Thursday, July 31, 2008

Answer: Writhing Chain
New Problem: Hourglass

Writhing Chain
The speed of sound in the chain depends on its tension, the same way it does in a violin string. You can raise the pitch of a violin string by tightening it. Similarly, the speed of traveling waves increases where the chain has increasing tension. The tension in the chain is high near the top because it's supporting the weight of all the chain below it. Near the bottom, the speed is low. The chain is set to rotate at just the speed of traveling waves near the bottom of the chain. That way, patterns can persist there even as the chain cycles around.

You sit an hourglass on a scale with the sand in the top, but blocked from falling. Then you remove the block and let the sand fall down the hourglass. Make a plot of the reading on the scale as a function of time.

Monday, July 28, 2008

Answer: Whisper Dish
New Problem: Writhing Chain

Whisper Dish
The answer is that they're parabolas, which I know because I read it on the little sign by the exhibit, which said, "these are parabolas." But several people pointed out that ellipses would also work, which is quite true. You could test this in person by seeing whether a person standing directly in front of you or directly behind you was more of a hindrance to hearing incoming messages.

There's a simple explanation for why parabolas focus incoming plane waves to a point. A parabola is defined as all the points equidistant from a focus (point) and a directrix (line, or plane in 3D). Since an incoming plane wave would hit a directrix evenly along its length, then in traveling the same distance all parts of the plane wave can meet at the focus together.

A person standing in between the two parabolas could potentially act as something of a low-pass filter. The speed of sound is about 300m/s, so a tone of 300Hz has a wavelength of about a meter. Tones lower than this will easily diffract around the person, but tones higher than this can scatter off. So if you were to use the whisper dish to broadcast music, the "color" might change when a person stands in the way because more high-frequency pitches could be blocked.

Writhing Chain
A chain hangs from a wheel. The wheel turns at a certain, fairly slow speed of about one meter per second. It pulls the chain around with it at the same rate. The wheel is about 3m off the ground, and the chain hangs from the wheel down to the ground. If you tap on the chain, a second or two later the bottom of the chain will twist and turn in slowly-changing patterns, snaking out from the plane of the rest of the chain. Why?

Thursday, July 24, 2008

Answer: Guitar
New Problem: Whisper Dish

The frets are designed so successive frets are a constant interval. An "interval" is a certain ratio of frequencies.
fn+1 = r*fn
fn+1 - fn = (r-1)fn

Where fn is the fundamental frequency of the string at the nth fret, and r is some number representing a ratio.
The frequency is inverse proportional to the length of the string. One way to see this is to find the differential equation describing waves on a string and solve it for the fundamental node. An easier method (although not a derivation) is to assume that the frequency of a vibrating string can depend only on the tension in the string, its linear mass density, and its length. The only way to combine these quantities and obtain a number whose dimension is frequency is

Then going back the equation for the change in frequency between frets, the change in lengths must also be geometric, but the factor is now a reciprocal (to double the frequency you must halve the length).
ln+1 = 1/r*ln
ln+1 - ln = (1/r - 1)ln = (1-r)/r * ln

So the distance between frets is proportional to the length of the string. As the string gets shorter, so does the distance between frets.

Whisper Dish

Two people can talk normally if they're positioned at large, hard dishes like the one in the picture. The dishes may be 30 meters apart, but the speakers can hear each other well if the dishes point towards each other and the speaker sit and speak at the right location.

What is the ideal shape for such a dish? Why does it still work if a third person stands in the middle to try to block the sound? Why can't that third person eavesdrop on the conversation?

Wednesday, July 23, 2008

Answer: Inverted Pendulum 2
New Problem: Guitar Frets

Inverted Pendulum 2
The longer the inverted pendulum, the easier it is to balance. The braindead way to see this is to notice that the torque on the pendulum is proportional to its length, while the moment of inertia is proportional to the square of the length. So the angular acceleration is inverse-proportional to length.

The period is longer, for the same reason. This difference between the pendulum oscillating through small angles with the weight near the bottom, and falling through a small angle with the pendulum near the top, is that the sine function becomes a hyperbolic sine. The time constant is unaffected.

