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Physics 16 Section 1, September 6-7, 2011

1 Getting Acquainted

On your sheet of scrap paper, please provide the following information.

1. What is your name?

2. What is your level of experience with math, what math courses are you currentlytaking, and what is the general state of your relationship with math?

3. Do you like to work with your hands? Have you built or fixed something recently?

4. How comfortable are you with computer programming, LaTeX, and Mathematica?

2 Getting Acquainted with Your Brain

On your sheet of scrap paper, please attempt to answer the following questions. Half-bakedideas, gut feelings, and wild guesses are welcome! You may skip questions that dont inspireyou.

1. Why are clouds white?

2. What is likelier to damage your microwave, a steel ball or a steel fork?

3. Why can you jump higher lowering yourself quickly before jumping than from a staticsquat position?

4. How do .zip files work? Why does .jpg format reduce the file size of a raw bitmap filemore than .zip compression?

5. Prove that a circle has the highest possible ratio of area to perimeter.

6. Why is it that you blow over a glass bottle with a steadystream of air it produces apitch, i.e. a vibratingcolumn of air inside the bottle?

3 Section, Office Hours, Homework

Section is a chance to review lecture, but this usually deserves at most fifteen minutesbecause you can review much more efficiently alone or with classmates. When I do review

in section, it is to summarize pithily ideas from lecture for your memory files and to givethem motivation and intuition. The real purpose of section is to solve problems. Usuallywe will do a warm-up problem and one or two harder problems. I will ask a lot of difficult

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questions on the way to a solution. The most obvious pitfall of easy questions is that they areboring. Furthermore, theres no glory in getting the right answer, while the wrong answer isembarassing. This causes awkward silences, which are a TFs greatest fear. It doesnt costany pride to get a hard question wrong and it feels great to get it right. If your answer has asliver of truth we can run with it and everyone is happy. The same applies if its wrong foran interesting reason. Its not for your ego that I reward partial answers its simply the

way science works. Finally, when you use ideas from lecture in unexpected ways, combinethem, and generalize them, you get a deeper and more confident grasp of them.

Atoffice hours(as well as before and after class) you can talk about anything. If some-thing was presented in class from a point of view that doesnt work well for you we canapproach it from a different angle. Or, you may have a few busy weeks and fall behind. Ifthis happens, dont be embarassed to come in for a detailed review of weeks of material. Myopinions about section do not imply a disdain for review in general. Nothing is too simplefor office hours, including, for example, line-by-line review of lecture slides. When you havethe time, office hours are great for oddball tangents such as rigorous proofs and other areasof physics.

Homework is mainly for your practice, but it is also a dialogue. It shows me how wellyou understand things and give comments accordingly. Sometimes I will just ask you to lookat the solutions, but if you make a very tempting error I try to explain why it doesnt quitework. I also note particularly elegant and original solutions. Basically, check the margins ofyour problem sets.

4 A Bit of Review

So far all we have seen is a bit ofF = ma and solutions to the differential equations thatthis implies.

4.1 Example: Air Resistance

As discussed in lecture, you can model air resistance with a frictional force proportional tothe square of speed. Hence

F =ma= mv2 v2 =dv

dt =

1

v2dv

dt, (1)

where we arbitrarily write the force constant as minstead ofjust to simplify our expres-sions. We have grouped all thev-dependent terms on the same side as the dv/dt becausethen we can integrate both sides using the change-of-variables formula. Integrating from

some initial time ti to some final time tfand changing variables from t to v(t)

tfti

dt=

tfti

1

v2dv

dtdt=

v(tf)v(ti)

1

v2dv =

1

v

v(tf)v(ti)

, (2)

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which reduces to

(tf ti) =

1

v(tf)

1

v(ti)

v(tf) =

v(ti)

1 +v(ti)(tf ti). (3)

4.2 Example: Harmonic Oscillator

As you probably saw in AP Physics, a useful model is a system whose degree of freedom isrestored to equilibrium with a force proportional to its displacement:

ma= md2x

dt2 =F = Kx. (4)

Note that for positiveKthe minus sign is essential. There are systematic ways of deriving thesolution to such an equation, but lets use of more venerable method of guessing a solutionof the form

x(t) =a cos(t) +b sin(t). (5)

Plugging this into the differential equation gives

m2 (a cos(t) +b sin(t)) = K(a cos(t) +b sin(t)) =

K/m (6)

We see that the differential equation has nothing to say abouta and b. They are determinedby the initial conditions. For example, we might have be given x(0) x0 andx

(0) v0. Interms ofa andb we find

x(0) =a a= x0

x(0) =b b= v0/. (7)

5 A Problem, if Time Permits

Problem: Up to some dimensionless prefactor, what is the frequency of the fundamentalmode of vibration of a circular drumhead?Solution: Frequency has dimensions T1. We need to figure out a few parameters thatought to affect the frequency. If were lucky, there will be a uniqueway of combining themso that the dimensions come out to inverse time.

The radiusr, with dimensionsL, ought to matter. Equivalently, we could express thingsin terms of the circumference or area. It seems intuitively clear that the density , withdimensions mass per area (ML2) should also affect the frequency something with lots of

inertia should vibrate slower.The last one is a bit trickier. In one dimension, you know that taut strings have higher

pitches. Thus we need the two-dimensional version of tension, that is, surface tension.Question: What are the dimensions of surface tension?

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Answer: Surface tension represents the amount of energy it takes to stretch a surface,that is, to increase its area. Thus we can define it as energy divided by (area), which hasdimensions

E

A=

ML2T2

L2 =M T2. (8)

So now we need to combine r, , and to get dimensions T1. Only surface tension

involves time, so the answer must contain 1/2, which has dimensions

1/2

= M1/2T1. (9)

Since frequency does not involve any length dimensions, we must combine r and so as tocancel out length. The combination that does this is

r2

= M. (10)

Finally, we take the square root of r2 to get dimensions of M1/2 that cancel the massdimensions in1/2:

r2

= T1. (11)

We conclude (1/r)

/.Question: Did some cheating occur here? Is there a variable we left out?Answer: Perhaps the height h of the vibration makes a difference. In fact, it does.

This ruins everything, because the ratio r/h is dimensionless, and as such you can act onit with any function, not just raise it to various powers. We would then have to include aprefactorf(r/h), wherefis an unknown function. This greatly reduces the predictive powerof dimensional analysis.

However, ignoringh actually isnt so bad.

Question: Why?Answer: We know from experience with the harmonic oscillator that for small oscilla-

tions, with restoring force approximately proportional to displacement, the frequency actu-ally doesnt depend on the amplitude. So in the case of small vibrations, our original answeris not so bad.

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Physics 16 Section 2, September 13, 2011

1 A Bit of Review

Last Thursdays lecture stressed two concepts. The first, conservation of momentum, is

familiar to you from high school. It will come up in todays problems. The second is notsimply vectors, which you know very well, but the subtler idea that it is often useful to ma-nipulate vectors independent of their components, that is, as objects with inherent meaning.The dot product and cross product are very useful tools for the so-called coordinate-freerepresentation of vectors. You can think of this as analogous to solving problems with allthe parameters as symbols and only plugging in values at the end, if at all. It saves timeand increases clarity to preserve generality as long as possible.

2 Problem 1: Sliding off a Hemisphere (Morin 5.53)

Problem: A point particle of mass m slides without friction, starting with infinitesimalinitial speed, at the top of a hemisphere of mass Mand radiusR that moves on a frictionlessplane. At what angle relative to the hemisphere does the particle fly off?Solution: First, note that we can treat this as a 2-D problem the particle moves along alongitude line, which we can say lies in the xy plane. Let vh = vhx and

vp = (vp,x, vp,y)denote the hemisphere and particle velocities. We will find it useful to express some things inthe lab frame of reference and some things in the hemispheres frame, so let vp

= (vp,x, v

p,y)be the particles velocity relative to the hemisphere. Let be the angle of the arc the particletravels relative to vertical, as in the diagram below.

Question: What is the condition for flying off?

Answer: The particle flies off when the normal force vanishes, that is, when gravity justbarely provides the centripetal force mv2/R for circular motion.Question: How do we express this in terms of our variables?

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while the vanishing normal force equation becomes

gR cos =

1 + m

M

2

1 + tan2

v2p,x. (8)

Define 1 +m/M for convenience and eliminate v2p,x to get

2gR(1 cos ) =(+2 tan2 ) gR cos

2 (1 + tan2 )

2(1 cos )

1 + tan2

=

1 + tan2

cos . (9)

Since tan2 = sin2 / cos2 = (1 cos2 )/ cos2 , this is a polynomial in cos :

(1 )cos3 + 3 cos 2= 0. (10)

Question: Are there any cases that we can check this results against?Answer: The one that first comes to mind is an infinitely heavy, i.e. immobile, hemi-

sphere. Then = 1 and we get 3 cos

2 = 0, so = cos

1

(2/3). Personally, I dont haveany reason to think this is intuitive, so lets try a different limit, perhaps slightly silly: aninfinitely massive particle or infinitely light hemisphere. Then and we have

cos3 + 3 cos 2= 0

3cos cos3 = 2, (11)

which implies = 0.Question: Why is this intuitive?Answer: If the hemisphere is infinitely light, it acquires an infinite recoil velocity in-

finitely quickly, hence it scoots out from under the particle when = 0.

