Feynman Simplified 1D: Angular Momentum, Sound, Waves, Symmetry & Vision [2 ed.]

Citation preview

Feynman Simplified 1D: Angular Momentum, Sound, Waves, Symmetry & Vision Everyone’s Guide to the Feynman Lectures on Physics by Robert L. Piccioni, Ph.D.

Copyright © 2014 by Robert L. Piccioni Published by Real Science Publishing 3949 Freshwind Circle Westlake Village, CA 91361, USA Edited by Joan Piccioni

All rights reserved, including the right of reproduction in whole or in part, in any form. Visit our web site www.guidetothecosmos.com

Everyone’s Guide to the Feynman Lectures on Physics Feynman Simplified gives mere mortals access to the fabled Feynman Lectures on Physics.

This Book Feynman Simplified: 1D covers about a quarter of Volume 1, the freshman course, of The Feynman Lectures on Physics. The topics we explore include: Angular Momentum Moments of Inertia Rotations in Three-Dimensions Sound & Beat Frequencies Modes & Harmonics Fourier Series & Transforms Complex Waves Symmetry Properties of Natural Laws To find out about other eBooks in the Feynman Simplified series, and to receive corrections and updates, click HERE. I welcome your comments and suggestions. Please contact me through my WEBSITE. If you enjoy this eBook please do me the great favor of rating it on Amazon.com or BN.com.

Table of Contents Chapter 39: Rotation & Angular Momentum Chapter 40: Centers & Moments of Rotation Chapter 41: 3D Rotation & Review of Rotation Chapter 42: Physics of Waves & Sound Chapter 43: Theory of Beats Chapter 44: Modes of Oscillation Chapter 45: Harmonics & Fourier Analysis Chapter 46: Complex Waves Chapter 47: Review of Physics of Waves Chapter 48: Physics of Vision Chapter 49: Symmetry & Physical Laws

Chapter 39 Rotation & Angular Momentum We have so far explored Newton’s laws as they apply to point particles, and objects that we can assume are equivalent to point particles. An example of the latter is the gravitational attraction of the Earth and Sun, which Newton proved can be calculated assuming all of each body’s mass is concentrated at its center. We can now move up to more complex objects and motions. As Feynman says in V1p18-1: “When the world becomes more complicated, it also becomes more interesting… the phenomena associated with the mechanics of a more complex object…are really quite striking.” We will start with the simplest cases, preparing ourselves for the “really quite striking” ones to come later. In this chapter, we will examine rotations of rigid bodies in two-dimensions moving at nonrelativistic velocities, so that we can employ Newtonian mechanics.

Parabolic path of center of mass From our prior studies, we know how to calculate the motion of a football thrown by a quarterback to a receiver. Assume the QB releases the ball at time t = 0, at location x=y=z=0, with velocity v= (v ,0,v ), where +z is straight up and +x is downfield. Ignoring air resistance, if the ball were a single point, its trajectory would be: x

z

x(t) = v t y(t) = 0 z(t) = v t – 1/2 g t x

2

z

This arcing curve is called a parabola, as illustrated in Figure 39-1.

Figure 39-1 Parabolic Path Of Thrown Ball

The motion is much more complicated if a more complex object is thrown, such as a gaucho’s bolo, three balls tied to a string. But even then, there is something that still moves in a parabolic arc — some central essence moves in a perfect parabola, while all else flails around it. That central essence is called the center of mass, a kind of average of the locations of all the particles within the complex object. The center of mass, often-abbreviated CM, is a mathematically determined point that doesn’t even have to be a material part of the object. For three balls attached to a loop of rope, the center of mass is a point in-between the balls where nothing exists. Let’s examine F, the sum of all forces on all particles within a complex object. Label the particles j = 1 to some enormous number N. The jth particle experiences a force F equal to its mass m times its acceleration, which equals the second derivative of its position r . Recall the linearity of differentiation: d(A+B)/dt = dA/dt + dB/dt. j

j

j

F = Σ F = Σ {d (m r ) / dt } F = d ( Σ {m r } ) / dt 2

j

j

j

2

j

2

j

2

j

j

j

Define: M = Σ {m } R = Σ {m r } / M j

j

j

j

j

Rewriting the prior equations with these definitions yields: F = M d (R) / dt 2

2

We have recovered the equation for the motion of a single point body of mass M and position R. As defined above, M equals the mass of the complex object and R is the position of its center of mass. Everything we learned before about the motion of a single point body applies directly to the motion of the CM of a complex body. In V1p18-2, Feynman suggests mentally separating the forces on each particle into two groups: internal and external. Newton’s third law (action begets reaction) ensures that the forces between particle j and particle k in the body are equal and opposite; hence they cancel one another in the sum F. Indeed all internal forces sum to zero in the calculation of F. This means the motion of the center of mass is determined only by external forces, such as Earth’s gravity. Imagine a rocket ship in outer space. If it is sufficiently isolated, we could assume it experiences no external forces and is stationary in an appropriate reference frame. More precisely, its center of mass is stationary. If the crew moves toward the front of the ship, the ship itself must move backward to keep the center of mass stationary. Indeed, this must happen on the International Space Station, although since the ISS is enormously more massive than its crew, this effect is imperceptible. Newton’s third law prevents a rocket ship from moving its own center of mass, but it can propel part of its mass (the ship) by expelling another part (the fuel). In whatever manner its rockets fire, the CM of all the ship’s original mass remains stationary in an appropriate reference frame. In any other reference frame, the ship’s CM moves at a constant velocity, regardless of how or whether its rockets

fire.

2D Rotation of Rigid Body V1p18-2 Since we know how to analyze the motion of the center of mass of a complex body, let’s now examine the motion of the rest of the body. We’ll make things simpler by considering only rigid bodies, those whose atoms are so strongly bound to one another that their relative orientations do not change. We thus avoid dealing with bending, twisting, and vibrating. Having agreed on all these constraints, the only motion left is for the body to rotate as one single object. What does “rotating” mean? A three-dimensional object rotates around an axis, whereas a twodimensional object rotates about a single point. Figure 39-2 illustrates rotation in two-dimensions. On the left side, the darker rectangle is rotated about the XY origin by a small angle, resulting in the lighter rectangle. The right side of this figure focuses on the rectangle’s upper right corner, labeled P. Initially, the radial line from the XY origin to P lies at angle ø relative to the x-axis. Rotation increases that angle by dø, moving the rectangle’s upper right corner P to position Q.

Figure 39-2 Rotation in 2D

In the figure, the line u is perpendicular to r, hence the angle between u and the vertical dotted line is also ø. Let the coordinates of P be (x,y) and of Q be (x+dx,y+dy). From trigonometry, we find these relationships: x = r cosø y = r sinø u = r tan(dø) u = dx = –r sinø dø u = dy = +r cosø dø x y

If we measure angles in radians, and take the limit of dø going to zero, we have: u = r dø u = –y dø u = +x dø x y

If dø occurs during an infinitesimal time interval dt, and we define ω = dø/dt, the equations become: v = dx/dt = –y ω v = dydt = +x ω v = v +v = ω (y +x ) v=ωr x y 2

2 x

2 y

2

2

2

The last equation jibes with P moving to Q, a distance u = r dø, in time dt, making v = r dø/dt = r ω. A quick note about measuring angles in radians. As mentioned earlier, in equations and calculations we always measure angles in radians rather than degrees. Radians are convenient: the arc length subtended by angle θ equals rθ, for radius r. For example, for θ = π/2, arc length = π r/2, both corresponding to 1/4 of a full circle. In the text, we sometimes use the more colloquial 90 degrees, rather than π/2, but in equations it’s always radians. To clarify the rotational parameters, Feynman stresses the analogy with linear motion. In linear motion we have position r, velocity v = dr/dt, acceleration a = d r/dt , and force F = ma. In rotational motion we have angular position ø, angular velocity ω = dø/dt, angular acceleration α = d ø/dt , and perhaps an unknown X. Isn’t there some angular X analogous to force? 2

2

2

2

Torques V1p18-4 Ask and ye shall receive: X is torque, from the Latin word torquere, to twist. We’ll follow Feynman’s lead and find quantitative expressions for torque by considering the work energy expended in rotating an object. Recall that for linear motion, work equals force times the distance through which the force acts. By analogy, work will equal torque times the angle through which the torque acts. Going back to our rectangle in Figure 39-2, pushing on the rectangle at point P and turning it by angle dø requires work W, according to: W = F • ds = (F dx + F dy) x

y

Using the expressions derived above for dx and dy, we derive the equation for torque τ in 2D: W = (–y dø F + x dø F ) W = τ dø τ=+xF –yF x

y

y

x

You might recognize the combination xy – yx; it’s the vector cross product. So, we can rewrite this equation in vector form, which we’ll discover later is also valid in 3D, as: τ=r×F If multiple torques act on a body, they simply add like vectors, as do forces in linear motion. Since an

object can have only one rotation angle dø (in 2D), the above equations generalized for multiple torques, labeled j = 1 through N, are: τ = Σ {x F – y F } W = τ dø j

j

yj

j

xj

Unlike a force, a torque is defined relative to the center of rotation in 2D, or the axis of rotation in 3D. The math shows this because r, x, and y are all measured from the center of rotation to the point at which the torque acts. One can calculate a torque about any point (or axis in 3D), but the values relative to different points will likely be quite different. If, and only if, the sum of all torques is zero, the body will be in rotational equilibrium — its rotational velocity ω = dø/dt will be constant. This is analogous to a body moving at a constant velocity if and only if the sum of all forces is zero. Complete dynamic equilibrium requires both the sum of all forces and the sum of all torques to be zero to ensure the body does not accelerate linearly or rotationally. Let’s take another look at the torque equation. Figure 39-3 shows force F acting at point P on a rigid body.

Figure 39-3 Torque From Force F at Position r

For convenience, choose coordinates which place the center of rotation at x=y=0, P on the x-axis, and define r to be the radial vector from the center of rotation to P. The light dotted arc shows the rotational path of P. The angle between F and r is µ. If we extend the line of F infinitely far in both directions (the dashed line), b is the distance of closest approach of that line to the center, and the radial line b is perpendicular to the dashed extension of F. If the body rotates slightly by an angle dø (counterclockwise for dø > 0), point P will move upward (+y-direction) a distance r dø, and the work done will be F r dø. Since work = F • ds, the xcomponent of F does not contribute because it is perpendicular to the motion. The torque equation for our current situation (x = r and y = 0) is: y

τ=+xF –0 τ = + r F = r F sinµ W = r F dø = τ dø y

y

y

Since only the component of force perpendicular to the radial vector contributes to torque, torque equals F r sinµ, which as you may recall from Chapter 6 is the definition of the cross product; hence,

this confirms τ = r × F. Now, let’s get back to b, the distance of closest approach of F to the center of rotation in Figure 39-3. This parameter is also called the lever arm of the force, or its impact parameter. Note that b is one side of a right triangle of which r is the hypotenuse. The interior angle of that triangle at P equals µ, so: τ = F r sinµ b = r sinµ τ=Fb The above give us several ways to calculate torque, the last being force multiplied by its lever arm. By now you might be wondering whether all this center-of-mass and torque stuff is actually useful or is merely esoteric physimath. If so, you’ll be surprised how enjoyable (and profitable) all this stuff can be. When Feynman taught me how to play pool, lesson #1 was where to hold the cue stick. If you grab it just anywhere, the stick is apt to turn unexpectedly when you shoot. This is because your stroke may apply a torque, rotating the stick and ruining your hopes for victory. To avoid torquing the cue stick, hold it at its center of mass — there’s no torque when a force is applied with zero lever arm. How can you conveniently find the CM of a tapered rod? Put your two index fingers a couple feet apart and point them straight ahead and level. Now balance the stick across your fingers, and slowly move your fingers together, while keeping the cue stick balanced. Where they meet is the CM. Why? Because, at any instant, the finger closer to the CM will bear most of the stick’s weight, causing greater friction on that finger than on the one farther from the CM. Hence, the farther finger will slide while the other is stuck by friction. As you move your fingers together, the stick will slide over one finger and then the other, alternating as the farther finger moves closer. When you master this, you’ll be ready for Feynman’s pool lesson #2: spin.

Angular Momentum We are ready to move beyond rigid bodies. We’ll now consider any collection of particles, even those that do not form a material body. Examples include: seed heads of dandelions; marbles tossed in the air; a chain; stars in a galaxy; or electrons in a particle beam. We will also no longer require that each particle turn in perfectly circular rotation. With this generalization our results will be applicable to elliptically orbiting planets, and particles following other irregular paths. At the beginning of this chapter, we discussed how to calculate the center of mass of an array of particles, and how to calculate its motion due to external forces. Here we will consider the twists and turns due to the torques associated with all the forces acting on the particles. Begin with just one such particle. As before, in 2D: τ=xF –yF τ = xm d y/dt – ym d x/dt τ = d {xm dy/dt – ym dx/dt } /dt y

2

x 2

2

2

Wow, how did I know how to do the last step? Some of you might be brilliant enough to figure that out by yourselves, but I read it in V1p18-5. We can show this is true by working backwards: τ = d {xm dy/dt – ym dx/dt } /dt τ = dx/dt m dy/dt + xm d y/dt – dy/dt m dx/dt – ym d x/dt 2

2

2

2

τ = xm d y/dt – ym d x/dt 2

2

2

2

So, torque is the rate of change of {xm dy/dt – ym dx/dt }, just as force is the rate of change of momentum. We therefore call {…} the angular momentum L. In 2D: L = {x d(my)/dt – y d(mx)/dt } L= x p – y p τ = dL/dt y

x

Everything we’ve done in this chapter employed Newton’s laws, but the last three equations are also correct relativistically. Just as we had several expressions for torque, we can write equivalent expressions for angular momentum. With b being the lever arm of p, and µ being the angle between r and p, we have: L= x p – y p L=r×p L = r p sinµ L= p b y

x

With our understanding of angular momentum, let’s re-examine Kepler’s second law. For a single planet orbiting a star, the force holding the planet in orbit is a radial vector originating in the center of the star. The force’s lever arm is therefore zero, hence no torque acts on the planet and its angular momentum cannot change (0 = τ = dL/dt). Using the above equations, and referring to Figure 39-4, we find in the limit that all changes are infinitesimal: L = r mv sinµ r dθ = v sinµ area swept = 1/2 r (r dθ) area = 1/2 r (v sinµ) area = 1/2 L/m d(area)/dt = dL/dt /(2m) = 0

Figure 39-4 Planet Sweeping Area

In calculating the area swept, we used the area of the large triangle and not that of the small triangle, because the latter has two sides proportional to the infinitesimal changes, making it negligible in our limit. Our result is that a planet sweeps equal areas in equal times, just as Kepler said. His second law is a direct consequence of the conservation of angular momentum in the absence of torques.

Conservation of Angular Momentum Recall that at the beginning of this chapter, we summed the forces and accelerations of every particle within a large body. We found that all the internal forces canceled one another, leaving only the external forces that accelerate the body’s center of mass. That simplicity results from the linearity of Newton’s laws and from his law that action begets reaction. We will now go through the same process for rotational motion: summing the torques and angular momenta of every particle in a body. Let L and τ be the angular momentum of the jth particle and the torque acting on it. Let L and τ be the total angular momentum of the body and the total torque acting on it. Then: j

j

L= Σ L τ=Σ τ τ = Σ dL /dt τ = dL/dt j

j j

j

j

j

Again, all the internal torques, those due to internal forces cancel one another because the forces of action and reaction are exactly equal in magnitude and exactly anti-parallel. Hence all that remains in τ are torques due to external forces. Therefore the rate of change of an object’s total angular momentum equals the total external torque acting upon it. The equation is: dL/dt = τ

external

This equation is valid regardless of what’s happening inside a large composite object, and whether the object is rigid or not. And it applies to motion about any axis. A very important special case arises when the sum of all external torques is zero: the body’s total angular momentum is constant. The total angular momentum of any closed system is therefore conserved.

