Arcsecond

Moving to Wordpress

2008-10-05T18:16:00.000-07:00

The LaTeX feature I installed here is no longer working, so I'm moving to Wordpress, where LaTeX support is built in. The new location for the blog is
http://arcsecond.wordpress.com/

Re-examing the Inner Product (Euclidean Space)

2008-09-29T19:45:00.000-07:00

I'm auditing Kip Thorne's "Applications of Classical Physics" this year. Today was the first day of class, and Kip gave a preamble which, if I may be allowed to paraphrase in a wild, inaccurate, and completely unfair manner, (which is basically to say every word of this is existed only in my head) went something approximately like this:

Welcome to class. Here is the website. Please call me "Kip". I'm 68 years old and my signature lucky ponytail is a thing of the past. There's not enough time left for ridiculous formalities. I am about to retire. I'm going to start writing books and making movies and various other things that famous retired physicists can do. I'm too old to waterski, so this is the next best option. This is will be last class I ever teach. There will be no grades. I would prefer you actually learn some stuff. Now, let's suppose the laws of physics are frame-independent, and see what restrictions this places on force law dealing with a classical field in Minkowski space..."

Awesome.

One thing that interested me was our classes' definition of the inner product between two vectors. Since we wanted to do everything in a frame-independent manner, we couldn't simply define the inner product by

because that presumes the existence of basis vectors. The other common definition is

which is good, but we wanted something that could generalize to the Minkowski space of special relativity. So the definition we came in two parts. First, we defined the inner product of a vector with itself. Then use that definition to bridge to the inner product of arbitrary vectors.

For a vector , define

where is the square of the physical invariant interval between two events. To measure this interval, have an unaccelerated clock move from the origin to the event. ( should be real, meaning the square of the interval is positive, and is negative. So the inner product of the vector between timelike events with itself is negative). In the case of spacelike events, you should instead find the inertial reference frame in which the events are simultaneous, and lay down a measuring stick between them to get the interval.

Now we know how to take the inner product of a vector with itself. Define the inner product of two vectors and by

This is more subtle than it might initially appear. You can't just go willy-nilly with the algebra and start simplifying that right hand side out. You don't know any properties of this inner product, so you cannot, for example, write

Instead you have to work with the actual sums. It was claimed in class that this definition of the inner product is bilinear in the arguments, which wasn't obvious to me. So I asked about it, and Kip suggested I try to work it out for myself in Euclidean space using the Pythagorean theorem. So here goes.

Break down the vector into a component parallel to and a component perpendicular to

The vector now becomes , where and are unit vectors in the directions parallel and perpendicular to .

Because these two vectors make a right angle, the Pythagorean theorem applies.

similarly,

This is great, because the right hand sides of those equations are just numbers. Expand that out, subtract the two equations, and divide by four to obtain

That's just the normal definition of the dot product, and is obviously linear in both vectors. Now, how about Minkowski space?

Why Am I Posting This Crap?

2008-09-29T18:21:00.000-07:00

Working with physical models inevitably leads to mathematical problems. In physics education, many professors build up enough tools and techniques to shove these problems aside, and then quickly move on to the real physics. That's simply a reflection of the fact that they care deeply about the physics itself, and I appreciate their efforts to keep students from getting bogged down in details. But taken to the extreme, this attitude produces calculating-maniacs with tons of knowledge and little mathematical intuition. Take at look at the canonical book on math methods for physics (Arfken), and you'll see that it's simply an encyclopedic dictionary of methods.

I'm only now becoming mature enough to go back and re-examine many of the things I learned earlier with an eye towards appreciation rather than computation. This practice intrigues me. The purpose of a blog is ostensibly to create content for public consumption. That's not what I'm doing. I'm working through things for myself. Not because I feel a dire need to understand. Because I can get enraptured by the absurd power of a little solid thought to unearth something I never saw before, and yet later becomes laughably obvious.

According to Google Analytics, in the last month, 58 total people have visited the site in 172 separate visits, and viewed 268 pages. They've spent a total of 5 hours, 41 minutes reading the posts. That's only a small fraction of the cumulative time it took me to write the posts. Also, a lot of the statistics I quoted are just me, logging in from various computers, trying to crank up the scores so I'll feel more important.

Nikita, Julia, and Kangway have actually been participating and giving me feedback, which is pretty much incredible, so thank you for doing that. But looking at the recent posts I've written, I realized a few things:

not that many people care about things such as the geometric interpretation of a vector product rule
people who do care were better students the first time around than I was, and mostly already know
people who would care, but don't already know this stuff, most likely don't have the assumed background to learn much from my posts.
even if the content of the posts is "just right" for you, I skip a bunch of steps and gloss over the parts of the arguments that are less interesting to me, because they're too tedious to draw/ for the blog. Hence, following the posts from scratch would be a rough road. It's not the sort of reading people are looking for when browsing through blog posts. They want quick, to-the-point, easily-comprehensible and entertaining content. This blog is the antithesis.

My goal is simply to explore nuances that interest me, however mundane. That explains a few of the points above. When I start writing a post, I have to think about how much background knowledge to assume the reader has. I want to write about the parts that I think are interesting, which means essentially the parts (and only the parts) that I didn't completely understand before I started working on the post. I generally assume that everything I am totally comfortable with, the reader is as well. That means people will frequently be flabbergasted by the strange logical jumps I take and the way I write certain details off as trivial. On the other hand, everything that I had to work out and think about for a while, I feel the need to explain in more detail than many readers would be interested in wading through.

Today I finished a post about vector products and determinants. I was aware of the geometric interpretations from a few sources, but I had never put the pieces together for myself. By the end, I was amazed. Not only had I built up the idea of a determinant, motivated geometrically, but I had also shown how to use the determinant to solve a system of equations. I didn't even know that would pop out of the argument. If you had asked me to demonstrate how a determinant works before I started writing, I probably wouldn't even have known. Additionally, the path that led me there was remarkably short, and since I had worked through it myself, it walks through a logical (to my mind) problem-solving process.

Math books tend to define and develop the determinant as a number characterizing a linear operator on an n-dimensional vector space. The treatment is more abstract. The proofs showing how to calculate it and how it solves systems of equations frequently rely on manipulating sums of indices and leave me feeling unsatisfied. I believe them, and see that they're correct, but I have difficulty visualizing them as well as the author of the book. I don't knock their approach. I think it's probably the best way. But this supplementary cogitation on the blog has bolstered my appreciation.

Physics books, on the other hand, will normally skip over the entire journey and roll the answer out right away, then tell you to start inverting some matrices. Save the math questions for your linear algebra class!

Writing this blog is my way of taking control of my own educational process. I think about whatever I want, at my own pace, in my own way. Why then post it on the internet? Anything I write has probably been written somewhere else, and probably with more insight and clarity than I'm providing. If my goals are personal education, why paste it all over this blog, advertising it for people to see?

If I were to keep these developments to myself, I wouldn't think them through fully. I'd work until I saw the gist of the argument, then say to myself, "the rest is just mopping up details from here. better move on to something more important."

If I don't take the time to mop up the details, the bits of understanding I did build flit away. So by assuming the responsibility to clean things up to the point where someone could potentially read through the post and work out the results for themself, I force myself to clean up the ideas in my own head. That's what I'm after.

What does a determinant have to do with a cross product? pt. 2

2008-09-29T14:20:00.000-07:00

In the previous post, I showed that the quantity

and the determinant

are both tests that the three vectors , , and are independent. Now we'll go about showing that they're actually the same test - that they're the same number.

You may be familiar with the geometric interpretation of as the volume of a parallelipiped (box) with sides , , and .

Just a quick explanation - is a vector perpendicular to both, whose length is the area of the parallelogram they form. See stolen image below:

If I take that and dot it with , I get the projection of into the perpendicular direction, times the area of the box. That is, I get the height of the box times the area of its base. That's the volume. Wikipedia visualization:

There's one subtlety, which is that the volume of the box records its "handedness" in the sign of . By handedness, I mean that if we looked at a mirror image of the box, its volume would be multiplied by negative one. If you do that funny thing with your right hand, and find that points the same sort of direction that does, you have a righthanded box. Otherwise it's left-handed.

This interpretation of as the volume turns out to be the same as the determinant. To see why, we'll make a list of properties that the area of the box has, and then show that those properties imply the normal calculation of the determinant.