3 Problem 2: Drag on a Sphere (Morin 5.86)

Problem: What is the drag force on a sphere of mass M, radius R, and speed v movingthrough a gas of particles with mass m and density (number per unit volume) , assumingthatm


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Question: In as coordinate-free a way as possible, what is the velocity of the gas particleafter collision, in terms of the position vector r of the point at which it hits the surface ofthe sphere?

Answer: Instead of a sphere, we could equally well imagine the particle bouncing off aplane tangent to the sphere and perpendicular to r.

Question: What is a coordinate-free way of saying that the angle of incidence equalsthe angle of reflection?

Answer: The component of v perpendicular to the plane (parallel to r) is reversed,

while the components parallel to the plane (perpendicular to r) are unaffected. We candivide the velocity vector into perpendicular and parallel components with the dot product:

v = (v r) r+ (v (v r) r) = parallel + perpendicular

vf = (

v r) r+ (v (v r) r) . (12)

Hence the momentum change is 2m (v r) r.Question: Can we simplify this by discarding extraneous information and/or exploiting

symmetry?Answer: Only the momentum change parallel to the direction of motion matters other

forces must balance out to zero. To isolate this component we take the dot product with v:

p= 2m (v r) r v= 2mv(r v)2 . (13)

Finally, we can delay the inevitable no longer and we institute a coordinate system. It isnot hard to see that v r = cos . In addition to , we have an angle that rotates aroundv.

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Question: We will find the average momentum transfer and then multiply by the totalrate of collisions. To get the average, do we simply integrate over the half sphere and divideby the surface area of the sphere?

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R2 sin d2

0 d2mvcos2

2R2 (14)

Answer: No! The incoming particles are not evenly distributed over the surface of thesphere. Rather, they are evenly distributed over the circular cross section of the sphere.Looking at the sphere head-on, this cross section can be divided in to annuli of infinitesimalwidthdr:

The area of an annulus at (cross-sectional) distance from the center of the circle is 2rdr,so what we want is R

0 (2r)(2mvcos2 )dr

R2 (15)

We must relate r to . From the diagram below, we see that R sin = r, and therefore

cos2 = 1 (r/R)2. (16)

Thus we have the average momentum loss per collision:

2mvR0 (2r)(1 (r/R)2)dr

R2

=mv. (17)

Question: Finally, what is the rate of collisions?

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Answer: The sphere moves through a distance vdt in time dt, hence it sweeps througha volumeR2vdt. This volume hasR2vdtgas particles, so the rate of collisions is R2v.Therefore, the rate of momentum loss (force) is

F =R2v2m. (18)

Question: Why does this start to fail for a denser medium?Answer: We assumed that each particle is hit once and then goes away forever. However,

it could be stopped by another gas particle and collide with the sphere again. For a liquid,our formula completely falls apart because the liquid molecules grip each other strongly.

Question: We have seen a reason why the drag is greater than our estimate. Is theresome effect that reduces the amount of drag?

Answer: Pressure in the medium mean that isolated collisions are a poor model and abetter approximation is sheets of laminar flow. The angle particles are deflected by wouldthen be much less. In effect, you could imagine an extreme case in which the sphere ispermanently attached to a streamlined narrow wedge of gas particles that it pushes in frontof itself, in which case we would solve the same problem with a wedge instead of a sphere tofind a smaller coefficient fo drag. Obviously, this is not realistic, but the effect exists.

4 Problem 3: Distribution of Velocities in a Gas1 Problem: What are the relative probabilities for molecules in a gas at equilibrium tohave velocity v = (vx, vy, vz)? That is, what is the function f(v) such that the probabilityto have velocity v0,x < vx < v0,x +dvx, v0,y < vy < v0,y +dvy,v0,z < vz < v0,z +dvz isf(v0,x, v0,y, v0,z)dvzdvydvz?

1Thanks go to Eric Kramer for suggesting how to solve this problem.

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Solution: This seems hopeless. When something in physics seems hopeless, a direct attackusually increases the hopelessness. We need something clever.

Question: What are some useful symmetries?Answer: First, no direction is more awesome than any other, so fcan only depend on

the magnitudev =

v2x+v2y+v

2z . Second, motion in every direction is independent, which

means that the probability distribution ofx-velocities is independent of the distribution ofy-velocities. When things are independent, probabilities simply multiply, so we must havefor some function g

f(v) =g(vx)g(vy)g(vz). (19)

Now this is actually a very significant constraint, because there are lots of ways to changethe components ofv in such a way that the magnitude v is unaffected. Then the left sideof this equation doesnt change, and so everything must also work out so that the left sidedoesnt change! To make this perhaps a little starker, take the natural logarithm:

ln f(v) = ln g(vx) + ln g(vy) + ln g(vz). (20)

Question: Now what? Hint: whats a mathematical way to talk about changes?Answer: At this point it is at most a hunch, but lets try taking a derivative. Specifically,

take the partial derivative of both sides with respect to vx, which just means a derivativein which we pretend vy andvz are constants. If you remember the chain rule, you get

1

f(v)

df

dv

dv

dvx=

1

g(vx)

dg

dvx(21)

Nowv =

v2x+v2y+v

2z , so

dv

dvx = 2vx

2

v2x+v2y+v

2z

=vx

v, (22)

whence

1

f(v)

df

dv

vxv

= 1

g(vx)

dg

dvx

1

f(v)v

df

dv =

1

g(vx)vx

dg

dvx(23)

Question: What is very peculiar about this equation?Answer: The left side is a function ofv alone. The right side is a function ofvx alone.

But these are two different variables! The left-side cant depend on vy orvz even though itdepends onv . This is only possible if it is a constant. Calling this constant 2, we get

1

f(v)v

df

dv = 2=

1

g(vx)vx

dg

dvx. (24)

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Question: How do we solve the left of the equations?Answer: Use the separation of variables method that you used in your first homework:

1

f(v)v

df

dv = 2

df

f =

2vdv ln f= v2 +c

f exp(v2) (25)

Question: What is the physical meaning of?Answer: When is large, large velocities are suppressed. Thus 1/ represents the

temperature. To see this more directly, we could use the distribution to calculate the averagekinetic energy of a molecule (which is proportional to the temperature) and we would findthat it is proportional to 1/.

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Physics 16 Section 3, September 20, 2011

1 Crash Course in Multivariable Differential Calculus

1.1 Partial Derivatives

Suppose we have some multivariable functionf(x, y) and want to know how much it changeswhenxorychanges. To do this, we simply differentiate with respect to x(say) in the familiarway and pretend that y is a constant. The definition of such of operation is

f

x(x0, y0) lim

x0

f(x0+ x, y0) f(x0, y0)

x . (1)

We define the partial derivative with respect to y analogously. For example

x

x2 +y2

= 2x,

x(xy) =y,

y(sin(xy)) = x cos(xy). (2)

1.2 Chain Rule

Now that we can take derivatives, lets invent a chain rule. First, a digression about fruitsales. Suppose I sellna apples and np pears at prices pa and pb. Further suppose that thenumber of apples and pears I can sell is proportional to the numberm of sunny days in thegrowing season, so that na(b) = a(b)m. Then the rate of change of my total revenue R withrespect to the number of sunny days is

R

m=apa+ppp. (3)

The way you reasoned this is as follows: each sunny day give a apples. And each one of

those apples gives revenue pa, for a total ofapa. Repeating the logic for pears gives theabove total. Now note that this is just another way of saying

R

m=

R

na

nam

+ R

np

npm

. (4)

This is exactly the same as the single-variable chain rule, but now the independent variablemhas two paths, one viana and one via np, through which it can cause a change in R. Thislogic generalizes: partial derivatives of composite functions are the sum of chain rule-likeexpressions for each path. For example, suppose we have some function f(g1(x, y), g2(x, y)).Then

f

y

= f

g1

g1

y

+ f

g2

g2

y

. (5)

Note that we often get lazy and write the partial derivative symbol when technically some-thing is a single derivative. Finally, you might think the apples and pears example wasunrigorous because it was a linear function. However, any well-behaved function looks linearlocally, so as far as derivatives were concerned our example did not suffer a loss of generality.

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1.3 Vector Operator # 1: Divergence

Quick definition: a vector field is something that assigns a vector to each point in space.Suppose we wanted a measure of how much a vector field is flowing outward. In thediagram below, the right face (colored yellow) of the infinitesimal box faces in the xdirection,so the amount of flow leaving this face is proportional to the x-component Ax of the

field, evaluated at (x+dx, y, z). The amount entering the left face is similarlyAx(x,y,z).Subtracting, we get a contribution

Ax(x+dx, y,z) Ax(x,y,z) Axx

(6)

Adding the contributions from the four other faces we can define the divergence

A Axx

+Ayy

+Azz

. (7)

The dot product notation is a slight abuse of notation but makes sense if we define thenabla operator

x

, y

, z

. (8)

Those of you who have seen the divergence theorem can now appreciate its intuitivemeaning. The flux across a surface is the amount flowing in minus the amount flowing out.If this surface encloses a volume V, then we can subdivided Vinto infinitesimal boxes, andthe flux is the total amount flowing in minus the total amount flowing out of these boxes.But that is simply the integral over the divergence over V.

1.4 Vector Operator # 1: Curl

The curl is also motivated by a tangible quantity: if you were to walk around an infinitesimalsquare, how much would the force field go with or against your motion? In other words,what is the integral of the component of the force field that is tangent to your path?