Moment of Inertia V1p18-7 We will next consider how a change of angular momentum affects an object’s speed of rotation. Imagine again a rigid body, all of whose particles rotate at the same rate about one point in 2D. The particles in this body will all move in circles about the center of rotation, and all will rotate at the same angular speed ω = dø/dt. Let the jth particle’s radial distance be r , its linear speed be v (v = r ω), and its angular momentum be L . Note that for circular motion, v is always perpendicular to r , for j

j

j

j

j

j

j

all j. L =m v r =m r ω L= Σ L = ω Σ m r L= I ω 2

j

j

j

j

j

j

2

j

j

j

j

j

Here we define the body’s moment of inertia I = Σ m r . Just as mass m times linear velocity v equals linear momentum p, moment of inertia I times angular velocity ω equals angular momentum L. 2

j

j

j

The key distinction between mass and moment of inertia is that the latter is proportional to the square of an object’s size. For linear motion, we learned that we could perform our analyses assuming all of an object’s mass was concentrated at a single point. Clearly that’s not true in rotational motion. Perhaps the most dramatic demonstration of this effect is the spectacular “scratch spin” of championship figure skating. Skaters turn in a small circle with limbs extended, as in the left side of Figure 39-5. When they then pull their arms and legs into a tight position, as in the image’s right side, their moment of inertia decreases and their rotational velocity increases to conserve angular momentum.

Figure 39-5 Figure Skater’s “Scratch Spin”

Some skaters can increase their angular velocities from 2.5 revolutions per second to 6.0 revolutions per second. That’s 2.4 times faster. Since L = Iω must be constant, their before I ω and their after I ω must be equal: I ω = I ω . This means the skaters must reduce their moment of inertia by a factor of 2.4, which requires reducing their average r to 64% of their original breadth. That seems like quite a feat, given that their heads and torsos don’t shrink. Each arm weighs about 6% of a person’s total weight, while each leg is about 20% — you do the math. B

A

A

A

B

B

A

B

Rotational Kinetic Energy V1p19-7 The one key quantity of rotational mechanics that we have yet to examine is rotational kinetic energy. We discussed the angular momentum L of a rotating object, the work done by torque W = τ dø, and how torques change angular momentum τ = dL/dt. Just as linear motion results in kinetic energy, so does rotational motion. What is the kinetic energy of an object rotating with angular velocity ω?

Since the rotational analog of mass m is moment of inertia I, and the analog of linear velocity v is angular velocity ω, it’s no surprise that the analog of 1/2 m v is 1/2 I ω . Let’s show this is indeed true. 2

2

Consider a collection of particles that are all rotating with angular velocity ω. Let the mass of the jth particle be m , and its distance to the rotational axis be r . Each particle’s linear velocity v equals ω r . Hence the total kinetic energy of all particles, according to the laws of Newtonian mechanics for linear motion, is: j

j

j

j

T = Σ {1/2 m (ωr ) } T = 1/2 ω Σ {m r } T = 1/2 ω I 2

j

j

j

2

2

j

j

j

2

Note that we must not double-count energy. The kinetic energy of the jth particle is m (ωr ) /2; we need to account for that once, either as linear kinetic energy or as rotational kinetic energy, but not as both. One could legitimately ignore rotational kinetic energy and do the sum of m (ωr ) /2, but it’s generally much easier to use the rotational kinetic energy equation Iω /2 and get the result in one calculation. Take your pick, but do not include both in any conservation of energy analysis. 2

j

j

2

j

j

2

In V1p19-8, Feynman raises an interesting question about spinning figure skaters. When skaters pull their arms and legs inward to spin faster, their angular momenta remain constant since no torques are present. Hence L = L , where the labels B and A refer to before and after. What about their rotational kinetic energies? A

B

L =L I ω =I ω ω =ω I /I T = 1/2 I (ω I / I ) T = 1/2 ω I / I T = T (I / I ) A

A

B

A

A

B

B

A

B

B

A

A

2

A A

B

B

B

B 2 B

B

2

A

A

A

To spin faster, skaters reduce their moments of inertia, I < I , this means kinetic energy increases as skaters pull in their arms and legs. Skaters must supply the added energy, exerting considerable force pulling mass inward against its natural tendency to fly outward. A

B

Chapter 39 Review: Key Ideas 1. For linear motion, the internal forces among a collection of particles cancel one another, leaving only external forces to act on the center of mass of the particles, as if it were a single object. 2. For rotational motion, we define angular position ø, angular velocity ω = dø/dt, angular acceleration α = d ø/dt , torque τ = r × F, the lever arm of force b, the angle µ between r and F or between r and p, angular momentum L, and moment of inertia I = Lω. These relationships 2

2

exist: Work W = τ dø τ = F r sinµ = r × F τ=Fb τ = dL/dt L=r×p L = r p sinµ L= p b I= mr Kinetic energy T = 1/2 I ω 2

2

Compared below are the linear and rotational variables of prime interest.

Chapter 40 Centers & Moments of Rotation

This chapter examines the centers of mass and moments of inertia of large bodies.

Properties of Center of Mass V1p19-1 In the prior chapter, we considered a large collection of particles with many forces acting on them. The particles could be molecules of a gas, atoms within a rigid body, or stars within a galaxy. We found that the internal forces amongst these particles summed to zero due to Newton’s third law of action and reaction. That left us to deal only with any external forces acting on these particles. We found it useful to define a center of mass (CM) of any group of particles as follows: R = Σ {m r } / M j

j

j

Here R is the position vector of the CM, m and r are the mass and position vector of the jth particle, M = Σ {m } is the total mass of all particles, and the sum extends over j = 1…N, with N being the number of particles in the group. As with any vector equation in 3D, the above is shorthand for three equations, one in each of any three mutually orthogonal spatial directions. j

j

j

j

Imagine initially that every particle has the same mass m. The equation for the center of mass is then: R = m Σ {r } / (Nm) = Σ {r } / N j

j

j

j

This means R is just the average position of all its parts, which is a perfectly reasonable definition of “center.” Now, let’s imagine we have n particles of type A that have mass m, and n particles of type B that have mass 2m. Our equation then becomes: R = [ Σ {mr } + Σ {2mr } ] /(nm+2nm) R = [ Σ {r } + 2 Σ {r } ] / (3n) jA jA

jA

jA

jB

jB

jB

jB

Above, we sum over all n A-particles in the first term and all n B-particles in the second term. While there are equal numbers of A- and B-particles, the B-particles have twice as much effect due to their greater mass. This also seems reasonable, particularly since we can imagine each B-particle being two particles of mass m glued together, thereby making R the average position of 3n particles, 2/3rds of which happen to be paired.

Let’s now put some bounds on R. Consider for a moment only the x-direction and let Xmin be the minimum value of the x components of all N particles. Now compute: R = Σ {m x } / M R = Σ {m (x – Xmin + Xmin} / M R = Xmin + Σ {m (x – Xmin)} / M x

j

j

x

j

j

j

j

x

j

j

j

To get the last equation, we pulled the constant Xmin out of the summation; its coefficient was Σ {m }/M = 1. Now, we know (x – Xmin) ≥ 0 for every j because Xmin is the minimum of all x . Hence R ≥ Xmin. We can similarly show that R ≤ Xmax, the maximum of all x . This means the x component of the center of mass must be in the range Xmin to Xmax; similarly for the y and z components. Therefore the center of mass of any collection of particles must lie within an envelope completely surrounding all particles, which is entirely reasonable. j

j

j

x

j

x

j

This doesn’t mean the center of mass is necessarily coincident with any of its particles. A donut, for example, has its center of mass in the hole in its middle. Next consider joining two objects A and B that have masses M and M and centers of mass R and R . What can we say about the center of mass of the combined object, whose mass is M = M + M ? A

B

B

A

A

B

R = Σ {m r } / M R = (Σ {m r } + Σ {m r }) / M R = (M R + M R )/ M R = R (M–M )/M + M R / M R = R + (R – R ) (M /M) j

j

j

jA

A

j

A

jB

B

A A

j

j

B

B

B

j

B

A

B

B

This means the CM of the combined object must lie somewhere along the line connecting the CMs of its two constituent parts. Clearly this can be extended to any number of combined objects. If the collection of particles is symmetric, its center of mass will be located along the line of symmetry. For example, the isosceles triangle shown on the left side of Figure 40-1 is symmetric about its vertical median, so its CM must lie along that line. This is because each point on the left side of the line of symmetry has a corresponding point on the right side. (Recall that a triangle’s three median lines run from each corner to the midpoint of the opposite side.)

Figure 40-1 Isosceles and Oblique Triangles

Now examine the oblique triangle on the right side of Figure 40-1. It has no symmetries, but if its mass is uniformly distributed, we know immediately that its center of mass is where its three medians

intersect. Why? Imagine dividing this triangle into a series of narrow strips parallel to the bottom side. The center of each strip is on the median line from the upper corner, because by definition half the points are to the left and half to the right of the median. Therefore, the CM of every strip lies along that median. As demonstrated above, the CM of all strips combined must lie along the line connecting their CMs, which is the vertical median. Similarly, the center of mass must lie along each of the other two medians. All this means is that the CM is at the intersection of the three medians.

Mathematical Methods This section explores more complex mathematical techniques for finding centers of mass, and provides another opportunity to sharpen one’s math skills. For a symmetric object, such as the isosceles triangle in Figure 40-1, we said symmetry ensures the center of mass is along its symmetry axis. Let’s now show that mathematically. Let x be the horizontal axis and let the line of symmetry be at x = 0. By “symmetric about x = 0”, we mean whatever exists at +x must also exist at –x. So in the sum Σ {m x }, for every j there must be a corresponding k such that m x = –m x . The summation must therefore equal 0, and the x component of the CM is at the symmetry axis x = 0. j

j

j

j

j

k k

Consider a body with multiple symmetries, such as a rectangle, which has both horizontal and vertical symmetry. Its center of mass must lie along both its horizontal and vertical symmetry axes. This means its center of mass must be at its midpoint. A useful tool for finding centers of mass is a theorem by Pappus of Alexandria, one of the last great ancient Greek geometers. The following statement of Pappus’ centroid theorem is somewhat clearer and more precise than that of V1p19-4. Pappus’ Centroid Theorem: any planar surface rotated about an external axis, sweeps through a volume V such that V = A d, where A is the area of the surface and d is the distance that its centroid moves. The centroid is the average position of all points within the surface, which for a uniform mass density is also the center of mass. Define XY coordinates with x=y=0 at the centroid, with the y-axis parallel to the axis of rotation, and D being the distance from the centroid to the axis of rotation, as shown in Figure 40-2.

Figure 40-2 Pappus Centroid Theorem

For a rotation angle dθ, the distance moved by a point at position (x,y) equals (D–x) dθ. We can calculate the volume V swept by the area by summing the distance moved by each point within the surface. One could write that as a sum Σ of small areas, each dx × dy, or equivalently one could write an integral equation for V: V = ∫ (D–x) dθ dx dy V = dθ ∫ D dx dy – dθ ∫ x dx dy V = dθ D A Above, we recognize that ∫dxdy equals the area A, and ∫(x)dxdy = A times the centroid’s x coordinate, which in our coordinate system is identically zero. Thus, the volume swept equals the area of the surface times dθ D, the distance moved by the centroid, proving Pappus’ theorem. This theorem applies to any sequence of rotations about any sequence of axes, as long as the axes are all external to the surface. Now let’s use this theorem to find a center of mass. In Figure 40-3, a right triangle of height H and base length D is rotated 2π radians around its vertical side, sweeping out the volume of a cone, while its CM, a distance x from the axis of rotation, rotates around the smaller circle.

Figure 40-3 Cone Swept by Rotating Triangle

The volume of this cone equals πHD /3, the distance the centroid travels equals 2πx, and the area of the triangle equals DH/2. Thus Pappus’ theorem states: 2

(d) (A) = V (2πx) (DH/2) = πHD /3 x = D/3 2

If we instead rotate the triangle about its horizontal side, we similarly find the y coordinate of the centroid: y = H/3. Another example: consider a semicircular disk, half a pizza. The area of the half pizza equals (π r /2). Rotating the half pizza 2π radians about its straight edge, we get a ball of volume 4πr /3. Let x again be the distance from CM to rotation axis. The CM moves a distance 2πx through the rotation. The 2

3

equation for x is: (2πx) (π r /2) = 4πr /3 x = 4r / 3π 2

3

Now let’s ask: what is the centroid of the crust of that half pizza? Let the crust’s width be w, making its area πwr. Rotating it as before sweeps a shell of volume 4πwr , yielding: 2

(2πx) (πwr) = 4πwr x = 2r / π

2

Insights on Scaling In V1p19-2, Feynman raises a subtle point. We have found that the motion of the center of mass of a collection of particles follows the same Newtonian laws that apply to the individual particles. This means Newton’s laws have the property that they scale: they apply equally well to pebbles, boulders, planets, stars, and galaxies. Remarkably, no new principles of mechanics emerge as we scale up from pebbles to galaxies. The universe didn’t have to be that way, but it is because, as Feynman says, “… the fundamental gears and wheels of the universe are of atomic dimensions.” Those fundamental “gears and wheels” actually follow the rules of quantum mechanics, which are far more complex than the rules of Newtonian mechanics. Newton’s laws are what emerge when the behaviors of trillions of particles are averaged — Newton’s laws are the asymptotic limit of the quantum behaviors of a vast number of subatomic particles. Scaling up from atoms to pebbles already brings us extremely close to the asymptotic limit; scaling up from pebbles to galaxies makes no perceptible change in basic mechanics. At least, that is our current understanding.

Moment of Inertia V1p19-5 We next turn to moment of inertia, the rotational analog of mass. Recall that the moment of inertia, I, depends on two things: what is rotating and how far “what” is from the axis of rotation. The moment of inertia of a collection of particles is the sum over all particles (j=1…N) of each particle’s mass m times the square of its distance to the rotational axis r : j

j

I=Σ m r j

j

2 j

If the rotation is about the z-axis, and the object has a uniform mass density µ, we can write this equation as an integral over the object’s volume: I = µ ∫ (x + y ) dx dy dz M = µ ∫ dx dy dz 2

2

Our first example is rotating a long bar around its end. Assume a bar of mass M and length L lies along the x-direction, from x=0 to x=L. (I’m using L here for length, not for angular momentum. Sorry, but that’s what physicists do: you’ll see this everywhere. With practice, you’ll immediately know

which meaning is intended.) Let’s rotate the bar about the y-axis at x=0. This is a bit like a singlebladed propeller. Let µ be the bar’s mass density per unit length (µ=M/L). The moment of inertia equals the integral of µx from x=0 to x=L. 2

I = µ ∫ x dx = µ x /3; from x=0 to x=L I = µ (L /3 – 0) I = M L /3 2

3

3

2

Now let’s rotate the bar about its midpoint, still around an axis perpendicular to its length. We use the same integral, but this time the limits are x = –L/2 to x = +L/2. I = µ {(+L/2) /3 – (–L/2) /3) I = µ {L /24 +L /24} I = M L /12 3

3

3

3

2

Equivalently, we could have computed the integral: I = µ ∫ (x–L/2) dx; from x=0 to x=L 2

which gives the same result. Try it. Feynman points out yet another approach. Since the summation Σ m r is linear, the moment of inertia of two objects around a common axis of rotation equals the sum of the objects’ individual moments of inertia. This is written mathematically as: 2

j

I =Σ m r I =Σ m r +Σ m r I =I +I

j

j

2

AB

jAB

AB

jA

AB

j

j

A

j 2

j

jB

j

2 j

B

This linearity property applies to both addition and subtraction: the moment of inertia of a ring equals the moment of inertia of a disk minus the moment of inertia of the missing center. This is true only if all parts rotate about the same axis. Now apply the linearity property to the bar of length L rotating about its midpoint: think of the bar consisting of two half-bars, each of mass M/2 and length L/2, which are joined end-to-end and are rotating about their joint. From an earlier result, the moment of inertia of two half-bars rotating about their ends is: I = 2{ (M/2) (L/2) /3 } I = M L /12 2

2

Whenever I can think of two different ways of calculating something, I do both, because even I make mistaeks. If you’re perfect, you needn’t waste time double-checking. Next, we discuss the parallel-axis theorem. It is extremely useful to separate the moment of inertia I of a large body about the Z axis into two parts: (1) the moment of inertia of its center of mass around

the Z axis, plus (2) the moment of inertia of all the body’s parts around its center of mass. Let’s see how this is done, starting again with the fundamental equation for the moment of inertia of a collection of particles about the Z axis (x=y=0). Let M be the body’s total mass and X and Y be the coordinates of its center of mass. CM

CM

I=Σ m r I = Σ m {x + y } 2

j

j

j

j

j

2

2

j

j

Examine the x part. x = [(x –X ) + X ] x = (x –X ) + 2(x –X )X +X 2

2

j

j

CM

2

CM

2

j

j

CM

j

CM

CM

2

CM

The last two terms are easy to sum: Σ m X =MX 2X Σ m {x –X } = 0 2

j

j

2

CM

CM

CM

j

j

j

CM

Putting those into the x sum yields: Σ m x =MX j

j

+ Σ m {(x –X ) }

2

2

j

2

CM

j

j

j

CM

Similarly for y. So, the moment of inertia becomes: I = M(X +Y ) +Σ m {(x –X ) +(y –Y ) } I = MR + I 2

2

CM 2 CM

2

CM

j

j

j

CM

2

j

CM

CM

The first term is the body’s moment of inertia of its center of mass about axis Z, and I is the body’s moment of inertia around its center of mass. CM

The benefit of the parallel-axis theorem is that once we calculate moments of inertia for a body of some shape around its CM, we need just one more term to obtain the moment of inertia of that body around any axis. We don’t need to start over every time summing mr for each particle. An essential requirement for using this theorem is that the axis of rotation about the center of mass and the axis about which the center of mass rotates must be parallel. Axes matter: the moment of inertia of the bar we discussed above is much greater around the z-axis, which we computed, than it is around the xaxis along the bar’s length. 2

Another useful theorem applies to any very thin flat object rotated about an axis perpendicular to its surface. Let x and y be the axes within the surface and z be the perpendicular axis, and let I , I and I be the moments of inertia about each axis. The theorem is: I =I +I . This is because the z coordinates of all the body’s particles are zero, or at least much smaller than the x and y coordinates. Here’s why: x

z

I + I = Σ m {(x +z ) + (y +z )} I + I = Σ m {(x ) + (y )} I +I =I 2

y

x

j

j

j

y

x

j

j

j

y

x

2

2

j

j

2

z

2

j

2

j

x

y

y

z

A table of moments of inertia for some common shapes is at the end of this chapter.

