 2 Vlm = ρs

Φlm Plm + r r

(r < R)

Because ∇2 δp = 0 and ∇2 δϕ = 0 when r ̸= R, we have { { Alm rl if r < R Clm rl if r < R Φlm (r) = P (r) = lm Blm r−l−1 if r > R 0 if r > R Note that

R + ∂δϕ = 4πρs sr , ∂r R−

R+ δϕ = 0. −

(36.3.19)

(36.3.20)

(36.3.21)

R

We have − (l + 1)Blm R−l−2 − lAlm Rl−1 = 4πρs Ulm (R),

Alm Rl = Blm R−l−1 .

(36.3.22)

The solution is Alm = −

4πρs Ulm (R) , (2l + 1)Rl−1

Blm = −

4πρs Ulm (R) l+2 R . (2l + 1)

(36.3.23)

Using equation 36.3.13, we have Clm Rl − The solution is Clm =

4π 2 ρ RUlm (R). 3 s

(36.3.24)

4πρ2s Ulm (R) . 3Rl−1

(36.3.25)

ρs ω 2 Vlm (r) = (ρs Alm + Clm )rl−1 .

(36.3.26)

Note that quation 36.3.19 and 36.3.20 gives ρs ω 2 Ulm (r) = (ρs Alm + Clm )lrl−1 , We can get ω2 =

8l(l − 1)πρs , 3(2l + 1)

Ulm (r) = Ulm (R)

( r )l−1 R

,

Vlm (r) =

( r )l−1 1 Ulm (R) . l R

It is compatible with 36.3.17. Now we have )( ) ∑( 1 r l−1 s= Ylm + Ψlm Ulm (R)e−iωt , l R

(36.3.27)

(36.3.28)

l,m

Define ξ as the root mean squared fractional departure of the surface point from the original sphere, average over the sphere. We have 2 ξpeak =

1 4π

∫ dΩ

s2r,peak (R) U 2 (R) = lm 2 2 R 4πR

(36.3.29)

The pulsing energy of the star 1 2 ω 2

∫ 2

( ) ∫ ( r )2l−2 ∫ 2 1 ρs Ulm (R)ω 2 R ∗ dΩ Ylm · Ylm dr r2 + 2 Ψlm · Ψ∗lm 2 R l 0 2 2 = πρs ω 2 R5 ξpeak l 16(l − 1) 2 2 5 2 = π ρs R ξpeak 3(2l + 1)

ρs |speak | dV =

368

(36.3.30) (36.3.31) (36.3.32)

The reduced second quadrupole moment of the star is ( ) ( ) ∫ ∫ 1 1 I = dV δρ r ⊗ r − r2 I = ρs R4 Ulm (R)e−iωt dΩ Ylm rˆ ⊗ rˆ − I ≡ ρs R4 Ulm (R)e−iωt X 3 3 (36.3.33) The radiation power of the gravitational wave is LGW =

1 ... ... 2π 2 6 10 2 ⟨ I ij I ij ⟩ = ρ ω R ξpeak Tr XX † . 5 5 s

We can show that Tr XX † =

(36.3.34)

8π δl2 15

(36.3.35)

So, for quadrupole oscillations, l = 2, we have ω2 =

16 πρs , 15

Epuls =

16 2 2 5 2 π ρs R ξpeak , 15

LGW =

They can also be written in terms of M and R, √ ( )2 4M 1 3 M 2 ω= , Epuls = Rξpeak , 5R R 5 R

65536 5 5 10 2 π ρs R ξpeak , 253125

LGW

192 = 3125

(

M R

)5 2 ξpeak ,

τ=

16875 −3 −3 −5 π ρs R . 4096 (36.3.36)

625 τ= 64

(

M R

)−3 R. (36.3.37)

Suppose M = 1.1 × 1030 g, R = 1.414 × 106 cm, ξpeak = 0.1. We have ω = 4.557 × 103 sec−1 ,

36.4

Epuls = 3.426 × 1050 erg,

LGW = 1.433 × 1050 erg s−1 ,

τ = 2.391s. (36.3.38)

PULSAR SLOWDOWN

The pulsar NP 0532 in the Crab Nebula has a period of 0.033 seconds and is slowing down at the rate dP /dt = 1.35 × 10−5 sec/yr. Assuming the pulsar is a typical neutron star, calculate the rate at which it is losing rotational energy. If this energy loss is due primarily to gravitational radiation reaction, what is the magnitude of the star’s nonaxial deformation?

Solution: For an ellipsoid with principal axes a, b and c, the reduced second quadrupole moment is  2  2a − b2 − c2 M  2b2 − c2 − a2 I ij = 15 2 2 2 2c − a − b

(36.4.1)

Applying the rotation operator R(z) around the z-axis and differentiating three times with respect to time, we obtain   sin 2ωt − cos 2ωt ... 4M 3 2  ω (a − b2 )− sin 2ωt − sin 2ωt (36.4.2) I ij = 5 0 The power of gravitational wave radiation will be LGW =

32 2 6 2 M ω (a − b2 )2 . 125

(36.4.3)

If we assume a ≈ b ≈ r and (a − b)/r = ϵ ≪ 1, then we have LGW =

32 2 2 6 128 2 6 4 2 M ω R ϵ = ϵ I ω , 125 5

369

(36.4.4)

where I = 2M R2 /5 is the moment of inertia of the ellipsoid around the z-axis. The energy conservation equation gives I dω 2 32 = − ϵ2 I 2 ω 6 (36.4.5) 2 dt 5 Putting back c and G, we have dω 32G = − 5 ϵ2 Iω 5 . dt 5c

(36.4.6)

Since ω=

dω 2π dP =− 2 = −2.47 × 10−9 sec−2 , dt P dt

2π = 190sec−1 , P

(36.4.7)

if we assume M = 1.1 × 1030 g, R = 1.414 × 106 cm, we can get ϵ = 0.025.

36.5

ENERGY AND ANGULAR MOMENTUM LOSSES DUE TO RADIATION REACTION

Derive equations (36.31) for the rate at which gravitational radiation damping saps energy and angular momentum from a slow-motion source. Base the derivation on equations (36.26b) and (36.30).

Solution: Firstly, we have 1 d5 I jk xj xk 5 dt5

Φreact = and dE =− dt

∫

dJj =− dt

ρΦreact vj d3 x , ,j

(36.5.1)

∫ ϵjkl xk ρΦreact d3 x . ,l

(36.5.2)

We can derive that Φreact = ,k

2 d5 I jk xk . 5 dt5

(36.5.3)

Thus, dE 2 d5 I jk =− dt 5 dt5

∫ ρxj vk d3 x = −

1 d5 I jk d 5 dt5 dt

(∫

) ρxj xk d3 x

=−

1 d5 I jk dI jk t. 5 dt5

(36.5.4)

After taking average, we can integrate by parts, and so we can get 1 dE =− dt 5

⟨

d3 I jk d3 I jk dt3 dt3

⟩ (36.5.5)

.

As for angular momentum, we have 2 d5 I li dJj =− dt 5 dt5

∫

2 d5 I li ϵjkl xk ρxi d3 x = − ϵjkl 5 I ki . 5 dt

(36.5.6)

After taking average, we can integrate by parts, and so we can get dJj 2 = − ϵjkl dt 5

⟨

d3 I li d2 I ki dt3 dt2

370

⟩ .

(36.5.7)

36.6

GRAVITATIONAL WAVES FROM BINARY STAR SYSTEMS

Apply the full formalism of section 36.7 and 36.8 to a binary star system with circular orbits. Calculate the angular distribution of the gravitational waves; the total power radiated; the total angular momentum radiated; the radiation-reaction forces; and the loss of energy and angular momentum due to radiation reaction. Compare the answers with the results quoted in section 36.6.

Solution: We choose the center of mass as the origin of the system. If the distance of between two stars is a, we have ) ( ) ( m m a cos ωt δ y − a sin ωt δ(z) ρ = Mδ x − M +m M +m ( ) ( ) M M + mδ x − a cos ωt δ y − a sin ωt δ(z). (36.6.1) M +m M +m The second momentum of the system is ∫ Iij = As a result, we have  cos 2ωt I¨ik = −2µω 2 a2  sin 2ωt 0

 cos 2ωt + 1 1 ρxi xj d3 x = µa2  sin 2ωt 2 0

sin 2ωt − cos 2ωt 0

 0 0 , 0

sin 2ωt 1 − cos 2ωt 0

 sin 2ωt ... I ik = 4µω 3 a2 − cos 2ωt 0

 0 0 . 0

− cos 2ωt − sin 2ωt 0

(36.6.2)

 0 0 . 0

(36.6.3)

The gravitational wave is given by hTT ij =

2 ¨TT I (t − r), r ik

(36.6.4)

where

1 xa xb TT TT TT I¨ik = Pja I¨ab Pbk − Pjk Pab I¨ab , Pab = δab − 2 . 2 r The total power radiated is 1 ⟨... ... ⟩ 32 2 6 4 LGW = I ik I ik = µ ω a . 5 5

(36.6.5)

(36.6.6)

Using ω 2 a3 = m1 + m2 ≡ M , we have − Note that

dE 32µ2 M 3 = LGW = . dt 5a5

  0 1 0 ⟨ ... ⟩ I¨ka I al = 8µ2 ω 5 a4 −1 0 0 . 0 0 0

(36.6.7)

(36.6.8)

The total angular momentum radiated is 2 jkl ⟨¨ ... ⟩ 32 2 5 4 ϵ I ka I al = µ ω a (0, 0, 1), 5 5

(36.6.9)

i.e., − Note that E=−

dLz a3/2 = 1/2 LGW dt M

µM , 2a

Lz = µM 1/2 a1/2 .

371

(36.6.10)

(36.6.11)

Equation 36.6.7 and 36.6.10 are compatible. Radiation reaction force is Fk /m = −Φreact ,k

  x sin 2ωt − y cos 2ωt 2 d5 I jk 32µω 5 a2  −x cos 2ωt − y sin 2ωt . =− xk = 5 dt5 5 0

(36.6.12)

The position and velocity of star are r = r(cos ωt, sin ωt, 0),

v = ωr(− sin ωt, cos ωt, 0).

We have Fk vk = − And so

32µmω 6 a2 r2 5

(36.6.14)

32µω 6 a2 32µ2 ω 6 a4 dE =− (m1 r12 + m2 r22 ) = − . dt 5 5

We also have Mz = xFy − yFx = − And so

(36.6.13)

32µω 5 a2 mr2 . 5

dJz 32µω 5 a2 32µ2 ω 5 a2 =− (m1 r12 + m2 r22 ) = − . dt 5 5

(36.6.15)

(36.6.16)

(36.6.17)

Equation 36.6.15 and 36.6.17 are identical with equation 36.6.7 and 36.6.10.

36.7

MAGNITUDE OF tµν

Consider a slow-motion source of gravitational waves. Show that far from the source, but in the near zone (R ≪ r ≪ λ) the components of the “stress-energy pseudotensor” tµν die out as 1/r4 , but in the radiation zone (r ≫ λ) they die out only as 1/r2 .

Solution: From the definition of tµν , we have t ∼ (∂h)2 . The gravitation wave originating from the source is ∫ [T µν + tµν ]ret 3 ′ µν j ¯ h (t, x ) = 4 d x . (36.7.1) |x − x′ | Note that ∂T ∼

T λ

and ∂(1/r) ∼ 1/r2 . So we have ∂h ∼

1 1 h + h where r λ

h∝

1 r

(36.7.2)

If r ≪ λ, we have ∂h ∼ h/r and so t ∼ 1/r4 . If r ≫ λ, we have ∂h ∼ h/λ and so t ∼ 1/r2 .

36.8

PROOF THAT THE TRANVERSE TRACELESS PARTS OF

Prove by direct computation that the TT parts of Ijk (36.42b) and I jk (36.46) are identical, no matter where the observer is who does the TT projection (i.e., no matter what the unit vector n in the projection operator may be).

372

Solution: Note that

1 I jk = Ijk − δjk I. 3

(36.8.1)

We have 1 I TT jk = Pja I ab Pbk − Pjk Pab I ab 2 ) ( ( ) 1 1 1 = Pja Iab − δab I Pbk − Pjk Pab Iab − δab I 3 2 3 1 1 TT = Ijk − IPja Pak + IPjk δab Pab 3 6 TT = Ijk .

36.9

(36.8.2)

ENERGY AND ANGULAR MOMENTUM RADIATED

(a) For the gravitational waves in asymptotically flat spacetime described by equation (36.47), calculate µν the smeared-out stress-energy tensor TGW ) of equation (35.23). (b) Perform the integrals of equations (36.23) and (36.25) to obtain the total power and angular momentum radiated.

Solution: (a) Firstly, we have hTT jk = and

TT (t − r) 2 d2 Ijk , r dt2

GW Tµν =

1 ⟨ TT TT ⟩ h h , 32π jk,µ jk,ν

1 TT Ijk = I TT jk = Pja I jk Pbk − Pjk Pab I ab . 2

We can derive that

)( ) 1 1 Pja I ab Pbk − Pjk Pab I ab Pjc I cd Pdk − Pjk Pcd I cd 2 2 1 = Pac Pbd I ab I cd − Pab Pcd I ab I cd 2 [ ] 1 = (δac − na nc )(δbd − nb nd ) − (δab − na nb )(δcd − nc nd ) I ab I cd 2 ( ) 1 1 = δac δbd − δab δcd − 2δac nb nd + δab nc nd + na nb nc nd I ab I cd 2 2 1 = I ab I ab − 2I ab I ad nb nd + na nb nc nd I ab I cd . 2

(36.9.1)

(36.9.2)

(

TT TT Ijk Ijk =

(36.9.3)

Note that TT hTT jk,0 = −hjk,r .

(36.9.4)

We have 1 ⟨ ...TT ...TT ⟩ 1 ⟨ TT TT ⟩ GW GW GW hjk,0 hjk,0 = T00 = −T0r = Trr = I jk I jk 32π 8πr2 ⟨ ⟩ ... ... ... ... 1 1 2 I I − 2n I I n + I n ) . (n = jk jk i ij jk k j jk k 8πr2 2

373

(36.9.5)

(b) ∫ LGW (t, r) =

GW 2 Trr r dΩ =

1 ⟨... ... ⟩ I ij I kl 8π

) ∫ ( 1 δik δjl − 2ni nk δjl + ni nj nk nl dΩ . 2

(36.9.6)

Using the fact that ∫ dΩ ni nj =

4π δij , 3

∫ dΩ ni nj nk nl =

we can derive that LGW (t, r) = Similarly, for Ji =

4π (δij δkl + δik δjl + δil δjk ), 15

1 ⟨... ... ⟩ I ij I ij . 5

⟩ ... ... 1 ijk ⟨ ¨km I mp np + 9nj I¨km nm np I pq nq , ϵ −6n I j 8πr2

(36.9.7)

(36.9.8)

(36.9.9)

we have ⟨ ⟩] [⟨ ∫ ⟨ ⟨ ... ⟩ ... ... ⟩ 2 ... ⟩ 1 3 ¨ ... 2 dΩ Jj r2 = ϵijk −2I¨km I mj + (I kq I jq + I¨kp I pj ) = − ϵijk I¨km I mj = ϵijk I¨jm I mk . 2 5 5 5 (36.9.10)

374

Chapter 37

DETECTION OF GRAVITATIONAL WAVES 37.1

GENERAL PLANE WAVE IN TT GAUGE

Show that the most general linearized plane wave can be described in the transverse-traceless gauge of linearized theory by expressions (37.1).

Solution: Under a gauge transformation generated by ξ µ , we have ¯ µν → h ¯ µν − ξµ,ν − ξν,µ + ηµν ξ ρ . h ,ρ

(37.1.1)

¯ µν . ξ µ,ν ν = h ,ν

(37.1.2)

¯ µν = 0, h ,ν

(37.1.3)

Suppose that ξ µ is the solution of After the gauge transformation, we have called Lorentz gauge. ¯ µν (t−z) can be expressed as a superposition In Lorentz gauge, a plane wave propagated along z direction h of monochromatic plane waves, Aµν exp(ikµ xµ ) where

Aµν kν = 0,

kµ = k(−1, 0, 0, 1),

(37.1.4)

or explicit Aµ0 = Aµz . Consider a gauge transformation generated by ξ µ = −iC µ exp(ikµ xµ ).

(37.1.5)

Note that ξµ,ν + ξν,µ = (Cµ kν + Cν kµ ) exp[−ik(t − z)]. We have the transformation Aµν → Aµν − Cµ kν − Cν kµ .

(37.1.6)

Now solve the equation for Cµ Ai0 − C i k − C 0 k i = 0,

375

Aµµ − 2C µ kµ = 0

(37.1.7)

In new gauges, we have Ai0 = 0, Aµµ = 0. Since transformation generated by 37.1 will keep the wave in Lorentz gauge, we have Aµ0 = Aµz . Now we can get A0µ = Azµ = 0 and Aµµ = 0. If we make gauge transformation for each mode of plane wave, the resulting wave must satisfy that ¯ µν = 0, h ,ν

¯ 0µ = h ¯ zµ = 0, h

¯ = 0. h

(37.1.8)

So the most general linearized plane wave can be described in the transverse-traceless gauge of linearized theory by TT TT hTT hTT (37.1.9) xx = −hyy = A+ (t − z), xy = hyx = A× (t − z) In the TT gauge, the time-space components Rj0k0 = R0j0k = −Rj00k = −R0jk0

(37.1.10)

of the Riemann curvature tensor have an especially simple form (see exercise 18.4) 1 Rj0k0 = − hTT 2 jk,00

(37.1.11)

So we have 1 Rx0x0 = −Ry0y0 = − A¨+ (t − z), 2

1 Rx0y0 = Ry0x0 = − A¨× (t − z). 2

In the TT gauge, the stress energy tensor of gravitational wave is ⟩ 1 ⟨ GW Tαβ = hµν,α hµν ,β . 32π So we can get GW GW GW T00 = Tzz = −T0z =

37.2

⟩ 1 ⟨ ˙2 A+ + A˙ 2× . 16π

(37.1.12)

(37.1.13)

(37.1.14)

TEST-PARTICLE MOTION IN PROPER REFERENCE FRAME

Show that a slowly moving test particle, falling freely through the proper reference frame of equation (37.2), obeys the equation of motion (geodesic equation) ˆ ( ) d2 xj kˆ = −a + O x . ˆ j dt2

(37.2.1)

Thus, one can interpret −aˆj as the “local acceleration of gravity”.

Solution: The metric in the “proper reference frame of the detector” is (

ˆ j

ds = − 1 + 2aˆj x 2

So we have

and

)( dx

)2 ˆ 0

( ) ˆ ˆ 2 + δˆj kˆ dx dx + O xj dxαˆ dxβ ˆ j

ˆ k

(37.2.2)

[ ( ][ )2 ( )] ˆ ˆ ˆ ˆ 0 j k ds = − dx + δˆj kˆ dx dx 1 + O xk

(37.2.3)

( ) ˆ ˆ Γi00 = aˆi + O xk

(37.2.4)

376

For a slowly moving test particle, we have dτ = − ds ≈ dt. So the equation of motion ˆ

ˆ

d2 xj dxαˆ dxβ ˆi + Γ =0 ˆ α ˆ β dτ dτ dτ 2 will become

37.3

(37.2.5)

ˆ ( ) d2 xj kˆ = −a + O x . ˆ j dt2

(37.2.6)

CONNECTION COEFFICIENTS IN PROPER REFERENCE FRAME

( ) ˆ j (a) Calculate Γαˆ βˆ for the metric 37.2.2, ignoring corrections of O |x | . ˆγ ˆ

(b) Calculate Rj ˆ0kˆˆ0 using the standard formula (8.44), and leaving spatial derivatives of the connection ) ( ˆ coefficients unevaluated because of the unknown corrections of O |xj | . ( ) ˆ ˆ (c) Use the answer to part (b) to evaluate the O |xj | corrections to Γj ˆ0ˆ0 .

Solution: (a) Firstly, we have

( ) ˆ gˆ0ˆ0,ˆi = −2aˆi + O |xj |

(37.3.1)

( ) ˆ j gαˆ β,ˆ ˆ γ = O |x |

(37.3.2)

and for other components. Using Γαˆ βˆ ˆγ = we can get

and

1 γˆ µˆ g (gµˆβ,ˆ ˆγ ,ˆ ˆ γ + gµ µ ), ˆγ ˆ ,βˆ − gβˆ 2

( ) ( ) 1 ˆ ˆ ˆ ˆ ˆ Γ0ˆ0ˆi = Γ0ˆiˆ0 = Γiˆ0ˆ0 = − gˆ0ˆ0,ˆi + O |xj | = aˆi + O |xj | 2 ( ) ˆ j Γαˆ βˆ ˆγ = O |x |

(37.3.3) (37.3.4) (37.3.5)

for other components. (b) ˆ

(c)

ˆ

ˆ

ˆ

ˆ

+ Γj αˆ kˆ Γαˆ ˆ0ˆ0 − Γj αˆ ˆ0 Γαˆ ˆ0kˆ Rj ˆ0kˆˆ0 = Γj ˆ0ˆ0,kˆ − Γj ˆ0k, ˆˆ 0 ) ( ˆ ˆ ˆ ˆ = Γj ˆ0ˆ0,kˆ − Γj ˆ0ˆ0 Γ0ˆ0kˆ + O |xj | ) ( ˆ ˆ = Γj ˆ0ˆ0,kˆ − aˆj akˆ + O |xj | .

(37.3.6)

) ( ˆ ˆ ˆ ˆ ˆ ˆ ˆ Γj ˆ0ˆ0 = Γj ˆ0ˆ0 (xi = 0) + Γj ˆ0ˆ0,kˆ (xi = 0)xk + O |xj |2 ) ( ˆ ˆ ˆ = aˆj + (Rj ˆ0kˆˆ0 + aˆj akˆ )xk + O |xj |2

(37.3.7)

377

37.4

WHY THE a · x ?

Explain the origin of the a · x correction in equation (5b) of Box 37.1.