If you have two pendulums, one with a longer arm but both with the same mass, and raise them to the same angle and let them go, the tension in the pendulum arms will be the same magnitude. The pendulums are going in circles, so their acceleration is v2/r. "r" is the length of the pendulum. "v2" is proportional to the kinetic energy. The kinetic energy comes from converting potential energy, and the potential energy is linear in the length of the pendulum. So the v2 term is also linear is the length. Dividing them, the acceleration and force are independent of the length.


Why do the frets on a guitar get closer together as you move down the neck?

Tuesday, July 22, 2008

Visualizing Complex Beats

Take two complex exponentials and add them.
ei*a*t + ei*b*t
If one is much faster than the other (b>>a), you get a circle that slowly "marches around". It runs around with angular frequency "b", while its center is also going in a slow circle with angular frequency "a". This is the picture above.

If the two frequencies are equal, they would just trace out a circle of radius 2. But if they are close but slightly different, they'll drift from being in phase to being out of phase and back again. They'll almost trace out a circle of radius 2, but just barely miss because by the time they get back to where they started, they're a little bit off from each other. The result is a gradual inward spiral, shown below.

You could see this analytically like so:
ei*a*t + ei*b*t
ei*t*(a+b)/2(ei*t*(a-b)/2 + ei*t*(b-a))

So the sum of two complex exponential runs around in a circle at the average of their frequencies, but the radius of that circle varies up and down sinusoidally in time. The projection of this complex exponential onto one of the axes describes beats - the "wa-wa-wa-wa-wa" sound you hear when two clarinet players tune.

Monday, July 21, 2008

Answer: Inverted Pendulum
New Problem: Still Inverted Pendulum

Inverted Pendulum
Remember you want to swing the pendulum all the way up, but your eyes are closed so you don't know where the pendulum is. You do know that you want to add energy to the system. From rest, slide the bar a little to get things started. Then wait until you feel a tug, and pull in the opposite direction. That way you can ensure that the force you exert and the motion of the slider bar are the same direction, and therefore the work is positive; you're adding energy. If you were to move the slider in the direction it "wants" to go based on the tug you feel, the pendulum would be doing work on you, pulling your hand along for the ride.

Inverted Pendulum 2
We're going a bit more conceptual and less quantitative for a while to encourage more participation in the problem of the day, which had been fairly low for things like the planet formation problem. So the question is, think about the same apparatus as yesterday, but make the pendulum longer while keeping the mass the same. How is the task different? Easier, harder, or the same difficulty to balance the pendulum in the inverted position? Are the tugs you feel while swinging the pendulum bigger, smaller, or the same? How about the period of oscillations?

Answer: Suspension Bridge
New Problem: Inverted Pendulum

Suspension Bridge
The cable takes the shape of a parabola. Consider the section of cable supporting a differential element of the road, length dx.

The physics that goes into this solution is:
  • the total force on this bit of cable is zero
  • the road pulls down on the cable due to its weight from gravity
  • the tension of the cable pulls from either side; this tension must be along the direction of the cable
We'll break the tension on each side into its x and y components. The x-components must cancel each other because they're the only forces in the x-direction. The equation stating that the total y-forces all cancel could be written

That last term is the linear mass density of the bridge times gravitational acceleration times the length of the segment. All together it's the weight of the road supported by that segment of cable. Finally, we bring in the fact that the tension must be along the direction of the cable

Now just mop it up

The equation gives a parabola. The constants "c1" and "c2" simply allow you to change the focus of the parabola. This is equivalent to saying that you are free to move the origin of the coordinate system around as you wish without changing the shape of the bridge. The third constant, Tx, does change the shape of the bridge. If you make Tx large, you can have a shallow parabola with low towers supporting the bridge, but you'll need a strong cable capable of supporting a lot of tension. On the other hand, you could get away with a weaker cable by making Tx small, but then the parabola would open up steeply, and you'd need a very long cable and tall towers to make it work.

Inverted Pendulum
The problem is basically, "how does this work"?

We went to the Exploratorium this weekend, and one exhibit was a pendulum attached to a slider bar that ran on a horizontal track. The idea was to slide the bar back and forth in such a way as to swing the pendulum to the vertical, then try to keep it there. Many people could get the pendulum swinging half way up, but then were stuck swinging the slider back and forth violently without getting the pendulum to rise any higher. However, it's possible to swing the pendulum around without using much force, without swinging the slider long distances, and done completely with your eyes closed. How? What algorithm would you use?