If the sides of the box are orthonormal (all perpendicular to each other, and all unit length), its volume is one (or minus one for a left-handed box).
The volume of the box is linear in a vector. By this I mean that if we multiplied any side of the box by some constant, we'd multiply the volume of the entire box by that constant. Also, if we make two boxes that share two vectors but differ in the third, the sum of the volumes is the same as the volume of the box whose third side is the sum of the vectors. Notationally:

and similarly for and

If you switch the roles of any two vectors, you'll multiply the volume of the box by minus one. This is obvious if you flip and , because then the angle between them gets measured the opposite way, so its sine is the opposite of what it used to be. The other possibilities you can visualize if you just play around with it a little. It works.

With these properties, we can uniquely determine the value of the volume of the box from its components, and show that it's the determinant. It's a little long for three dimensions, so you can work through that on your own if you want, after seeing it in two dimensions.

Imagine we have two 2-D vectors in component form, and we put them in a matrix. Then we ask for the determinant. The first property says the determinant of the identity is one.

Using linearity in the vectors, we find

Now we switch the vectors in the identity and multiply the determinant by -1.

and

Finally use the linearity within a vector, which states

where we could drop the determinants of matrices with an entire column zero because they are dependent vectors.

That is the determinant for a 2-by-2 matrix. If you want to do it for 3-by-3, you can use the exact same process of breaking it down into 27 matrices with three nonzero entries each, and then see by permutations of the identity which ones are zero, and where the plus and minus signs go. Out will fall the normal expression for a 3-by-3 determinant.

So that's where that comes from.

Why it Matters

In the previous post, we were trying to solve the problem of taking three vectors and finding a linear combination of them that yielded a fourth vector. That is, find , , such that

where , , , and are all arbitary vectors.

We worked far enough just to test if the problem had a solution, but we found this result:

From there, we need to find by dotting it with a unit vector perpendicular to both and .

Plug this into the equation for , and the stuff about cancels, leaving

Those are determinants. That's why they matter. They let you solve systems of linear equations, such as the system of linear equations that results from the problem we just solved.

Debate Debate

2008-09-27T02:14:00.000-07:00

Relying Exclusively of Volatile Eruptions of the Reptile Brain Mark: You should watch the debate.

Preposterously Idealized Self SErving Rational Mark: Will it be interesting?

REVERB Mark: You better be interested! It's the Presidential Debate!

PISSER Mark: I don't follow that.

REVERB Mark: You're a functioning, adult member of the society now. You live in a democracy, and that's a great privilege. It's your responsibility to take part in the governance of this nation, and you didn't even vote in the last election! Now you better turn on that debate.

PISSER Mark: Given my demographic, occupation, community, and personal interests it's not difficult to guess which candidate I support. I doubt that watching the debate will change that.

REVERB Mark: That's not the point. Even if you know who you're going to vote for, you still need to stay well informed.

PISSER Mark: Why?

REVERB Mark: Because you're more privileged than any but the tiniest fraction of human beings alive, and yet you take it all for granted. You live in a nation where you personally can have real, meaningful participation in the political process, and it's your responsibility to exercise that right.

PISSER Mark: I admit I frequently take the perks of life in America for granted, but I don't see why following politics is a responsibility, rather than a personal choice. I'd rather do something else.

REVERB Mark: What do you think politics is, then?

PISSER Mark: A hobby.

REVERB Mark: A what?

PISSER Mark: A hobby. A pastime. Like being a sports fan. You can read about sports in the newspaper, listen to experts talk about it on the radio. Then you can choose which team you like, and "support" them by buying bumper stickers and paying for admission to their games. All the same applies with politics, except the bumper sticker becomes a sign on your front lawn and the ticket to the game becomes a vote. They're both just ways for people to have something to get riled up about without doing much harm.

REVERB Mark: Not an apt analogy at all. Sports are just games. It doesn't really matter who wins the World Series. But it sure matters who wins the Presidency!

PISSER Mark: A little bit, maybe. My job prospects might be a little dimmer with one candidate in charge than the other. Their differences on social issues could affect how pissed off the people around me are. The foreign policy they enact will cause me to read different headlines in the news.

REVERB Mark: That's not just news items. All that stuff is really happening out there, to real people. It's not just all about you. To some people it makes a huge difference who's President.

PISSER Mark: But we're discussing my personal decision on whether to watch the debate.

REVERB Mark: Look, I know what you're thinking. You're thinking, "my vote is just one gumball in Candy Land. Doesn't make a bit of difference. The vote in my state always goes to the same party, and one more vote cast either way can't change the result." But you know what? One hundred thousand votes can change the result. And what if one hundred thousand people just like you are using the same logic?

PISSER Mark: Are you saying that if I go to vote it will change one hundred thousand lazy people's minds?

REVERB Mark: You know that's not what I'm saying. I'm saying you shouldn't vote because you think your particular vote is going to make all the difference. You should vote because it's a statement that you give a rat's ass about the future of this country! Not just for yourself, but for everyone you share this country with. And since the US is a leading world superpower, everyone you share this Earth with. And before you vote, you need to be informed. So you need to watch the debate.

PISSER Mark: Watching just one debate won't make me informed. Becoming a competent participant in democracy is a significant commitment.

REVERB Mark: One that every adult American ought to make.

PISSER Mark: Have you met many adult Americans? On the whole, they're not the ones I want making any important decisions. Basically, ability to choose a good Presidential candidate will undoubtedly fall on a curve like this:

So let the people who are innately interested in the whole process, and can do a good job of it, pour their efforts into politics.

REVERB Mark: All the more reason for you personally to get involved. You're way at the top of that curve. Remember high school? You don't want to common man making the decisions - then YOU better get involved and do something about it.

PISSER Mark: That's not really my point. I don't want to invest huge amounts of time and energy into becoming an "informed citizen", which by the way makes the dubious presupposition that it's possible to make a good choice from behind a computer screen, because it doesn't interest me much. Even if I did watch the debate, I don't know enough about foreign policy or economics to decide who I agree with. I'd inevitably fall back on judging demeanor and rhetoric. I could do more good by putting that effort into something I'll enjoy.

REVERB Mark: Like what? What can you do that's more important than your civic duty?

PISSER Mark: I don't know... plant a tree or something, I guess.

REVERB Mark: Oh come on. You've never planted a tree in your life.

PISSER Mark: I did, actually. I planted the free pine tree seedling they gave me third grade on Arbor Day. I think dad accidentally ran over it with the lawn mower.

REVERB Mark: Not the point. That's not what you will do with that extra time. What you will do is read stupid websites or play flash games, or something inane like that.

PISSER Mark: Or maybe I'll study for a while. Or work on that project I had planned, or...

REVERB Mark: Oh please. You've already got three tabs open and they're Pacman, SomethingAwful.com, and then something more awful than that.

PISSER Mark: sigh. Where is it streaming from?

What does a determinant have to do with a cross product? pt. 1

2008-09-26T21:19:00.000-07:00

You're probably aware of a nifty mnemonic for calculating a cross product.

If you take a dot product of the above expressions with another vector , you'll get

What a crazy magic coincidence that vector products and determinants can be calculated the same way!

I won't say magic coincidences don't exist, but I will say that when you find one it may behoove you to look a little deeper. There is a deeper connection.

I'll hint at it today by posing a certain problem, then showing that the problem has a solution whenever either expression, the determinant or the vector product, is nonzero. So is it two different conditions that show the problem can be solved, or the same condition in different guises? We'll deal with that later.

Imagine you are given four random vectors in space, . Then you are asked to express the fourth as a linear combination of the first three. That is, find such that

I won't solve the problem. I'm just asking for a test to see if it's soluble (besides throwing water on it). Especially, I want to see if it's soluble for all possible , not just special cases.

Using A Basis

Set up a standard Cartesian basis in three dimensions, and write out the vector equation in components

which is a system of linear equations that can be written in matrix form as

(Aside: If you know how to tell to make the vertical columns the same size in those matrices, drop me a comment.)

If you've taken linear algebra, you'll recognize that the problem can be solved only if the above matrix has a non-zero determinant. That same determinant is the one we were interested in earlier, since it's also the value of

Without a Basis

Vectors and define a plane. They're two independent vectors (I'm assuming they're independent - if they're dependent, you might as well just get rid of one of them for the purposes of solving this problem, but it's clear that in general that won't work) in a two-dimensional space, so they form a basis for that space.

