Introduction to Quantum Field Theory: Classical Mechanics to Gauge Field Theories: Solutions Manual for Teachers

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Solutions Manual for Teachers Introduction to Quantum Field Theory: Classical Mechanics to Gauge Field Theories ETHAN N. CARRAGHER AND ANTHONY G. WILLIAMS

Contents

Preface to Solutions Manual

iii

page vi

1 Lorentz and Poincar´ e Invariance 1.1 Problem 1 1.2 Problem 2 1.3 Problem 3 1.4 Problem 4 1.5 Problem 5 1.6 Problem 6 1.7 Problem 7 1.8 Problem 8 1.9 Problem 9 1.10 Problem 10 1.11 Problem 11 1.12 Problem 12 1.13 Problem 13 1.14 Problem 14 1.15 Problem 15 1.16 Problem 16

1 1 2 3 5 5 7 8 8 9 10 11 12 13 13 14 15

2 Classical Mechanics 2.1 Problem 1 2.2 Problem 2 2.3 Problem 3 2.4 Problem 4 2.5 Problem 5 2.6 Problem 6 2.7 Problem 7 2.8 Problem 8 2.9 Problem 9 2.10 Problem 10 2.11 Problem 11 2.12 Problem 12 2.13 Problem 13 2.14 Problem 14 2.15 Problem 15

18 18 19 20 22 24 29 30 34 35 36 37 40 42 43 45

Contents

iv

3 Relativistic classical fields 3.1 Problem 1 3.2 Problem 2 3.3 Problem 3 3.4 Problem 4 3.5 Problem 5 3.6 Problem 6 3.7 Problem 7 3.8 Problem 8

48 48 52 53 57 60 61 61 64

4 Relativistic Quantum Mechanics 4.1 Problem 1 4.2 Problem 2 4.3 Problem 3 4.4 Problem 4 4.5 Problem 5 4.6 Problem 6 4.7 Problem 7 4.8 Problem 8 4.9 Problem 9 4.10 Problem 10 4.11 Problem 11 4.12 Problem 12

68 68 69 72 75 76 77 79 81 82 86 88 89

5 Introduction to Particle Physics 5.1 Problem 1 5.2 Problem 2 5.3 Problem 3 5.4 Problem 4 5.5 Problem 5 5.6 Problem 6 5.7 Problem 7 5.8 Problem 8 5.9 Problem 9 5.10 Problem 10 5.11 Problem 11 5.12 Problem 12

91 91 92 94 95 97 98 100 102 103 104 108 110

6 Formulation of Quantum Field Theory 6.1 Problem 1 6.2 Problem 2 6.3 Problem 3 6.4 Problem 4 6.5 Problem 5

117 117 118 120 121 123

Contents

v

6.6 6.7 6.8 6.9 6.10 6.11 6.12 6.13

Problem Problem Problem Problem Problem Problem Problem Problem

6 7 8 9 10 11 12 13

127 130 133 135 147 150 151 154

7 Interacting Quantum Field Theories 7.1 Problem 1 7.2 Problem 2 7.3 Problem 3 7.4 Problem 4 7.5 Problem 5 7.6 Problem 6 7.7 Problem 7 7.8 Problem 8 7.9 Problem 9 7.10 Problem 10

163 163 164 164 165 166 167 168 170 171 173

8 Symmetries and Renormalization 8.1 Problem 1 8.2 Problem 2 8.3 Problem 3 8.4 Problem 4 8.5 Problem 5 8.6 Problem 6 8.7 Problem 7 8.8 Problem 8

176 176 177 179 180 181 182 189 191

9 Gauge Field Theories 9.1 Problem 1 9.2 Problem 2 9.3 Problem 3 9.4 Problem 4 9.5 Problem 5 9.6 Problem 6 9.7 Problem 7 9.8 Problem 8 9.9 Problem 9

192 192 194 198 200 202 205 208 220 227

References

231

Preface to Solutions Manual

This solutions manual has been provided to assist teachers who adopt the textbook as a teaching resource for their classes. It is not intended for broad distribution as that would defeat the purpose of the problem sets at the end of each chapter. Some of the problems are challenging for students and such problems could be assigned to a small group or as small projects for individual students. As can be seen from the length of this manual, it took some time and effort for us to prepare and typeset these solutions and it is almost certain that we have not caught every error. So we ask the reader’s forgiveness for any mistakes found and we would be very grateful if these could be reported at the website below. In that way corrections can be made and an appropriate acknowledgement given. Ethan N. Carragher and Anthony G. Williams Adelaide, May 28, 2022

Numbering of equations Two-part equation numbers, such as Eq. (1.42) and Eq. (9.51), refer to equations in this Solutions Manual. Three-part equation numbers, such as Eq. (3.1.16) and Eq. (9.4.23), refer to equations in the book.

Corrections to this book As is the case for the book itself, the current list of corrections for this Solutions Manual along with the names of those who suggested them can be found at: www.cambridge.org/WilliamsQFT. It would be greatly appreciated if anyone finding additional errors in the book itself or in this Solutions Manual could please report them using the relevant corrections link provided on this website.

vi

1

Lorentz and Poincar´e Invariance 1.1 Problem 1 Problem: A muon is a more massive version of an electron and has a mass of 105.7 MeV/c2 . The dominant decay mode of the muon is to an electron, an electron antineutrino and a muon neutrino, µ− → e− + ν¯e + νµ . If we have N (t) unstable particles at time t then the fraction of particles decaying per unit time is a constant, i.e., we have dN/N = −(1/τ )dt for some constant τ . This gives dN/dt = −(1/τ )N , which has the solution N (t) = N0 e−t/τ where we have N0 unstable particles at t = 0. The fraction of particles decaying in the interval t to t + dt −t/τ is −dN/N dt. So the mean lifetime (or lifetime) 0 = (−dN/dt)dt/NR 0 = (1/τ )e R∞ ∞ −x is 0 t (−dN/dt) dt/N0 = τ 0 xe dx = τ , where x = t/τ . The half-life, t1/2 , is the time taken for half the particles to decay, e−t1/2 /τ = 21 , which means that t1/2 = τ ln 2. The decay rate, Γ, is defined as the probability per unit time that a particle will decay, i.e., Γ = (−dN/dt)/N = 1/τ is the inverse mean lifetime. A muon at rest has a lifetime of τ = 2.197 × 10−6 s. Cosmic rays are high-energy particles that have traveled enormous distances from outside our solar system. Primary cosmic rays are particles that have been accelerated by some extreme astrophysical event and secondary cosmic rays are those resulting from collisions of primary cosmic rays with interstellar gas or with our atmosphere. Most cosmic rays reaching our atmosphere will be stable particles such as photons, neutrinos, electrons, protons and stable atomic nuclei (mostly helium nuclei). Muon cosmic rays therefore are secondary cosmic rays produced when primary or secondary cosmic rays collide with out atmosphere. A typical height in the atmosphere for the production of cosmic ray muons is ∼ 15 km. What is the minimum velocity that this muon be produced with in order that it have a 50% chance of reaching the surface of the Earth before decaying? Solution: A stationary observer on Earth will see the time experienced by a muon traveling at speed v to be dilated by a factor of γ = (1 − v 2 /c2 )−1/2 compared to their own. So according to the observer, the half-life of the muon will be γ t1/2 . If the muon travels at a speed that allows it to traverse the L = 15 km of the atmosphere in this time, it will therefore have a 50% chance of reaching the Earth’s surface. That

1

Lorentz and Poincar´e Invariance

2

is, the speed of the muon must satisfy q L = γ t1/2

v=

v2 c2

1−

L

t1/2

(1.1)

to have a 50% chance of reaching the surface before decaying. This can be rearranged to give L v=q t21/2 +

L2 c2

.

(1.2)

Now from the given information, the half-life of the muon is t1/2 = τ ln 2 = (2.197 × 10−6 s) ln 2 = 1.523 × 10−6 s,

(1.3)

so the necessary speed is v=q

15 × 103 m (1.523 × 10−6 s)2 +

(15×103 m)2 (3×108 ms−1 )2

= 299, 653, 700 ms−1 = 0.9995c.

(1.4)

An inertial observer traveling with the muon will see the height of the atmosphere contracted by a factor of γ, which at this speed gives a height of r L v2 = 1− 2L γ c p = 1 − 0.99952 (15 × 103 m) = 456 m.

(1.5)

In other words, from the muon’s point of view, it only needs to travel 456 m to reach the Earth’s surface.

1.2 Problem 2 Problem: The diameter of our Milky Way spiral galaxy is approximately 100,000 - 180,000 light years and our solar system is approximately 25,000 light years from the center of our galaxy. Recalling the effects of time dilation, approximately how fast would you have to travel to reach the center of the galaxy in your lifetime? Estimate how much energy would it take to accelerate your body to this speed. Solution:

3

Problem 3

From the solution to Problem 1.1, the speed needed for an observer to travel a proper distance L in a time T is L v=q T2 +

L2 c2

.

(1.6)

So for a human with a lifetime of, say, T = 80 years, to travel to the center of the Milky Way L = 25,000 light years = 25,000c years away, a speed of 25,000c years v=p (80 years)2 + (25,000 years)2 = 0.999995c

(1.7) (1.8)

would be required. The kinetic energy possessed by a human of mass m = 70 kg at this speed is Ekin = (γ − 1)mc2   1 = √ − 1 (70 kg)(3 × 108 ms−1 ) 1 − 0.9999952 = 2 × 1021 J.

(1.9)

For reference, it would take humanity slightly more than 3 years to consume this much energy at current rates.

1.3 Problem 3 Problem: Consider two events that occur at the same spatial point in the frame of some inertial observer O. Explain why the two events occur in the same temporal order in every inertial frame connected to it by a Lorentz transformation that does not invert time. Show that the time separation between the two events is a minimum in the frame of O. (Hint: Consider Figs. 1.2 and 1.5.) Solution: Since the two events E1 and E2 occur at the same spatial point in frame O, the displacement vector E2 − E1 between them will point entirely along the time axis in this frame, in the positive direction, say. Under any given Lorentz transformation, displacement vectors will be moved along hyperbolae in a spacetime diagram, as illustrated in Fig. 1.1. Further, Lorentz transformations that do not invert time will only move displacement vectors along branches of those hyperbolae. Our vector of interest E2 − E1 lies on a positive-time branch, so under an orthochronus Lorentz transformation it will remain on that positive-time branch. That is, the temporal order of the events remains the same. And the point on such a positive-time branch that has the smallest time component is the one that lies on the time axis, as E2 −E1

4

Lorentz and Poincar´e Invariance

1 1 2

2 2

1

t

Figure 1.1

A two-dimensional representation of spacetime displacement with displacements represented by vectors. Vectors labeled by the same number (1 or 2) are related by Lorentz transformations. y 0

does in frame O. So under an orthochronus Lorentz transformation, E2 − E1 will be moved to a point on the spacetime diagram with a time component at least as 0 the time separation between the events is large as that in frame O, showing that minimized in this frame. To see this algebraically, label the two events in frame00O as T

✓E1 = (ct1 , x) , 00 E2✓= (ct2 , x)T ,

(1.10) (1.11)

where t2 > t1 . Then under a Lorentz transformation Λ, the events go to E1 → E1′ = ΛE1 = Λ(ct1 , x)T , E2 →

E2′

x

(1.12)

T

= ΛE2 = Λ(ct2 , x) .

(1.13)

The difference between theseyevents is y0 T

E2′ − E1′ = Λ (c(t2 − t1 ), 0) .

(1.14)

In particular, the temporal component of this difference is Λ00 c(t2 − t1 ), which is at least as great as c(t2 −t1 ), because Lorentz transformations that do not reverse time satisfy Λ00 ≥ 1. Again we conclude the time separation is minimized in frame O.

x0 ✓ x

Problem 4

5

1.4 Problem 4 Problem: Consider any two events that occur at the same time in the frame of an inertial observer O. Show that by considering any Lorentz transformation there is no limit to the possible time separation of the two events and that the smallest spatial separation of the two events occurs in the frame of O. Solution: Label the two events in frame O as E1 = (ct, x1 )T , T

E2 = (ct, x2 ) .

(1.15) (1.16)

Assume without loss of generality that the events are separated in the x-direction by an amount |∆x| = ̸ 0, and consider a boost in the x-direction with speed v. The difference in the boosted temporal coordinates will be given by Eq. (1.2.89),   v v (1.17) c∆t′ = γ c∆t − ∆x = − γ∆x, c c and so the magnitude of the time separation in the boosted coordinates will be v |∆x|. (1.18) |∆t′ | = q 2 c2 1 − vc2 The factor |∆x| is a non-zero constant, so this expression approaches infinity as v → c. Hence, there is no limit to the possible time separation between the two events. Next, the spatial separations between the events along each direction in the boosted frame, ∆x′ = γ (∆x − v∆t) = γ∆x,

(1.19)

′

(1.20)

′

(1.21)

∆y = ∆y, ∆z = ∆z,

imply a total spatial separation of p |∆x′ | = (γ∆x)2 + (∆y)2 + (∆z)2 p ≥ (∆x)2 + (∆y)2 + (∆z)2 = |∆x|.

(1.22) (1.23)

Hence, the smallest spatial separation between the two events occurs in frame O.

1.5 Problem 5 Problem:

6

Lorentz and Poincar´e Invariance

y

γ1

θ/2

x

θ/2

t

Figure 1.2

γ2

Coordinate system being used for Problem 1.5, where γ1 and γ2 label the two beams of light. Two narrow light beams intersect at angle θ, where θ is the angle between the outgoing beams. The beams intersect head on when θ = 180◦ . Using the addition of velocities formula in Eq. (1.2.126) show that for any angle θ there is always an inertial frame in which the beams intersect head on. Solution: Set up the coordinate system such that the beams of light lie in the x − y plane, with the origin at their intersection and the x axis bisecting the angle θ made by the beams, as in Fig. 1.2. Then the velocity vectors of the beams will be   T   θ θ , c sin ,0 , 2 2      T θ θ u2 = c cos , −c sin ,0 . 2 2 

u1 =

c cos

(1.24) (1.25)

We will boost with speed v in the +x-direction, hoping to find a speed such that the beams are directed entirely in the y ′ -direction and moving oppositely. The velocities in the x′ -direction, given by Eq. (1.2.126) in the text, must therefore vanish: u′x = =⇒

ux − v =0 1 − ucx2v v = ux = c cos

(1.26)   θ . 2

(1.27)

If cos(θ/2) = 1 then the beams intersect head on in the original frame, and it is not necessary to perform a boost. If not, then Eq. 1.27 shows that v < c and therefore our proposed boost is valid. The velocities in the y ′ -direction are u′y =

1 uy γ 1 − ucx2v

Problem 6

7

y0

y

` θ `x

t

Figure 1.3

boost

`0

`y

`0y = `y

θ0

x

`0x =

`x γ

x0

Diagram showing the length contraction of the ruler in a boosted frame. uy =q 1−

v2 c2

±c sin θ2 =q 1 − cos2


θ 2




= ±c

(1.28)

where the ± refers to the first and the second beam, respectively. We see that the beams are oppositely directed in the y ′ -direction and therefore intersect head on in this frame.

1.6 Problem 6 Problem: A ruler of rest length ℓ is at rest in the frame of inertial observer O and is at an angle θ with respect to the +x-direction. Now consider an inertial observer O′ in an identical inertial frame except that it has been boosted by speed v in the +xdirection. What is the length ℓ′ of the ruler and what is the angle θ′ with respect to the +x′ -direction that will be measured by O′ ? Solution: As a result of Eq. (1.2.89), the lengths measured by observer O′ will be contracted compared to the lengths in frame O by a factor of γ in the x′ -direction, and remain the same in the perpendicular directions, as illustrated in Fig. 1.3. If the ruler lies in the x − y plane, O′ will therefore measure the lengths in the x′ and y ′ directions to be r ℓx v2 ′ ℓx = = 1 − 2 ℓ cos θ, (1.29) γ c ℓ′y = ℓy = ℓ sin θ,

(1.30)

Lorentz and Poincar´e Invariance

8

and therefore the total length to be ′

ℓ =

q

r (ℓ′x )2

+

(ℓ′y )2

=ℓ 1−

v2 cos2 θ. c2

(1.31)

The angle θ′ in the boosted frame will satisfy ℓ′y ℓy tan θ′ ≡ ′ = γ ≡ γ tan θ ℓx ℓx



 tan θ . =⇒ θ′ = tan−1  q 2 1 − vc2

(1.32)

1.7 Problem 7 Problem: If a spaceship approaches earth at 1.5×108 ms−1 and emits a microwave frequency of 10 GHz, what frequency will an observer on Earth detect? Solution: The ratio of the frequencies in the source frame and observer frame is given by Eq. (1.2.123) in the text, s 1+β fs = . (1.33) fo 1−β Since the ship is moving towards Earth at a speed of 1.5 × 108 ms−1 = c/2, we write v = −c/2 and so β = v/c = −1/2. An observer on Earth will then detect a frequency of s 1−β fo = fs (1.34) 1+β s 1 + 21 = (10 GHz) (1.35) 1 − 21 = 17.3 GHz.

(1.36)

Note that fo > fs , so the beam has been blueshifted as expected when the source is moving towards the observer.

1.8 Problem 8 Problem: An inertial observer observes two spaceship moving directly toward one another. She measures one to be traveling at 0.7 c in her inertial frame and the other at 0.9

Problem 9

9

y γ θ x v

t

Figure 1.4

A particular ray of light emitted by the moving light source in frame O. c in the opposite direction. What is the magnitude of the relative velocity that each spaceship measures the other to have? Solution: Take the +x-direction to be the direction of motion of the spaceship moving at 0.9c in the frame of the inertial observer. Boost by v = 0.9c in this direction to go to the rest frame of the spaceship. In this frame, the other spaceship will have a speed u′x =

ux − v , 1 − ucx2v

(1.37)

where ux = −0.7c is the speed of that spaceship in the original frame along the +x-direction, by Eq. (1.2.126). Substituting in the given values, we find u′x =

−0.7c − 0.9c

1−

(−0.7c)(0.9c) c2

= −0.9816c.

(1.38)

That is, each spaceship measures the other to be moving towards them at 0.98c.

1.9 Problem 9 Problem: A light source moves with constant velocity v in the frame of inertial observer O. The source radiates isotropically in its rest frame. Show that in the inertial frame of O the light is concentrated in the direction of motion of the source, where half of the photons lie in a cone of semi-angle θ, where cos θ = v/c. Solution:

Lorentz and Poincar´e Invariance

10

In frame O, assume without loss of generality the source of light is moving in the +x-direction at speed v. Consider the ray of light emitted by the source in the x − y plane making an angle θ with the x-axis, shown in Fig. 1.4. It will have velocity vector u = (c cos θ, c sin θ, 0)T .

(1.39)

Boost by speed v in the +x-direction to move to the rest frame of the light source. Here, the x′ -component of the ray’s velocity vector will be given by Eq. (1.2.126): ux − v 1 − ucx2v c cos θ − v = 1 − vc cos θ

u′x =

≡ c cos θ′ ,

(1.40)

which implies cos θ =

 v  v + 1 − cos θ cos θ′ . c c

(1.41)

Now the source emits photons uniformly in all directions in this boosted frame, so half of the photons it emits will lie in the region −90◦ < θ′ < +90◦ . The boundary of this region satisfies cos θ′ = 0, which corresponds in the original frame to cos θ = v/c by Eq. 1.41. Since the boost transformation is continuous, the interior of the original region (satisfying cos θ′ > 0) will be mapped to the interior of the boosted region (satisfying cos θ > v/c). That is, half of the emitted photons will lie in a cone of semi-angle θ = cos−1 (v/c) < 90◦ , showing that the light is concentrated in the direction of motion of the source.

1.10 Problem 10 Problem: Write down the Lorentz transformation rule for an arbitrary (3, 2) tensor Aµνρστ . Hence show that any double contraction of this tensor leads to a contravariant vector, i.e., to a (1, 0) tensor. Solution: Under a Lorentz transformation Λ, a (3, 2) tensor Aµνρστ will transform as ′ ′ ′

Aµνρστ → A′µνρστ = Λµµ′ Λν ν ′ Λρρ′ Aµ ν

′

ρ

σ′ τ ′

′

(Λ−1 )σ σ (Λ−1 )τ τ .

Define T µ = Aµνρνρ . Then using Eq. 1.42, T µ will transform as T µ → T ′µ = A′µνρνρ

′ ′ ′

= Λµµ′ Λν ν ′ Λρρ′ Aµ ν

′

ρ

σ′ τ ′

′

(Λ−1 )σν (Λ−1 )τ ρ

(1.42)

Problem 11

11

h ih i ′ ′ ′ ′ ′ = (Λ−1 )σν Λν ν ′ (Λ−1 )τ ρ Λρρ′ Λµµ′ Aµ ν ρ σ′ τ ′ ′

′ ′ ′

′

= δ σν ′ δ τ ρ′ Λµµ′ Aµ ν =

ρ

σ′ τ ′

′ ′ ′ Λµµ′ Aµ ν ρ ν ′ ρ′ ′

= Λµµ′ T µ .

(1.43)

This is the transformation rule for a (1, 0) tensor, so T µ is a (1, 0) tensor. Of course, the same steps apply if any other pairs of indices were chosen in the double contraction of Aµνρστ to form T µ - our choice here is equivalent to any other choice under a change of basis. Therefore, any double contraction of Aµνρστ produces a (1, 0) tensor.