Lets say your path were in the x-y plane (this is defined as the z-component of thecurl, since paths in this plane wrap around the z-axis). Then the component of the force

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along the path c4 is Fx(x, y+ y), while along c2 it is Fx(x, y). This gives a total ofFx(x, y)Fx(x, y+ y) Fx/y. Adding in the sides c2 and c4, the total amount theforce field F(x, y) pushes you as you go around the path is

F Fyx

+Fyx

, (9)

which defines the curl. As with the divergence, the notation can be taken literally the curl

can be computed as

F=

x y z

x

y

zFx Fy Fz

. (10)

Using this geometric definition of the curl, we can give a non-rigorous but compellingproof of Stokes Theorem, which says that the line integral of a vector field along the bound-ary of a surface (the surface is two-dimensional, possibly curved like the surface of a hemi-sphere, and the boundary is one-dimensional) equals the integral of the curl of the vectorfield over the surface itself.

The proof follows from the picture above. If we add (integrate) the curl over all theinfinitesimal squares, we see that contributions from an edge shared by two squares cancels;

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the clockwise flow is going left for one square and right for the other, or up for one and downfor the other. The only thing that doesnt cancel are edges that arent shared, that is, theboundary of the surface.

A corollary of Stokes Theorem is that a field with zero curl does zero net pushingaround any closed path, not just infinitesimal ones. This explains why conservative forcesmust have zero curl. If not, there would exist some path that would give you more and more

energy the more you walked around it.Another corollary is that the work a conservative force does between two points doesnt

depend on the path. To prove this, suppose we have two paths,1 and 2, from A to B.Then reversing 2 and adding it to 1 gives a closed path that starts and end at A. Thusthe work along 1 minus the work along 2 has to equal zero, which implies that the workalong the two paths is the same. This lets us define the potential energy at pointXas the(path-independent) amount of work it takes to get to X from some arbitrary starting point.The arbitrary starting point only affects the potential by an additive constant.

2 Resonance

Resonance is simple: as we have seen, many systems like to oscillate at a particular frequency.If you push it back and forth at that frequency, it will keep absorbing energy without limitunless some damping force keeps the amplitude in check. If you push it near the naturalfrequency, you will generate a large amplitude.

In section we also talked about the bottle problem posed in section 1. However, it wouldtake too long to code the diagrams. Its easier to explain in person.

3 Conceptual Problem

Complete the following sentence profoundly and concisely:

Non-conservative forces appear to exists even though the fundamental forcesare conservative because...

Answer: Think about friction. An object slows down because it transfers momentum to theatoms of the surface its moving on. If there were only a few of these atoms (and if quantumphysics were turned off), we could keep track of them as a billiards problem. However, thereare unspeakably many atoms, which in turn transfer energy to unspeakably more atoms inthe earths crust. Clearly, its much easier to pretend there is some frictional force acting onthe moving object than to solve for the motion of gondwillions (one gondwillion is defined

as the number of years ago that tectonic activity produced Gondwanaland) of atoms. Henceour answer is

...we cant keep track of all the degrees of freedom.

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4 Problem: Collisions and Vectors

Problem: A moving mass collides elastically with a pair of other masses at rest, somethinglike figure 5.44 in Morin. Assume that the three masses are equal. Show that in a twodimensional collision, it is impossible to have the same non-zero angle between the velocitiesof each pair of particles after the collision. Note that the statement of the problem implicitly

assumes that none of the velocities in the final state vanish, because at zero velocity, thatangles are not defined.Solution: The idea is to use energy and momenum conservation efficiently in vector nota-tion. Again call the initial velocity of mass 1 v. Call the final velocity of massj vj. Thenconservation of energy is (cancelling a factor ofm/2)

v v= v1 v1+ v2 v2+ v3 v3 (11)

and conservation of momentum is

v= v1+ v2+ v3 (12)

Taking the dot product of both sides of (12) with itself gives

v v= (v1+ v2+ v3)(v1+ v2+ v3) =v1 v1+v2 v2+v3 v3+2 (v1 v2+ v1 v3+ v2 v3)(13)

Subtracting (11) from (13) gives

v1 v2+ v1 v3+ v2 v3 = 0 (14)

This is the mathematics we need for both parts. If all the angles are equal, (14) becomes

(v1v2+v

1v3+v

2v3)cos = 0 (15)

wherevj |vj|is the length ofvj.Plane geometry implies that three vectors in a plane can all be at the same angle from

one another only if = 120 or 0. But in both cases, cos = 0, and so (15) implies

v1v2+v1v3+v2v3= 0 (16)

which is impossible if all the masses are moving in the final state. Thus two dimensionsdoesnt work.

Q: Show that this ispossible in three dimensions, and compute .A: This is now easy from (15). We can satisfy it only if cos = 0. This is possible in

three dimensions. The three velocities in the final state are perpendicular to one another.

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5 Problem: More Springs

Problem: A mass is attached to a two-dimensional collection of springs, all with differentspring constants and non-zero relaxed lengths. Describe the motion after it is bonked fromequilibrium.Solution: As we have seen, the relaxed lengths need to be zero for the force to simplify

nicely. What about the potential? The potential of one spring is (k/2)|r r0|2, which isproportional to (r r0) (r r0). Hence the total potential is a quadratic function of thecoordinates. This means that near equilibrium (choose coordinates where this isx = y = 0),we can write

U(x, y) =c0+c1x2 +c2y

2 +c3xy. (17)

Note that there are no terms x or y, since these have non-zero partial derivative and thusyield a non-zero force at x = y = 0. A new feature is the cross termxy, which in generalappears in any quadratic function ofx andy . Differentiating, we get

F(x, y) = U

x

,U

y= (2c1x+c3y,2c2y+c3x) =

2c1 c3

c3 2c2

x

y . (18)

Q: Is it true as it was in the homework that the system oscillates with simple harmonicmotion along the axis of the bonk?

A: No. As we can see, a bonk in the x-direction produces a force in the y-direction.Q: Do there exist any directions in which motion is simple harmonic along a single axis?A:Lets check. We need for some direction nthat the force of motion along nis parallel

ton. That is 2c1 c3c3 2c2

nxny

=

nxny

(19)

for some number. Those of you who know linear algebra will recognize this as an eigenvector

equation. It yields two perpendicular vectorsn1 and n2, the eigenvectors, that are the axesof simple harmonic motion.

Q: What are the eigenvalues 1,2?A:Along the directions n1,2, the force equals the displacement times 1,2. Thus they are

effective spring constants for the two special directions.We have found two types of motion of the system that have a definite frequency. These

are its normal modes. A general motion is a superposition of the two of these, which is aLissajous Figure (see Wikipedia) along axes n1,2.

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Physics 16 Section 4, September 27-28, 2011

1 Review

1.1 Euler-Lagrange Equations

If you can write the kinetic energy T and the potential energy U in terms of some set ofcoordinates {qi} and their time derivatives, then, defining the LagrangianL = T U, theequations of motion are

d

dt

L

qiL

qi= 0 i. (1)

In Morin it is proved that these are equivalent to Newtonian mechanics. The proof isunilluminating and there is no point recapitulating it here. This may seem like pulling arabbit out of a hat, and historically Lagrangian mechanics was just a convenient coincidence.From the modern point of view, the Lagrangian is more fundamental than force. Physicistslike to start with a few symmetries and logical properties that the laws of physics ought to

exhibit and then deduce the laws from these properties. It turns out, as perhaps we may seelater in the semester, that the Lagrangian is an excellent way to encode symmetries.

By the way, you might be skeptical whether taking a partial derivative with respect toq (while holding qconstant) is a legal operation. If you retrace Morins proof, you will seethat what this really means is that you write the Lagrangian in terms ofq and q, pretendthat they have nothing to do with each other, and then take the derivative as a formal(and perfectly well-defined) operation. Since this is the operation that the proof refers to,the Euler-Lagrange equations you obtain this way are valid. Your skepticism is about thenotation, not the equations.

1.2 Conservation Laws and Noethers TheoremThere are two kinds of conservation laws that follow from the Lagrangian method. Theeasy one is that if the Lagrangian depends on qi but not qi for some coordinate qi, then theEuler-Lagrange equations immediately tell us that

d

dt

L

qi= 0. (2)

Hence L/qi is conserved. Conservation laws of this easy variety include conservation oflinear and angular momentum.

These are a subcategory of laws that follow from Noethers Theorem, which goes asfollows. Suppose the Lagrangian (and hence the physics) is unchanged under some transfor-mation law

qj qj() =qj+j+. . . (3)

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3 Waves on a String

Problem: Derive the equation of motion of a string of length L, tension T, and density with its endpoints held in place. You may assume that all motion occurs in a single plane.Solution: The string is essentially a continuous object, while we have only formulatedLagrangian mechanics in terms of finitely many degrees of freedom. To get around this

difficulty, we could concentrate all the mass of the string atNevenly-spaced points and taketheN limit: The mass isL = mN, wherem is the mass of each point, so m = L/N.

The natural choice of coordinates is are the heights yi of each mass, where y0 =yN= 0 arenot free.