Break down into two parts, the part in the plane of and , which we'll call and the part perpendicular to the plane, .

and are irrelevant to the perpendicular part of the equation, so we're left with

We need to find , so we better dot with a unit vector in the perpendicular direction.

Combining the previous two equations lets us solve for .

Unless is by extraordinary chance in the plane of and (and we're interested in the general case, not special ones), we need to be finite, meaning needs to be nonzero. This condition also has built into it the requirement that and be independent, so it's a sufficient condition for the problem to have a solution.

Summary for today

The equation

can be solved for any whatsoever if

also, the same is true if

That doesn't prove the determinant and the vector product are identical, but it's hinting at why it's true. Tomorrow I'll probably post a little about what determinants are, and the geometrical interpretation of the vector product.

A X (B X C)

2008-09-22T14:23:00.000-07:00

Any intro to physics course will teach you the identity

It's easy to prove by explicitly writing out each vector in component form, cranking through both sides of the equation, finally getting the same thing on each. But there's some insight to be gained in looking at the identity from a purely geometric point of view. Warning: the following is a bit dense, but also short. Go one sentence at a time and you should be fine.

Together, and are a basis for a plane. is perpendicular to the plane. Take that perpendicular vector product, cross it with , and you get a vector perpendicular to the perpendicular to the plane. That's back in the plane again. So we know that the double cross product is in the plane somewhere. But any vector in the plane can be written as a sum of and (that's what it means to be a basis), so we only need to find the right coefficients. Those coefficients come from requiring that we find a combination of and perpendicular to . Just dot the expression
with , and you'll get zero.

That proves both sides of the equation

point in the same direction, but doesn't prove they have the same length. It's not crazy difficult to prove the they have the same length without using a coordinate system, but I don't have a neat insightful trick for it, either.

I started by noting that you could break down into perpendicular and parallel components to the plane defined by the other two vectors, and that as far as the identity is concerned the perpendicular part of doesn't matter (plug it in and you'll see). Then, because both sides of the equation are proportional to the lengths of all three vectors, you might as well divide by their lengths to make them unit vectors. All that's left is to find the lengths of both sides of the equation in terms of trig functions of the angles between the vectors. You have tree angles between the coplanar vectors. Write one as a sum of the other two, expand out all the trig functions, and fiddle around with it a bit until you see that it's actually a trig identity. That gets the final part of the theorem in there.

Space Elevators

2008-09-22T10:29:00.001-07:00

Space elevators are in the news again, so I'll write a post covering the "intro to physics" aspects of what a space elevator is and why it's hard to build. This doesn't address the actual difficulties of building the elevator - which are in the engineering. It's just to get the overall conceptual framework.

Geosynchronous Orbits

A space elevator is a theoretical long tether that reaches all the way up into outer space and just hangs there all by itself. In theory, once you put one up, you can use it to climb up into outer space very cheaply compared to chemical rockets.

How could a long tether "just hang there?" Because it's in orbit around the Earth. A space elevator is a special type of satellite, specifically, one in a geosynchronous orbit.
The idea here is that it's possible to have a satellite always directly above the same spot on Earth. This isn't the normal state of things. Most satellites are in low-Earth orbit and whip around the planet in fast circles every few hours. But the higher up you put a satellite, the longer it will take to orbit. If you put it high enough, it will take 24 hours to orbit. That way, the Earth can spin underneath it at just the same pace it's orbiting, and by happy coincidence the satellite appears to stay in the same spot, to an observer on the surface of the Earth.

There are a lot of restrictions on where you can place such an orbit. First off, the shape and size of an orbit, along with the gravity of the planet and the laws of physics, determine how long an orbit takes. You can't just blast your rockets a little softer and expect to orbit slowly. If you do that, you won't be going fast enough to orbit any more, and gravity will pull you back to the ground.

If you want an orbit that takes 24 hours and is "smooth" (always has the same angular velocity), you're restricted to making it a circular orbit at a height that works out to be 36,000 km above the surface of the Earth. The radius of the Earth is only 6,000 km, so a space elevator is long enough to wrap around the Earth multiple times, and together they look roughly like a flying sparrow with a two-meter long hair trailing out its butt. Hold a hula hoop around a soccer ball. That is roughly the correct scale for how high up this circular orbit would be. By comparison, most orbits for satellites and space shuttle trips and things like that would be loops hovering about half a centimeter from the ball's surface.

Second, a geosynchronous orbit has to be over the equator. A satellite in a circular orbit doesn't have to be over the equator, of course. Take that soccer ball and hula hoop from before. Spin and twirl the hula hoop however you like. As long as the Earth is still at the center, you have a valid orbit. But it will only stay over the same spot if it's at the equator. Otherwise, the satellite wanders north and south again over the course of its day. If you try to cheat by simply sliding the hula hoop up some, so it's parallel to the equator but over say, the 45^o line of latitude, it's no longer a possible physical orbit. The soccer ball isn't at the center any more, so its gravity would inevitably drag the orbit back down.

Tethers

Imagine a satellite in geosynchronous orbit. Now imagine you drop a little fishing line from the bottom of the satellite, reaching down towards Earth. In response, the rest of the satellite will have to rise up a tiny little bit to keep the center of mass at the same height. Now drop that line lower and lower. The rest of the satellite continues creeping up a little bit higher, and the line keeps reaching lower, until eventually you have a tether that reaches all the way from Earth to geosynchronous orbit. That's a space elevator.

The first problem that creates is that the fishing line is under enormous tension. If it's made of any ordinary material, it will rip apart. To see why, imagine a spot on the line a few thousand kilometers above the Earth. If you cut the line there, the stuff below it would clearly not be in geosynchronous orbit - it isn't high enough up. So all that stuff would fall.

Since that material below the cut point isn't falling in an operational space elevator, something must be holding it up. That something is tension in the line. The higher up the space elevator you go, the more material there is to hold up, so the tension gets greater and greater. By the time you reach geosynchronous orbit, the line has to support the weight of tens of thousands of kilometers of line beneath it. That's a lot of weight, and any normal material would imply rip apart. So to build a space elevator you need something really strong. So far, there are no long ropes that are anywhere near strong enough, although some people speculate that it's possible to build one out of carbon nanotubes.

Geosynchronous Orbits II

The gravitational potential of two point masses is given by

is a constant, and are the objects' masses, and is the distance between them. We're giving the Earth (which in real life takes up lots of space) a single definite location at one point. Gauss' law allows us to do this for a body with spherical symmetry, but really it doesn't matter for the calculation I have in mind, because the distance for a space elevator will turn out to be much greater than the radius of the Earth.

Using a reference frame in which the center of mass of the Earth does not move, and assuming this is an inertial frame, the kinetic energy of the system is

Where is the magnitude of the satellite's velocity in this frame. We'll use ordinary spherical coordinates, so that the position of the satellite is given by . We're only interested in circular orbits, where is a constant, and . Then the velocity is

where

.

But we need a circular orbit, for which

The Lagrangian is

Then the Euler-Lagrange equations give

which yields

where is a constant depending on the gravitational constant and the mass of the Earth.

Setting and plugging through the numbers gives r=4.2*10²⁴ km. That's the height of a geosynchronous orbit from the center of the Earth.

Tension II

The force on a differential element of the cable, , is

where is the force of gravity on the segment and is the change in tension from below to above that point. is the mass of the length element, and is its acceleration. The positive direction is defined to be away from the center of the Earth. The element is accelerating in a circular orbit, with acceleration

This gives

If the mass of the element, , is written in terms of the cross-sectional area of the cable at a given length , and the density , we can then write the tension as

The difference between r and l is that l is a dummy variable for integrating over. It begins at r_e, the radius of the Earth, and goes to whatever radius we're considering.

The tension is zero at the surface of the Earth, then increases until the integrand becomes zero at geosynchronous orbit. Then the tension slowly slinks back down to zero at the far end of the elevator.

To determine whether or not the cable will rip, it's not the total tension that matters, but whether the tension is greater than the cable's tensile strength. And the cable's strength is proportional to it's cross section . This suggests a strategy for building the elevator - down near the ground where the tension is small, make the cable very thin. It doesn't need to be strong there. Up near geosynchronous orbit where the tension is very high, make the cable thick to deal with the high stress. This plan could be called "tapering" the cable.