1.11 Problem 11 Problem: A rocket of initial rest mass M0 has a propulsion system that accelerates it by converting matter into light with negligible heat loss and directs the light in a collimated beam behind it. The propulsion system is turned on for some period of time during which the speed of the rocket is boosted by speed v. Use energy and momentum p conservation to show that the final rest mass of the rocket is given by Mv = M0 (c − v)/(c + v). Solution: In the frame where the rocket is initially at rest, the total momentum is p = 0 and the total energy is the rest energy of the rocket, E = M0 c2 . Once the propulsion is completed, the rocket will have momentum of magnitude γMv v and energy γMv c2 where γ = (1 − v 2 /c2 )−1/2 , while the emitted light will have some momentum pγ directed oppositely to that of the rocket, and an energy pγ c. Momentum conservation then imposes the condition γMv v − pγ = |p| = 0, =⇒

pγ = γMv v,

(1.44)

and energy conservation the condition γMv c2 + pγ c = E = M0 c2 . Putting this information together, γMv c2 + γMv vc = M0 c2  v = M0 γMv 1 + c q =⇒

Mv = M0

1−

1+

v2 c2 v c

(1.45)

Lorentz and Poincar´e Invariance

12

q = M0

1+

v c




1+ s

1− 1+

r

c−v , c+v

= M0 = M0

1−

v c




v c

v c v c

(1.46)

as required.

1.12 Problem 12 Problem: Two identical spaceships approach an inertial observer O at equal speeds but from opposite directions. A second observer O′ traveling on one of the spaceships measures the length of the other spaceship to be only 60% of its length at rest. How fast is each space ship traveling with respect to O? Solution: In frame O, say that each spaceship has speed v and that observer O′ is moving in the +x-direction, so the other spaceship has velocity ux = −v in this direction. The rest frame of O′ is obtained by a boost of speed v along the +x-direction, so in this frame the speed of the other spaceship will be given by the velocity addition formula Eq. (1.2.126), u′x =

2v ux − v 2 , ux v = − 1 − c2 1 + vc2

(1.47)

Now O′ measures the other spaceship to be contracted by a factor 1 = γ

s

 1−

u′x c

2 = 0.6,

(1.48)

so its speed must be |u′x | = 0.8c. With the previous information, this implies 2v 2 = 0.8c, 1 + vc2

or

 c 0 = c2 − 2.5cv + v 2 = v − (v − 2c) . 2

(1.49)

Discounting the superluminal solution, it must be that the spaceships are traveling at speed v = c/2 with respect to O.

Problem 13

13

1.13 Problem 13 Problem: An observer on earth observes the hydrogen-β line of a distant galaxy shifted from its laboratory measured value of 434 nm to a value of 510 nm. How fast was the galaxy receding from the Earth at the time the light was emitted? [You may neglect the effects of the cosmological redshift due to the expansion of the universe during the time of flight of the photons]. Solution: The relation between the frequency fs in the source frame (the frequency that would be measured in a laboratory) and the frequency fo in the observer’s frame on Earth is given by Eq. (1.2.123), s 1−β fs . (1.50) fo = 1+β Since c = f λ, the analogous equation between wavelengths is s 1+β λo = λs . 1−β

(1.51)

With this, we get 

λo λs

2 =

c+v , c−v

(1.52)

and rearranging for the relative speed between the source and the observer,  2 λo −1 λs c. (1.53) v =  2 λo + 1 λs For the case at hand we have λo /λs = (510 nm)/(434 nm) = 1.175, meaning the galaxy is receding from Earth at a speed of v=

1.1752 − 1 c = 0.16c. 1.1752 + 1

(1.54)

1.14 Problem 14 Problem: An inertial observer O is midway between two sources of light at rest in her frame with the sources 2 km apart. Each source emits a flash of light that reach O simultaneously. Another inertial observer O′ is moving parallel to the line that

Lorentz and Poincar´e Invariance

14

passes through the light sources and inertial observer O. In her frame O′ measures the time of each flash and finds that they are spaced 10 ns apart in time. What is the speed of O′ with respect to O? Solution: Say without loss of generality that the sources of light lie on the x-axis in the frame of observer O. Then according to O, the events at which the flashes originated are separated in time by an amount ∆t = 0 and in space by an amount ∆x = 2 km. The temporal separation of the events in the frame of O′ , who is boosted by speed v in the x-direction, is given by the Lorentz transformation rule Eq. (1.2.89),   v v c∆t′ = γ c∆t − ∆x = − q ∆x. (1.55) 2 c c 1 − vc2 Rearranging for the speed, we get v=p

c∆t′ (∆x)2 + (c∆t′ )2

c.

(1.56)

Now we are told that ∆t′ = 10 ns (and so c∆t′ = 3 m), meaning observer O′ must be traveling at a speed of v=p

3m (2 ×

103

m)2 + (3 m)2

c = 0.0015c = 4.5 × 105 ms−1

(1.57)

with respect to observer O.

1.15 Problem 15 Problem: A particle with integer or half-integer spin s has 2s+1 values of spin with respect to any arbitrary spin quantization axis n in the particle’s rest frame. If parity is a good symmetry, a spin eigenstate for a massless particle is a helicity eigenstate with helicity eigenvalues ±s. If parity is not conserved then a single helicity eigenstate is possible with eigenvalue either +s or −s. As succinctly as you can, summarize the key reasons for this result. Solution: Particles are classified according to the representations of the Poincar´e group under which they transform. For the classification of the internal states of the particle, i.e. the intrinsic spin, it helps to consider the little group of Poincar´e transformations that leave the four-momentum of the particle invariant. For massive particles, we can always consider the rest frame with zero momentum, in which case the little group corresponds to the group SO(3) of 3D rotations. We have seen that the representations of SO(3) are labeled according to an integer or half-integer s,

15

Problem 16

and are 2s + 1 dimensional. This corresponds to the 2s + 1 possible spin states for a massive particle of spin s. For massless particles, on the other hand, there is no frame in which the momentum vanishes. The little group is instead the group SE(2) of translations in a plane orthogonal to the momentum vector, and rotations in that plane. For this group, requiring a finite number of helicity degrees of freedom in fact forces there to be only a single possible helicity λ, and the opposite helicity for the antiparticle. This is because the ladder operators that raise or lower the helicity do not give rise to a terminating tower of helicity states (there is no analog to the maximal spin ms = s state that occurs in the massive case), unless there is only a single valid state that they both annihilate. Now under a parity transformation, helicity reverses sign: Shelicity = Sspin · np → −Sspin · np = −Shelicity .

(1.58)

So if a theory is parity invariant, it must be possible for a massless particle to have both helicity +λ and −λ. However, if parity is not a good symmetry, then negating the helicity of a massless particle may not result in a valid particle state, and only a single helicity would be possible.

1.16 Problem 16 Problem: The Poincar´e group is a subgroup of a larger Lie group, the conformal group C(1, 3). This fifteen-parameter group consists of the ten-parameter subgroup of Poincar´e transformations, the one-parameter subgroup of scale transformations (dilatations), and the four-parameter subgroup of special conformal transformations (SCT). The generators of the unitary representation of the group are: P µ = iℏ ∂ µ (translations); M µν = iℏ (xµ ∂ ν −xν ∂ µ ) = xµ P ν − xν P µ (Lorentz
transformations); D = iℏ xµ ∂µ = x · P (dilatations); and K µ = iℏ 2xµ x·∂ −x2 ∂ µ = 2xµ D − x2 P µ (SCT). Using [P µ , xν ] = iℏg µν we have already seen that P µ and M µν satisfy the Lie algebra of the Poincar´e group. Show that the remaining commutators of the Lie algebra of the conformal group are: [D, P µ ] = −iℏ P µ , [D, K µ ] = iℏ K µ , [D, M µν ] = 0, [K µ , P ν ] = −iℏ 2(M µν +g µν D), [K µ , K ν ] = 0 and [K µ , M ρσ ] = iℏ (g µρ K σ −g µσ K ρ ). Solution: The following derivations will make extensive use of the commutator identities [AB, C] = A[B, C] + [A, C]B,

(1.59)

[A, BC] = [A, B]C + B[A, C],

(1.60)

which may readily be verified by expansion of both sides of the equations.

Lorentz and Poincar´e Invariance

16

(i) Using the first identity, we have [D, P µ ] = [xν Pν , P µ ] = xν [Pν , P µ ] + [xν , P µ ]Pν = 0 − iℏg νµ Pν = −iℏP µ . (1.61) (ii) For this part it helps to first calculate [D, xµ ] = [xν Pν , xµ ] = xν [Pν , xµ ] + [xν , xµ ]Pν = iℏg νµ xν + 0 = iℏxµ (1.62) and [D, x2 ] = [D, xµ xµ ] = [D, xµ ]xµ + xµ [D, xµ ] = iℏxµ xµ + iℏxµ xµ = 2iℏx2 . (1.63) Note that a commutator containing a covariant index is just the covariant version of that same commutator with a contravariant index, and vice versa, because the metric components are constant and thus commute with all operators. Now we use these findings to calculate [D, K µ ] = [D, 2xµ D − x2 P µ ]

= 2[D, xµ ]D − [D, x2 ]P µ − x2 [D, P µ ]

= 2iℏxµ D − 2iℏx2 P µ + iℏx2 P µ

= iℏ(2xµ D − x2 P µ ) = iℏK µ .

(1.64)

(iii) Next we see [D, xµ P ν ] = [D, xµ ]P ν + xµ [D, P ν ] = iℏxµ P ν − iℏxµ P ν = 0,

(1.65)

and so [D, M µν ] = [D, xµ P ν − xν P µ ] = 0.

(1.66)

(iv) First, [x2 , P µ ] = xν [xν , P µ ] + [xν , P µ ]xν = −iℏxν g νµ − iℏδνµ xν = −2iℏxµ . (1.67) Hence, [K µ , P ν ] = [2xµ D − x2 P µ , P ν ]

= 2xµ [D, P ν ] + 2[xµ , P ν ]D − x2 [P µ , P ν ] − [x2 , P ν ]P µ

= −2iℏxµ P ν − 2iℏg µν D − 0 + 2iℏxν P µ

= −2iℏ(xµ P ν − xν P µ + g µν D)

= −2iℏ(M µν + g µν D)

(1.68)

(v) Using our earlier results, we have [K µ , xν ] = [2xµ D − x2 P µ , xν ] = 2xµ [D, xν ] − x2 [P µ , xν ] = 2iℏxµ xν − iℏx2 g µν

(1.69)

Problem 16

17

and [K µ , x2 ] = [K µ , xν ]xν + xν [K µ , xν ] = 2iℏxµ xν xν − iℏx2 g µν xν + 2iℏxν xµ xν − iℏxν x2 δνµ

= 2iℏx2 xµ ,

(1.70)

since components of the position vector commute amongst themselves. It follows that [K µ , K ν ] = [K µ , 2xν D − x2 P ν ]

= 2[K µ , xν ]D + 2xν [K µ , D] − [K µ , x2 ]P ν − x2 [K µ , P ν ]

= 4iℏxµ xν D − 2iℏx2 g µν D − 2iℏxν K µ − 2iℏx2 xµ P ν + 2iℏx2 (M µν + g µν D)

= 2iℏ(2xµ xν D − xν K µ − x2 xµ P ν + x2 M µν )

= 0.

(1.71)

(vi) Finally, let us calculate [xµ , M ρσ ] = [xµ , xρ P σ − xσ P ρ ]

= xρ [xµ , P σ ] − xσ [xµ , P ρ ]

=⇒ =⇒

µ

ρσ

2

ρσ

xµ [x , M [x , M

= −iℏxρ g µσ + iℏxσ g µρ , ρ σ

(1.72)

σ ρ

] = −iℏx x + iℏx x = 0 ] = 0.

(1.73)

Then [K µ , M ρσ ] = [2xµ D − x2 P µ , M ρσ ]

= 2[xµ , M ρσ ]D + 2xµ [D, M ρσ ] − [x2 , M ρσ ]P µ − x2 [P µ , M ρσ ]

= 2iℏ(xσ g µρ − xρ g µσ )D + 0 − 0 − iℏx2 (g µρ P σ − g µσ P ρ )
= iℏ g µρ (2xσ D − x2 P σ ) − g µσ (2xρ D − x2 P ρ ) = iℏ(g µρ K σ − g µσ K ρ ),

(1.74)

given we already know P µ and M µν satisfy the Lie algebra of the Poincar´e group, Eq. (1.2.167).

2

Classical Mechanics 2.1 Problem 1 Problem: A cylinder of mass M , moment of inertia I about the cylindrical axis and radius R rolls on a horizontal surface without slipping: (a) Express the no-slip constraint in differential form. Is it holonomic? Explain. (b) Write down a Lagrangian for the system. Solution: (a) To not slip, the distance traveled by the center of mass of the cylinder must equal the distance rolled along the circumference. So if a particular point on the circumference turns through a differential angle dθ while the center of the cylinder moves a horizontal distance dx, the no-slip constraint is given by dx = Rdθ,

(2.1)

x − Rθ − constant = 0.

(2.2)

or equivalently

This is the form of a holonomic constraint since there is no dependence on the ˙ generalized velocities x˙ and θ. (b) The cylinder, being a rigid body, has kinetic energy equal to the sum of its translational kinetic energy and the rotational kinetic energy about its center of mass: 1 1 (2.3) T = M x˙ 2 + I θ˙2 . 2 2 Assuming the cylinder is constrained to always be touching the ground, its potential energy will be constant and can therefore be regarded as zero. So an appropriate Lagrangian for the system is L=T −V =

1 1 M x˙ 2 + I θ˙2 . 2 2

(2.4)

Lagrange multipliers will be needed to enforce the constant-height constraint in the equations of motion. 18

Problem 2

19

x2 m x1 − `

t

Figure 2.1

M

x=0 Coordinates used to analyse the sliding blocks of Problem 2.2. The natural length of the spring is denoted by ℓ.

2.2 Problem 2 Problem: A small block of mass m slides without friction on top of a larger block of mass M and is attached to a pin in this block by a massless spring with spring constant k. All motion is in one dimension. The larger block slides without friction on a flat table. Construct a Lagrangian for the system and obtain the equations of motion. Identify any conserved quantities. Relate these to the symmetries of the system. Solution: Let us use the coordinates shown in Fig. 2.1: x1 is the horizontal coordinate of the pin in the larger block plus the natural length of the spring, and x2 is the horizontal coordinate of the smaller block, so that the end of the spring is displaced from its equilibrium position by an amount (x2 − x1 ). Then the total kinetic energy of the system is simply T =

1 1 M x˙ 21 + mx˙ 22 , 2 2

(2.5)

while the potential energy is that stored in the spring: V =

1 k(x2 − x1 )2 . 2

(2.6)

In total, the Lagrangian for the system is L=T −V =

1 1 1 M x˙ 21 + mx˙ 22 − k(x2 − x1 )2 . 2 2 2

(2.7)

The equation of motion for the coordinate x1 is d ∂L ∂L − dt ∂ x˙ 1 ∂x1 d = (M x˙ 1 ) − k(x2 − x1 ) dt = Mx ¨1 − k(x2 − x1 ),

0=

(2.8)

Classical Mechanics

20

and that for x2 is d ∂L ∂L − dt ∂ x˙ 2 ∂x2 d = (mx˙ 2 ) + k(x2 − x1 ) dt = m¨ x2 + k(x2 − x1 ).

0=

(2.9)

Notice that summing Eq. 2.8 and Eq. 2.9 yields the equation d (M x˙ 1 + mx˙ 2 ) = 0, dt

(2.10)

which is simply a statement of the conservation of momentum. This is to be expected because momentum is the conserved quantity related to spatial translations x1 → x1 + s and x2 → x2 + s, which are clearly a symmetry of our Lagrangian. Further, we note that the system is natural and the Lagrangian is time independent, so the total energy E = T + V should be conserved too. This is simple to check, using the above equations of motion:   dE d 1 1 1 2 2 2 = M x˙ 1 + mx˙ 2 + k(x2 − x1 ) dt dt 2 2 2 = M x˙ 1 x ¨1 + mx˙ 2 x ¨2 + k(x2 − x1 )(x˙ 2 − x˙ 1 )

= x˙ 1 [M x ¨1 − k(x2 − x1 )] + x˙ 2 [m¨ x2 + k(x2 − x1 )] = 0,

(2.11)

as expected.

2.3 Problem 3 Problem: A uniform circular wire of radius R is forced to rotate about a fixed vertical diameter at constant angular velocity ω. A bead of mass m experiences gravity, is smoothly threaded on the wire and can slide without friction. Is the system holonomic? Construct a Lagrangian for the system. Show that there is just one solution for the off-axis motion of the bead where it does not oscillate. Does the system conserve energy? Solution: Let us use the cartesian coordinate system of Fig. 2.2 with origin at the center of the wire circle. When the bead is at an angle θ from the bottom of the wire, its position can be written as x = (R sin θ cos(ωt), R sin θ sin(ωt), −R cos θ)T ,

(2.12)

Problem 3

21

z

y

x θ

t

Figure 2.2

R

ω Coordinates used to analyse the rotating wire and bead system of Problem 2.3. given an appropriate time coordinate. Note that this implies a constraint of the form x · x − R2 = x2 + y 2 + z 2 − R2 = 0,

(2.13)

which does not depend on the generalized velocities and is thus holonomic. The kinetic energy of the bead is 1 ˙ 2, m|x| 2 while the potential energy is entirely gravitational: T =

V = mgz = −mgR cos θ.

(2.14)

(2.15)

Now the velocity of the bead can be decomposed into perpendicular components, ˙ and the other tangential to the one tangential to the wire circle (of magnitude Rθ), circle of radius R sin θ swept out by the bead (of magnitude (R sin θ)ω). It follows that the speed of the bead satisfies ˙ 2 = R2 θ˙2 + R2 ω 2 sin2 θ, |x|

(2.16)

as can be verified explicitly from Eq. 2.12. Hence, the Lagrangian of the system is given by 1 mR2 (θ˙2 + ω 2 sin2 θ) + mgR cos θ. 2 The equation of motion is therefore L=T −V =

d ∂L ∂L − dt ∂ θ˙ ∂θ 2¨ = mR θ − mR2 ω 2 sin θ cos θ + mgR sin θ.

(2.17)

0=

(2.18)

Classical Mechanics

22

Now suppose the bead is off-axis (so sin θ ̸= 0) and does not oscillate (so θ¨ = 0). Then the equation of motion dictates −mR2 ω 2 sin θ cos θ + mgR = 0.

(2.19)

Therefore there is only one solution, cos θ = g/(Rω 2 ), for off-axis motion of the bead with no oscillations. It might be thought that energy is conserved in this system because the Lagrangian has no explicit time dependence. However, this is incorrect because the ˙ so the Hamilsystem is not natural (the kinetic energy is not homogeneous in θ), tonian is not equal to the energy of the system. In fact, it is conceptually clear that energy E = T + V is generically not conserved in this system: the wire is driven by an external force so that it rotates with a constant angular speed ω independently of the motion of the bead, so the energy associated to this rotation, supplied by the external driving system, does not need to balance the other forms of energy of the bead. Checking this algebraically:   dE d 1 2 2 ˙2 2 = mR (θ + ω sin θ) − mgR cos θ dt dt 2 ˙ 2 sin θ cos θ + mgRθ˙ sin θ = mR2 θ˙θ¨ + mR2 θω ˙ 2 sin θ cos θ = 2mR2 θω

(by the equation of motion)

̸= 0,

(2.20)

as was reasoned above.

2.4 Problem 4 Problem: A bead of mass m is threaded without friction on a massless wire hoop of radius R which is forced to oscillate vertically in a fixed vertical plane at angular frequency ω and with amplitude a. Is the system holonomic? Construct a Hamiltonian for the system and obtain Hamilton’s equations. Does the system conserve energy? Solution: Let us use a cartesian coordinate system with origin at the average position of the center of the wire hoop, as in Fig. 2.3. Then the position of the center can be parameterized as xcenter = (0, a sin ωt)T .

(2.21)

The position of the bead, x = (x, y)T , is constrained to lie on a circle of radius R around this point: |x − xcenter |2 = R2 ,

(2.22)

Problem 4

23

y

a a

x θ

t

Figure 2.3

R

Coordinates used to analyse the oscillating hoop and bead system of Problem 2.4. and therefore the coordinates obey the holonomic constraint x2 + (y − a sin ωt)2 − R2 = 0.

(2.23)

Now whether the system is natural depends on the form of the kinetic energy T = mx˙ 2 /2. To calculate this, we first parameterize the coordinates by the angle θ of the bead from the bottom of the hoop as x = xcenter + (R sin θ, −R cos θ)T = (R sin θ, a sin ωt − R cos θ)T ,

(2.24)

leading to x˙ = (Rθ˙ cos θ, aω cos ωt + Rθ˙ sin θ)T ,

(2.25)

x˙ 2 = R2 θ˙2 cos2 θ + a2 ω 2 cos2 ωt + 2Raω θ˙ sin θ cos ωt + R2 θ˙2 sin2 θ = R2 θ˙2 + a2 ω 2 cos2 ωt + 2Raω θ˙ sin θ cos ωt.

(2.26)

and therefore

˙ it contains both θ˙ and θ˙2 terms. The This expression is not homogeneous in θ: ˙ and so the system is not natural. kinetic energy is therefore not homogeneous in θ, To construct a Hamiltonian for the system we must first write a Lagrangian. We already have an expression for the kinetic energy, and the potential energy of the bead is entirely gravitational: V = mgy = mg(a sin ωt − R cos θ), so our Lagrangian is L=T −V =

1 1 mR2 θ˙2 + ma2 ω 2 cos2 ωt + mRaω θ˙ sin θ cos ωt 2 2

(2.27)

Classical Mechanics

24

− mg(a sin ωt − R cos θ).