The kinetic energy is easy:

KE = m2j

y2j . (10)

The potential energy is the tension times the amount of stretch, which is

PE =Ti

x2 + y2 x

= T

i

x2 + (yi+1 yi)2 x

(11)

We could leave the potential energy like this, but then the equations of motion would benon-linear. In a future section we might discuss the effects of non-linearity. For now, wenote that for small displacements we can take a Taylor approximation to get

PE = T

2xi

(yi+1 yi)2

. (12)

Putting it together,

L=m

2

j

y2j T

2x

i

(yi+1 yi)2. (13)

To write, the Euler-Lagrange equations, the easy part is

L

yj=myj =xyj. (14)

Q:In evaluatingL/yj, what do we have to careful of?A: We have to be careful because we get a contribution when the dummy index i equals

j but also wheni + 1 =j . Thus we have

L

yj=

T

2x

yj

(yj+1 yj)

2 + (yj yj1)2

= T

x[(yj+1 yj) (yj yj1)] . (15)

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Hence the equations of motion are

xyj T

x[(yj+1 yj) (yj yj1)] = 0 (16)

Q: What should we do next?A:Dividing everything by x, the right side of the equation becomes a second derivative

as x 0:

yk T

yk+1ykx

ykyk1x

x = 0

2y

t2 T

2y

x2 = 0. (17)

Q: How do we solve this equation?A:If you guessed that the time dependence of the solution should involve cos(t) due to

the presence of the second derivative, you are correct. But lets digress a little to show whythis is so. This equation clearly has the property of time translation invariance, which

means that ify(x, t) is a solution, then so is y(x, t+ t). Since the equation is also linear,all solutions can be written as linear combinations of some basis yi(x, t) of solutions. Thusif we time-translate some solution yj, we get

T[yj] =k

Mjkyk, (18)

whereTis the time translation operator T :f(x, t) f(x, t+ t) and M is some matrix.Taking the eigenvectors ofM, we get solutions that map to constant multiples of themselveswhen time-translated:

T[y](x, t) =y(x, t+ t) y(x, t). (19)

The only well-behaved functions with this property are exponentials, so we are guaranteeda basis of solutions

y(x, t) =eitf(x). (20)

Q: Furthermore. . .A: . . . the same logic applies to x, so we get solutions

y(x, t) =eiteikx. (21)

Plugging this in, we find that it is a solution iff

2 +T k2 = 0 = kT/. (22)

Taking linear combinations of complex exponentials we get a real-valued solution

y(x, t) = cos(nt+) sin(knx). (23)

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Q: What is one more contraint that fixes the different values kn, and in turn, n?A: We need y(t, 0) = y(t, L) = 0, so kn = n/L. Thus we have determined the normal

modes of the string and their frequencies.Q: What quantity is conserved by virtue ofhorizontaltranslational invariance?A:Before we take the limit N , this is a discrete symmetry and Noethers Theorem

technically doesnt apply. However, it is a continuous symmetry in reality so we expect

the result to be forgiving of a lack of rigor. In terms of the continuous function y(x, t) atranslation is

y(x, t) y(x+, y) =y(x, t) +y

x+ . . . (24)

Back in discrete language, this says that

yi yi+x(yi+1 yi), (25)

from which we read off yi = x(yi+1 yi). Thus we have the consevred quantity

xi (yi

+1

yi)

L

yi =mxi (yi

+1

yi)yi L0

y

x

y

t dx. (26)

If it seems redundant to switch back and forth from continuous to discrete so much, it should.A tool called the functional derivative allows one to carry out the whole analysis withoutdiscretizing.

4 Conceptual Questions

4.1 Bathtubs

Question: When the water in a bathtub drains, it forms a whirlpool. How is this possiblewithout violating Noethers Theorem?

4.2 Throat Singing

Question: It is possible to simultaneously sing a low note and a very high overtone. Whydoesnt changing the low note (essentially, reducing tension in your vocal cords to reduce, which should affect all modes, not just the fundamental one) also change the high note?Note: in section I will try to demonstrate this. If I fail, look it up on YouTube.

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Physics 16 Section 5, October 4, 2011

1 Review

1.1 Partial Derivatives and Their Notation

Suppose we had a function f(x) that depends on x via x2and x3. For example

f(x) = sin(x2 +x3) =g(x2, x3), where g(a, b) = sin(a+b). (1)

We could directly obtain the derivative

df

dx= cos(x2 +x3)(2x+ 3x2). (2)

Alternatively, we could differentiate with respect to g. As discussed earlier, we apply thechain rule to all paths of intermediate variables by which a change inxcan changeg(x2, x3).

Here they arex

2

and x

3

and there are two paths. The formula for what we are doing isdg

dx=

g

1st argument of g

d 1st argument of g

dx +

g

2nd argument of g

d 2nd argument of g

dx , (3)

where we evaluate the partial derivates of g at (1st argument, 2nd argument) = (x2, x3).Henceforth lets call the arguments arg1 and arg2. Then

g

arg1=

arg1sin(arg1 + arg2) = cos(arg1 + arg2)

g

arg2=

arg2sin(arg1 + arg2) = cos(arg1 + arg2). (4)

We can introduce notation that is precise, well-suited for computers, and perhaps somewhatsoulless as follows. Define

g(1,0)(arg1, arg2) garg1

g(0,1)(arg1, arg2) garg2

. (5)

Then, for example,g(2,0)is the second partial derivative with respect to its first argument.The precise statement of the chain rule is then

dg

dx=g(1,0)(x2, x3)

d(x2)

dx +g(0,1)(x2, x3)

d(x3)

dx = cos(x2 +x3)(2x+ 3x2), (6)

as obtained by the direct method. The important thing to take away from this is that wewere in no sense pretending that x2 and x3 were independent of each other. All we said is

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that change inx affects bothx2 andx3, which in turns affects sin(x2 + x3). Confusion comeswhen we get sloppy and write

g(1,0)(x2, x3) g(x2)

g

(0,1)

(x

2

, x

3

) g

(x3) . (7)

On the LHS, it is clear that we differentiate g with respect to its first argument and thenplug in the value x2. On the RHS,it seems that x2 and x3 are two distinct independentvariables.

Incidentally, this lets us make sense of many of the partial derivatives in the textbook.For example, for Lagrangian L = L(q, q, t)

L

t= dL

dt??? (8)

Heres what these two quantities refer to. The partial derivative really means the partialderivative ofL with respect to its third argument. That is

L(0,0,1)(q, q, t)sloppiness L

t (9)

The normal derivative,on the other hand is the derivative of the number L,which dependson t via its first argument q(t), its second argument q(t),and its third argument t. By thechain rule we have

dL

dt =L(1,0,0)(q(t), q(t), t)

dq

dt+L(0,1,0)(q(t), q(t), t)

dq

dt+L(0,0,1)(q(t), q(t), t)

dt

dt

sloppiness Lq dqdt + Lq dqdt + Lt. (10)

1.2 Stationary Action and Extrema of Functionals

Lets review how stationary action leads to the Euler-Lagrange equations using the goodnotation introduced above. We define the actionS: C[t1, t2] R

S[q] t2t1

L(q(t), q(t), t) dt. (11)

This is a very general kind of functional that comes up in all sorts of contexts. Now supposewe want to find the analog of a critical point minimum, maximum, or saddle point. Justas a single-variable function has a critical point when its derivative vanishes,that is, whenyou can change its independent variable by a small amount and get no change up to first

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order in, the condition for some functionq(t) to be a stationary value, given fixed boundaryconditionsq(t1(2)) =q1(2), is

d

d

t2t1

L(q(t) +(t), q(t) +(t), t) dt= 0 s.t. (t1) = (t2) = 0. (12)

We need (t) to vanish at the boundaries because otherwise we would be comparing a nearbyfunction with different boundary conditions. This would be a valid question, but its not thequestion we are asking! We are only maximizing or minimizing or finding a saddle pointwithin the set of functions that have the same boundary conditions at t1(2). There is no apriori reason for this except that a lot of useful problems are of this form. Of course, we knowa posteriori that finding the stationary action subject to the boundary conditions yields theEuler-Lagrange equations.

Anyway, the math proceeds and we cavalierly take the derivative inside the integral sign.Then we apply the chain rule, noting that t does not depend on. The condition forq(t) tobe stationary is (the condition s.t. (t1) = (t2) = 0 is implicit throughout)

t2

t1

dd

L(q(t) +(t), q(t) +(t), t) dt= 0 (13)

t2t1

L(1,0,0)(q(t), q(t), t)d

d(q(t) +(t)) +L(0,1,0)(q(t), q(t), t)

d

d

q(t) +(t)

+. . .

+L(0,0,1)(q(t), q(t), t) dtd

dt = 0 (14) t2t1

L(1,0,0)(q(t), q(t), t)(t) +L(0,1,0)(q(t), q(t), t)(t) dt= 0 (15)

(16)

Now we integrate by parts (

u dv = uv

v du with u = L(0,1,0)(q(t), q(t), t) and dv =

(t) dt, hence v = (t)):

t2t1

L(0,1,0)(q(t), q(t), t)(t) =

L(0,1,0)(q(t), q(t), t)(t)t2t1 t2t1

d

dt

L(0,1,0)(q(t), q(t), t)

(t) dt

(17)

= 0 t2t1

d

dt

L(0,1,0)(q(t), q(t), t)

(t) dt, (18)

where we used the fact that (t1(2)) = 0. Substituting this the stationarity (is that a word?)condition becomes

t2t1

L(1,0,0)(q(t), q(t), t) d

dtL(0,1,0)(q(t), q(t), t)

(t) dt= 0 (19)

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Now this is true for all (t). The only thing that can be multiplied by any function suchthat the product is always zero is the zero function (you should be able to convince yourselfof this hand-wavingly), so we conclude that the thing in brackets is zero:

L(1,0,0)(q(t), q(t), t) ddt

L(0,1,0)(q(t), q(t), t) = 0 (20)

sloppiness

Lq ddt Lq = 0. (21)Thus we have obtained the Euler-Lagrange equation. There are two nifty consequences ofthis derivation. First, we have shown that a lot of interesting optimization problems reduceto the Euler-Lagrange equation. Second, we have shown that Newtonian mechanics, whichis equivalent to the Euler-Lagrange equations, is equivalent to the principle of stationaryaction.