Sunday, July 20, 2008

Answer: Spinning Fish Tank
New Problem: Suspension Bridge

Spinning Fish Tank
The surface of the spinning fish tank takes the shape of a paraboloid. It's undergoing uniform circular motion, so its acceleration is directly proportional to its distance from the center of the tank. By the equivalence principle, this is just like having a component of gravity pointing outwards, in addition to normal gravity. The surface of the water must be perpendicular to this artificial gravity, so its tangent must have a vertical component directly proportional to its distance from the center. Integrate this and you get a paraboloid. (Parabola rotated around its axis of symmetry).

If you dropped a little rubber duck in the water, it would stay right where you dropped it, despite having an apparent "hill" to fall down. It feels the same artificial gravity the water does. The pressure of the water is increased compared to the non-spinning case. This is because the water weighs more. The further out you go from the center, the stronger the artificial gravity and the greater the pressure. This creates a pressure gradient in the horizontal direction, which in turn produces a force on the water. This is just the centripetal force necessary to keep the water rotating in a circle. A fish swimming in the tank would not feel thrown out to the outside or sucked towards the middle. It would not sense that it was twirling around, either, except that it would see the room were spinning. The fish would feel a bit "squished" due to the higher water pressure, and would (in theory) be able to detect the change in the pressure gradient from its head to its tail.

People who wish to create a perfect parabola, say for the reflecting mirror of a telescope, might care. In fact, there are projects that seek to use this effect to make reflecting mirrors for telescopes.

Philosophers of science, starting with Newton, have also taken interest in this problem. The basic idea is: how do you know the bowl is spinning in the room instead of the room spinning around it (imagine the bowl spinning freely - no friction and therefore no forces)? The kinematics are the same, but the physics is different. If this sounds silly, you haven't thought about it hard enough yet.

Suspension Bridge
A suspension bridge has a cable that is very light compared to the weight of the road it supports. Assume each bit of the cable supports the weight of the bridge directly beneath it, neglect the mass of the cables, and assume the cables do not stretch (have infinite Young's modulus). What shape is formed by the arc of the cables from one tower to the next?

Friday, July 18, 2008

Playing Catch in a Rotating Spaceship (revisited)

My Mathematica Demonstration is now online.

Answer: Planet Formation
New Problem: Spinning Fish Tank

Planet Formation

The dust cloud has a great deal of potential energy due to the particles' mutual gravitational attraction. As the cloud collapses, all this potential energy is converted to heat (random kinetic energy of the particles). Find the (negative) gravitational potential of the planet by integrating
-GMm/r from 0 to R, where M = 4/3*pi*r3*density, m = 4*pi*r2*dr*density. You'll find the total heat released was proportional to M2/R. Plugging in some reasonable values for Earth, the temperature comes to 105K. Clearly, the planet cannot be this hot (random thermal motions would keep it from condensing that much), so the planet must cool as it forms.

Spinning Fish Tank

You take a fish tank and spin it (at constant angular frequency). What shape does the surface of the water settle down to? Who cares?

Cos(x) = i*Sin(x)

Several students asked me today whether, by some trick they didn't know, Cos(x) = i*Sin(x). (Not random coincidence, it popped out of a question on their problem set).

The answer is no, of course, since Cos(0) = 1, i*Sin(0) = 0. Further, there are clearly no solutions to this equation for real x, since the cosine will be real, and the sine imaginary. But are there solutions for complex x?

Cos(a+bi) = i*Sin(a+bi)

let's start with the cosine term, and break it up using the angle addition law

Cos(a + bi) = Cos(a)Cos(bi) - Sin(a)Sin(bi)

We now need the sines and cosines of imaginary angles. We can get them using Euler's equation.

eix = Cos(x) + i*Sin(x)
e-ix = Cos(-x) + i*Sin(-x) = Cos(x) - i*Sin(x) using the even and odd properties of sine and cosine

now add these

eix + e-ix = 2Cos(x)

let x be an imaginary number i*b, b is real

e-b + eb = 2Cos(i*b) = 2Cosh(b)

where the last bit about the hyperbolic cosine is just from the definition Cosh(x) = [ex + e-x]/2

do the same to find the sine of an imaginary angle. you'll obtain

Sin(i*b) = i*Sinh(b)

go back and plug this into the expansion of Cos(a+bi)