(2.28)

Now we must calculate the conjugate momentum ∂L = mR2 θ˙ + mRaω sin θ cos ωt ∂ θ˙ in order to finally construct the Hamiltonian pθ :=

(2.29)

H := pθ θ˙ − L

1 1 = mR2 θ˙2 + mRaω θ˙ sin θ cos ωt − mR2 θ˙2 − ma2 ω 2 cos2 ωt 2 2 − mRaω θ˙ sin θ cos ωt + mg(a sin ωt − R cos θ) 1 1 = mR2 θ˙2 − ma2 ω 2 cos2 ωt + mg(a sin ωt − R cos θ). (2.30) 2 2 Importantly, the Hamiltonian is a function of position and momentum only, and not of the generalized velocities. This means we must use Eq. 2.29 to substitute out the generalized velocity θ˙ and write the Hamiltonian instead as H=

1 (pθ − mRaω sin θ cos ωt)2 − ma2 ω 2 cos2 ωt + mg(a sin ωt − R cos θ). 2 2mR 2 (2.31)

Hamilton’s equations of motion are ∂H pθ − mRaω sin θ cos ωt θ˙ = , = ∂pθ mR2

(2.32)

which is simply a restatement of Eq. 2.29, and ∂H (pθ − mRaω sin θ cos ωt) = (−mRaω cos θ cos ωt) − mgR sin θ. (2.33) ∂θ mR2 A rather involved calculation would show that the total energy E = T + V is not conserved over time when these equations of motion are satisfied. This does not follow from the explicit time dependence of the Lagrangian, because the system is not natural and so the Hamiltonian is not equal to the total energy. However, it is clear on physical grounds that the energy will generically not be conserved: imagine the bead sitting at the bottom of the hoop, driven at a very high frequency ω with a very small amplitude a. The kinetic energy of the bead would vary wildly, while the potential energy would be almost constant, so the total energy would be changing over time. The energy is not conserved because the system is not isolated; the external system driving the oscillation is supplying extra energy to the bead. p˙θ = −

2.5 Problem 5 Problem: Two point masses m and M are connected by a massless rod of length ℓ and placed

Problem 5

25

y x

M

b

θ ` x

m

t

Figure 2.4

Diagram showing the generalized coordinates x and θ for the constrained rod and spring system of Problem 2.5. on a horizontal table. The mass m is also connected to a fixed point P on the table by a spring with spring constant k and zero natural length; it can slide without friction in any direction on the table. The mass M is constrained to slide along a straight line a perpendicular distance b < ℓ from P . Set up a Lagrangian for the system and obtain the Euler-Lagrange equations. In the limit of small oscillations, solve for the characteristic angular frequencies and normal modes. Explain whether or not energy is conserved? Solution: Let us use a cartesian coordinate system with origin located at P , as in Fig. 2.4. The position of mass M will then be written xM = (x, b)T .

(2.34)

If θ is the angle the rod makes from the vertical (going clockwise), the mass m at the end of the rod will be located at the point xm = xM + (−ℓ sin θ, −ℓ cos θ)T = (x − ℓ sin θ, b − ℓ cos θ)T .

(2.35)

Our two generalized coordinates here are the horizontal position x of the mass M , and the angle θ of the rod. The velocities of the two masses are x˙ M = (x, ˙ 0)T , x˙ m = (x˙ − ℓθ˙ cos θ, ℓθ˙ sin θ)T , so the kinetic energy of the system is T =

1 1 M x˙ 2M + mx˙ 2m 2 2

(2.36)

Classical Mechanics

26

1 M x˙ 2 + 2 1 = M x˙ 2 + 2 =

1 m(x˙ 2 − 2xℓ ˙ θ˙ cos θ + ℓ2 θ˙2 cos2 θ + ℓ2 θ˙2 sin2 θ) 2 1 m(x˙ 2 − 2xℓ ˙ θ˙ cos θ + ℓ2 θ˙2 ). 2

(2.37)

The potential energy of the system is that stored in the spring, which is extended by a length |xm |: 1 2 kx 2 m
1 = k (x − ℓ sin θ)2 + (b − ℓ cos θ)2 2 1 ∼ = k(x2 − 2xl sin θ − 2bl cos θ), 2

V =

(2.38)

where we have dropped inconsequential constant terms in the last line. In total, an appropriate Lagrangian for the system is L=T −V =

1 1 1 M x˙ 2 + m(x˙ 2 − 2xℓ ˙ θ˙ cos θ + ℓ2 θ˙2 ) − k(x2 − 2xl sin θ − 2bl cos θ). 2 2 2 (2.39)

The Euler-Lagrange equation for the coordinate x is d ∂L ∂L − dt ∂ x˙ ∂x d = (M x˙ + mx˙ − mℓθ˙ cos θ) − (−kx + kℓ sin θ) dt = (M + m)¨ x − mℓθ¨ cos θ + mℓθ˙2 sin θ + k(x − ℓ sin θ),

0=

(2.40)

and that for the coordinate θ is d ∂L ∂L − dt ∂ θ˙ ∂θ d = (mℓ2 θ˙ − mxℓ ˙ cos θ) − (mxℓ ˙ θ˙ sin θ + kxℓ cos θ − kbℓ sin θ) dt = mℓ2 θ¨ − m¨ xℓ cos θ + kℓ(b sin θ − x cos θ).

0=

(2.41)

Now to find the behavior of small oscillations, we must first identify the stable equilibrium points of the system, about which oscillations can take place. At such points, the gradient of the potential energy must vanish:  ∂V      kx − kℓ sin θ 0 ∂x = = . (2.42) ∂V −kxℓ cos θ + kbℓ sin θ 0 ∂θ There are two classes of solutions to these equations: 1) 2)

x = 0, p x = ± ℓ2 − b2 ,

sin θ = 0, x sin θ = , ℓ

and and

cos θ = ±1; b cos θ = . ℓ

(2.43) (2.44)

Since b < ℓ, both of these solutions are valid. As for which of these is a stable

Problem 5

27

equilibrium, we look to the matrix of second derivatives !   ∂2V ∂2V k −kℓ cos θ 2 ∂x ∂x∂θ K(x, θ) ≡ = . ∂2V ∂2V −kℓ cos θ kxℓ sin θ + kbℓ cos θ ∂θ∂x ∂θ 2

(2.45)

Stability occurs if this matrix is positive-definite, which, since k > 0, will be the case if and only if det K(x, θ) > 0. Now in the first class of solutions we have cos θ = ±1, so the second derivative matrix will have the form   k −kℓ(±1) K= , (2.46) −kℓ(±1) kbℓ(±1) and therefore a determinant of k 2 ℓ(±b − ℓ) < 0, so this solution does not represent a stable equilibrium point. For the second class of solutions, where cos θ = b/ℓ, the second derivative matrix is   k −kb K= . (2.47) −kb kℓ2 The determinant here is k 2 (ℓ2 − b2 ) > 0, so this solution does represent a stable equilibrium setup. To summarize, the only stable equilibria in this system are those √ configurations where x = ± ℓ2 − b2 , sin θ = x/ℓ, and cos θ = b/ℓ. Let us denote the coordinates at such a point by x0 and θ0 , so then we can expand our coordinates in terms of small deviations x ˜ and θ˜ about that point as x =: x0 + x ˜, ˜ θ =: θ0 + θ.

(2.48)

Small oscillations are found by expanding the Lagrangian to second order in x ˜ and ˜ and their derivatives. In fact we already know the second order expansion of the θ, potential energy: the only important terms are the quadratic terms, dictated by the matrix K in Eq. 2.47. So only the kinetic energy needs to be expanded further. To do so, we note the Taylor expansion ˜ ˜ = cos θ0 − θ˜ sin θ0 + O(θ˜2 ) = b + O(θ) cos θ = cos(θ0 + θ) ℓ allows us to expand the kinetic energy as 1 M x˙ 2 + 2 1 = M x˙ 2 + 2  1 ˙ = (x˙ θ) 2

T =

1 m(x˙ 2 − 2xℓ ˙ θ˙ cos θ + ℓ2 θ˙2 ) 2   1 b 2 2 ˙2 ˙ m x˙ − 2xℓ ˙ θ + ℓ θ + O(θ3 ) 2 ℓ   x˙ M + m −mb + O(θ3 ). −mb mℓ2 θ˙

(2.49)

(2.50)

In an abuse of notation we have replaced the symbols x ˜ and θ˜ by simply x and θ from the second line onwards. We recognize the matrix   M + m −mb 0 M = , (2.51) −mb mℓ2

Classical Mechanics

28

that dictates the frequency ω of small oscillations through the relation 0 = det(K − ω 2 M 0 )   k − ω 2 (M + m) −kb + ω 2 mb = det −kb + ω 2 mb kℓ2 − ω 2 mℓ2

= ω 4 [M mℓ2 + m2 (ℓ2 − b2 )] − ω 2 k[M ℓ2 + 2m(ℓ2 − b2 )] + k 2 (ℓ2 − b2 )  
k 2 =m ω − [M ℓ2 + m(ℓ2 − b2 )]ω 2 − k(ℓ2 − b2 ) . (2.52) m

The two solutions are r ω1 =

k , m

s ω2 =

k(ℓ2 − b2 ) . M ℓ2 + m(ℓ2 − b2 )

(2.53)

Now the normal modes v1 and v2 corresponding to these frequencies are those vectors that satisfy (K − ω12 M 0 )v1 = 0,

(K − ω22 M 0 )v2 = 0.

(2.54)

If the top component of one of these vector equations is satisfied then the bottom 2 component will be satisfied too, since det(K − ω1,2 M 0 ) = 0. So let us solve the top equation for the first normal mode:   M +m v11 = 0, =⇒ v11 = 0. (2.55) k−k m Really we only need to solve for v1,2 up to a scalar multiple C1,2 , so we can write   0 v1 = C1 . (2.56) 1 As for the normal mode v2 corresponding to ω2 , we get the equation (K − ω22 M 0 )v2 = 0,   k(ℓ2 −b2 )mb kb − 1 2 2 2 M ℓ +m(ℓ −b ) ℓ2 v2  = , =  =⇒ 2 2 2 −b )(M +m) v2 b k − k(ℓ M ℓ2 +m(ℓ2 −b2 ) ! 2 and therefore

v2 = C2

ℓ b

1

.

The normal mode oscillations are then simply     x(t) x(t) = v1 cos(ω1 t + ϕ1 ) and = v2 cos(ω2 t + ϕ2 ), θ(t) θ(t) for some arbitrary constants ϕ1 and ϕ2 .

(2.57)

(2.58)

Problem 6

29

2.6 Problem 6 Problem: A mass-m particle moves along the x-axis subject to the potential energy V (x) = λ[(x2 − a2 )2 + 2ax3 ], where λ and a are positive constants: (a) Find all equilibrium points and determine which are stable. (b) Find the angular frequency of small oscillations about the point with the lowest energy. Solution: (a) Equilibrium points are the stationary points of the potential, i.e. the points that satisfy
∂V = λ 2(x2 − a2 ) · 2x + 6ax2 = 0 ∂x =⇒ 2x(2x − a)(x + 2a) = 0.

(2.59)

Hence, there are three stationary points: x = 0, x = a/2, and x = −2a. Whether these are stable depend on the second derivative of the potential:  2  −4λa for x = 0 ∂2V 2 2 2 = λ(12x − 4a + 12ax) = (2.60) 5λa for x = a2  ∂x2 2 20λa for x = −2a. The first of these is negative, so x = 0 is an unstable equilibrium point, while the others are positive, so x = a/2 and x = −2a are stable equilibrium points. (b) By the second derivative test, the point with the lowest energy is either x = a/2 or x = −2a. Direct calculation reveals

13 4 λa > −7λa4 = V (−2a), (2.61) 16 so x = −2a is the point with the lowest energy. To find the frequency of small oscillations about this point, let us change coordinates to η := x + 2a. Equilibrium occurs at η = 0. The kinetic energy is simply V (a/2) =

1 1 mx˙ 2 = mη˙ 2 , (2.62) 2 2 while Taylor expansion of the potential gives ∂V 1 ∂ 2 V V (x) = V (−2a) + (x + 2a) + (x + 2a)2 + O((x + 2a)3 ) ∂x x=−2a 2 ∂x2 x=−2a 1 = −7λa4 + 0 + 20λa2 η 2 + O(η 3 ). (2.63) 2 Using the general procedure outlined in Section 2.3, we recognize the matrices T =

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30

M 0 = [m] and K = [20λa2 ] from the above equations. The frequency ω of small oscillations about η = 0 then satisfies det(K − ω 2 M 0 ) = 0,

(2.64)

giving r ω=

20λa2 . m

(2.65)

In this simple case where the system is one-dimensional, it is easy to obtain this same result by recognizing the Lagrangian 1 mη˙ 2 − 2 is, for small enough η, the Lagrangian of above frequency. L=T −V =

1 20λa2 η 2 + O(η 3 ), (2.66) 2 a simple harmonic oscillator with the

2.7 Problem 7 Problem: A uniform thin rod of length ℓ and mass m is suspended from fixed points A and B by two identical massless springs with zero natural length and spring constant k. At equilibrium (as shown in the figure), the angle between the springs and the vertical is ϕ and the length of the springs is a. All motion is restricted to this vertical plane. (a) Construct a Lagrangian for the system and obtain the Euler-Lagrange equations. (b) Find the normal modes of small oscillations of the system about equilibrium and their angular frequencies. (c) Construct the Hamiltonian, obtain Hamilton’s equations and show equivalence to the Euler-Lagrange equations. Solution: (a) Let us use cartesian coordinates x = (x, y)T to denote the position of the center of the rod, with origin such that equilibrium occurs at (x, y) = (0, 0), and also let θ be the angle the rod makes with the horizontal. These will be our generalized coordinates, depicted in Fig. 2.5. With these coordinates the kinetic energy takes on a simple form, for it is the sum of the translational kinetic energy of the rod and the rotational kinetic energy about its center of mass: 1 1 T = m(x˙ 2 + y˙ 2 ) + I θ˙2 , (2.67) 2 2

Problem 7

31

y A

B `/2 xL

t

Figure 2.5

x

`/2

xR

θ x

Diagram showing our generalized coordinates x and θ for the system of Problem 2.7. where I = mℓ2 /12 is the moment of inertia of the rod about its center of mass. The potential energy of the system is more involved. It is the sum of the potential energy stored in the each spring, and also the gravitational potential energy of the rod. We need to find the length of each spring in terms of x, y, θ, and the given constants. Now the left endpoint of the rod will be at position T  ℓ ℓ , (2.68) xL = x − cos θ, y − sin θ 2 2 and the right at position xR =

 T ℓ ℓ x + cos θ, y + sin θ . 2 2

(2.69)

Meanwhile, the positions of the fixed endpoints of the left and right springs (A and B, say) are  T ℓ xA = − − a sin ϕ, a cos ϕ , 2  T ℓ xB = + a sin ϕ, a cos ϕ , (2.70) 2 as may be ascertained from the geometry of the equilibrium position. The potential energy will then be 1 1 k|xL − xA |2 + k|xR − xB |2 + mgy 2 2  2  2 1 ℓ ℓ ℓ = k x − cos θ + + a sin ϕ + y − sin θ − a cos ϕ 2 2 2 2  2  2 ! ℓ ℓ ℓ + x + cos θ − − a sin ϕ + y + sin θ − a cos ϕ + mgy 2 2 2

V =

(2.71) The identity (a + b)2 + (a − b)2 = 2(a2 + b2 ) is useful here, ultimately leading

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32

to V =

    1 ℓ 2 k 2x2 + 2 (y − a cos ϕ) − 2ℓ + a sin ϕ cos θ + mgy, 2 2

(2.72)

where inconsequential constant terms have been discarded. In total, our Lagrangian for the system is   1 1 2 ˙2 2 2 L = T − V = m x˙ + y˙ + ℓ θ 2 12     1 ℓ 2 2 − k 2x + 2 (y − a cos ϕ) − 2ℓ + a sin ϕ cos θ − mgy. 2 2 (2.73) The Euler-Lagrange equations for the x coordinate are d ∂L ∂L − dt ∂ x˙ ∂x = m¨ x + 2kx,

0=

(2.74)

for the y coordinate are 0=

d ∂L ∂L − dt ∂ y˙ ∂y

= m¨ y + 2k(y − a cos ϕ) + mg,

(2.75)

and for the θ coordinate are d ∂L ∂L − 0= dt ∂ θ˙ ∂θ   1 ℓ 2¨ = mℓ θ + kℓ + a sin ϕ sin θ. 12 2 (b) Let us find the equilibrium points of this system, where potential energy vanishes:  ∂V     2kx ∂x  ∂V  =  2k(y − a cos ϕ) + mg  =  ∂y
∂V kℓ 2ℓ + a sin ϕ sin θ ∂θ

(2.76) the gradient of the  0 0 . 0

(2.77)

Now by definition of our coordinates, (x, y, θ) = (0, 0, 0) is an equilibrium point. The middle equation then amounts to a relation between given constants: −2ka cos ϕ + mg = 0.

(2.78)

We see that the equilibrium points are those such that x = 0, y = 0, and sin θ = 0. Whether they are stable depends on the matrix of second derivatives  ∂2V    ∂2V ∂2V 2k 0 0 ∂x2 ∂x∂y ∂x∂θ 2 2  ∂2V  ∂ V . K(x, y, θ) ≡  ∂y∂x ∂∂yV2 =  0 2k 0 ∂y∂θ 
ℓ ∂2V ∂2V ∂2V 0 0 kℓ 2 + a sin ϕ cos θ 2 ∂θ∂x

∂θ∂y

∂θ

(2.79)

Problem 7

33

At an equilibrium point we have sin θ = 0 and so cos θ = ±1. Now for stability, K needs to be positive-definite, which in this case means all entries on the diagonal are positive. This requires cos θ = +1, so in fact the only stable equilibrium point is (x, y, θ) = (0, 0, 0). Furthermore, we see from Eq. 2.79 that our Lagrangian is already diagonal up to second order in our coordinates:     1 1 1 1 L= mx˙ 2 − 2kx2 + my˙ 2 − 2ky 2 2 2 2 2     1 1 1 ℓ + mℓ2 θ˙2 − kℓ + a sin ϕ θ2 + constant + O(θ3 ). (2.80) 2 12 2 2 That is, it is the Lagrangian of three non-interacting harmonic oscillators of frequencies s r r
kℓ 2ℓ + a sin ϕ 2k 2k , ωy = , and ωθ = , (2.81) ωx = 1 2 m m 12 mℓ so the three normal modes of oscillation are    x(t) cos(ωx t + ϕx )  y(t)  = Cx  0 θ(t) 0    x(t) 0  y(t)  = Cy  cos(ωy t + ϕy ) θ(t) 0    x(t) 0 and  y(t)  = Cθ  0 θ(t) cos(ωθ t + ϕθ )

 ,  ,  .

(2.82)

(c) To construct the Hamiltonian, we must calculate the conjugate momenta ∂L = mx, ˙ ∂ x˙ ∂L py := = my, ˙ and ∂ y˙ ∂L 1 ˙ mℓ2 θ. pθ := = 12 ∂ θ˙

px :=

(2.83)

Then the Hamiltonian is H := px x˙ + py y˙ + pθ θ˙ − L   1 2 ˙2 1 2 2 = m x˙ + y˙ + ℓ θ 2 12     1 ℓ 2 2 + k 2x + 2 (y − a cos ϕ) − 2ℓ + a sin ϕ cos θ + mgy. 2 2

(2.84)

To make sure this is a function of position and momentum coordinates only,

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34

we substitute out the generalized velocities using Eq. 2.83, yielding H=

p2y p2x p2 + + 6 θ2 2m 2m mℓ    ℓ 1 2 2 + k 2x + 2 (y − a cos ϕ) − 2ℓ + a sin ϕ cos θ + mgy. 2 2

(2.85)

Now half of Hamilton’s equations of motion just reproduce the relations in Eq. 2.83: ∂H px = , ∂px m ∂H py y˙ = = , ∂py m pθ ∂H = 12 2 , θ˙ = ∂pθ mℓ

x˙ =

(2.86)

while the other half of the equations are ∂H = −2kx, ∂x ∂H p˙y = − = −2k(y − a cos ϕ) − mg, ∂y   ∂H ℓ p˙θ = − = −kℓ + a sin ϕ sin θ. ∂θ 2

p˙x = −

(2.87)

Clearly, these contain the same content as the Euler-Lagrange equations Eqs. (2.74), (2.75), and (2.76).

2.8 Problem 8 Problem:
2
4 Show that the Lagrangian L = q1 q˙2 − q2 q˙1 − a q1 − q2 is invariant under the transformations      q1 1 + s −s q1 → q2 s 1−s q2 and construct the corresponding conserved quantity. Solution: Under the given transformation, we have q1 q˙2 − q2 q˙1 → ((1 + s)q1 − sq2 )(sq˙1 + (1 − s)q˙2 )

− (sq1 + (1 − s)q2 )((1 + s)q˙1 − sq˙2 )

= (1 + s)sq1 q˙1 + (1 − s2 )q1 q˙2 − s2 q2 q˙1 − s(1 − s)q2 q˙2

− s(1 + s)q1 q˙1 + s2 q1 q˙2 − (1 − s2 )q2 q˙1 + s(1 − s)q2 q˙2

Problem 9

35

= q1 q˙2 − q2 q˙1 ,

(2.88)

and q1 − q2 → (1 + s)q1 − sq2 − (sq1 + (1 − s)q2 ) = q1 − q2 .