Solution: Take the viewpoint of an observer at rest at the spatial origin who watches two freely falling particles respond to the inertial force. Suppose the force is at zˆ direction. At time tˆ = 0, put the particle A at the origin and the particle B at (0, 0, z0 ). As time passes, the separation of the particles in their common Lorentz frame remains fixed; so there develops a Lorentz contraction in zˆ direction from the viewpoint of the observer fixed at the origin. So we have ( )1/2 zB − zA = z0 1 − v 2 ,

(37.4.1)

where v = at. Take the derivative with respect to time. We have d2 zA z0 a2 d2 zB − = − ≈ −z0 a2 (1 − v 2 )3/2 dtˆ2 dtˆ2

when tˆ ≈ 0.

(37.4.2)

Because particle A is at the origin, we have d2 zA = a. dtˆ2

(37.4.3)

d2 zB = −a(1 + az0 ). dtˆ2

(37.4.4)

As a result,

Or more generally, we have ˆ

d2 xj ˆ = −aj (1 + a · x). 2 ˆ dt

37.5

(37.4.5)

ORIENTATION OF POLARIZATION DIAGRAM

Derive equation (2) of Box 37.2.

Solution: New coordinates relate with old coordinates through x ˆ=x ˆ′ cos ϕ0 − yˆ′ sin ϕ0 ,

yˆ = x ˆ′ sin ϕ0 + yˆ′ cos ϕ0 .

(37.5.1)

The equation of motion in will be ] 1[¨ d2 x ˆ′ cos ϕ0 − yˆ′ sin ϕ0 ′ ′ ′ ′ ¨× (ˆ A (ˆ x cos ϕ − y ˆ sin ϕ ) + A x sin ϕ + y ˆ cos ϕ ) = + 0 0 0 0 2 dtˆ2

(37.5.2)

] 1[ ¨ d2 x ˆ′ sin ϕ0 + yˆ′ cos ϕ0 ′ ′ ′ ′ ¨× (ˆ = − A (ˆ x sin ϕ + y ˆ cos ϕ ) + A x cos ϕ − y ˆ sin ϕ ) . + 0 0 0 0 2 dtˆ2

(37.5.3)

and

378

It can be simplified as

and

] d2 x ˆ′ 1[ ¨ ′ ′ ¨× sin 2ϕ0 )ˆ ¨+ sin 2ϕ0 + A¨× cos 2ϕ0 )ˆ ( A cos 2ϕ + A x + (− A y = + 0 2 dtˆ2

(37.5.4)

] d2 yˆ′ 1[ ¨ ′ ′ ¨× sin 2ϕ0 )ˆ ¨+ sin 2ϕ0 + A¨× cos 2ϕ0 )ˆ = −( A cos 2ϕ + A y + (− A x + 0 2 dtˆ2

(37.5.5)

We will get the same pattern in new coordinates if − A¨+ sin 2ϕ0 + A¨× cos 2ϕ0 = 0. We have the solution 1 ϕ0 = arctan 2

37.6

(

A¨× A¨+

(37.5.6)

) .

(37.5.7)

RELATIVE MOTION OF FREELY FALLING BODIES AS A DETECTOR OF GRAVITATIONAL WAVES

Consider two test bodies initially at rest with respect to each other in flat, empty spacetime. (The case where other, gravitating bodies are nearby can be treated without too much more difficulty; but this exercise concerns only the simplest example!) A plane, nearly monochromatic gravitational wave, with angular frequency ω and polarization e+ , impinges on the bodies, coming from the −z direction. As shown in exercise 35.5, the bodies remain forever at rest in those TT coordinates that constituted the bodies’ global inertial frame before the wave arrived. Calculate, for arbitrary separations (∆x, ∆y, ∆z) of the test bodies, the redshift and the round-trip travel time of photons going back and forth between them. Compare the answer, for large ∆x, ∆y, ∆z, with the answer one would have obtained by using (without justification!) the equation of geodesic deviation. Physically, why does the correct answer oscillate with increasing separation? Discuss the feasibility and the potential sensitivity of such a detector using current technology.

Solution: We can decompose the geodesic of the photon into a background path plus a perturbation, xµ (λ) = xµ(0) (λ) + xµ(1) (λ),

(37.6.1)

where xµ(0) solves the geodesic equation in the background. For convenience we denote the wave vector of the background path as kµ and the derivative of the deviation vector as lµ . The condition that a path be null is of course (ηµν + hµν )(lµ + k µ )(lν + k ν ) = 0. (37.6.2) At zeroth order we simply have ηµν k µ k ν = 0. We can obtain that k µ = k(1, sin θ cos ϕ, sin θ sin ϕ, cos θ), where θ and ϕ are spherical coordinates of (∆x, ∆y, ∆z). At first order we obtain 2ηµν k µ lν + hµν k µ k ν = 0.

(37.6.3)

We now turn to the perturbed geodesic equation. The zeroth-order geodesic equation simply tells us that xµ(0) (λ) is a straight trajectory, while at first order we have dlµ = −Γµρσ k ρ k σ . dλ

379

(37.6.4)

Particularly, we have for µ = 0 dl0 = −(Γ0xx k x k x + Γ0yy k y k y ), dλ where Γ0xx = −Γ0yy =

−iω A+ e−iω(t−z) . 2

(37.6.5)

(37.6.6)

Note that We have

xµ(0) (λ) = k µ λ.

(37.6.7)

z dl0 iω = A+ e−iωλ(k−k ) [(k x )2 − (k y )2 ] dλ 2

(37.6.8)

The integration gives ∆l0 = −

A+ (k x )2 − (k y )2 −iω∆λ(k−kz ) kA+ sin2 θ cos 2ϕ −iω∆r(1−cos θ) iϕ0 (e −1)eiϕ0 , (37.6.9) (e −1)e = − 2 k − kz 2 1 − cos θ

where ϕ0 = −ω(t0 − z0 ) is the initial phase. The redshift of the photon is z1 = −

A+ l0 = cos 2ϕ(1 + cos θ)(e−iω∆r(1−cos θ) − 1)eiϕ0 . k 2

(37.6.10)

For photon propagate back, we have z2 = −

l0 A+ = cos 2ϕ(1 − cos θ)(e−iω∆r(1+cos θ) − 1)e−iω∆r(1−cos θ) eiϕ0 . k 2

(37.6.11)

The total redshift is z = z1 + z2 . If ω∆r ≪ 1, we have

z = A+ cos 2ϕ sin2 θω∆rei(ϕ0 −π/2) .

(37.6.12)

The correct answer oscillate with increasing separation because the spacetime is oscillate when the photon propagates.

37.7

EARTH-MOON SEPARATION AS A GRAVITATIONAL-WAVE DETECTOR

In the early 1970’s one can monitor the Earth-moon separation using laser ranging to a precision of 10 cm, with successive observations separated by at least one round-trip travel time. Suppose that no oscillations in round-trip travel time are observed except those (of rather long periods) to be expected from the Earth-moon-sun-planets gravitational interaction. What limits can one then place on the energy flux of gravitational waves that pass the Earth? The mathematical formula for the answer should yield numerically Flux ≲ 1018 erg/cm2 sec for 0.3 cycle/sec ≲ ν ≲ 1 cycle/sec

(37.7.1)

corresponding to a limit on the mass density in gravitational waves of Density ≲ 10−13 g/cm3 . Why is this an uninteresting limit?

Solution:

380

(37.7.2)

The Earth-moon separation is about 3.84 × 1010 cm. With a 10 cm precision, the upper limit of the amplitude of gravitational wave is 10 h= = 2.6 × 10−10 . (37.7.3) 3.84 × 1010 The corresponding energy flux is 1 πν 2 h2 c3 ⟨hjk,µ hjk,ν ⟩ ∼ ∼ 1018 erg/cm2 sec. 32π 16G

(37.7.4)

The corresponding density is πν 2 h2 ∼ 10−13 g/cm3 . (37.7.5) 16G The energy flux of solar radiation is 1.368×106 erg/cm2 sec. So this upper limit is too high to be interesting. ρ∼

37.8

POWER RERADIATED

The idealized gravitational wave detector of Figure 37.4 vibrates with angular frequency ω. Show that the power it radiates as gravitational waves is given by equation (37.37).

Solution: Put the detector on the x axis. We have ρ = M δ(x − ξ − L)δ(y)δ(z) + M δ(x + ξ + L)δ(y)δ(z).

(37.8.1)

The second moment of mass distribution is Ixx = 2M (L + ξ)2

(37.8.2)

and all other components vanish. The reduced quadruple is I xx =

4 M (L + ξ)2 , 3

2 I yy = − M (L + ξ)2 , 3

2 I zz = − M (L + ξ)2 . 3

(37.8.3)

The power of radiation is 1 ⟨... ... ⟩ 8M 2 P = I ij I ij = 5 15

⟨

d3 (L + ξ)2 d3 (L + ξ)2 dt3 dt3

⟩ (37.8.4)

If ξ is mush smaller of L, we have 32M 2 L2 P = 15

37.9

⟨

d3 ξ d3 ξ dt3 dt3

⟩ =

32M 2 L2 ω 6 ⟨ 2 ⟩ ξ . 15

(37.8.5)

CROSS SECTIONS CALCULATED BY DETAILED BALANCE

Use the principle of detailed balance to derive the cross sections (37.41) for a vibrating, resonant detector of any size, shape, or mass (e.g., for the vibrating Earth, or Weber’s vibrating cylinder, or the idealized detector of Figure 37.4).

Solution:

381

Let the detector be in thermal equilibrium with a bath of blackbody gravitational waves. Then it must be losing energy by reradiation as rapidly as it is absorbing it from the waves. Internal damping can be ignored because, in true thermal equilibrium, energy loss by internal damping will match energy gain from random internal Brownian forces. In detail, the balance of energy in and out reads ∫ [4πIν (ν0 )]blackbody × ⟨σ⟩all directions dν = AGW × (Energy in normal mode of detector). (37.9.1) Using the familiar form of the Planck spectrum and the fact that gravitational waves have two independent states of polarization, we have 2hν 3 . (37.9.2) Iν = exp[hνi /kT ] − 1 On the other hand, the average number of quanta (phonon) for a normal mode with frequency νi of detector is given by 1 . (37.9.3) ni = exp[hν/kT ] − 1 Thus, the energy in normal mode with frequency νi of detector is Ei =

hνi . exp[hνi /kT ] − 1

Put them together, we can obtain ∫ ⟨σ⟩all directions dν =

37.10

(37.9.4)

AGW π = λ20 AGW . 2 8πν0 2

(37.9.5)

NORMAL-MODE ANALYSIS OF VIBRATING, RESONANT DETECTORS

Derive all the results for vibrating, resonant detectors quoted in Box 37.4. Pattern the derivation after the treatment of the idealized detector in section 37.6.

Solution: (a) Let the detector be driven by the polarized waves of equation hjk = A(t − n · x)ejk ,

ejk nk = 0,

ejj = 0,

ejk ejk = 2,

(37.10.1)

and let it be wave-dominated (Evibration ≪ kT ). The displacements δx = ξ(x, t) of its mass elements satisfy that ξ˙ 1¨ ξ¨ + + ω02 ξ = Ae · x. (37.10.2) τ0 2 We decompose the displacements into normal modes of the detector as ∑ ξ= Bn (t)un (x). (37.10.3) n

Using the orthonormalization condition

∫ ρum · un d3 x = M δmn ,

multiplying 37.10.2 by ρun and integrating over the whole space, we can get ∫ B˙ n 1¨ ρ i 2 ¨ Bn + + ωn Bn = A u eij xj d3 x . τn 2 M n

382

(37.10.4)

(37.10.5)

Note that ω0 = ωn and τ0 = τn for nth mode. Using the definition of the moment of inertia factor for the nth normal mode ∫ ∫ I(n)jk ≡ −(ρuln ),l xj xk d3 x = ρ(ujn xk + ukn xj ) d3 x , (37.10.6) and corresponding reduced quadrupole factor for the nth normal mode 1 I (n)jk ≡ I(n)jk − I(n)ll δjk , 3

(37.10.7)

˙ ¨n + Bn + ωn2 Bn = 1 AI ¨ (n)jk eij = 1 AI ¨ (n)ij eij . B τn 4M 4M

(37.10.8)

equation 37.10.5 can be written as

(b) Fourier-analyze the amplitudes of the detector and waves, ∫ ∞ ∫ −1/2 −iωt −1/2 ˜ Bn (t) = (2π) Bn (ω)e dω , A(t) = (2π) −∞

∞

−iωt ˜ A(ω)e dω ,

(37.10.9)

−∞

and substitute 37.10.9 into 37.10.8. We can derive that 1 2˜ ˜n − i ω B ˜n + ω 2 B ˜ − ω2 B ω AI (n)ij eij . n n =− τn 4M

(37.10.10)

The solution is

˜ (n)ij eij ω 2 AI . 2 4M (ω + iω/τn − ωn2 ) If |ω ± ωn | ≪ ωn , we have the approximation ˜n = B

˜n = B

(37.10.11)

˜ (n)ij eij ωn AI . 8M (|ω| + i/2τn − ωn )

(37.10.12)

(c) The total energy deposited in the detector can be calculated by integrating ∫ (energy deposited) = (Force per unit volume) · (Velocity) d3 x dt .

(37.10.13)

For nth mode, the force per unit volume is fi =

1¨ Aρeij xj . 2

(37.10.14)

The velocity of a small volume is vi = B˙ n uin . It follows that 1 (energy deposited in nth normal mode) = 2

∫

¨ ij xj B˙ n ui d3 x dt = 1 I (n)ij eij Aρe n 4

(d) Apply Parseval’s theorem to 37.10.16, we have ∫ ∫ ˜ dω . A¨B˙ n dt = −ω 2 A˜∗ (−iω B)

(37.10.15) ∫ A¨B˙ n dt . (37.10.16)

(37.10.17)

Using 37.10.12, we can obtain (energy deposited in nth normal mode) ∫ ˜ 2 I (n)ij eij ωn |A| 1 dω = I (n)ij eij iω 3 4 8M (|ω| + i/2τn − ωn ) ∫ ˜ 2 (|ω| − ωn − i/2τn ) 1 ωn |A| dω = |I (n)ij eij |2 iω 3 4 8M [(|ω| − ωn )2 + (1/2τn )2 ] ∫ ˜ n )|2 (|ω| − ωn − i/2τn ) ω 4 |A(ω 1 dω . ≈ |I (n)ij eij |2 i n 4 8M [(|ω| − ωn )2 + (1/2τn )2 ]

383

(37.10.18)

∫

Since

|ω| − ωn dω (|ω| − ωn )2 + (1/2τn )2

(37.10.19)

must vanish because of symmetry consiseration, we have ∫ ˜2 π |I (n)ij eij |2 ωn2 /τn ω 2 |A| (energy deposited in nth normal mode) = dν , 2 2 4 M (|ω| − ωn ) + (1/2τn ) 8 (37.10.20) where ν = ω/2π. (e) The energy flux of the gravitational wave is T tn = The total energy per unit area is

1 ⟨ ˙ 2⟩ A . 16π time avg

∫ E=

T

tn

(37.10.21)

∫

1 dt = 16π

A˙ 2 dt .

(37.10.22)

˜2 ω 2 |A| dν . 8

(37.10.23)

Using Parseval theorem, we have 1 E= 16π

∫

∫ ˜ 2 dω = ω |A| 2

Thus the total energy per unit area per unit frequency carried by the waves past the detector is Fν =

˜2 ω 2 |A| . 8

(37.10.24)

From the definition of scattering cross section, we have ∫ (energy deposited in nth normal mode) =

σn (ν)Fν dν .

(37.10.25)

Using equation 37.10.20 and 37.10.24, we can conclude that σn (ν) =

π |I (n)ij eij |2 ωn2 /τn . 4 M (|ω| − ωn )2 + (1/2τn )2

(37.10.26)

It follows that ∫

1 |I (n)ij eij |2 ωn2 σn (ν) dν ≈ 4 M resonance

∫

dx π |I (n)ij eij |2 ωn2 = . x2 + 1 4 M

(37.10.27)

(f) For simplicity, we suppose the n is in the z direction. For unpolarized radiation (random mixture of polarizations) with propagation direction n, we have ( ) cos θ sin θ eij = , (37.10.28) sin θ − cos θ with θ uniformly distributed within [0, 2π). Thus, ⟩ ⟨ ⟩ ⟨ |I (n)ij eij |2 = |I (n)xx cos θ − I (n)yy cos θ + 2I (n)xy sin θ|2 = Recall that

|I (n)xx − I (n)yy |2 + 4|I (n)xy |2 . 2

( ) 1 I TT (n) = P I (n) P − P Tr P I (n) , 2

It follows that TT I TT (n)xx = −I (n)yy =

Pjk = δjk − nj nk .

1 (I (n)xx − I (n)yy ), 2

384

I TT (n)xy = I (n)xy ,

(37.10.29)

(37.10.30)

(37.10.31)

and all other components vanish. Thus, we have ⟨ ⟩ 2 |I (n)ij eij |2 = |I TT (n)ij |

(37.10.32)

and so π |I (n)ij | ωn2 /τn , 4 M (|ω| − ωn )2 + (1/2τn )2 TT

σn (ν) =

2

∫ σn (ν) dν = resonance

TT 2 2 π |I (n)ij | ωn . 4 M

(g) From equation 37.10.30, we have )( ) ( 1 1 TT I TT I I P − P P I P I P − P P I = P ij ab ab ic cd dj ij cd cd ia ab bj ij ij 2 2 1 = Pac Pbd I ab I cd − Pab Pcd I ab I cd 2 [ ] 1 = (δac − na nc )(δbd − nb nd ) − (δab − na nb )(δcd − nc nd ) I ab I cd 2 ( ) 1 1 = δac δbd − δab δcd − 2δac nb nd + δab nc nd + na nb nc nd I ab I cd 2 2 1 = I ab I ab − 2I ab I ad nb nd + na nb nc nd I ab I cd . 2 When averaged over all directions, we have ) ∫ ( ⟨ TT TT ⟩ 1 1 I ij I ij direction avg = I ij I kl δik δjl − 2ni nk δjl + ni nj nk nl dΩ . 4π 2 Using the fact that ∫ 1 1 dΩ ni nj = δij , 4π 3

1 4π

we can derive that

∫ dΩ ni nj nk nl = ⟨

⟩

1 (δij δkl + δik δjl + δil δjk ), 15

2 I ij I ij 5 For unpolarized radiation, averaged over all directions, cross sections are σn (ν) =

TT I TT ij I ij

direction avg

=

ωn2 /τn π |I (n)ij |2 . 10 M (|ω| − ωn )2 + (1/2τn )2

(37.10.33)

(37.10.34)

(37.10.35)

(37.10.36)

(37.10.37)

(37.10.38)

For nth normal mode, we have δIjk = I(n)jk Bn (t),

δx = Bn (t)un .

(37.10.39)

The power of radiated gravitational waves are LGW =

⟨ ⟩ 1 1 ⟨ ... ... ⟩ δ I ij δ I ij time avg = |I (n)ij |2 ωn6 Bn2 time avg . 5 5

The kinetic energy of the detector is ∫ ∫ 1 2 2 1 1 2 3 ˙ d x = ωn Bn ρun · un d3 x = M ωn2 Bn2 . ρ(δ x) Ekin = 2 2 2

(37.10.40)

(37.10.41)

The total energy of the detector is ⟨ ⟩ Edet = 2 ⟨Ekin ⟩ = M ωn2 Bn2 time avg .

(37.10.42)

The Einstein A coefficients for gravitational wave radiation is AGW =

1 |I (n)ij |2 4 LGW = ωn . Edet 5 M

385

(37.10.43)

The Einstein A coefficients for internal damping is 1 . τ0

(37.10.44)

π 2 AGW Adiss λ . 2 (|ω| − ωn )2 + (Adiss /2)2

(37.10.45)

Adiss = Now we can write down σn (ν) = ∫

and

π 2 λ AGW . 2

σn (ν) dν = resonance

37.11

(37.10.46)

SPECTRUM OF ENERGY RADIATED BY A SOURCE

Derive the results quoted in the last section of Box 37.4.

Solution: The component of polarization eij of a gravitational wave hTT ij is 1 Tr(hTT e)eij . 2

(37.11.1)

Noting that 1 hTT = P hP − P Tr(P h), 2

Pjk = δjk − nj nk

(37.11.2)

ejk ejk = 2,

(37.11.3)

and ejk nk = 0,

ejj = 0,

we have Tr(hTT e) = Tr(P hP e) −

1 1 Tr(P h) Tr(P e) = Tr(P he) − Tr(P h) Tr e = Tr(eP h) = Tr{he}. (37.11.4) 2 2

The gravitational wave generated by the source is given by hij =

2¨ I ij . r

(37.11.5)

So the energy flux of gravitational wave with polarization tensor eij is − Ttr = −

⟩e e 1 ⟨ 1 ⟨ (... ) (... )⟩ ij ij Tr I e Tr I e , Tr(he),t Tr(he),r = 32π 4 16πr2

(37.11.6)

i.e.,

⟩ dE 1 ⟨ ... = | I ij eij |2 . dt dΩ 16π Using Parseval’s theorem, we can obtain ∫ ∫ ... dE 1 1 = | I ij eij |2 dt = |I˜ij eij |2 ω 6 dω . dΩ 16π 16π

(37.11.7)

(37.11.8)

So the total energy per unit frequency (ν > 0) radiated over all time, into a unit solid angle about the direction n, and with polarization tensor e, is dE 1 ∑ ˜ = |Iij eij |2 ω 6 . dν dΩ 8 ω=±2πν

386

(37.11.9)

Summed over polarizations, we have − Ttr = −

1 ⟨ TT TT ⟩ 1 ⟨...TT ...TT ⟩ hij,t hij,r = I ij I ij . 32π 8πr2

(37.11.10)

By similar method, we have ∑ dE 1 1 1 ∑ ˜TT 2 6 TT = |I˜ij |2 ω 6 = = |I | ω . dν dΩ 4 ω=±2πν,i,j 2 2 i,j ij

(37.11.11)

Using equation 37.10.5, the total energy radiated per unit frequency, integrated over all directions, is dE 1∑ 2 4π ∑ ˜ 2 6 = 4π × |I˜ij |2 ω 6 = |I ij | ω . dν 2 i,j 5 5 i,j

37.12

(37.11.12)

PATTERNS OF EMISSION AND ABSORPTION

The elementary dumbbell oscillator of Figure 37.4, initially unexcited, has a cross section for absorption of unpolarized gravitational radiation proportional to sin4 θ, and when excited radiates with an intensity also proportional to sin4 θ (Chapter 36). The patterns of emission and absorption are identical. Any other dumb bell oscillator gives the same pattern, apart from a possible difference of orientation. Consider a nonrotating oscillator of general shape undergoing free vibrations in a single nondegenerate (and therefore nonrotatory) mode, or excited from outside by unpolarized radiation. (a) Show that its pattern of emission is identical with its pattern of absorption. [Hint: Make the comparisons suggested in the last few parts of Box 37.4.] (b) Show that this emission pattern (absorption pattern), apart from three Euler angles that describe the orientation of this pattern in space, and apart from a fourth parameter that determines total intensity, is uniquely fixed by a single (“fifth”) parameter. (c) Construct diagrams for the pattern of intensity for the two extreme values of this parameter and for a natural choice of parameter intermediate between these two extremes. (d) Define the parameter in question in terms of a certain dimensionless combination of the principal moments of the reduced quadrupole tensor.