Cos(a+bi) = Cos(a)Cosh(b) - i*Sin(a)Sinh(b)
Sin(a+bi) = Sin(a)Cosh(b) + i*Sinh(b)Cos(a)

our original equation was iSin(x) = Cos(x), with "x" complex ("a" and "b" above are real). Taking the two equations above and equating their real and imaginary parts

Cos(a)Cosh(b) = -Sinh(b)Cos(a)
-Sin(a)Sinh(b) = Sin(a)Cosh(b)

Both of these equations yield the same condition:

Cosh(b) = -Sinh(b)
Tanh(b) = -1

But Tanh(b) approaches this value asymptotically in the limit b ==> -Infinity
So there are no solutions to the equation.

Answer: Ice Cube New Problem: Planet Formation

Ice Cube
You should wait first, then add the ice cube. It's fairly straightforward to be quantitative about this, but the thing to notice is that the ice cube has some latent heat of fusion. To melt the ice cube, you will have to put in a certain minimum amount of heat, no matter what (as opposed to adding cold water, where the water will take in more heat if the coffee is hotter). Because the coffee loses heat faster when it's hotter, let it cool a bit, then drop in the ice cube.

Planet Formation
An Earth-like planet forms from a cloud of cold, diffuse dust. How hot is the planet?

Wednesday, July 16, 2008

Melting Ice in Coffee

I'm sponsoring a daily physics problem contest for the students. Today's is:

You have a cup of hot coffee and a small ice cube (which is in a freezer). Should you drop the ice cube in the coffee first, then wait for it to finish cooling enough to drink, or let it cool some first, then drop in the ice cube? (No, you cannot just put the coffee in the freezer).

(Answer to appear with tomorrow's problem).

Tuesday, July 15, 2008

What I've Learned About Lecturing

Goals for My Afternoon Lectures
  • 15 - 20 minutes long
  • focus on a single idea
  • forces student involvement. not just by the audience questions, but by assigning actual problems and calculations in the middle of lecture
  • makes the student feel like he/she is brilliant, not like the lecturer is
  • write big (bigly?)

My quantum mechanics students arrived today (scattering through the open door - no tunneling was required). I walked into our first house meeting alone, began playing the James Bond theme song on my computer (since it's the theme of our dorm decorations), and told the students I was going to deliver a "short lecture, no more than ninety minutes" on the history of the James Bond film series. I talked for about 10 seconds until the other counselors simultaneously burst into the room from all corners, dressed up in Bond-like tuxes, and shot me down with water guns.

But soon I'll get to deliver some real lectures. I enjoy this. I enjoy sitting down and trying to work out the order I want to make all my points. I enjoy trying to simulate what the students would and would not be able to understand. I enjoy being on the stage presenting the material, because I can choose what to talk about and I talk about whatever I think is interesting. But simply lecturing however I feel like it is something of a disservice to the students, because I also need to do everything I can to keep the lecture digestible.

I think I learned the most about this after watching one afternoon while one of the other counselors lectured. He did an excellent job. His delivery was smooth and well planned. Everything was technically correct, even when I asked asinine technical questions from the audience. But the presentation was too long and covered too much material. I realized that I was doing all the same things in my own lectures.

My goal then is to make all my afternoon lectures fit a sort of mold. They should be compact, conceptual, and unified. Each part should be there as a subservient to the goal of the lecture as a whole. No little tangents on things I personally think are interesting, since for the most part these just break the students' concentration. I used to do these starting at 1:00PM, and finishing between 1:15 and 2:00 depending. Now I'm going to start at 2:00 to let them think of the lecture as a short break in their 1:00 - 3:30 study time, to let them work on the problems they're solving first, and to let the food coma that comes on after lunch wear off.

Possibly the most effective lecture I gave was a guided exercise on the binomial theorem. I first asked the students for a crazy number, then told them I would take its square root. I did this using the binomial theorem (to first order, obtaining 3 sig figs). Then I actually wrote problems on the board such as finding the square root of 101 and made the students work through them. After four or five, they had the hang of using the binomial theorem to estimate square roots, so I gave them some cube roots and fourth roots. Then I showed them how to do otherwise difficult division problems with the same theorem and made them work through a few of those. Then finally I gave them an example of how using the binomial theorem to first order fails in some cases, using a series that converges to e.