(2.89)

It follows that the Lagrangian is invariant under these transformations. Now the identity transformation is at the point s = 0, so the conserved quantity corresponding to this symmetry is given by Eq. (2.2.13) in the text, C =η·

∂L , ∂ η˙

(2.90)

where d ds



q1′ q2′

d = ds



1 + s −s s 1−s

η :=



s=0



q1 q2



 = s=0

q1 − q2 q1 − q2

 .

(2.91)

Explicitly, C = (q1 − q2 )[2(q1 q˙2 − q2 q˙1 )(−q2 )] + (q1 − q2 )[2(q1 q˙2 − q2 q˙1 )(q1 )] = 2(q1 q˙2 − q2 q˙1 )(q1 − q2 )2 .

(2.92)

It is straightforward but tedious to check that C is indeed conserved, i.e. dC/dt = 0, when the equations of motion are satisfied.

2.9 Problem 9 Problem: A system with a single degree of freedom is governed by the Lagrangian L = βt 1 e ( 2 mq˙2 − 21 kq 2 ): (a) Write down the equation of motion. What sort of motion does it describe?; (b) Show that the explicit time dependence of L can be removed by introducing a new coordinate variable Q = eβt/2 q; and (c) Construct the first integral whose conservation is implied by this property. Solution: (a) The equation of motion is d ∂L ∂L − dt ∂ q˙ ∂q d βt = (e mq) ˙ − (−eβt kq) dt

0=

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36

= βeβt mq˙ + eβt m¨ q + eβt kq =⇒

0 = m¨ q + mβ q˙ + kq.

(2.93)

This is recognized as the equation of motion of a damped harmonic oscillator. (b) If we define Q = eβt/2 q, then we have β Q˙ = eβt/2 q + eβt/2 q, ˙ 2

(2.94)

or q˙ = e

−βt/2



 β ˙ Q− Q . 2

In these new coordinates, the Lagrangian becomes !  2 β 1 −βt 2 1 −βt ˙ βt me Q − Q − ke Q L=e 2 2 2   1 ˙2 1 1 mβ 2 ˙ = mQ − mβ QQ − k− Q2 . 2 2 2 4

(2.95)

(2.96)

The explicit time dependence has vanished. (c) As the Lagrangian in Eq. 2.96 has no explicit time dependence, the energy function ∂L −L h := Q˙ ∂ Q˙     1 1 ˙2 1 1 mβ 2 ˙ ˙ ˙ = Q mQ + mβQ − mQ + mβ QQ + k− Q2 2 2 2 2 4   1 mβ 2 1 k− Q2 (2.97) = mQ˙ 2 + 2 2 4 will be conserved over time. Note, however, that since the system is not natural, h is not equal to the energy of the system. The energy of a damped harmonic oscillator will not be conserved over time.

2.10 Problem 10 Problem: A particle of mass m moves in one dimension in a potential V (q) = −Cqe−λq where C and λ are positive constants. Write down the Lagrangian and find the point of stable equilibrium. In the small-oscillation approximation find the angular frequency of the normal mode. Solution:

Problem 11

37

With the given potential, the Lagrangian is 1 2 1 mq˙ − V = mq˙2 + Cqe−λq . (2.98) 2 2 Points of stable equilibrium are local minima of the potential energy. These satisfy L=

dV = −Ce−λq + Cλqe−λq = 0 dq

(2.99)

d2 V = Cλe−λq + Cλe−λq − Cλ2 qe−λq > 0. dq 2

(2.100)

and

There is only one solution to Eq. 2.99, given by q = λ−1 . At this point the second derivative Eq. 2.100 is equal to Cλ/e > 0, so this is indeed a point of stable equilibrium. Just as in Problem 2.6, we solve for the small perturbations about this minimum by expanding the Lagrangian in terms of η := q − λ−1 . The kinetic energy is T =

1 2 1 mq˙ = mη˙ 2 , 2 2

(2.101)

while the potential is

∂V 1 ∂ 2 V V λ−1 + η = V λ−1 + η 2 + O(η 3 ) η + ∂q q=λ−1 2 ∂x2 q=λ−1 =−

C 1 Cλ 2 +0+ η + O(η 3 ). λe 2 e

(2.102)

We recognize the matrices M 0 = [m] and K = [Cλ/e] that dictate the frequency ω of small oscillations about η = 0 through the relation det(K − ω 2 M 0 ) = 0,

(2.103)

giving r ω=

Cλ . em

(2.104)

2.11 Problem 11 Problem: The Lagrangian of a particle of mass m and charge e in a uniform, static magnetic ˙ = 12 mx˙ 2 + (e/2m)L · B, where L = x × mx˙ is the field B is given by L(x, x) mechanical angular momentum: ¯ 2 /2m, where p ¯ = p − (e/2)B × x; (a) Show that the Hamiltonian is H(x, p) = p (b) Obtain the Poisson bracket result {¯ pi , p¯j } = e ϵijk Bk ; and ¯ · B is conserved using Poisson brackets; and (d) Are components (c) Show that p ¯ perpendicular to B conserved? of p

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38

Solution: (a) In index notation, the Lagrangian is given by L=

1 e mx˙ i x˙ i + εijk xi x˙ j Bk , 2 2

(2.105)

where repeated indices are summed over and we do not distinguish between upper and lower indices. To find the Hamiltonian, we must first compute the conjugate momenta pn :=

∂L 1 1 e = mδin x˙ i + mx˙ i δin + εijk xi δjn Bk ∂ x˙ n 2 2 2 1 e 1 = mx˙ n + mx˙ n − εikn xi Bk 2 2 2 e = mx˙ n − (x × B)n , 2

(2.106)

or in vector notation, e p = mx˙ − (x × B). 2

(2.107)

Then the Hamiltonian is H := p · x˙ − L e 1 e ˙ ·B = mx˙ 2 − (x × B) · x˙ − mx˙ 2 − (x × x) 2 2 2 1 e e ˙ ·B = mx˙ 2 − (x˙ × x) · B − (x × x) 2 2 2 1 = mx˙ 2 , 2

(2.108)

where we have used the cyclic property of the scalar triple product in going from the second to the third line, and the antisymmetry of the cross product between the third and the fourth. But the Hamiltonian must be expressed in terms of position and momentum only, so we use Eq. 2.107 to substitute out the velocity and obtain  2 p + 2e (x × B) ¯2 p 1 ≡ , (2.109) H= m 2 m 2m ¯ := p + 2e (x × B) = p − 2e (B × x). where p (b) Using the bilinearity of Poisson brackets, we can write e e {¯ pi , p¯j } = {pi − εikl Bk xl , pj − εjmn Bm xn } 2 2 e e = {pi , pj } − εjmn Bm {pi , xn } − εikl Bk {xl , pj } 2 2 e2 + εikl εjmn Bk Bm {xl , xn }, 4

(2.110)

Problem 11

39

because the components of B are constant. Now we use the fundamental Poisson brackets Eq. (2.4.28), {xi , pj } = δij ,

{xi , xj } = 0 = {pi , pj },

and the antisymmetry of the brackets to simplify e e {¯ pi , p¯j } = 0 − εjmn Bm (−δin ) − εikl Bk δlj + 0 2 2 e = (εjmi Bm − εikj Bk ) 2 = e εijk Bk .

(2.111)

(2.112)

Here, the total antisymmetry of εijk was used and we relabeled the dummy index m → k. ¯ · B is given, from Eq. (2.4.24), by (c) The time derivative of p ∂(¯ p · B) d(¯ p · B) = {¯ p · B, H} + dt ∂t = {¯ p · B, H}.

(2.113)

Going back to index notation and using the product rule of the Poisson bracket, we calculate Bi d(¯ p · B) = {¯ pi , p¯j p¯j } dt 2m Bi = ({¯ pi , p¯j }¯ pj + p¯j {¯ pi , p¯j }) 2m Bi p¯j = {¯ pi , p¯j } m Bi p¯j = e εijk Bk m e ¯ · (B × B) = p m = 0. (2.114) ¯ · B is conserved. Thus, p ¯ perpendicular to B are given by (d) Those components of p p¯⊥ ¯i − i =p

(¯ p · B) Bi , B2

(2.115)

¯ ⊥ · B = 0. These evolve in time as may readily be verified by observing that p as ∂ p¯⊥ d¯ p⊥ i i = {¯ p⊥ i , H} + dt ∂t = {¯ p⊥ i , H} Bi = {¯ pi , H} − 2 {¯ p · B, H} B

Classical Mechanics

40

e εijk p¯j Bk − 0 m e = (¯ p × B)i , m =

(2.116)

using the results of Part (c). Generically this time derivative is non-zero, show¯ perpendicular to B are not necessarily conserved, ing the components of p as might reasonably be expected from the Lorentz force law. In general, the perpendicular components are all conserved only when all of them vanish.

2.12 Problem 12 Problem: 2

(a) Show that the relativistic Lagrangian L = −(1/γ)mc2 eU/c reduces to the Newtonian Lagrangian for a particle of mass m in a gravitational potential U in the slow motion, weak field limit v ≪ c, |U | ≪ c2 . (b) Obtain the Euler-Lagrange equations of motion, and show that they can be written in the relativistic form dV µ /dτ = (g µν − (1/c2 )V µ V ν )∂ν U , where V is the four-velocity and τ is the proper time of the particle. Verify that not all four component equations are independent. Solution: (a) We find the slow motion, weak field limit of the Lagrangian by Taylor expanding the various factors in terms of v/c and U/c2 , and keeping only the first-order terms: L=−

mc2 U/c2 e γ r

v 2 U/c2 e c2   4    2  1 v2 v U U 2 = −mc 1 − +O 1+ 2 +O 2 c2 c4 c c4  2  1 v U = −mc2 + mv 2 − mU + O . 2 c4 = −mc2

1−

(2.117)

Ignoring the higher-order terms, and also the constant −mc2 , which does not affect the equations of motion, we have L=

1 mv 2 − mU = T − V, 2

(2.118)

where V := mU is the Newtonian gravitational potential energy. This is the standard Newtonian Lagrangian for a particle of mass m in a gravitational field.

Problem 12

41

(b) To find the equations of motion, it is first helpful to calculate ! r     d 1 γv i d 1 2v i v2 q = − . = = − 1 − 2 dv i γ dv i c2 c2 c2 2 1 − vc2

(2.119)

Then the equations of motion are d ∂L ∂L − i i dt ∂v ∂x      mc2 U/c2 1 ∂U d −γv i U/c2 2 e − − = −mc e dt c2 γ c2 ∂xi   2 2 d 0=γ γv i eU/c + eU/c ∂i U. (2.120) dt 0=

=⇒

Recall the 4-velocity is given by V µ = (γc, γv), so γv i = V i . Also recall that γ≡

dt , dτ

so

γ

d d = . dt dτ

With these, the equations of motion Eq. 2.120 become 2 d  i U/c2  0= V e + (∂i U )eU/c dτ 2 1 dU 2 dV i U/c2 = e + V i eU/c 2 + (∂i U )eU/c dτ c dτ V i dU dV i + 2 + ∂i U. =⇒ 0= dτ c dτ

(2.121)

(2.122)

To simplify further, note that dU ∂U dxν = = (∂ν U )V ν , dτ ∂xν dτ

(2.123)

∂i U = −(−δ iν )∂ν U = −g iν ∂ν U,

(2.124)

and that

where g µν is the Minkowski metric tensor. We are following the convention that Greek letters denote spacetime indices, while Latin letters denote only spatial indices. With these identities, Eq. 2.122 reduces to   dV i 1 i ν iν 0= − g − 2V V ∂ν U. (2.125) dτ c The tensor on the right-hand side of this equation evidently has vanishing spatial components in all inertial frames. The only tensor with this property is the zero tensor, so the above equation must also remain true if we replace i → µ. We arrive at the desired equations of motion   dV µ 1 µ ν µν = g − 2V V ∂ν U. (2.126) dτ c Now we deduced the µ = 0 equation from the three spatial equations, so it

Classical Mechanics

42

is clear that not all of these equations are independent. To see this explicitly, consider the fact that V µ Vµ is the constant c2 . This implies 0=

1 d dV 0 dV i dV µ (V µ Vµ ) = Vµ = V0 + Vi , 2 dτ dτ dτ dτ

(2.127)

leading to dV 0 Vi dV i =− , dτ V0 dτ

(2.128)

and therefore showing the equations are not independent. Checking that we do indeed arrive at the desired temporal equation, we substitute in the known spatial equations and use the fact that there is no numerical difference between covariant and contravariant temporal components of a tensor:   Vi 1 i ν dV 0 iν ∂ν U =− g − 2V V dτ γc c   1 1 i i ν =− Vi ∂ U − 2 Vi V V ∂ν U γc c   1 1 ν 0 i ν =− V ∂ν U − V ∂0 U − 2 Vi V V ∂ν U γc c   1 1 = ∂0 U − 1 − 2 Vi V i V ν ∂ν U γc c   1 1 = ∂0U − 1 − 2 (Vµ V µ − V0 V 0 ) V ν ∂ν U γc c   1 1 = g 0ν ∂ν U − 1 − 2 (c2 − γcV 0 ) V ν ∂ν U γc c   1 (2.129) = g 0ν − 2 V 0 V ν ∂ν U, c which is the desired form.

2.13 Problem 13 Problem: Consider a differentiable function V (q) of a generalized coordinate q. Consider the Lagrangian L = (1/12)m2 q˙4 + mq˙2 V (q) − V 2 (q). Show that this system is equivalent to a much simpler Lagrangian for a particle moving in a potential. Solution: Two Lagrangians are equivalent if their equations of motion have the same solutions. The equation of motion of the given Lagrangian is 0=

d ∂L ∂L − dt ∂ q˙ ∂q

Problem 14

43

    d 1 2 3 dV dV m q˙ + 2mqV ˙ − mq˙2 − 2V dt 3 dq dq     dV dq dV dV = m2 q˙2 q¨ + 2m¨ q V + 2mq˙ − mq˙2 − 2V dq dt dq dq    1 2 dV mq˙ + V , = 2 m¨ q+ dq 2

=

(2.130)

where we have made extensive use of the chain rule. For the above equation to be satisfied, at least one of the following equations must hold: dV = 0, dq

(2.131)

1 2 mq˙ + V = 0. 2

(2.132)

m¨ q+

Suppose the second equation holds. Then also the first equation holds, for   d 1 2 mq˙ + V = 0 dq 2 dV dq˙ dt + =0 mq˙ dt dq dq dV =⇒ m¨ q+ = 0. (2.133) dq In other words, the motion of the system is entirely dictated by the equation m¨ q=−

dV . dq

(2.134)

But this is the equation of motion of the more familiar Lagrangian L′ :=

1 2 mq˙ − V (q) 2

(2.135)

for a particle moving in a potential V , and therefore the two systems are equivalent.

2.14 Problem 14 Problem: (a) A particle with rest mass M phas kinetic energy T . Show that the magnitude of its momentum is p = (1/c) T (T + 2M c2 ). (b) A particle A decays at rest into a lighter particle B and a photon γ. Use fourvector methods to determine the energies of B and γ in terms of the rest masses mA and mB of A and B. Solution:

Classical Mechanics

44

(a) The magnitude of the particle’s momentum is given by p = γM v ≡ q

Mv 1−

v2 c2

,

(2.136)

while the particle’s kinetic energy is equal to the total energy, minus the rest mass energy: T = γM c2 − M c2 = (γ − 1)M c2 .

(2.137)

Performing some algebra, we find    T (T + 2M c2 ) = (γ − 1)M c2 (γ + 1)M c2 = (γ 2 − 1)M 2 c4   1 = 1 − 2 γ 2 M 2 c4 γ    v2 = 1− 1− 2 γ 2 M 2 c4 c = γ 2 M 2 v 2 c2 = p 2 c2 p T (T + 2M c2 ) , p= c

=⇒

(2.138)

as desired. (b) Denote the four-momenta of the three particles by pA , pB , and pγ . By the conservation of four-momentum, we have pA = pB + pγ .

(2.139)

Consider now the quantity p2γ = (pA − pB )2

= p2A − 2pA · pB + p2B .

(2.140)

Recall the following properties of a particle’s four-momentum vector conveyed by Eq. (2.6.4) in the text: (i) The temporal component is equal to the particle’s energy, divided by c. (ii) The length of the vector is equal to the particle’s mass, multiplied by c. The above equation therefore reduces to 0=

m2A c2

 − 2 (mA c)

EB c



+ m2B c2 ,

(2.141)

where we have used the fact that in the rest frame of particle A, its fourmomentum is pA = (EA /c, 0)T = (mA c, 0)T . Rearranging, we find the energy of particle B EB =

m2A + m2B 2 c . 2mA

(2.142)

Problem 15

45

γ1 y

θ1 e+

t

Figure 2.6

e−

x θ2

γ2

Geometry of the e+ + e− → 2γ process in the rest frame of the electron. Of course, the electron and positron will not exist at the same times as the two photons. Similarly, the energy of the photon is found through the following calculation: p2B = (pA − pγ )2

= p2A − 2pA · pγ + p2γ   Eγ 2 2 2 2 +0 mB c = mA c − 2 (mA c) c m2 − m2B 2 =⇒ Eγ = A c . 2mA

(2.143)

Note that for this decay process to be physical we must have Eγ > 0, which enforces mB < mA .

2.15 Problem 15 Problem: A relativistic positron e+ with kinetic energy T annihilates a stationary electron − e and produces two photons e+ + e− → γ + γ: (a) At what angles are the photons emitted when they have equal energies?. (b) Is it possible for the photons to have unequal energies? Explain. Solution: (a) We will use the conservation of four-momentum to answer this question. In the rest frame of the electron, the four-momenta of the initial particles are T  T  M c2 E e− , 0 = , 0 , pe− = c c

Classical Mechanics

46

 pe+ =

Ee+ , p, 0, 0 c

T =

T + M c2 , c

p

T (T + 2M c2 ) , 0, 0 c

!T ,

(2.144)

where we have taken the +x-direction to be the direction of motion of the positron, and we have used the result of Problem 2.14 (a) to express the momentum of the positron in terms of its kinetic energy. The total four-momentum is then !T p T (T + 2M c2 ) T + 2M c2 , , 0, 0 , (2.145) ptot = pe− + pe+ = c c and this must be equal to the total four-momentum of the two photons. Now it is easy to see that the three-momenta of the photons and of the positron all lie in a single plane: the momentum vectors of the two photons define a plane, and the momentum of the positron in the direction normal to this plane must vanish in order for momentum to be conserved. For convenience, we shall take this to be the x − y plane, as illustrated in Fig. 2.6. Say the two photons make respective angles θ1 and θ2 to the x-axis. Then under the assumption the photons have equal energies Eγ , their four-momenta will be Eγ T (1, cos θ1 , sin θ1 , 0) , c Eγ T = (1, cos θ2 , sin θ2 , 0) . c

pγ1 = pγ2

(2.146)

(The magnitude of a photon’s three-momentum is always Eγ /c). Therefore, the total four-momentum may equivalently be written ptot = pγ1 + pγ2 =

Eγ T (2, cos θ1 + cosθ2 , sin θ1 + sin θ2 , 0) . c

(2.147)

Comparing with Eq. 2.145, we get the following relations: 2Eγ = T + 2M c2 , p Eγ (cos θ1 + cosθ2 ) = T (T + 2M c2 ), sin θ1 + sin θ2 = 0.

(2.148) (2.149) (2.150)

The last of these tells us θ1 = −θ2 ≡ θ. Then combining this with the first two, we get p 2Eγ cos θ = (T + 2M c2 ) cos θ = T (T + 2M c2 ). (2.151) Hence, the photons make angles −1

θ = cos

r

T T + 2M c2

!

on either side of the line of motion of the positron.

(2.152)

47

Problem 15

(b) Consider now the collision in the center-of-mass frame of the electron and positron, so the magnitude of the total three-momentum is zero. To conserve momentum, the photons must therefore be directed oppositely with equal energies:  T Eγ pγ1 = ,p , c  T Eγ pγ2 = , −p . (2.153) c Clearly, any direction p is consistent with the requirement of energy and momentum conservation, and is therefore allowed. Now to return to our original frame in part (a), let us boost by some speed v in the x-direction so that the electron becomes stationary. Four-momenta transform as tensors, so in this boosted frame the temporal components of the photons’ four-momenta will be   Eγ′ 1 Eγ v =γ − px , c c c   Eγ′ 2 Eγ v =γ − (−px ) . (2.154) c c c These energies are not equal in general, so it is possible for the photons to have unequal energies in the stationary frame of the electron.