Chapter 40 Review: Key Ideas 1. A body’s center of mass must lie within an envelope containing all its parts. 2. The CM of a two-part object lies along the line connecting the CM of each individual part. 3. The center of mass of a symmetric body lies along its line of symmetry. 4. Pappus’ centroid theorem is: the volume swept by a surface rotating about an external axis equals its area times the distance that its centroid moves. 5. The moment of inertia of a multi-part object equals the sum of the moments of each part, provided all rotate about the same axis. 6. The parallel-axis theorem is: I = M R + I A body’s moment of inertia I about an axis Z, equals the moment of inertia of the body about its own center of mass, I , plus M R , its mass times the square of the distance of its center of mass to Z, the axis of rotation. 2

CM

CM.

2

CM

Some Common Moments of Inertia

CM

Chapter 41 Rotations: 3D & Review In this chapter, we will examine rotations in three-dimensions, using Newtonian mechanics and vector algebra. In V1p20-1, Feynman says that the behavior of the rotating wheel is “…one of the most remarkable and amusing consequences of mechanics.” He saved that amusement for later in the lecture. While looking forward with anticipation, we must first prepare by extending our 2D knowledge of rotational motion to three-dimensions. That is best done using vectors. What we learned about 2D rotations remains valid, but must be expanded. Recall our 2D equations for angular momentum L and torque τ: L= x p – y p τ = dL/dt τ=xF –yF y

x

y

x

These equations describe activity in the xy-plane. If we correctly guess the signs using the right-hand rule, we can easily write similar equations for activity in the zx-plane and the yz-plane. In V1p20-1, Feynman asks us to ponder if we need to continue writing new equations for countless other planes beyond these three. Fortunately, that is unnecessary. We already know that three orthogonal axes are sufficient to describe all possible linear motions in 3D. After all, what “3D” means is that there are three independent dimensions of space, no more, no less. Rotational motion is no different in this regard: three orthogonal planes are necessary and sufficient to describe all possible rotational motion in 3D. In fact, three orthogonal planes are really the same thing as three orthogonal axes. The xy-plane is defined as that surface which is everywhere perpendicular to the z-axis — a vector in the +z direction, such as (0,0,1), defines the xy-plane. Similarly, the vector (0,1,0) defines the zx-plane, while (1,0,0) defines the yz-plane. The angular momentum L = x p – y p is in fact a vector in the z-direction; +z if L>0, and –z if L 0. Using your right hand, aim your fingers along that object’s path, turning them counterclockwise. Your thumb is now pointed toward you in the +z-direction. Perhaps surprisingly, like the linear velocity v, the angular velocity ω is a vector: it has a direction, the axis of rotation, and a magnitude, the rate of rotation, dθ/dt. Some vector equations involving ω are: τ•ω = power expended by torque

L=Iω If the sum of all torques in one direction equals zero, the angular momentum in that same direction will not change — it will be conserved. This is exactly analogous to linear momentum being conserved in one direction if all forces sum to zero in that direction. When τ=0, the sum of all torques is zero in every direction, and angular momentum is conserved about every axis in every direction. Another useful vector equation for rotational motion relates the linear velocity v of a particle at position r rotating with angular velocity ω: v = ω × r. Indeed, an even more general equation holds: dq/dt = ω × q, for any vector q rotating with angular velocity ω. In V1p20-4, Feynman lists rules for manipulating vectors in 3D using the dot and cross products. I added a few more. Here a, b, and c are vectors and h is any constant. Keep in mind that when a = b × c, a is perpendicular to both b and c. 1. 2. 3. 4. 5. 6. 7. 8. 9.

(ha)•b = h(a•b) (ha)×b = h(a×b) b•a = a•b b×a = –a×b a×a = 0 a×(b+c) = a×b + a×c a•(b×c) = (a×b)•c a×(b×c) = b(a•c) – c(a•b) a•(a×b) = 0

Gyroscopes V1p20-5 We now come to Feynman’s “most remarkable and amusing” spinning wheel: the gyroscope. Imagine a person holding a spinning wheel while sitting on a chair that swivels freely. The wheel’s axis is initially horizontal. To demonstrate the conservation of angular momentum, the person (a physicist, who else?) turns the wheel’s axis upward so that it now points vertically. Initially, the total angular momentum of the entire system — wheel, chair, and physicist — about the vertical axis was zero. The only external torque is the floor pushing upward to support the system’s weight. Since a vertical torque can’t change vertical angular momentum, the total angular momentum about the vertical axis must be conserved. To achieve this, the chair and physicist must rotate in the opposite direction of the wheel. That states the result we expect — angular momentum conservation — but exactly how are the chair and physicist forced to rotate? Figure 41-1 illustrates a gyroscope with forces +F and –F being applied to the ends of its axle. The force is in the +x-direction on the left and in the –x-direction on the right. Initially, at time A, the gyroscope is spinning about the y-axis with angular velocity ω and angular momentum L . A

A

Figure 41-1 Gy roscope with Torques

The two forces F produce a torque τ in the +z direction, which causes the gyroscope to rotate about the x-axis with angular velocity Ω. At time B, a short time interval dt after time A, the gyroscope’s axis has rotated about the x-axis by angle dθ. Both ω and L rotate about the x-axis toward the z-axis to their new directions ω and L . Although its not labeled in the figure, dθ is the angle between L and L . Let’s go through the math to see why each part of this description is correct. A

B

A

B

A

B

Let b equal the lever arm of force F acting on each end of the axle. On the left, b is in the –y-direction and F is in the +x-direction. For clarity, let’s do this for the left side in both component and vector notation: τ = xF – yF = 0 – (–b)F τ = r×F = (–b) F (times –1 as zyx is odd) z

y

x

We get the same result for the right side of the figure, where both b and F flip polarities. The total torque is then τ = 2bF. This total torque rotates angular momentum L, changing its direction but not its magnitude. At time A shown in the figure, the axle is horizontal, the torque is vertical, and L is in the +y-direction. In the infinitesimal time interval dt, the vertical torque increases the vertical component of L, by: z

dL = dt τ z

z

Since dθ is the angle through which L rotates in time dt, we also have: dL = L dθ z

Combining these yields: dt τ = L dθ τ = L dθ/dt = L Ω z

z

Recalling the equation dq/dt = ω × q, and that Ω = dθ/dt, we can write this in vector notation as: τ = dL/dt = Ω × L Thus applying force F to each end of the axle rotates the axis of rotation about the x-axis at the rate Ω = 2bF/L. To turn the spinning wheel’s axle from horizontal to vertical, the physicist must apply such

forces to the axle. By Newton’s third law, the axle applies equal and opposite forces to the physicist making his head spin, along with all the rest of him and the chair. One might wonder what happened to the original horizontal angular momentum. When the axle was turned vertically, L became zero. Why didn’t the physicist and chair start rotating about the y-axis? The answer is the floor. L isn’t conserved because the floor applies an external torque by pushing up against the chair and preventing it from flipping and rotating about the y-axis. y

y

The same ideas apply to spinning tops. As shown in Figure 41-2, gravity exerts force F, pulling the top downward. We treat that force as acting on the top at R , the top’s center of mass. This creates a horizontal torque that changes the direction of the top’s axis of rotation. CM

Figure 41-2 Spinning Top with Precession Ω

The top spins rapidly about its axis, while its axis of rotation slowly precesses, turning about the vertical direction of gravity’s pull. The equations for the precession rate Ω are: τ=R ×F τ=Ω×L τ=Ω×ωI CM

The direction of precession is horizontal, perpendicular to the applied force.

Math or Miracle? In V1p20-6, Feynman muses about the extent to which we really understand gyroscopes. Yes, we know all the equations and can describe their behavior mathematically. But, do we really understand it in a visceral sense? Who would be comfortable saying: “Of course the gyroscope’s motion is perpendicular to the applied force”? Feynman adds this very important guidance:

“It will turn out, as we go to more and more advanced physics, that many simple things can be deduced mathematically more rapidly than they can be really understood in a fundamental or simple sense. This is a strange characteristic, and as we get into more and more advanced work there are circumstances in which mathematics will produce results which no one has really been able to understand in any direct fashion.” Can we better understand why a gyroscope moves perpendicular to the applied force? Figure 41-3 shows that the particles of the gyroscope are not moving entirely within the xz-plane.

Figure 41-3 Gy roscope with Torques

The two F forces are trying to turn the gyroscope about the z-axis. The forces move the spinning particles very slightly in the +y-direction on the +x side, as shown. On the –x side, the particles move in the –y-direction (not shown). This out-of-plane motion turns the gyroscope, rotating it slightly about the x-axis, as explained above.

Nutation What we’ve described in the prior section is the steady precession of a gyroscope subject to a constant torque. This circumstance would occur if one end of the axle were supported on a gimbal bearing, and the other end were lovingly put into motion at exactly the right precession rate. In other words, smooth uniform precession is an equilibrium state in which the precession rate precisely matches the applied torque. But, what if precession and torque don’t match? For example, what if we hold the free end of the axle stationary, and then let go? While we were holding it stationary, we had to support its weight, thereby balancing the gravitational torque. But when we release the axle, the gyroscope is suddenly exposed to the torque of gravity. Is it really possible that the downward pull of gravity, which we all know so well, pushes the gyroscope sideways instead of down? As Feynman says in V1p20-7: “Anyone in his right mind would think that the [gyroscope] would fall.” What really happens is that the free end of the axle does start to fall and it also starts to precess. Like a mass on a string that is stretched and then released, the gyroscope will overshoot. Its axle will drop below its equilibrium height and precess too fast, faster than the rate that matches the gravitational torque. The excess precession rate will drive the axle upward, again overshooting equilibrium, whereupon the cycle repeats. This oscillating process, shown in Figure 41-4, is call nutation.

Figure 41-4 Nutating Gy roscope

Without friction or other damping, the path of the free end of the axle would be a cycloid, the path of a stone stuck in the tread of an automobile tire. More realistically, friction damps the nutation and the oscillations diminish. The gyroscope eventually settles into equilibrium, with its precession rate matching the gravitational torque. At equilibrium, the axle end will be somewhat below its starting point. This lower height causes a slight tilt that reduces the vertical component of angular momentum of the spinning motion, which balances the increased vertical angular momentum of precession. The detailed equations involve third-order polynomials and are quite complex, hence Feynman’s amusement.

Principal Axis Theorem The principal axis theorem states: every rigid body, however irregular, has three mutually orthogonal axes through its center of mass such that when rotating about any one of these axes, the body’s angular momentum is parallel to its angular velocity. In addition, the two axes that correspond to the body’s maximum and minimum moments of inertia are both principal axes. We’ll examine below a situation involving rotation about a non-principal axis. But first, let’s finish discussing principal axes. If a body has a symmetry property, as does an isosceles triangle, the line of symmetry is a principal axis. Consider any body, define coordinates such that the body’s principal axes are along the x, y, and z directions, and call the moments of inertia about these axes I , I , and I . We can write any angular velocity vector ω and angular momentum L as: x

y

z

ω = (ω , ω , ω ) L = (L , L , L ) L = (I ω , I ω , I ω ) x

y

x

x

y

x

z

z

y

y

z

z

The kinetic energy of rotational motion is then: T = 1/2 L • ω Finally, we arrive at the promised example of rotation along a non-principal axis. As shown in Figure 41-5, a disk is mounted askew on a rod that passes through the disk’s CM. The axis of rotation is the centerline of the rod, but that is not a principal axis of this disk.

Figure 41-5 Rotation About Non-Principal Axis

Choose the z-axis to be perpendicular to the disk face, passing through the CM, and pick any two orthogonal diameters of the disk to be the x and y axes. By symmetry, our coordinate axes are the disk’s three principal axes. As shown at the end of this chapter, the moments of inertia of a disk of mass M and radius R are MR /2 about the z- axis and MR /4 about both the x and y axes. As above, we can write the angular velocity vector ω and angular momentum L as: 2

2

L = (I ω , I ω , I ω ) ω = (ω , ω , ω ) L = (ω , ω , 2ω ) MR /4 x

x

y

x

y

x

y

y

z

z

z

2

z

The key point here is that angular momentum L is not parallel to ω, the axis of rotation. This means that as the rod turns, L is constantly changing, requiring continual torques in ever changing directions. Whatever bearings hold this rod in place will be substantially stressed. This is one reason it pays to get your tires balanced and aligned.

Moments of Inertia of Disk & Ring We’ll do the calculation for a ring with outer radius R, inner radius r, total mass M, and area mass density µ (M = µπ[R –r ]). A disk is the special case of r = 0. First find the moment of inertia I for the axis of rotation through the CM and perpendicular to the ring’s plane by integrating over the disk’s radial coordinate u, from u=r to u=R. The distance from the disk point at (u,θ) to the axis of rotation equals u. At each value of u, the arc length around the disk equals 2πu. The integral sums (mass/area) × (distance to axis) × (area). 2

2

2

I = ∫ µ u 2πu du; u from r to R I = 2πµ ∫ u du I = 2πµ u /4; u from r to R I = πµ (R –r )/2 I = πµ (R –r )(R +r )/2 I = M(R +r )/2 2

3

4

2

4

4

2

2

2

2

2

It may seem surprising that a larger r, the hole radius, seems to increase I. A larger r does increase the average distance from the rotational axis, but it also reduces M. From the prior equation I = πµ (R – r )/2, we see that the net effect of larger r is actually to reduce I, as expected. Now calculate the moment of inertia for any diameter across the ring; call that the y-axis. The distance of the disk point at (x,y) to the axis of rotation equals x. 4

4

I = ∫ µ x 2πu du; u from r to R 2

y

This looks like a tough integral; it’s time for a valuable clever trick. If we compute I about the x-axis, we get the same equation except x would be replaced by y . Since the ring is circularly symmetric, I must equal I . It is much easier to calculate I +I and then divide by 2. x

2

2

x

y

x

y

I = 1/2 ∫ µ (x +y ) 2πu du; u from r to R I = 1/2 ∫ µ (u ) 2πu du; u from r to R I = πµ u /4; u from r to R I = πµ (R –r )/4 I = πµ (R –r )(R +r )/4 I = M(R +r )/4 2

2

2

4

2

4

4

2

2

2

2

2

Rotational Motion Review Chapters 39, 40 & 41 For linear motion, the internal forces among a collection of particles cancel one another, leaving only external forces to act on the center of mass of the particles, as if it were a single object. Primary linear and rotational variables are compared below.