Solution: (a) For unpolarized radiation with propagation direction n, the cross section of absorption of gravitational waves is TT 2 π |I (n)ij | ωn2 /τn σn (ν) = . (37.12.1) 4 M (|ω| − ωn )2 + (1/2τn )2 The total energy per unit frequency (ν > 0) radiated over all time, into a unit solid angle about the direction n, and summed over polarizations, is 1 TT dE = |I˜ij |2 ω 6 . dν dΩ 2 Since

δIjk = I(n)jk Bn e−iωn t−t/(2τn )

we have I˜ij = I(n)jk Bn

(37.12.2)

for t > 0,

1 . 1/(2τn ) − i(ω − ωn )

387

(37.12.3) (37.12.4)

and so TT

2 2 |I˜ij |2 = |I TT (n)ij | |Bn |

1 . (ω − ωn )2 + (1/2τn )2

(37.12.5)

Thus, its pattern of emission is identical with its pattern of absorption. (b) The emission pattern is determined by I (n)jk . Adjusting three Euler angles that describe the orientation of this pattern in space properly, we can bring I (n)jk into diagonal form as  I (n)jk ∝ I = 

−x

 

x−1

(37.12.6)

1 where 0 ≤ x ≤ 1/2. Apart from a fourth parameter that determines total intensity, the pattern is uniquely fixed by a single parameter a. 2 (c) The pattern of the emission is determined by |ITT ij | . Suppose n = (sin θ cos ϕ, sin θ sin ϕ, cos θ). We have

) ] 1 [( 4 cos4 ϕ − 4 cos2 ϕ + 1 sin4 θ − 4 sin2 θ + 4 x2 2 [( ) ( ) ] − 2 cos4 ϕ − 5 cos2 ϕ + 2 sin4 θ + 6 cos2 ϕ − 5 sin2 θ + 2 x ) ( ) 1( 4 1 + cos ϕ − 4 cos2 ϕ + 4 sin4 θ + 3 cos2 ϕ − 2 sin2 θ + 2 2

2 |ITT ij | =

(37.12.7)

For x = 0, we have 2 |ITT ij | =

) ( ) 1( 4 1 cos ϕ − 4 cos2 ϕ + 4 sin4 θ + 3 cos2 ϕ − 2 sin2 θ + . 2 2

(37.12.8)

9 sin4 θ. 8

(37.12.9)

For x = 1/2, we have 2 |ITT ij | =

(d) The parameter can be defined as x=

min(|Ii |) , max(|Ii |)

(37.12.10)

where I1 , I2 and I3 is the principal moments of the reduced quadrupole tensor.

37.13

MULTIMODE DETECTOR

Consider a cylindrical bar of length very long compared to its diameter. Designate the fundamental mode of end-to-end vibration of the bar as n = 1, and call the mode with n − 1 nodes in its eigenfunction the “nth mode”. Show that the cross section for the interception of unpolarized gravitational waves at the nth resonance, integrated over that resonance, and averaged over direction, is given by the formula [Ruffini and Wheeler (1971b)] ∫ 32 v 2 M σ(ν) dν = for n odd (zero for even n), (37.13.1) 15π c2 n2 nth resonance, random where v is the speed of sound in the bar expressed in the same units as the speed of light, c; and M is the mass of the bar (geometric units; multiply the righthand side by the factor G/c = 2.22 × 10−18 cm2 Hz/g ∫ when employing conventional units). Show that this expression gives σ dν = 1.0 × 10−21 cm2 Hz for the lowest mode of Weber’s bar. Multimode detectors are (1973) under construction by William Fairbank and William Hamilton, and by David Douglass and John A. Tyson.

388

Solution: For nth mode, we have uxn

=

√

[( 2 cos

x 1 + L 2

)

] nπ ,

ωn =

πnv . L

The “moment of inertia factor for the nth normal mode” is { [( ) ] ∫ −1 √ for n odd 2M L/2 √ x 1 2 2 + nπ x dx = 4 2M L n π I(n)xx = 2 cos L −L/2 L 2 0 for n even So for n odd, we have ∫

(37.13.2)

(37.13.3)

2

σ(ν) dν = nth resonance, random

π |I jk |2 2 π I(n)xx 2 32M v 2 ωn = ωn = . 10 M 15 M 15πn2

(37.13.4)

For Weber’s bar, we have M = 1.4 × 106 g, So we obtain

v = νλ = 153cm × 2 × 1660Hz = 5.0796 × 105 cm/s. ∫ σ dν =

37.14

32 GM v 2 = 6 × 10−22 cm2 Hz. 15π c3

(37.13.5)

(37.13.6)

CROSS SECTION OF IDEALIZED MODEL OF EARTH FOR ABSORPTION OF GRAVITATIONAL RADIATION

The observed period of quadrupole vibration of the earth is 54 minutes [see, e.g., Bolt (1964) or Press (1965) for survey and bibliography]. To analyze that mode of vibration, with all due allowance for elasticity and the variation of density in the earth, is a major enterprise. Therefore, for a first estimate of the cross section of the earth for the absorption of quadrupole radiation, treat it as a globe of fluid of uniform density held in the shape of a sphere by gravitational forces alone (zero rigidity). Let the surface be displaced from r = a to r = a + aαP2 (cos θ), (37.14.1) where θ is polar angle measured from the North Pole and a is the fractional elongation of the principal axis. The motion of lowest energy compatible with this change of shape is described by the velocity field 1 ξ x = − αx, 2

1 ξ y = − αy, 2

ξ z = αz

(37.14.2)

(zero divergence, zero curl). (a) Show that the sum of the kinetic energy and the gravitational potential energy is ( ) 3 M2 α2 3 E=− 1− + M a2 α˙ 2 . 5 a 5 20 (b) Show that the angular frequency of the free quadrupole vibration is ( )1/2 16π ω= ρ1/2 . 5

(37.14.3)

(37.14.4)

(c) Show that the reduced quadrupole moments are I xx = I yy = −

M a2 α , 5

389

I zz =

2M a2 α . 5

(37.14.5)

(d) Show that the rate of emission of vibrational energy, averaged over a period, is ⟨ ⟩ dE 3 2 − = M 2 a4 ω 6 αpeak . dt 125

(37.14.6)

(e) Show that the exponential rate of decay of energy by reason of gravitational wave dampIng, or “gravitational radiation line broadening”, is 4 AGW = M a2 ω 4 . (37.14.7) 25 (f) Show that the resonance integral of the absorption cross section for radiation incident from random directions with random polarization is ∫ ⟨σ(ν)⟩ dν = resonance

2π M a2 . 25 λ2

(37.14.8)

(g) Evaluate this resonance integral.

Solution: (a) The kinetic energy is ) ∫ ∫ ( ∫ a 1 ρ 2 1 2 1 2 ρ 1 3 2 3 2 ˙ K= ρξ d x = α˙ x + y + z d3 x = α˙ 2 2πr4 dr = πρa5 α˙ 2 = M a2 α˙ 2 . 2 2 4 4 2 5 20 0 (37.14.9) The potential energy of the earth without viberation is ( ) ∫ ∫ 1 1 a 2π 2 16 3 M2 V0 = ρϕ0 d3 x = ρ ρr − 2πρa2 4πr2 dr = − π 2 a5 ρ2 = − . (37.14.10) 2 2 0 3 15 5 a From Exercise 36.3, we have δρ = ρξ · rˆδ(r − a) where

αr ξ · rˆ = 2

(

3z 2 −1 r2

)

I can not figure out the correct answer for the variation of potential energy. I thought ∫ 1 δρ (r) δρ (r ′ ) 3 3 ′ δV = − d xd x , 2 |r − r ′ |

(37.14.11)

(37.14.12)

but only to find that δV = − The correct answer should be δV =

8 2 2 2 2 π α a ρ . 25

(37.14.13)

16 2 2 2 2 π α a ρ . 75

(37.14.14)

Any help is appreciated. (b) Since E = V0 + we have

( ω=

3M 2 20 25a 3M a2

3 M2 2 3 α + M a2 α˙ 2 , 25 a 20 (

)1/2 =

12M 5a3

390

(

)1/2 =

16π 5

(37.14.15) )1/2 ρ1/2 .

(37.14.16)

(c)

∫ δI =

∫ δρ r ⊗ r d3 x = ρa4

 −1 1 (ξ · rˆ)ˆ r ⊗ rˆ dΩ = M a2 α 5

 .

−1

(37.14.17)

2

It is easy to see that δI ij = δI ij . (d) The rate of emission of vibrational energy, averaged over a period, is ⟩ ⟨ ⟨ ⟩ 1 ⟨... ... ⟩ 6 3 dE 2 − = I ij I ij = M 2 a4 ω 6 α2 = M 2 a4 ω 6 αpeak . dt 5 125 125 (e) The average viberating energy of the earth in a period is ⟨ ⟩ 3 3 2 2 2 ⟨E⟩ = 2 M a α˙ = M a2 ω 2 αpeak . 20 20

(37.14.18)

(37.14.19)

So “gravitational radiation line broadening” is AGW =

− ⟨ dE/dt ⟩ 4 2 = M a2 ω 4 αpeak . ⟨E⟩ 25

(37.14.20)

(f) The resonance integral of the absorption cross section for radiation incident from random directions with random polarization is ∫ π 2π 2π M a2 ⟨σ(ν)⟩ dν = λ2 AGW = . (37.14.21) M a2 ω 2 = 2 25 25 λ2 resonance (g) This resonance integral is 2π GM a2 16πGρ 24πG2 M 2 = = 5.58cm2 Hz. 3 25 c 5 125ac3

391

(37.14.22)

392

Chapter 39

OTHER THEORIES OF GRAVITY AND POST-NEWTONIAN APPROXIMATION 39.1

ORDERS OF MAGNITUDE IN GRAVITATIONALLY BOUND SYSTEMS

Use Newtonian theory to derive conditions (39.7) for any gravitationally bound system. [Hint: Such concepts as orbital velocities, the speeds of sound and shear waves, the virial theorem, and hydrostatic equilibrium are relevant.]

Solution: For circular orbits in solar system, we have 1 2 1 v = |U | 2 2

(39.1.1)

and so ϵ2 ≳ v 2 . From hydrostatic equilibrium, we have ∇p = ρ0 ∇U,

(39.1.2)

and so ϵ2 ≳ p/ρ0 . Stress-energy tensor of gravity is tij ∼ U,i U,j ∼ U U,ij ∼ U ρ0 ,

(39.1.3)

and so ϵ2 ≳ |tij |/ρ0 . Finally, we have Π≡

ρ − ρ0 p |tij ∼ v2 , , , ρ0 ρ0 ρ0

(39.1.4)

and so ϵ2 ≳ Π.

39.2

PATTERN OF TERMS IN POST-NEWTONIAN EXPANSION

Verify the statements in the paragraph following / equation (39.11). In particular, suppose that one wishes to evaluate the coordinate acceleration, d2 xj dt2 , to accuracy ϵ2N U,j for some integer N . Show that this undertaking requires a knowledge of g00 to accuracy ϵ2N +2 , of g0k to ϵ2N +1 , and of gjk to ϵ2N . Also

393

suppose that one knows T 00 to accuracy ρ0 ϵ2N , T 0j to ρ0 ϵ2N +1 , and Tjk to ρ0 ϵ2N +2 Show that to calculate T 0α;α with accuracy ϵ2N +1 ρ0,j and T jα;α with accuracy ϵ2N +2 ρ0,j , one must know g00 to ϵ2N +2 , g0k to ϵ2N +1 , and gjk to ϵ2N . This dictates the pattern of Box

Solution: Firstly, from equation (39.11), we have d2 xj = U,j + terms of order {ϵg0k,j ; ϵ2 gkl,j } + terms of higher order in g00,j . dt2

(39.2.1)

/ Notice that U,j is of the order ϵ2 . If one wishes to evaluate d2 xj dt2 to accuracy ϵ2N U,j . We must require a knowledge of g00 to accuracy ϵ2N +2 , of g0k to ϵ2N +1 , and of gjk to ϵ2N . Secondly, we have T 0α;α = T 0α,α + Γ0βα T βα + Γαβα T 0β = T 00,0 + T 0i,i + Γ000 T 00 + 2Γ00i T 0i + Γ0ij T ij + Γ000 T 00 + Γ0j0 T 0j + Γi0i T 00 + Γiji T 0j ∼ (g00,0 + g0i,i + gij,0 )ρ0 + (g00,i + gi0,0 + gij,k )ϵρ0 + (g0i,j + gij,0 )ϵ2 ρ0 ∼ (ϵg00 + g0i + ϵgij )ρ0,i + (g00 + ϵgi0 + gij )ϵρ0,i + (g0i + ϵgij )ϵ2 ρ0,i .

(39.2.2)

To calculate T 0α;α with accuracy ϵ2N +1 ρ0,j , one must know g00 to ϵ2N , g0k to ϵ2N +1 , and gjk to ϵ2N . Thirdly, we have T jα;α = T jα,α + Γj βα T βα + Γαβα T jβ (39.2.3) Notice that Γj 00 T 00 ∼ (g00,j + g0j,0 )ρ0 ∼ (g00 + ϵg0j )ρ0,j . To calculate T jα;α with accuracy ϵ2N +2 ρ0,j , one must know g00 to ϵ2N +2 and g0j to ϵ2N +1 . In conclusion, to calculate T 0α;α with accuracy ϵ2N +1 ρ0,j and T jα;α with accuracy ϵ2N +2 ρ0,j , one must know g00 to ϵ2N +2 , g0k to ϵ2N +1 , and gjk to ϵ2N .

39.3

NEWTONIAN APPROXIMATION

(a) Derive equations (39.13) for the components of the stress-energy tensor in the PPN coordinate frame. (b) Show that, in the PPN coordinate frame, T 0α;α reduces to equation (39.15a), and T jα;α , when combined with (39.15a), reduces to equation (39.15b).

Solution: (a) In the rest frame of the matter, we have Tˆ0ˆ0 = ρ, Tˆ0ˆi = 0 and Tˆiˆj = tˆiˆj . Lorentz-transform these components by a pure boost with ordinary velocity −vj to obtain Tαβ . The boost matrix is (

γ γvi

γvi δij + (γ − 1)ni nj

394

) (39.3.1)

So we have ( )( )( ) ρ 0 γ γvni γ γvni αβ T = 0 tˆiˆj . γvni δij + (γ − 1)ni nj γvni δij + (γ − 1)ni nj ( ) 2 2 γ (ρ + v ni tˆiˆj nj ) γ 2 ρvni + γvnk tkˆˆi + γ(γ − 1)vnk tkˆˆl nl ni = γ 2 ρvni + γvnk tkˆˆi + γ(γ − 1)vnk tkˆˆl nl ni γ 2 ρv 2 ni nj + tˆiˆj + 2(γ − 1)ni nk tkˆˆj + (γ − 1)2 nk tkˆˆl nl ni nj (39.3.2) Notice that

( ) ρ = ρ 0 + O ρ 0 ϵ2 ,

v ∼ O(ϵ),

( ) 1 γ = 1 + v 2 + O ϵ4 , 2

( ) tˆiˆj ∼ O ρ0 ϵ2 .

(39.3.3)

We can get ( ) T 00 = ρ0 + O ρ0 ϵ2 ,

( ) T 0i = ρ0 vi + O ρ0 ϵ3 ,

( ) T ij = tˆiˆj + ρ0 vi vj + O ρ0 ϵ4 .

(39.3.4)

( ) (b) To the order of O ϵ2 , we have T 0α;α = T 00,0 + T 0i,i + 2Γ000 T 00 + 2Γ00i T 0i + Γ0ij T ij + Γ0j0 T 0j + Γi0i T 00 + Γiji T 0j ( ) = T 00,0 + T 0i,i + 2Γ000 T 00 + +Γi0i T 00 + O ϵ2 ( ) ∂ρ0 ∂ρ0 vi (39.3.5) = + + O ϵ2 . i ∂t ∂x ( ) To the order of O ϵ3 , we also have T jα;α = T jα,α + Γj βα T βα + Γαβα T jβ ( ) = T j0,0 + T ji,i + Γj 00 T 00 + O ϵ3 ( ) 1 = T j0,0 + T ji,i − g00,j T 00 + O ϵ3 2 ∂tˆjˆi + ρ0 vi vj ( ) ∂ρ0 vj ∂U = + − ρ0 j + O ϵ3 ∂t ∂xi ∂x ) ( ∂tˆjˆi ( ) ∂ρ0 vi ∂vj ∂vj ∂U ∂ρ0 + + ρ0 + ρ 0 vi i + − ρ 0 j + O ϵ3 = vj i i ∂t ∂x ∂t ∂x ∂x ∂x ∂t ( ) ˆ dvj ∂U jˆi = ρ0 + − ρ 0 j + O ϵ3 . dt ∂xi ∂x

39.4

(39.3.6)

A USEFUL FORMULA

Derive from equations (39.15) the following useful formula, valid for any function f (x, t): d dt

∫

∫ ρ0 (x, t)f (x, t) d3 x =

ρ0 (x, t)

( ) df (x, t) 3 d x + fractional errors of O ϵ2 . dt

(39.4.1)

Here both integrals are extended over all of space; and df /dt is the derivative following the matter (39.16).

Solution:

395

d dt

∫ ∫

= ∫ = ∫ = ∫ = ∫ =

39.5

ρ0 (x, t)f (x, t) d3 x [ ] ∂ρ0 (x, t) ∂f (x, t) 3 f (x, t) + ρ0 (x, t) d x ∂t ∂t [ ] ( ) ∂ρ0 (x, t)v i ∂f (x, t) 3 − f (x, t) + ρ (x, t) d x + fractional errors of O ϵ2 0 i ∂x ∂t ] [ i ( ) ∂v i ∂f (x, t) 3 ∂ρ0 (x, t)f (x, t)v + ρ0 (x, t)f (x, t) i + ρ0 (x, t) d x + fractional errors of O ϵ2 − i ∂x ∂x ∂t I ( ) df (x, t) 3 ρ0 (x, t) d x− ρ0 (x, t)f (x, t)v · dS + fractional errors of O ϵ2 dt rS →∞ ( ) df (x, t) 3 d x + fractional errors of O ϵ2 . ρ0 (x, t) (39.4.2) dt

STRESS TENSOR FOR NEWTONIAN GRAVITATIONAL FIELD

Define a “stress tensor for the Newtonian gravitational field U ” as follows: ( ) 1 1 U,j U,k − δjk U,l U,l . tjk = 4π 2

(39.5.1)

Show that the equations of motion for the matter (39.15b) can be rewritten in the forms ∂tjk + tˆjˆi ( ) dvj =− + fractional errors of O ϵ2 k dt ∂x ( ) (ρ0 vj ),t + (tjk + tˆj kˆ + ρ0 vj vk ),k = 0 + fractional errors of O ϵ2 . ρ0

(39.5.2) (39.5.3)

Solution: Firstly, we have tjk,k =

1 1 (U,jk U,k + U,j U,kk − δjk U,lk U,l ) = U,j U,kk = −ρ0 U,j . 4π 4π

(39.5.4)

So

∂tˆj kˆ ( ) dvj ∂U + − ρ0 j = fractional errors of O ϵ2 k dt ∂x ∂x is equivalent to 39.5.2. From exercise 39.3, we have ρ0

∂tˆjˆi + ρ0 vi vj ∂tˆjˆi ( ) ∂ρ0 vj dvj = ρ0 + O ϵ3 . + + i i ∂t ∂x dt ∂x

(39.5.5)

(39.5.6)

So we can get 39.5.3.

39.6

NEWTONIAN VIRIAL THEOREMS

(a) From equation 39.5.3 show that ∫ ( ) d2 Ijk = 2 (tjk + tˆjˆi + ρ0 vj vk ) d3 x + fractional errors of O ϵ2 , 2 dt

396

(39.6.1)

where Ijk is the second moment of the system’s mass distribution, ∫ Ijk = ρ0 xj xk d3 x .

(39.6.2)

This is called the “time-dependent tensor virial theorem”. (b) From this infer that, if ⟨ ⟩long time longtime denotes an average over a long period of time, then ⟨∫

⟩

(∫ =O

3

(tjk + tˆjˆi + ρ0 vj vk ) d x

) 4

3

ρ0 ϵ d x .

(39.6.3)

long time

This is called the “tensor virial theorem”. (c) By contraction of indices and use of equations (39.18), (39.14a), and (39.5e), derive the (ordinary) virial theorems: ∫ ∫ ∫ ( ) 1 d2 I 1 3 2 3 = ρ0 v d x − ρ0 U d x + 3 p d3 x + fractional errors of O ϵ2 , (39.6.4) 2 dt2 2 where I is the trace of the second moment of the mass distribution ∫ I = Ijj = ρ0 r2 d3 x ; and

⟨∫

∫ ρ0 v 2 d3 x −

1 ρ0 U d3 x + 3 2

(39.6.5) (∫

⟩

∫

=O

p d3 x

) ρ0 ϵ4 d3 x .

(39.6.6)

long time

Solution: (a) Firstly, we have d2 Ijk d = dt2 dt

∫ ρ˙0 xj xk d3 x .

Secondly, we have ∫ ∫ ∫ ∫ ∂ρ0 vi ∂xj ∂xk ρ˙0 xj xk d3 x = − xj xk d3 x = ρ0 vi xk d3 x + ρ0 vi xj d3 x ∂xi ∂xi ∂xi ∫ ∫ 3 = ρ0 vj xk d x + ρ0 vk xj d3 x Thus we have

d2 Ijk =2 dt2

∫

(39.6.7)

(39.6.8)

∫ 3

(ρ0 vj ),t xk d x + 2

(ρ0 vk ),t xj d3 x .