The ideal taper for the cable is one in which the tension per unit area is constant all along the cable. To make the problem workable, assume the tension doesn't drop to zero at the surface of the Earth (in which case the "ideal" cable would be zero thickness there, and all the way up), but instead goes to some finite value .

Then we have:

this can be combined with the expression for the tension to give a differential equation for the cross-section length

which you can solve to find

where C is just a constant set so that the cable has the appropriate width at the surface of the Earth.

Depth Perception, part II

2008-09-20T23:18:00.000-07:00

The goal is to understand why a mirror allows depth perception, but a television or painting does not. In the first part of the answer, I demonstrated that depth perception relies on seeing the same object with two eyes. Eyes detect to angle to the object, but the two eyes will detect different angles. The bigger the difference between these angles, the closer the object.

Here's a diagram of a television and a mirror. A pair of eyes looks at each, and sees a purple ball.

Notice that for the television, the light rays converge at the screen. But the light rays bounce off the mirror in different places, converging only at the ball.

Although the ball is actually in front of the mirror, there's a problem. The light you detect is the light right at your eyes. You can't detect the light's entire path. So if you want to reconstruct the path of the light to see where light beams intersect, you just have to guess. That's what your brain does, and it simply guesses that the light doesn't bend or bounce or anything like that. It just goes back straight to the source. So the situation for the mirror is that the light actually travels the solid line, but your brain thinks it travels the dotted line.

An object can appear to have a depth well back behind the mirror. Objects on a television screen will always appear to be right at the television screen, since that's where the lines leading to those objects converge.

The essential difference is that the television or painting is generating the light you see. The light from any given point of the painting radiates out in all directions, always showing the same object.

The mirror doesn't generate its light - it reflects it. So if two observers are in different places and look at the same spot on a mirror, they see different objects there, not the same.

Depth Perception, part 1

2008-09-19T00:53:00.000-07:00

This post will take a look at how depth perception works. The next post will build on this background to answer yesterday's question.

First experiment:

Close your eyelids.
Now open them.

If you did the first part of that experiment right away, you shouldn't have known what to do next, because you wouldn't be able to read the instructions. Close your eyes and you can't see any more. Conclusion: you see with your eyes. (Actually, this just shows that eyes are "necessary but not sufficient". But hey, let's not split corneas here.)

Second experiment: Cut open a cow's eye and diagram it. You should get this:

The salient point is that your eye is designed to give your brain the information "lots of light is coming from this direction, not so much from that direction, and none at all from a third direction". Overall, there are lots and lots of directions. When you put them all together, you get an image.

That's what one eye does. The way it does it is optics. First, light comes into the eye and passes through the lens. All light that comes in parallel, regardless of whether it enters the left or right side of the eye, then gets focused to the same part of the retina. That's the glory of the lens. So all the light falling on a certain part of the retina came from the same direction in space, not from entering the same part of your eye.

This directional knowledge lets your build an image. Here is a picture you can look at with your eyes:

To see each pixel, you must look in a slightly different direction. Putting each pixel with its appropriate direction allows you to build the image back up in your mind.

So eyes only care about the direction light comes in. However, you have two eyes, and they're in different places. That means light from the same source will enter from different angles. Here's a gratuitous graphic:

Okay, so light from an object, like a baseball you want to catch (actually, depth perception is not the tool baseball outfielders use to flag down a fly ball, but that's another story), enters your left eye from a direction (&theta_l, &phi_l), and enters your right eye from a different direction (&theta_r, &phi_r). How far away is the object? Here, &theta is the polar angle, the angle between the line from the top of your head to your feet and the path of the light. &phi is the azimuthal angle, telling how far to the right or left the direction is. Some people use a different convention. Mine is like this:

You stand with your head pointed towards z and your face towards x.

Your eyes are separated (obviously) by a distance d. Also, we'll make the approximation that the object is far enough away from your face that it's essentially the same distance from each eye. In that case, the polar angle &theta will be the same for each eye.

The three lines, one connecting your eyes, the other two from each eye to the object, form a triangle like this:

Boom! The math is right in there! I actually just spent ten minutes searching for and installing this thing that should let me start 'ing up my blog, and then I didn't even use it (except to make that sexy logo).

Anyway, yup. where d is the distance to the object, and w is the width of your eyes. &Delta is the difference between the azimuthal angles, and &phi is that azimuthal angle (either one. the difference is small). I made two simplifying assumptions. The first is that the difference in the angles is small, which just means the thing isn't right up in your face. The second is that the object is close to right in front of you, which is necessary to be able to use depth perception on it.

Nice.

Now there's a problem, which is that you only get so much accuracy in determining the angles. As &delta becomes smaller and smaller, small errors result in big problems for d. So depth perception only works for objects close to you, or a medium distance away. For very far objects, you need to use other clues. The same problem occurs for astronomers attempting to find the distance to a star by parallax.

New Problem: Depth Perception

2008-09-18T02:48:00.000-07:00

No matter how good your cameraman, or the equipment he's using, it's impossible to create the illusion of depth on a television. You might be able to tell some things are further away than others, but they won't ever look truly 3-D the way the real world does when you have two eyes open.

On the other hand, a mirror can give you a perfect illusion of depth (I am not using "perfect" in a technical sense here). A mirror image world is just as depth-rich to your binocular vision as the genuine article.

So why is that? Why can't a television gives depth perception? And what can a mirror do that a television can't?

Meditation

2008-09-16T01:12:00.001-07:00

A lot of people have asked me about my vipassana meditation retreat last month. The only ones who've received answers are the people who asked me in person, since they're much harder to shirk.

The meditation camp was a psychologically intense experience. It was important enough to me that if I want to explain it, I think I should put the effort into explaining it well.

I think it will take me about two weeks to write up all my thoughts as a memoir of my time there. I'll post the entire story at once when I'm done. In the mean time, thanks to everyone for your interest - it shows me that putting the effort into writing the story will be worthwhile.

Answer: Circling Bugs

2008-09-15T23:02:00.000-07:00

The mathematical answer to yesterday's problem about the circling bugs is much prettier than the previous one about the demon and lake. We might as well just solve the problem for "n" bugs, then let n=4 for the square. It is a bit of a special case, though, which you might be able to solve completely by intuition.

We'll let n bugs be distributed uniformly around a unit circle, and set the units of time so their walking speed is one. Each begins chasing its nearest neighbor. Because the situation is exactly the same for every bug, the shape cannot distort. If it starts out a square, it stays a square. It can only get larger or smaller, translate, or rotate.

Also by symmetry, it can't translate. Which way would it go? If you let it translate in the direction of one bug, you're being unfair to all the others, but the problem is perfectly equable.

The shape will rotate and change size. Begin by imagining the case of infinitely many bugs, all point-sized, that form a complete circle. Then if they chase the bug in front of them, they'll simply march around the circle without it ever changing size at all.

If there are finite bugs, the shape must shrink as it rotates. In the case of a circle, the bug can never get any closer to the next guy he's chasing, because whoever he's chasing is running directly away from him at the same speed. But for a finite-sided object, the chased bug isn't running directly away any more. He's running at an angle, so the distance between the bugs decreases. Hence, the shape shrinks.

Down to business: find the equation for the motion of a bug.

Let's write an equation for its velocity in polar coordinates. If the position of the bug is (r,&theta), then the position of the next bug is (r,&theta+2&pi/n). Skipping a step or two, the velocity of the bug is (-sin[&pi/n],cos[&pi/n]). You get that by taking the geometric fact that the exterior angle of a regular n-gon is 2&pi/n, and noticing the deviation of the bug's path from a pure circular motion is exactly half that.

The bugs all start a distance of one from the center, so they will all meet there after a time (and distance walked) 1/sin(&pi/n). If there are lots of bugs, we can make a small-angle approximation to get n/&pi. More bugs, longer to meet.

Integrating the velocity above (and remembering that things are a little more complicated in polar coordinates, because d&theta/dt = v_&theta/r(t)), we have, for a bug that begins at (1,0)
p(t) = (1-sin[&pi/n]*t,-cot[&pi/n]*ln[1-sin(&pi/n)t])
A logarithmic spiral.

For the case n=4, our formula says the bugs will walk a distance 2^1/2 before colliding. In this special case, the motion of the bug "running away" is at a right angle. He isn't running away at all, so much as trying to execute a circular orbit. So, from a bug's perspective, the distance between the bugs shrinks just as fast as it would if the one he were chasing were simply stationary.

