Monday, December 17, 2007

Heisenberg's uncertainty principle

Well, it’s getting to be about time to wrap up the semester, and I’ve saved the most important thing for last. Perhaps you’ve gotten the sense in the last few weeks that quantum mechanics is one incredibly bizarre thing after another, and this is certainly true. However, in a very real way, all that unbelievable weirdness actually emanates from a single source: the uncertainty principle. More than that, in fact: everything I’ve told you so far about quantum mechanics can actually be derived from the uncertainty principle.

A friend of mine once referred to uncertainty as “the beating heart of quantum mechanics”. This may be overly poetic, but it is true that the uncertainty principle is one of the two pillars on which all of quantum mechanics is based (the other being Pauli’s exclusion principle). In spite of being so important, it’s usually misquoted, and it’s very often misunderstood.

So what exactly is uncertainty? Actually, when Heisenberg originally wrote his paper in 1925, he didn’t really explain it very well. (I personally think he was hedging his bets. Remember, this was long before the debate about the realist vs. orthodox positions had been resolved.) Basically, what he said was this: the more precisely the momentum of a particle is known, the less precisely its position can be known, and vice versa.

To be a little more mathy about it, let’s talk about the quantities Δx and Δp. Strictly, these are the standard deviations of the position and momentum, x and p, though people typically just call them the uncertainties. (I’d rather not actually define these, just because I haven’t mentioned a couple of things that go into the definition. However, if you don’t remember what a standard deviation is, it’s safe to think of Δx and Δp as the “spread” in the position and momentum. Of course, Rebecca might be upset with you if you don’t remember what standard deviations are…) Anyway, what Heisenberg said was this:

ΔxΔp ≥ hbar/2

That is, the product of the two uncertainties is greater than some constant. This equation may seem fairly innocuous, but it’s actually an unbelievable result: it doesn’t matter what the actual constant is, the fact that the uncertainties must be greater than zero is the incredible thing. I’m really not sure that there’s anything in human experience which might prepare us for this. What Heisenberg had shown, even though he himself may not have realized it at the time, was the incompatibility of position and momentum.

When a lot of authors (including Heisenberg himself) talked about uncertainty, they made it sound as if it was somehow the experimenter’s fault. For instance, one way to measure a particle’s position is to hit it with a beam of light. If you hit it with low-energy light, you can do your measurement without disturbing the particle too much, so its momentum can be fairly well known. The trade off is that low-energy light isn’t good at resolving the particle, so you don’t know much about where exactly it is. Conversely, you could pummel it with high-energy light, in which case you’d get a great sense of where exactly it is, but the high-energy light would send the particle skittering off to wherever, so we’d have no idea what its momentum is.

This is not only silly, it’s downright misleading. The only conclusion that we would draw from that story is that this particular way of measuring a particle may not be very good. Sheesh, maybe the people doing these experiments just aren’t very smart; it sounds like a clever person would just find a less obtrusive way of measuring the particle’s position. For that matter, maybe the problem is even simpler: perhaps we just need to spend more money to buy a better machine.

But this isn’t it at all! What the uncertainty principle really says is much deeper than this. Recall from the orthodox position that a particle doesn’t always have an exact position or momentum. According to uncertainty, the more definitely a particle has a position, the less definitely it has a momentum, and vice versa. It’s not that an experiment that’s good at finding a particle’s position necessarily has to be bad at finding the particle’s momentum, it’s that when you measure the particle’s position, it doesn’t really have a momentum, so there’s nothing to measure!

At the two extremes, the uncertainty relation is actually even more surprising: if Δx = 0 (that is, the position is known exactly), then Δp → ∞ (that is, the particle has literally every momentum at once). On the other hand, if the particle has an exact momentum (Δp = 0), it exists everywhere simultaneously (Δx → ∞).

Alright, no doubt about it, that’s weird. But what makes uncertainty so important? What makes it the source of all weirdness in quantum mechanics? Actually, uncertainty is more general than I’ve let on: I started off talking about position and momentum, but the wonderful thing about uncertainty is that it applies to any two physical observables, be they position, momentum, energy, spin, or whatever else we can dream up that we might want to measure. Recall that every physical observable has its own operator in quantum mechanics. What the generalized uncertainty principle says is this: if A and B are any two operators, then

(ΔA)2(ΔB)2[A, B]2/4

This second, more general form of the uncertainty principle is the real engine here: we know that physical observables in quantum mechanics are represented by operators, and we know that sometimes two operators don’t commute. What uncertainty does is take this purely mathematical fact and turn it into something physical: because of uncertainty it is now impossible for some observables to have definite values at the same time; it’s why quantum mechanics has wave functions in the first place, and therefore probabilities, and by extension it’s the cause of the realist/orthodox/agnostic debate. So, no fooling, uncertainty really is the motivating force behind everything we’ve talked about in quantum mechanics.