In evaluating terms in a cross product, such as L = +x p –y p , the sign rule is: even combinations xyz, yzx, and zxy have plus signs, while odd combinations xzy, yxz, and zyx, have minus signs. The number of swaps from xyz determines if a combination is even or odd. z

For rotational motion, we define: b, the lever arm of force F µ, the angle between r & F or r & p These additional relationships exist:

y

x

Work W = τ dø τ = F r sinµ = r × F τ=Fb τ = dL/dt L=r×p L = r p sinµ L= p b I= mr Kinetic energy T = 1/2 I ω 2

2

Centers of Mass (CM) & Moments of Inertia A body’s CM lies within an envelope containing all its parts. The CM of a two-part object lies along the line connecting the CM of each individual part. The CM of a symmetric body lies along its line of symmetry. The moment of inertia of a multi-part object equals the sum of the moments of each part, provided all rotate about the same axis. The Pappus centroid theorem says the volume swept by a surface rotating about an external axis equals its area times the distance its centroid moves. The parallel-axis theorem states I = MR + I A body’s moment of inertia I about an axis Z, equals the moment of inertia of the body about its own center of mass, I , plus MR , its mass times the square of the distance of its center of mass to Z, the axis of rotation. 2

CM

CM.

2

CM

CM

The principal axis theorem states every rigid body has three mutually orthogonal axes through its CM such that when rotating about any one of these axes the body’s L is parallel to ω. Also, the two axes that correspond to the maximum and minimum moments of inertia of that body are both principal axes. 3D Rotations that are “Quite Striking” Angular velocity ω is a vector; its direction is the axis of rotation and its magnitude is the rotation rate. Some equations involving ω are: τ•ω = power expended by torque L = I ω, definition of angular momentum dq/dt = ω×q, for any q rotating at ω When a gyroscope is subjected to a torque, it precesses with angular velocity Ω, according to: τ = dL/dt = Ω × L = Ω × ω I

Some Common Moments of Inertia

Chapter 42 Physics of Waves & Sound In this chapter, we begin exploring the physics of waves, a phenomenon that in V1p47-1 Feynman says: “appears in many contexts throughout physics. …Waves are related to oscillating systems… wave oscillations appear not only as time-oscillations at one place, but propagate in space as well.” We have often encountered oscillations and waves in our prior studies. Harmonic oscillators execute repetitive cycles, moving about fixed locations (Feynman Simplified 1B Chapters 12 through 14), while light waves (Feynman Simplified 1C Chapters 30 through 38) oscillate in both time and space. We discovered how the interference of light waves explains reflection, refraction, and other intriguing phenomena. We now consider some additional aspects of wave behavior. One such behavior is interference in time, in which the interference of two combining waves changes with time rather than position. A second behavior arises when waves are confined within limited regions, and interfere due to reflections from boundaries. Other aspects of wave behavior arise when wave velocities change due to differences in the media that the waves traverse. Yet another wave behavior involves wave velocities that vary with wavelength, which we discuss in the next chapter. In this chapter, we confine our attention to waves whose velocities are the same at all wavelengths. We know from our everyday experience that neither the speed of sound nor the speed of light changes appreciably over the range of wavelengths to which humans are sensitive. If for example, high frequency sound traveled faster than low frequency sound, a concert would sound quite different in the front row than in the rear balcony. The bass vocal would appear late in the balcony. Similarly, if red light traveled faster than blue light, a distant flash of white light would appear red initially, would rapidly become white, and then would finally turn blue. The absence of such effects confirms that the speeds of the sound and light we perceive do not change with

wavelength.

Wave Motion Waves are motion. They move through space and change over time. Let’s briefly review the wave terminology presented in Chapter 31. As illustrated in Figure 42-1, waves oscillate in identically repeating cycles. Wavelength λ is the spatial extent of one complete cycle, the distance between consecutive crests or troughs. Amplitude A is how high the wave goes up above and goes down below its average.

Figure 42-1 Wave Terminology

If the above wave were moving to the left, its entire shape would move left in unison as if it were a solid object. If we focus on a specific point, such as point Q, we would see the wave height going up and down as time passes, oscillating between +A and –A. If the wave goes through 9 full cycles per second, its frequency f is 9 cycles/second, or 9 Hertz, which is abbreviated 9 Hz. The product of wavelength and frequency equals the wave’s velocity v: λf=v; (meters/cycle) times (cycles/second) = (meters/second). Frequency can also be expressed in radians/second: the angular frequency ω=2πf. Another important quantity is the wave number k, which equals 2π/λ and has units of radians/meter. Figure 42-2 shows a hypothetical example of an electric field from an accelerating charge.

Figure 42-2 Electric Field E vs. Time t and Distance r/c

The upper part of the image plots field E versus time t, at two fixed positions. The solid line represents E(t) near the accelerating charge, and the dashed line is E(t) at a greater distance r. Since electric fields travel at the speed of light, the time shift between the solid and dashed curves equals r/c. As often discussed earlier, a remote observer sees E(t*), the field at retarded time t*=t–r/c. The lower part of Figure 42-2 plots the electric field seen at various positions at a fixed time. The horizontal axis is distance r from the source divided by the wave speed c. Here we see that A, the field at an earlier time, has traveled farther than B, the field at a later time. In general, the electric field’s wave height is most simply represented as a function of r–ct: f(r–ct). To demonstrate this, compare the field E at a position r and time t with E at position r+Δr and time t+Δt: f(r+Δr –c[t+Δt]) = f(r–ct + [Δr–cΔt]) f(r+Δr –c[t+Δt]) = f(r–ct), if Δr = cΔt This means every part of the wave is reproduced identically at all values of Δr and Δt for which Δr=cΔt. Hence the entire waveform moves in unison through space at velocity c. The above analysis can be applied to other types of waves. Any wave phenomenon governed by a linear differential equation and with the same velocity v at all frequencies can be represented by a function of the form f(r–vt).

Propagation of Sound Unlike light, we know that sound needs a medium to travel through — sound cannot travel through vacuum. This is because sound is the organized motion of the atoms and molecules of which the medium is composed. Its motion is determined by the properties of those atoms and molecules. As such, Feynman says in V1p47-2: “In short, sound is a branch of mechanics, and so it is to be understood in terms of Newton’s laws.” He provides this important insight: “We shall give a derivation of the properties of the propagation of sound between the source and the receiver as a consequence of Newton’s laws, and we shall not consider the interaction with the source and the receiver. Ordinarily we emphasize a result rather than a particular derivation of it. In this chapter we take the opposite view. The point here, in a certain sense, is the derivation itself. This problem of explaining new phenomena in terms of old ones, when we know the laws of the old ones, is perhaps the greatest art of mathematical physics. The mathematical physicist has two problems: one is to find solutions, given the equations, and the other is to find the equations which describe a new phenomenon. The derivation here is an example of the second kind of problem.” Consider the simplest example of sound propagation: motion in one dimension through air. We must first understand the physics of what happens when an object moves through air. As it moves, an object pushes air molecules out of its way. If the object moves very slowly, air will simply flow

around the object with only minor disturbance. From Chapter 15, we know the average molecular velocity due to thermal energy is given by: mv

avg

2

/ 2 = 3 kT / 2

Here T is temperature measured in Kelvin, and k is Boltzmann’s constant (not the wave number which is also denoted by k). For nitrogen molecules (N ) at 293K (20ºC, 68ºF): 2

v ≈ 509 m/s. avg

When an object moves rapidly through air, faster than air can gently flow out of its way, the air becomes compressed and its pressure rises. This high-pressure air compresses neighboring air molecules, resulting in a pressure wave that spreads outward through space, as illustrated in Figure 42-3.

Figure 42-3 Sequential Views of Pressure Wave In Air

In the above image, the dots represent air molecules and the black rectangle is a piston. The left quarter of the image shows an initially static condition. In the second quarter, the piston expands, compressing the adjacent air molecules, raising their pressure. In the third and fourth quarters, air molecules compress adjacent molecules, resulting in a pressure wave moving to the right. What variables do we need to analyze this process? Since this process is dynamic, changing in both space and time, we must know key variables as functions of both position and time. Those key variables are density, pressure, and displacement, the latter being how much the air molecules have moved. Molecular velocities and accelerations are also important, but these are calculable by taking time derivatives of displacements. If the original source of sound is very far away, the wavefront, the surface of a wave crest, is nearly flat. If the direction of motion is along the x-axis, the flat wavefront comprises a yz-plane perpendicular to the x-axis. By symmetry, the values of all variables must be the same at all y and all z. This reduces the problem to one spatial dimension, and we can write the displacement as D(x,t). The Newtonian analysis of sound that we will present relates the displacement, density, and pressure of small gas volumes — in V1p47-3 Feynman calls these elements of gas. This analysis makes certain assumptions about the size of these gas elements and the wavelengths of the sound waves that propagate through them. Firstly, in order to apply the principles of thermodynamics, the gas elements in our analysis must be much larger than atoms. From Chapters 15 through 24, we know that concepts like pressure and density are valid on macroscopic scales, but not on atomic scales. An individual molecule does not have a well-defined pressure, but millions of gas molecules do. For our macroscopic descriptors — density, pressure, and displacement — to be valid, our gas elements must be much larger than the

mean free path of individual molecules. Secondly, the sound waves of interest must have wavelengths much longer than that same molecular mean free path. For a pressure-density wave to move through a gas, energy must flow from molecules in one region to molecules in adjacent regions. Molecules moving away from higher-pressure regions must collide with molecules in neighboring regions and thereby transfer energy to them. For pressure-density waves with wavelengths much longer than the mean free path, molecules undergo many collisions in moving from high-pressure to low-pressure regions. Myriad collisions efficiently transfer energy, thus maintaining the wave’s driving force and keeping the wave moving forward. Conversely, for pressure-density waves with wavelengths much shorter than the mean free path, molecules can undergo few if any collisions while moving far enough to equalize pressure differences. Here, energy is not efficiently transferred, the driving force is not sustained, and the wave rapidly fades away. The result of this effect is that sound travels farther when its wavelength is longer and its frequency is lower. A particularly impressive example is the infrasonic communication of elephants. Infrasonic denotes frequencies below 20 Hz, the limit of human hearing. Elephants can communicate via infrasound across distances of up to 17 km (11 miles). By comparison, human vocalizations have minimum frequencies of 70 Hz for men and 140 Hz for women. These assumptions are entirely valid for typical sound waves. Under normal conditions, the mean free path in air is typically about 0.1 microns, while the wavelength of 3 kHz sound is about one million times larger at 0.1 meters.

The Sound Wave Equation The physics of sound involves the interaction of three effects: 1. Gas motion creates density gradients. 2. Density gradients create pressure gradients. 3. Pressure gradients drive gas motion. Note that the gas density we are discussing here is its mass per unit volume, not the number of molecules per unit volume. We begin with the second effect. From Chapter 15, recall the ideal gas law: PV = NkT Here P is pressure, N is the number of molecules in volume V, T is temperature in Kelvin, and k is Boltzmann’s constant. Assuming that T is constant, pressure P will be a function of only (N/V). In an

ideal gas at constant temperature, pressure is: P = ρ kT/m Here m is the average molecular mass, and ρ is the mass density. While air is a nearly ideal gas, let’s be more general and allow pressure to be an unknown function of density: P = g(ρ) It is convenient to define equilibrium values of key variables with a subscript zero: at equilibrium, without sound waves, the gas pressure is P and its density is ρ , with P = g(ρ ). At sea level, the average atmospheric pressure P is 1.01325 bars, where 1 bar = 10 newton/m = 10 kg/m-sec . 0

0

0

5

0

2

5

2

0

Experimentally, we find that the pressure changes caused by sound waves are quite small, typically parts per million of P . Sound intensity is measured using a logarithmic scale, since that nearly matches the response curve of human hearing. An rms (root-mean-square) pressure change of ΔP, corresponds to a sound intensity in db (decibels) of: 0

Sound Intensity I = 20 log (ΔP/P ) db 10

ref

Here P = 2×10 bar. To calculate rms, take the square root of the average of the square of the pressure change throughout a complete cycle. For a sinusoidal wave, the rms amplitude equals the peak amplitude divided by √2. –10

ref

Some examples of sound intensity are: 60 dB = 10 P = 2×10 bar: conversation 80 dB = 10 P = 2×10 bar: screaming child 100 dB = 10 P = 2×10 bar: chainsaw 120 dB = 10 P = 2×10 bar: rock concert 3

–7

ref

4

–6

ref

5

–5

ref

6

–4

ref

We see that sound waves change air pressure by only parts per ten-thousand to parts per ten-million. Hearing loss can result from excess exposure to loud sounds. Audiologists recommend these maximum continuous exposure limits: 10 seconds at a rock concert, 15 minutes operating a chainsaw, and 12 hours with screaming children. Other issues may reduce one’s tolerance to much shorter durations. Millions of years of human evolution have resulted in an intense parental sensitivity to their child’s scream. We will later consider much louder sounds, such as explosions. Here, we will consider only pressure changes much smaller than 1 bar. Let: ρ = ρ + dρ, with dρ 0 f(x,t) = Σ A sin(ω t+ø ) sin(nπx/L) n

n

n

n

n

We noted that any string displacement function f(x,t), however complex, can be represented by the above sum. Let us now restrict our attention to one value of x and consider f(t) at that x. The factor sin(nπx/L) is then constant, and can be absorbed into the A ‘s. Let’s also define ω=vπ/L, which is the frequency at which all modes repeat: ω=2π/T=2π/(2L/v). The most general waveform is then represented by: n

f(t) = Σ A sin(nωt+ø ) n

n

n

Alternatively, we can replace the phase angles ø with cosine functions, by using: n

sin(nωt+ø ) = sin(nωt) cos(ø ) + sin(ø ) cos(nωt) n

n

n

Cos(ø ) and sin(ø ) are constants that can be absorbed into the coefficients of sin(nωt) and cos(nωt). This provides another expression for the most general waveform: n

n

f(t) = Σ { a cos(nωt) + b sin(nωt) } n

n

n

The above expression is called the Fourier series for f(t), a function that repeats with frequency ω. In musical terms, the n=1 term is called the first harmonic, n=2 is the second harmonic, etc. Waves are almost always expressed with an average value of zero, such as sin(x), but the above sums can accommodate waves with offsets, such as 1+sin(x), by including an n=0 term, which is simply the constant a . 0

Figure 45-2 shows how increasing the number of terms in a Fourier series better approximates a square wave. A true square wave is shown in Figure 45-3.