Circling Bugs

2008-09-14T04:01:00.000-07:00

Here's another problem I liked from "How Would You Move Mount Fuji". I'll state the original problem, then the generalization, and hope that when I try to solve it tomorrow I can do a little better than I did yesterday with the boat and demon. I think this one is actually a famous problem.

There are four beetles located the the four corners of a square. At the same moment, they all start crawling directly towards the next beetle clockwise around from them. That beetle they're aiming for is also crawling, so each beetle continually adjusts its course so that it aims straight towards the next beetle.

Will they collide in finite time? How far will they walk before they collide?

Generalizations: Let there be "n" beetles at the corners of a regular n-gon. Same questions. Also, if you drew a path marking the route of a single beetle, how many degrees around the center of the square would it subtend (i.e. how many times does the beetle wrap around the center)? What is their exact path?

(Kind of) Answer: Lake and Demon

2008-09-13T20:42:00.001-07:00

This post is about yesterday's problem. I have to admit - the question is tougher than I thought it was (it's another of those I asked before solving). The question in the book is simply, "Imagine the demon goes four times as fast as your boat, how can you escape?" That question is purely conceptual. I'll answer it, then go as far as I can right now towards, "what is the minimum ratio of the speed of your boat to speed of the demon for you to be able to guarantee escape?" (Obviously less than 1/4).

I'll set the units of distance so the radius of the lake is 1, then set the units of time so the speed of your boat is also 1. The speed of the demon is D.

At first glance, you can realize that if D < Pi, you can just shoot for the shore opposite of where the demon is standing, and you'll get there first. But you can escape even if your ratio is worse than this.

Qualitatively, here is the plan: You start rowing in the direction directly away from the demon. He picks a direction and starts running that way, anticipating where you'll land. But it's to your advantage to keep the demon 180 degrees around the lake from you - as far as possible. To do this, just angle your boat a bit so whichever way the demon runs, you stay directly opposite him.

As long as you're still close to the center of the lake, this is easy to do. Your angular speed can be quite high because it's a small circle around the center. The demon's angular speed is low because he's so far out. Hence, you can keep him 180 degrees away from you.

What's the maximum distance from the center of the lake where you can still force the demon to be 180 degrees away from you? That happens when your maximum angular velocities match, which is at a distance 1/D.

After that, you still have to be able to make it to shore. If you were going on a straight shot, your speed would have to be great enough so that the time for you to get to shore is less than the time for the demon to run halfway around the lake. Algebraically,
(1 - 1/D) < Pi/D
D < 1 + Pi

The new strategy takes us from D < Pi to D < 1+Pi (slightly greater than 4), but is there something even better? Maybe it's worth your time to curve away from the demon despite the fact you can't keep him a full 180 degrees away from you.

A moment's thought will tell you undoubtedly yes - you can do better than moving out to the cutoff circle and then sprinting straight for shore. Suppose you are making that beeline for shore. The demon has picked a direction and is running towards your expected landing spot. If you turn your heading a very slight amount away from his side of the lake, you now have two components to your velocity - a "radial" component towards the edge of the lake and a "circumferential" component around in a circle. By the Pythagorean theorem, the squares of these speeds add up to the maximum speed of your boat. So if the component of circumferential speed is small, the change in the radial speed is second-order. Explicitly:

S_r² + S_c² = 1
S_r = (1 - S_c²)^1/2

using the binomial theorem, for small S_c, this is

S_r = 1 - 1/2*S_c²

So you can pay a very small price to your radial velocity in return for some circumferential velocity. Losing radial velocity means it takes you longer to get to shore, while gaining circumferential velocity means it takes the demon longer to get to your landing spot. There's a positive effect to veering off course, and a negative effect. But since you can make the ratio of the negative effect to the positive effect arbitrarily small, it'll always be worth your while to veer away from a straight shot to shore at least a little bit, assuming your goal is to land with the demon as far away from you as possible.

Let's go back to the beginning and try a more quantitative approach.

If we want to escape the demon, we should always go at full speed. The demon's strategy is simple - he also goes at full speed in whatever direction takes him closer to you. If both directions are equal (because you're 180 degrees apart), he picks one at random. So the only thing we have to choose is our bearing, based on our current location and the demon's current location. What should the criterion be? I don't have a proof this is optimal, but if your goal is to land on shore as far from the demon as possible, let's make the criterion:

Choose the heading that maximizes the quantity:
(dI/dt) + (dr/dt*I)/(1-r)
Where "I" indicates the distance from the demon's position to the "intercept point", the point on the shore closest to your boat, and "r" is your distance from the center of the lake.

I derived this criterion with the following reasoning:
Imagine you were going to make a beeline to shore from wherever you currently are. If we divide the demon's distance to the intercept point "I" by his speed "D", we get the demon's expected time to intercept, I/D. On the other hand, your expected time to get there is just your distance from shore, 1-r, because your speed is 1. Finally, if we divide those two times by each other, we get a ratio I/(D*(1-r)). If that ratio is bigger than one, you've made it - you could escape with a bee-line trajectory now. Unfortunately, if D>Pi, that ratio at the beginning of the problem is less than one. But by choosing a clever trajectory, we can try to push that ratio up to one, and then escape. So at every moment, we're going to choose the heading so that this ratio is as high as we can get it a short time "dt" from now. To do that, just take the derivative of the ratio with respect to time and maximize it.

If you take the derivative, you'll noticed I dropped the constant "1/(D*(1-r))". This yields the condition above. As long as the derivative of that ratio is positive, we'll assume you still have a chance. Once you can't push the ratio any higher he's got you, since if you were going to escape, the ratio would go to infinity right before you got away, and if he's going to catch you, the ratio goes to zero.

Choose "@" to be the angle between your heading and the heading straight for shore. I'll let you work through the calculus for yourself should you so desire. The goal is to get "@" as a function of "I" and "r". That is, choose the best course whatever the current situation is. I found:

tan(@) = (1-r)/(I*r)

Reality check: I expect that as you near the shore, you should aim more and more towards it. As r->1, @->0.

I have to be careful with this formula - it only applies when the demon is NOT directly opposite you. If he's directly opposite you, then angling your boat brings him closer no matter which way you turn - a factor not accounted for in my calculation. So this formula only applies once the boat is already past the cutoff circle and racing for shore.

Now we've worked out a strategy, all that's left is integration. Unfortunately, the integration isn't very easy. We'll start things off by introducing a polar coordinate system (r,@). The center of the lake is (0,0). The demon starts out at (1,0), and the boat, having already completed the first stage of its maneuvers, begins at (1/D,Pi).

Let's say the demon chooses to run up towards the top of the circle (increasing @). Then his position as a function of time is (1,D*t).

Your position is trickier. It depends on your bearing, which depends on both your own position again and the demon's position. If you write out these equations, you'll see that they're pretty hopelessly coupled. I get

dr/dt = I*r/((1-r)^2 + (I*r)^2)^1/2
d@/dt = (1-r)/(r*((1-r)^2 + (I*r)^2)^1/2)

where I = @ - D*t, and "@" and "r" are your polar coordinates as functions of time.

So from there, I'm basically analytically screwed. If you'd like to do the numerical analysis and figure something out, please let me know what you find.

The sad part is, even doing the integration doesn't actually find an upper bound for D. It only finds an upper bound assuming my particular strategy is the best possible strategy. Just because it seems like a good one in my head doesn't mean there isn't one that's better, and I honestly have no idea how to go about proving either:
A) such-and-such a strategy is optimal for escaping the demon
B) no strategy can do better than such-and-such

Moral: These toy problems, though simple and well-defined, can be tougher than they look when you actually start to bite into them. I was originally just attracted to the little trick of realizing you could force the demon away from you by making small circles about radii close to the center of the pond. But after I got that I naturally wanted to go a step further. Turned out the next step was biting off more than I could chew.

Lake and Demon

2008-09-12T23:26:00.000-07:00

This is a slightly-modified form of one of the most interesting problems from a book I just finished, called How Would You Move Mount Fuji? The book is about using such problems in job interviews, but the context is irrelevant to how clever the problem is. So without further ado:

You're on a boat in the middle of a circular lake. On the outside edge of the lake is a demon who wants to kill you to death. He is bigger than you and has pointy teeth. With poison on them. Death poison. He can't go in the water, so he runs around the edge of the lake, waiting to intercept you when you land.