It was short and interactive. The kids loved it because they were getting results they could understand without using anything very advanced or mystical. It was much better than showing them a proof by induction, or even worse a calculus-based proof of the theorem. And it was relevant, because many times during their three-week course they had to do calculations where the result was a number very close to one, and their calculator would fail them while the binomial theorem would not. So that lecture is a model for what I want to try to do here.

There's no rush to get a certain amount of material covered, because no one's even requiring me to give lectures. I'm doing it just for me own amusement. I can break a difficult concept up over three days if I want. It's a wonderful blank slate. One big challenge is that the students come from a variety of backgrounds, since some do not know calculus and others have taken AP calculus and physics.

Saturday, July 12, 2008

Susskind's High School Black Hole Lecture

On the last day of relativity class, our instructor took us to meet for almost 90 minutes with Leonard Susskind, who delivered a lecture based around a simple calculation. The goal was to demonstrate the connection between the mass, entropy/information, and temperature of a black hole. This is a summary of what he said, as nearly as I remember it.

First: entropy vs. information
I asked Susskind what the difference was. The qualitative description of entropy and information seem different. Entropy is how "messed up" something is, information is how much you worked at setting it up.
But when you get a quantitative definition, it suddenly appears they're the same. Entropy is the log of the number of microstates that correspond to a given macrostate. (You need a log because suppose you have a system with two independent subsystems. The first subsystem has N possible microstates. The second has M. The whole system then has N*M microstates. Its entropy is log(M*N) = log(M) + log(N). Defined as logarithms, entropies add.)

Information is the number of bits required to specify the exact (micro)state you've prepared. If you multiply the number of possible microstates by 2, you need one extra bit to describe which of those microstates you picked. So information is also a logarithm of the number of microstates.

Imagine I have a chess board and some pieces. The squares of the board are numbered. (This is done so that two boards that are 180 degree rotations of each other will count as separate microstates. You wouldn't actually have to number every square. You could mark one of the black corners as being the "A1" and the rest would be determined from there.)

Now suppose I have one pawn to place on the board. I can place it on any of the 64 squares, so there are 64 microstates (specific configurations) that correspond to the macrostate "one pawn on a chessboard". That means the entropy is log(64) = 8. (I'm using base 2. It doesn't really matter what base you choose because they are all related to each other by a constant.)

The information is the number of bits needed to describe the state. It is also 8. You must specify a number between 0 and 63 to indicate which square the pawn is on. That takes 8 bits.

If I had the entire slew of 32 chess pieces, nothing much would change. There would be many more possible configurations. Say N of them. Then the entropy would be log(N). But to find the information, I could make a long numbered list of every possible configuration of the pieces. To specify the exact position on the board, I find that position on the list and write write down its corresponding number. Since there are N numbers, I'd need log(N) bits to say which number I wanted. It's the same number as the entropy.

This is called the information because I could easily work the process the other way. If I wanted to send a message containing log(N) bits, I could put those bits in order to make a number, see which chess position corresponded to that number, set up the board that way, and send someone the board instead of the binary number. Once they got the board, they could work backwards (assuming they had the same list) to get the bits I originally intended to send.

The information a system has is how long a message it could send. This turns out to be mathematically the same as the entropy. So why use the term "information" at all? We already had the term "entropy" in place. Why not just keep it?

Of course my question was much shorter, and just asked what the difference is. Susskind said that entropy is "hidden information". His example was a bath tub of hot water. If there are waves on the surface, that's information because I could take a snapshot of them, analyze them, and see what sort of message they held (about where water is dripping in from a faucet, for example. (Many geologists are in this business.)) But the entropy of the bathtub is much higher than just the amount of information you could store in waves. The bathtub has 10^30 or so molecules of water. The log of the number of possible quantum states consistent with a tub of 10^30 water molecules at 300K would be the entropy and, in theory, the information. It's a huge number. But there's no point in saying the tub has that much information, because there's no way we could ever measure the state of all those water molecules. No one would ever try to send a message across enemy lines by setting up a bunch of water molecules in an exact quantum state to be decoded by a quantum-bathtub-reader in the general's field headquarters. (Here I'm deviating a bit from Susskind's exact analogy.) Even if you did, the bathtub would immediately interact with its environment, and decoherence would set in almost immediately. Although there's technically information there, we can't use it. It's still entropy, but doesn't earn the tag of information.