Anyway, now that we know what commutators are (see last week’s blog), we have all the math we need to evaluate this relation. Well, almost. Actually the “absolute value” brackets mean something slightly different when we’re talking about imaginary numbers, but for the problems we’re doing you’ll be fine if you just remember to make sure the square of the commutator is positive. With this, you should be in good shape for this week’s homework.

Monday, December 3, 2007

Compatibility in quantum mechanics

Last week I claimed (and in the Bell’s theorem blog I proved) that particles generally don’t have physical properties until those properties are specifically measured. This week I’d like to explain a little more about how measurement actually works. We already have some sense of how to “do” quantum mechanics: we know what we’re looking for (the wave function Ψ) and, at least in principle, how to get it (solve the Schrodinger equation).

But what do we do once we have Ψ? Well, we start applying operators to it, and see what happens. To review, if we let an operator A act on Ψ, there are generally two possible outcomes:

(1) We get back the original wave function times a constant: AΨ = aΨ, where a is just some number. In this case the particle already had a definite value for that property. For example, take the momentum operator p. If we apply it to a particle’s wave function and find that pΨ = 4Ψ, then the particle was already in a state with a definite momentum (of 4, though we’d need to know the units before this would mean much).

(2) We get something else. Then the particle does not have a definite value for this property, and the wave function can only tell us the probability of getting a particular value if we make a measurement.

So what’s the big deal? We find Ψ and then start applying operators to it. If we get outcome (1), fantastic, we get a value for that physical property without doing any more work. If we get outcome (2), well, we’ll just go measure that property! Then it will have a definite value, and we can move on to the next operator. When we get done with all the operators we care about, we’ll have a long list of values representing the properties the particle has.

Well, this is tempting, but unfortunately there’s a teensy little problem. Recall from last week that a measurement actually changes the wave function to something else. In many cases, measuring one property will destroy any information you had about another property, because these two properties cannot exist simultaneously. If this is true, the two measurements are said to be “incompatible”. Incompatibility is yet another unbelievably weird feature of quantum mechanics.

As an example, let’s consider the spin of an electron. I can measure the electron’s spin in the x-direction or the y-direction, but I can never know both simultaneously. If I measure the spin in the x-direction I will always find it pointing exactly left or exactly right. Say I measure it pointing left. If I measure it again in the x-direction then I will definitely see it pointing left again. If I measure it in the y-direction, however, then I will have 50/50 odds of seeing it point either way. Once I do that measurement, however, I will have fixed it to be pointing either up or down, and a left/right measurement will once again yield 50/50 odds. Conclusion: the electron cannot have a definite spin in more than one direction at a time.

If you happen to have a more experimental turn of mind, you might be starting to wonder how we can ever get any work done at all. If making a measurement can cause you to lose the information you got from your previous measurement, why bother? Fortunately, not all measurements are incompatible. The trick, as it turns out, is figuring out which measurements can be made while still keeping the information from previous measurements.

So how do we determine which observables (physical properties) are compatible? We use an operator called a “commutator”. If A and B are any two operators, then the commutator of A and B is

[A, B] = AB – BA

Uh-oh. AB minus BA? Isn’t this always going to be equal to zero? Actually, no, not always. It would be, of course, if A and B were numbers, but they’re not; they’re operators. In fact, if you’re in EM this semester, you’ve probably already run into an example of something like this: the cross product. If instead we have two vectors a and b, then when you take their cross product the order matters quite a bit, since a x b is not the same as b x a.

Anyway, In the homework this week we’ll look at examples of two operators that have a nonvanishing commutator, but for now you’ll just have to take my word for it that the commutator is not always zero. (By the way, to use one more bit of jargon, if the commutator of A and B does happen to be zero, then A and B are said to “commute”.)