If you can make it to shore without the demon being right where you land, you'll be okay. You're fast enough over land to run away and escape. But if the demon is right exactly there when you hit shore, you lose. Your wife and kids grieve. Your life insurer goes broke.

Assume the demon and your boat both have a maximum speed, which they can maintain indefinitely. What is the minimum ratio of the speed of your boat to the speed of the demon for you to have an infallible strategy for escape? What is that strategy, and how long does it take you to get to the edge of the lake evading the demon (as a function of the ratio of your speeds, and the ratio of your speed to the size of the lake)?

Interview with Brian Greene

2008-09-12T17:47:00.000-07:00

Brian Greene gave me a half hour interview earlier this week. I asked him some obligatory questions about his new book, Icarus at the Edge of Time. After reading the book (twice. It's very short.), I think Greene is a bit out of his league in writing fiction. Nonetheless, he gave some good insight on the relationship between thinking like a scientist and thinking like a writer, and how one could do both.

I was nervous enough during the interview to have a hard time afterwards remembering exactly what we had talked about (I recorded it, of course), but not so nervous as to be completely incapable of creating follow-up questions on the spot after getting his answers to my prepared ones.

What's more - Brian Greene appeared a little bit nervous, too. His tapped his foot throughout the interview and began every single response with the word "well" (which I deleted from the transcription, along with about two hundred of my own "um"s, "uhh"s, and "so"s.)

This should come out in the school paper, in some shorter form, once the new school year starts, but for the time being check out the link to Ideotrope at the top of the post.

Conjuring Trees From Thin Air?

2008-09-08T10:32:00.000-07:00

At TED.COM, Jonathan Drori gave a talk claiming that graduates from MIT could not correctly answer the following four questions:

An apple seed is small, but an apple tree is large. Where did all that extra mass come from?
It's easy to light a flashlight bulb with a 1.5V battery and two wires, but can it be done with just one wire? (Presumably, no "lateral thinking" solutions, such as cutting one long wire to make two wires, are allowed.)
Why is it hotter in summer than in winter?
Draw a diagram of the solar system.

I believe that most students from Caltech would get all four mostly correct, although of course no one would be able to draw a perfect diagram of the solar system off the top of their heads. (However, I did recently have a conversation with one Caltech graduate who, although he could answer question one correctly, tripped up on its inverse. That is, he thought that when a human loses weight, that weight comes out in their poo.)

Question 1 is the one I want to focus on, because I think that the guy who was giving a talk about how people get it wrong, was himself wrong. In his talk, Drori claims that 99% of the mass of a piece of wood comes from the air.

Eighth grade biology gives us the tools to figure this out. Trees are made of cells, and cells bags holding organic molecules floating around in water. So the mass of a tree is mostly organic molecules and water. The organic molecules come in different varieties, but lots of them are carbohydrates having the empirical formula CH₂O. The series of reactions that makes these carbohydrates is photosynthesis, summarized by:

CO₂ + H₂O -> CH₂O + O₂

So the mass of the tree comes primarily from carbon dioxide and water. There are other things sucked up by the roots - the tree needs plenty of nitrogen to build amino acids, various other chemicals such as phosphorus and iron, and these along probably add up to more than 1%, but let's just think about the carbon dioxide and water.

The carbon dioxide comes from the air - that was Drori's point. The most important element in life is carbon -sometimes people refer to terrestrial beings as carbon-based life forms. Trees get their carbon from the air. Fine. But that's not the whole story. Trees need lots of water, too. And they get it from the ground.

How do I know they get their water from the ground? Because in my new apartment, there are a bunch of trees on the back patio. They're all dead, because the previous tenants believed what Drori said, and thought the trees would suck up water from the air around them. Nope. Empirical evidence in the form of very dead trees now suggests that you have to pour water on the ground around the trees, and then they can suck up that water through the ground.

PS - when you lose weight, you're going the other way from photosynthesis, turning carbohydrate and oxygen into water and carbon dioxide. So the weight you lose comes out partially as water, which has lots of ways to exit the body, and partially as CO₂, which is exhaled. Every time you breathe out, you're losing weight - at rate of about a pound every 50,000 breaths (based on breathing 15 times a minute and burning 2000 calories a day).

End of the Problem-of-the-Day Contest, On to New Adventures

2008-08-12T01:14:00.000-07:00

The problem of the day contest ended one week ago. Our winner at camp was Lucas Brown, followed closely by Jonathan Lilley. Tyler Fitzgerald and Ben Zinberg were third and fourth. Fantastic prizes were awarded to all.

Now that I'm going on to new things and new ways of life, the problem of the day will probably become the "occasional problem when I see an interesting one, supplemented by other various things that are on my mind". That's sort of what the blog was intended for anyway. It covers various things I'm thinking about, but only has a one-arcsecond window on the breadth of human experience.

My next adventure is to travel to the California Vipassana Center in North Fork, to learn a certain form of meditation. It's a ten-day course that involves a rather extreme schedule of 10 hours of meditation a day, no talking, reading, writing, exercising, doing physics, looking at females, contacting the outside world, or in general doing anything that would distract from the task at hand (you wouldn't want to be distracted from doing nothing, would you?) So I'll be completely off the radar until Aug 25 or so.

Answer: Square Root Calculator

2008-08-08T17:39:00.000-07:00

Pick a number. Set the ball at that number on the ramp and let it go. Watch where it falls. The number where it falls is the square root of the number on the ramp.

This works because the velocity of the ball as it leaves the ramp is proportional to the square root of its kinetic energy. Its kinetic energy comes from the potential energy it originally had while sitting on the ramp. That's linear in the height.

Incidentally, not all the ball's initial potential energy gets converted into kinetic energy of translation. Some is also converted into kinetic energy of rotation. However, the fraction of kinetic energy that goes into rotation is constant, so this doesn't affect our ability to build a calculator. It just means the marks on the landing zone have to be closer together than they would for a frictionless, sliding ball.

The thing about the square root calculator is that it uses two linear scales. That's why it's a great machine - it doesn't take any difficult work to decide where the marks go. The laws of physics do the actual computation.

There were a variety of answers to the challenge to create your own analog calculator. Most of them missed the mark of what I was shooting for.

For example, an abacus can do multiplication. It's a physical system, as well. So I suppose it's a physical system that does a useful computation. But it's not interesting from a physics perspective. It's interesting only from a logical perspective. The abacus is probably made of wood, but doesn't depend essentially on the properties of wood to work. I could make the abacus out of stones sitting in from of me, or fancy metal rings, or little holes that I dig or fill in with sand at the beach. The abacus only cares about the positional relationship between its parts, not their physical interactions. I could even, if my powers of imagination were strong enough, picture an abacus with all its beads in my head, move them around mentally, and do abacus-style multiplication completely inside my head. If you imagine a different universe, in which the laws of physics work differently, the abacus wouldn't care. Multiplication would still be the same, and you could still do it with any device whose pieces could move in relation to each other the same way they can in an abacus.

So slide rules are out, too. Electronic computers can certainly compute useful quantities and solve differential equations numerically, and they are physical systems. But again, the magic of the computer is in its logical gates rather than its physics. Some other answers, like tree rings and sundials counting the passage of time, were not really doing a mathematical computation.

If you're good at circuit analysis, you could build an analog computer to calculate all kinds of interesting things. In fact, people used to do this all the time before digital computers got so good. So I won't talk about circuit analysis.

Instead, here's a fairly simple system to do multiplication. It's a tube of dilute gas, say argon. There's a piston allowing you to change the volume to whatever you want. The volume is marked off on the sides like it is in graduated cylinders or Nalgene bottles. You adjust the piston up or down with a stack of weights. If you want the cylinder to have less volume, stack on more weight. By counting the weights you have on there, you can get the pressure. There's a thermometer inside the tube positioned so you can easily read it. Finally, there's a bunsen burner and some cold packs lying around, so you can add or remove heat at will.

To a good approximation, the gas obeys the equation of state PV = nRT. So choose the amount of gas in your cylinder, along with your units pressure, volume, and temperature, so that nR = 1. Then PV = T. You can now do multiplication. Start by setting the weight stack so that the pressure inside is the first number you want to multiply. If the volume is lower than the second number you want to multiply, add heat. If it's greater, suck heat out. Keep adding or subtracting heat until the volume is equal to the second number you want to multiply. Now read off the temperature from the thermometer, and that is the multiple you sought.