The heat death of the universe would be the extreme example, I suppose. In this theoretical state long in the future, everything has come into thermal equilibrium. The universe is a uniform temperature throughout, with no stars and galaxies and certainly no life. It's "as boring as possible", meaning there's almost no information because if you wanted to tell someone what the universe was like, you could basically give its size and temperature and maybe one or two other things. That would be all they needed to know. This is the state in which the universe's entropy is at a maximum, while its information is minimal.

Next we went into a calculation to demonstrate the importance of entropy for a black hole. The problem we're trying to solve is that John Wheeler used General Relativity to prove the "no hair" theorem for black holes. If black holes were bathtubs, they would have no ripples on their surface. There is nothing about a black hole that can give you detailed information. Black holes have only mass, electric charge, and angular momentum. Any two black holes with the same mass, charge, and angular momentum are just as identical as any two electrons. That means the entropy and information of a black hole are exceedingly tiny. In fact there is just one microstate corresponding to the macrostate we observe, so the entropy is zero! (I think. Don't quote me on that. Besides, this is the internet, meaning I'm basically supposed to get stuff wrong.)

If you toss a bucket of hot water into the black hole, you've added to it something with a great deal of entropy. But the black hole still has almost no entropy. So the entropy seems to have disappeared. This violates the second law of thermodynamics, which states that entropy always increases.

The calculation Susskind next presented he attributed to Jakob Bekenstein. I won't pretend to understand how to justify all the steps. I'll just repeat the general idea. I don't remember quite what all his steps were anyway. Somehow he avoided doing an integral, and never mentioned Boltzmann's constant. So my calculation is a bit different, but it starts and ends in the same places.

We'll imagine a procedure for adding mass and information to a black hole one step at a time. This is done by shooting a photon into the black hole. If we shoot a high-frequency photon, we're adding a lot of information. A high-frequency photon has high resolving power, so if we shot a high-frequency photon we'd be able to tell where it entered the black hole, which counts as information. We'd be adding a great deal of entropy. Instead, we decide to shoot a photon whose wavelength is comparable to the radius of the black hole. That way the photon cannot resolve the black hole at all. We add exactly one bit of information/entropy: whether or not the photon entered. The entropy is measured in units of Boltzmann's constant. (It turns out that if you shoot a photon with wavelength greater than the black hole, it'll probably just bounce off. Also, when I say, "it turns out", that means "I don't understand this")

We consider the photon to be an incremental energy addition, dE. The energy of a photon is

E = h*c/L

with h Planck's constant, c the speed of light, and L the wavelength. Therefore we can write

dE = h*c/L*dN

dE is the change in energy of the black hole. N is the number of photons we've shot in, so dN is just one. This term is there so we have a differential equaling a differential. (I made this up, too. I really wish I remember what Susskind wrote exactly.)

We're setting the wavelength L equal to the radius of the black hole.

dE = h*c/R*dN

The relationship between temperature and entropy is

dE = T*dS

with dE the change in energy of a system, T its temperature, and dS the change in entropy. For a black hole, we already know dE, because it's the energy of the photon, which we just calculated. We know dS, too (one times Boltzmann's constant). Plugging in what we already know about dE and dS we get

h*c/(R*k) = T
R = h*c/(k*T)

Here, k is Boltzmann's constant. This gives the temperature of the black hole given its radius, or the radius given the temperature. They're inverse proportional. Incidentally, the temperature of the universe is about 3K. Plugging in the constants, that temperature corresponds to a black hole of radius 5mm. Since real black holes are much larger than this, they are also much colder. Black holes are colder than their surroundings, so heat flows into them and they are all getting bigger. They'll continue to do so until the universe is much, much colder. Then they will begin to lose mass as their blackbody radiation emits mass in the form of photons faster than the black hole absorbs mass (also mostly photons).