1.3 Some Remarks on Noethers Theorem

1. Land Sdo not really have an intuitive meaning.

2. Noethers Theorem is sort of intuitive within quantum mechanics.

3. Noethers Theorem is really not an important computational tool in classical physics.However, it is very useful in quantum field theory. Furthermore, it provides a new wayofdefining momentum and energy as the quantities that are conserved by virtue oftranslation invariance in space and time. This lets us apply conservation of momentumwhen matter interacts with electromagnetic fields, for example. It also tells us to thinkof energy as the time component of momentum, which is exactly what we will do inspecial relativity.

2 Problem 1: Lagrangian and Small Oscillation

Problem: A particle of mass m slides along a hoop of radius R. The hoop is alignedvertically and rotates at angular speed around a vertical axis running through its center.What are the equilibria of the particles position, are they stable, and what are the frequenciesof small oscialltions about those equilibria if they are stable?Solution: The easiest coordinate to use is the angle along the hoops circumference thatthe particle is displaced from the bottom. There are two perpendicular components of thevelocity. They areR along the hoop and R sin in a plane parallel to the ground due thethe hoops rotation. The height relative to the center of the hoop isR cos , so

L=mR2

2

2 sin2 + 2

mgR cos . (22)

The corresponding Euler-Lagrange equation is

mR2 mR22 sin cos +mgR sin = 0. (23)

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The points of equilibrium have constant , so = 0. We find that either sin = 0, in whichcase = 0 or = , or cos = g/(R2). = is obviously unstable. What about = 0.Then expanding to first order in small we get

2+g/R= 0. (24)

By now we can read off that the frequency of small oscillations is =

g/R 2

, which isreal (stable) ifg/(R2)1. Whats interesting about this is that this is the condition forthe other equilibrium not to exist at all, since cos = g/(R2) 1 is impossible.

Now lets look at the other equilibrium. We have already found a condition for itsexistence; now we must check its stability. Expand to first order in the deviation of from0 = cos

1(g/(R2)): = 0+. Then

sin = sin(0+) = sin(0)cos() + cos(0)sin() sin(0) + cos(0) (25)cos = cos(0+) = cos(0) cos() sin(0)sin() cos(0) sin(0). (26)

Substituting these into the Euler-Lagrange equation gives

2 (sin(0) + cos(0)) (cos(0) sin(0)) +g/R (sin(0) + cos(0)) = 0 (27)Now the part of the LHS that doesnt involve must equal zero, since this is the definitionof an equilibrium (you could of course check this). The first order remainder is

+

2(sin2 0 cos2 0) + gR

cos 0

= 0 (28)

+

2(1 2cos2 0) + gR

cos 0

= 0 (29)

+

2 g

2

R22

= 0 (30)

The freqeuncy is then

2 g

2

R22, which is real (stable) iff

2 g2

R22 0 1 g

2

R24 0 g

2

R24 1 g

R2 1. (31)

This is the same condition as before, so if this equilibrium exists, it is stable.

3 Problem 2: 2-D bubbles

Problem: Show that a two-dimensional bubble subject to pressure and surface tensionforms a circle.Solution: We need to find some energy functional for the system. Perhaps it is

E= circumference surface tension () area pressure (P) (32)

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Q: How accurate is this? Does it matter?A: That the pressure decreases to zero as the circle expands is the glaring error. This

shouldnt matter because the essential qualitative feature is that there exists some isotropictendency to expand. It is not hard to generalize this to handle pressure as a function of area.

Q: Are we looking for a global minimum?A: In general we would do this, but with our caricature of pressure we can expand the

shape by some factor x so that the circumference, which goes as x, is dwarfed by thearea, which goes as x2. Thus the energy becomes infinitely low as x increases. However,we hope that there will be a local minimum where area and circumference are balanced incompetition.

Q: How do we formulate the problem? What coordinate system? What boundary con-ditions?

A: The simplest way to represent a closed curve is in polar coordinates with the angle going from 0 to 2. For the curve to close we need r(0) = r(2) = R for some R. We arethen hoping that our Euler-Lagrange equation can be solved by r() =R = constant.

Q: Whats the problem with this?

A: The physics will be controlled by the ratio /P. For some value of this ratio andfixed boundary condition R, there is no guarantee that the solution has radius R. It couldbe a circle of a different radius, and then we would have to try a solution that is a circle notcentered at the origin.

Q: Is there a better way?A: Yes there is. We know that for some value of the ratio /P, the radius will be R.

Thus we can plug r() = R and then solve for the /P that makes this work. If we want,we could then invert this relationship to get the radius R in terms of the parameters.

Now we can start. You may recall that the area in polar coordinates is

A= 2

0

1

2r2() d (33)

Q: What is the circumference?A: The infinitesimal length can be found from the Pythagorean theorem with one per-

pendicular direction along increasing r (dl= dr) and another along increasing (dl= rd).Then

dl=

dr2 +r2d2 =

r()2 +r2d L= 20

r()2 +r2d (34)

Then

E 20

P

r()2 +r2 1

2r2()

d (35)

The Lagrangian is simply the integrand, and the Euler-Lagrange equation is

P

d

d

r()r()2 +r2

P

r()r()2 +r2

+r() = 0 (36)

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Plugging inr() =R andr (= 0), we get

R P

R

R= 0

P =R (37)

We have shown that the shape is a circle and we have found how the radius depends on thephysical parameters.

4 Challenge Problem

Find the function that maximizes (absolute maximum) the functional

V[f] =

10

x0

f(t) dt

2dx (38)

subject to the constraint

1

0 f2

(t) dt= 1. (39)

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Physics 16 Section 6, October 11-12, 2011

1 Review

Unlike previous weeks, the new material is not mathematically difficult but is conceptually

bizarre. Thus we will do something new and integrate conceptual questions into the review.Anyway, relativity starts from two postulates:

1. The laws of physics are the same in all non-accelerating reference frames.

2. The speed of light (in a vacuum) is a law of physics.

Now this seems very nice, but something really ought to bother you about the secondpostulate.

Q:What is it?A:Whats so special about light that its speed is a law of physics? After all, an electrons

speed is not a law of physics electrons move at all sorts of speeds. If thats too abstract,

yourspeed is not a law of physics. Hence,Q:Why is the speed of light a fundamental law of the universe? You should think about

it for a few minutes, after which the hint is: think hand-wavingly about the Newtonianmechanics of massless objects.

A1: If you apply F =ma to a massless object, you will find that a massless object thatinteracts with other stuff with any force at all experiences an infinite acceleration. This isuntenable and requires that we somehow change the laws of physics so that speeds dontbecome infinite. We could do this in three ways give every massless object its own speedlimit with its own ad hoc reason for having that speed, invent some fancy new speed limitlaws, or simply impose the same universal speed limit on all objects. The third is the mostaesthetically satisfying.

A2: Using only Maxwells equations, one can derive a wave equation for light which isidentical to the equation of a string we found two weeks ago:

2

x2+

1

c22

t2 = 0, (1)

where is some component of the elctric or magnetic field. The actual equation is the 3-Dgeneralization of this, but thats not important. We solved this and found standing waves,but you can equally well get travelling waves. Try a solution (x, t) = f(x vt) for somefunctionf. This represents a shape f(x) travelling at velocity v . Plugging in, we get

f(x vt)v

2

c2 f(x vt) = 0. (2)

Hencev =c. Since Maxwells equations are laws of physics, the corollary that electromag-netic waves travel at the fixed speed c, which can be computed from the permeability andpermittivity of free space, must also be a law of physics.

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frame counts the number of black ticks and vice versa. The problem is that this relies onthe right blocks and left blocks lining up simultaneously, which can be true in one frame andfalse in the other. Thus there is no paradox, althought there is still weirdness.

Another thing we introduced was the idea of spacetime coordinates and a spacetimeinterval. If we have a point xand a timet, then we can combine them into a four-dimensionalobject (t, x). This is just convenient notation for a theory where space and time get mixed

up. Then the spacetime interval between two events (t1, x1) and (t2, x2) is just the difference(t2, x2) (t1, x1). In Morin, the way to switch between different reference frames is derived.Suppose reference frame S moves at velocity vxwith respect to reference frame S. Then ifframeSmeasures some two events separated by the interval (x, t), then frameS measuresit occurring at coordinates (x, t), where in units c= 1

x =(x vt) (3)

t =(t vx) (4)

y = y (5)

z = z (6)

This is called a Lorentz transformation and can be expressed as the matrix equation

t

x

y

z

=

v 0 0v 0 0

0 0 1 00 0 0 1

txyz

. (7)

Q:Without doing any calculation, what is the inverse of this matrix?A: This matrix gives coordinate moving at velocityvxrelative to our original coordinates.

Moving with velocity vx restores the original frame, so the inverse is given by takingv v above.

Q:If the spacetime interval between two events is zero, the linearity of the transformationimplies that it is zero in every reference frame. Whats the common sense reason for this?