Once we compute the commutator of two operators, we have a mathematical way of determining whether their observables are compatible (that is, can be measured at the same time): if the two operators do commute, then their observables are compatible, and vice versa. (If they don’t commute, then their observables aren’t compatible.)

As an example, the energy and momentum of a free particle are represented by the operators H and p, respectively. As it turns out, these two operators commute ([H, p] = 0), so it’s possible to measure a particle’s energy without damaging what you know about that particle’s momentum, because the two quantities are compatible.

What about an example of operators that don’t commute? I’m glad you asked, because that brings us to this week’s homework. (As always, if you get stuck, feel free to ask me for help.)

Thursday, November 29, 2007

Bell's theorem

Hey folks,

For those of you who are interested, I went ahead and wrote up the proof for the "orthodox" position. Let me know what you think.

Monday, November 26, 2007

The Schrodinger equation

I just know you’ve been in suspense for two weeks now, so I think it’s time I told you which of the three viewpoints (realist, orthodox, or agnostic) is correct. It’s worth mentioning that only recently did physicists discover there even was a right answer to this question. (Well, not that recently; the news is over forty years old by now.) Before the 60s you could basically pick whichever viewpoint appealed most to you, and each one had a large group of fans.

In 1964 though, an Irish physicist named John Bell demonstrated that it makes a real, observable difference whether the particle had a definite (though unknown) position before the measurement. As Griffiths puts it: “Bell’s discovery effectively eliminated agnosticism as a viable option, and made it an experimental question whether the realist or orthodox position is the correct choice.” So if any of you picked agnostic thinking it was the safest route, you might be surprised to learn it was the first one to bite the dust.

I’ve struggled a lot with the question of whether to show you Bell’s proof, but I finally decided against it. I realize this probably means that Taylor will never believe me, but unfortunately the proof requires a fair bit of background information, as well as being somewhat complicated in its own right.

But let’s say for the sake of argument that you believe me, and agree that there is a right answer. So what is it? Alright, fine, I’ll tell you: it turns out that the correct position is… the orthodox one. Believe it or not, the particle did not have a position before you measured it. This is why quantum mechanics is so insanely weird. Before the measurement, that particle wasn’t entirely anywhere; however, it was somewhat everywhere. It’s a safe bet that no human on earth can really “comprehend” this, and it’s difficult to even communicate this fact through language, but there it is anyway: particles just don’t have positions.

And this isn’t just true of positions! Get down small enough, and you’ll find that particles don’t have definite momenta either, or angular momenta, or energy, or anything else that can be measured. The act of measurement gives these things definite values. If you thought relativity was weird, get a load of that.

In the comments last week, Taylor made an extremely good point. The complaint was this: suppose you measure a particle’s position and find it was at point P (from the graph in the “wave function” blog post). Then, immediately after that, you make a second measurement. Who’s to say you won’t find it far away in, say, Atlanta? If you make the measurement fast enough, wouldn’t the particle have had to travel faster than the speed of light?

That’s a legitimate objection, and if quantum mechanics allowed things like that to happen, we’d have a real problem. But wait! We’re now asking a very different question! We’re now asking what happens when we perform a second measurement on the same particle. The trick is, the first measurement changes the particle’s wave function, so now it looks like this:

As you can see, there’s now zero chance of finding the particle anywhere except at point P. So if you make a second measurement (and third, forth, etc.) you will continue to get the same value back. That is, before the first measurement there was no telling where it would be, but once you’ve made that measurement, any subsequent measurement will give the same value. (To throw around a bit of jargon, the first measurement is often said to “collapse” the wave function into the function above, which is called a “Dirac delta function”.)

Alright, time to get back to work. In the “wave function” blog post I did a sort of comparison between classical and quantum mechanics. In both cases, we consider a particle of mass m, and for simplicity, we only let it move in the x-direction. Then we subject it to some kind of force, F(x, t), and we’re ready to roll. In classical mechanics, the goal is to figure out the particle’s position at any particular time, x(t), and in quantum mechanics (where particles often don’t have positions), the goal is to figure out the particle’s wave function at any particular time and place, Ψ(x, t).

But how do you get these in the first place? In classical mechanics, the answer is easy: that’s what Newton’s second law is for! Once you know that F = ma, your worries are pretty much over. Sometimes it’s a teeny bit more complicated, because you’ll be given a potential energy instead of a force. But that’s no big deal: you know from the definition of work that F = -dU/dx (in this case, I’m using U to mean potential energy). If we put this in, Newton’s second law reads: md2x/dt2 = -dU/dx. Plug stuff in with the right initial conditions and you’re done.