The radius and mass of a black hole are directly related via

R = 2Gm/c2

This is because the radius, called the Schwarzchild radius, is whatever distance from a point source of mass m the escape velocity is that of light. The escape velocity depends on the kinetic energy needed to escape, which is equal to the potential well you're in. So the Schwarzchild radius depends on the shape of the potential.

The potential is the integral of the force, which is GMm/R2 ("M" the test object mass, "m" the black hole mass). So the potential falls off as GMm/R. The kinetic energy is 1/2*M*v2. Set v=c, set the potential and kinetic energies equal and solve for R. You get R = 2Gm/c2. This feels like it shouldn't work, since the kinetic energy of something moving at speed c is actually infinite (if it has mass), but somehow the answer pops out regardless. The real calculation a bit more involved.

The energy that's gone into a black hole can tell us its mass.

E = mc2
m = E/c2

Plugging this into the equation for the radius

R = 2G*E/c4
E = R*c4/(2G)
dE = dR * c4/(2G)

Finally we return to the definition of entropy and write everything in terms of R. We'll use the expression above for dE and the earlier expression for the temperature in terms of the radius.

dE = T*dS

dS = dE * 1/T
dS = [dR*c4/2G]*[(R*k)/(h*c)]

Integrating over dR from zero to the final radius

S = k*R2c3/(4G*h)

This is the result we were shooting for. The entropy of a black hole is proportional to its area. A better calculation gives slightly different constants, but the idea is the same. The number G*h/c3 must be an area to make the units work out. It comes to 10-70m2. Its square root is the Planck length, 10-35m2. This is a length built out of natural constants, and it characterizes this particular process.

What's interesting (according to Susskind. To me it's not very interesting, because I don't understand it well enough) is that this stuff about the thermodynamics of black holes involves an apparent paradox. General relativity says black holes don't emit radiation. But the black hole has temperature, and so it must emit black body radiation like anything else with temperature. When you look at a process whose characteristic length scale is the Planck length, you have to consider the interactions between relativity and quantum mechanics.

By the end of all this, I thought the calculation was straightforward enough, but I didn't really believe it meant anything. This was a calculation for one idealized process of adding information a bit at a time. But shouldn't the relationship between entropy and mass depend on the procedure for adding the mass? I could dump a bucket of hot water into the black hole, or freeze the water and dump in an ice cube. I'm adding the same mass, but it seems like the entropy I'm adding is very different. Intuitively to me, the radius of the black hole, which depends only on the mass, should increase by the same amount in both cases, but the increase in entropy should be different.

Apparently, this is not so. I have no idea why not, but the explanation came from Hawking, who generalized the simple procedure discussed here. Hawking worked out exactly what the constants should be in the relation between the entropy and area of a black hole, and showed that's it's a general relation. Maybe someday before I die I'll be smart enough to catch a glimmer of how he did it.

I used these wikipedia pages:

Wednesday, July 9, 2008

Playing Catch in a Rotating Spaceship

It's a standard physics problem, but you don't really get it until you work through it for yourself.

You're standing in a circular spaceship, which is rotating to provide artificial gravity. You throw a ball straight up in the air. What path do you see it follow?

The longish version of the answer is here.

What was interesting was that as I went along, more and more little things fell out of the math. Once I had the equations for the trajectory of the ball, for example, I saw that I could take a Taylor expansion about any point in the trajectory, and the Coriolis and centripetal accelerations would fall right out - even though I made no reference to them in the solution (which is purely kinematical).

It also turns out that the shape of the trajectory is specified completely by the angular velocity of the ship's rotation and the ratio of the velocity of the ball to the size of the ship. If you make the ship twice as big, and throw the ball twice as hard, the trajectory will have exactly the same shape.

I also made a Mathematica Demonstration plotting the trajectory as a function of angular velocity, size of ship, and how hard you throw the ball. If they publish it it'll be online soon, but in the mean time there are some sample plots on the pdf.

Friday, July 4, 2008

Triangles on A Sphere

Here's a problem that was on the homework set my students were supposed to complete yesterday:

Prove that the sum of the angles of a triangle on a sphere is more than 180 degrees.

A related and easy problem is to prove that on a sphere the ratio of circumference of a circle to its radius is less than 2π. However, the one about triangles is trickier.