A:If you are walking through Harvard Square, and a fish falls from the sky and hits you,then in your reference frame the events of you being in Harvard Square and a fish being inHarvard Square had zero spacetime interval between them. But one would hope that thefish hits you in all reference frames, which requires zero spacetime interval.

There is an odd generalization of distance in Euclidean space, which is invariant underrotations, to four-dimensional spacetime. It is defined as

s2 c2t2 x2 (8)

and you can check that it is invariant under Lorentz transformations. In words, s measuresthe time surplus relative to how long it takes light to travel a certain distance in space. Thenifs2 is positive it means light could travel the spatial interval in less than the time interval.If two events are separated by such an interval, this means that the earlier can affect thelater one.

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Q:How does this logic work?A:All forces are transmitted by particles travelling at or slower than c, so if two events

are too far separated in space for light to traverse the interval in time, one cannot influencethe other. conversely, if the distance in space is not too great, one can influence the other.

Q:What worries you about something I said recently?A:The earlier one can affect the later one. Which event is earlier might depend on the

reference frame, in which case the order of cause an deffect is switched! Fortunately, you canshow that ifs2 is positive then Lorentz transformations do not affect which event is earlierand which is later.

In lecture it was pointed out that time and space arent as different as we perceive. Forexample, we only experience one instant of time at once, while space seems to be spread outbefore us all at once. However, really we only experience one pointin space at once as well.Photons (sound waves, wafting scents, etc.) emitted at different points in space and previoustimes travel through time and space to reach us, so we see the past. however, because lighttravels so quickly we think we are seeing an instantaneous snapshot of all points in space.This brings up a conundrum.

Q: If you can only observe one point in spacetime, then how is a sense of self possiblewhen our brains are extended objects and not localized to a single point?A:I have no idea!Finally, since a lot of people mentioned this on their QAs, lets talk a bit about rapidity.

Unlike some things we have dealt with recently, rapidity has a very clear physical meaning.In Newtonian mechanics, if an object is pushed by a constant force, its velocity increaseslinearly with time. In relativity, an object pushed by a constant force (that is, it experiencesa constant acceleration in its rest frame) has a rapidity that increases linearly with time.Basically, rapidity measures how much something has been pushed, but because there is afundamental speed limit c, to get a speed you have to smush the interval [0,) of possi-ble rapidities into the interval [0, c] of possible speeds. As you can check for yourself, the

hyperbolic tangent is such a smushing function.To prove that its a hyperbolic tangent, let be the total amount at that an object has

accelerated according to its rest frame and let it accelerate in its rest frame an infinitesimalamount . Then by velocity addition

v(+ ) = v() +

1 +v() =v() + (1 v()2)) +O(2). (9)

Thus

v() = lim0

v(+ ) v()

= 1 v2 v= tanh (10)

I uploaded a document that derives Lorentz transformations from the invariant intervalonto the website, and it shows the mathematical significance of the rapidity.

Lets do some problems.

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2 Problems

2.1 Problem 1: Lorentz invariance of wave equation

Problem: Show that the wave equation is Lorentz-invariant. That is, if

2x2

2

t2

= 0 (11)

is true in one reference frame, then the same equation is true if we replace (t, x) (t, x).Solution: This calls for the chain rule, which we will apply to second derivatives for thefirst time. We can say that depends on (t, x) via the intermediate variables

x =(x vt) (12)

t =(t vx) (13)

Then the chain rule says that

f

x=

f

xx

x +

f

tt

x (14)

=f

x v

f

t. (15)

As an operator equation, this says that

x=

x v

t

. (16)

Likewise, we find that

t

=v

x+

t

. (17)

Hence

2

x2

2

t2 =2

x v

t

2

v

x+

t

2 (18)

=2

1 v2 2

(x)2

2

(t)2

+ (2v 2v)

2

xt

(19)

=2(1 v2) 2

(x)2

2

(t)2 (20)=

2

(x)2

2

(t)2

. (21)

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2.2 Problem 2: Taste of Relativistic Electromagnetism

Problem: One of the postulates of relativity is that the laws of physics are the same in allinertial reference frames. It would certainly contradict this if a force existed in one frameand not in another. But wait... do you remember magnetism from AP physics?

Q:Whats the problem?

A:Magnetic fields are produced by and act on moving charges. A charged particle thatis moving in one frame and experiences a magnetic force is motionless in some other frame.The force is no more.

Q:Whats a possible resolution?A:As we will see, the forces remain the same, but in one frame they are called magnetic

and in another they are called electric. In other words, given relativity and electrostatics,magnetism must exist. If thats not compelling enough for you, I dont know what is. Letsnow state the quantitative part of our problem.

Problem: Suppose we model an electric current as a very long line with density ofpositive charges moving at velocity vzand a density of negative charges moving at velocityvz, for a net current ofI= 2v. This produces a magnetic field

B(r) = 0I

2r. (22)

If a charge q is a distance r from the wire and moves at a small velocity uz, it experiencesa magnetic force quB, which you can check points toward the wire for positive u and q.How can we understand this in the rest frame of the charged particle?Solution: We have no magnetic field, so somehow the wire must produce an electric forceacting on the charged particle. However, the wire is neutral, so it seems there can be noelectric force.

Q:Whats going on?

A:Perhaps the wire is not so neutral after all. Consider the positive and negative currentcarriers separately, since they are moving with different velocities. We can use the velocityaddition formula to obtain the velcitiesv of the positive and negative carriers in the frameof the particle (the lab frame has velocity -u wrt. the particle, to which we add velocitiesv wrt the lab frame):

v=u v

1 uv (23)

Common sense and the above formula tell us that for u >0 the negative carriers, which aremoving opposite the direction of the particle in the lab frame, have a larger velocity in theparticles frame. Thus|v|> |v+|.

Q:Whats the next step?

A: Consider some length-L chunk of the wire, and consider the positive and negativecharges separately. They are length-contracted by factors

L= L/= L

1 v2 (24)

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This means that their densities are increasedby the same factors

= /= /

1 v2 (25)

Since the negative carriers move faster, > + and their density in enhanced more. Thusthe charged particle sees a negatively charged wire with net charge density(+). This

confirms that forq >0 andu >0 the particle is attracted to the wire.

2.3 Problem 3: Trains, Conductors, etc.

Problem: A train has length L in its rest frame and is moving with speed vtxwith respectto a station, wherev >0. A conductor walks to the right end at speed vcrelative to the train.His dog also starts at the left end and runs to the right end, back to the conductor, back tothe right end, etc. at speedvd > vc relative to the train. The conductor has an extremelyaccurate biological clock to measure the dogs age at both ends of the train. According tothe conductors clock, how much does his dog age?

Solution: The first thing to realize is that the station frame is irrelevant. The secondthing to do is to translate this as a physics question.Q:What is the question?A:It is, how much time elapses in the dogs reference frame?Q:How can we obtain this?A: We could find the time in the trains reference frame and then use the fact that

the dogs clock runs slowly. Its essentially the twin paradox with the dog replacing thespace-traveling twin and the train replacing the Earth.

Q:In the trains reference frame, how much time elapses?A:Considering the conductors path, it is L/vc, so the time elapsed in the dogs frame is

Lvc

= L

1 v2d

vc. (26)

Q:Does it matter that the conductors frame is not the same as the trains frame?A:No. The clock is moving with respect to the train , but its measurements are perfectly

valid spacetime events in the train frame. Of course, we could solve the problem in theconductors frame too, but it would be more complicated.

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1 Review

1.1 4-vectors

A 4-vector is defined as a collection of four numbers A = (A0, A1, A2, A3) = (A0,A) thatmix up with the same Lorentz transformation that mixes up position and time as

A0= (A0 vA1) (1)A1= (A1 vA0) (2)

when you switch between reference frames with relative velocity v. This is analogous toregular vectors in two and three dimensions. For example, in two dimensions the positionvector (x, y) transforms as

x

= cos x sin y (3)y = sin x+ cos y (4)

when you rotate your coordinates by an angle . Then velocity and acceleration are vectorsbecause the same rule works if you replace (x, y) by (vx, vy) or (ax, ay). In contrast, (Moose,Squirrel) is not a 4-vector, since when you rotate your coordinates you do not get (Moosecos + Squirrel sin ...)

We can divide both sides of the above equations by some number that does not dependon the reference frame to get an identical transformation law for A/, which shows thatA/is a 4-vector. This seems a little useless it tells us that 2Ais also a 4-vector. However, letx() = (t,x) be the 4-vector describing a particles spacetime coordinates in a laboratory

frame as a function of the particles proper time . Thenx(+ d) x() is also a 4-vectorfor any d. Now dis the same in every frame it is defined that way so we can choose= dand take the limit 0 to get that

v x.d

=dx

dt

dt

d =(1,v) (5)

is a 4-vector. Multiplying by the particles mass m we get another 4-vector

P (m,mv). (6)

Taylor expanding gives

P0m+12mv2, (7)

while Plooks like regular momentum with a factor ofslapped on to let it become infiniteeven thoughv is bounded byc. Furthermore, Morin shows that both of these quantities are

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conserved in simple collisions. Thus heuristically it seems like they ought to be the enrgyand momentum, so that P = (E, p). More on that later.