So that’s how you get x(t) in classical mechanics, but how do you get Ψ(x, t) in quantum mechanics? As it turns out, we get it by solving something called the Schrodinger equation. So the Schrodinger equation does the exact same thing in quantum mechanics that Newton’s second law does in classical mechanics. This may not seem like a big point, but I went through an entire year of quantum mechanics before anyone bothered to tell me this. So in case you ever take a course in QM, maybe this will save you some grief.

Anyway, maybe you’re wondering why I haven’t actually shown you what the Schrodinger equation looks like yet. That’s because it’s your job to tell me, which brings us to this week’s homework.

Thursday, November 8, 2007

The wave function

Okay, now I’m going to sum up in one paragraph almost every problem you can get in your mechanics class. Let’s say we find a particle of mass m and we let it move back and forth in the x-direction. While we’re at it, I’m going to apply a force to the particle, F(x, t). Basically the whole idea of classical mechanics is to figure out, given that force, what the particle’s position will be at any particular time, x(t). Pretty much everything else follows from that. Yours for the asking are the velocity, dx/dt; the momentum, mv; the kinetic energy, (1/2)mv2; or whatever else Rebecca decides to ask you for.

The point of this is simply to give you something to compare quantum mechanics to, to help you get oriented. So let’s compare: suppose we have the exact same setup (a particle of mass m moving in the x-direction) in quantum mechanics. The problem is that at the microscopic level, particles don’t necessarily have positions, so it doesn’t make sense to ask about x(t). Instead, what we’d like to find out is something called the wave function of the particle: Ψ(x, t). This is the interesting thing, and once we’ve got it we can figure out anything else we want.

So the wave function, Ψ, does much the same thing for you in quantum mechanics that the position, x, does for you in classical mechanics. It’s the big thing, from which pretty much anything else can be figured out. You say you want to know about the particle’s momentum? Simply apply the momentum operator to Ψ and see what happens. The kinetic energy? We’ve got an operator for that, too. Potential energy? Oh, jeez, we’ve got dozens of operators for that. (Of course, you can’t actually apply any of these operators yet, because I haven’t actually told you what they are.)

Alright, so that’s what the wave function does, but what exactly is it? The best answer was provided by a German dude named Max Born, who provided the “statistical interpretation”:

If Ψ(x, t) is the wave function of a particle, then Ψ2 is the probability of finding the particle at point x, at time t.

To be a teeny bit more mathy about it:

ab Ψ(x, t)2 dx = {probability of finding the particle between a and b, at time t}

That is, if you’ve got the wave function of a particle Ψ, and you’d like to know the probability of finding it between two particular points, just take the area under the curve of Ψ2 from a to b. For example, the picture is a graph of Ψ2, so we would be much more likely to find the particle near the middle than out at the edges.

If you’re thinking all of this is a little bit hokey, I can hardly blame you. What kind of a theory is this? We get the wave function, the most important thing we can learn about this particle, and the best we can do with it is find the probability that a particle will be in a certain place? Why can’t we say beforehand exactly where it’s going to be when we measure it? Is there just something wrong with quantum mechanics?

Actually, all this can be summed up in one question. Let’s say a particle has the wave function in the picture, and I actually do measure the particle’s position. It so happens that I discover the particle was somewhere right around point P. The question is: where was the particle the instant before we measured it?

There are three general opinions you might have on this (courtesy of David Griffiths):

1. The realist position: “Well, obviously, the particle had to have been at point P. That’s where we found it after all. If you find your socks under your bed, clearly they were under you bed right before you found them, you just didn’t know it.” It certainly seems hard to argue with this logic, and Einstein himself took this position. If it’s true, the theory of quantum mechanics is a bit of a letdown, because it must be incomplete. That is, the particle had a position before we measured it, and quantum mechanics just couldn’t tell us what it was.

2. The orthodox position: “Believe it or not, the particle wasn’t really anywhere until you measured it. The measurement itself caused the particle to have a position.” So how about that? In addition to being really, really weird, this claim requires us to believe some pretty farfetched things. In particular, it means there really is such a thing as a “random” event: it was not possible, even in principle, to know where the particle was going to be before you measured it. Even if you knew every true thing in the entire universe, you still could not have said where that particle would be.