(39.6.9)

Using 39.5.3, we have ∫ ∫ ∫ (ρ0 vj ),t xk d3 x = − (tji + tˆjˆi + ρ0 vj vi ),i xk d3 x = (tji + tˆjˆi + ρ0 vj vi )xk,i d3 x ∫ = (tjk + tˆj kˆ + ρ0 vj vk ) d3 x (39.6.10) Now we obtain equation 39.6.1. (b) Since tjk + tˆjˆi + ρ0 vj vk ∼ ρ0 ϵ2 , equation 39.6.1 is equivalent to d2 Ijk =2 dt2

∫ [

( )] tjk + tˆjˆi + ρ0 vj vk + O ρ0 ϵ4 d3 x .

397

(39.6.11)

So we have ⟨∫ ⟩ (tjk + tˆjˆi + ρ0 vj vk ) d3 x

⟨ = long time

d2 Ijk +O dt2

(∫

⟩ ) ρ0 ϵ4 d3 x

(∫ =O

) ρ0 ϵ4 d3 x

long time

(39.6.12) (c)Firstly, we have ( ) 1 1 1 1 1 1 1 tii = U,i U,i − δii U,l U,l = − U,i U,i = U U,ii − (U U,i ),i = − U ρ0 − (U U,i ),i (39.6.13) 4π 2 8π 8π 8π 2 8π and tˆiˆi = 3p.

(39.6.14)

By contracting equation 39.6.1, we have d2 I =2 dt2

∫

( ) 1 (− U ρ0 + 3p + ρ0 v 2 ) d3 x + fractional errors of O ϵ2 , 2

(39.6.15)

i.e., equation 39.6.4. By contracting equation 39.6.12, we have 39.6.6.

39.7

PULSATION FREQUENCY FOR NEWTONIAN STAR

Use the ordinary, time-dependent virial theorem 39.6.4 to derive the following equation for the fundamental angular frequency of pulsation of a nonrotating, Newtonian star: |star’s self-gravitational energy| ; (trace of second moment of star’s mass distribution) ∫ 3 ¯ 1 = pressure-weihted average of adiabatic index ≡ ∫Γ1 p d x . Γ p d3 x

¯ 1 − 4) ω 2 = (3Γ

(39.7.1) (39.7.2)

In the derivation assume that the pulsations are “homologous” -i.e., that a fluid element with equilibrium position xj (relative to center of mass xj = 0) gets displaced to xj + ξ j (x, t), where ξ j = (small constant)xj e−iωt .

(39.7.3)

Assume nothing else.

Solution: Suppose ξ = sr cos ωt. For an mass element at r, it will move to r(1 + s cos ωt). The density of the mass element will be ρ0 (1+s cos ωt)−3 . The gravitational potential at the mass element will be U (1+s cos ωt)−1 . For adiabatic process, we have p ∝ ρΓ1 . So the pressure at the mass element will become p(1+s cos ωt)−3Γ1 . / For a static star, we have d2 I dt2 = 0. Thus ∫ 3

∫ p d3 x =

1 ρ0 U d3 x ≡ |Ω|, 2

(39.7.4)

where Ω is star’s self-gravitational energy. For a pulsing star, the trace of second moment of star’s mass distribution is ∫ ′ I = ρ0 (1 + s cos ωt)−3 (1 + s cos ωt)5 r4 dr dΩ = (1 + s cos ωt)2 I ≈ (1 + 2s cos ωt)I

398

(39.7.5)

Star’s self-gravitational energy is Ω′ =

∫

ρ0 (1+s cos ωt)−3 U (1+s cos ωt)−1 (1+s cos ωt)−3 r2 dr dΩ = −(1+s cos ωt)−1 |Ω| ≈ −(1−s cos ωt)|Ω|. (39.7.6)

The pressure integral is ∫ 3

¯ 1 )s cos ωt]|Ω|. (39.7.7) p(1+s cos ωt)−3Γ1 (1+s cos ωt)3 r2 dr dΩ = (1+s cos ωt)3−3Γ1 |Ω| ≈ [1+(3−3Γ ¯

Note that kinetic energy

∫

ρ0 v 2 d3 r vanishes to the first order of s. So we have ¯ 1 )s cos ωt. − ω 2 Is cos ωt = |Ω|s cos ωt + (3 − 3Γ

(39.7.8)

It follows that ¯ 1 − 4) ω 2 = (3Γ

39.8

|Ω| . I

(39.7.9)

ABSENCE OF “METRIC-GENERATES-METRIC” TERMS IN POSTNEWTONIAN LIMIT

In writing down the post-Newtonian metric corrections, one might be tempted to include terms that are generated by the Newtonian potential acting alone, without any direct aid from the matter. After all, general relativity and other metric theories are nonlinear; so the two-step process (matter) → U → (postNewtonian metric corrections) seems quite natural. Show that such terms are not needed, because the equations (39.14)-(39.16) of the Newtonian approximation enable one to reexpress them in terms of direct integrals over the matter distribution. In particular, show that ∫

∂ 2 U (x′ , t) d3 x′ = 2π[Vj (x, t) − Wj (x, t)], ∂x′j ∂t |x − x′ |

(39.8.1)

where Vj and Wj are defined by equations (39.23c,d); also show that ∫

∂U (x′ , t) ∂U (x′ , t) 1 d3 x′ = −2π[U (x, t)]2 + 4π ′ ′ ∂xj ∂xj |x − x′ |

∫

ρ0 (x′ , t)U (x′ , t) 3 ′ d x . |x − x′ |

(39.8.2)

Note that the terms on the righthand sides of 39.8.1 and 39.8.2 are already included in the expressions (39.23e,f) for g0j and g00 .

Solution: (a) Firstly, we have ∫

(∫ ) ρ0 (x′′ , t) 3 ′′ d3 x′ ∂2 d x ′ ′ ′′ ∂xj ∂t |x − x | |x − x′ | ( ) ∫ 1 1 ∂ρ0 (x′′ , t)vk (x′′ , t) ∂ =− d3 x′′ d3 x′ ∂x′′k ∂x′j |x′ − x′′ | |x − x′ | ( ) ∫ ∂2 1 1 ′′ ′′ = − ρ0 (x , t)vk (x , t) ′ ′ d3 x′′ d3 x′ ′ ′′ ∂xk ∂xj |x − x | |x − x′ |

∂ 2 U (x′ , t) d3 x′ = ∂x′j ∂t |x − x′ |

∫

(39.8.3)

399

Secondly, we notice that ∫

R

r2 dr dΩ 0

∫

ra

=

dr dΩ 0 ∞ ∑

l ∑

∫

1 r|r − ra |

r + |r − ra | ra

∫

R

dr dΩ ∫

ra

r |r − ra |

rl+1 4π ∗ dΩ Ylm (θ, ϕ)Ylm (θa , ϕa ) l+1 2l + 1 r a l=0 m=−l 0 ∫ R ∫ ∞ ∑ l ∑ rl 4π ∗ + dr al Ylm (θ, ϕ)Ylm (θa , ϕa ) dΩ r 2l + 1 ra =

dr

l=0 m=−l

∫ R ∞ ∑ l ∑ rl+1 rl 4πδ + dr al 4πδl0 l0 l+1 r ra l=0 m=−l 0 l=0 m=−l ra ∫ R ∫ ra r + 4π dr = 4π dr r a ra 0 =

∫ ∞ ∑ l ∑

ra

dr

= 4πR − 2πra . So we have

(39.8.4)

) ) ( ( ∫ 1 1 ∂2 ∂2 1 1 3 ′ d x = d3 x′ ∂x′k ∂x′j |x′ − x′′ | |x − x′ | |x′ − x′′ | ∂x′k ∂x′j |x − x′ | ∫ ∫ 1 ∂2 1 ∂2 1 1 3 ′ d x = d3 x′ = ∂xk ∂xj |x′ − x′′ | |x′ − x| ∂xk ∂xj |x′ | |x′ − (x − x′′ )| ( ) (xj − x′′j )(xk − x′′k ) ∂2 δij ′′ − = (−2π|x − x |) = −2π ∂xk ∂xj |x − x′′ | |x − x′′ |3 ∫

(39.8.5)

It follows that ∫ 2 ∫ ∫ ρ0 (x′′ , t)vk (x′′ , t)(xk − x′′k )(xj − x′′j ) 3 ′′ ∂ U (x′ , t) d3 x′ ρ0 (x′′ , t)vj (x′′ , t) 3 ′′ = 2π d x − 2π d x ∂x′j ∂t |x − x′ | |x′ − x′′ | |x − x′′ |3 = 2π[Vj (x, t) − Wj (x, t)]

(39.8.6)

(b) ∫

∂U (x′ , t) ∂U (x′ , t) 1 d3 x′ ′ ′ ∂xj ∂xj |x − x′ | ( ) ∫ ∂U (x′ , t) 1 ∂ ′ = − U (x , t) ′ d3 x′ ∂xj ∂x′j |x − x′ | ( )] ∫ [ 2 ′ ′ ∂ U (x , t) 1 ∂ 1 ∂U (x , t) =− U (x′ , t) + U (x′ , t) d3 x′ ∂x′j ∂x′j |x − x′ | ∂x′j ∂x′j |x − x′ | ( )] ∫ [ 1 1 1 ∂U 2 (x′ , t) ∂ ′ ′ =− U (x , t)(−4πρ0 (x , t)) + d3 x′ |x − x′ | 2 ∂x′j ∂x′j |x − x′ | ( )] ∫ [ 1 4πρ0 (x′ , t)U (x′ , t) 1 2 ′ ∂2 d3 x′ = + U (x , t) ′ ′ |x − x′ | 2 ∂xj ∂xj |x − x′ | ] ∫ [ 4πρ0 (x′ , t)U (x′ , t) 2 ′ ′ = − 2πU (x , t)δ(x − x ) d3 x′ |x − x′ | ∫ ρ0 (x′ , t)U (x′ , t) 3 ′ = −2π[U (x, t)]2 + 4π d x . |x − x′ |

400

(39.8.7)

39.9

REMOVAL OF Σ TERM FROM g00

Show that the coordinate transformation (39.25) removes the Σ term from the metric coefficient g00 of equation (39.24), as claimed in the text.

Solution: After coordinate transformation, we have gα′ β ′ (xnew ) = gαβ (xold )

∂xα ∂xβ . ∂xα′ ∂xβ ′

(39.9.1)

From (39.25), we have 1 x0new = x0old − Σ 2 And so ∂x0old = ∂x0new

(

∫

ρ0 (x′ , t)vj (x′ , t)(xj − x′j ) 3 ′ d x . |x − x′ |

1 1− Σ 2

∫

(ρ0 vj ),t (xj − x′j ) 3 ′ d x |x − x′ |

(39.9.2)

)−1 .

(39.9.3)

Notice that ∫

∫ (tji + tˆjˆi + ρ0 vj vi ),i (xj − x′j ) 3 ′ (ρ0 vj ),t (xj − x′j ) 3 ′ d x = − d x |x − x′ | |x − x′ | ( ) ∫ ( ) ∂ (xj − x′j ) = (tji + tˆjˆi + ρ0 vj vi ) ′ d3 x′ ∼ O ϵ4 . ′ ∂xi |x − x | ( 4) To the order of O ϵ , we have new old g00 (xnew ) = g00 (xold )

∂x0old ∂x0old ∂x0new ∂x0new

(39.9.4)

∫

( ) (ρ0 vj ),t (xj − x′j ) 3 ′ d x + O ϵ6 ′ |x − x | ∫ ∫ ′ ′ ( ) (ρ0 vj ),t (xj − xj ) 3 ′ ρ0 vj (xj − xj ) 3 ′ d x −Σ d x + O ϵ6 = g00 + U,t Σ (39.9.5) |x − x′ | |x − x′ | = g00 (xnew ) + g00,0 (xnew )(told − tnew ) + g00 (xnew )Σ

Notice that

( ∫ ) ( ) ρ0 vj (xj − x′j ) 3 ′ ρ0 vj (xj − x′j ) 3 ′ ∂ d x ∼ U Σ d x ∼ O ϵ6 ′ ′ |x − x | ∂t |x − x |

∫ U,t Σ

Thus we have

∫ new g00 (xnew ) = g00 − Σ

( ) (ρ0 vj ),t (xj − x′j ) 3 ′ d x + O ϵ6 . ′ |x − x |

(39.9.6)

(39.9.7)

Using ∂tˆjˆi + ρ0 vi vj ∂U ∂ρ0 vj − ρ0 j = 0, + ∂t ∂xi ∂x

(39.9.8)

we have ∫

new g00

∫ (tˆj kˆ + ρ0 vk vj ),k (xj − x′j ) 3 ′ ( ) ρ0 U,j (xj − x′j ) 3 ′ = −Σ d x + Σ d x + O ϵ6 |x − x′ | |x − x′ | ( ) ∫ ∫ ( ) (xj − x′j ) ρ0 U,j (xj − x′j ) 3 ′ ∂ old = g00 − Σ d x − Σ (tˆj kˆ + ρ0 vj vk ) ′ d3 x′ + O ϵ6 ′ ′ |x − x | ∂xk |x − x | ∫ ′ ρ U (x − x ) 0 ,j j j old = g00 −Σ d3 x′ |x − x′ | ] [ ∫ ( ) (xj − x′j )(xk − x′k ) 3 ′ δjk d x + O ϵ6 (39.9.9) + Σ (tˆj kˆ + ρ0 vj vk ) − ′ ′ 3 |x − x | |x − x | old g00

401

Notice that the last term can be absorbed into (39.23g,h,i), and ∫ ∫ ∫ ρ0 (x′ , t)U,j (x′ , t)(xj − x′j ) 3 ′ ρ (x′ , t)(xj − x′j ) ρ (x′′ , t)(x′j − x′′j ) 3 ′ 0 3 ′′ 0 −Σ d x = Σ d x d x |x − x′ | |x − x′ | |x′ − x′′ |3 ∫ ∫ ′ ′′ ′ ′ ρ0 (x , t)ρ0 (x , t)[(x − x ) · (x − x′′ )] 3 ′ 3 ′′ =Σ d x d x . |x − x′ ||x′ − x′′ |3 (39.9.10) We conclude that the coordinate transformation (39.25) removes the Σ term from the metric coefficient g00 of equation (39.24).

39.10

VERIFICATION OF FORMS OF POST-NEWTONIAN CORRECTIONS

Verify the claims in the text immediately preceding equations (39.23a,b,c,f).

Solution: The elements we can use to construct post-Newtonian correction of metric is ( ) ( ) ( ) ρ0 , ρ0 Π ∼ O ρ0 ϵ2 , tˆiˆj ∼ O ρ0 ϵ2 , p =∼ O ρ0 ϵ2 , v ∼ O(ϵ) and ∫ X≡

( ) ( 2 −1 ) d3 x′ 1 ∼ O ϵ ρ ∼ O , 0 ′ |x − x | r

∫ Y≡

d3 x′

(39.10.1)

( ) ( 2 −1 ) 1 (x − x′ )(x − x′ ) ∼ O ϵ ρ ∼ O . 0 ′ 3 |x − x | r (39.10.2)

Claims preceding equations (29.23a) ( ) The corrections to the spatial components are O ϵ2 . The only choice is the combination of X or Y with ρ0 . So we have equation (39.23a). Claims preceding equations (29.23b) For coordinate transformation xinew = xiold + Γ

∂χ ∂xi

∫ where

χ(x, t) = −

ρ0 (x′ , t)|x − x′ | d3 x′ .

(39.10.3)

We have ∂xiold ∂xjnew

( ) ∂2χ = δij − Γ i j + O ϵ4 = δij + Γ ∂x ∂x ( ) = δij + Γ(U δij − Uij ) + O ϵ4 .

∫

′

ρ0 (x , t)

(

(xi − x′i )(xj − x′j ) δij − |x − x′ | |x − x′ |

)

( ) d3 x′ + O ϵ4 (39.10.4)

It follows that new old gij (xnew ) = gkl (xold )

∂xkold ∂xlold ∂xjnew

∂xjnew

( ) old = gij (xold ) + 2Γ(U δij − Uij ) + O ϵ4 .

Notice that old old old gij (xold ) − gij (xnew ) ∼ gij,k (xkold − xinew ) ∼ U,i

So we have

( ) ∂χ ∼ O ϵ4 . j ∂x

( ) new old gij (xnew ) = gij (xnew ) + 2Γ(U δij − Uij ) + O ϵ4 .

402

(39.10.5)

(39.10.6) (39.10.7)

old The term 2ΓUij in gij will be eliminated.

Claims preceding equations (29.23c) ( ) The corrections to the time-spatial components are O ϵ3 . The only choice is the combination of X or Y with ρ0 v. So we have equation (39.23cd). Claims preceding equations (29.23f) ( ) The corrections to the 00 components are O ϵ4 . The possible choice can be the combination of v 2 , ρΠ, p, U with ρ0 X or the combination of vi vj , tˆiˆj with Yij . One more possible chose is U 2 .

39.11

TRANSFORMATION TO MOVING FRAME

Show that the change of coordinates (39.29) changes the PPN metric coefficients from the form (39.23) to the form (39.32).

Solution: From the definition of χ(x, t), we have ∫ ( ) ∂ρ0 (x′ , t) xj − x′j 3 ′ ∂ρ0 v i xj − x′j 3 ′ d x = d x + fractional errors of O ϵ2 ′ ′i ′ ∂t |x − x | ∂x |x − x | ( ∫ ′ ) ( ) xj − xj ∂ = − ρ0 v i ′i d3 x′ + fractional errors of O ϵ2 ∂x |x − x′ | [ ] ∫ ( ) (xi − x′i )(xj − x′j ) 3 ′ δij i − = ρ0 v d x + fractional errors of O ϵ2 ′ ′ |x − x | |x − x | ( ) (39.11.1) = V j − W j + O ϵ5 .

∂2χ =− ∂xj ∂t

∫

Similarly, we also have ∂2χ = −U δjk + Ujk . ∂xj ∂xk

(39.11.2)

From equation (39.23), we have ( ) ( ) 3 1 ∂told 1 ∆2 + ζ − 1 wj (Vj − Wj ) + O ϵ6 = 1 + w2 + w4 + ∂tnew 2 8 2 ( ) ( ) 1 2 ∂told 1 = (1 + w )w + ∆ + ζ − 1 (Ujk wk − U wj ) + O ϵ5 j 2 j 2 2 ∂xnew j ( ) ∂xold 1 = (1 + w2 )wj + O ϵ5 ∂tnew 2 ( ) ∂xjold 1 = δkj + wj wk + O ϵ4 . k ∂xnew 2

(39.11.3)

The transformation of the metric is new old gαβ (xnew ) = gµν (xold )

∂xµold ∂xνold . β ∂xα new ∂xnew

(39.11.4)

Keep firmly in mind the fact that the potentials U , Vj , Wj , A and D are not scalar fields. Each coordinate

403

system possesses its own potentials. We have [

] ( ) 1 j k Uold (xold , told ) = Unew − w − + w w χ,jk + O ϵ6 2 xnew ,tnew ( new ) ( 5) old Vj (xold , told ) = Vj + wi Unew x ,t + O ϵ new new ) ( ) ( old new i new Wj (xold , told ) = Wj + w Uij x ,t + O ϵ5 new new ( ) ( ) Φold (xold , told ) = Φnew + 2β1 wi Vinew + β1 w2 Unew xnew ,tnew + O ϵ6 ( ) ( ) new Aold (xold , told ) = Anew + 2wi Winew + wj wk Ujk + O ϵ6 xnew ,tnew ( ) Dold (xold , told ) = Dnew (xnew , tnew ) + O ϵ6 j

(Vjnew

Wjnew )

(39.11.5)

The metric in new coordinates will be ∂xj ∂xkold ∂told ∂xjold ∂told ∂told old old + 2g0j (xold ) + gjk (xold ) old ∂tnew ∂tnew ∂tnew ∂tnew ∂tnew ∂tnew [ ( ) ] 1 2 2 4 j new new = (−1 + 2Uold − 2βUold + 4Ψold − ζAold − ηDold ) × 1 + w + w + 2 ∆2 + ζ − 1 w (Vj − Wj ) 2 ( ) ( ) 7 1 + 2 − ∆1 Vjold − ∆2 Wjold wj + δjk (1 + 2γUold )(1 + w2 )wj wk + O ϵ6 2 2 ) ] [ ( 1 j new new 2 4 ∆2 + ζ − 1 w (Vj − Wj ) + 2Unew (1 + w2 ) =− 1+w +w +2 2 [ ] ( ) 1 j k j new new 2 + 2 −w (Vj − Wj ) + w w χ,jk − 2β 2 Unew + 4 Φnew + 2β1 wi Vinew + β1 w2 Unew 2 ) ( ( ) 1 7 new new i new j k new − ∆2 W j wj − ζ Anew + 2w Wi + w w Ujk − ηDnew + 2 − ∆1 Vj 2 2 ( ) ( ) 1 7 k new + 2 − ∆1 wj Unew − ∆2 w Ukj wj + (1 + w2 )w2 + 2γUnew w2 + O ϵ6 2 2

new old g00 (xnew ) = g00 (xold )

2 = (−1 + 2Unew − 2βUnew + 4Ψnew − ζAnew − ηDnew ) + (2 − 2ζ + 8β1 − 7∆1 − ∆2 )wj Vjnew ( ) new j k + (1 + 2γ + 4β1 − 7∆2 )w2 Unew + (1 − ζ − ∆2 )Ujk w w + O ϵ6 . (39.11.6)

∂told ∂xkold ∂told ∂xkold ∂told ∂told ∂xk ∂xl old old old + g0k (xold ) + g0k (xold ) j + gkl (xold ) old jold j j ∂tnew ∂xnew ∂tnew ∂xnew ∂tnew ∂tnew ∂xnew ∂x ( )[ ( ) new ] 1 2 1 2 1 new k = (−1 + 2Uold ) 1 + w (1 + w )wj + ∆2 + ζ − 1 (Ujk w − Unew wj ) 2 2 2 ( ) ( ) ( ) 7 1 1 1 + − ∆1 Vkold − ∆2 Wkold δjk + δkl (1 + 2γUold )(1 + w2 )wk δjl + wl wj + O ϵ5 2 2 2 2 [ ( ) ] ( ) 1 7 1 new k = − (1 + w2 )wj + ∆2 + ζ − 1 (Ujk w − Unew wj ) + 2Unew wj + − ∆1 Vjnew − ∆2 Wjnew 2 2 2 ( ) ( ) ) ( 7 1 1 1 new + − ∆1 wj Unew − ∆2 wk Ukj + wk δjk + wk wj + 2γUnew wj + w2 wj + O ϵ5 2 2 2 2 ( ) ( ) ( ) 7 1 1 7 new new k new = − ∆1 Vj − ∆2 W j + (1 − ζ − ∆2 )w Ukj + 1 + 2γ + ζ + ∆2 − ∆1 wj Unew + O ϵ5 . 2 2 2 2 (39.11.7)

new old g0j (xnew ) = g00 (xold )

404

( ) ∂told ∂told ∂xi ∂xlold old + gil (xold ) jold + O ϵ4 j k k ∂xnew ∂xnew ∂xnew ∂xnew )( ) ( ( ) 1 1 = −wj wk + δil (1 + 2γUold ) δji + wi wj δkl + wl wk + O ϵ4 2 2 ( 4) = δjk (1 + 2γUnew ) + O ϵ .

new old gjk (xnew ) = g00 (xold )

(39.11.8)

Now we obtain the components for the metric in the new PPN coordinates.