Figure 45-2 Fourier Series Fit To Square Wave Upper Left: f(ωt) and f(3ωt) Lower Left: f(5ωt) and Fit with n=1,3 Upper Right: Fit with n=1,3,5 Lower Right: Fit with n=1,3,5,7

Musical Quality & Consonance Fourier series help us describe the musical concepts of quality and consonance. Even when the tones from a violin and an oboe have the same pitch (the same ω), the differing characteristics of these instruments produce sounds with differing harmonic contributions — differing coefficients in their Fourier series representations. Musicians say violins and oboes have different tone qualities. Sounds that contain only one harmonic are called pure tones. Sounds composed of many strong harmonics are called rich tones. In V1p50-3, Feynman explains that electric organs have keys that select pitch and stops that control the contributions of higher harmonics. With different stops, an electric organ can approximate the sound of a flute, an oboe, or a violin. I say “approximate” because electric organs don’t have a Stradivarius stop, at least so far. Feynman notes that our ears detect the intensity of sound at various frequencies, but not the phase differences — human hearing is not sensitive to phase angles, the ø in the above equations. Therefore instruments such as electric organs need only generate either sin(nωt) or cos(nωt) but not both. Other applications are more demanding, requiring both the sine and cosine of each frequency in the Fourier series. n

It is also interesting that we distinguish vowels, such as “a” and “e”, by the quality of their sounds, their harmonic compositions. We produce the different vowel sounds by changing the shape of our mouths to change the contribution of various harmonics. Feynman notes that the shape of our mouths emphasize certain specific frequencies, and that these frequencies do not change when our vocal

cords produce a different pitch. Thus, tone quality, the ratios of various frequencies, of a spoken “e” changes at different pitches (different ω’s). We seem to distinguish vowels more by specific frequencies rather than frequency ratios. In V1p50-4, Feynman returns to Pythagoras’ discovery that two strings produce a pleasant sound when their lengths are in the ratio of two small integers. Perhaps we can now understand why this is so. Each string produces sounds at several harmonics: ω, 2ω, 3ω, …. If the strings have a length ratio of 3:2, for example, they produce sounds at these frequencies: L=3: modes at (vπ/3) × (1, 2, 3, …) L=2: modes at (vπ/2) × (1, 2, 3, …) Note that the third harmonic of the L=3 string matches the second harmonic of the L=2 string; both have frequencies of vπ. When the higher harmonics match, beats are eliminated. Recall from Chapter 43 the equation for the intensity I of the sum of two waves of frequencies ω and ω : 1

2

cos(ω t) + cos(ω t) = 2 cos(ωt) cos(Δωt) 1

2

with ω = (ω +ω )/2 and Δω=(ω –ω )/2 1

1

2

2

I = 1 + cos(2Δωt) Here 2Δω is the beat frequency, the frequency at which the sound slowly waves and wanes. If the two frequencies are exactly equal, Δω=0, we hear a pure tone without beats. But if the two frequencies are close but unequal, we hear an annoying beat: OOOoooOOOooo. A small-number ratio eliminates annoying higher harmonic beats, making the sound more pleasant. This is called consonance. When string lengths differ from a small-number ratio, their combined sound has beats and is called dissonant. Stringed instruments can be accurately tuned by adjusting string tension until beats vanish. Using a piano as an example, Feynman offers a charming illustration of harmonic consonance. Let's label three successive C’s near the middle of the keyboard as C, C*, and C** and similarly the three next higher G’s as G, G*, and G**. Recall that frequencies double for each octave. The frequencies of these keys are in these ratios: C, C*, C** : 2, 4, 8 G, G*, G**: 3, 6, 12 Feynman says:

“press C* slowly—so that it does not sound but we cause the damper to be lifted. If we then sound C, it will produce its own fundamental [the frequency 2 above] and some second harmonic [4]. The second harmonic will set the strings of C* into vibration. If we now release C (keeping C* pressed) the damper will stop the vibration of the C strings, and we can hear (softly) the note C* as it dies away. In a similar way, the third harmonic of C [3×2=6] can cause a vibration of G* [6]. Or the sixth of C [6×2=12] (now getting much weaker) can set up a vibration in the fundamental of G** [12]." “A somewhat different result is obtained if we press G quietly and then sound C*. The third harmonic of C* [3×4=12] will correspond to the fourth harmonic of G [4×3=12], so only the fourth harmonic of G will be excited. We can hear (if we listen closely) the sound of G** which is two octaves above the G we have pressed! It is easy to think up many more combinations for this game.” He notes that the major scale has three major chords (FAC, CEG, and GBD) that are each in the frequency ratio 4:5:6. That fact, plus the doubling of frequency in each octave, fully defines the “ideal” scale called just intonation. For practical reasons, Feynman says keyboard instruments are not tuned this way, but are tempered with 12 equal frequency ratios of 1: 2 ≈ 1:1.0595. A fifth is no longer in the ratio 3:2 but rather 2 ≈ 1.4983; apparently, that’s close enough for most of us. Near middle C, that corresponds to a beat frequency of 1 cycle per 2.2 seconds. Since most musical notes have shorter durations, most of us don’t hear the beats. 1/12

7/12

Calculating Fourier Coefficients Fourier analysis is a powerful tool for linear systems because any periodic function f is a sum of sine and cosine functions, with some set of Fourier coefficients. We can often solve difficult equations for the special case of simple sinusoidal functions. The solution for the complex function f is then simply the sum of the sinusoidal solutions, with the appropriate Fourier coefficients. All we need is a procedure to compute the Fourier coefficients. Not surprisingly, the person who developed this procedure was Jean-Baptiste Joseph Fourier. Consider again the Fourier series: f(t) = Σ { a cos(nωt) + b sin(nωt) } n

n

n

Integrate f(t) over one full period T = 2π/ω. The integral of sine and cosine over a full cycle is zero. Hence only the cosine part of the n=0 term remains on the right hand side after integration. ∫ f(t) dt = ∫ a cos(0) dt = T a a = (1/T) ∫ f(t) dt T

0

T

0

0

T

Fourier discovered that calculating the other coefficients isn’t much harder. Multiply the Fourier series by cos(jωt) for some integer j>0, and then integrate over T:

∫ f(t) cos(jωt) dt = Σ { ∫ a cos(nωt) cos(jωt) dt } + Σ { ∫ b sin(nωt) cos(jωt) dt } T

n

n

T

T

n

n

Note that: 2 cos(nωt) cos(jωt) = cos([n+j]ωt) + cos([n–j]ωt) 2 sin(nωt) cos(jωt) = sin([n+j]ωt) + sin([n–j]ωt) 2 sin(nωt) sin(jωt) = cos([n–j]ωt) – cos([n+j]ωt) We will use the first two equations now and the third equation later. Each term on the right side of each equation is a sinusoidal function of frequency mω, for m either n+j or n–j. If m is non-zero, the integral over period T of either sine or cosine is zero. Therefore we have shown that after integrating from t=0 to t=T=2π/ω: ∫ cos(nωt) cos(jωt) dt = T/2 if n=j, else = 0 ∫ sin(nωt) cos(jωt) dt = 0 for any n,j ∫ sin(nωt) sin(jωt) dt = T/2 if n=j, else = 0 The prior integral becomes: ∫ f(t) cos(jωt) dt = ∫ a cos(jωt) cos(jωt) ∫ f(t) cos(jωt) dt = a T/2 a = (2/T) ∫ f(t) cos(jωt) dt T

T

T

j

j

j

T

We can repeat this logic multiplying f(t) by sin(jωt), and integrating over T. The result is: b = (2/T) ∫ f(t) sin(jωt) dt j

T

We have been very successful analyzing many repetitive motion problems using exponentials with complex exponents. We can employ that technique here as well and rewrite the Fourier equations as: f(t) = Real part of Σ { z exp(inωt) }, with z = a + ib = (2/T) ∫ f(t) exp(–inωt) dt, for n≥1 a = (1/T) ∫ f(t) dt, and b = 0 n

n

0

n

n

n

T

T

0

Fourier Series of Square Wave Feynman next examines the simple example of the Fourier series for a square wave. Let the square wave be a periodic function, repeating with period T and defined by: f(t) = +1 for 0 ≤ t < T/2 f(t) = –1 for T/2 ≤ t < T f(t+T) = f(t) for any t The square wave function is shown in Figure 45-3.

Figure 45-3. Square Wave Function

Clearly a , the average value of f(t), is zero. To compute the other coefficients, we must separate each integral into two parts: (1) the integral from t=0 to t=T/2, where f(t)=1; and (2) the integral from t=T/2 to t=T, where f(t)=–1. Recall ωT=2π. 0

a = (2/T) { ∫ cos(jωt) dt – ∫ cos(jωt) dt } b = (2/T) { ∫ sin(jωt) dt – ∫ sin(jωt) dt } j

(1)

j

(1)

(2)

(2)

a = (2/Tjω) { sin(jωT/2) – 0 – sin(jωT) + sin(jωT/2) } b = (2/Tjω) {–cos(jωT/2)+1 + cos(jωT)–cos(jωT/2) } j

j

a = (1/jπ) { 2sin(jπ) – sin(j2π) } b = (1/jπ) { 1 –2cos(jπ) + cos(j2π) } j

j

a = (1/jπ) { 0 – 0 } b = (1/jπ) { 2 –2cos(jπ) } j

j

a = 0 for all j b = 0 for j even b = (4/jπ) for j odd j

j j

f(t) = (4/π) {sin(ωt) + sin(3ωt)/3 + sin(5ωt)/5 + … } Feynman shows that we can obtain an equation for the sum of an interesting infinite series by setting ωt=π/2. f(T/4) = +1 = (4/π) {1 –1/3 + 1/5 –1/7 + … } Feynman also notes that the Fourier series of a square wave cannot exactly match the square wave at its discontinuity (t=T/2 in this case). Here we find: f(T/2) = (4/π) {sin(π) + sin(3π)/3 + sin(5π)/5 + …} = 0 The Fourier series yields the value half way between the square wave’s value at tT/2. This seems reasonable. Natural phenomena are almost never discontinuous. Any physically realistic function that goes from +1 to –1 must pass through zero.

Fourier Transform of Gaussian Feynman doesn’t do this, but let’s find the Fourier representation of a Gaussian distribution. Gaussians are very important because many natural phenomena follow such distributions. The equation for a Gaussian distribution, G(x), centered at x=0, with standard deviation σ (mean square variance = σ ), is: 2

G(x) = exp{–x /2σ } / √(2πσ ) 2

2

2

Rather than compute the Fourier series, which is a sum of discrete frequencies, we will instead perform a Fourier transform that is an integral over all frequencies. The Fourier series is most appropriate for waves confined to a finite space, whereas the Fourier transform corresponds to the limit when that space grows toward infinity. The Fourier transform, S(k), is a function of wave number k given by an integral over x, from x=–∞ to x=+∞: S(k) = ∫ f(x) exp{–ikx} dx / √(2π) Here S(k) is the Fourier transform of function f(x). For a Gaussian: S(k) = ∫ exp{–x /2σ –ikx} dx / (2πσ) 2

2

This exponent isn’t the prettiest we’ve seen, but it is integrable with a neat trick called “completing the square.” We can make the exponent a perfect square by adding the right constant. –(x/σ√2 + A) = –x /2σ – 2xA/σ√2 – A 2

2

2

2

To complete the square we want: 2xA/σ√2 = ikx A = ikσ/√2 We can therefore rewrite the exponent as: –x /2σ –ikx = –(x/σ+ikσ) /2 –k σ /2 2

2

2

2

2

The integral then becomes: S(k) = ∫ exp{–(x/σ+ikσ) /2} exp{–k σ /2} dx / (2πσ) 2

2

2

We next substitute u = x/σ+ikσ, and du = dx/σ. S(k) = exp{–k σ /2} ∫ exp{–u /2} σdu / (2πσ) S(k) = exp{–k σ /2} √(2π) / (2π) S(k) = exp{–k σ /2} / √(2π) 2

2

2

2

2

2

2

We see that the Fourier transform of a Gaussian, G(x), is also a Gaussian, S(k). Also note that the standard deviation of G(x) is σ, while the standard deviation of S(k) is 1/σ. For any Gaussian distribution, about 50% of the population is contained within 1/√2 standard deviations of the mean. In the context of a wave packet, we can view 1/√2 standard deviations as being the uncertainties Δx and Δk in the values of x and k, respectively. The product of these two uncertainties is: Δx Δk = (σ/√2) (1/σ√2) = 1/2 This analysis proves that reducing Δx increases Δk, and vice versa, demonstrating the unavoidable tradeoff between the uncertainty of a wave packet’s location and the uncertainty of its wave number. Since quantum mechanics equates momentum p with ħk, we have proven: Δx Δp = ħ/2, for Gaussian distributions which is the Heisenberg Uncertainty principle of quantum mechanics.

Fourier Series & Energy The energy carried by a wave is proportional to its amplitude squared. For a wave, f(t), that repeats with period T, the energy it carries is given by an integral over one full period T. E ~ ∫ f(t) dt 2

If f(t) is represented by a Fourier series, with coefficients a and b , as above, we can expand the energy equation as follows: n

n

E ~ ∫ { Σ [a cos(nωt) + b sin(nωt)] } dt E ~ ∫ { Σ [a cos(nωt) + b sin(nωt)] } × { Σ [a cos(jωt) + b sin(jωt)] } dt 2

j

n

n

n

n

n

n

n

j

Here, we have changed the summation index in the second { } to j. This allows us to rewrite the integrand as: Σ [a cos(nωt) a cos(jωt) + a cos(nωt) b sin(jωt) + b sin(nωt) a cos(jωt) + b sin(nωt) b sin(jωt) ] nj

n

j

n

n

j

n

j

j

As shown earlier, when integrated over a full period T, cos(nωt) sin(jωt) is zero for all n and j. Also cos(nωt) cos(jωt), and sin(nωt) sin(jωt) are nonzero only when n=j. Eliminating these terms reduces the summation to: Σ [ a cos (nωt) + b sin (nωt) ] n

2 n

2

2 n

2

Putting this back into the integral we get:

E ~ ∫ { Σ [ a cos (nωt) + b sin (nωt) ] } dt 2

n

2

2

n

2

n

For n=0, cos (0)=1 and sin (0)=0. For n>0, cos (nωt) and sin (nωt) each average 1/2. 2

2

2

2

The result is: E ~ ∫ f(t) dt = T a + (T/2) Σ [a + b ] 2

2

0

n>0

2 n

2 n

In V1p50-8, Feynman calls this equation the energy theorem. It says a wave’s energy equals the sum of the energies of each of its Fourier components. We can obtain the sum of another infinite series by applying this theorem to the Fourier series for a square wave, which we derived above. For a square wave, f(t) is always 1. 2

∫ f(t) dt = T = (T/2) (4/π) {1+ 1/3 + 1/5 + 1/7 + … } 2

2

2

2

2

Feynman says using the energy theorem on the Fourier series for f(t)=(t–T/2) proves the equation for another infinite sum: 2

π /90 = 1 + 1/2 + 1/3 + 1/4 + … 4

4

4

4

Do you want to try to prove that? I provide the solution at the end of this chapter.

Nonlinear Systems So far, we have analyzed systems that were assumed to be linear. This is an idealization that may only be approximately correct for many real-world systems. While the most general nonlinear systems cannot be analyzed simply, we can gain insight into the interesting behaviors of slightly nonlinear systems. Let’s now consider the effects of nonlinear responses, such as a current in an electrical circuit that is not strictly proportional to voltage. Consider a device subject to an input x(t) that produces a response to that input of y(t). Examples include a voltage x that produces a current y, or a force x that produces a displacement y. A linear device has a relationship between x and y of the form: y(t) = K x(t) Here K is a proportionality constant that never changes — it is the same for all x and all t. By contrast, in a nonlinear device, the relationship may have the form: y(t) = K [x(t) + εx (t)] 2

If ε is small enough, the nonlinear term εx (t) is small compared with x(t). 2

Consider a sinusoidal input, x, and the response y, as shown in Figure 45-4. x(t) = cos(ωt) y(t) = K[cos(ωt) + εcos (ωt)] 2

Figure 45-4 Linear (light) vs. Nonlinear (dark) Responses

If ε were zero, the response would be linear, resulting in the lighter curve in Figure 45-4. If ε is small but greater than zero, the response is slightly nonlinear, resulting in the darker curve. We can rewrite the prior equation using 2cos (θ) = 1–cos(2θ). 2

y(t) = K cos(ωt) + Kε/2 – (Kε/2) cos(2ωt) The first of the three terms above is the normal linear response at the same frequency ω as the input x. The second term adds a constant offset to y(t), shifting its average value. This shifting of the entire response curve is called rectification. The third term adds a higher frequency harmonic to y(t). Feynman notes that this is the second harmonic (twice the input frequency) because we assumed a nonlinearity proportional to x . He says nonlinearities proportional to x or x would add third and fourth harmonics respectively. The most general nonlinearity would introduce an entire spectrum of harmonics. 2

3

4

The addition of harmonics, exemplified by the third term, is called modulation, a process that we studied in Chapter 43. Now consider the response of a nonlinear device to an input composed of two components of different frequency and amplitude. x(t) = A cos(ωt) + B cos(Ωt) y(t) = K [x(t) + εx (t)] y(t) = K x(t) + Kε [A cos(ωt) + B cos(Ωt)] 2

2

y(t) = K x(t) + Kε [A cos (ωt) + B cos (Ωt)] + 2KεAB cos(ωt) cos(Ωt) 2

2

2

2

The term in [ ]’s is the same as the prior example; it produces second harmonics. In Chapter 43, we often dealt with terms containing the product of cosines of different frequencies, as in the last term above. Such terms produce sidebands with cosines of the sum and difference of ω and Ω. Making the familiar substitutions yields: y(t) = KA cos(ωt) + KB cos(Ωt) + (Kε/2) [A +B ] – (Kε/2) [A cos(2ωt) + B cos(2Ωt)] + KεAB {cos([ω+Ω]t) + cos([ω–Ω]t)} 2

2

2

2

We have here a combination of interesting effects: linear response: KA cos(ωt) + KB cos(Ωt) rectification: (Kε/2) [A +B ] harmonics: A cos(2ωt) + B cos(2Ωt) sidebands: cos([ω+Ω]t) + cos([ω–Ω]t) 2

2

2 2

As in Chapter 43, if ω≈Ω, the sidebands include a term with a frequency of about 2ω and another at ω–Ω. If ω>>Ω, the two sidebands are at nearly the same frequency. An entirely equivalent way of looking at this term comes from considering its prior form: cos(ωt) cos(Ωt). If ω≈Ω, this term produces beats, as we found in Chapter 43. If ω>>Ω, we could say the signal oscillates at ω, while being slowly modulated at frequency Ω. Both descriptions are perfectly correct. Note that all the nonlinear effects are proportional to the second power of amplitudes: A , B , or AB. This means nonlinear effects will be more important for larger inputs, for example for stronger electrical signals. 2

2

These nonlinear effects — rectification, harmonics, modulation, sum and difference frequencies — have many practical implications. In V1p50-9, Feynman notes that the human ear is thought to be somewhat nonlinear. Very loud sounds give us the sensation of harmonics and sum and difference frequencies even when the input is monotonic. Additionally, audio components, including amplifiers and speakers, are never perfectly linear. Many people find these nonlinearities objectionable enough to warrant paying more for higher fidelity (less nonlinear) equipment. Feynman notes that, for unknown reasons, people seem to object less to the nonlinearities of their ears than to the nonlinearities of their loudspeakers. Finally, Feynman recalls our prior study of light causing electrons to oscillate and emit radiation that interferes with the incident light, resulting in refraction. He says our assumption that electrons respond linearly to light’s electric field is a very good approximation, but not perfectly accurate. With the development of high-powered lasers, scientists have observed small nonlinearities. Red lasers passing through glass produce faint blue light, the second harmonic of red, due to electrons’ slightly nonlinear response.