The above machine is also a division machine, because you could equally well set the pressure to the denominator, add/subtract heat until the temperature is the numerator, and read off the volume using V = T/P. Similarly the square root calculator is also a square calculator. Take a guess at the square of a number, drop the ball from there and see where it lands. If it lands in front of the number you're squaring, drop it from higher. Repeat until you find the number on the ramp such that when you drop the ball from there, it hits your target number on the landing zone. That's the square.

Any system that satisfies the differential equation df/dx = a*f is a potential logarithm/exponential calculator, since its solution is f = e^ax = (e^a)^x. An example is a system whose friction is proportional to its amplitude. This occurs in the viscous flow of a fluid, for example. A runaway chemical reaction might obey this law, at least over some regime. Population growth of organism does this. The expansion of the universe due to a nonzero vacuum energy does this, since the rate of growth is proportional to the amount of vacuum, but growth is itself the change in vacuum.

Trig functions come from any physical system that obeys d²f/dx² = -a*f, where "a" is positive. A mass on a spring is the canonical example, all sorts of things will exhibit this same behavior. The reason is that particles will minimize their potential energy. When you're near a potential energy minimum, you can expand the potential as a Taylor series, but the constant term is not physically relevant, and the linear term will be zero because you're at a minimum. Then the quadratic term dominates over a certain regime, and in that regime the system obeys the differential equation for trig functions.

Answer: Lifeguard New Problem: Square Root Calculator

2008-08-05T13:44:00.000-07:00

Lifeguard
Let the guard be a distance "b" from the edge of the beach, a horizontal distance "h" from where the drowning guy is, and let the drowning guy be a distance "d" from the shoreline.

Finally, let the lifeguard run to a point "x" on the shore before entering the water. See picture:

Then the time it takes him to get out there is

We want to find "x" that minimizes this time, so take a derivative and set it equal to zero.

Now take another look at the geometry of the problem. The expression above can be rewritten as

which is our answer: Snell's Law. Notice that all the actual distances fall out - only two of their proportions matter.

Square Root Calculator
The Exploratorium in San Francisco has a device they call a "square root calculator". It's just a ramp, a ball, a jump, and a landing zone, like this:

How is this a square root calculator? How does it work? Can you suggest another simple physical system that would operate as an multiplication calculator, logarithm calculator, trigonometric calculator, etc?

Answer: To Swim or Not to Swim New Problem: Lifeguard

2008-08-04T01:11:00.000-07:00

To Swim or Not to Swim
The lake is a hyperbola.
Actually, it's a section of a hyperbola on one side of the axis of symmetry, then the mirror image of that section on the other side.

This is good knowledge to have if you're building optical equipment. Light follows paths of stationary time, so light passing through a lens is very similar to this problem. Most lenses are built with edges that are sections of sphere, not hyperbolas. This is just because it's easier to manufacture them that way. Then if you're doing some fancy optics, you'll have to correct for the "spherical aberration" of your lens.

Proving the the hyperbola is straightforward, but algebra-y. Just set up a coordinate system where the middle of the lake is (0,0) and the edge of the lake is given by x(y). Then the edge of the make must be long enough to satisfy the condition
((d-x)² + y²)^.5 + 2x = c
where c is some constant, equal to (d² + (w/2)²)^.5.

I got this by letting "y" be the height at which you cross the lake. Then I counted the distance to the point where you enter the lake, and the twice the distance to the axis of symmetry (since you go twice as slow in water). This must be equal to the same number for all "y" by the specification of the problem. If it is, the rest of the problem (you're going from A to B, not A to the axis of symmetry) follows because the second half is the same as the first half backwards, and also takes time "c".

The equation you get is a hyperbola. This makes some sense. Consider the limit as the width of the lake is much greater than your initial distance from it. Then in that limit, the additional path length added by crossing the lake one unit higher is one unit (for a trip to the axis of symmetry). So in that same limit, the width of the lake must be directly proportional to height of the lake, so that it slows you down the right amount. That's what a hyperbola does - goes asymptotically to a line. Ah well, I just realized that random babbling probably doesn't get the idea across very well. But I also don't care that much because if you're reading this I'm sure you can figure it out for yourself anyway. I just thought it was neat.

Also, note that you can quickly get into situations where the lake extends back behind your starting point, so if the lake is too wide, it's impossible. It'll be faster to swim across. This happens when (w/2)² + d²>4d². If that happens, you'll need the start farther away from the lake, or make it out of something that's harder to swim through.

New Problem: Lifeguard
Today's and yesterday's problems were inspired by Feynman's popular book "QED", which in turn is based on these lectures.

You're a lifeguard who sees a swimmer drowning somewhere off shore. You need to get out to the swimmer by a combination of running on the beach and swimming in the water. See the picture:

One particular path is fastest. Assume you run at speed n₁ and swim at speed n₂. You take the fastest path. What is the relationship between the angle of your path on the beach relative to the normal, and your angle into the water relative to the normal?

Answer: Square Wheel New Problem: To Swim or Not To Swim

2008-08-02T22:30:00.000-07:00

Square Wheel
Sorry to be lame, but I'm not going to prove this one. It's a catenary.

To Swim or Not to Swim
There's a lake in the middle of the field, as shown. You want to go from point A to point B as quickly as possible. You can go over land or over water, but you're twice as slow in the water as one the land.

Assume (or prove) the best strategy is to head straight to some spot on the shore, swim straight across, and then head on from the far shore to point B by land. The shape of the lake is such that it doesn't matter where you choose to enter - all paths take the same amount of time. What is the shape of the lake?

Assume: points A and B are equidistant from the center of the lake. The lake is symmetric about a vertical line through its center.

Answer: Hourglass New Problem: Square Wheel

2008-08-01T13:18:00.000-07:00

Hourglass
It's hard to say exactly what the plot would look like, if you try to account for things like the varying pressure on the sand, the varying height of the pile in the bottom of the hourglass, etc. We can say generally what the weight-vs-time graph looks like by finding an important property. Start with

F = dp/dt
Fdt = dp

Integrating, we find that the total force-seconds is equal to the change in momentum. But the hourglass has zero momentum when the sand is all up top, and again zero when the sand is all in the bottom. So the integral of force over time for the whole experiment must be zero.

The force on the hourglass is from gravity and from the scale. Because the force of gravity isn't changing, that means the force read on the scale must, on time-average, be equal to the weight of the hourglass. Any "bumps" in the scale's readings must be canceled out by "valleys".

That tells us about the entire outline of the plot. As for the details, it depends on the acceleration of the center of mass of the hourglass (F=ma). Just as the sand starts falling, the center of mass is certainly accelerating down. The force of gravity is stronger than the force exerted by the scale, and so the reading goes down for a time. When the sand hits the bottom, it accelerates up. If the rate of sand flowing were constant, and all the sand fell the same distance, and the sand in the top weren't sinking, then the total acceleration of the hourglass would be zero. There are complications so that while the sand is falling, the center of mass might have some nonzero acceleration, but it would be small.

As the last bits of sand fall, the acceleration of the center of mass must be upwards - it's going towards a state where it's no longer falling. So the basic idea of the outline should be something like this:

Square Wheel
A square wheel rolls on a bumpy road smoothly, and without slipping. What is the shape of the road?

Answer: Writhing Chain New Problem: Hourglass

2008-07-31T12:19:00.000-07:00

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.

Hourglass
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.

Answer: Whisper Dish New Problem: Writhing Chain

2008-07-28T01:26:00.000-07:00

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?

Answer: Guitar New Problem: Whisper Dish

2008-07-24T02:47:00.001-07:00

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

Where f_n is the fundamental frequency of the string at the n^th 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
Tension^1/2*density^-1/2*length^-1

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).
l_n+1 = 1/r*l_n
l_n+1 - l_n = (1/r - 1)l_n = (1-r)/r * l_n

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?

Answer: Inverted Pendulum 2 New Problem: Guitar Frets

2008-07-23T15:31:00.000-07:00

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.

Guitar

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

Visualizing Complex Beats

2008-07-22T14:44:00.000-07:00

Take two complex exponentials and add them.
e^i*a*t + e^i*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:
e^i*a*t + e^i*b*t
e^i*t*(a+b)/2(e^i*t*(a-b)/2 + e^i*t*(b-a))
e^i*t*(a+b)/22Cos[t*(a-b)/2]

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.