If this surprises you, maybe you can relate to this: when I was a kid, I thought everything in the universe was completely determined. A coin flip, for instance, only seems random; if you knew exactly how it was flipped, and exactly how the air moved around it, you could figure out whether it would come up heads or tails every time. A computer claims to generate “random” numbers, but it really just takes a seed number from its internal clock and performs calculations on it. Even humans would just be machines, following deterministic paths (though they might think they had free will). In short, I figured if you knew all the starting conditions of the universe, you could figure out everything that would happen, from a coin flip to what a person would say at a particular moment. However, if the orthodox position is right, even with a perfect knowledge of the universe, there are some things that can’t possibly be known beforehand.

3. The agnostic position: “I refuse to answer the question. I don’t want to, and you can’t make me.” Actually, this isn’t as ridiculous as it sounds. The point is, it makes no sense to ask about where the particle was before a measurement: the only way to know that would be to make a measurement! In that case, it wouldn’t be before the measurement anymore, would it? It’s just goofy philosophy to ask about something that can’t, by definition, be known. It’s like asking “what if time suddenly started to run at half-speed?” There’d be no way to check. It may sound like you’re describing two different things, but you’re actually just describing the same thing in two different ways.

This question bugged me for years, but I always thought it was basically a matter of opinion. Believe it or not, though, there actually is a correct answer! Incredibly, it makes an observable, testable difference which position is right. I’ll give it away next week, but I’d like to know: what do you think? Position 1, 2, or 3? Or do you have a different position altogether? Write a comment and let me know.

In the meantime, here's the problem for this week. Feel free to write me if you get stuck, too.

Friday, October 26, 2007

Quantum mechanics

Offhand, I can’t think of any physics theory that was discovered just by one person. Often, though, there’s one go-to guy who wrapped the whole thing up in a paper somewhere, and whenever we think of that theory we think of them. Newton’s mechanics, for example, or Maxwell’s electrodynamics, or even Einstein’s relativity. But quantum mechanics isn’t like that; it was put together by a collection of extremely clever folks over the span of maybe twenty years or so. Many of them didn’t even like each other. Heisenberg, for example, couldn’t stand Schrodinger’s work, even though it turned out they were doing the same thing.

As a result, even today no two physicists completely agree on what quantum mechanics is. No one knows what its fundamental principles are, or how it should be taught. (Although a friend of mine likes to say that “quantum mechanics is just Heisenberg’s uncertainty plus Pauli’s exclusion.” I’m not sure what that means, but I like it because it’s short and snappy.) I heard a story once about a kid who misses the very first day of kindergarten, when the teacher explains what “numbers” are. He gets there the second day and everyone already seems to know, so he’s too embarrassed to ask. He doesn’t want anyone to catch on that he’s got no idea what a number is, so he works incredibly hard. He learns arithmetic and algebra, and later trigonometry and calculus. Eventually he becomes a famous mathematician and wins the Fields Medal. But inside he always feels like a fraud, because the truth is he still doesn’t know what a number is, just how it works. That’s sort of how I feel when I “do” quantum mechanics.

Over the next few weeks, I’d like to crack the QM door and let you peek in. Unfortunately, a peek is really all that’s possible, because there’s a fair amount of math that goes along with the subject. Things like spherical harmonics, Fourier transforms, and Hilbert spaces, along with a host of others. Probably you’ve never heard of any of that stuff before, but you’ll run into some of it if you’re interested in technical careers like engineering or the sciences.

Okay, so in my opinion, the very first thing to know about quantum mechanics is the idea of operators. In physics there are all kinds of observable quantities that might interest us; things like position, velocity, momentum, angular momentum, potential energy, kinetic energy, and so forth. In classical mechanics (what you’ve been studying) these quantities are just variables, which hopefully you’re familiar with by now. In quantum mechanics, each of these quantities gets an upgrade—they’re turned into operators.

So what is an operator? Well, I suppose it doesn’t help to say they operate on something, so maybe an example is the way to go here. Let’s say I invent an operator called D. What D does is take the derivative of something with respect to x. That is,

D = d/dx

Sitting out there, all by itself, D doesn’t mean very much. It’s a derivative, sure, but of what? To be useful, D has to operate on something, say a function f(x). Suppose f(x) is something simple, like, oh, 3x4. If we let D operate on it, we get

Df(x) = 12x3

I just made up the operator D, but I should actually mention that it’s not the first operator you’ve run into. In trigonometry, things like sine and cosine are operators. Obviously a “cos” just sitting around doesn’t mean anything, but everyone’s seen cos θ. It doesn’t mean you should multiply θ by cos, rather the cosine operates on θ.