Solution to Infinite Sum

Let’s find the Fourier series expansion for f(t)=t . Feynman suggested f(t)=(t–T/2) but that’s needlessly messy; simply shifting the time axis gives us an cleaner equation. Since t is an even function t =(–t) , the Fourier series will not have any sine function contributions (b =0 for all n). The coefficients we need are: 2

2

2

2

2

n

a = (1/T) ∫ f(t) dt a = (2/T) ∫ f(t) cos(jωt) dt 0

T

j

T

The integrals cover the range t=–T/2 to t=+T/2. a = (1/T) ∫ t dt = t / (3T) evaluated at the limits a = {T /8 – [–T] /8} / (3T) = T /12 2

0

3

T

3

3

2

0

Next, calculate a : j

a = (2/T) ∫ t cos(jωt) dt 2

j

T

Make the substitution x=jωt. The integration limits become x=±jωT/2=±jπ. a = (2/T) ∫ x cos(x) dx / (jω) 2

j

3

T

We find the integral we need from tables (or you can derive this using integration by parts): ∫ x cos(x) dx = 2x cos(x) + (x –2) sin(x) 2

2

Evaluating this at the limits yields: 2 (jπ) cos(jπ) – 2 (–jπ) cos(–jπ)} + 0 = 4jπ (–1) j

The final 0 in the first line arises from terms proportional to sin(jπ)=0. Note that cos(jπ) equals –1 when j is odd and +1 when j is even, which we have written as (–1) . j

Plugging this into the equation for a gives us: j

a = (2/T) 4jπ (–1) / (jω) a = (2/T) (T/j2π) 4jπ (–1) a = (T/jπ) (–1) j

3

j

3

j

j

2

j

j

Finally, we use the energy theorem: ∫ f(t) dt = T a + (T/2) Σ [a + b ] ∫ t dt = T (T /12) + (T/2) Σ [ (T/jπ) (–1) ] 2

2

0

4

2

2 n

n>0

2

2 n

2

j

2

n>0

(1/5) {T /32–[–T] /32} = T /144 + (T/2) Σ [ (T/jπ) ] (1/5) = (1/9) + 8 Σ (1/jπ) 5

5

5

4

n>0

4

n>0

9/45 – 5/45 = 8 Σ (1/jπ) 1/90 = Σ (1/jπ) π / 90 = 1 + 1/2 + 1/3 + 1/4 + … 4

n>0

4

n>0

4

4

4

4

Chapter 45 Review: Key Ideas In each section below, ∫ denotes integration over one full period T. T

1. Musical tones are characterized in terms of loudness, pitch, and quality. Loudness is the amplitude of a sound wave Pitch is the frequency of a tone’s most basic pattern Quality corresponds to the multiplicity of frequencies from which a waveform is composed 2. The intensity I of the sum of two musical tones of frequencies ω and ω is: 1

2

I = 1 + cos(2Δωt) Here 2Δω = ω –ω is the beat frequency at which sound intensity slowly waves and wanes. When instrument string lengths have a small-number ratio, their harmonics match with Δω=0 and beats are eliminated, producing the pleasing effect called consonance. At other length ratios, the combined sound has beats and is called dissonant. 1

2

3. Any function f(t) that repeats with period T has a Fourier series representation that is a sum of sine and cosine functions. f(t) = Σ { a cos(nωt) + b sin(nωt) } n

n

n

where T=2π/ω, and a = (1/T) ∫ f(t) dt b =0 a = (2/T) ∫ f(t) cos(jωt) dt b = (2/T) ∫ f(t) sin(jωt) dt 0

T

0

j

T

j

T

Here we have employed these relationships: ∫ cos(nωt) cos(jωt) dt = T/2 if n=j>0, else = 0 ∫ sin(nωt) cos(jωt) dt = 0 for any n,j ∫ sin(nωt) sin(jωt) dt = T/2 if n=j>0, else = 0 T T T

4. We can rewrite the Fourier equations using exponentials with complex exponents as:

f(t) = Real part of Σ { z exp(inωt) } z = a +ib = (2/T) ∫ f(t) exp(–inωt) dt, for n≥1 a = (1/T) ∫ f(t) dt, and b = 0 n

n

n

n

0

n

T

T

0

5. Any function G(x) has a Fourier transform S(k) given by: S(k) = ∫ G(x) exp{–ikx} dx / √(2π) where the integral is from x=–∞ to x=+∞. If G(x) is a Gaussian distribution, S(k) will also be Gaussian: G(x) = exp{–x /2σ } / √(2πσ ) S(k) = exp{–k σ /2} / √(2π) 2

2

2

2

2

The standard deviation of G(x) is σ, while the standard deviation of S(k) is 1/σ. Let Δx and Δk be the uncertainties in the values of x and k of a wave packet. Define these uncertainties to be the limits that include 50% of the population of each Gaussian distribution, and note that quantum mechanics equates momentum p with ħk. The resulting Δx and Δp satisfy the uncertainty principle: Δx Δp = ħ / 2. 6. The energy of a wave is proportional to its amplitude squared, averaged over a full period T, which equals the sum of the average amplitudes squared of the wave’s Fourier components. E ~ ∫ f(t) dt = T a + (T/2) Σ [a + b ] 2

2

T

0

n>0

2 n

2 n

7. Fourier analysis helps evaluate some interesting infinite series: π/4 = 1 –1/3 + 1/5 –1/7 + … π /90 = 1 + 1/2 + 1/3 + 1/4 + … 4

4

4

4

Chapter 46 Complex Waves In V1p51-1, Feynman says this lecture is about “some of the more complex phenomena associated with waves.” He adds that he will discuss these phenomena qualitatively rather than quantitatively because they are “too complicated to analyze in detail here.” Indeed this chapter contains fewer equations than normal.

Bow Waves The first topic concerns waves created by a source moving faster than the wave velocity. Consider the example of a jet flying faster than the speed of sound in air. Let the jet’s velocity be v in the +xdirection. At each point P along the jet’s path, the jet compresses the air in front of it, creating a pressure wave that spreads outward in all directions at the speed of sound, c . s

Figure 46-1 shows the crests of pressure waves emitted at equally spaced times. All such waves will be tangent to a common line, thus creating a bow wave, as we will demonstrate.

Figure 46-1 Jet (grey ) Creating Bow Wave An Intense Conical Wave With Half-Angle θ

The image in Figure 46-1 corresponds to time t=0, when the tip of the jet is at x=0. Consider the largest wave on the left side of the image. Define T to be the time that wave was emitted, and define X to be the position of the jet at time T; X is also the position of the wave’s emission point. Both T and X are negative quantities. From the jet’s velocity, we know that X=vT, which is the length of hypotenuse of the right triangle shown in the image with dotted lines. From the wave’s speed, we know that the radius of the largest wave equal Tc , which is the length of the side of the right triangle farthest from the jet. Therefore: s

sinθ = Tc /Tv = c /v s

s

We chose the left most wave as an example, but the same analysis applies to all other waves created by the jet at other times. Thus all wave crests are tangent to the same line at angle θ. They interfere constructively and produce a high intensity wavefront — a bow wave. The figure shows a 2D cross-section. In all three dimensions, the bow wave is conical with θ being one-half the cone angle. Recall that we assumed v>c , that the speed of the jet, v, is faster than the speed of sound, c . There is no bow wave if v0

n

2

n

In complex number notation, the Fourier equations are: f(t) = Real part of Σ { z exp(inωt) } z = a +ib = (2/T) ∫ f(t) exp(–inωt) dt, for n≥1 a = (1/T) ∫ f(t) dt, and b = 0 n

n

n

n

0

n

T

T

0

10. Any function G(x) has a Fourier transform S(k) given by this integral from x=–∞ to x=+∞: S(k) = ∫ G(x) exp{–ikx} dx / √(2π) If G(x) is a Gaussian distribution, S(k) will also be Gaussian: G(x) = exp{–x /2σ } / √(2πσ ) 2

2

2

S(k) = exp{–k σ /2} / √(2π) 2

2

The standard deviation of G(x) is σ, while the standard deviation of S(k) is 1/σ. 11. An object moving with velocity v through a medium with wave speed c produces bow waves with a conical shape if v>c . The cone half-angle θ is given by: s

s

sinθ = c /v s

12. Shock waves occur when trailing waves move faster than the waves in front of them. High intensity sound waves heat the air they pass through thereby increasing the speed of the sound waves that follow. Shock waves always travel faster than low intensity sound waves. 13. Tidal bores occur in strong rising tides, in shallow narrow channels, as incoming waves “pile up” creating a shock wave effect. 14. Solid materials support two types of waves: longitudinal waves, also called compression waves transverse waves, also called shear waves Compression waves in solids are analogous to sound waves in air. Transverse waves occur only in solids, not in gases or liquids. Wave oscillations are perpendicular to the wave’s direction of motion in transverse waves and along the direction of motion for longitudinal waves. 15. Seismology, the study of waves traveling through Earth’s interior, has enabled us to determine

our planet’s internal structure. 16. Water waves do not transport water. As a wave passes, each molecule moves in a roughly circular path, approximately returning to its starting point. This circular path lies in a vertical plane that is perpendicular to the wavefront. The phase velocity of waves in open, deep water is given by: v = √ [2πτ/ρλ + gλ/2π] P

Here τ is water’s surface tension, ρ its density, g is the acceleration of gravity, and λ is the wavelength. The first term dominates for short waves, where surface tension (the capillary effect) is the primary restoring force. The second term dominates for long waves, where gravity is the primary restoring force.

Chapter 48 The Physics of Vision Vision is a complex process involving physics and physiology. As this is a physics course, we will focus more on the former. In V1p35-1, Feynman begins this topic by looking into the eye.

The Human Eye A stationary human right eye has a field of view that extends from the central line of sight: 95º to the right 60º to the left 75º downward and 60º upward The left eye covers the same range on the opposite side. Each eye rotates about ±45º. The combination of eye rotation and field of view gives us an angular range of 280º horizontally and 225º vertically, without turning our head. That leaves us blind to only an 80º-wide zone directly behind us. In terms of an analog clock, with 12 o’clock being straight ahead and 3:00 to the right, our blind zone is between 4:40 and 7:20, hence fighter pilots’ concern about what’s on their “6.” (I wonder why evolution didn’t produce a creature with a third eye at the back of its head.) The full range of light intensities through which our vision operates effectively is 100 trillion to 1. The iris controls the eye’s entrance aperture, the pupil, whose diameter ranges from 2 mm to 8 mm, but is typically 4 mm. In terms of intensity discrimination, we can detect stationary objects with as little as a 1% intensity contrast. Figure 48-1 shows a horizontal cross section of a right human eye. Light rays enter through the cornea, refract, and are focused on the retina.

Figure 48-1 Diagram of Human Ey e 1. Conjunctiva, 2. Scelra, 3. Cornea, 4. Aqueous humor, 5. Lens, 6. Pupil, 7. Uvea, 8. Iris, 9. Ciliary body, 10. Choroid, 11. Vitreous humor, 12. Retina, 13. Macula, 14. Blind spot, 15. Optic nerve

Optics of Vision The eye features a two-element optical system composed of the cornea and the lens. Its total focusing power is 43 diopters, meaning that it can focus from infinity to as close as 1m/43 = 2.3 cm (0.9 inches). The aqueous humor fills the space between cornea and lens, and the vitreous humor fills the space between lens and retina; both are almost entirely water. The cornea provides about 2/3rds of the eye’s focusing power due to its shape and the change of refractive index at its exterior surface. The cornea’s index of refraction is 1.37, which is substantially larger than the index of air (1.0003), but only slightly larger than the index of water (1.33). Since the bending of light at a surface is determined by the change in refractive index across that surface, our vision performs well in air but not in water. Underwater, we lose most of our focusing power and our vision becomes blurry. This can be remedied with a diving mask that maintains air in front of our eyes. The shape of the cornea is superior to that of normal man-made optics. Mass-produced lenses are spherical, and therefore suffer from spherical aberration that limits their performance (see Chapter 30). (Eyeglass lenses correct astigmatism with the addition of a cylindrical surface on the opposite side). By comparison, the cornea is less curved at its sides than a sphere, having a shape that reduces spherical aberration. The lens consists of multiple layers, called laminae, whose arrangement is similar to an onion. To enhance transparency, mature lens cells have no nerves, blood vessels, connective tissues,

mitochondria, or nuclei. Lens cells are nourished by the aqueous humor. The refractive index of the lens varies, being 1.406 in the center and 1.386 at its edges. This variation gives the lens greater focusing power than is possible with a uniform index. The shape of the lens is adjusted to optimally focus objects of interest. For distant objects, the ciliary muscles relax and the lens flattens. For nearby objects, the ciliary muscles contract, making the lens more curved and thicker at its center.

The Retina The eye’s light sensor is the retina, which contains a variety of photoreceptors and covers about 2/3rds of the eyeball’s inner surface. At the periphery of the visual field, photoreceptors are sparsely distributed. Here, neural density declines much more rapidly, because up to 100 receptors feed into a single nerve fiber. Peripheral acuity is therefore quite limited. Receptor density and acuity increase steadily toward the center of the visual field. The highest acuity is provided by the 0.2 mm-wide foveola that lies at the center of the macula. Each receptor in the foveola is connected to an individual nerve fiber, with the result that the foveola drives about half of the fibers in the optic nerve. The foveola enables reading and other precise visual activities. Vision employs two types of photoreceptors: rods and cones. In the 1990’s, it was discovered that a small percentage of the eye’s ganglion cells are also photosensitive. The ganglion receptors sense overall light intensity for the control of the pupil, and possibly to inform the circadian rhythm. The 120 million rod receptors in a human eye cover the bulk of our visual field, and provide black and white vision. Rods are extremely sensitive, up to 10,000 times more sensitive than cones. In optimal conditions, a rod cell can be triggered by a single photon of light. However, a neural response requires 5 to 10 photons detected within 0.1 seconds. Rods cells are about 2 microns in diameter. The 6 million cone receptors provide color vision, and require much higher light intensity. This is why we see only black and white at low light levels. Cone cells are typically 6 microns in diameter, but are thinner in the densely packed foveola. Humans have three types of cones, each type being sensitive to a different range of wavelengths of light. Some other species have four or five types of cones, and are able to see infrared and ultraviolet light that we cannot. Cones are sharply concentrated in the macula, as shown in Figure 48-2.

Figure 48-2 Density of Rods & Cones Across Visual Field Dark Line Marks 10 Million Photoreceptors Per Square cm

In V1p36-2, Feynman explains that the retina is effectively part of the brain. In embryonic development, he says: “a piece of the brain comes out in front, and long fibers grow back, connecting the eyes to the brain. The retina is organized in just the way the brain is organized and, as someone has beautifully put it, ‘The brain has developed a way to look out upon the world’.” After photoreceptors detect light, signal processing starts immediately in the retina itself. The retina has ten layers of cells that process an estimated 9 million bits of visual information per second, before sending the results on to the brain. Retinal processing provides rapid edge and motion detection, which are essential survival traits. One puzzling feature of our visual system is that our retinas seem “inverted.” In all vertebrates, the photoreceptors are on the backside of the retina, the side farthest from the pupil. Light must therefore travel through the entire retina to reach its photoreceptors. Nerves run across the front side of the retina, the side nearest the pupil. On their way to the brain, these nerves pass through a void in the retina called the blind spot. The blind spot contains no photoreceptors, which results in each eye being blind to a small portion of its visual field that is typically 15º to the temporal side of the primary line of sight. Since the left and right eyes’ blind spots are in different parts of our field of view, each spot is noticeable only when the other eye is closed. By contrast, the retinas of cephalopods, such as octopus and squid, are not “inverted.” Their photoreceptors are on the light-facing side of their retinas, and their nerves run across the backside. As a result, cephalopods have no blind spots.