3

Relativistic classical fields 3.1 Problem 1 Problem: √ We define a free complex scalar field as ϕ(x) = (1/ 2)[ϕ1 (x) + iϕ2 (x)] ∈ C in terms of two real free scalar fields ϕ1 (x), ϕ2 (x) ∈ R and the Lagrangian density as the sum of the two free Lagrangian densities. (a) Show that the Lagrangian density in terms of ϕ(x) and ϕ∗ (x) is as given in Eq. (3.2.48), L = (∂µ ϕ)∗ (∂ µ ϕ) − m2 ϕ∗ ϕ. Using the Euler-Lagrange equations for ϕ1 and ϕ2 show that the complex scalar field and its complex conjugate satisfy (∂µ ∂ µ + m2 )ϕ(x) = (∂µ ∂ µ + m2 )ϕ∗ (x) = 0. R (b) Using the action S[ϕ] = d4 x L and the Wirtinger calculus results in Sec. A.2.5, show that these equations of motion follow from Hamilton’s Principle, δS[ϕ]/δϕ∗ (x) = δS[ϕ]/δϕ(x) = 0. (c) Using Eq. (3.1.15) evaluate the conjugate momenta π(x) = ∂L/∂(∂0 ϕ)(x) and π ∗ (x) = ∂L/∂(∂0 ϕ∗ )(x). Use these R results Rto construct the Hamiltonian using Eq. (3.1.16) and show that H = d3 x H = d3 x [π ∗ (x)π(x) + ∇ϕ∗ (x) · ∇ϕ(x) + m2 ϕ∗ (x)ϕ(x)]. (d) Obtain Hamilton’s equations using Eq. (3.1.21) in terms of ϕ and ϕ∗ and show that these reproduce the Klein-Gordon equations for ϕ and ϕ∗ . (e) The complex scalar field Lagrangian in terms of ϕ1 and ϕ2 has a well-defined Legendre transform since the Hessian matrix ML is positive definite. This follows from the straightforward generalization of Eq. (3.1.48), δ2 L = δij δ 3 (x − y) , (3.1) (ML )ij (x, y)|x0 =y0 = δ ϕ˙ i (x)δ ϕ˙ j (y) 0 0 x =y

for i, j = 1, 2. Show that the Wirtinger calculus version of the nonsingular Hessian matrix has the four elements 2 ˙ ˙ ˙ ∗ (x)δ ϕ˙ ∗ (y) δ 2 L/δ ϕ(x)δ ϕ(y) = δ L/δ ϕ = 0, (3.2) x0 =y 0 x0 =y 0 ˙ ˙ δ 2 L/δ ϕ(x)δ ϕ˙ ∗ (y) 0 0 = δ 2 L/δ ϕ˙ ∗ (x)δ ϕ(y) = δ 3 (x − y) . 0 0 x =y

Solution: 48

x =y

49

Problem 1

(a) For two free real scalar fields ϕ1 and ϕ2 of mass m, the Lagrangian density is     1 1 1 1 L= (∂µ ϕ1 )(∂ µ ϕ1 ) − m2 ϕ21 + (∂µ ϕ2 )(∂ µ ϕ2 ) − m2 ϕ22 2 2 2 2 1 1 = [(∂µ ϕ1 )(∂ µ ϕ1 ) + (∂µ ϕ2 )(∂ µ ϕ2 )] − m2 (ϕ21 + ϕ22 ). (3.3) 2 2 √ √ Now with the definitions ϕ := (ϕ1 + iϕ2 )/ 2 and ϕ∗ := (ϕ1 − iϕ2 )/ 2, we note the identity    ϕ1 − iϕ2 ϕ1 + iϕ2 1 ∗ √ √ ϕ ϕ= = (ϕ21 + ϕ22 ), (3.4) 2 2 2 and similarly for the derivatives,   µ  ∂µ ϕ1 − i∂µ ϕ2 ∂ ϕ1 + i∂ µ ϕ2 √ √ (∂µ ϕ)∗ (∂ µ ϕ) = 2 2 1 = [(∂µ ϕ1 )(∂ µ ϕ1 ) + (∂µ ϕ2 )(∂ µ ϕ2 )] . 2

(3.5)

From these it follows that the Lagrangian density can equivalently be written as L = (∂µ ϕ)∗ (∂ µ ϕ) − m2 ϕ∗ ϕ.

(3.6)

Looking back to the first expression of the Lagrangian density, and using the identities ∂(∂µ ϕi ) = δ ν µ δ ji , ∂(∂ν ϕj )

∂(∂ µ ϕi ) = g µν δ ji , ∂(∂ν ϕj )

(3.7)

the Euler-Lagrange equation for the field ϕ1 is seen to be   ∂L ∂L 0 = ∂ν − ∂(∂ν ϕ1 ) ∂ϕ1   1 ν µ = ∂ν (δ µ ∂ ϕ1 + ∂µ ϕ1 g µν ) − (−m2 ϕ1 ) 2 = ∂ν ∂ ν ϕ1 + m2 ϕ1 = (∂ 2 + m2 )ϕ1 .

(3.8)

The equation for ϕ2 is then obviously (∂ 2 + m2 )ϕ2 = 0, because the Lagrangian density is symmetric with respect to ϕ1 and ϕ2 . Now we can combine these two equations to see  1  0 = √ (∂ 2 + m2 )ϕ1 ± i(∂ 2 + m2 )ϕ2 2 ϕ1 ± iϕ2 = (∂ 2 + m2 ) √ , 2 or in other words, (∂ 2 + m2 )ϕ = 0 and (∂ 2 + m2 )ϕ∗ = 0.

(3.9)

Relativistic classical fields

50

(b) Hamilton’s Principle that the equations of motion will be satisfied when δS = 0 under variations of ϕ and of ϕ∗ will lead to the Euler-Lagrange equations of motion, where ϕ and ϕ∗ are the fields:     ∂L ∂L ∂L ∂L . (3.10) 0 = ∂ν − and 0 = ∂ν − ∂(∂ν ϕ) ∂ϕ ∂(∂ν ϕ∗ ) ∂ϕ∗ The Wirtinger calculus results tell us that these derivatives, with ϕ and ϕ∗ defined as above, act in the expected way as long as ϕ and ϕ∗ are thought of as completely independent fields. So the first of these Euler-Lagrange equations, for example, gives by Eq. 3.6   ∂L ∂L 0 = ∂ν − ∂(∂ν ϕ) ∂ϕ = ∂ν (∂ µ ϕ)∗ δ νµ − (−m2 ϕ∗ )

= (∂ 2 + m2 )ϕ∗ ,

(3.11)

and the second gives  0 = ∂ν

∂L ∂(∂ν ϕ∗ )

 −

∂L ∂ϕ∗

= ∂ν (∂ ν ϕ) − (−m2 ϕ) = (∂ 2 + m2 )ϕ,

(3.12)

where we have used the fact that (∂µ ϕ∗ ) = (∂µ ϕ)∗ . We see that we have derived the equations of motion from Part (a) by using Hamilton’s Principle, with ϕ and ϕ∗ as the variational fields. (c) Using Eq. 3.6, we calculate π=

∂L ∂L = ∂ 0 ϕ∗ , and π ∗ = = ∂ 0 ϕ. ∂(∂0 ϕ) ∂(∂0 ϕ∗ )

(3.13)

Note that since ∂ 0 = ∂0 , we can also write π = ∂0 ϕ∗ and π ∗ = ∂0 ϕ. The Hamiltonian density is then by definition H := π∂0 ϕ + π ∗ ∂0 ϕ∗ − L   = ππ ∗ + π ∗ π − g µν (∂µ ϕ)∗ (∂ν ϕ) − m2 ϕ∗ ϕ

= 2ππ ∗ − g 00 (∂0 ϕ)∗ (∂0 ϕ) − g ij (∂i ϕ)∗ (∂j ϕ) + m2 ϕ∗ ϕ = 2ππ ∗ − (∂0 ϕ)∗ (∂0 ϕ) + δ ij (∂i ϕ)∗ (∂j ϕ) + m2 ϕ∗ ϕ = 2ππ ∗ − ππ ∗ + δ ij (∇i ϕ)∗ (∇j ϕ) + m2 ϕ∗ ϕ = ππ ∗ + (∇ϕ∗ ) · (∇ϕ) + m2 ϕ∗ ϕ.

(3.14)

Here we have used the fact that ∂µ = (∂0 , ∇)T . Or in terms of the full Hamiltonian, Z Z   3 H = d x H(x) = d3 x π(x)π ∗ (x) + (∇ϕ∗ (x)) · (∇ϕ(x)) + m2 ϕ∗ (x)ϕ(x) . (3.15)

Problem 1

51

(d) With the Hamiltonian density Eq. 3.14, the Hamilton’s equations that are relevant to us are ∂H = π ∗ , and ∂π ∂H ∂H ∂0 π = − +∇· = −m2 ϕ∗ + ∇ · (∇ϕ∗ ), ∂ϕ ∂(∇ϕ) ∂0 ϕ =

(3.16)

and of course the conjugate counterparts ∂H = π, and ∂π ∗ ∂H ∂H ∂0 π ∗ = − ∗ + ∇ · = −m2 ϕ + ∇ · (∇ϕ). ∂ϕ ∂(∇ϕ∗ ) ∂0 ϕ∗ =

(3.17)

Combining these two sets of equations yields ∂0 ∂0 ϕ = ∂0 π ∗ = −m2 ϕ + ∇ · (∇ϕ)

=⇒

(∂0 ∂ 0 − ∇2 + m2 )ϕ = 0, (3.18)

and similarly ∂0 ∂0 ϕ∗ = ∂0 π = −m2 ϕ∗ + ∇ · (∇ϕ∗ )

=⇒

(∂0 ∂ 0 − ∇2 + m2 )ϕ∗ = 0. (3.19)

These are simply the Klein-Gordon equations (∂ 2 +m2 )ϕ = 0 and (∂ 2 +m2 )ϕ∗ = 0 that we derived previously. (e) For a functional of the form F [σ] := the property that

R

dx f (σ(x)), functional derivatives have

∂f (σ(x)) δF = . δσ(x) ∂σ(x)

(3.20)

Importantly, this holds as long as F is not a functional of the derivatives of σ. Applied to the case at hand, we can identify F → L, f → L, and σ → ϕ˙ (∗) ≡ ∂0 ϕ(∗) . Then, for example, δL ∂L(y) = ˙ ˙ δ ϕ(y) ∂ ϕ(y)  ∂  ˙∗ ˙ = ϕ (y) ϕ(y) + ∂i ϕ∗ (y) ∂ i ϕ(y) − m2 ϕ∗ (y)ϕ(y) ˙ ∂ ϕ(y) = ϕ˙ ∗ (y).

(3.21)

It follows that δ2 L δ ϕ˙ ∗ (y) δ2 L δ ϕ˙ ∗ (y) = = 0, and = = δ 4 (x − y). ˙ ϕ(y) ˙ ˙ ˙ δ ϕ(x) δ ϕ(x) δ ϕ˙ ∗ (x)ϕ(y) δ ϕ˙ ∗ (x) Similarly, we have δL ∂L(y) = ∗ ˙ δ ϕ (y) ∂ ϕ˙ ∗ (y)

(3.22)

Relativistic classical fields

52

  ∂ ˙ ϕ˙ ∗ (y) ϕ(y) + ∂i ϕ∗ (y) ∂ i ϕ(y) − m2 ϕ∗ (y)ϕ(y) ∂ ϕ˙ ∗ (y) ˙ = ϕ(y), (3.23)

=

and so ˙ ˙ δ2 L δ ϕ(y) δ2 L δ ϕ(y) = = δ 4 (x − y), and = = 0. (3.24) ˙ ϕ˙ ∗ (y) ˙ δ ϕ(x) δ ϕ(x) δ ϕ˙ ∗ (x)ϕ˙ ∗ (y) δ ϕ˙ ∗ (x) Restricted to equal-time spacetime points (x0 = y 0 ), these reduce to the desired equations with factors of δ 3 (x − y) (the common infinite factors of δ(x0 − y 0 ) = δ(0) should be understood to be implicit).

3.2 Problem 2 Problem: Let ϕ be a Lorentz-scalar field.  (a) Show that the quantities Tµν = ∂µ ϕ∂ν ϕ − 12 gµν (∂ϕ)2 − λ2 ϕ2 with λ ∈ R are components of a Lorentz (0, 2) tensor field. (b) Show that ∂µ T µν = 0 if ϕ(x) obeys the wave equation (∂ 2 + κ2 )ϕ = 0. Solution: (a) First, we note that since ϕ is a Lorentz-scalar field, it transforms under a Lorentz transformation Λ as ϕ(x) → ϕ′ (x′ ) = ϕ(x).

(3.25)

Obviously ϕ2 will also be a Lorentz-scalar, transforming as ϕ2 (x) → (ϕ′ )2 (x′ ) = ϕ2 (x). From Eq. (1.2.134), the derivatives of ϕ will form a Lorentz four-vector, ′

∂µ ϕ(x) → ∂µ′ ϕ′ (x′ ) = (Λ−1 )µ µ ∂µ′ ϕ(x),

(3.26)

from which it follows that (∂ϕ)2 ≡ (∂µ ϕ)(∂ µ ϕ) is also a Lorentz-scalar. Then knowing that gµν is a (0, 2) tensor, we see that Tµν will transform as ′ Tµν (x) → Tµν (x′ ) ′

′

= (Λ−1 )µ µ ∂µ′ ϕ(x) (Λ−1 )ν ν ∂ν ′ ϕ(x)

′ ′ 1 − (Λ−1 )µ µ (Λ−1 )ν ν gµ′ ν ′ {(∂ϕ)2 (x) − λ2 ϕ2 (x)} 2   ′ ′ 1 = (Λ−1 )µ µ (Λ−1 )ν ν ∂µ′ ϕ(x) ∂ν ′ ϕ(x) − gµ′ ν ′ {(∂ϕ)2 (x) − λ2 ϕ2 (x)} 2 ′

′

= (Λ−1 )µ µ (Λ−1 )ν ν Tµ′ ν ′ (x).

(3.27)

This is the transformation rule for a (0, 2) tensor, so Tµν are the components

Problem 3

53

of a (0, 2) tensor. (b) By direct calculation we have
1 ∂µ T µν = ∂µ ((∂ µ ϕ)(∂ ν ϕ)) − g µν ∂µ (∂ϕ)2 − λ2 ϕ2 2
1 2 ν µ = (∂ ϕ)(∂ ϕ) + (∂ ϕ)(∂µ ∂ ν ϕ) − ∂ ν (∂µ ϕ)(∂µ ϕ) − λ2 ϕ2 2
1 2 ν µ ν = (∂ ϕ)(∂ ϕ) + (∂ ϕ)(∂ ∂µ ϕ) − 2(∂ µ ϕ)(∂ ν ∂µ ϕ) − 2λ2 ϕ∂ ν ϕ 2   = ∂ 2 ϕ + λ2 ϕ ∂ ν ϕ. (3.28) We see that ∂µ T µν will vanish if ϕ obeys the wave equation (∂ 2 + λ2 )ϕ = 0.

3.3 Problem 3 Problem: Consider n real equal-mass scalar fields ϕ1 , . . . , ϕn with a quartic interaction of the form Lint = −(λ/4!)(ϕ21 + · · · + ϕ2n )2 then
Pn Pn L = i=1 Lfree + Lint = i=1 12 ∂µ ϕi ∂ µ ϕi − 21 m2 ϕ2i − (λ/4!)(ϕ21 + · · · + ϕ2n )2 i = 12 ∂µ ϕT ∂ µ ϕ − 21 m2 ϕT ϕ − (λ/4!)(ϕT ϕ)2 .

(3.29)

(a) Obtain the Euler-Lagrange equations for this theory; (b) Explain why this theory is invariant under O(n) transformations. Derive the conserved Noether current and explain why it corresponds to the SO(n) subgroup and not O(n); and (c) For n complex equal-mass scalar fields write a corresponding Lagrangian density invariant under U (n). Obtain the Euler-Lagrange equations. Construct the conserved Noether current. Solution: (a) Our fields are all independent, so we have ∂ϕi = δij , ∂ϕj

and

∂(∂µ ϕi ) = δ αµ δij . ∂(∂α ϕj )

(3.30)

Then since ∂µ ϕi ∂ µ ϕi = g µν ∂µ ϕi ∂ν ϕi , it follows that ∂ (∂µ ϕi ∂ν ϕi ) = g µν [(δ αµ δij )∂ν ϕi + ∂µ ϕi (δ αν δij )] ∂(∂α ϕj ) = 2∂ α ϕi δij . With these, we calculate the Euler-Lagrange equations of motion as   ∂L ∂L 0 = ∂α − ∂(∂α ϕj ) ∂ϕj

(3.31)

Relativistic classical fields

54

= ∂α

n X

! α

∂ ϕi δij

i=1

−

n X λ −m ϕi δij − 2(ϕ21 + · · · + ϕ2n ) 2ϕi δij 4! i=1 i=1 2

n X

!

λ 2 (ϕ + · · · + ϕ2n )ϕj . (3.32) 3! 1 For each field ϕj this is simply the free equation of motion, with an added interaction term. = ∂ 2 ϕj + m2 ϕj +

(b) Inspection of Eq. 3.29 reveals the Lagrangian density, rather than depending on the individual field values, is actually a function of only the “lengths” in field space q p (3.33) |ϕ| := ϕT ϕ = ϕ21 + · · · + ϕ2n , and similarly for the derivatives. Any transformation of the fields that preserves these lengths is therefore a symmetry of the theory. But the group of lengthpreserving transformations acting on n elements is precisely O(n), so this theory is invariant under O(n). Algebraically, O(n) consists of those n×n matrices A with the property AT A = I, as seen in Section 2.1.1. Then under such a transformation ϕ → Aϕ,

∂µ ϕ → ∂µ (Aϕ) = A ∂µ ϕ,

(3.34)

the Lagrangian density transforms as 1 λ 1 (A ∂µ ϕ)T (A ∂ µ ϕ) − m2 (Aϕ)T (Aϕ) − [(Aϕ)T (Aϕ)]2 2 2 4! 1 1 λ = ∂µ ϕT (AT A)∂ µ ϕ − m2 ϕT (AT A)ϕ − [ϕT (AT A)ϕ]2 2 2 4! 1 2 T λ T 2 1 T µ = ∂µ ϕ ∂ ϕ − m ϕ ϕ − (ϕ ϕ) 2 2 4! = L.

L→

(3.35)

That is, the Lagrangian density is invariant under O(n) transformations, as was reasoned earlier. This is an internal symmetry, so the associated conserved Noether current is given by Eq. (3.2.18) from the text, ⃗ j µ = ⃗π µ · Φ,

(3.36)

⃗ is the vector of field changes under an infinitesimal O(n) transformawhere Φ tion, and ⃗π is the vector of conjugate momenta  T ∂L ∂L T µ ⃗π := , ... , = (∂ µ ϕ1 , . . . , ∂ µ ϕn ) . (3.37) ∂(∂µ ϕ1 ) ∂(∂µ ϕn ) ⃗ i.e. we must determine the behavior of ϕ under an infinitesIt remains to find Φ, imal O(n) transformation. Now we know from Section 2.1.1 that the matrices of O(n) are grouped into those that satisfy det(A) = +1, which constitute

Problem 3

55

the group SO(n), and those that satisfy det(A) = −1. Infinitesimal transformations must have determinants infinitesimally close to det(I) = +1, so any infinitesimal transformation of O(n) must lie in SO(n). The conserved current coming from an O(n) symmetry is therefore the same as that from an SO(n) symmetry1 . Now SO(n) is the group of rotations of n-dimensional space. A rotation is specified by a plane of rotation, and a rotation angle. The number of linearly independent planes in n-dimensional space is equal to the number of ways of choosing 2 dimensions out of n, which is   n(n − 1) n = . (3.38) 2 2 We expect this many degrees of freedom in parameterizing arbitrary SO(n) transformations, along with the freedom afforded by a rotation angle. Infinitesimal SO(n) transformations A can then be parameterized as A(⃗ α) = I + iαa Ma ,

(3.39)

where αa (a = 1, . . . , n(n − 1)/2) are a set of infinitesimal rotation angles, and Ma are as-yet unspecified transformations that each rotate about an independent plane. Such matrices Ma are known as generators of SO(n). To make sure that A is indeed an element of SO(n), we must impose AT A ≈ I + iαa (Ma + MaT ) = I,

(3.40)

giving the condition MaT = −Ma . In other words, the generators must be antisymmetric. The vector space of antisymmetric n×n matrices indeed has dimension n(n − 1)/2, so everything will be consistent if we take the generators to be any basis of the antisymmetric matrices. Now under this transformation, our fields transform as ϕ → Aϕ = ϕ + iαa Ma ϕ,

or ϕi → ϕi + iαa

n X (Ma )ij ϕj .

(3.41)

j=1

There is actually an independent symmetry transformation here for each Ma , so there will be a conserved current associated to each of these, which we also index by a. Our vectors of infinitesimal field changes are then given by  T n n X X d(Aϕ) ⃗ a := = i (Ma )1j ϕj , . . . , i (Ma )nj ϕj  , (3.42) Φ dαa |⃗α=⃗0 j=1 j=1 1

More generally, the conserved current is associated to the component of the symmetry group that is continuously connected to the identity element. For O(n) this connected component is SO(n), because on any continuous path in O(n) the determinant map is continuous, so it cannot jump from +1 to −1 at any point. Since det(I) = +1, the only elements of O(n) continuously connected to I are those with determinant +1, i.e. those in SO(n).

Relativistic classical fields

56

allowing us to finally arrive at the conserved currents, ⃗a = i jaµ = ⃗π µ · Φ

n X n X (∂ µ ϕi )(Ma )ij ϕj = i(∂ µ ϕ)T Ma ϕ.

(3.43)

i=1 j=1

It might seem strange that we have not specified the matrices Ma exactly, but in fact any choice of an antisymmetric basis will lead to current conservation under the equations of motion. (c) If we now have a collection of n complex scalar fields ϕ1 , . . . , ϕn , we can generalize our previous result Eq. 3.6 to write an appropriate Lagrangian density with interactions as L=

n X i=1

Lfree + Lint i

n X   λ = (∂µ ϕi )∗ (∂ µ ϕi ) − m2 ϕ∗i ϕi − (ϕ∗1 ϕ1 + · · · + ϕ∗n ϕn )2 4! i=1

= (∂µ ϕ)† (∂ µ ϕ) − m2 ϕ† ϕ −

λ † 2 (ϕ ϕ) . 4!

(3.44)

Now the group U (n) is similar to O(n), only it preserves terms of the form ϕ† ϕ instead of ϕT ϕ; complex matrices U of U (n) are those that satisfy U † U = I. Clearly our Lagrangian density is invariant under U (n), then, for much the same reason as in Part (b): L → (U ∂µ ϕ)† (U ∂ µ ϕ) − m2 (U ϕ)† (U ϕ) − = ∂µ ϕ† (U † U )∂ µ ϕ − m2 ϕ† (U † U )ϕ − = ∂µ ϕ† ∂ µ ϕ − m2 ϕ† ϕ −

λ [(U ϕ)† (U ϕ)]2 4!