Groovy? Hopefully you’re happy with this, but if not you can always write me with questions. While we’re at it, I should also mention a couple of pitfalls so you’ll know to avoid them. Let’s go back to my made up operator D again. What if this time it operates on the function g(x) = e2x? In that case, we get

Dg = 2e2x = 2g

Does this mean we can cancel out the g’s and conclude that D = 2? Of course not! Remember, we’re not multiplying by g here, we’re operating on g. In this case, g was an exponential, so we got it back when we used D on it. If you’re still not convinced, consider this equation:

cos θ = sin θ

Is it okay to cancel out the θ’s and say cos = sin? Nope.

Here’s another thing you should be aware of. What does it mean if I have D2? Should you take the derivative of something and then square it? You might think so, but that’s not it. When I say D2, I’m really saying you should apply the operator D twice. For example, going back to the f(x) from earlier,

D2f = DDf = D(Df) = D(12x3) = 36x2

And there you go. To be a bit more explicit, since D is a derivative, D2 is a second derivative. (If you applied them to a particle’s position, D would give the velocity, and D2 would give the acceleration.) Armed with this knowledge, you should be more than a match for this week’s homework.

Thursday, October 25, 2007

G’bye to special relativity

So that about wraps up the special relativity I wanted to cover with you guys. I suppose it’s time we move on to other things. If you survived (and possibly even enjoyed) the problems from the last few weeks, you should be fairly impressed with yourselves. At many universities, the material we covered would be part of a sophomore-level course, so you’re seeing it some two years early. Perhaps you breezed through them, but I’ve found from my own experience that the biggest challenge to working problems like this is just learning not to be intimidated by them. They use new language and deal with very new ideas, and it can be very tempting to give up without actually trying them. Personally, it took me years to learn that lesson, and it wasn’t until midway through my undergrad that I stopped freaking out every time I saw something new in physics. Well, mostly. I still freak out every now and again.

If you really liked special relativity, you might be happy to hear that we really only scratched the surface. There are deeper and more beautiful ideas in relativity, like spacetime structure, covariant and contravariant vectors, relativistic energy and momentum, field transformations, and tensor potentials. So if this stuff really grabbed you—and you’re willing to learn a little more math—you might consider taking a course on relativity in college sometime.

Thursday, October 18, 2007

Length contraction problems

Well, since no one left a comment, I’m assuming that no one had any thoughts the length problem (or, surprisingly, the Veyron). But as you might anticipate, the explanation is similar to the one for time dilation. So, if A says that B’s measuring stick is short, and B says that A’s measuring stick is short, who is right? Answer: they both are. It makes more sense if we think about how length is actually measured.

Let’s say, once again, that A is driving the Maserati, and we’d like to measure its length on the ground. If A happens to be parked, this is easy: we just lay our ruler on the ground, check the positions of the front and back bumpers, and subtract the two. No problem.

For that matter, it’s not hard to check the length of the car if it happens to be moving, either. Same procedure, but this time obviously we have to be sure to check the two positions at the same time. If you don’t, then clearly the car will move while you’re measuring and you’ll get the wrong answer.

Hmmm. Just maybe, this could be starting to sound a little familiar to you. The thing that bails us out of the paradox, once again, is the first consequence of relativity: two events that are simultaneous in one frame are not going to be simultaneous in another. We say we were careful to measure both points at the same time, but no matter how careful we are, the dude in the car is never going to agree. He’s going to complain that we read the front end of the Maserati first, and then read the back end after the car had already moved forward, so of course we got a number that was too small.

Well, A has always been something of a complainer, and we’re thinking he’s really just full of it, but it’s important to remember that neither of us is “actually” right. In his reference frame, it really is true that we measured the two points at different times. In our reference frame, we really did measure at the same time, and his car really is shorter. For that matter, someone moving in a third frame would tell us we were all wrong: they would claim that we mismeasured, and his car is shorter (exactly how short would depend on how fast they were going, but they’d get a different length than us).