Just as t2 x2 is invariant as a consequence of how Lorentz transformations work, thesame must be true for any 4-vector:

A20

A2 = constant. (8)

In particular, for the energy-momentum 4-vector

P20 P2 =E2 p2 =m22(12 v2) =m22

2 =m2. (9)

This is actually just a special case of the invariance of the 4-vector inner product. Defining

A B A0B0 A B, (10)

one can show that A B = A B is the same in any inertial reference frame. Then forenergy-momentum we have

P2 P P =m2. (11)For a single particle this is a useful algebraic relation between E, p, andm, but we can useit for the total momentum of a system, since the sum of 4-vectors is a 4-vector. Then we getthe mass of the system

m=P2tot, (12)

which says that the mass of a system is the energy in its zero-momentum frame (P0 = mwhen v = 0 = 1). Said another way, mass is defined as the energy something has

just for existing, and not for doing anything. The unsettling consequence of this is that themass of a system of particles is notsimply the sum of the masses of its constituents. If this

doesnt bother you, you are the most sanguine physics student in history. Lets try to makeit bother you less.Q: Basically the problem is that you think of mass as essentially stuff. Masses should

add together in the same way that 4 ducks and 5 ducks make 9 ducks (or 4 atoms plus 5atoms etc). How do you get around this?

A:You skirt the issue completely by stating that stuff does not exist! On a subatomiclevel there is no such thing as a solid lump of stuff, just point particles and empty spaceor fuzzy wavefunctions, depending on your point of view. Thus there is no sense in whichtwo electron lumps make a lump thats twice as big. So we have gotten over this issue,with the price that now we realize that in some sense only abstract mathematical objects,notthings, actually exist.

Now maybe it bothers you that something has energy just by existing, without doinganything. For a composite particle its not a big deal, since in the zero-momentum frameof a proton there are a lot of internal motions of the quarks, hence the proton is doingsomething. But for a fundamental particle like an electron, it seems weird that somethinghas energy in the absence of dynamics.

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and time, so the machinery we developed earlier gives

E=i

L

xixi L (14)

=i

mx2i1

x2

+m

1 x2 (15)

=mv2+m/ (16)

= (m/)(1 +v22) (17)

= (m/)

1 +

v2

1 v2

(18)

= (m/)1 v2 +v2

1 v2 =m2

=m (19)

and

pi = L

xi=mxi (20)

These are the energy and momentum that we found earlier. The relativistic Lagrangian hasnow served its purpose.

2 Problems

2.1 Pair Formation

Problem: Two photons of energy E1 and E2 collide at an angle . What is the minimumpossible E1 to produce of two particles of massm?Solution: Instead of immediately crunching a lot of kinematics, lets make our task easier.

Q: What is the least possible total energy of the two emitted particles? Hint: theirtotal momentum is fixed by momentum conservation, so the question is: givenp, how do we

minimizeE?A: For fixed total momentum, energy is minimized by having the least possible internal

dynamics of the two-particle system, that is, the least possible energy in the zero-momentumframe. This is achieved if the two particles have the same velocity. Then in this best-casescenario, we essentially produce one particle of mass 2m. LetPbe the 4-momentum of thisfictitious particle. Then

1+2 = P, (21)

where1(2) are the photon 4-momenta.

Q: What are these?A: Since E= |p| for photons, we can choose coordinates where

1= (E1, E1, 0, 0) (22)

2= (E2, E2cos , E2sin , 0) (23)

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Q: Now what?A: Square both sides of 4-momentum conservation and use 2 = 0 to get

(1+2)2 =P2 (24)

21 2 = (2m)2 (25)12 = 2m

2 (26)

E1E2(1 cos ) = 2m2 (27)

E1 = 2m2

E2(1 cos ) . (28)

2.2 Photon Rocket

Problem: A rocket propels itself by emitting photons. When it starts from rest it has massm0 and eventually reaches 4-momentum (E, p) and then decelerates back to rest, again byemitting photons. What is its final mass?Solution: By conservation of momentum, the photons emitted during the acceleration

phase have total momentump, and hence have 4-momentum (p,p). Thus 4-momentumconservation gives

(m0,0) = (p,p) + (m1,p) (29)m1 = m0 p, (30)

where m1 is the mass after acceleration. By the same logic, the deceleration photons havemomentum p and carry off energy p, which is reflected in the rockets loss of mass, so

m2= m0 2p. (31)

2.3 Morin 13.11

Problem: Three particles go off at equal speedsv at angle 2/3 with respect to each other.What is the angle between any two particles velocities in the rest frame of the third?Solution: The strategy for problems like this is to (1) write out relevant 4-vectors, in thiscase the velocity 4-vectors, in terms of unknowns in the desired frame; (2) take inner productsof 4-vectors in this frame and relate them to the desired unknown quantities; (3) calculatethe inner products in whatever frame is easiest. In the rest frame of particle 3 (frame 3),

V 1 =(1, v, 0, 0) (32)

V

2 =

(1, v

cos , v

sin , 0) (33)V 3 = (1, 0, 0, 0), (34)

wherev and are the speed and associatedfactor of particles 1 and 2 in frame 3. Thereare only two independent unknowns, since ()2 = 1/(1 (v)2). Next we take some dot

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products there are only two interesting ones:

V 1 V 2 = ()2(1 (v)2 cos ) (35)V 1 V 3 =. (36)

The second of these is really convenient since we have an expression for without any

algebra. Using also the relation

(v)2 = 1

()2(()2 1), (37)

we get

V 1 V 2 = (V 1 V 3)2

1 1(V 1 V 3)2

(V 1 V 3)2 1

cos

(38)

Now we exploit invariance of the inner product to replace the primed inner products byunprimed ones. Furthermore, symmetry lets us say

V

1 V

2 =V1 V2 = V

2 V

3 =V

2V3 . (39)then we must compute V1 V2 and solve

= 2

1 122 1 cos

(40)

cos =

+ 1 (41)

So finally, what is the inner product? As in the tetrahedron problem from a problem set,the sum of velocity 3-vectors is zero and the square of a velocity 4-vector is 1. Thus

(V1+V2+V3)2 = 3V21 + 6V1 V2= (3, 0, 0, 0)2 (42)3 + 6= 92 (43)

=32 1

2 (44)

(45)

Hence

cos =321

2321

2 + 1

=321

232+1

2

=32 132 + 1

(46)

=3

1v2 1

31v2

+ 1=

31+v2

1v2

3+1v2

1v2

=2 +v2

4 v2 (47)

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1 Review

Theres not much new this week. We have the definition

F= dpdt

(1)

and its corollarydE= F dr. (2)

We learned a bit about relativistic rocket motion, which is essentially just conservation ofmomentum. Relativistic strings are a just a matter of connecting objects with a constanttension. That doesnt imply that your homework this week is easy, just that a review ofconcepts wont be as useful as going straight to the problems.

2 Problems

2.1 Three Quarks

Problem: Three massless quarks are in the same place and go off with equal energies Eatequal angles from each other. They are connected by massless strings of tensionT. What isthe period of the resulting motion?Solution: The time it takes to stop, which for particles travelling at speedc = 1 is equal tothe distance it takes to turn around, is one quarter of a period. Thus the period is four timesthe distanced from the center of an equilateral triangle of side lengthLto one of its vertices,whereL is the length of each string when the quarks first turn around. By conservation of

energy, we need 3E= 3LT, so L = E/T. Some geometry gives d = L/3, so

P = 4E

T

3. (3)

2.2 Laser-Driven Rocket

Problem: A rocket ship is driven, starting from rest, by a laser beam of powerw shot fromEarth. The ship varies the fraction of photons it absorbs by spreading photon-absorbingsails so that it experiences a constant acceleration g in its own frame. What is(t) accordingto the Earths frame?

Solution: First, we need a plan of what information we need to calculate in what order.Lets work backwards. To get(t), we need to know the rate at which the ship consumesenergy. To get energy consumption, we need to know its energy as a function of time. FromE = m, we need to know its speed v(t) and mass m(t) as a function of time. Since itsacceleration is known, in one frame at least, it seems like v(t) cant be too hard to obtain.

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Perhaps m(t) will be a bit harder, but maybe once we have v(t) we will have a chance. So,a loose plan is

1. Getv(t) somehow.

2. Getm(t) somehow.

3. GetE(t) and dE/dtstraightforwardly.

4. Findthe rate of laser energy available to ship.

5. Take the ratio.

Q: To get you warmed up, tell me whats wrong with this argument: The rapidity

is just the integral of acceleration in the ships frame, so (t) =

t

0

mg dt = mgt, so

v(t) = tanh(mgt).A: The problem is that the upper limit of integration that defines rapidity is the ships

proper time, not the Earth time t. And you cant just slap on a factor because the speedis not constant.

Heres an efficient way to get v(t). First, convince yourself that if you know the acceler-ation as a function of time, then you know the velocity as a function of time, independentof the mass. Thus to get v(t), we can pretend we have an object of constant mass m thatfeels acceleration g in its own frame. Then the force is F =mg in its frame, and since therelative vecloity of the frames is collinear with the force, this is also the Earth frame force.Now apply F =dp/dtto get

F =mg =dp

dt p= mgt = mv (t)v(t) =gt. (4)

Hencev(t)

1 v2(t) =gt v(t) = gt

g2t2 + 1. (5)

Now we need to find m(t).Q: How can we get m(t) quickly. Hint: all the momentum and energy comes from

photons.A: Since energy and momentum come from photons, the increase in one equals the

increase in the other. the ship starts from rest with E= m0,p = 0, so we must always haveE=p + m0. Now we plug this into the definitions ofEandp, which involve m(t) andv(t):

E(t) =(t)m(t) =p(t) +m0= (t)m(t)v(t) +m0 (6)

m(t) = m0(t)(1 v(t)) (7)

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The next step is rote plugging in:

E(t) =m(t)(t) = m0

1 v(t) (8)dE

dt =

m0dv/dt

(1 v(t))2 (9)

Finally, we divide this by the rate at which energy reaches the ship in the Earth frame.This is not simply w.