39.12

THE TRANSFORMATION BETWEEN COMOVING FRAME AND PPN FRAME

Carry out the details of the derivation of the transformation matrix (39.41); and in the process calculate ( ) ˆ the correction of O ϵ4 to A0ˆ0 .

Solution: Firstly, we have the ˆ

ω α˜ = Aα˜ βˆ ω β . (

The transformation matrix is Aα˜ βˆ =

(39.12.1)

) γ˜ γ˜ v˜i , γ˜ v˜i δij + (˜ γ − 1)n˜i n˜i

(39.12.2)

( ) where v˜i = vi (1 + U + γU ) + O ϵ5 , γ˜ = (1 − v˜i2 )−1/2 and n˜i = v˜i /|˜ v |. From equation (39.38), we have ˜

dxα = Aαβ˜ ω β , where Aαβˆ =

(

( 4) 1 + U (+ O ϵ ) O ϵ5

(39.12.3)

( ) ( )) − 27 ∆1 Vj − 12 ∆2 Wj − 12 α1 − α2 wj(U )− α2 wk Ukj + O ϵ5 (1 − γU )δjk + O ϵ4

(39.12.4)

ˆ

So for dxα = Aαβˆ ω β , we have ( ) ( ) 1 ˜ A0ˆ0 = A0β˜ Aβ ˆ0 = (1 + U )˜ γ + O ϵ4 = 1 + U + v 2 + O ϵ4 . 2

(39.12.5)

( ) ˜ ˜ A0ˆi = A0β˜ Aβˆi = (1 + U )˜ γ v˜i + A0˜j Ajˆi + O ϵ5 ( ) ( ) ( ) 1 2 7 1 1 = 1 + v + γU + 2U vi − ∆1 Vi − ∆2 Wi − α1 − α2 wi U − α2 wj Uji + O ϵ5 (39.12.6) 2 2 2 2 ( ) ( ) ˜ ˜ γ v˜i + O ϵ5 = Aiˆ0 = Aiβ˜ Aβ ˆ0 = Ai˜j Aj ˆ0 + O ϵ5 = (1 − γU )˜

(

1 1 + v2 + U 2

)

( ) + O ϵ5

[ ] ( ) ( ) 1 ˜ Aiˆj = Aiβ˜ Aβˆj = (1 − γU )δik δkj + (˜ γ − 1)nk˜ n˜j + O ϵ4 = (1 − γU ) δij + vi vj + O ϵ4 2

(39.12.7)

(39.12.8)

( ) ˆ The calculation of the correction of O ϵ4 to A0ˆ0 is similar but more tedious. I do not perform the calculation here.

405

39.13

EQUATIONS OF MOTION

Carry out the details of the derivation of the equations of motion (39.44), (39.46), and (39.47). As part of the derivation, calculate the following values of the Christoffel symbols in the PPN coordinate frame: ) ( ( ) Γ000 = −U,t + O U,j ϵ3 , Γ00j = −U,j + O U,j ϵ2 , ( ) ( ) 7 1 1 0 Γ jk = γU,t δjk + ∆1 V(j,k) + ∆2 W(j,k) + α1 − α2 w(j U,k) + α2 wi Ui(j,k) + O U,j ϵ3 2 2 2 [ ] 1 1 1 1 1 j 2 2 Γ 00 = −U,j + (β + γ)U − 2Ψ + ζA + ηD + (α1 − α2 − α3 )w U + (α1 − 2α3 )wi Vi + α2 wi wk Uik 2 2 2 2 2 ,j ( ) ( ) 7 1 1 − ∆1 Vj,t − ∆2 Wj,t + α2 − α1 wj U,t − α2 wi Uij,t + O U,j ϵ4 2 2 2 ( ) ( ) 7 1 1 Γj 0k = γU,t δjk − ∆1 + ∆2 V[j,k] − α1 w[j U,k] + O U,j ϵ3 2 2 2 ( ) j 2 (39.13.1) Γ kl = −γ(U,j δkl − 2U,(k δl)j ) + O U,j ϵ . Here square brackets on tensor indices denote antisymmetrization, and round brackets denote symmetrization. As part of the derivation, it may be useful to prove and use the relations ∫ χ(t, x) =

ρ0 (t, x′ )|x − x′ | d3 x′ ,

(39.13.2a)

χ,jk = −δjk U + Ujk , ( ) χ,it = Vi − Wi + O ϵ5 ,

(39.13.2b) (39.13.2c) (39.13.3)

W[k,j] = V[k,j] . Here χ is the function originally defined in equation (39.29c).

Solution: The metric is ( ) g00 = −1 + 2U − 2βU 2 + 4Ψ − ζA − ηD + (α2 + α3 − α1 )w2 U + (2α3 − α1 )wj Vj − α2 Ujk wj wk + O ϵ6 , ( ) ( ) 1 1 7 g0j = − ∆1 Vj − ∆2 Wj − α2 wk Ukj + α2 − α1 wj U + O ϵ5 , 2 2 2 ( 4) (39.13.4) gjk = δjk (1 + 2γU ) + O ϵ . The inverse of the metric is ( ) g 00 = −1 − 2U − 4U 2 + 2βU 2 − 4Ψ + ζA + ηD − (α2 + α3 − α1 )w2 U − (2α3 − α1 )wj Vj − α2 Ujk wj wk + O ϵ6 , ( ) ( ) 7 1 1 0j k g = − ∆1 Vj − ∆2 Wj − α2 w Ukj + α2 − α1 wj U + O ϵ5 , 2 2 2 ( ) (39.13.5) g jk = δ jk (1 − 2γU ) + O ϵ4 . Christoffel symbols of the first kind are 1 1 1 g00,0 ∼ O(U,j ϵ), Γ00j = g00,j ∼ O(U,j ), Γ0jk = g0(j,k) − gjk,0 2 2 2 1 1 = gj0,0 − g00,j ∼ O(U,j ), Γj0k = g0[j,k] + gjk,0 ∼ O(U,j ϵ), Γjkl = 2 2

Γ000 =

∼ O(U,j ϵ)

Γj00

1 (gjk,l + gjl,k − gkl,j ) ∼ O(U,j ) 2 (39.13.6)

406

Christoffel symbols of the second kind are ( ( ) ) 1 Γ000 = g 00 Γ000 + g 0i Γi00 = −Γ000 + O U,j ϵ3 = − g00,0 + O U,j ϵ3 , 2 ( ) ( ) 1 0 00 0i 2 Γ 0j = g Γ00j + g Γi0j = −Γ00j + O U,j ϵ = − g00,j + O U,j ϵ2 , 2 ) 1 ( 0 00 0i 3 Γ jk = g Γ0jk + g Γijk = −Γ0jk + O U,j ϵ = gjk,0 − g0(j,k) , 2 ) ( ) ( 1 j j0 ji Γ 00 = g Γ000 + g Γi00 = (1 − 2γU )Γj00 + O U,j ϵ4 = gj0,0 − (1 − 2γU )g00,j + O U,j ϵ4 , 2 ( ) ( ) 1 j j0 ji 3 Γ 0k = g Γ00k + g Γi0k = Γj0k + O U,j ϵ = g0[j,k] + gjk,0 + O U,j ϵ3 , 2 ( ) 1 ( ) j j0 ji 2 Γ kl = g Γ0kl + g Γikl = Γjkl + O U,j ϵ = (gjk,l + gjl,k − gkl,j ) + O U,j ϵ2 . (39.13.7) 2 Substituting equation 39.13.4 into 39.13.7, we can derive the equation 39.13.1 easily. The only trick lies in the simplification of g0[j,k] . We need to derive equation 39.13.2 and 39.13.3 first. We have already derived the equation 39.13.2 in Exercise 39.11. As for 39.13.3, we have ∫ xk − x′k 3 ′ d x (39.13.8) Vj,k = − ρ0 (t, x′ )vj (t, x′ ) |x − x′ |3 and Wj,k

[

] δik (xj − x′j ) + δjk (xi − x′i ) (xi − x′i )(xj − x′j )(xk − x′k ) 3 ′ = ρ0 (t, x )vi (t, x ) −3 d x |x − x′ |3 |x − x′ |5 [ ] ∫ (xj − x′j )(xk − x′k ) 3 ′ (xi − x′i ) (39.13.9) = −Vk,j + ρ0 (t, x′ )vi (t, x′ ) δ − 3 d x . jk |x − x′ |3 |x − x′ |2 ∫

′

′

It follows that W[k,j] = V[k,j] . Now we have ( ) ( ) 1 1 7 g0[j,k] = − ∆1 V[j,k] − ∆2 W[j,k] − α2 wi Ui[j,k] + α2 − α1 w[j U,k] + O U,j ϵ3 2 2 2 ( ) ( ) 7 1 1 =− ∆1 + ∆2 V[j,k] − α1 w[j U,k] − α2 wi (Ui[j,k] − δi[j U,k] ) + O U,j ϵ3 2 2 2 ( ) ( ) 7 1 1 =− ∆1 + ∆2 V[j,k] − α1 w[j U,k] − α2 wi χ,i[jk] + O U,j ϵ3 2 2 2 ( ) ( ) 7 1 1 =− ∆1 + ∆2 V[j,k] − α1 w[j U,k] + O U,j ϵ3 2 2 2 (39.13.10) And then we can get the expression for Γj 0k . Using the law of conservation of baryon number, we have 1 √ (ρ0 uα );α = √ ( −gρ0 uα ),α = 0. −g

(39.13.11)

Since

( ( ) ) √ ( 4) ( ) 1 2 1 2 3 ρ0 u −g = ρ0 1 + v + U (1 − 2U )(1 + 2γU ) + O ρ0 ϵ = ρ0 1 + v + 3γU + O ρ0 ϵ4 , 2 2 (39.13.12) we would define ( ) 1 ρ∗ ≡ ρ0 1 + v 2 + 3γU , (39.13.13) 2 0√

and so we have ) ( √ −gρ0 u0 = ρ∗ + O ρ0 ϵ4 ,

√

−gρ0 ui =

√

407

) ( −gρ0 u0 v i = ρ∗ v i + O ρ0 ϵ5

(39.13.14)

The law of conservation of baryon number would be ( ) ρ∗,t + (ρ∗ v i ),i = O ρ0,j ϵ5 . The stress-energy tensor in the PPN coordinate frame is ( ) T 00 = ρ0 (1 + Π + v 2 + 2U ) + O ρ0 ϵ4 , ( 5) T 0j = ρ0 (1 + Π + v 2 + 2U )v j + tˆj m ˆ vm + O ρ 0 ϵ , ( 6) 1 T jk = tˆj kˆ (1 − 2γU ) + ρ0 (1 + Π + v 2 + 2U )vj vk + (vj tkˆm jm ˆ vm ) + O ρ 0 ϵ . ˆ vm + vk tˆ 2

(39.13.15)

(39.13.16)

The law of conservation of local energy-momentum gives T αβ;β = T αβ,β + Γαµβ T µβ + Γβ µβ T αµ = 0.

(39.13.17)

Notice that Γβ µβ = (ln

√

−g),µ =

( ) ( ) 1 [ln(1 − 2U )(1 + 2γU )3 ],µ + O U,µ ϵ2 = (3γ − 1)U,µ + O U,µ ϵ2 . 2

(39.13.18)

For α = 0, we have T 0β;β = T 00,0 + T 0j,j + Γ0µβ T µβ + Γβ µβ T 0µ = 0.

(39.13.19)

Γ0µβ T µβ = Γ000 T 00 + 2Γ00j T 0j + Γ0jk T jk ( ) = −U,t ρ0 − 2U,j ρ0 v j + O ρ0,j ϵ5

(39.13.20)

Since

and ( ) Γβ µβ T 0µ = (3γ − 1)U,µ T 0µ + O ρ0,j ϵ5

( ) = (3γ − 1)U,t ρ0 + (3γ − 1)U,j ρ0 v j + O ρ0,j ϵ5 ,

(39.13.21)

equation 39.13.19 becomes ( ) j 5 [ρ0 (1 + Π + v 2 + 2U )],t + [ρ0 (1 + Π + v 2 + 2U )v j + tˆj m ˆ vm ],j + (3γ − 2)U,t ρ0 + (3γ − 3)U,j ρ0 v = O ρ0,j ϵ . (39.13.22) Using ( ) ρ0,t + (ρ0 v j ),j = O ρ0,j ϵ3 , (39.13.23) we can get ( ) j O ρ0,j ϵ5 = ρ0,t + (ρ0 v j ),j + ρ0 (Π + v 2 + 2U ),t + ρ0 v j (Π + v 2 + 2U ),j + (tˆj m ˆ vm ),j + (3γ − 2)U,t ρ0 + (3γ − 3)U,j ρ0 v ( ) 1 dΠ = 1 + v 2 + 3γU [ρ0,t + (ρ0 v j ),j ] + ρ0 + ρ0 (v 2 + 3γU ),t + ρ0 v j (v 2 + 3γU − U ),j + (tˆj m ˆ vm ),j . 2 dt (39.13.24) Using

( ) 1 dv 2 j 5 ρ0 = ρ0 U,j v j − tˆj k, ˆk ˆ v + O ρ0,j ϵ , 2 dt

(39.13.25)

we can get ( ) O ρ0,j ϵ5 =

(

1 1 + v 2 + 3γU 2

) [ρ0,t + (ρ0 v j ),j ] + ρ0

dΠ + ρ0 (v 2 + 3γU ),t + ρ0 v j (v 2 + 3γU − U ),j dt

1 dv 2 j + tˆj m ˆ vm,j + ρ0 U,j v − ρ0 dt ( ) 2 ( ) ( ) 1 2 dΠ 1 2 1 2 j j = 1 + v + 3γU [ρ0,t + (ρ0 v ),j ] + ρ0 + ρ0 v + 3γU + ρ0 v v + 3γU + tˆj m ˆ vm,j 2 dt 2 2 ,t ,j = ρ∗,t + (ρ∗ v i ),i + ρ0 = ρ0

dΠ + tˆj m ˆ vm,j dt

dΠ + tˆj kˆ vj,k . dt

(39.13.26)

408

For α = j, we have T jβ;β = T j0,0 + T jk,k + Γj µβ T µβ + Γβ µβ T jµ = 0.

(39.13.27)

Using equation 39.13.22, we have T j0,0 + T jk,k = [ρ0 (1 + Π + v 2 + 2U )v j + tˆj m ˆ vm ],t

( ) 1 6 + [tˆj kˆ (1 − 2γU ) + ρ0 (1 + Π + v 2 + 2U )vj vk + (vj tkˆm jm ˆ vm )],k + O ρ0,j ϵ ˆ vm + vk tˆ 2 dv j k = ρ0 (1 + Π + v 2 + 2U ) − v j [(tkˆm ˆ vm ),k + (3γ − 2)U,t ρ0 + (3γ − 3)U,k ρ0 v ] dt ( ) 1 6 + [tˆj kˆ (1 − 2γU ) + (vj tkˆm (39.13.28) ˆ vk ),t + O ρ0,j ϵ . jm ˆ vm )],k + (tˆ jk ˆ vm + vk tˆ 2

Using tˆj kˆ vk,t + tˆj kˆ vm vk,m = tˆj kˆ U,k −

tˆj kˆ tkˆm,m ˆ ρ0

( ) + O ρ0,j ϵ5 ,

(39.13.29)

we have dv j + [tˆj kˆ (1 − 2γU )],k − v j [(3γ − 2)U,t ρ0 + (3γ − 3)U,k ρ0 v k ] + tˆj kˆ U,k dt tˆj kˆ tkˆm,m ( ) 1 1 ˆ + tˆj k,t + O ρ0,j ϵ6 ˆm ˆˆ ˆ vk + (vj,k tk jm ˆ vm,k vk ) + [vm (tm ˆˆ j vk ),k − vj (tk ˆ vm − tˆ l vk ),l ] − ∗ 2 2 ρ (39.13.30)

T j0,0 + T jk,k = ρ0 (1 + Π + v 2 + 2U )

Also, we have Γ

j

00 T

00

[ ] 1 1 1 1 = −ρ0 (1 + Π + v + 2U )U,j − ρ0 2Ψ − ζA − ηD − α2 wi wk Uik + ρ0 (α1 − 2α3 )wk Vk,j 2 2 2 2 ,j ] [ 1 − ρ0 U,j −2(β + γ)U + (α2 + α3 − α1 )w2 2 ( ) ( ) 7 1 1 − ρ0 ∆1 Vj,t − ρ0 ∆2 Wj,t + α2 − α1 ρ0 wj U,t − α2 ρ0 wi Uij,t + O ρ0,j ϵ6 (39.13.31a) 2 2 2 2

( 2Γj 0k T 0k = 2γρ0 U,t vj −

) ( ) 7 1 1 1 ∆1 + ∆2 (Vj,k − Vk,j )ρ0 vk − α1 wj U,k ρ0 vk + α1 ρ0 wk vk U,j + O ρ0,j ϵ6 2 2 2 2 (39.13.31b)

( ) Γj kl T kl = −γU,j (3p + ρ0 v 2 ) + 2γU,k (tˆj kˆ + ρ0 vj vk ) + O ρ0,j ϵ6 .

(39.13.31c)

Adding equations 39.13.31 together and using the fact that ( ) Uij,t − δij U,t = χ,ijt = χ,itj = (Vi − Wi ),j + O ϵ5 ,

(39.13.32)

we have Γ

j

µβ T

µβ

[ ] 1 1 1 = −ρ0 (1 + Π + v + 2U )U,j − ρ0 2Ψ − ζA − ηD − α2 wi wk Uik + α2 wi (Vi − Wi ) 2 2 2 ,j [ ] 1 1 1 + ρ0 (α1 − 2α3 )wk Vk,j − ρ0 U,j γv 2 − α1 vi wi + (α2 + α3 − α1 )w2 − 2(β + γ)U + 3γp/ρ0 2 2 2 ( ) 7 1 1 7 1 1 − ρ0 ∆1 Vj,t − ρ0 ∆2 Wj,t − α1 ρ0 wj U,t + 2γρ0 U,t vj − ∆1 + ∆2 Vj,k ρ0 vk − α1 wj U,k ρ0 vk 2 2 2 2 2 2 ( ) ) ( 7 1 + 2γU,k (tˆj kˆ + ρ0 vj vk ) + ∆1 + ∆2 Vk,j ρ0 vk O ρ0,j ϵ6 . (39.13.33) 2 2 2

409

/ Using d/dt = ∂/∂t + v k ∂ ∂xk , we have j

Γ

µβ T

µβ

[ ] 1 1 1 = −ρ0 (1 + Π + v + 2U )U,j − ρ 2Ψ − ζA − ηD − α2 wi wk Uik + α2 wi (Vi − Wi ) 2 2 2 ,j [ ] 1 1 − ρ∗ U,j γv 2 − α1 vi wi + (α2 + α3 − α1 )w2 − (2β − 2)U + 3γp/ρ∗ + (2γ + 2)ρ0 U U,j 2 2 [ ( ) ] d 7 1 1 1 + ρ∗ − ∆1 + ∆2 Vj − α1 wj U + ∆2 ρ∗ (Vj − Wj ),t + 2γρ∗ vj (U,t + vk U,k ) dt 2 2 2 2 ) ( 1 ∗ (39.13.34) + ρ [(7∆1 + ∆2 )vk + (α1 − 2α3 )wk ]Vk,j + 2γU,k tˆj kˆ + O ρ0,j ϵ6 . 2 2

∗

Finally, we have ) ( Γβ µβ T jµ = (3γ − 1)U,t ρ0 v j + (3γ − 1)U,k tˆj kˆ + ρ0 vj vk

(39.13.35)

Now focus on the expression dv j + [tˆj kˆ (1 − 2γU )],k − v j [(3γ − 2)U,t ρ0 + (3γ − 3)U,k ρ0 v k ] + tˆj kˆ U,k dt − ρ0 (1 + Π + v 2 + 2U )U,j + (2γ + 2)ρ0 U U,j + 2γρ∗ vj (U,t + vk U,k ) + 2γU,k tˆj kˆ ( ) ( ) + (3γ − 1)U,t ρ0 v j + (3γ − 1)U,k tˆj kˆ + ρ0 vj vk + O ρ0,j ϵ6 . (39.13.36)

C ≡ρ0 (1 + Π + v 2 + 2U )

We have dv j − ρ0 (1 + Π + v 2 − 2γU )U,j + (1 − 2γU )tˆj k,k ˆ dt ( ) + 3γtˆj kˆ U,k + (2γ + 1)ρ0 v j U,t + (2γ + 2)ρ0 v j v k U,k + O ρ0,j ϵ6 .