Some physiologists suggest that our “inverted” retinas facilitate blood flow to our energy-intensive retinas, more than compensating for what seems to be a poor optical design. In V1p36-4, Feynman is not so kind, calling the inverted design of our retinas “apparently stupid.”

Color & Light Intensity In V1p35-2, Feynman remarks on a striking feature of our vision: dark adaptation. When light levels diminish rapidly, we see very little initially. Then slowly our eyes become increasingly sensitive, and eventually, we may be able to see quite well — but only in black and white. Our cone receptors do not function in the dark, so our dark vision comes only from rods. In Feynman’s day, astronomers observing faint nebulas by eye often saw them only in black and white, whereas long-exposure photographic images of the same objects demonstrated beautiful colors. The difference, of course, was due to our visual insensitivity to color at low light intensity. In modern professional astronomy, naked-eye viewing and film photography have been largely replaced by highperformance CCD imaging due to its much greater sensitivity. This effect is even more interesting because the colors we perceive change with light intensity. Figure 48-3 illustrates how the sensitivity of each type of photoreceptor varies with wavelength. Here each curve is normalized to the same peak sensitivity.

Figure 48-3 Spectral Response of Various Receptors

Cones are designated blue, green, or red according to the colors of their peak sensitivity — or by short, medium, or long according to the corresponding wavelengths. Note that rods are reasonably sensitive to blue light but not to red light. Therefore, as Feynman notes, when comparing pieces of red and blue paper, red might seem brighter in daylight but darker at night. In daylight, we see the red paper very well with our red cones, while our rods-only night vision is very insensitive to red. This is called the Purkinje effect.

Astronomers take advantage of the Purkinje effect. Using red lights that activate cone receptors, astronomers can find their way in the dark without ruining the dark adaptation of their rod cells. This is important because astronomers spend so much time in the dark. Another interesting visual effect involves what astronomers call averted vision. Our high-acuity foveola contains few if any rods, making it useless in dim light. We are accustomed to aligning our foveola with objects of special interest, but that is counterproductive in darkness. Naked-eye astronomers train themselves to concentrate on areas outside the center of their field of view. They avert their vision to place objects where the retina has a higher density of rods and provides the best night vision. Two other peculiar features of our vision arise at the periphery of our visual field. Since our cones are concentrated along our primary line of sight, we do not see colors peripherally even in bright light. An object that crosses our field of view becomes colorful as it crosses our primary line of sight. Our peripheral vision does excel, however, at detecting motion. The retina’s image processing layers promptly alert us to objects moving into our field of view; clearly this is a survival mechanism.

Measuring Color Response In V1p35-3, Feynman addresses the question of what does “color” mean in our visual system? We all know that our instruments can precisely measure light intensity as a function of wavelength. But the question is: what color do our eyes perceive when viewing various combinations of wavelengths? If green light enters our eyes we will see green — that is simply the definition of “green.” But are there other combinations of light that also look “green”? The answer turns out to be: Yes. Many different combinations of wavelengths of various intensities will look “green” to us, because our three-color vision senses only a subset of all possible combinations of light wavelengths. While the human eye can distinguish 10 million different colors, there are an infinite number of wavelengths and an infinite number of ways of combining them. So how can we measure and quantify what we mean by “green”? It would be enormously difficult, perhaps impossible, to determine what each person senses when they see certain combinations of wavelengths or to determine which stimuli look “green.” Instead, Feynman says, the most effective approach is a null experiment. In general, null experiments consist of finding circumstances where two entities are equal — where their difference is null. Such experiments often achieve astonishing precision. A prime example is the Michael-Morley null experiment confirming the equality of the speed of light in different directions. In a color null experiment, a subject views two light sources, each a combination of wavelengths. The composition of one source is adjusted until the subject says two sources appear identical. By repeating the experiment with many subjects, one discovers the extent to which these wavelength

combinations are universally indistinguishable. This technique avoids the conundrum of whether or not people experience the same sensation from the same light. In his lecture, Feynman demonstrated the technique using four light projectors. Three projectors produced colored spots of any desired intensity, with one each for red, green, and blue. The fourth projector produced a large white circle with a black spot at its center. Initially, Feynman turned on only the green and red lights, producing a shade of yellow. Different projector intensities produce various shades of green, yellow, orange, and red. Feynman showed that any specific color can be produced in other ways as well. For example, some combination of green and red make a shade of orange, but that same shade can be made from orange and white, or yellow and red. Let’s represent this mathematically. Denote the projector colors using R for red, G for green, and B for blue. Define r, g and b to be the intensities of the red, green, and blue projectors. The resultant color Z for any combination of three projector intensities is: Z = rR + gG + bB The question is: can we make all the colors visible to humans from just red, green, and blue? Feynman demonstrated that equal intensities of RGB made a “fairly nice white.” But when the three colors were projected onto the central black spot from the fourth projector, its annular white ring surrounding the color spots no longer seemed white, but was somewhat yellowish. Feynman then tried to make brown, which, in V1p35-4, he says is very difficult. “People who give lectures on color make all the ‘bright’ colors, but they never make brown, and it is hard to recall ever having seen brown light. As a matter of fact, this color is never used for any stage effect, one never sees a spotlight with brown light; so we think it might be impossible to make brown … we point out that brown light is merely something that we are not used to seeing without its background. As a matter of fact, we can make it by mixing some red and yellow. To prove that we are looking at brown light, we merely increase the brightness of the annular background against which we see the very same light, and we see that that is, in fact, what we call brown! Brown is always a dark color next to a lighter background.”

Mixing Colors The first important principle of color mixing is that color combinations can be indistinguishable to human eyes, even if their spectral compositions are different. Mathematically, let X and Z be two different combinations of light wavelengths. If X and Z are indistinguishable to our eyes, we will say for the present purposes that they are equal: X = Z, if we can’t see the difference

If we now add any Q, some other combination of light wavelengths, then: X+Q=Z+Q Let’s be clear about what this equation represents. If we shine light X at one spot and shine light Z at another spot, and then shine the additional light Q on both spots, the resultant colors of the two spots are as indistinguishable to human eyes as are X and Z. This means that the colors of emitted light are additive. Paint and ink colors are different. Both paint and ink absorb rather than emit light, hence their colors are subtractive. In V1p35-5, Feynman says it is very important to note that the indistinguishablility of two colors does not change when the state of our eyes changes. If for example, we stare at a bright red light for a long time, and then look at white paper, that paper will appear greenish. This is because our photoreceptor response to red diminishes under intense exposure and takes some time to recover. Nonetheless, any X and Z that are equal, by the above definition, remain equal even during such changes in our visual response. Each might appear different as our eyes change, with less red in this example, but they will still appear identical. This principle of color mixing holds as long as the light intensity adequately activates our cone cells. The second important principle of color mixing is that any visible color can be made by mixing three primary colors, such as red, green, and blue. (There is a caveat coming shortly.) The only restriction on the choice of primary colors is that they must be linearly independent: none of the three can equal any sum of the other two. In V1p36-2, Feynman points out how surprising this principle might be to most non-physicists. “The total sensation that is associated with the absorption characteristics of the three [cone types] acting together is not necessarily the sum of the individual sensations. We all agree that yellow simply does not seem to be reddish green; … presumably the sensation of light is due to some other process than a simple mixture like a chord in music, where the three notes are there at the same time and if we listen hard we can hear them individually. We cannot look hard [at yellow] and see the red and the green.” Combining colors can be elegantly represented using vectors, as first demonstrated by Schrödinger who is more famous for quantum mechanics and bizarre cats. Let our three primary colors be: A, B, and C. And let colors X and Z be represented by: X= a A+ b B + c C Z= a A+ b B + c C x

x

x

Z

Z

Z

Then the color Q = X + Z is: Q = (a +a ) A + (b +b ) B + (c +c ) C x

Z

x

Z

x

Z

This is equivalent to the vector equation: Q=X+Z We can perform color analysis using our vector analysis skills. The primary colors A, B, and C are the three axes of the color space. Any color X is equivalent to a vector in that space with coordinates a , b , and c . Adding colors is just a matter of adding their vectors. x

x

x

Here is the caveat: producing some colors from three primary colors may require negative intensities. For example, if red, yellow, and blue are the primary colors, no combination of those colors from three projectors shining on the same spot will match green from a fourth projector shining on an adjacent spot. But we can get a color match with some trickery. If we add red to the green spot, we will be able to adjust the three primary color projectors to match the green-red spot. In fact, we will only need to project some yellow and some blue. We can write this mathematically: G + rR = yY + bB G = –rR +yY +bB This means we can match any visible color with any set of linearly independent primary colors, but we might need a negative intensity. While theoretically satisfying, this isn’t usually a practical solution. Some sets of primary colors provide a greater range of color combinations than others, without using negative intensities. RGB is the most common choice for additive colors. In printing, where colors are subtractive, the standard choice is CMYK: cyan (blue-green), magenta (blue-red), yellow, and black.

Chromaticity Diagram Three primary colors form the axes of a 3-dimensional space. Normalizing the three primary intensities to produce a constant combined intensity reduces the color space to two dimensions. This is equivalent to reducing 3D space to the surface of a 2D sphere by normalizing all vectors to the same radius. On that basis, colors are characterized using a 2D chromaticity diagram. The diagram plots the limits of human sensitivity to various wavelength combinations, as determined experimentally by examining many individuals. One standard format for this diagram is CIE 1931 shown in Figure 48-4. The diagram has a horseshoe shape, with a straight line across the bottom from blue (420 nm) on the left to red (680 nm) on the right, and a curved end at green (520 nm) in the upper left.

Figure 48-4 CIE 1931 Chromaticity Diagram Numbers on Periphery are Wavelengths in nm

Those with black and white ereaders can view this color image on their computer at http://www.guidetothecosmos.com/feynman/F1D-color.html. On the CIE diagram, I have superimposed the position of R, G, and B, along with the triangle that connects them. Any color within the triangle can be represented by a linear combination of these primary colors, with positive coefficients. Colors outside the triangle can be represented, but only with some negative coefficients. The main areas that are lost with positive-only coefficients are green. Any specific wavelength of light (any pure monochromatic light of only one frequency) can be represented by a linear sum of primary colors, using both positive and negative coefficients. In the 1920’s this was carefully done for all wavelengths in the visible range by W. David Wright and John Guild. Their results, shown in Figure 48-5, are the basis of the CIE diagram.

Figure 48-5 RGB Coefficients of Monochromatic Light vs. Wavelength λ

Monochromatic colors (single wavelength light) are on the CIE diagram’s edges, while mixed wavelength light is in its interior. All points in the interior are linear combinations, with coefficients between 0 and 1, of the monochromatic light on the perimeter.

Chapter 48 Review: Key Ideas 1. The human visual system has a field of view that spans 280º horizontally and 225º vertically. Its focusing power is 43 diopters, enabling us to focus from infinity to 2.3 cm (0.9 inches). 2. The photoreceptors in each retina include 120 million rods that see the world in black and white, and 6 million cones of three types that detect color. The three types of cones correspond to peak sensitivities at blue, green, and red wavelengths. The density of cones is very strongly peaked in the foveola, which provides high-acuity vision. The density of rods is nearly zero in the foveola. Rod density peaks in an annular ring surrounding the foveola, and decreases rapidly toward the edge of the visual field. 3. Rods are highly sensitive to light, up to 10,000 times more sensitive than cones. In dim light, our cones are useless and we see only in black and white. 4. The retina is an extension of the brain that originates in the embryonic brain. Processing of the 9 million bits of visual information per second begins in the retina, with the results transferred through the optic nerve to the brain. 5. The first principle of color mixing is that color combinations can be indistinguishable to human eyes even if their spectral compositions are different. Two indistinguishable colors remain indistinguishable even when the state of our eyes changes. 6. The second principle of color mixing is that any visible color can be made from three primary

colors, such as red, green, and blue (possibly with some negative coefficients), provided those colors are linearly independent. Color mixing is elegantly represented with vectors. For primary colors A, B, and C, let colors X and Z be written: X= a A+ b B + c C Z= a A+ b B + c C x

x

x

Z

Z

Z

Then the color Q = X + Z is: Q = (a +a ) A + (b +b ) B + (c +c ) C x

Z

x

Z

x

Z

This is equivalent to the vector equation: Q=X+Z 7. A chromaticity diagram shows all the colors humans can perceive. The diagram is horseshoe shaped, with a straight line across the bottom from blue on the left to red on the right, and a curved end at green in the upper left. Monochromatic colors (single wavelength light) are on the diagram’s edges, while mixed wavelength light is in its interior.

Chapter 49 Symmetry & Physical Laws Symmetry has become become one of the most powerful concepts in theoretical physics. In trying to identify the laws governing new phenomena, physicists are able to exclude a wide variety of potential but incorrect solutions on the basis of symmetry. In V1p52-1, Feynman begins his lecture on symmetry and physical laws with this poetic and insightful introduction: “Symmetry is fascinating to the human mind, and everyone likes objects or patterns that are in some way symmetrical. It is an interesting fact that nature often exhibits certain kinds of symmetry in the objects we find in the world around us. Perhaps the most symmetrical object imaginable is a sphere, and nature is full of spheres—stars, planets, water droplets in clouds. The crystals found in rocks exhibit many different kinds of symmetry, the study of which tells us some important things about the structure of solids. Even the animal and vegetable worlds show some degree of symmetry, although the symmetry of a flower or of a bee is not as perfect or as fundamental as is that of a crystal. “But our main concern here is not with the fact that the objects of nature are often symmetrical. Rather, we wish to examine some of the even more remarkable symmetries of the universe—the symmetries that exist in the basic laws themselves which govern the operation of the physical world.” Feynman attributes the best definition of symmetry to Herman Weyl. In my own words: An entity E is symmetric under process S if S leaves E indistinguishable from its original state. For example, rotating a sphere about the vertical axis leaves it indistinguishable from its original state; this makes a sphere symmetric with respect to rotation about that axis. More generally, a sphere is symmetric under the process of rotation by any angle about any axis. A more interesting example is that kinetic energy is symmetric with respect to time reversal, which is the replacement of t by –t. Let’s see what this means. K.E. (t) = m v /2 = m (dx/dt) /2 K.E.(–t) = m (dx/d[–t]) /2 K.E.(–t) = m (dx/dt) /2 = K.E.(t) 2

2

2

2

To understand what this means, consider taking a video of an object moving with velocity +v, and later play the video backwards. In reverse, the video shows the object moving with velocity –v, now

going in the opposite direction. But in both forward and reverse, the object has the same kinetic energy. This is because the kinetic energy equation contains v , which doesn’t change values under time reversal. If time reversal symmetry is a universal property of nature, kinetic energy must be the same in forward and reverse time. That means the kinetic energy equation cannot contain terms like v. 2

3

The following symmetry processes are universally valid. Translation in Space Translation in Time Spatial Rotation Motion at Constant Velocity Exchange of Identical Particles Change of Quantum Phase CPT Symmetry The following symmetry processes are valid for all but the weak interaction. Spatial Reflection Time Reversal Exchange of Matter & Antimatter As Feynman stresses, the natural laws as we now know them have these symmetries. We can only teach what we know now, and remain open to new discoveries that could teach us all something more.

Translations in Space & Time By translation in space or time physicists mean moving to some other location in space or some other moment in time. Translational symmetry means that the laws of nature are invariant under any translations of the following form: x is replaced by x + Δx y is replaced by y + Δy z is replaced by z + Δz t is replaced by t + Δt A simple example is: F = m dv/dt Let’s make the above translation, marking the translated v and F with asterisks. v* = d(x+Δx)/d(t+Δt) F* = m dv*/d(t+Δt)

Because Δx is a constant, any change in (x+Δx) must equal the change in x. Similarly for (t+Δt) and t. We therefore have: d(t+Δt) = dt d(x+Δx) = dx v* = d(x+Δx)/d(t+Δt) = dx/dt = v F* = m dv*/d(t+Δt) = m dv/dt = F Science might well be impossible if natural laws were not symmetric under space and time translations. For example, how could we make sense of our world if natural phenomena behaved differently in my lab compared with your lab, or if observations we make today are not reproducible tomorrow? Some things, of course, do change with position and time. Atmospheric pressure is not the same atop Mt. Everest as it is at the Dead Sea. Atmospheric pressure also changes as the weather changes. When we look for symmetries, we must exclude varying external factors. So, by translational symmetry we mean natural laws are invariant when everything is translated, or at least everything that might influence an observation of interest. Ultimately, what this really means is that space and time are homogeneous: all locations in space are intrinsically identical and indistinguishable, as are all instants in time. The laws of nature may specify time differences or distances between points, but they do not reference specific instants in times or points in space. The mass of the electron and the strength of its electric field are meaningful concepts only because competent measurements yield the same values everywhere and always. The laws of nature, as currently known, break down at the center of black holes and at the instant of the Big Bang. Therefore, these are exceptions to the universal principle of space and time homogeneity and translational symmetry.