λ † † [ϕ (U U )ϕ]2 4!

λ † 2 (ϕ ϕ) 4!

= L.

(3.45)

We construct the corresponding conserved current(s) in the same way as above. First we note that our vector of conjugate momenta is now  T ∂L ∂L ∂L ∂L µ , , ... , , ⃗π := ∂(∂µ ϕ1 ) ∂(∂µ ϕ∗1 ) ∂(∂µ ϕn ) ∂(∂µ ϕ∗n ) T

= (∂ µ ϕ∗1 , ∂ µ ϕ1 , . . . , ∂ µ ϕ∗n , ∂ µ ϕn ) ,

(3.46)

because we need to treat ϕi and ϕ∗i as independent fields, as per Problem 3.1. ⃗ of infinitesimal field transformations must also include Similarly, the vector Φ those of the conjugate fields. For infinitesimal unitary transformations U (⃗ α) = I + iαa Ha ,

(3.47)

Problem 4

57

where Ha are the generators of U (n), we require U † U ≈ I + iαa (Ha − Ha† ) = I.

(3.48) Ha†

Since α ⃗ is an arbitrary infinitesimal, this gives the condition = Ha . That is, the generators of U (n) must be hermitian. Any basis of the n×n hermitian matrices will be suitable for our purposes. Hermitian matrices must have a symmetric real component and an antisymmetric imaginary component, so in total there are [n(n + 1)/2] + [n(n − 1)/2] = n2 real degrees of freedom available. Our index a therefore ranges from 1 to n2 . Under this infinitesimal transformation our fields transform as ϕ → U ϕ = ϕ + iαa Ha ϕ,

ϕ† → (U ϕ)† = ϕ† − iαa ϕ† Ha† = ϕ† − iαa ϕ† Ha ,

(3.49)

or in other words, ϕi → ϕi + iα

a

ϕ∗i → ϕ∗i − iαa

n X

(Ha )ij ϕj ,

j=1 n X

ϕ∗j (Ha )ji .

(3.50)

j=1

Our vectors of infinitesimal field transformations are then   n n n n X X X X ⃗ a = i ϕ∗j (Ha )jn  . (Ha )nj ϕj , −i ϕ∗j (Ha )j1 , . . . , i Φ (Ha )1j ϕj , −i j=1

j=1

j=1

j=1

(3.51) These give conserved currents ⃗a jaµ = ⃗π µ · Φ n X  µ ∗ =i (∂ ϕ1 )(Ha )1j ϕj − (∂ µ ϕ1 )ϕ∗j (Ha )j1 + · · · j=1

 +(∂ µ ϕ∗n )(Ha )nj ϕj − (∂ µ ϕn )ϕ∗j (Ha )jn n n X X  µ ∗  (∂ ϕi )(Ha )ij ϕj − (∂ µ ϕi )ϕ∗j (Ha )ji =i i=1 j=1

  = i (∂ µ ϕ)† Ha ϕ − ϕ† Ha (∂ µ ϕ) .

(3.52)

Since U (n) is connected, these n2 currents are indeed associated with U (n) symmetry, and not a smaller symmetry group.

3.4 Problem 4 Problem:

Relativistic classical fields

58

A Lagrangian density with two complex scalar fields, ϕ and χ, is given by L = ∂µ ϕ∂ µ ϕ∗ + ∂µ χ∂ µ χ∗ − m2ϕ |ϕ|2 − m2χ |χ|2 − λϕ |ϕ|4 − λχ |χ|4 − g[χ∗ ϕ2 + χ(ϕ∗ )2 ],

(3.53)

where mϕ , mχ , λϕ , λχ , g are real constants. (a) Using the transformations ϕ → eiα ϕ and χ → eiβ χ choose α, β such that L is invariant. Construct the Noether current j µ ; and (b) use the Euler-Lagrange equations to verify that ∂µ j µ (x) = 0. Solution: (a) Under the phase transformations ϕ → eiα ϕ and χ → eiβ χ, the first line of Eq. 3.53 is clearly invariant because all of its terms depend only on the magnitudes |ϕ|, |χ|, |∂ϕ|, |∂χ| of the fields and their derivatives. Only the term in the second line is not necessarily invariant, for it transforms as χ∗ ϕ2 + χ(ϕ∗ )2 → (eiβ χ)∗ (eiα ϕ)2 + (eiβ χ)(eiα ϕ∗ )2

= ei(2α−β) χ∗ ϕ2 + e−i(2α−β) χ(ϕ∗ )2 .

(3.54)

Evidently, it is only when ei(2α−β) = 1 that this term is invariant, as would happen if we fixed β = 2α for example. In this case the fields and conjugate fields transform as ϕ → eiα ϕ ≈ ϕ + iαϕ,

χ → ei2α χ ≈ χ + 2iαχ,

ϕ∗ → e−iα ϕ∗ ≈ ϕ∗ − iαϕ∗ ,

χ∗ → e−2iα χ∗ ≈ χ∗ − 2iαχ∗

(3.55)

for infinitesimal α. This tells us that the vector of changes in the fields under this symmetry is Φ = (iϕ, −iϕ∗ , 2iχ, −2iχ∗ )T . Because the symmetry is internal, we know from Eq. (3.2.18) in the text that the associated conserved Noether current is given by ⃗ j µ = ⃗π µ · Φ,

(3.56)

where ⃗π is the vector of conjugate momenta  T ∂L ∂L ∂L ∂L ⃗π µ : = , , , ∂(∂µ ϕ) ∂(∂µ ϕ∗ ) ∂(∂µ χ) ∂(∂µ χ∗ ) T

= (∂ µ ϕ∗ , ∂ µ ϕ, ∂ µ χ, ∂ µ χ∗ ) .

(3.57)

We therefore have the current ⃗ = i(ϕ∂ µ ϕ∗ − ϕ∗ ∂ µ ϕ) + 2i(χ∂ µ χ∗ − χ∗ ∂ µ χ). j µ = ⃗π µ · Φ

(3.58)

Note that if we had instead used β as our infinitesimal transformation parameter and fixed α = β/2, then the current we would have calculated would be scaled by a factor of 1/2 compared to the one above. This scaling would have no effect on the conserved nature of the current.

Problem 4

59

(b) Noting that |ϕ|2 = ϕ∗ ϕ and |ϕ|4 = (ϕ∗ )2 (ϕ)2 , and the same for χ, the EulerLagrange equations can easily be found by using the Wirtinger calculus. We have in particular   ∂L ∂L 0 = ∂µ − ∂(∂µ ϕ) ∂ϕ
µ ∗ 2 ∗ = ∂µ ∂ ϕ − −mϕ ϕ − 2λ(ϕ∗ )2 ϕ − 2gχ∗ ϕ , (3.59) and  0 = ∂µ

∂L ∂(∂µ χ)

 −

∂L ∂χ


= ∂µ ∂ µ χ∗ − −m2χ χ∗ − 2λ(χ∗ )2 χ − g(ϕ∗ )2 ,

(3.60)

and since the Lagrangian is real-valued, the equations for the conjugate fields are simply the conjugates of these equations:
0 = ∂µ ∂ µ ϕ − −m2ϕ ϕ − 2λϕ∗ ϕ2 − 2gχϕ∗ ,
0 = ∂µ ∂ µ χ − −m2χ χ − 2λχ2 χ∗ − gϕ2 . (3.61) These results will allow us to show the current Eq. (3.58) is indeed conserved, for we calculate ∂µ j µ = i(∂µ ϕ∂ µ ϕ∗ + ϕ∂µ ∂ µ ϕ∗ − ∂µ ϕ∗ ∂ µ ϕ − ϕ∗ ∂µ ∂ µ ϕ)

+ 2i(∂µ χ∂ µ χ∗ + χ∂µ ∂ µ χ∗ − ∂µ χ∗ ∂ µ χ − χ∗ ∂µ ∂ µ χ)

= i(ϕ∂µ ∂ µ ϕ∗ − ϕ∗ ∂µ ∂ µ ϕ) + 2i(χ∂µ ∂ µ χ∗ − χ∗ ∂µ ∂ µ χ).

(3.62)

By the equations of motion, the first term here is i(ϕ∂µ ∂ µ ϕ∗ − ϕ∗ ∂µ ∂ µ ϕ) = iϕ −m2ϕ ϕ∗ − 2λ(ϕ∗ )2 ϕ − 2gχ∗ ϕ




− iϕ∗ −m2ϕ ϕ − 2λϕ∗ ϕ2 − 2gχϕ∗




= 2igχ(ϕ∗ )2 − 2igχ∗ ϕ2 ,

(3.63)

and similarly the second term is 2i(χ∂µ ∂ µ χ∗ − χ∗ ∂µ ∂ µ χ) = 2iχ −m2χ χ∗ − 2λ(χ∗ )2 χ − g(ϕ∗ )2
− 2iχ∗ −m2χ χ − 2λχ2 χ∗ − gϕ2 = 2igχ∗ ϕ2 − 2igχ(ϕ∗ )2 .




(3.64)

Together, these give ∂µ j µ = 2igχ(ϕ∗ )2 − 2igχ∗ ϕ2 + 2igχ∗ ϕ2 − 2igχ(ϕ∗ )2 = 0,

(3.65)

so the current is conserved as expected. Note that the factor of 2 multiplying the χ terms in the current, which was so essential for all of the cancellations in ∂µ j µ , comes directly from the relation β = 2α in accordance with the symmetry.

Relativistic classical fields

60

3.5 Problem 5 Problem: Let F µν be the electromagnetic tensor and consider the symmetric stress-energy tensor defined in Eq. (3.3.9), T¯µν = −F µτ Fντ + 14 δ µν F στ Fστ . In the presence of an external conserved four-vector current density j µ show that ∂µ T¯µν = 0 becomes ∂µ T¯µν = (1/c)jµ F µν . Solution: First of all, we can raise an index on T¯µν to get 1 1 T¯µν = g νρ T¯µρ = −g νρ F µτ Fρτ + g νρ δ µρ F στ Fστ = −F µτ F ντ + g µν F στ Fστ . 4 4 (3.66) Now Maxwell’s equation’s can be expressed in Lorentz covariant form as ∂µ F µν =

1 ν j , c

(3.67)

together with the Jacobi identity ∂ ν F αβ + ∂ α F βν + ∂ β F να = 0.

(3.68)

With these, we calculate 1 ∂µ T¯µν = −∂µ (F µτ F ντ ) + g µν ∂µ (F στ Fστ ) 4 1 = − (∂µ F µτ ) F ντ − F µτ (∂µ F ντ ) + ∂ ν (Fστ F στ ) 4   1 τ 1 =− j F ντ − Fµτ (∂ µ F ντ ) + Fστ (∂ ν F στ ) c 2 1 1 = − jτ F ντ − Fµτ (∂ µ F ντ ) + Fστ (−∂ σ F τ ν − ∂ τ F νσ ) , c 2 and now recognizing we can relabel dummy indices,   1 1 σ τ ν 1 τ νσ νµ µν σ ντ ¯ ∂µ T = − jµ F − Fστ ∂ F + ∂ F + ∂ F c 2 2   1 1 1 ∂ σ F ντ + ∂ τ F νσ , = jµ F µν − Fστ c 2 2

(3.69)

(3.70)

where we have also used the antisymmetry of F µν . Now the term in brackets is symmetric in σ and τ , so its contraction with the antisymmetric Fστ vanishes. We are left with ∂µ Gµν = as desired.

1 jµ F µν , c

(3.71)

Problem 6

61

3.6 Problem 6 Problem: Let Aµ be the electromagnetic four-vector potential and define the four-vector Kµ = ϵµνρσ Aν ∂ ρ Aσ . Show that ∂µ K µ = −(2/c)E · B. (Hint: consider F˜ µν .) Solution: By direct calculation, we have ∂µ K µ = ϵµνρσ (∂µ Aν )(∂ρ Aσ ) + ϵµνρσ Aν (∂µ ∂ρ Aσ ).

(3.72)

Now partial derivatives commute, so ∂µ ∂ρ Aσ is symmetric in µ and ρ. Its contraction with the antisymmetric tensor ϵµνρσ therefore vanishes, giving ∂µ K µ = ϵµνρσ (∂µ Aν )(∂ρ Aσ ).

(3.73)

We can further exploit the antisymmetry of ϵµνρσ and the arbitrariness of dummy index labels to write 1 ∂µ K µ = (ϵµνρσ − ϵνµρσ − ϵµνσρ + ϵνµσρ ) (∂µ Aν )(∂ρ Aσ ) 4 1 µνρσ = ϵ [(∂µ Aν )(∂ρ Aσ ) − (∂ν Aµ )(∂ρ Aσ ) 4 −(∂µ Aν )(∂σ Aρ ) + (∂ν Aµ )(∂σ Aρ )] 1 = ϵµνρσ (∂µ Aν − ∂ν Aµ )(∂ρ Aσ − ∂σ Aρ ) 4 1 µνρσ = ϵ Fµν Fρσ 4 1 = Fµν F˜ µν , (3.74) 2 using the definitions of the field strength and dual field strength tensors Fµν and F˜ µν . But we have seen in Eq. (2.7.26) that the contraction of these two tensors evaluates to −4(E · B)/c , so in total 2 ∂µ K µ = − E · B. c

(3.75)

3.7 Problem 7 Problem: A general real scalar Lagrangian density at most quadratic in the four-vector potential is L = − 12 (∂µ Aν ∂ µ Aν + ρ ∂µ Aν ∂ ν Aµ ) + λ Aµ Aµ , where ρ, λ ∈ R: (a) Obtain the Euler-Lagrange equations for this system. (b) Assume plane wave solutions with the general form Aµ (x) = Cϵµ (k)e−ik·x and express the Euler-Lagrange equation in terms of ϵµ (k) and k µ .

Relativistic classical fields

62

(c) For k 2 ̸= 0 we can define transverse and longitudinal projectors as T µν ≡ g µν − (k µ k ν /k 2 ) and Lµν ≡ (k µ k ν /k 2 ). Verify the properties T µρ Tρν = T µν , Lµρ Lρν = Lµν and T µν + Lµν = δ µν . Define ϵµℓ ≡ Lµν ϵν and ϵµt ≡ T µν ϵν and so verify that k·ϵℓ = k·ϵ, k·ϵt = 0 and ϵµt + ϵµℓ = ϵµ . (d) Show that there is a plane wave solution to the Euler-Lagrange equations with k 2 = m2ℓ ≡ 2λ/(1 + ρ) that is longitudinal ϵµ = ϵµℓ . (e) Similarly show that there is a plane wave solution with with k 2 = m2t ≡ 2λ that is transverse. (f) We say that the ℓ and t modes propagate with masses mℓ and mt respectively. Choosing ρ = −1 gives mℓ → ∞ so that the ℓ modes cannot propagate, leaving only the t modes. Show that in this limit we recover the Proca Lagrangian density for a massive vector field given in Eq. (6.5.1) with mt becoming the mass of the vector field. Solution: (a) Using the identities ∂Aν ∂(∂µ Aν ) = δ αµ δ βν , = δ βν and ∂Aβ ∂(∂α Aβ )

(3.76)

we get for the given Lagrangian density the equations of motion   ∂L ∂L 0 = ∂α − ∂(∂α Aβ ) ∂Aβ   1 α β µ ν µ ν α β α β ν µ µ ν α β = ∂α − δ µδ ν ∂ A + ∂ A δ µδ ν + ρ δ µδ ν ∂ A + ρ ∂ A δ ν δ µ 2 − λ(δ βµ Aµ + Aµ δ βµ )

= −∂α (∂ α Aβ + ρ ∂ β Aα ) − 2λAβ ,

(3.77)

or in a more conventional form, ∂ 2 Aµ + ρ ∂ µ (∂ · A) + 2λAµ = 0.

(3.78)

(b) If we assume a plane wave ansatz Aµ (x) = Cϵµ (k)e−ik·x , the derivatives of the four-vector potential will be   α ∂ν Aµ = Cϵµ ∂ν e−ikα x = Cϵµ e−ik·x ∂ν (−ikα xα ) = −ikν Cϵµ e−ik·x = −ikν Aµ ,

(3.79)

∂ µ (∂ · A) = −ikν ∂ µ Aν = (−ikν )(−ik µ Aν ) = −k ν (k · A),

(3.80)

from which it follows

Problem 7

63

and ∂ 2 Aµ = ∂ ν ∂ν Aµ = −ikν ∂ ν Aµ = (−ikν )(−ik ν Aµ ) = −k 2 Aµ .

(3.81)

The Euler-Lagrange equations found in Part (a) then reduce to −k 2 Aµ − ρk ν (k · ϵ) + 2λAµ = 0

(2λ − k 2 )ϵµ − ρk µ (k · ϵ) = 0,

=⇒

(3.82)

having factored out Ce−ik·x , which is nowhere vanishing. (c) To avoid redundant calculations, we shall prove the operator identities in the reverse order given. First, we show T µν = gνα T µα = gνα g µα − gνα kµ kν k2 − Lµν

kµ kα k2

= δ µν − = δ µν =⇒

T µν + Lµν = δ µν .

(3.83)

Next, k µ k ρ kρ kν k 2 k µ kν k µ kν = = = Lµν , k2 k2 k4 k2 and finally, using the above, Lµρ Lρν =

(3.84)

T µρ Tρν = (g µρ − Lµρ ) (gρν − Lρν )

= δ µν − Lµν − Lµν + Lµρ Lρν = δ µν − Lµν = T µν .

(3.85)

The required dot products are straightforward to evaluate: kµ kν k2 ν ϵ = k ϵν = k · ϵ, ν k2 k2 k · ϵt = kµ T µν ϵν = kµ (g µν − Lµν )ϵν = k ν ϵν − kµ Lµν ϵν = 0,

k · ϵℓ = kµ Lµν ϵν = kµ

(3.86) (3.87)

and hidden in the second line here is the last required identity ϵµℓ + ϵµt = (T µν + Lµν )ϵν = g µν ϵν = ϵµ .

(3.88)

(d) If we substitute ϵ → ϵℓ and k 2 → m2ℓ in the equation of motion Eq. 3.82, we can simplify using the results of Part (c): 0 = (2λ − m2ℓ )ϵµℓ − ρk µ (k · ϵℓ ) kµ kν = (2λ − m2ℓ ) 2 ϵν − ρk µ (k · ϵ) mℓ   2λ − 1 − ρ kµ . = (k · ϵ) m2ℓ

(3.89)

64

Relativistic classical fields

Discounting the case where k · ϵ = 0, it must be that

2λ 2λ − 1 − ρ = 0, i.e. m2ℓ = . m2ℓ 1+ρ

(3.90)

Therefore, there is a valid longitudinal solution to the Euler-Lagrange equations with the given mass. (e) Similarly, if we substitute ϵ → ϵt and k 2 → m2t in the equation of motion Eq. 3.82, the results of part (c) allow us to simplify considerably: 0 = (2λ − m2t )ϵµt − ρk µ (k · ϵt ) = (2λ − m2t )ϵµt ,

(3.91)

so discounting the case where ϵµt = 0, it must be that m2t = 2λ. Hence, there is a valid transverse solution to the Euler-Lagrange equations with this mass. (f) With ρ = −1, the given Lagrangian density is

1 L = − (∂µ Aν ∂ µ Aν − ∂µ Aν ∂ ν Aµ ) + λAµ Aµ 2 1 m2 = − (∂µ Aν )(∂ µ Aν − ∂ ν Aµ ) + t Aµ Aµ 2 2 1 1 2 µν µ = − (∂µ Aν )F + mt Aµ A , 2 2

(3.92)

where we have substituted λ → m2t /2 from part (e). Now since the field strength tensor F µν is antisymmetric, we can use the same trick as in Problem 3.6 to write 1 1 L = − (∂µ Aν )(F µν − F νµ ) + m2t Aµ Aµ 4 2 1 1 2 µν = − (∂µ Aν − ∂ν Aµ )F + mt Aµ Aµ 4 2 1 1 2 µν µ = − Fµν F + mt Aµ A , (3.93) 4 2 by relabeling dummy indices to arrive at the second line. This is exactly the Proca Lagrangian density for a free massive vector field, where mt is the mass of the vector field (cf. Eq. (6.5.1)).

3.8 Problem 8 Problem: Scalar electrodynamics is the gauge invariant minimal coupling of a charged scalar field and the electromagnetic field. It has the Lagrangian density L = − 14 Fµν F µν + (Dµ ϕ)∗ (Dµ ϕ) − m2 ϕϕ∗ , where the covariant derivative is defined as as Dµ ≡ ∂µ + iqAµ and where q is the electric charge of the scalar particle:

Problem 8

65

(a) The system is invariant under global phase transformations ϕ → e−iα ϕ. Use this to derive the Noether current that replaces Eq. (3.2.49), j µ = iqc [ϕ∗ (Dµ ϕ)−(Dµ ϕ)∗ ϕ] = iqc [ϕ∗ (∂ µ ϕ)−(∂ µ ϕ∗ )ϕ+2iqAµ ϕ∗ ϕ] (b) Show that the Euler-Lagrange equations can be written as ∂µ F µν = j ν /c and (Dµ Dµ + m2 )ϕ = 0; and (c) Show that Aµ (x) → Aµ (x) + ∂ µ α(x), ϕ(x) → e−iqα(x) ϕ(x) leaves the system invariant. This is an example of U (1) gauge invariance. Solution: (a) The given symmetry transformation is internal, so the corresponding conserved current is given by ⃗ j µ = ⃗π µ · Φ,

(3.94)

⃗ is the vector of field changes under an infinitesimal symmetry transwhere Φ formation, and ⃗π is the vector of conjugate momenta  T ∂L ∂L µ ⃗π := , . (3.95) ∂(∂µ ϕ) ∂(∂µ ϕ∗ ) Now with the covariant derivatives written out explicitly, the Lagrangian density is 1 L = − Fµν F µν + (∂µ ϕ∗ − iqAµ ϕ∗ )(∂ µ ϕ + iqAµ ϕ) − m2 ϕϕ∗ , 4

(3.96)

and so its derivatives are ⃗π µ = (∂ µ ϕ∗ − iqAµ ϕ∗ , ∂ µ ϕ + iqAµ ϕ)T T

= ((Dµ ϕ)∗ , Dµ ϕ) .