C = ρ0 (1 + Π + v 2 + 2U )

(39.13.37)

As dU − ρ0 v j v k U,k dt ( ) dU v j dv j = (2γ + 2)ρ∗ − (2γ + 2)ρ0 U − ρ∗ v j U,t + O ρ0,j ϵ6 , dt dt

(2γ + 1)ρ0 v j U,t + (2γ + 2)ρ0 v j v k U,k = (2γ + 2)ρ0 v j

(39.13.38)

we have ( j ) j ) ( dv ∗ j ∗ dU v C = ρ0 (1 + Π + v 2 − 2γU ) − U,j + (1 − 2γU )tˆj k,k U − ρ v U + (2γ + 2)ρ + O ρ0,j ϵ6 ˆ + 3γtˆ ˆ ,k ,t jk dt dt ( j ) ( )( ) ( ) j dv 1 2 dv 1 2 = ρ∗ − U,j + v + Π − 5γU ρ0 − ρ0 U,j + tˆj k,k − v + Π − 5γU tˆj k,k ˆ ˆ dt 2 dt 2 j ( ) ∗ j ∗ dU v + (1 − 2γU )tˆj k,k + O ρ0,j ϵ6 ˆ + 3γtˆ ˆ U,k − ρ v U,t + (2γ + 2)ρ jk ( j ) ( ) dt j ( ) dv 1 ∗ j ∗ dU v = ρ∗ − U,j + [tˆj kˆ (1 + 3γ)U ],k − v 2 + Π tˆj k,k + O ρ0,j ϵ6 ˆ − ρ v U,t + (2γ + 2)ρ dt 2 dt (39.13.39)

Now bring equations 39.13.30, 39.13.34, 39.13.35, 39.13.36 and 39.13.39 together, we have the post-

410

Newtonian Euler equation ) ( j ( ) tˆj kˆ tkˆm,m 1 2 dv ˆ ∗ ∗ j ρ − U,j + [tˆj kˆ (1 + 3γ)U ],k − v + Π tˆj k,k ˆ − ρ v U,t + tˆ ˆ vk − j k,t dt 2 ρ∗ [ ( ) ] 7 1 1 1 j ∗ d (2γ + 2)U v − ∆1 + ∆2 Vj − α1 wj U + ∆2 ρ∗ (Vj − Wj ),t +ρ dt 2 2 2 2 [ ] 1 1 1 1 − ρ∗ 2Ψ − ζA − ηD − α2 wi wk Uik + α2 wi (Vi − Wi ) + ρ∗ [(7∆1 + ∆2 )vk + (α1 − 2α3 )wk ]Vk,j 2 2 2 2 ,j [ ] 1 1 2 ∗ ∗ 2 − ρ U,j γv − α1 vi wi + (α2 + α3 − α1 )w − (2β − 2)U + 3γp/ρ 2 2 ( ) 1 1 6 + (vj,k tkˆm (39.13.40) ˆˆ jm ˆ vm,k vk ) + [vm (tm ˆˆ j vk ),k − vj (tk ˆ vm − tˆ l vk ),l ] = O ρ0,j ϵ . 2 2

39.14

POST-NEWTONIAN APPROXIMATION TO GENERAL RELATIVITY

Perform a post-Newtonian expansion of Einstein’s field equations, thereby obtaining the values cited in Box 39.2 for the PPN parameters of general relativity. The calculations might best follow the approach of Chandrasekhar (1965a): Set gαβ = ηαβ + hαβ and assume ( ) ( ) ( ) ( ) h00 = O ϵ2 + O ϵ4 , h0j = O ϵ3 , hjk = O ϵ2 . (39.14.1) Choose the space and time coordinates so that the four “gauge conditions” ( 4 ) ( 5 ) ϵ ϵ 1 1 hjk,k − h,j = O , h0k,k − hkk,0 = O with h = hαβ η αβ = −h00 + hll 2 R⊙ 2 R⊙

(39.14.2)

are satisfied. (a) Show that the spatial gauge conditions are the post-Newtonian approximations to those (35.1a) used in the study of weak gravitational waves, but that the temporal gauge condition is not. (b) Use these gauge conditions and the post-Newtonian limit in equations (8.24) and (8.47) to obtain for the Ricci tensor, accurate to linearized order, ( 4 ) ( 4 ) ( 5 ) 1 ϵ 1 ϵ 1 1 ϵ R00 = − h00,mm + O , Rjk = − hjk,mm + O , R0j = − h0j,mm − h00,0j + O . 2 2 2 2 R⊙ 2 R⊙ 2 4 R⊙ (39.14.3) (c) Combine these with the Newtonian form (39.13) of the stress-energy tensor, and with equation (39.27), to obtain the following metric coefficients, accurate to linearized order: ( ) h00 = 2U + k00 + O ϵ6 ,

( ) 7 1 h0j = − Vj − Wj + O ϵ5 , 2 2

( ) hjk = 2U δjk + O ϵ4 .

(39.14.4)

Here U , Vj , and Wj are to be regarded as defined by equations (39.34a,b,c). By comparing these metric coefficients with equations 39.13.4, discover that γ = 1,

∆1 = 1,

∆2 = 1

(39.14.5)

for general relativity. (d) With this knowledge of the metric in linearized order, one can carry out the analysis of section 39.10 to obtain the post-Newtonian corrections to the stress-energy tensor.

411

(e) Calculate, similarly, the post-Newtonian corrections to the Ricci tensor component R00 , using gαβ = ηαβ + hαβ , using hαβ as given in equations (39.54), and using the gauge conditions (39.52). The answer should be ( ) ( 6 ) 1 ϵ 2 R00 = −U − k00 − U + 4U U,mm + O . (39.14.6) 2 2 R ⊙ ,mm (f) Evaluate the Einstein equation R00 = 8π(T00 − g00 T /2), accurate to post-Newtonian order, and solve it to obtain the post-Newtonian metric correction k00 = −2U 2 + 4Ψ,

(39.14.7)

where Ψ is given by equation (39.43d) with β1 = β2 = β3 = β4 = 1. By comparing with equations (39.32c) and (39.34d), discover that β = β1 = β2 = β3 = β4 = 1,

ζ=η=0

(39.14.8)

for general relativity. (g) Knowing the full post-Newtonian metric, and the full post-Newtonian stress-energy tensor, one can carry out the calculations of section 39.11 to obtain the post-Newtonian equations of motion for the matter [equations (39.44), (39.46), and (39.47)].

Solution: (a) Gauge conditions used in the study of weak gravitational waves is ¯ µα = 0. h ,α

(39.14.9)

The temporal component is ( ¯ 00 + h ¯ 0k = h ,0 ,k

) 1 1 1 − h0k,k = h00,0 + hkk,0 − h0k,k = 0. h00 + h 2 ,0 2 2

(39.14.10)

( ) Since h00,0 ∼ O ϵ3 /R⊙ , temporal component is different from those in 39.14.1. The spatial components are

Since hj0,0

) ( 1 ¯ j0 + h ¯ jk = −hj0,0 + hjk − 1 δjk h = hjk,k − h,j − hj0,0 = 0. h ,0 ,k 2 2 ,k ( 4 ) ∼ O ϵ /R⊙ , spatial components are the same as those in 39.14.1.

(39.14.11)

(b) To calculate Ricci tensor to linearized order, we are free to use η to lower and raise indices. We have ( ) 1 ( ) Rµν = Γαµν,α − Γαµα,ν + O Γ2 = (hµα,να + hν α,µα − hµν,αα − h,µν ) + O Γ2 . 2 So

( 4 ) ( 4 ) 1 1 ϵ ϵ = − (2h0α,0α − h00,αα − h,00 ) + O h + +O 00,mm 2 2 2 R⊙ 2 R⊙ ( 5 ) 1 ϵ = (h0α,jα + hj α,0α − h0j,αα − h,0j ) + O 2 2 R⊙ ( 5 ) 1 ϵ = (−h00,j0 + h0k,jk + hjk,0k − h0j,kk − h,0j ) + O 2 2 R⊙ ( 5 ) 1 1 1 ϵ = (−h00,j0 + hkk,0j − h0j,kk − h,0j ) + O 2 2 2 2 R⊙ ( 5 ) 1 1 ϵ = − h0j,mm − h00,0j + O . 2 2 4 R⊙

R00 = R0j

412

(39.14.12)

(39.14.13)

(39.14.14)

( 4 ) 1 α ϵ (hj ,kα + hkα,jα − hjk,αα − h,jk ) + O 2 2 R⊙ ( 4 ) 1 ϵ = (hjm,km + hkm,jm − hjk,mm − h,jk ) + O 2 2 R⊙ ( 4 ) 1 1 1 ϵ = ( h,jk + h,kj − hjk,mm − h,jk ) + O 2 2 2 2 R⊙ ( 4 ) 1 ϵ . = − hjk,mm + O 2 2 R⊙

Rjk =

Notice that

(

R = ηαβ R

αβ

ϵ4 +O 2 R⊙

)

( 4 ) 1 1 ϵ = h00,mm − hkk,mm + O . 2 2 2 R⊙

(c)The Newtonian form (39.13) of the stress-energy tensor is ( ) ( ) ( ) T 00 = ρ0 + O ρ0 ϵ2 , T 0i = ρ0 vi + O ρ0 ϵ3 , T jk = tˆj kˆ + ρ0 vj vk + O ρ0 ϵ4 . Since

) ϵ2 ρ0 ∼ O , 2 R⊙

(39.14.15)

(39.14.16)

(39.14.17)

(

(39.14.18)

so we have ( 4 ) 1 ϵ 1 − h00,mm − hkk,mm = 8πρ0 + O 2 4 4 R⊙ ( 4 ) ϵ 1 1 1 − hjk,mm − δjk h00,mm + δjk hll,mm = O 2 2 4 4 R⊙ ( 5 ) ϵ 1 1 . − h0j,mm − h00,0j = −8πρ0 vi + O 2 2 4 R⊙ It is easy to find that

( ) h00 = 2U + O ϵ4 ,

( ) hjk = 2U δjk + O ϵ4 .

(39.14.19) (39.14.20) (39.14.21)

(39.14.22)

It follows that h0j,mm = 16πρ0 vi − U,0j .

(39.14.23)

The solution is ∫ h0j (x, t) = −4

1 ρ0 (x′ , t)vi (x′ , t) 3 ′ d x + ′ |x − x | 4π

∫

( ) ∂ 2 U (x′ , t) d3 x′ + O ϵ5 . ′ ′ ∂xj ∂t |x − x |

(39.14.24)

Using equation 39.8.1, we have ( ) ( ) 1 7 1 h0j (x, t) = −4Vi (x, t) + [Vj (x, t) − Wj (x, t)] + O ϵ5 = − Vi (x, t) − Wj (x, t) + O ϵ5 . (39.14.25) 2 2 2 Now we conclude that γ = ∆1 = ∆2 = 1 in general relativity. (d) Substitute γ = ∆1 = ∆2 = 1 into equation (39.42). We obtain the stress-energy tensor to PPN order in general relativity. ( ) T 00 = ρ0 (1 + Π + v 2 + 2U ) + O ρ0 ϵ4 , ( 5) T 0j = ρ0 (1 + Π + v 2 + 2U )v j + tˆj m ˆ vm + O ρ 0 ϵ , ( 6) 1 T jk = tˆj kˆ (1 − 2U ) + ρ0 (1 + Π + v 2 + 2U )vj vk + (vj tkˆm (39.14.26) jm ˆ vm ) + O ρ0 ϵ . ˆ vm + vk tˆ 2

413

(e)Suppose the unknown post-Newtonian correction to h00 is k00 , i.e., ( ) h00 = 2U + k00 + O ϵ6 .

(39.14.27)

Using equations 39.13.1 and 39.13.7, we have R00 = Rα0α0 = Rk0k0 = Γk00,k − Γk0k,0 + Γkµk Γµ00 − Γkµ0 Γµ0k = Γk00,k − Γk0k,0 + Γk0k Γ000 + Γklk Γl00 − Γk00 Γ00k − Γkl0 Γl0k [ ] ( 6 ) 1 ϵ 2 = hk0,0 − (1 − 2U )h00,k − 3U,tt − 3U,l2 − U,k +O . 2 2 R ⊙ ,k

(39.14.28)

Using the gauge condition, we have hk0,0k = So we obtain

( 6 ) ( 6 ) 1 ϵ ϵ hkk,00 + O = 3U + +O ,tt 2 2 2 R⊙ R⊙

( 6 ) 1 ϵ 2 R00 = − k00,kk − U,kk − 2U,k + 2U U,kk + O 2 2 R⊙ ( ) ( 6 ) 1 ϵ . = −U − k00 − U 2 + 4U U,mm + O 2 2 R ⊙ ,mm

(39.14.29)

(39.14.30)

(f)Firstly, we have ( ) 2 T00 = g00 T 00 + 2g00 g0i T i0 + g0i g0j T ij = (−1 + 2U )2 T 00 + O ρ0 ϵ4 ( ) = ρ0 (1 + Π + v 2 − 2U ) + O ρ0 ϵ4

(39.14.31)

( ) T = g00 T 00 + 2g0i T 0i + gij T ij = (−1 + 2U )T 00 + δij T ij + O ρ0 ϵ6 ( ) = −ρ0 (1 + Π + v 2 ) + 3p + ρ0 v 2 + O ρ0 ϵ6 .

(39.14.32)

and

So we obtain the equation (

R00

i.e.,

) ( ) 1 = 8π T00 − g00 T = 4π[ρ0 (1 + Π + 2v 2 − 2U ) + 3p] + O ρ0 ϵ6 2 ) ( ( ) 1 2 + 4U U,mm + O ρ0 ϵ6 , = −U − k00 − U 2 ,mm

) ( ( ) 1 = 4π[ρ0 (1 + Π + 2v 2 + 2U ) + 3p] + O ρ0 ϵ6 −U − k00 − U 2 2 ,mm

We have the solution 1 U + k00 + U 2 = U + 2 2

∫

( ) 1 3p 1 ρ0 v 2 + U + Π + d3 x′ . 2 2ρ0 |x − x′ |

(39.14.33)

(39.14.34)

(39.14.35)

We can conclude that β = β1 = β2 = β3 = β4 = 1,

ζ = η = 0.

(39.14.36)

(g) The post-Newtonian equations of motion for the matter is ( ) ρ∗,t + (ρ∗ v i ),i = O ρ0,j ϵ5 ρ0

where

( ) 1 ρ∗ ≡ ρ0 1 + v 2 + 3U 2

( ) dΠ + tˆj kˆ vj,k = O ρ0,j ϵ5 dt

(39.14.37) (39.14.38)

414

and

) ( ) tˆj kˆ tkˆm,m dv j 1 2 ˆ ∗ j − U,j + [tˆj kˆ (1 + 3U )],k − v + Π tˆj k,k ˆ − ρ v U,t + tˆ ˆ vk − j k,t dt 2 ρ∗ ) 1 d( j + 4ρ∗ U v − Vj + ρ∗ (Vj − Wj ),t − 2ρ∗ Ψ,j + 4ρ∗ vk Vk,j − ρ∗ U,j (v 2 + 3p/ρ0 ) dt 2 ( ) 1 1 6 + (vj,k tkˆm (39.14.39) ˆˆ ˆˆ j vk ),k − vj (tk jm ˆ vm,k vk ) + [vm (tm ˆ vm − tˆ l vk ),l ] = O ρ0,j ϵ . 2 2

ρ∗

39.15

(

MANY-BODY SYSTEM IN POST-NEWTONIAN LIMIT OF GENERAL RELATIVITY

Consider, in the post-Newtonian limit of general relativity, a system made up of many gravitationally interacting bodies with separations large compared to their sizes (example: the solar system). Idealize each body to be spherically symmetric, to be free of internal motions, and to have isotropic internal stresses, tˆj kˆ = pδjk . Let the world line of the center of body A, in some chosen PPN coordinate frame, be xA (t); and let the (coordinate) velocity of the center of body A be vA =

dxA . dt

(39.15.1)

The total mass-energy of body A as measured in its neighborhood (rest mass-energy plus internal energy plus self-gravitational energy) is given by ( ) ∫ ( ) 1 MA = = 1 + Π − Uself d(rest mass) + O MA ϵ4 , (39.15.2) 2 VA where Uself is the body’s own Newtonian potential (no contributions from other bodies), and VA is the interior of the body. (a) Show that, when written in the chosen PPN coordinate frame, this expression for MA becomes ( ) ∫ ( ) 1 2 1 MA = ρ0 1 + Π + vA + 3U − Uself d3 x + O MA ϵ4 . (39.15.3) 2 2 VA Use equations 39.13.13, 39.13.15, and 39.13.26 to show that MA is conserved as the bodies move about, dMA /dt = 0. (b) Pick an event (t, x) outside all the bodies, and at time t denote rA ≡ xA − x,

rAB ≡ xA − xB ,

rA ≡ |rA |,

rAB ≡ |rAB |.

(39.15.4)

Show that the general-relativistic, post-Newtonian metric 39.13.4 at the chosen event has the form ( ) ∑ MA ( ) gjk = δjk 1 + 2 + O ϵ4 , rA A [ ] ∑ MA 7 ( ) 1 (vA · rA )rAj vAj + + O ϵ5 , g0j = − 2 rA 2 2 rA A )2 ( ∑ MA v 2 ∑ MA ∑ ∑ M A MB ∑ MA ( ) A +3 (39.15.5) g00 = −1 + 2 −2 −2 + O ϵ6 . rA rA rA rA rAB A

A

A

(c) Perform an infinitesimal coordinate transformation, 1 ∑ MA (vA · rA ) told = tnew − , 2 rA A

415

A B̸=A

xold = xnew ,

(39.15.6)

to bring the metric 39.15.5 into the standard form originally devised by Einstein, Infeld, and Hoffman (1938), and by Eddington and Clark (1938): ( gjk = δjk g0j = −4

1+2

∑ MA

∑ MA A

g00 = −1 + 2

rA

rA

A

( ) + O ϵ4 ,

( ) vAj + O ϵ5 , (

∑ MA A

)

−2

rA

∑ MA A

)2 +3

rA

∑ MA v 2

A

A

rA

−2

∑ ∑ MA MB ( ) ∂2χ − 2 + O ϵ6 , rA rAB ∂t

(39.15.7)

A B̸=A

where χ [equation (39.49a)] is given by χ=−

∑

MA rA .

(d) The equations of motion for the bodies can be obtained in either of two ways: by performing a volume integral of the Euler equation 39.13.40 over the interior of each body; or by invoking the general arguments of section 20.6. The latter way is the easier. Use it to conclude that any chosen body K moves along a geodesic of the metric obtained by omitting the terms A = K from the sums in 39.15.7. Show that the geodesic equation for body K reduces to  ) ∑ MB ∑ ∑ MC ( d2 xK dvA MA  rAK · rCA 1−4 ≡ = rAK 3 − 1− 2 dt2 dt rAK rBK rCA 2rCA B̸=K A̸=K C̸=A ( )2 ] 3 vA · rAK 2 2 + vK + 2vA − 4vA · vK − 2 rAK ∑ MA MC MA rAK · (3vA − 4vK ) 7 ∑ ∑ + rCA − (vA − vK ) 3 3 . rAK 2 rAK rCA

(39.15.8)

A̸=K C̸=A

A̸=K

Equations 39.15.7 and 39.15.8 are called the Einstein-Infeld-Hoffman (“EIH”) equations for the geometry and evolution of a many-body system. They are used widely in analyses of planetary orbits in the solar system.

Solution: (a) The rest mass energy in volume d3 x is [ ] [ ( )] ( ) √ 1 d(rest mass) = ρ0 u0 −g d3 x = ρ∗ 1 + O ϵ4 d3 x = ρ0 1 + v 2 + 3U + O ϵ4 d3 x . 2

(39.15.9)

The total mass energy of the star is ( MA = ρ0 1 + Π − V ∫ A ( = ρ0 1 + Π + ∫

VA

)[ ] ) ( 4) 3 ( 1 2 1 Uself 1 + v + 3U + O ϵ d x + O M A ϵ4 2 2 ) ) ( 1 2 1 vA + 3U − Uself d3 x + O MA ϵ4 . 2 2

(39.15.10)

I have not figured out how to prove MA is conserved as the bodies move about. Any help is appreciated.