Q: Why?A: Because the laser beam travels at speed c = 1, while the ship travels at speed v(t).

Thus energy reaches the ship at a reduced speed w(1 v(t)), so we have

(t) = dE/dt

w(1 v(t))=m0

w

dv/dt

(1 v(t))3 =m0g

w

gt+

g2t2 + 1

3

. (10)

2.3 Why Field Theory is Necessary

Problem: Construct the Lagrangian of two interacting relativistic particles.Solution: Thats easy, just take the free particle kinetic part and add an interaction:

S=

i

mi

1 x2

idt

U(|x1 x2|) dt (11)

Q: Whats wrong with this?A: Two things. First, the distance between the particles is not Lorentz-invariant. Second,

this says that particles separated by a spacelike interval can influence one another. To solveit, we could replace the spatial distance by the invariant interval, and then require thepotential to vanish for spacelike intervals:

S=

i

mi

1 x2

idt

U((t1 t2)2 (x1 x2)2) dt1dt2 (12)

Unfortunately, it seems that we must have a double integral over time, since different times t1andt2 interact. We no longer have an action expressed as the time integral of a Lagrangian.Worse, by making the interaction Lorentz-invariant, we have lost the Lorentz invariance ofthe action, sincedt1and dt2are not invariant. To fix the first problem, we could say that weonly want an interaction that applies when t1= t2, but if we need the condition t1 t2= 0to hold in every frame, we also need x1 x2= 0, so that we need something like

Sint= L(0, 0(ifx1= x2)

0 (otherwise) (13)

In addition to our queasiness with the discontinuity inherent in having an interaction thatsuddenly turns on for particles that are in exactly the same place, we still dont have Lorentzinvariance since dt is not invariant.

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Q: What could we do instead of

dt? Hint: what is the determinant of a Lorentz

matrix?

A: the quantity dtdxdy dz is Lorentz invariant, so we could integrate over space andtime. To verify this, recall that

dt dx dy dz =dtdxdy dz

dt

dt

dx

dt

dy

dt

dz

dtdt

dx

dx

dx

dy

dx

dz

dxdt

dy

dx

dy

dy

dy

dz

dydt

dz

dx

dz

dy

dz

dz

dz

(14)

Because the Lorentz transformation is linear, the matrix of derivatives (the Jacobian) issimply the matrix of Lorentz transformation coefficients. For boosts along the x-axis, thiscomes outs to:

v 0 0v 0 00 0 1 00 0 0 1

=2(1 v2) = 1. (15)

Hencedt dx dy dz =dtdxdy dz. The interaction part of the action now looks like

Sint=

Lint((x), (x)) d

4x, (16)

wherex = (t,x,y,z) andd4x dtdxdydz.We now have a theory with a degree of freedom defined at each point in spacetime.

The spatial coordinates are just dummy variables. This means we have a field theory. Forexample, an electric field does not have a position. Rather, it has some value at every pointin space, and its Lagrangian is obtained by integrating over all of space.

Finally, the form we gave above isnt quite right because /tis not invariant. However,the combination

2

t2

2

x2

2

y2

2

z2 (17)

is invariant. Then one simple action for a relativistic field is

S= 2(x) +

2

t2

2

x2

2

y2

2

z2

d4x. (18)

2.4 Relativistic Bucket

See Morin problem 12.19.

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Physics 16 Section 10, November 8, 2011

1 Review

1.1 Tensors

By explicit calculation, one finds that the angular momentum of an object is related to itsangular velocity by

L= I, (1)

where I is a symmetric 3x3 matrix that depends on the geometry of the object and withrespect to which point L is measured. In general, you calculate it by doing tedious integrals.

Okay, I lied a little. Technically, I is a tensorwhich is represented as a matrix when wechoose a particular coordinate system. Just as the vector one meter in length that points infront of you could be (3, 3, 1) or (4, 6, 7) etc, but always points in front of you, a tensor hasan abstract meaning independent of any coordinate system. Lets explore what makes it atensor. One definition of a vector is that it transforms in a certain way. In special relativitywe saw that 4-vectors transform by Lorentz matrices when we change coordinate systems.Similarly, a 3-vector is an object that transforms via rotation matrices, which I assume mostof you have encountered before. That is, for each rotation there exists a matrix R such that

x =Rx, xi =j

Rijxj (2)

relates the vectors coordinates in the original and rotated frames. Now L and are vectors,and

L= I, Lj =j

Iijj (3)

ought to be true in any coordinate system. Thus we must have

L =I (4)

RL= I(R) (5) L= (R1IR)) =I (6) I =R1IR (7)

We can actually do a little better using the fact that rotation matrices leave angles andmagnitudes invariant, and hence do not affect the inner product. This is analogous to the4-vector inner product of special relativity, which normal human beings learn about after

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covering 3-vectors. the conseuence of invariance is

a b =a b (8) (Ra)T(Rb) =aTb (9)

aTRTRb= aTb (10) R

T

R= Id3x3 (11) R1 =RT (12)

Hence

I =R1IR (13)

=RTIR (14)

Ijk = RTIRjk =m,n

RTjmImnRnk (15)

= m,n

ImnRmjRnk (16)

This looks like two separate matrix multiplications by the rotation matrix, one for thefirst index and one for the second. This transformation defines a tensor. If you have everencountered the tensor product in your math classes, this definition is equivalent, for supposeyou have a tensor product x = v w, xij viwj. Then to get its value under a change ofbasis, you would transform the two vectors separately:

xij =viw

j =

m,n

RimRjnvmwn=m,n

RimRjnxmn (17)

In short, a tensor transforms like the outer product of two vectors.

1.2 Inertia Tensor Meaning

Mathematically, the existence of three orthogonal principal axes, where angular momentumand angular velocity are parallel to one another, follows from the spectral theorem, whichsays that any real symmetric matrix (or complex matrix equal to its own conjugate transpose,or under certain conditions, self-adjoint infinite-dimensional linear operators) has a completeset of eigenvectors. The unintuitive existence ofnon-principal axes follows from the fact thatnot everything is an eigenvector! Now lets try to understand things more physically. Whatfollows is my personal interpretation, which I hope you will find helpful.

Since angular momentum is conserved while angular velocity is not, you can think of as a temporary conditions that holds only as long as rotation around a certain axis isimposed. A system may want to rotate around one axis based on its angular momentumbut may have to rotate around another axis, for example if a rod is stuck through it. Ifthe rod is removed instantaneously, L does not change but certainly may. Thus, at least

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one of the angular velocities is not parallel to the angular momentum. Consider just aboutthe simplest example: a globe. Suppose the globe is a perfect sphere and has a rod goingthrough the (geographic, not magnetic) poles. By symmetry the rod is a principal axis. Nowlets ruin this by adding two really big identical mountains at antipodal points, for exampleLake Baikal and Tierra del Fuego.

Q: How do these two mountains want to move?

A:You know in your heart that centrifugal forces will try to push the mountains towardthe equator, which the globe could achieve by rotating. If the rod were removed, that isexactly what the globe would try to do. Thus we see a tangible case where a torque has tobe applied to maintain the direction of angular velocity.

Now suppose these two mountains are in the x zplane, os that their coordinates are(x, z) and (x, z). Then their contribution to the inertia tensors off-diagonal element Ixzis mxz+m(x)(z) = 2mxz. The off-diagonal element is non-zero, which represents thatthese are not principal axes.

Q: How could we locate the mountains so that the off-diagonal inertia tensor elementscancel?

A:If the mountains were related by reflection symmetry, that is, at the same latitude butopposite longitude, the contribution would be mxz+ m(x)z= 0. In fact, if we had a lot ofmountains, but every mountain at (x,y,z) had a clone at (x,y,z), the off-diagonal elementwould vanish. Physically, our centrifugal force argument doesnt work because you cantmove them both closer to the equator simultaneously, so they remain at equal distancesfrom the equator. The off-diagonal elements of the inertia tensor cancel to zero due tosymmetry. We can turn this around and say The off-diagonal elements of the inertia tensormeasure the amount of asymmetry of an object. We can further say that the principal axesdefine rotational motion where all the centrifugal forces balance each other.

1.3 Addition of Angular Velocities

The proof that angular velocities add is quite simple. Lets go over it.Theorem:Suppose object A is rotating around some point R with angular velocity A

and object B is rotating with angular velocity B with respect the same point R on objectA. Then object B is (instantaneously) rotating around Rwith angular velocityA+B.

Proof: When object B is at point r, it is moving with velocity B (r R) in the restframe of object A. The same point r on object A has velocity A (r R). Since relativevelocities simply add, the net velocity is (A+B) (r R), which describes rotationalmotion with angular velocity A+B.

Morin first discusses Chasles theorem because we had to know that the most generalmotion of a rigid body is a translation plus a rotation, which means that the only in-stantaneous motion about some fixed point is a rotation. This let us say that the veloc-ity (A+B) (r R) must equal (r R) for some ,which lets us conclude that= A+B.

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One final note, this only applies to instantaneous angular velocity, which in general isalways changing. It does not mean that the motion over time is a simple rotation!

1.4 Principal Axis Guessing

Often the princiapl ax

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