(3.97)

⃗ Under the given symmetry transformation, the Now all we need to find is Φ. two fields transform as ϕ → e−iα ϕ = ϕ − iαϕ + O(α2 ), ∗

ϕ → (e

−iα

∗

∗

∗

(3.98) 2

ϕ) = ϕ + iαϕ + O(α ).

(3.99)

From these, we see that the vector of field changes under an infinitesimal sym⃗ = (−iϕ, +iϕ∗ )T . Hence, the conserved current is metry is Φ j µ = (Dµ ϕ)∗ (−iϕ) + (Dµ ϕ)(iϕ∗ ) = i [ϕ∗ Dµ ϕ − ϕ(Dµ ϕ)∗ ]

= i [ϕ∗ (∂ µ ϕ + iqAµ ϕ) − ϕ(∂ µ ϕ∗ − iqAµ ϕ∗ )]

= i [ϕ∗ ∂ µ ϕ − ϕ∂ µ ϕ∗ + 2iqAµ ϕ∗ ϕ] .

(3.100)

66

Relativistic classical fields

Of course, any constant multiple of this current will also be conserved, so we may rescale j µ → qc j µ to arrive at the normalization given in the question, j µ = iqc [ϕ∗ ∂ µ ϕ − ϕ∂ µ ϕ∗ + 2iqAµ ϕ∗ ϕ] . (b) For the equations of motion of the field Aµ , we note the result   ∂L ∂ 1 αβ = − Fαβ F = −F µν ∂(∂µ Aν ) ∂(∂µ Aν ) 4

(3.101)

(3.102)

from Eq. (3.3.5) of the text. The Euler-Lagrange equations then reduce to   ∂L ∂L 0 = ∂µ − ∂(∂µ Aν ) ∂Aν
= −∂µ F µν − iqϕ∂ ν ϕ∗ − iqϕ∗ ∂ µ ϕ + 2q 2 Aν ϕ∗ ϕ   1 ν µν = −∂µ F − − j , (3.103) c or ∂µ F µν = j ν /c. And for the scalar field, the equation of motion is   ∂L ∂L 0 = ∂µ − ∂(∂µ ϕ∗ ) ∂ϕ∗
= ∂µ (Dµ ϕ) − −iqAµ (∂µ ϕ + iqAµ ϕ) − m2 ϕ = ∂µ (Dµ ϕ) + iqAµ Dµ ϕ + m2 ϕ

= (∂µ + iqAµ )Dµ ϕ + m2 ϕ = Dµ Dµ ϕ + m2 ϕ = (Dµ Dµ + m2 )ϕ.

(3.104)

Of course, the equation of motion of the conjugate field is simply the conjugate of this equation. (c) Under a local phase transformation, the field strength tensor transforms as Fµν = ∂µ Aν − ∂ν Aµ → ∂µ (Aν + ∂ν α) − ∂ν (Aµ + ∂µ α)

= (∂µ Aν − ∂ν Aµ ) + (∂µ ∂ν α − ∂ν ∂µ α)

= Fµν ,

(3.105)

because partial derivatives commute. That is, the field strength tensor is invariant under a local phase transformation. Next, we look at the covariant derivative: Dµ ϕ = ∂µ ϕ + iqAµ ϕ → ∂µ (e−iqα ϕ) + iq(Aµ + ∂µ α)(e−iqα ϕ) = −iq(∂µ α)e−iqα ϕ + e−iqα ∂µ ϕ

+ iqAµ e−iqα ϕ + iq(∂µ α)(e−iqα ϕ)

= e−iqα (∂µ ϕ + iqAµ ϕ) = e−iqα Dµ ϕ.

(3.106)

67

Problem 8 We see that Dµ ϕ transforms in the same way as ϕ. It follows that (Dµ ϕ)∗ (Dµ ϕ) will transform in the same way as the mass term ϕ∗ ϕ. Now the mass term is invariant under the local transformation: ϕ∗ ϕ → (e−iqα ϕ)∗ (e−iqα ϕ) = e+iqα ϕ∗ e−iqα ϕ = ϕ∗ ϕ,

(3.107)

and therefore all terms in the Lagrangian density are unchanged under a local phase transformation. In other words, such a transformation leaves the whole system invariant.

4

Relativistic Quantum Mechanics 4.1 Problem 1 Problem: Show that the symmetric n-boson basis states in Eq. (4.1.149) are normalized, i.e., show ⟨bi1 · · · bin ; S|bi1 · · · bin ; S⟩ = 1. Solution: The symmetric n-boson basis states are defined by rQ X m nm ! |bσ(i1 ) bσ(i2 ) · · · bσ(in ) ⟩, |bi1 · · · bin ; S⟩ = n! σ

(4.1)

where the sum is over all distinct permutations σ of the elements i1 , i2 , · · · , in , and nm is the number of times bm occurs in the list bi1 bi2 · · · bin . The squared norm of such a state is then !2 rQ XX n ! m m ⟨bi1 · · · bin ; S|bi1 · · · bin ; S⟩ = ⟨bτ (i1 ) · · · bτ (in ) |bσ(i1 ) · · · bσ(in ) ⟩ n! σ τ Q nm ! X X = m δτ (i1 )σ(i1 ) δτ (i2 )σ(i2 ) · · · δτ (in )σ(in ) , n! σ τ (4.2) where we have used Eq. (4.1.149) from the text. Note that the terms in this sum vanish unless σ(im ) = τ (im ) for all m = 1, · · · , n, i.e. unless σ and τ are the same permutation, in which case the terms are unity. So the double sum reduces to Q nm ! X ⟨bi1 · · · bin ; S|bi1 · · · bin ; S⟩ = m 1. (4.3) n! σ We see this sum is counting the number of distinct permutations of our symbols bi1 , bi2 , · · · , bin . The number of permutations of n distinct elements is n!, but in fact this overcounts the number of permutations we have by a factor of nm ! for each bm in our list, because permuting these identical symbols does not result in a different ordering. That is to say, X σ

68

n! , m nm !

1= Q

(4.4)

Problem 2

69

so in total, Q ⟨bi1 · · · bin ; S|bi1 · · · bin ; S⟩ =

nm ! n! Q = 1. n! m nm !

m

(4.5)

The states are correctly normalized.

4.2 Problem 2 Problem: ˆ in quantum mechanics we have [Vˆ i , Jˆj ] = iℏϵijk Vˆ k For any vector operator V from Eq. (4.1.114). It will be helpful to also recall Eq. (1.2.64). (a) Use these results to show that ⟨jb mb |Vˆz |ja ma ⟩ = 0 unless ma = mb . (b) Define Vˆ± ≡ Vˆx ± iVˆy in the usual way and show that ⟨jb mb |Vˆ± |ja ma ⟩ = 0 unless mb = ma ± 1. (c) Evaluate [Jˆ+ , Vˆ+ ] and use this result for the special case j = ja = jb to obtain the dependence of ⟨j (m + 1)|Vˆ+ |jm⟩ on m; and ˆ ˆ (d) Extending this result show that ⟨jmb |V|jm a ⟩ = Cj ⟨jmb |J|jma ⟩, where Cj is a constant independent of mb and ma . Solution: (a) From [Vˆ i , Jˆj ] = iℏϵijk Vˆ k , we have that [Vˆz , Jˆz ] = 0, or Vˆz Jˆz = Jˆz Vˆz . Now since Jˆz |jm⟩ = ℏm|jm⟩, and since Jˆz is self-adjoint, we can calculate ℏma ⟨jb mb |Vˆz |ja ma ⟩ = ⟨jb mb |Vˆz Jˆz |ja ma ⟩ = ⟨jb mb |Jˆz Vˆz |ja ma ⟩

= ℏmb ⟨jb mb |Vˆz |ja ma ⟩

(4.6)

Hence, if ma ̸= mb , we must have ⟨jb mb |Vˆz |ja ma ⟩ = 0. (b) Using the linearity of the commutator, we have [Vˆ± , Jˆz ] = [Vˆx , Jˆz ] ± i[Vˆy , Jˆz ] = −iℏVˆy ± i(iℏVˆx ) = ∓ℏ(Vˆx ± iVˆy ) = ∓ℏVˆ± .

(4.7)

This allows us to perform a similar calculation to that in Part (a): ℏma ⟨jb mb |Vˆ± |ja ma ⟩ = ⟨jb mb |Vˆ± Jˆz |ja ma ⟩ = ⟨jb mb |(Jˆz Vˆ± ∓ ℏVˆ± )|ja ma ⟩ = ℏ(mb ∓ 1)⟨jb mb |Vˆ± |ja ma ⟩.

(4.8)

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Hence, if mb ̸= ma ± 1, we must have ⟨jb mb |Vˆ± |ja ma ⟩. (c) With our known commutation relations, noting that [Vˆ i , Jˆj ] = −[Jˆj , Vˆ i ], we have [Jˆ+ , Vˆ+ ] := [Jˆx + iJˆy , Vˆx + iVˆy ] = [Jˆx , Vˆx ] + i[Jˆx , Vˆy ] + i[Jˆy , Vˆx ] − [Jˆy , Vˆy ] = 0 + i(iℏVˆz ) + i(−iℏVˆz ) − 0 = 0,

To continue, we must know that p Jˆ± |jm⟩ = ℏ j(j + 1) − m(m ± 1)|j (m ± 1)⟩. † Then since Jˆ+ = Jˆx + iJˆy = (Jˆx − iJˆy )† = Jˆ− , it follows that †  p ⟨jm|Jˆ+ = Jˆ− |jm⟩ = ℏ j(j + 1) − m(m − 1) ⟨j (m − 1)|,

(4.9)

(4.10)

(4.11)

and so 1 ⟨j (m + 1)|Vˆ+ Jˆ+ |j (m − 1)⟩ ⟨j (m + 1)|Vˆ+ |jm⟩ = p ℏ j(j + 1) − m(m − 1) 1 = p ⟨j (m + 1)|Jˆ+ Vˆ+ |j (m − 1)⟩ ℏ j(j + 1) − m(m − 1) s j(j + 1) − m(m + 1) = ⟨j m|Vˆ+ |j (m − 1)⟩. (4.12) j(j + 1) − m(m − 1) The inner product on the right-hand side here is of the same form as the one on the left-hand side, only with m → m − 1. We may therefore continually repeat this process on the right-hand side until m reaches its minimum possible value, m = −j. Fortunately the product telescopes, leaving us with s j(j + 1) − m(m + 1) ⟨j (m + 1)|Vˆ+ |jm⟩ = ⟨j (−j + 1)|Vˆ+ |j (−j)⟩. j(j + 1) − (−j)(−j + 1) (4.13) We may gather the terms that depend on j, but not on m, into ⟨j (−j + 1)|Vˆ+ |j (−j)⟩ Cj := p , j(j + 1) − (−j)(−j + 1) to see that the m-dependence of our expression is given by p ⟨j (m + 1)|Vˆ+ |jm⟩ = Cj j(j + 1) − m(m + 1).

(4.14)

(4.15)

(d) Note that the above result can be written as ⟨j (m + 1)|Vˆ+ |jm⟩ = Cj ⟨j (m + 1)|Jˆ+ |jm⟩,

(4.16)

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71

and therefore more generally ⟨j mb |Vˆ+ |jma ⟩ = Cj ⟨j mb |Jˆ+ |jma ⟩

(4.17)

because both sides vanish whenever mb ̸= ma + 1, by Part (b). In order to arrive at the analogous equation for Vˆz and Jˆz , we must relate Vˆz to Vˆ+ , for this is used in the definition of Cj . The only relations at our disposal are [Vˆ i , Jˆj ] = iℏϵijk Vˆ k , which tell us 1 Vˆz = [Vˆx , Jˆy ] iℏ i Vˆz = [Vˆy , Jˆx ] ℏ

=⇒ =⇒

1 1 Vˆz = [Vˆx + iVˆy , Jˆy ] = [Vˆ+ , Jˆy ], iℏ iℏ 1 1 Vˆz = [Vˆx + iVˆy , Jˆx ] = [Vˆ+ , Jˆx ], ℏ ℏ

(4.18) (4.19)

and therefore 1 Vˆz = 2ℏ



1 [Vˆ+ , Jˆx ] + [Vˆ+ , Jˆy ] i

 =

1 ˆ ˆ [V+ , J− ]. 2ℏ

(4.20)

Using this, we have 1 ⟨j mb |(Vˆ+ Jˆ− − Jˆ− Vˆ+ )|jma ⟩ ⟨j mb |Vˆz |jma ⟩ = 2ℏ 1  p ℏ j(j + 1) − ma (ma − 1)⟨j mb |Vˆ+ |j (ma − 1)⟩ = 2ℏ  p −ℏ j(j + 1) − mb (mb + 1)⟨j (mb + 1)|Vˆ+ |j ma )⟩ C j p = j(j + 1) − ma (ma − 1)⟨j mb |Jˆ+ |j (ma − 1)⟩ 2  p − j(j + 1) − mb (mb + 1)⟨j (mb + 1)|Jˆ+ |j ma )⟩ p C j p j(j + 1) − ma (ma − 1) j(j + 1) − ma (ma − 1)δma mb = 2  p p − j(j + 1) − mb (mb + 1) j(j + 1) − ma (ma + 1)δma mb Cj [j(j + 1) − ma (ma − 1) − j(j + 1) + ma (ma + 1)] δma mb 2 = Cj ma δma mb = Cj ⟨j mb |Jˆz |jma ⟩. (4.21) =

For compactness, we used δma mb to denote 1 if ma = mb , and 0 otherwise. Finally, we need to derive the analogous relation for Vˆ− and Jˆ− , for then the result follows. Using the same technique, we have Vˆx = [Jˆy , Vˆz ]/(iℏ) and Vˆy = −[Jˆx , Vˆz ]/(iℏ), so 1 1 Vˆ− = Vˆx − iVˆy = [Jˆx − iJˆy , Vˆz ] = [Jˆ− , Vˆz ]. ℏ ℏ Hence, ⟨j mb |Vˆ− |jma ⟩ =

1 ⟨j mb |(Jˆ− Vˆz − Vˆz Jˆ− )|jma ⟩ ℏ

(4.22)

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1 p ℏ j(j + 1) − mb (mb + 1)⟨j (mb + 1)|Vˆz |j ma ⟩ ℏ  p −ℏ j(j + 1) − ma (ma − 1)⟨j mb |Vˆz |j (ma − 1)⟩ p = Cj j(j + 1) − mb (mb + 1)⟨j (mb + 1)|Jˆz |j ma ⟩  p − j(j + 1) − ma (ma − 1)⟨j mb |Jˆz |j (ma − 1)⟩ p = Cj j(j + 1) − mb (mb + 1)ma δma (mb +1)  p − j(j + 1) − ma (ma − 1)mb δ(ma −1) mb p j(j + 1) − ma (ma − 1)ma = Cj  p − j(j + 1) − ma (ma − 1)(ma − 1) δ(ma −1) mb p = Cj j(j + 1) − ma (ma − 1)δ(ma −1) mb = Cj ⟨j mb |Jˆ− |jma ⟩. (4.23) =

Combining these results, we see   1 ⟨j mb |Vˆx |jma ⟩ = ⟨j mb | Vˆ+ + Vˆ− |jma ⟩ 2   Cj ⟨j mb | Jˆ+ + Jˆ− |jma ⟩ = 2 = Cj ⟨j mb |Jˆx |jma ⟩,

(4.24)

and similarly   1 ⟨j mb |Vˆy |jma ⟩ = ⟨j mb | Vˆ+ − Vˆ− |jma ⟩ 2i   Cj = ⟨j mb | Jˆ+ − Jˆ− |jma ⟩ 2i = Cj ⟨j mb |Jˆy |jma ⟩.

(4.25)

ˆ ˆ Therefore in total, ⟨j mb |V|jm a ⟩ = Cj ⟨j mb |J|jma ⟩, where Cj is defined in Eq. 4.14.

4.3 Problem 3 Problem: A scalar particle with mass m and charge q interacts with a time-independent electromagnetic potential given by qA0 = 0 for z < 0 and qA0 = V with V > 0 for z ≥ 0. A particle plane wave is moving in the +z-direction with energy E > 0. Calculate the reflection (R) and transmission (T ) coefficients, where these are the √ probability of reflection and transmission respectively, R + T = 1. Define p ≡ E 2 −m2 and p 2 2 K ≡ (E −V ) −m .

73

Problem 3

(a) Show that if |E − V | < m, then there is no transmission, R = 1 and T = 0. (b) Show that if E−V > m, then R = |(p−K)2 /(p+K)2 | < 1 and T = 1−R and there is partial transmission and partial reflection. (c) Show that if E−V < −m, then R = |(p+K)2 /(p−K)2 | > 1 and T = 1 − R, which corresponds to a negative current being transmitted and more positive current reflected than was incident. This is the Klein paradox and shows that |V | > 2m can lead to particle-antiparticle pair creation. This illustrates the breakdown of relativistic quantum mechanics in the presence of strong fields. (Hint: Use continuity of the wavefunction and its first derivative.) Solution: In the presence of an electromagnetic field, the behavior of a scalar particle ϕ is governed by the modified Klein-Gordon equation     mc 2  q q ϕ = 0, (4.26) ∂ µ + i Aµ ∂µ + i Aµ + ℏc ℏc ℏ from Eq. (4.3.57) in the text. In our case, this reduces to  h
i  0 = ∂ µ ∂µ + mc 2 ϕ for z < 0, ℏ i h


2  0 = ∂ 0 + i V ∂0 + i V − ∇2 + mc ϕ for z ≥ 0. ℏc ℏc ℏ

(4.27)

In the region z < 0 we have the usual Klein-Gordon equation, which has the familiar plane wave solutions  iEt  ipz ipz (z < 0), (4.28) ϕ(x) = Ae ℏ + Be− ℏ e− ℏ where A and B are arbitrary constants, and 1p 2 p := E − m2 c4 . c

(4.29)

The term proportional to A represents a wave moving in the +z-direction, i.e. the incident wave, while the term proportional to B represents a wave moving in the −z-direction, i.e. the reflected wave. Now to solve the equation in the region z ≥ 0, let us assume a solution of the form ϕ(x) = Ceiκz/ℏ e−iEt/ℏ , and find conditions for C and κ. Substituting this in, the equation becomes 2  2   mc 2 E V iκ 0 = −i +i ϕ(x) − ϕ(x) + ϕ(x), (4.30) ℏc ℏc ℏ ℏ which is easily seen to be solved by 1p κ=± (E − V )2 − m2 c4 ≡ ±K c

(4.31)

for arbitrary C. Here we run into a subtle point that lies at the heart of this exercise: are we to take the positive or negative solution for κ? The answer lies in considering

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the group velocity of this wave; on physical grounds it must point in the positive zdirection, because this solution represents the transmitted wave. The group velocity is found from Eq. (4.2.21): vg =

dE Kc ˆz = ± ˆz , e ce dκ E−V

(4.32)

and therefore we must take the positive solution when E − V > 0, and the negative solution when E − V < 0. That is,  +K if E − V > 0, κ= (4.33) −K if E − V < 0. This will be important later. In total, our wavefunction is  iEt  ipz ( ipz ϕ− (x) ≡ Ae ℏ + Be− ℏ e− ℏ ϕ(x) = iEt iκz ϕ+ (x) ≡ Ce ℏ e− ℏ

for z < 0 for z ≥ 0.

(4.34)

Since this wavefunction is non-normalizable, in order to find the reflection and transmission coefficients we need only find the relative probabilities of reflection and transmission, given by the strengths of the reflected and transmitted current densities relative to the incident current density. The current densities associated to each of the wave components are found using Eq. (4.3.24): j µ = iqc[ϕ∗ ∂ µ ϕ − (∂ µ ϕ)∗ ϕ].

(4.35)

The vector component of the current for the incident wave, for example, is h ipz ipz iEt iEt jA = −iqc (Ae ℏ e− ℏ )∗ ∇(Ae ℏ e− ℏ ) i ipz ipz iEt iEt −(∇(Ae ℏ e− ℏ ))∗ (Ae ℏ e− ℏ ) qc ˆz = + (p + p∗ )|A|2 e ℏ 2qc ˆz ≡ jA e ˆz , p|A|2 e (4.36) =+ ℏ and similarly 2qc ˆ z ≡ jB e ˆz , p|B|2 e (4.37) ℏ qc ˆ z ≡ jC e ˆz . jC = + (κ + κ∗ )|C|2 e (4.38) ℏ Note that p is always real for an on-shell particle, which allowed us to simplify the expressions for jA and jB , but the same is not necessarily true of κ. The reflection and transmission coefficients are therefore jB = −

R≡

−jB |B|2 = , jA |A|2

T ≡

jC |C|2 κ + κ∗ = . jA |A|2 2p

(4.39)

Now we must impose some relations between A, B, and C so that we may calculate these coefficients. Since the Klein-Gordon equation is second order, we have two

Problem 4

75

conditions to impose: ϕ(x) must be continuous everywhere, and ∂µ ϕ(x) must be continuous everywhere. If we consider the boundary plane z = 0, these conditions respectively require ϕ− (x)|z=0 = ϕ+ (x)|z=0

=⇒

A + B = C,

(4.40)

∂z ϕ− (x)|z=0 = ∂z ϕ+ (x)|z=0

=⇒

p(A − B) = κC.