416

∫

(b) Define mA ≡ U (x, t) =

VA

∑∫ VA

A

Ψ1 (x, t) =

∑∫ A

Ψ2 (x, t) =

Ψ4 (x, t) = Vj (x, t) = Wj (x, t) =

A

A

A

′

∑

rA

∫

VA

∫

( ) + O ϵ6

ρ0 (x′ , t)Uself d3 x′ rA

+

∑ ∑ MA MB ( ) + O ϵ6 rA rAB A B̸=A

ρ (x′ , t)Π d3 x′ VA 0

( ) + O ϵ6

VA

ρ0 (x , t)Π 3 ′ d x = |x − x′ |

VA

p(x′ , t) 3 ′ ∑ d x = |x − x′ |

VA

ρ0 (x , t)v 3 ′ ∑ d x = |x − x′ |

VA

j ( ) ρ0 (x , t)[v · (x − x )(xj − x′j )] 3 ′ ∑ MA (vA · rA )rA d x = + O ϵ5 . ′ 3 |x − x | rA

∑∫ A

A

∑ MA v 2

VA

∑∫ A

2

ρ0 (x′ , t)U 3 ′ ∑ d x = |x − x′ |

∑∫ A

A

′

VA

∑∫ A

∑ MA ( ) ρ0 (x′ , t) 3 ′ ∑ mA d x = = + O ϵ4 ′ |x − x | rA rA ρ0 (x , t)v 3 ′ d x = |x − x′ |

∑∫ A

Ψ3 (x, t) =

ρ0 d3 x. For an event (t, x) outside all the bodies, we have

rA

A

∫

p(x′ , t) d3 x′

VA

rA

A

′

j MA v A

rA

A ′

′

( ) + O ϵ6

( ) + O ϵ5 (39.15.11)

A

From the Newtonian virial theorem 39.6.6, applied to body A by itself in its own rest frame, we can conclude that ∫ ∫ ( ) 1 3 − ρ0 Uself d x + 3 p d3 x = O MA ϵ4 . (39.15.12) 2 In General relativity, we have ( ) g00 = −1 + 2U − 2U 2 + 4Ψ1 + 4Ψ2 + 2Ψ3 + 6Ψ4 + O ϵ6 We also have

∑ MA A

rA

( ) 1 = U + Ψ3 + Ψ1 + 3Ψ2 − 3Ψ4 + O ϵ6 . 2

(39.15.13)

(39.15.14)

So we can get g00 = −1 + 2

∑ MA A

= −1 + 2 −2

∑

∑ MA A

∫

−2

( )2 ∑ MA A

rA

rA ∑ MA

+ 12

( )2 ∑ MA

∑ MA v 2

A

+3 ∑

∫

A

rA

−2

A B̸=A

p(x′ , t) d3 x′ VA rA

A

∑ ∑ MA MB rA rAB

−2

+3

∑ MA v 2

( ) + O ϵ6

∑ ∑ MA MB ( ) + O ϵ6 ; rA rA rA rA rAB A A A A B̸=A [ ] ∑ MA 7 ( ) ( ) 7 1 1 (vA · rA )rAj = − Vj − Wj + O ϵ5 = − vAj + + O ϵ5 ; 2 2 2 rA 2 2 rA A ) ( ∑ MA ( 4) ( ) = δjk (1 + 2U ) + O ϵ = δjk 1 + 2 + O ϵ4 . rA = −1 + 2

gjk

rA

( ) − 2U 2 + 3Ψ1 − 2Ψ2 + 12Ψ4 + O ϵ6

ρ (x′ , t)Uself d3 x′ VA 0

A

g0j

rA

A

−2

(39.15.15a) (39.15.15b) (39.15.15c)

A

(c) Firstly, we have ( ) ∂2χ = Vj − Wj + O ϵ5 , j ∂t∂x

417

told = tnew +

1 ∂χ . 2 ∂t

(39.15.16)

The transformation matrix is ∂told 1 ∂2χ = 1+ , ∂tnew 2 ∂t2

∂told ∂xjnew

=

( ) 1 ∂2χ 1 = (Vj − Wj ) + O ϵ5 , 2 ∂t∂xj 2

∂xjold = 0, ∂tnew

∂xjold = δkj . (39.15.17) ∂xknew

Thus, (

)2 ( ) ( ) ∂told ∂2χ old = = g00 (xnew , told ) 1 + 2 + O ϵ8 ∂tnew ∂t 2 ( ) ( ) 1 ∂χ ∂ χ ∂2χ old old old = g00 (xnew , tnew ) + g00,0 (xnew , tnew ) − 2 + O ϵ6 = g00 (xnew , tnew ) − 2 + O ϵ6 2 ∂t ∂t ∂t ∂told ∂xkold ∂told ∂told old new old + g0k (xold , told ) g0j (xnew , tnew ) = g00 (xold , told ) ∂tnew ∂xjnew ∂tnew ∂xjnew ( ) 1 old = − (Vj − Wj ) + g0k (xnew , tnew ) + O ϵ5 2 ∂told ∂told ∂told ∂told old old gjk (xnew , tnew ) = g00 (xold , told ) j + g0k (xold , told ) j + gj0 (xold , told ) k + gjk (xold , told ) ∂xnew ∂xnew ∂xknew ∂xnew ( ) old (39.15.18) = gjk (xnew , tnew ) + O ϵ4 .

new g00 (xnew , tnew )

old g00 (xold , told )

So we derive the equation 39.15.7. (d) The geodesic equation is duα + Γαβγ uβ uγ = 0. dτ

(39.15.19)

So we have dv j dτ d = dt dt dτ

(

uj u0

) =

1 duj uj du0 − = −(Γj 00 +2Γj 0k v k +Γj kl v k v l )+v j (Γ000 +2Γ00k v k +Γ0kl v k v l ) (u0 )2 dτ (u0 )3 dτ (39.15.20)

Using equations 39.13.7, 39.15.7, and omitting the terms A = K from the sums in 39.15.7, we have 0

Γ

00

Γ00j

( 5) ∑ MA (rAK · vA ) ϵ = +O 3 rAK R A̸=K ( ) ∑ MA r j ϵ4 AK =− +O 3 rAK R

(39.15.21a)

(39.15.21b)

A̸=K

( 5) ∑ MA (rAK · vA ) ∑ MA (v j rk + v k rj ) ϵ A AK A AK δ + 2 + O jk 3 3 rAK rAK R A̸=K A̸=K   ( 6) ∑ MB ∑ MA r j ∑ MC 1 3 2 ϵ AK   1−4 + χ,ttj + gj0,0 + O =− + vA − 3 rAK rBK 2 rAC 2 R B̸=K A̸=K C̸=A ( ) ∑ MA (rAK · vA ) ∑ MA (v j rk − v k rj ) ϵ5 A AK A AK =− δ − 2 + O jk 3 3 rAK rAK R A̸=K A̸=K ( ) ∑ MA ϵ4 j k l = (rAK δjl + rAK δkj − rAK δkl ) + O 3 rAK R

Γ0jk = −

(39.15.21c)

Γj 00

(39.15.21d)

Γj 0k Γj kl

(39.15.21e) (39.15.21f)

A̸=K

Using

( 4) ∑ MC rCA dvA ϵ = + O , 3 dt rCA R C̸=A

418

(39.15.22)

we find that

    ( 6) j 7 ∂  ∑ MA j  1 ∂  ∑ MA (vA · rAK )rAK 1 ϵ  χ,ttj + gj0,0 = − v − +O 3 2 2 ∂t rAK A 2 ∂t rAK R A̸=K

A̸=K

j 7 ∑ ∑ MA MC rCA 7 ∑ MA (rAK · vA ) j =− + vA 3 3 2 rAK rCA 2 rAK A̸=K C̸=A A̸=K ) j j ∑ ( MC 1 ∑ MA rAK 1 ∑ MA (vA · rAK )vA − r · r − CA AK 3 3 3 2 rA rCA 2 rAK A̸=K C̸=A A̸=K ( 4) j 2 1 ∑ M A vA 3 ∑ MA (vA · rAK )2 rAK ϵ 3 − rAK + +O 5 j 2 2 r R rAK AK A̸=K A̸=K

(39.15.23)

We then have

  ( ) ( )2 j ∑ ∑ ∑ M r M M r · r 3 v · r A B C CA AK A AK 2 AK   Γj 00 = − 1−4 + 2vA − 1− − 3 2 rAK rBK rAC 2rCA 2 rAK A̸=K B̸=K C̸=A ( 6) j ∑ MA (rAK · vA ) j ∑ ∑ MA MC rCA 7 ϵ vA − +O (39.15.24) +3 3 3 rAK 2 rAK rCA R A̸=K

A̸=K C̸=A

We can also figure out that j k − 2Γj 0k vK + Γ000 vK ( 6) ∑ MA [v j (vK · rAK ) − (vA · vK )rj ] ∑ MA (rAK · vA )v j ∑ MA (rAK · vA ) j ϵ A AK K v + 4 + + O =2 K 3 3 3 rAK rAK rAK R A̸=K A̸=K A̸=K ) ( ∑ MA (rAK · vA ) j ∑ MA (vK · rAK ) j ∑ MA r j ϵ6 AK =3 (39.15.25) v + 4 v − 4 v · v + O A K K A 3 3 3 rAK rAK rAK R A̸=K

A̸=K

A̸=K

and j k k l − Γj kl vK vK + 2Γ00k vK vK ( 6) ∑ MA ∑ MA (rAK · vK ) j ϵ j j 2 = [−2(r · v )v + r v ] − 2 v + O AK K K AK K K 3 3 rAK rAK R A̸=K A̸=K ( ) ∑ MA r j ∑ MA (rAK · vK ) j ϵ6 AK 2 v − 4 v + O = . K K 3 3 rAK rAK R A̸=K

(39.15.26)

A̸=K

( ) We also notice that Γ0kl v k v l v j ∼ O ϵ6 /R . Putting equations 39.15.20, 39.15.24, 39.15.25 and 39.15.26 together, we can get the equation 39.15.8.

419

420

Chapter 40

SOLAR-SYSTEM EXPERIMENTS 40.1

PPN METRIC FOR IDEALIZED SUN

Show that for an isolated, static, spherical sun at rest at the origin of the PPN coordinate system, the PPN metric (39.32) reduces to expressions (40.3), (40.3’). As part of the reduction, show that the sun’s total mass-energy is given by ∫ M⊙ =

R⊙

ρ0 (1 + 2β2 U + β3 Π + 3β4 p/ρ0 )4πr2 dr .

(40.1.1)

0

Solution: For an isolated, static, spherical sun at rest at the origin of the PPN coordinate system, we have ∫ U (x, t) = ∫ Ψ(x, t) =

1 ρ0 (x′ , t) 3 ′ d x = ′ |x − x | r

∫

R⊙

ρ0 4πr′2 dr′

0

ρ0 (x′ , t)ψ(x′ , t) 3 ′ 1 d x = |x − x′ | r

∫

R⊙

0

) ( 3p 1 ρ0 β2 + β3 Π + 4πr′2 dr′ 2 2ρ0

V i = Wi = A = D = 0

(40.1.2)

The PPN metric will be ( ) g00 = −1 + 2U − 2βU 2 + 4Ψ + O ϵ6 ,

( ) g0j = O ϵ5 ,

( ) gjk = (1 + 2γU )δjk + O ϵ4 .

(40.1.3)

If we define M⊙ as 40.1.1, it will be easy to find that ( ) 2M⊙ = 2U + 4Ψ = 2U + O ϵ4 . r

(40.1.4)

So we have g00

( ) 2M 2 2M⊙ = −1 + − 2⊙ + O ϵ5 , r r

g0j

( ) = O ϵ5 ,

gjk

( ) ( ) 2γM⊙ = 1+ δjk + O ϵ4 . r

(40.1.5)

In order to calculate the metric for a theory with a preferred “universal rest frame”, we have to know ∫ Ujk (x, t) =

ρ0 (x′ , t)(xj − x′j )(xk − x′k ) 3 ′ d x |x − x′ |3

421

(40.1.6)

Recall that

∂2χ = Uij − U δjk ∂xj ∂xk ∫

where χ(x, t) = − Notice that ∫

ρ0 (x′ , t)|x − x′ | d3 x′ =

∫

R⊙

ρ0 (x′ , t)|x − x′ | d3 x′ .

ρ0 (r′ )r′2 dr′

0

∫

∫

(

1

√

−1

′

∫

where I⊙ =

′2

(40.1.8)

∫

2π

r2 + r′2 − 2rr′ cos θ d cos θ

r′2 = ρ0 (r )4πr r+ 3r 0 ( ) I⊙ = M⊙ r + + O ϵ4 r 2 , 3r R⊙

(40.1.7)

)

dϕ 0

dr′

ρ(x′ , t)r′2 d3 x′ .

(40.1.9)

(40.1.10)

So we have Ujk

( ) ( ) ( 4 ) M⊙ ( ) M⊙ I⊙ I⊙ 1 2 = δjk − M⊙ r + + O ϵ = 3 xj xk − 5 xj xk − r δjk + O ϵ4 . r 3r ,jk r r 3

(40.1.11)

Now we have g00 g0j gjk

( [ )] 2 ( ) 2M⊙ I⊙ 1 2 2M⊙ M⊙ 2 M⊙ − 2 + (α2 + α3 − α1 )w − α2 wj wk xj xk − 5 xj xk − r δjk = −1 + + O ϵ6 ; 3 r r r r r 3 ) [ ( )] ( ( 5) M⊙ I⊙ 1 2 1 M⊙ − α2 wk xj xk − 5 xj xk − r δjk +O ϵ ; = α2 − α1 wj 2 r r3 r 3 ( ) ( ) 2γM⊙ = 1+ δjk + O ϵ4 . (40.1.12) r

40.2

TRAJECTORY OF LIGHT RAY IN SUN’S GRAVITATIONAL FIELD

Derive equation (40.6) for the path of a light ray in isotropic coordinates (40.3) in the sun’s “equatorial plane”. Use one or more of three alternative approaches: (1) direct integration of the geodesic equation (the hardest approach!); (2) computation based on the three integrals of the motion k · k = 0,

k·

∂ = k0 , ∂t

k·

∂ = kϕ = −bk0 , ∂ϕ

k≡

d = tangent vector to geodesic dλ

(40.2.1)

(see section 25.2 and 25.3); (3) computation based on the Hamilton-Jacobi method (Box 25.4), which for photons (zero rest mass) reduces to the “eikonal method” of geometric optics (see section 22.5).

Solution: I would use Hamilton-Jacobi method to derive the trajectory. The metric of the spacetime is [ ] [ ] ( 4) ( 4) [ 2 ] 2M⊙ 2γM⊙ 2 2 ds = − 1 − +O ϵ dt + 1 + +O ϵ dr + r2 (dθ2 + sin2 θ dϕ2 ) . r r

(40.2.2)

For massless photons, we have g αβ pα pβ = 0

422

(40.2.3)

Suppose the photon is in plane θ = π/2. Hamilton-Jacobi equation would be ( ) ( )2 ( ) [ ( )2 ( )2 ] 2M⊙ ∂S 2γM⊙ ∂S 1 ∂S − 1+ + 1− + 2 = 0. r ∂t r ∂r r ∂ϕ

(40.2.4)

Assume S = −Et + Lϕ + Sr (r).

(40.2.5)

We can obtain the solution ∫

r

√(

S = −Et + Lϕ ±

2(γ + 1)M⊙ 1+ r

) E2 −

L2 dr . r2

(40.2.6)

The trajectory is given by ∂S = ϕ0 . ∂L ∫

We can obtain

r

ϕ = ϕ0 ∓ r2

(40.2.7)

L

√( 1+

2(γ+1)M⊙ r

)

(40.2.8)

dr E2 −

L2 r2

Far from the sun, the ray trajectory is ϕ = b/r. So we have L = bE and ∫

∞

ϕ= r

r2

b

√( 1+

2(γ+1)M⊙ r

∫ )

dr = −

b2 r2

0

b/r

√ −t2 +

1 2(γ+1)M⊙ t b

) ( 2) M⊙ (γ + 1)M⊙ 1 t √ − dt + O = 2 )3/2 2 b b2 (1 − t −t + 1 0 [ ] ( ) ( 2) M⊙ b (γ + 1)M⊙ 1 = arcsin + 1− √ +O r b b2 1 − b2 /r2 ∫

b/r

It can be simplified into

40.3

dt +1

(

( 2) M⊙ b (γ + 1)M⊙ = sin ϕ − (cos ϕ − 1) + O . r b b2

(40.2.9)

(40.2.10)

FERMAT’S PRINCIPLE

Prove Fermat’s principle for a static gravitational field.

Solution: The geodesic equation is

duα + Γαβγ uβ uγ = 0. (40.3.1) dτ Using t as the parameter of the geodesic, we have ( ) 1 duj dv j dτ d uj uj du0 = = − = −(Γj 00 +2Γj 0k v k +Γj kl v k v l )+v j (Γ000 +2Γ00k v k +Γ0kl v k v l ) dt dt dτ u0 (u0 )2 dτ (u0 )3 dτ (40.3.2) Using the fact that g0j = 0 and gαβ,0 = 0, we can get dv j + Γj 00 + Γj kl v k v l − 2Γ00k v k v j = 0. dt

423

(40.3.3)

Lowering the indices, we have gjk

gjk dv k + Γj00 + Γjkl v k v l − 2 Γ00l v l v k = 0. dt g00

(40.3.4)

For photons, we also have (g00 + gkl v k v l )u0 u0 = 0.

(40.3.5)

) k l ( dx dx d2 xk gkl gjk gjk 2 + Γjkl − Γj00 −2 Γ00l dt g00 g00 dt dt

(40.3.6)

Now the geodesic equation becomes

Define γjk = gjk /g00 . We have gjk,l = g00 γjk,l + γjk g00,l

(40.3.7)

So 2Γjkl = gjk,l + gjl,k − gkl,j = g00 (γjk,l + γjl,k − γkl,j ) + γjk g00,l + γjl g00,l − γkl g00,j .

(40.3.8)

Notice that 2Γj00 = −g00,j ,

2Γ00l = g00,l .

(40.3.9)

The geodesic equation will become γjk

1 dxk dxl d2 xk + (γ + γ − γ ) = 0. jk,l jl,k kl,j dt2 2 dt dt

(40.3.10)

Then notice that this is a geodesic equation with affine parameter t in a three dimensional manifold with metric γjk . The familiar extremum principle for this geodesic is ∫

∫

bj

bj

(γjk dxj dxk )1/2 = δ

δ aj

dt = 0,

(40.3.11)

aj

which is precisely Fermat’s principle.

40.4

DERIVATION OF PERIHELION SHIFT IN PPN FORMALISM

Derive equation (40.17) for the shape of any bound orbit of a test particle moving in the equatorial plane of the PPN gravitational field (40.3). Keep only “first-order” corrections beyond Newtonian theory (first order in powers of M⊙ /r).

Solution: Hamilton-Jacobi equation, referred to a test body of unit mass, is [

2M⊙ −1 = g αβ S˜,α S˜,β = − 1 + + (4 − 2β) r

(

M⊙ r

 ( )2  )2 ] ( ˜ )2 [ ] ( )2 M⊙  ∂ S˜ 1 ∂ S˜  ∂S + 1 − 2γ + 2 . ∂t r ∂r r ∂ϕ (40.4.1)

˜ + Lϕ ˜ + S˜r . We have Assume S˜ = −Et [

Notice that

2M⊙ 1+ + (4 − 2β) r

(

M⊙ r

)2 ]

  ] ( ˜ )2 [ 2 ˜ ∂ S L M r ˜ 2 − 1 = 1 − 2γ ⊙  + 2 . E r ∂r r

( ) M⊙ 2 ˜ E −1∼O , r

424

( ) ˜2 M⊙ L ∼O . r2 r

(40.4.2)

(40.4.3)

We have ∫

r

{

Sr = ±

] [ 2 ˜2 ˜ 2 − 1) + 2M⊙ E ˜ 2 + γ(E˜2 − 1) + M⊙ (4 − 2β + 4γ) − L (E r r2 r2

}1/2 dr

(40.4.4)

/ ˜ = 0: The shape of the orbit is determined by the “condition of constructive interference”, ∂ S˜ ∂ L ∫

r

ϕ=±

{

]}−1/2 ( ) [ )2 ( ] ˜2 1−E 1 1 2M⊙ [ M⊙ ˜ 2 − + 1 − (1 + γ)(1 − E ) − 2 1 − 2 (2 − β + 2γ) d 2 2 ˜ ˜ ˜ r r L rL L (40.4.5)

Define

( A≡1−2

M⊙ ˜ L

)2 (2 − β + 2γ),

B≡

] M⊙ [ 1 − (1 + γ)(1 − E˜2 ) , ˜2 L

˜ 2 < 1, we have the solution For the case E ϕ=A

−1/2

( arcsin

A/r − B √ B 2 − AC

C≡

˜2 1−E ˜2 L

(40.4.6)

) (40.4.7)

.

By redefine the position of ϕ = 0, we can get r= Define AC 1−e ≡ 2 = B 2

(

˜ L M⊙

)2

(

1−E

˜2

)

1+ [

√

A/B

( ). 1 − AC/B 2 cos A1/2 ϕ (

1 + 2(1 + γ) 1 − E

˜2

)

( −2

( )] M⊙ [ B ˜2 = 1 − (1 + γ) 1 − E ˜2 C 1−E )2 ( 2) ( M⊙ M⊙ (2 − β + 2γ) + O . δϕ0 ≡ 2π(1 − A1/2 ) = 2π ˜ r2 L

M⊙ ˜ L

(40.4.8)

)2

]

( 2) M⊙ (2 − β + 2γ) + O r2

a≡

The solution can be rewritten as r=

a(1 − e2 ) . 1 + e cos [(1 − δϕ0 /2π) ϕ]

The perihelion shifts is ( )2 ( 2) ( 2) M⊙ M⊙ M⊙ 2 − β + 2γ 6πM⊙ δϕ0 = 2π (2 − β + 2γ) + O = + O ˜ r2 3 a(1 − e2 ) r2 L

40.5

(40.4.9)

(40.4.10)

(40.4.11)

PERIHELION SHIFT FOR OBLATE SUN

(a) The Newtonian potential for an oblate sun has the form ( ) M⊙ R2 3 cos2 θ − 1 U= 1 − J2 2 , r r 2

(40.5.1)

where J2 is the “quadrupole-moment parameter”. One knows that J2 ≲ 3 × 10−5 . Show that if an oblate sun is at rest at the origin of the PPN coordinate system, the metric of the surrounding spacetime [equations (39.32)] can be put into the form [ [ ( )2 ] ] ] 2γM⊙ [ 2 M⊙ R2 3 cos2 θ − 1 M⊙ M⊙ 2 2 dt + 1 + dr + r2 (dθ2 + sin2 θ dϕ2 ) + 2J2 + 2β ds = − 1 − 2 3 r r 2 r r + corrections of post-post-Newtonian magnitude.

425

(40.5.2)

(b) Let a test particle move in a bound orbit in the equatorial plane. Use Hamilton-Jacobi theory to show that its orbit is a precessing ellipse [equation (40.17)] with a precession per orbit given by δϕ0 =

2 − β + 2γ 6πM⊙ 3πR2 + J2 2 . 2 3 a(1 − e ) a (1 − e2 )2

(40.5.3)

For the significance of this result, see Box 40.3.