Answer: Inverted Pendulum New Problem: Still Inverted Pendulum

2008-07-21T19:37:00.000-07:00

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

2008-07-21T11:50:00.000-07:00

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, T_x, does change the shape of the bridge. If you make T_x 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 T_x 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?

Answer: Spinning Fish Tank New Problem: Suspension Bridge

2008-07-20T02:58:00.000-07:00

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?

Playing Catch in a Rotating Spaceship (revisited)

2008-07-18T18:02:00.001-07:00

My Mathematica Demonstration is now online.

Answer: Planet Formation New Problem: Spinning Fish Tank

2008-07-18T15:32:00.000-07:00

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)

2008-07-18T13:38:00.000-07:00

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.

e^ix = 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

e^ix + e^-ix = 2Cos(x)

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

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

where the last bit about the hyperbolic cosine is just from the definition Cosh(x) = [e^x + 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

2008-07-18T13:35:00.001-07:00

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?

Melting Ice in Coffee

2008-07-16T09:49:00.000-07:00

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).

What I've Learned About Lecturing

2008-07-15T01:53:00.001-07:00

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.

Susskind's High School Black Hole Lecture

2008-07-12T23:50:00.001-07:00

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:
http://en.wikipedia.org/wiki/Schwarzschild_radius
http://en.wikipedia.org/wiki/Entropy
http://en.wikipedia.org/wiki/Hawking_radiation
http://en.wikipedia.org/wiki/Planck_length
http://en.wikipedia.org/wiki/Black_hole_thermodynamics

Playing Catch in a Rotating Spaceship

2008-07-09T15:42:00.000-07:00

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.

Triangles on A Sphere

2008-07-04T19:46:00.001-07:00

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

2008-07-04T19:23:00.000-07:00

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

2008-07-04T00:59:00.000-07:00

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.

First Thoughts

2008-06-09T23:18:00.000-07:00

It is impossible to begin. Every sentence must contradict itself.

Every thought that rushes up is beaten back down immediately by others, equally vaporous. I need to think clearly. When I learned I had failed quantum mechanics, and will not graduate, I wrote to the professor the most honest statement I was capable of producing.

Professor C-,
[The registrar] informed me I haven't done enough work to pass your class.  I'm
sure this is true.  The work I turned in was:
Midterm (I got about 25%)
HW 1
HW 2
Final (I answered six questions before time ran out).

I've attempted HW 3 a few times, but I still don't understand it.  I've
tried reading the lecture notes and other homework assignments but have
not made any significant headway.  I admit I did not learn more than a
fraction of the material for the course, and that this has become a
pattern for me over the last two years.  The total work I turned in for
professor M-'s course was similar, except I was unable to do any of
the problems on his final.

I don't understand the fundamentals from ph125a and ph125b, or peripheral
knowledge like ph106ab or ACM95.  Ph125c is mostly incomprehensible to me.
I should have attended the lectures earlier this year, studied
independently while I wasn't enrolled, and entered third term prepared.  I
don't have a good explanation for why I did not.  The only poor
explanation I have is that at the time, I mistakenly (perhaps
delusionally) believed I could catch up in a week or two of concerted
effort.  On other occasions I had the feeling, on opening the text to an
advanced chapter, that the magnitude of elementary material I would have
to study before I could make any progress on the assigned work was so
great that the effort was futile, and so I set it aside and did something
else.

Ph125c is my last requirement for graduation.  I no longer have
aspirations towards a career in physics, but I do realize that any career
path will become nearly unnavigable if I never finish my degree.

I'm trying to think of a reason why, if you were to allow me to make up
the missing work over the summer, I would actually get the work done.  It
would involve radically different behavior patterns than any I've
exhibited in the past for longer than a couple of weeks.

I believe I would get the work done, but I have always, in the past,
believed I was going to get work done, and I cannot explain why I believe
this.  I've learned that drawing up schedules and making commitments for
myself provides me with no psychological incentive if I do not believe
they are fundamentally important.  Only a complete shift in my mentality
could suffice to change my habits.

I'm unable to come up with a reason why this summer should be different
than last that is convincing to me.  The closest I can come is that this
summer I am more jaded and less inquisitive, and therefore less likely to
abandon my work in favor of other diversions.

If you think it could be productive, I'd like to meet with you in person
to discuss Ph125c and whether a viable option remains to finish the
course.

I apologize for the long and personal nature of this email, but I thank
you for your time and attention.

Mark

There was the hope that, if he knew failing the class was going to hold me here another year, he might decide 30% or so was sufficient, after all. But I didn't actually believe this email would accomplish anything. Nothing material. I felt a need to simply state the case. To acknowledge a professor as a human, and to acknowledge my own academic struggles as those of a human.
I have not collaborated on homework more than a handful of times over the course of my entire Caltech career. I have hardly ever sought help from TA's or professors. I never understood learning as an interpersonal process. I only understood academics as something I must do.

Am I, now, by stating it so plainly and so publicly, crying a pathetic mewl in hopes of sympathy? Am I "coping"? I hope not. I hope I am beginning.

I need a radical and deep alteration of the fundamental patterns of my cognition. In these first moments it is impossible to speculate. When I think, my only thoughts are how pointless all endeavors are. They only end in failure. We're six billion pinwheels flapping in the breeze. We dream of giant wind turbines, but we generate no power. Only a little noise.

Of course I was angry - the was enough work turned in it wouldn't have been immoral to pass me, by any stretch. But this anger was impossible to sustain under the waves of self-loathing, washing up regularly in a rising tide, each crest bowling me over with the force of its impact.

I watch them walking, driving, talking, yelling. What is their point? What do they think they can ever accomplish with such a trivial thing as existence? I cannot understand how this enduring pattern began. How I got to where I am now, tottering on the edge of the nest, realizing I have wings but no feathers.

I don't know in what direction even to wander. My belief, my hypothesis, is that I am at war with my ego. In moments of quiet solitude, when I am not thinking about myself, I am the happiest person I know (also, at that moment, the only person, but that is another thing). My mind can fly off in incommunicable flights of joy. I can sit by myself, laughing uncontrollably at the idea of a joke I do not know how to articulate, nor imagine, nor describe. It's simply a shape and a rhythm bouncing back and forth between my neurons. It emerges unbidden and enraptures me. Sometimes.

But I kill it. An innate, insane irrationality consumes me as soon as I think about the future, and my place in it. I'm important. Therefore, I must do important things. I can't even tell you right now what one important thing is. This will pass, in a few days. But will there be any lasting impression? Will I cease defeating myself with my life of fits and starts?

I am even watching myself typing, laughing at the predictability of my petty emotional response. I am mocking myself for being so crestfallen by something so trivial. Pitying myself for being a simple cognitive automaton, no different from any of millions of others of the same mold.

And I wonder why I recognize intrinsic meaning in such an image. Why should a sentient being be downtrodden simply because it is insignificant? Aren't the glories of life made more vibrant when reflected off the shining friendly faces of others?

The Harmonic Series

2008-06-09T02:39:00.001-07:00

I've been daydreaming about how I would teach an introduction to calculus in just two hours. It's hard because when I start thinking about calculus, which I generally take for granted, I begin to see holes and gaps in my understanding. (What precisely is an "infinitesimal" differential element supposed to mean, anyway?)

Just now, instead of another hole, I found this nice smooth pebble, my first for this blog. It's a simple proof that the harmonic series diverges. The harmonic series is this one:

The sum "goes to infinity", meaning that if you take any number at all, after enough terms the harmonic series grows bigger than it. The proof proceeds via reductio ad absurdum.

Assume the limit is some number S. Then break the sum into two smaller sums, one of the odd terms and one of the even terms, like so:

All the terms from the original sum are there in one of those two smaller guys. Also, s₁,the sum of odd terms, must be greater because each individual element is greater (i.e. 1> 1/2, 1/3 > 1/4, etc.)
So algebraically:

Now for the trick. Just multiply s₂ by 2.

If s₂ is half of S, then s₁ must be the other half.

Wait! s₁ is supposed to be bigger, but now they're equal to the same thing. To check, plug (8) and (9) into (6), and you're left with:

Which is a contradiction. Therefore, the hypothesis is wrong, and the sum is not a finite number. It is positive, since each of its terms is, and so the sum simply gets bigger and bigger without bound.

After a quick googling, it turns out my proof is number 8 on a long list (continued here) of proofs, but that no one found it until 1979!