Rotational Symmetry In addition to being homogeneous, space is isotropic, which means it is the same in all directions. One might think a homogeneous space is necessarily isotropic, but that’s not true. One can imagine a space that is expanding in the x direction, but not in the y or z directions. That space is homogeneous because the same rule applies everywhere, but it is not isotropic because x is different from y and z. The converse, though, is true: if a space is isotropic, it is necessarily homogenous. The proof is: if a space were not homogeneous, at least one point would be different, hence the direction to that point would be different from other directions; thus proving the space is not isotropic. This is an example of proving a proposition by proving its contrapositive. Because all directions in space are identical and indistinguishable, rotating everything of interest by any angle about any axis leaves the laws of nature invariant.

Constant Velocity Symmetry The Special Theory of Relativity (Chapter 25) states that the laws of nature are invariant under a transformation of constant velocity. This means that adding a constant velocity of any amount in any direction to everything of interest has no physically observable consequences. Recall that constant velocity requires constant speed in a constant direction; Earth does not orbit the Sun with a constant velocity since its direction of motion is continuously changing. We interpret this symmetry to say that absolute velocities are physically meaningless; there is no universal standard reference relative to which all other velocities are defined. Natural laws can depend only on relative velocities — on differences in velocity, or velocity changes — but not on absolute velocities. In V1p52-2, Feynman says: “As a matter of fact, it was the study of the relativity problem [by Einstein] that concentrated physicists’ attention most sharply on symmetry in physical laws.”

Identical Particle Exchange Symmetry The above are all symmetries of space and time. A symmetry of an entirely different sort is identical particle exchange symmetry. This symmetry states that, if any two identical particles are exchanged, there will be no physically observable consequences. As Feynman notes, some might object that the prior statement is a tautology — the definition of “identical” is consequence-free interchangeability. Yes, perhaps the statement could be better phrased, but arguing semantics won’t help us learn physics. The full context of this principle is enlightening. Our observable universe contains 10 particles, all of which belong to one of only 10 types (that’s 89 fewer digits). Furthermore, all the particles of each type are absolutely identical in a manner completely unlike any macroscopic entities. In everyday conservation, we might say two things are identical, such as two coins from the same mint, but in truth, they never are. Two macroscopic entities can never have exactly the same number of neutrons, protons, and electrons. By stark contrast, even nature cannot distinguish the intrinsic properties of one electron from those of any of the other 10 electrons. To say that they are interchangeable with absolutely zero physically observable consequences is indeed a profound statement. 90

1

80

Invalid Symmetries By now you may be wondering: is everything symmetric in every imaginable way? No. Indeed, there are many possible symmetries that nature chooses not to obey. Nature is not symmetric with respect to constant angular velocities, although it is symmetric with respect to constant linear velocities. If an initially stationary object is later rotated at a constant

angular velocity, very distinct observable changes result. There is an absolute standard of angular velocity. By using a Foucault pendulum, or similar devices, we can easily determine when a body’s angular velocity is zero radians per second, without reference to any external entity. Additionally, nature is not scale-invariant. Feynman notes that a 1 m-wide container of sodium atoms radiates light at λ=589 nm, and a 10 m-wide container of sodium radiates at the same wavelength, not at λ=5890 nm. This is true for many reasons, including the fact that sodium atoms come in only one size. Feynman provides another example of scale-non-invariance. Someone made the news by gluing matchsticks together to build a tiny model of a great cathedral. If we imagined enlarging his model to equal the size of the real cathedral, it might immediately collapse; the wooden beams wouldn’t be strong enough. If every dimension is enlarged by a factor of 100, the weight of every object would increase by a factor of 100 = 1 million. At the same time, the cross-section of each vertical beam would increase by 100 = 10,000. This means the weight per unit area on vertical beams increases 100-fold. At some scale factor, what used to be a strong wooden structure is crushed under its own weight. This is why skyscrapers are made of steel. This is also why the leg bone of an elephant has vastly different proportions compared with the leg bone of a bird or an insect. 3

2

Sometimes scale-non-invariance works in favor of large things. Tigers lose much less body heat per unit mass than do shrews who must eat constantly to overcome their heat losses. In water, whales and even people can swim with far less effort than can bacteria for whom water seems as viscous as honey does to us. Ultimately, natural phenomena are not scale invariant because particles and atoms cannot be scaled. As an aside, an American physicist went to do an experiment at CERN, the famous high-energy research center near Geneva, Switzerland. He knew Europeans use metric units, so he prepared the drawings for his equipment with every dimension stated in centimeters. He had no idea that European engineering drawings all use millimeters. Eventually, the CERN machine shop delivered a 1/10thscale model of what he intended, with a note saying the job had been delayed because he hadn’t used standard screw sizes. They had hand-machined every screw precisely per his incorrect drawings. The experiment was delayed, but the model was a work of art, perhaps reflecting the heritage of Swiss watch making.

Is Time Reversal Symmetric? Does reversing the “arrow of time”, by playing a video backwards for example, result in any violations of natural laws? Are the laws of nature equally satisfied by the actions we observe regardless of whether the video is played forward or backward? From everyday experience the answer seems to be obvious: time reversal is not a symmetry of nature. Imagine a video of a decanter of wine falling off a table and crashing onto the carpet, shattering the decanter and splattering the wine. The reversed video shows wine un-soaking from the carpet, glass fragments and wine droplets flying together to form a decanter that fills with wine and

jumps upward back onto the table. How ridiculous! Yet, at an atomic level, (almost) all the laws of nature are precisely time-reversal-invariant. Time reversal is a symmetry of nature at its fundamental level. If we examine the microscopic motions shown in the reversed video, each atom’s actions would be entirely consistent with all natural laws, including the conservation of energy and momentum as well as Newton’s laws of motion. (I said “almost” because of certain weak force interactions that we will discuss later and that are beyond everyday experience.) Why the stark distinction between the obvious irreversibility of time macroscopically, and the fundamental reversibility of time microscopically. While physicists debate the importance of this distinction, it is agreed that the difference is explained by entropy (Chapters 21 and 23). The second law of thermodynamics states that entropy increases in all irreversible processes (all macroscopic processes are irreversible). Entropy thus distinguishes forward and backward time: entropy grows as time runs forward, and diminishes as time runs backward. The “arrow of time” points toward greater entropy. Recall that thermodynamics describes the large-scale properties — temperature, pressure, entropy, etc. — of systems with vast numbers of atoms. Individual atoms do not have a definable pressure, only vast collections of atoms do. The laws of thermodynamics, including the second law, are statements about what’s probable. The probability of a system’s entropy decreasing isn’t exactly zero, but it does drop exponentially with the number of particles in that system. For a system of 10 particles, the probability of the system’s entropy decreasing is absurdly small. One atom of wine might bounce off the carpet and land back on top of the table, but an entire decanter filled with wine will not — not in a trillion times the age of the universe.

20

Thus the “arrow of time” is important but is not as fundamental as the conservation of electric charge. We believe electric charge is conserved always, everywhere, and in every possible interaction — no ifs, ands, ors, or buts. By comparison, the “arrow of time” is very well defined in large enough systems, but is meaningless at the fundamental level, where nature is (almost) completely symmetric with respect to time reversal.

Symmetry & Conservation Laws The symmetries of natural laws have profound consequences. As Feynman says in V1p52-3: “A fact that most physicists still find somewhat staggering, a most profound and beautiful thing, is that, in quantum mechanics, for each of the rules of symmetry there is a corresponding conservation law.” This profound principle is named Noether’s theorem for the famed mathematician Amalie Emmy Noether. Her theorem actually does apply in Newtonian physics as well as in modern physics. In quantum mechanics, Noether’s theorem becomes more expansive, as we will discover later in this course. The proof of Noether’s theorem isn’t overly complex, but requires introducing more

mathematics than is merited at this point. Examples of Noether’s theorem are: 1. Translation symmetry in the x-direction implies the conservation of the x-component of momentum. Similarly for y and z. 2. Spatial rotational symmetry implies the conservation of angular momentum. 3. Translation symmetry in time implies the conservation of energy. An example from quantum mechanics is invariance under a change of phase angle. In quantum mechanics, physical observables are determined by the squares of the magnitudes of probability amplitudes: Prob(A) = |Ψ(A)| . Shifting the phase of all probability amplitudes by a constant angle θ has no observable effect: 2

|Ψ(A)exp{iθ}| = |Ψ(A)| |exp{iθ}| = |Ψ(A)| 2

2

2

2

Feynman says: “The conservation law which is connected with the quantum-mechanical phase seems to be the conservation of electrical charge. This is altogether a very interesting business!”

Spatial Reflection Spatial reflection is the process of inverting one spatial dimension, such as replacing x with –x. Spatial reflection is the same as substituting one phenomenon with its mirror image. In V1p52-4, Feynman considers two identical clocks, physical clocks made of real gears and so forth, but with one being the exact mirror image of the other — all the parts in clock L are the mirror image, and are assembled in the mirror image fashion, to those of clock R. So while clock R runs clockwise, clock L runs counterclockwise. If we start both clocks at the same time, will they forever keep the same time? That seems reasonable. We can think of no reason why they shouldn’t keep the same time (with their hands moving in opposite directions). But do they? Let’s assume for now that the clocks do run identically. If so, they could not be used to distinguish “right” from “left” by any purely physical means. If we sent both clocks to a scientist on Mars who is unfamiliar with “right” and “left”, nothing we could say would help him identify which clock is “left.” Feynman suggests we look to nature for phenomena that might possibly be fundamentally right-handed or left-handed, phenomena with distinct chirality. Many complex molecules have left-handed and right-handed versions. In Chapter 33, we discussed optical activity, the ability of a molecule to rotate the polarization of light. Right-handed molecules rotate light’s polarization counterclockwise when viewed along light’s velocity vector. Left-handed molecules rotate the polarization clockwise.

The form in which I’ve defined the rotation direction is observer-independent: from every vantage point, the polarization rotates about the velocity vector in the same manner. This invariant definition is how particle physicists describe the spin of elementary particles. However, optical activity was originally analyzed by looking at light that had passed through a liquid of chiral molecules. Here the observer’s line of sight is anti-parallel to light’s velocity vector (light is coming toward the observer). In this specific observer orientation, the polarization rotation direction is reversed: righthand molecules rotate polarization clockwise. This observer-specific definition is standard in chemistry. In some cases, both chiralities of a given molecule occur with equal abundance; those can’t help us define “right” and “left.” However, it turns out that some molecules occur predominately or entirely in one chirality. For example, all organically produced protein molecules are left-handed, and all organically produced sugar molecules are right-handed. In V1p52-5, Feynman relates an amusing scenario. A solution of organically produced sugar molecules (all right-handed) rotates light polarization clockwise. A solution of the same sugar molecules produced artificially (equally right- and left- handed) does not rotate light’s polarization. But bacteria added to this solution will eat half the sugar (the right-handed half), after which the remaining sugar molecules (now all left-handed) rotate light’s polarization clockwise with respect to light’s velocity vector. He says: “It seems very confusing, but is easily explained.” Biochemists can easily produce the opposite-handed versions of such molecules. These opposite versions are chemical equivalent to the naturally occurring versions. Detailed analysis shows both have the same energy levels, chemical reaction rates, and in purely artificial reactions (using no organic compounds) equal numbers of both chiralities are produced. Yet, all life on Earth uses only one chirality. Scientists believe this is evidence that life on Earth had one common origin that by random chance selected the chiralities that all subsequent life employs. Feynman muses that if we were able to construct a frog with all its molecules of the opposite chirality, call it a left-hand frog, it would function like any real right-hand frog. The only problem is that our frog wouldn’t find any left-hand food. Earth has only right-hand flies that he can’t digest. Thus, we could tell our Martian scientist that “left” is the direction in which his DNA spirals. That would work if life on Mars had a common origin with life on Earth. If not, the odds are 50-50 that a completely independent life form has our chirality. The chirality of molecules, therefore, is not fundamentally one-handed. As a universal rule, chemical processes seem left/right symmetric.

Reflection of Vectors & Vector Equations In V1p52-6, Feynman explores some subtleties associated with the mirror reflection of vectors and equations involving vectors. The essential point here is that not all of the vectors that physicists employ have the same reflection properties. Vectors belong to two distinct categories: polar and axial.

Polar vectors are simpler and more familiar. Shown in Figure 49-1 are examples of polar vectors: the vector AW from point A to point W; and the vector from the origin of a coordinate system to point T.

Figure 49-1 Mirror Reflection of Polar Vectors

Polar vectors change under reflection; if the mirror lies in the yz-plane, just the x-component of a polar vector changes sign under reflection. Other polar vectors include: velocity, acceleration, momentum, and force. Axial vectors, often called pseudovectors, are more complex. Consider the example of a disk spinning in the yz-plane. The velocity of each atom in the disk lies entirely within the yz-plane; nothing is moving in the x-direction. Yet, we define the angular velocity ω and the angular momentum L as vectors that are parallel to the x-axis (see Chapters 39 and 41). We can define axial vectors this way because, in three dimensions, every plane defines a direction perpendicular to its surface. Motion entirely within the yz-plane identifies x as a special direction: only lines parallel to the x-axis are perpendicular to the plane in which the motion occurs. Mathematically, the perpendicular to a plane is given by the cross product of two non-parallel vectors that lie entirely within that plane. Unlike polar vectors, axial vectors are defined with cross products that explicitly employ the righthand rule. In V1p52-7, Feynman uses the right-hand rule in one paragraph and the left-hand rule in another paragraph, without being as clear about this as he might be. I will be more explicit. The angular velocity vector is a typical axial vector, with a cross product whose sign is set by convention according to the right-hand rule: ω=v×r Writing out all the components of this vector equation yields three equations (right-hand rule):

ω = +v r – v r ω = +v r – v r ω = +v r – v r x

y

z

z

y

y

z

x

x

z

z

x

y

y

x

Exchanging x with –x changes v to –v , and r to –r . The resultant ω, call it ω*, is (in the right-hand rule): x

x

x

x

ω* = +v r – v r ω* = +v (–r ) – (–v ) r ω* = +(–v ) r – v (–r ) x

y

y

z

z

z

y

x

z

x

x

y

z

y

x

ω* = +ω ω* = –ω ω* = –ω x

x

y

y

z

z

We see that the sign changes in axial vectors are opposite to those of polar vectors. For a reflection of the x-axis (a mirror in the yz-plane), the sign changes are in a polar vector’s x-component and an axial vector’s y- and z-components. This is illustrated in Figure 49-2; note that ω changes sign when it is parallel to the mirror and not when it is perpendicular to it.

Figure 49-2 Mirror Reflection of Axial Vectors

Above, we just reversed the x-axis. If we instead switched from a right-handed universe to a lefthanded universe, every “right” would be replaced by a “left.” In addition to reversing the x-axis, we would also need to switch to the left-hand rule. That would reverse the sign of the cross product, inverting the sign of every component of ω*. Reversing x and using the left-hand rule yields ω**: ω** = –v r + v r ω** = –v (–r ) + (–v ) r x

y

y

z

z

z

x

y

x

z

ω** = –(–v ) r + v (–r ) z

x

y

y

x

ω** = –ω* ω** = –ω ω** = +ω ω** = +ω x

x

y

y

z

z

Upon reversing x and using the left-hand rule, axial vectors change the same way that polar vectors do: only the x-components change sign. Let’s consider another example, an electric charge moving in a magnetic field. We will thoroughly explore electromagnetism later in this course. For now, know that magnetic fields are created by moving electric charges. A bar of iron wrapped with a current-carrying wire makes a simple magnet. By convention, we draw magnetic field lines B emanating from a magnet’s north pole and converging on its south pole. The Lorentz force F exerted by such a magnet on an electron with charge q (q