(4.41)

Together, these imply p(A − B) = κ(A + B)

and p(A − (C − A)) = κC,

(4.42)

|C|2 κ + κ∗ 2p (κ + κ∗ ). = 2 |A| 2p |p + κ|2

(4.43)

allowing us to conclude R=

|B|2 |p − κ|2 = , |A|2 |p + κ|2

T =

Notice that since p is real, it is always the case that R+T =

(p − κ)(p − κ∗ ) + 2p(κ + κ∗ ) (p + κ)(p + κ∗ ) = = 1. 2 |p + κ| |p + κ|2

(4.44)

(a) If we assume that |E − V | < mc2 , then from its definition κ will be purely imaginary. As a result, κ + κ∗ = κ + (−κ) = 0, and so T = 0: there is no transmission. Furthermore since R + T = 1, we must have R = 1. (b) If we assume that E − V > mc2 , then κ will be real. From our considerations above, we must have in this case κ = K. Then our coefficients reduce to R=

(p − K)2 , (p + K)2

T =

4pK . (p + K)2

(4.45)

Now clearly R > 0 and T > 0, so since R + T = 1 it must be that R < 1 and T < 1: there is partial transmission and partial reflection. (c) Finally, if we assume that E − V < −mc2 , then (E − V )2 > m2 c4 and once again κ is real, though now we must have κ = −K. Then κ∗ = −K, so we may write R=

(p + K)2 , (p − K)2

T =−

4pK . (p − K)2

(4.46)

Notice that T < 0. Then since R + T = 1, we must have R > 1 in this regime.

4.4 Problem 4 Problem: Using the properties of the Dirac γ-matrices verify that the matrices ( 12 σ µν ) satisfy the Lie algebra of the Lorentz transformations in Eq. (1.2.179).

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Solution: We wish to find the commutation relations among the matrices M µν := 21 σ µν = i µ ν 4 [γ , γ ]. For calculating these, note that the anticommutation relations of the γ matrices imply  −γ ν γ µ for µ ̸= ν µ ν γ γ = (4.47) ±I for µ = ν, which in turn implies [γ µ γ ν , A] = −[γ ν γ µ , A] for all values of µ and ν and any matrix A. This simplifies our calculation greatly, for then i µ ν [γ γ , M ρσ ] − 4 i = [γ µ γ ν , M ρσ ] 2 i2 = [γ µ γ ν , γ ρ γ σ ] 4 Then using the commutator identities [M µν , M ρσ ] =

i ν µ [γ γ , M ρσ ] 4

by the same logic.

[A, BC] = [A, B]C + B[A, C]

and

[AB, C] = A{B, C} − {A, C}B,

(4.48)

(4.49) (4.50)

we get 1 [M µν , M ρσ ] = − ([γ µ γ ν , γ ρ ]γ σ + γ ρ [γ µ γ ν , γ σ ]) 4 1 µ ν ρ σ = − (γ {γ , γ }γ − {γ µ , γ ρ }γ ν γ σ + γ ρ γ µ {γ ν , γ σ } − γ ρ {γ µ , γ σ }γ ν ) 4 1 = − (g νρ γ µ γ σ − g µρ γ ν γ σ + g νσ γ ρ γ µ − g µσ γ ρ γ ν ) . (4.51) 2 Now we note that 1 γ µ γ ν = ([γ µ , γ ν ] + {γ µ , γ ν }) 2 = −2iM µν + g µν , (4.52) and so, using the symmetry of g µν and antisymmetry of M µν , 1 [M µν , M ρσ ] = − (g νρ (−2iM µσ + g µσ ) − g µρ (−2iM νσ + g νσ ) 2 +g νσ (−2iM ρµ + g ρµ ) − g µσ (−2iM ρν + g ρν )) = i (g νρ M µσ − g µρ M νσ − g νσ M µρ + g µσ M νρ ) .

(4.53)

This is exactly the Lie algebra of the Lorentz transformations, as was to be shown.

4.5 Problem 5 Problem:

Problem 6

77

Explain why the hermiticity of γ 0 and γ 5 and the antihermiticity of γ are preserved in any representation unitarily equivalent to the Dirac representation. Solution: Suppose H+ is a hermitian matrix such as γ 0 or γ 5 , and suppose H− is an antihermitian matrix such as γ i (working in the Dirac representation). Then we † can write H± = ±H± . In any representation unitarily equivalent to the Dirac representation, these matrices will be related to those in the original representation through a similarity transformation ¯ ± ≡ U H± U −1 = U H± U † , H

(4.54)

for some unitary matrix U . We see that ¯ † = (U H± U † )† H ±

† † = (U † )† H± U

= U (±H± )U † ¯ ±. = ±H

(4.55)

¯ + is hermitian and H ¯ − is antihermitian; (anti)hermiticity is preIn other words, H served in any representation unitarily equivalent to the Dirac representation. Note that the unitarity is important here because it allowed us to express the similarity transformation in terms of U † rather than U −1 , enabling the later manipulations.

4.6 Problem 6 Problem: Provide detailed proofs of the trace identities: (a) traces of products of γ matrices (with no γ µ† ) are representation independent (same for γ µ and γ¯ µ ≡ Sγ µ S −1 ). (b) tr (I) = 4, tr (γ µ γ ν ) = 4g µν , tr (γ µ ) = tr (odd # of γ µ′ s) = tr (σ µν ) = 0, tr (γ µ γ ν γ ρ γ σ ) = 4 (g µν g ρσ −g µρ g νσ +g µσ g νρ ).


(c) tr γ 5 γ µ γ ν = tr γ 5 (odd # of γ µ′ s) = 0 and tr γ 5 γ µ γ ν γ ρ γ σ = −4iϵµνρσ . Solution: (a) Moving from one representation of γ matrices to another amounts to a change of basis. Since the trace of a matrix is the sum of its eigenvalues, it does not depend on the basis in which the matrix is expressed. Therefore, the trace of any product of γ matrices is representation independent. To see this algebraically, suppose we have any collection of γ matrices γ µi ,

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i = 1, . . . , N , in a certain representation. In any other representation, these matrices will be of the form γ¯ µi = Sγ µi S −1 for some invertible matrix S. Then
tr (¯ γ µ1 γ¯ µ2 · · · γ¯ µN ) = tr Sγ µ1 S −1 Sγ µ2 S −1 · · · Sγ µN S −1
= tr Sγ µ1 γ µ2 · · · γ µN S −1
= tr γ µ1 γ µ2 · · · γ µN S −1 S = tr (γ µ1 γ µ2 · · · γ µN ) ,

(4.56)

where we have used the cyclic property of the trace1 in the third step. So we see the trace of any collection of γ matrices is independent of representation. (b)

(i) The 4×4 identity matrix is I = diag(1, 1, 1, 1), so tr(I) = 1 + 1 + 1 + 1 = 4. (ii) Using the anticommutation relations {γ µ , γ ν } = 2g µν I and the linearity and cyclic properties of the trace, we have 1 [tr (γ µ γ ν ) + tr (γ ν γ µ )] 2 1 = tr ({γ µ , γ ν }) 2 1 = 2g µν tr(I) 2 = 4g µν .

tr (γ µ γ ν ) =

(4.57)

(iii) Consider an odd number of γ matrices γ µi , i = 1, . . . , 2N + 1. Since γ 5 anticommutes with each of these matrices, we have γ 5 γ µ1 γ µ2 · · · γ µ2N +1 = (−1)2N +1 γ µ1 γ µ2 · · · γ µ2N +1 γ 5 .

(4.58)

Then since (γ 5 )2 = I, using the cyclic property of the trace we calculate
tr (γ µ1 γ µ2 · · · γ µ2N +1 ) = tr γ 5 γ 5 γ µ1 γ µ2 · · · γ µ2N +1
= (−1)2N +1 tr γ 5 γ µ1 γ µ2 · · · γ µ2N +1 γ 5 = −tr (γ µ1 γ µ2 · · · γ µ2N +1 ) .

(4.59)

We conclude the trace of any odd number of γ matrices must vanish. Note this includes the case for a single γ matrix. (iv) It follows from the cyclic property of the trace that the trace of any commutator vanishes (at least for finite-dimensional matrices). So then since σ µν = i[γ µ , γ ν ]/2, we must have tr(σ µν ) = 0. (v) Using the anticommutation relations along with our previous results, we calculate tr (γ µ γ ν γ ρ γ σ ) = 2g µν tr (γ ρ γ σ ) − tr (γ ν γ µ γ ρ γ σ )

= 2g µν · 4g ρσ − 2g µρ tr (γ ν γ σ ) + tr (γ ν γ ρ γ µ γ σ )

= 8g µν g ρσ − 2g µρ · 4g νσ + 2g µσ tr (γ ν γ ρ ) − tr (γ ν γ ρ γ σ γ µ )

1

This is the easily-proven property that tr(AB) = tr(BA) for any square matrices A and B of the same size.

Problem 7

79

= 8g µν g ρσ − 8g µρ g νσ + 2g µσ · 4g νρ − tr (γ µ γ ν γ ρ γ σ ) , (4.60) having used the cyclic property of the trace in the last step. Solving for tr(γ µ γ ν γ ρ γ σ ) yields the desired relation tr (γ µ γ ν γ ρ γ σ ) = 4 (g µν g ρσ − g µρ g νσ + g µσ g νρ ) .

(4.61)

Obviously, this approach generalizes to any even number of γ matrices. (c)

(i) Since γ 5 = iγ 0 γ 1 γ 2 γ 3 , any odd number of gamma matrices times γ 5 will be the product of an odd number of gamma matrices, and this has a vanishing trace by an earlier result. (ii) Now consider tr(γ 5 γ µ γ ν ). Let γ ρ be distinct from both γ µ and γ ν . Then γ ρ appears exactly once in the product γ 5 γ µ γ ν = iγ 0 γ 1 γ 2 γ 3 γ µ γ ν , and it anticommutes with the other five matrices in the product, meaning that γ ρ γ 5 γ µ γ ν = (−1)5 γ 5 γ µ γ ν γ ρ . Since (γ ρ )2 = ±I depending on ρ, we may calculate

tr γ 5 γ µ γ ν = ±tr γ ρ γ ρ γ 5 γ µ γ ν
= ±tr −γ ρ γ 5 γ µ γ ν γ ρ
= −tr γ 5 γ µ γ ν , (4.62) using the cyclic property of the trace in the last step. This implies the result tr(γ 5 γ µ γ ν ) = 0. (iii) Finally, consider the quantity tr(γ 5 γ µ γ ν γ ρ γ σ ). If any two of γ µ , γ ν , γ ρ , and γ σ are equal, then they can be brought together by some number of commutations with the other γ matrices. Since (γ µ )2 ∝ I, this leaves us with the trace of the product of γ 5 with only two other γ matrices. But this vanishes by the previous result, so we see that all of the γ matrices in the product must be distinct for the trace to be non-zero. Distinct γ matrices all anticommute, so this means tr(γ 5 γ µ γ ν γ ρ γ σ ) must be proportional to the antisymmetric tensor with four indices, ϵµνρσ . We can evaluate the constant of proportionality by considering the special case

tr γ 5 γ 0 γ 1 γ 2 γ 3 = −i tr γ 5 γ 5 = −i tr(I) = −4i = −4iϵ0123 , (4.63) implying that tr(γ 5 γ µ γ ν γ ρ γ σ ) = −4iϵµνρσ .

4.7 Problem 7 Problem: Prove the following identities: (a) γµ γν γρ = gµν γρ − gµρ γν + gνρ γµ + iϵµνρσ γ σ γ 5 .

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(b) γ µ γµ = 4, γ µ γ 5 γµ = −4γ 5 , γ µ ̸ kγµ = −2̸ k, γ µ γ 5 ̸ kγµ = 2γ 5 ̸ k, γ µ ̸ k̸ ℓγµ = 4k · ℓ, γ µ ̸ k̸ ℓ̸ pγµ = −2̸ p̸ ℓ̸ k, γ µ ̸ k̸ ℓ̸ p̸ qγµ = 2 (̸ q̸ k̸ ℓ̸ p + ̸ p̸ ℓ̸ k̸ q). Solution: (a) We begin by performing a series of commutations: γµ γν γρ = 2gµν γρ − γν γµ γρ

= 2gµν γρ − 2gµρ γν + γν γρ γµ

= 2gµν γρ − 2gµρ γν + 2gνρ γµ − γρ γν γµ ,

(4.64)

giving 1 (γµ γν γρ + γρ γν γµ ) = gµν γρ − gµρ γν + gνρ γµ . 2 The required identity therefore follows as long as

(4.65)

1 (γµ γν γρ + γρ γν γµ ) = γµ γν γρ − iϵµνρσ γ σ γ 5 . (4.66) 2 It is easy to see that this relation holds if any two of µ, ν, and ρ are equal, because the square of any γ matrix is proportional to the identity and therefore commutes with everything, and also ϵµνρσ will vanish. It remains to check the cases where µ, ν, and ρ are all distinct. Since distinct γ matrices anticommute, in these cases the above condition simplifies to γµ γν γρ = iϵµνρσ γ σ γ 5 .

(4.67)

Both sides here are completely antisymmetric, so it suffices to check the four cases (µ, ν, ρ) = (0, 1, 2), (0, 1, 3), (0, 2, 3), (1, 2, 3). This is a tedious but rather straightforward application of the identities γ0 = γ 0 , γi = −γ i , (γ 0 )2 = I, (γ i )2 = −I, and the the anticommutivity of distinct γ matrices: iϵ012σ γ σ γ 5 = −ϵ0123 γ 3 γ 0 γ 1 γ 2 γ 3 = (−1)3 γ 0 γ 1 γ 2 γ 3 γ 3 = γ0 γ1 γ2 , σ 5

2 0 1 2 3

σ 5

1 0 1 2 3

σ 5

0 0 1 2 3

2 0 1 2 2 3

iϵ013σ γ γ = −ϵ0132 γ γ γ γ γ = −(−1) γ γ γ γ γ = γ0 γ1 γ3 , 1 0 1 1 2 3

iϵ023σ γ γ = −ϵ0231 γ γ γ γ γ = (−1) γ γ γ γ γ = γ0 γ2 γ3 , 1 2 3

iϵ123σ γ γ = −ϵ1230 γ γ γ γ γ = −γ γ γ = γ1 γ2 γ3 .

(4.68) (4.69) (4.70) (4.71)

We have now verified all distinct cases of our relation, proving that γµ γν γρ = gµν γρ − gµρ γν + gνρ γµ + iϵµνρσ γ σ γ 5 ,

(4.72)

as required. (b)

(i) With a slick application of the anticommutation relations, we have γ µ γµ = 1 µ ν µν µ 2 {γ , γ }gµν = g gµν I = δ µ I = 4I. Clearly, generalizing to d-dimensional µ spacetime will give γ γµ = dI. (ii) Since γ 5 anticommutes with all other γ matrices, we have from above γ µ γ 5 γµ = −γ µ γµ γ 5 = −4Iγ 5 = −4γ 5 .

Problem 8

81

(iii) Another application of the anticommutation relations gives γ µ ̸ kγµ = γ µ kν γ ν γµ = kν (2g µν − γ ν γ µ )γµ = 2̸ k − ̸ kγ µ γµ = 2̸ k − 4̸ k = −2̸ k. (iv) Similarly, γ µ γ 5 ̸ kγµ = −γ 5 γ µ ̸ kγµ = 2γ 5 ̸ k. (v) The same technique yields γ µ ̸ k̸ ℓγµ = kν ℓρ γ µ γ ν γρ γµ = kν ℓρ (2g µν I − γ ν γ µ )(2gρµ I − γµ γρ ) = (2k µ I − ̸ kγ µ )(2ℓµ I − γµ ̸ ℓ) = 4k · ℓI − 4̸ k̸ ℓI + ̸ kγ µ γµ ̸ ℓ = 4k · ℓ, using the first result. (vi) Going further, γ µ ̸ k̸ ℓ̸ pγµ = (2k µ − ̸ kγ µ )̸ ℓ̸ pγµ = 2̸ ℓ̸ p̸ k − ̸ k(4ℓ · p), from the previous result. Now ̸ ℓ̸ p = ℓµ pν γ µ γ ν = ℓµ pν (2g µν I − γ ν γ µ ) = 2ℓ · p − ̸ p̸ ℓ, so this reduces to γ µ ̸ k̸ ℓ̸ pγµ = −2̸ p̸ ℓ̸ k as desired. (vii) Finally, γ µ ̸ k̸ ℓ̸ p̸ qγµ = γ µ ̸ k̸ ℓ̸ p(2qµ −γµ ̸ q) = 2̸ q̸ k̸ ℓ̸ p−(−2̸ p̸ ℓ̸ k)̸ q = 2(̸ q̸ k̸ ℓ̸ p+ ̸ p̸ ℓ̸ k̸ q), again using the previous result.

4.8 Problem 8 Problem: Prove each of the identities: γ µ σ ρσ γµ = 0, σ µν σµν = 12, γ µ σ ρσ γ λ γµ = 2γ λ σ ρσ , µ λ ρσ γ γ σ γµ = 2σ ρσ γ λ , σ µν γ ρ σµν = 0, σ µν γ ρ γ σ σµν = 4 (4g ρσ −γ ρ γ σ ) and σ µν σ ρσ σµν = −4σ ρσ . Solution: The following identities concern the quantities σ µν := 2i [γ µ , γ ν ]. For proving them, it helps to note the implications of the identities in Problem 4.5: γ µ ̸ kγµ = −2̸ k µ

γ ̸ k̸ ℓγµ = 4k · ℓ µ

γ ̸ k̸ ℓ̸ pγµ = −2̸ p̸ ℓ̸ k

=⇒ =⇒ =⇒

γ µ γ ν γµ = −2γ ν , µ ρ σ

(4.73)

ρσ

γ γ γ γµ = 4g , µ ρ σ λ

(4.74) λ σ ρ

γ γ γ γ γµ = −2γ γ γ .

(4.75)

(a) It follows from Eq. 4.74 that γ µ (γ ρ γ σ − γ σ γ ρ )γµ = 4(g ρσ − g σρ ) = 0, and therefore γ µ σ ρσ γµ = 0. (b) With a judicious relabeling of dummy indices, we can write σ µν σµν = (i2 /4)(γ µ γ ν − γ ν γ µ )(γµ γν − γν γµ ) = (−1/2)(γ µ γ ν γµ γν − γ µ γ ν γν γµ ).

(4.76)

Using Eq. 4.73 along with the result γ µ γµ = 4I, this reduces to σ µν σµν = (−1/2)(−8I − (4I)(4I)) = 12I. (c) From Eq. 4.75, we have γ µ σ ρσ γ λ γµ = (i/2)(γ µ γ ρ γ σ γ λ γµ − γ µ γ σ γ ρ γ λ γµ ) = −i(γ λ γ σ γ ρ − γ λ γ ρ γ σ ) = 2γ λ σ ρσ ,

(4.77)

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82

as required. (d) Similarly, γ µ γ λ σ ρσ γµ = (i/2)(γ µ γ λ γ ρ γ σ γµ − γ µ γ λ γ σ γ ρ γµ ) = −i(γ σ γ ρ γ λ − γ ρ γ σ γ λ ) = 2σ ρσ γ λ ,

(4.78)

as required. (e) The easiest way to calculate σ µν γ ρ σµν is to consider the form it must have. It is a linear combination of products of five γ matrices, with only γ ρ left uncontracted. These products can be simplified, of course, by repeated application of the anticommutation relations, which would introduce factors of g µν into the expression and nothing else. The only terms that might exist in the expression, then, are proportional to either g µν gµν γ ρ ∝ γ ρ , g µν γµ γν γ ρ ∝ γ ρ or g ρν γν = γ ρ . That is, σ µν γ ρ σµν = kγ ρ for some constant k. Now multiplying by γρ from the right gives σ µν γ ρ σµν γρ = 4kI. The left-hand side vanishes by the result in Part (a), so we get k = 0 and therefore σ µν γ ρ σµν = 0. (f) Again by relabeling dummy indices, we can calculate σ µν γ ρ γ σ σµν = (i2 /4)2[γ µ γ ν γ ρ γ σ γµ γν − γ µ γ ν γ ρ γ σ γν γµ ] = (−1/2)[(−2γ σ γ ρ γ ν )γν − γ µ (4g ρσ )γµ ] = (−1/2)[−2γ σ γ ρ (4I) − 4g ρσ (4I)] = 4(2g ρσ + γ σ γ ρ ),

(4.79)

where we have used Eqs. 4.74 and 4.75. Now using the anticommutation relations, we arrive at the desired identity σ µν γ ρ γ σ σµν = 4(4g ρσ − γ ρ γ σ ). (g) Finally, we calculate with the previous result σ µν σ ρσ σµν = (i/2)[σ µν γ ρ γ σ σµν − σ µν γ σ γ ρ σµν ]

= 2i[(4g ρσ − γ ρ γ σ ) − (4g σρ − γ σ γ ρ )] = −2i[γ ρ , γ σ ] = −4σ ρσ ,

as required.

4.9 Problem 9 Problem:

(4.80)

Problem 9

83

Consider the sixteen matrices {Γ1 , Γ2 , . . . , Γ16 } ≡ {I, γ µ , σ µν |µ