Solution: The equation 40.5.2 (40.25) in the textbook is incorrect. (a) Notice that g00 = −1 + 2U + 2βU 2 = −1 + 2

2 ( ) M⊙ M⊙ R2 3 cos2 θ − 1 M⊙ − 2J2 + 2β + O ϵ6 . 3 2 r r 2 r

(40.5.4)

So we have the metric 40.5.2. (b) Hamilton-Jacobi equation, referred to a test body of unit mass at plane θ = π/2, is [ −1 = g αβ S˜,α S˜,β = − 1 +

2M⊙ + (4 − 2β) r

(

M⊙ r

)2 + J2

M⊙ R r3

2

](

∂ S˜ ∂t

 )2 [ ( )2  ] ( ˜ )2 M⊙  ∂ S 1 ∂ S˜  + 1 − 2γ + 2 . r ∂r r ∂ϕ (40.5.5)

˜ + Lϕ ˜ + S˜r . We have Assume S˜ = −Et [ 1+

2M⊙ + (4 − 2β) r

(

M⊙ r

)2 + J2

M⊙ R r3

2

]

  [ ] ( ˜ )2 2 ˜ L ˜ 2 − 1 = 1 − 2γ M⊙  ∂ Sr E + 2 . r ∂r r

(40.5.6)

The solution for Sr is }1/2 ∫ r{ [ ] M2 2 ˜2 2M L M R ⊙ ⊙ ⊙ 2 2 ˜ − 1) + ˜ + γ(E˜2 − 1) + dr Sr = ± (E E (4 − 2β + 4γ) − 2 + J2 r r2 r r3 (40.5.7) / ˜ ˜ The shape of the orbit is determined by the “condition of constructive interference”, ∂ S ∂ L = 0: ] }−1/2 ( ) [ ( )2 ] ˜2 1−E 2M⊙ [ 1 M⊙ M⊙ R 2 1 ˜ 2 ϕ=± − + 1 − (1 + γ)(1 − E ) − 2 1 − 2 (2 − β + 2γ) + J2 d 2 2 3 2 ˜ ˜ ˜ ˜ r r L rL L r L (40.5.8) ˜ 2 /M⊙ r. We have Define x = L ] ( ( )2 [ )2 [ ] 2 ˜ 2 )L ˜2 J2 R2 M⊙ M⊙ dx (1 − E + 1−2 = 0. (2 − β + 2γ) x2 − 2 1 − (1 + γ)(1 − E˜2 ) x − x3 + 2 ˜ ˜4 dϕ M⊙ L L (40.5.9) We take the derivative of with respect to ϕ to get ∫

r

{

d2 x + (1 − 2ϵ1 ) x − (1 − ϵ0 ) − ϵ2 x2 = 0, dϕ2 where

( ϵ0 = (1 + γ)(1 − E˜2 ),

ϵ1 =

M⊙ ˜ L

)2 (2 − β + 2γ),

ϵ2 =

2 3J2 R2 M⊙ . 2L˜4

( ) Suppose x = x0 + x1 + O ϵ2 , where x0 = 1 + e cos ϕ. So we have ( ) d2 x1 ϵ2 e 2 1 2 + x1 = cos 2ϕ + 2(ϵ1 + ϵ2 )e cos ϕ + 1 + e ϵ2 + 2ϵ1 − ϵ0 . dt2 2 2

426

(40.5.10)

(40.5.11)

(40.5.12)

The solution is

( ) ϵ2 e 2 1 2 x1 = − cos 2ϕ + (ϵ1 + ϵ2 )eϕ sin ϕ + 1 + e ϵ2 + 2ϵ1 − ϵ0 6 2

(40.5.13)

The three terms here have different characters. The first oscillates around zero, while the third is simply a constant displacement. The important effect is thus contained in the second term, which accumulates over successive orbits. We therefore combine this term with the zeroth-order solution to write x = 1 + e cos ϕ + (ϵ1 + ϵ2 )eϕ sin ϕ

(40.5.14)

This is not a full solution, even to the perturbed equation, but it encapsulates the part that we care about. In particular, this expression for x can be conveniently rewritten as the equation for an ellipse with an angular period that is not quite 2π: ] [ δϕ0 1 + e cos (1 − )ϕ (40.5.15) 2π where ( δϕ0 = 2π(ϵ1 +ϵ2 ) = 2π(2−β +2γ)

40.6

M⊙ ˜ L

)2 +

2 ( ) 3πJ2 R2 M⊙ 3πR2 2 − β + 2γ 6πM⊙ +J +O ϵ2 . = 2 2) 2 (1 − e2 )2 ˜ 4 3 a(1 − e a L (40.5.16)

PRECESSIONAL ANGULAR VELOCITY

Derive equations (40.33) for the precession of a gyroscope in the post-Newtonian limit. Base the derivation on equations (40.29)-(40.32).

Solution:  dSˆj dτ

= S · ∇u eˆj = gαβ S α  (

deˆβj dτ

 + Γβ µν eˆµj uν 

( ( ) ) ( ) d 1 + δab Saˆ −γU δjb + vj vb + Γbµν eˆµj uν + O ϵ3 U,j Sk dτ 2 ( ) ) ( ( ) 1 1 = −vk Skˆ aj + U,j + Γ0l0 eˆlj u0 + Sˆb −γU,α uα δjb + (U,j + aj )vb + (U,b + ab )vj + Γbµν eˆµj uν + O ϵ3 U,j Sk 2 2 ( ) 1 1 = −vk Skˆ aj − γSˆj (U,t + U,k vk ) + (U,j + aj )Sˆb vb + (U,b + ab )Sˆb vj + Sˆb (−U,b vj + Γbj0 + Γbjc vc ) + O ϵ3 U,j Sk 2 2 ( 1 1 = (U,j − aj )Skˆ vk − γSˆj (U,t + U,k vk ) + (−U,k + ak )Skˆ vj + Sˆj γU,t + g0[b,j] Sˆb + γ(δbj U,c + 2δc[b U,j] )vc Sˆb + O ϵ3 U,j 2 2 ( ) 1 1 = (U,j − aj )Skˆ vk + (−U,k + ak )Skˆ vj + g0[b,j] Sˆb + 2γv[b U,j] Sˆb + O ϵ3 U,j Sk 2 [ 2 ] ( ) = Skˆ (2γ + 1)v[k U,j] − v[k aj] + g0[k,j] + O ϵ3 U,j Sk . (40.6.1) = −vk Skˆ

dvj + Γ0µν eˆµj uν dτ

)

It is straightforward to bring the equation above into three-dimensional vector form, which is equation (40.33) in the textbook.

427

40.7

OFF-DIAGONAL TERMS IN METRIC ABOUT THE EARTH

Idealize the Earth as an isolated, rigidly rotating sphere with angular momentum J. Use equations (39.34b,c) and (39.27) to show that (in three-dimensional vector notation) V ≡ Vj eˆj = W ≡ Wj eˆj =

1 J × r/r3 2

(40.7.1)

outside the Earth, in the Earth’s PPN rest frame. From this, infer equation (40.36).

Solution: For an isolated, rigidly rotating sphere, we have ∫ U (x, t) = Equations (29.27)

ρ0 (x′ , t) 3 ′ M d x = ′ |x − x | r ∫

∫ where

M=

ρ0 (x′ , t) d3 x′ ,

r = |x|.

( ) ∂ 2 U (x′ , t) d3 x′ = 2π[Vj (x, t) − Wj (x, t)] + O ϵ5 ′ ′ ∂xj ∂t |x − x |

then implies that

( ) Wj = Vj + O ϵ5 .

We also have

∫ V =

ρ0 (x′ , t)v(x′ , t) 3 ′ d x =Ω× |x − x′ |

(40.7.2)

(40.7.3)

(40.7.4) ∫

ρ0 (x′ )x′ 3 ′ d x . |x − x′ |

(40.7.5)

And ∫

ρ0 (x′ )z ′ 3 ′ ∑ 4πYlm (θ, ϕ) d x = |x − x′ | (2l + 1)rl+1

∫

′

′

dr ρ(r )r

′l+3

∫

Ylm (θ′ , ϕ′ ) cos θ′ dΩ′

l,m

√ ∫ ∑ 4πYlm (θ, ϕ) ∫ π ′ ′ ′l+3 ′ ′ = dr ρ(r )r Ylm (θ , ϕ )2 Y10 (θ′ , ϕ′ ) dΩ′ l+1 (2l + 1)r 3 l,m ∫ ∫ √ Y10 (θ, ϕ) π ′ ′ ′4 dr ρ(r )r 2 = dΩ′ 3r2 3 ∫ z = 3 ρ(x′ , t)r′2 d3 x′ . 3r

So generally, we have r V =Ω× 3 3r

∫

ρ(x′ , t)r′2 d3 x′ .

If we notice that the inertial tensor for the earth is ∫ ∫ 2 ′ ′2 ′ ′ 3 ′ Iij = ρ(x , t)(r δij − xi xj ) d x = δij ρ(x′ , t)r′2 d3 x′ . 3

(40.7.6)

(40.7.7)

(40.7.8)

The angular momentum of the earth is 2 Ji = Iij Ωj = Ωi 3

∫

ρ(x′ , t)r′2 d3 x′

(40.7.9)

Now we obtain that V =

1 r J × 3. 2 r

428

(40.7.10)

40.8

SPIN-CURVATURE COUPLING

Consider a spinning body (e.g., the Earth or a gyroscope or an electron) moving through curved spacetime. Tidal gravitational forces produced by the curvature of spacetime act on the elementary pieces of the spinning body. These forces should depend not only on the positions of the pieces relative to the center of the object, but also on their relative velocities. Moreover, the spin of the body, ∫ S ≡ (ρr × v) d(volume) (40.8.1) is a measure of the relative positions and velocities of its pieces. Therefore one expects the spin to couple to the tidal gravitational forces-i.e., to the curvature of spacetime-producing deviations from geodesic motion. Careful solution of the PPN equations of Chapter 39 for general relativity reveals [Papapetrou (1951), Pirani (1956)] that such coupling occurs and causes a deviation of the worldline from the course that it would otherwise take; thus, Duα D2 uβ αµνβ 1 λµρτ αν = −Sµ uν ϵ + ϵ R λµ uν Sρ uτ . (40.8.2) dτ dτ 2 2 Evaluate, in order of magnitude, the effects of the supplementary term on planetary orbits in the solar system. m

Solution: The magnitude of ϵλµρτ Rαν λµ uν Sρ uτ is about 3 M⊙ M⊙ v × M Rv ∼ ∼ (2 × 10−6 )3 ∼ 10−17 . (40.8.3) ⊙ R3 R3 So the effects of the supplementary term on planetary orbits in the solar system is too small to be detected.

ϵλµρτ Rαν λµ uν Sρ uτ ∼ ϵjkl0 Ri0jk Sl ∼

40.9

CAVENDISH CONSTANT FOR IDEALIZED SUN

Idealize the sun as a static sphere of perfect fluid at rest at the origin of the PPN coordinates. Then its external gravitational field has the form 40.1.5, with M⊙ given by 40.1.1. Consequently, a test body of mass m, located far away at radius r, is accelerated by a gravitational force Force = −

mM⊙ r2

(40.9.1)

(a) Calculate the mass of the sun, M , in the sense of the amount of energy required to construct it by adding one spherical shell of matter on top of another, working from the inside outward. (b) Use the virial theorem to rewrite equation 40.1.1 in the form ( ) ] ∫ R⊙ [ 1 M⊙ = ρ0 1 + β3 Π + 2β2 + β4 U 4πr2 dr . 2 0

(40.9.2)

(c) Combine the above equations with the definition GC mM r2 of the Cavendish constant for r far outside the sun, to obtain Force = −

(40.9.3)

mass of sun as defined by its effect in bending world line of a faraway test particle mass-energy as defined by applying law of conservation of energy [ ] ∫ ρ0 1 (40.9.4) =1+ (β3 − 1)Π + (4β2 + β4 − 6γ + 1)U 4πr2 dr M0 2

GC =

429

Unless β3 = 1, and 4β2 + β4 − 6γ + 1 = 0 (as they are, of course, in Einstein’s theory), GC will depend on the sun’s internal structure! Specialize equation 40.9.4 to conservative theories of gravity, and explain why the result is what one would expect from equation (40.43).

Solution: (a) The rest mass and internal energy required to construct one spherical shell of matter with thickness dr is ( ) √ (40.9.5) dE 1 = ρu0 −g4πr2 dr + O ρ0 r2 dr ϵ4 , where

( ) ( ) √ 1 u0 −g = 1 + v 2 + 3γU + O ϵ4 = 1 + 3γU + O ϵ4 , 2

So we have

∫

∫

R⊙

dE 1 =

ρ = ρ0 (1 + Π).

ρ0 (1 + Π + 3γU )4πr2 dr .

(40.9.6)

(40.9.7)

0

The work done to overcome gravity to add one spherical shell of matter on top of another is ( ) ˜ 4πr2 dr + O ρ0 r2 dr ϵ4 , dE 2 = ρ0 U where

( ) ˜ (r) = M (r) + O ϵ4 U r

(40.9.8)

(40.9.9)

After construction, we have ∫

( ) M (r′ ) ′ dr + O ϵ4 r′2

R⊙

˜ (R⊙ ) + U (r) = U r

(40.9.10)

Notice that ∫

R ( ) ∫ R⊙ ( 4) ( ) M (r) M 2 (r) ⊙ M (r) dE 2 = dM (r) + O M ϵ = − M (r) d + O M ϵ4 r r r 0 0 0 ∫ ∫ R⊙ 2 2 ( ) M M (r) dr + O M ϵ4 . (40.9.11) = − dE 2 + 2 R⊙ r 0 ∫

R⊙

We have

∫ 2

dE 2 =

M2 + R⊙

∫

R⊙

0

( ) M 2 (r) dr + O M ϵ4 2 r

(40.9.12)

On the other hand, we have ∫ 0

R⊙

] ′ ( ) M (r ) ˜ (R⊙ ) + dr′ dM (r) + O M ϵ4 ρ0 U 4πr2 dr = U ′2 r 0 r R⊙ ∫ [ ] ∫ R⊙ R⊙ ′ ( ) −M (r) M (r ) ′ ˜ (R⊙ ) + M (r) − = U dr M (r) dr + O M ϵ4 ′2 2 r r 0 r 0 ∫ R⊙ 2 2 ( ) M M (r) = dr + O M ϵ4 . + (40.9.13) 2 R⊙ r 0

So we conclude that

∫

R⊙

[

∫

∫

1 dE 2 = 2

The mass of the sun is ∫ ∫ M ≡ (dE 1 + dE 2 ) = 0

R⊙

R⊙

∫

R⊙

( ) ρ0 U 4πr2 dr + O M ϵ4 .

(40.9.14)

0

[ ( ) ] ( ) 1 ρ0 1 + Π + 3γ − U 4πr2 dr + O M ϵ4 2

430

(40.9.15)

(b) From equation 40.1.1, we have ∫ M⊙ =

R⊙

ρ0 (1 + 2β2 U + β3 Π + 3β4 p/ρ0 )4πr2 dr .

(40.9.16)

0

From the Newtonian virial theorem , ⟨∫ ⟩ (∫ ) ∫ ∫ 1 2 3 3 3 4 3 ρ0 v d x − ρ0 U d x + 3 p d x =O ρ0 ϵ d x . 2 long time applied to the sun by itself in its own rest frame, we can conclude that ∫ ∫ ( ) 1 − ρ0 U d3 x + 3 p d3 x = O M ϵ4 . 2 So we have

∫

R⊙

M⊙ = 0

(40.9.17)

(40.9.18)

) ] [ ( 1 ρ0 1 + β3 Π + 2β2 + β4 U 4πr2 dr . 2

(40.9.19)

(c) ( ) ] ∫ R⊙ [ ( ) ρ0 1 + β3 Π + 2β2 + 12 β4 U 4πr2 dr M⊙ 0 GC = = + O ϵ4 ) ] ( ∫ R⊙ [ 1 2 M ρ0 1 + Π + 3γ − 2 U 4πr dr 0 [ ( ) ] ∫ R⊙ ( ) 1 + 0 ρ0 /M0 β3 Π + 2β2 + 12 β4 U 4πr2 dr + O ϵ4 = ) ] [ ( ∫ R⊙ 1 + 0 ρ0 /M0 1 + Π + 3γ − 12 U 4πr2 dr ] [ ∫ ( ) ρ0 1 =1+ (40.9.20) (β3 − 1)Π + (4β2 + β4 − 6γ + 1)U 4πr2 dr + O ϵ4 , M0 2 ∫ where M0 = ρ0 4πr2 dr. For a conservative theory, we have ζ1 = ζ2 = ζ3 = ζ4 = α3 = 0, i.e., ζ = 0,

2β + 2β2 − 3γ − 1 = 0,

β3 − 1 = 0,

β4 − γ = 0,

4β1 − 2γ − 2 − ζ = 0.

(40.9.21)

So ∆Gc ∼ (4β2 + β4 − 6γ + 1)U = [4β2 + γ − (2β + 2β2 − 1) − 3γ + 1]U = −2(β + γ − β2 − 1)U (40.9.22) Gc

40.10

CAVENDISH CONSTANT FOR ANY BODY

Extend the analysis of exercise 40.9 to a source that is arbitrarily stressed and has arbitrary shape and internal velocities (subject to the constraints v 2 ≪ 1, |tˆj kˆ |/ρ0 ≪ 1, U ≪ 1, Π ≪ 1 of the post-Newtonian approximation). Assume that the body is at rest relative to the universal rest frame. Show that GC depends on the internal structure of the source unless 2β1 − β4 = 1,

4β2 + β4 − 6γ = −1,

β3 = 1,

ζ = 0,

η = 0.

(40.10.1)

Of course, these PPN constraints are all satisfied by Einstein’s theory.

Solution: The mass of the star, M , in the sense of the amount of energy required to construct it by adding one spherical shell of matter on top of another, working from the inside outward is ( ) ] ∫ R⊙ [ ( ) 1 2 M= ρ0 1 + Π + v + 3γ − U 4πr2 dr + O M ϵ4 (40.10.2) 2 0

431

In order to have Newtonian potential far from the star, g00 must be isotropic and so terms A and D must vanish, i.e., ζ = η = 0. (40.10.3) The mass of sun as defined by its effect in bending world line of a faraway test particle is ∫ M⊙ =

R⊙

ρ0 (1 + 2β1 v 2 + 2β2 U + β3 Π + 3β4 p/ρ0 )4πr2 dr .

(40.10.4)

0

Using 40.9.17, we have ∫ M⊙ = 0

If M = M⊙ , i.e.,

R⊙

[ ( ) ] 1 2 ρ0 1 + (2β1 − β4 )v + 2β2 + β4 U + β3 Π 4πr2 dr . 2 2β1 − β4 = 1,

4β2 + β4 − 6γ = −1,

GC will be independent of the internal structure of the source.

432

β3 = 1,

(40.10.5)

(40.10.6)

Chapter 41

SPINORS 41.1

ELEMENTARY FEATURES OF THE ROTATION MATRIX

Write equation (41.3) in the form ( ) ( ) θ θ R(θ) = cos − i sin (σ · n) 2 2

(41.1.1)

and establish the following properties: (a) (σ · n)2 = 1 ≡ unit matrix; (b) tr(σ · n) = 0 (tr means “trace”, i.e., sum of diagonal elements); (c) [R, (σ · n)] ≡ R(σ · n) − (σ · n)R = 0; (d) dR i = − (σ · n)R. dθ 2

(41.1.2)

Solution: (a) Using {σi , σj } = 2δij , we have ∑ ∑ ∑ ∑ (σ · n)2 = ni nj σi σj = n2i σi2 + {σi , σj }ni nj = n2i = 1. i,j

i

i0 s=R,I

Suppose the length scale of the space is L. We can get H=

1 2π

(

2πℏ L

)3 ∑ ∑ ∑ ˜˙ 2 Ar,s + k 2 A˜2r,s 2 r=1,2 s=R,I

where

kx >0

ki =

2πℏ ni . L

(43.4.6)

Using 43.3.1, the quantum amplitude will be ′′

′′

′

′

⟨A , t ; A , t ⟩ =

∏

∏ ∏ (

s=R,I r=1,2 kx >0

Mk 2πiℏ sin k(t′′ − t′ )

)1/2

′ ′′ ′ ′′ 2 ) cos ˛(t′′ − t′ ) − 2A˜r,s A˜r,s ] iM k[(A˜r,s2 + A˜r,s ×exp , 2ℏ sin k(t′′ − t′ )

(43.4.7) where M=

1 2π

(

2πℏ L

)3 (43.4.8)

.

Notice that A is completely determined by B in Coulomb’s gauge.

43.5

HAMILTON-JACOBI FORMULATION OF MAXWELL ELECTRODYNAMICS

Regard the four components Aµ of the electromagnetic 4-potential as the primary quantities; split them into a space part Ai and a scalar potential ϕ. (1) Derive from the action principle (in flat spacetime) ∫ 1 I= (E 2 − B 2 ) d4 x , (43.5.1) 8π by splitting off an appropriate divergence, an expression qualitatively similar in character to (43.7). (2) Show that the appropriate quantity to be fixed on the initial and final spacelike hypersurface is not really Ai itself, but the magnetic field, defined by B = ∇ × A. (3) Derive the Hamilton-Jacobi equation for the dynamic phase or action S(B, S) in its dependence on the choice of hypersurface S, and the choice of magnetic field B on this hypersurface, δS 1 2 (4π)2 − = B + δΩ 8π 8π

(

δS δA

)2 .

(43.5.2)

The quantity on the left is Tomonaga’s “bubble time” derivative [Tomonaga (1946); see also Box 21.1].

441

Solution: (1) The Lagrangian density for EM field is L=

(A˙ i + ϕ,i )2 − (ϵijk Ak,j )2 1 (E 2 − B 2 ) = . 8π 8π

(43.5.3)

The canonical momentum for Ai is i πture =

∂L A˙ i + ϕ,i Ei = =− . 4π 4π ∂ A˙ i

(43.5.4)

i Define π i ≡ 4ππture = −Ei . We have ∫ [ 2 ] 1 I= πi − (ϵijk Ak,j )2 d4 x 8π } ∫ { ] 1 1[ 2 = π i A˙ i − πi + (ϵijk Ak,j )2 + π i ϕ,i d4 x 4π 2 } ∫ { ] 1 1[ 2 i ˙ 2 i = π Ai − π + (ϵijk Ak,j ) − π,i ϕ d4 x . 4π 2 i

(43.5.5)

(2) 1 δI = 4π =

1 4π

∫

∫

t2

d3 x [−Ei (δ A˙ i + δϕ,i ) − Bi ϵijk δAk,j ]

dt t1 ∫ t2

∫ dt

d3 x [(E˙ k + ϵijk Bi,j ) δAk + Ei,i δϕ] −

t1

1 4π

If δAi = χ,i , using the fact that Ei,i = 0, we can get ∫ ∫ 3 d x Ei δAi = d3 x (Ei χ),i = 0.

∫

t2 d3 x Ei δAi

(43.5.6)

t1

(43.5.7)

So a transformation Ai → Ai + χ,i on the space sheet t1 and t2 will not change the value of action. The appropriate quantity to be fixed on the initial and final spacelike hypersurface is not really Ai itself, but the magnetic field, defined by B = ∇ × A. (3) From equation 43.5.6, we have δS Ei πi i =− = = πture δA 4π 4π From equation 43.5.5, we can obtain the Hamiltonian density H=

(43.5.8)

] 1 [ 2 π + (ϵijk Ak,j )2 8π i

(43.5.9)

Since δS/δΩ = −H, the Hamilton-Jacobi equation is −

δS 1 2 (4π)2 = B + δΩ 8π 8π

442

(

δS δA

)2 .

(43.5.10)