I'm sure that if you know a bit about non-euclidean geometry you can simply prove the theorem in all spaces with positive curvature. Then the sphere falls out as a special case. I don't know how to do that.

Here is another approach, which is not very rigorous, but pretty much good enough to convince me.

Take the longest side of the triangle and draw the perpendicular through it that crosses the opposite vertex. I should prove such a perpendicular exists, but won't. Now you have two right triangles. If you add up all the interior angles of the two right triangles, you get the interior angles of the big triangle, plus the two extra right angles. (draw your own picture. I don't feel like it.)

Now if the two right triangles both have the sum of their angles more than 180 degrees, added together they make something more than 360. Subtract the two right angles. You find that the big triangle also has its interior angles more than 180 degrees.

How do I know if one of those two right triangles has its interior angles more than 180 degrees? Drop a perpendicular from the hypotenuse to the right angle vertex, breaking the right triangle into two smaller right triangles. Now the same reasoning holds as before, and I only have to show that the smaller right triangles have interior angles that sum to more than 180. I can keep breaking the triangles down smaller and smaller, so all I have to show is that a very small right triangle has the sum of its interior angles greater than 180, and I'm done.

Now introduce coordinates on the sphere so that the small right triangle's right angle lines up with the intersection of the prime meridian and the equator. Then the length of the side on the equator is dφ, while the length of the side on the prime meridian is dθ. Finally, the metric on a sphere is ds2 = dθ2 + sin2θ *dφ2, so that is also the length of the hypotenuse.

If I let θ = π/2, I have a normal right triangle in euclidean space. But θ = π/2 only right along the equator. Along the length of the hypotenuse, sin2θ < 1, meaning the hypotenuse is actually shorter than that of a right triangle.

The law of sines shows that the shorter the side of a triangle, the bigger the angles on at its endpoints. So the angles at the end of the hypotenuse are large than those of a triangle in euclidean space. The right triangle has a sum of its interior angles more than 180 degrees, and then so do all other triangles on the sphere.

I admit the last step is a bit of a stretch. The law of sines is really useless because it's true only in euclidean space, but for a very small triangle the space is a better and better approximation of euclidean.

Balls Rolling Down Planes

I'm a TA for a 3-week course in relativity for high school students. I've been putting together short (and sometimes not-too-short) lectures for them in the afternoons, but last weekend and this upcoming one I've been tackling a more ambitious project: teaching them calculus. (How does one conjugate a verb referring to an action you both did in the past and will do in the future, but aren't doing right now?)

It took me several hours to put together a simple example of modeling a physical problem using calculus - a ball rolling down a plane without slipping. It's in the category of simple problems used to drill students. There are hundreds of examples like it in every introductory physics book. For ten bucks you can buy a giant practice book with 3000 of them. Yet somehow it took me hours to put the notes together and they're monstrously long (7 pages with 3 diagrams).

Of course, most of the reason for this is that I do a fair amount of extraneous stuff. I solve the problem without using forces at all. I derive the expression for the kinetic energy of a rolling sphere by integrating over the kinetic energy of a differential volume element, just as an example of how to do such an integral.

The idea is that I don't want things like "moment of inertia" to be black boxes used to make quick numerical calculations, the way introductory texts invite their students to do.

On the other hand, everything I did this afternoon is pretty much worthless in terms of my original goals. Someone who's spent only three hours of their life studying calculus has no prayer of understanding the notes I wrote. Probably, the only people who would understand them are the people who already know everything they have to say.

Students complain about hard-to-follow lectures and obscure notes. But flip yourself over to the other side for a while and you'll begin to appreciate how difficult it is to give a clear explanation, even for something that's crystal clear inside your own head.

Constant Acceleration in Special Relativity

I admit I am completely powerless to make the internet do what I want. Instead of posting here, I'm putting a pdf up at scribd.

Also, I'm not going to put in all the effort to change it, but at the end I somehow added a magical power of 10^-1 to the value of Earth's gravity. A spaceship accelerating away from Earth at a constant 1g would only need about a one light-year head start to keep permanently ahead of a photon chasing it.

For an observer riding on the ship, this essentially means that objects more than a light year back disappear - light from them will never reach the ship. Objects very near one light year back become more and more red-shifted and time-dilated. These are the same effects that occur as an objects falls towards the event horizon of a black hole.