Is there really a paradox hiding somewhere in length contraction? Once again, it sure feels like it, but I’ve been looking for years and I’ve never actually found one. Still, no one’s ever accused me of being too clever, so maybe you’ll have better luck. I encourage you to roll this one around in your heads for a while, see if you can come up with something that looks like a contradiction. To help get you started, this week’s problems involve a couple of entertaining “paradoxes” about length contraction. No worries; you won’t need to figure them out to answer the questions, but if you have any guesses on how to fix the paradox, leave me a comment. (Seriously. I get lonely out here in these French villages.)

Here are this week's problems:

and here are a couple of pictures to help you visualize the ladder paradox:

Wednesday, October 10, 2007

Length contraction

Alright, maybe you found the discussions about simultaneity and time dilation a little difficult to picture. In that case, you’ll probably appreciate the third consequence of relativity a little more. Length contraction simply claims:

Moving objects are shortened.

Creepy, I know. To be a little more specific, moving objects are shortened in the direction of motion. So if a Maserati MC12 goes zipping by you at near light speed, its front bumper will be closer to its back bumper, but its height and width will stay the same. Just as with simultaneity and time dilation, I want to remind you that this is not some trick of observation: it’s not just that the car looks shorter, it’s that it actually is shorter (in the reference frame of the ground). While we’re on the subject, take a guess: by what factor do you predict the length changes?

Okay, well, I hate to keep you in suspense. If you guessed that the factor was γ, just like for time dilation, then you have a devious mind. But you’re also correct. If the length of an object at rest is L’, then the length that someone on the ground measures is L = L’/γ. If you don’t believe it, you can check out my proof:

So, huh. That’s weird. Why γ? Why not, I dunno, ? Or some entirely different function of velocity? If you read the proof, you’ll see that the way γ crops up is quite different from the way it came about in time dilation, so what gives? Actually, it’s no coincidence that the factor by which time is increased is exactly the same as the factor by which length is decreased.

You’ve probably just started studying vectors in physics, but of course you know that they’re basically just a magnitude and a direction. If I want to, I’m free to rotate a vector however I like (by multiplying by something called a “rotation matrix”). Of course, the rotation can’t possibly change the magnitude of the vector, just which way it’s pointing. The secret is, special relativity is just a very abstract kind of rotation: instead of rotating a normal three-dimensional vector, say, out of the x-direction and into the y-direction, it rotates a four-dimensional vector out of space entirely, and into time. The magnitude of the “spacetime” vector doesn’t change, which is why the amount that length decreases has to exactly match the amount that time increases.

Of course, the guy driving the Maserati doesn’t think his car is shortened. To measure it, he’d have to use some kind of measuring stick, but all his measuring sticks are also shortened by the same factor! As a result, no matter how carefully he measures, he’s never going to agree that it’s not its normal length.

In fact, just as with time dilation, remember that as far as he’s concerned, he’s stationary, and it’s the road that’s moving underneath him. Actually, if he sees anyone on the ground whip out a measuring stick, he’s going to think they’re the ones that are shortened. So this raises a similar problem to the one in the time dilation section: if Al is the guy in the car and Bob is a scientist on the ground, Al says Bob’s measuring sticks are short, and Bob says Al’s measuring sticks are short. Which one is right? Write me a comment and let me know what you think. In the meantime, here are a couple of questions about length contraction to entertain you. (Remember, if you get stuck, ask me questions; I’m always happy to help out.)

Friday, September 28, 2007

Working on the Compact Muon Solenoid

It sounds like the end of the six weeks is coming up, so I thought this week I’d cut you guys a break and share with you some of what it’s like to work at CERN as a grad student. In a way, all of these physics puzzles are a little misleading, because they really have nothing to do with what professional physicists actually “do”. At the edges of particle physics things are fairly ragged, and nobody’s making up nifty little problems because nobody knows quite how things work yet. Instead, a typical day for me will consist of coding at a computer, or working with hardware and circuits, or occasionally sitting through seven hours of meetings. (As I write this, I’m in a meeting about something called “level 1 trigger efficiencies”, which is exactly as exciting as it sounds.)

To take a small step back, I work at a detector called the Compact Muon Solenoid. If you remember my earlier post, the Large Hadron Collider is responsible for accelerating protons up to almost the speed of light and then colliding them together, and at the actual collision points we stick detectors to see what comes out. At startup, there will be two detectors: CMS, and at almost the opposite end of the ring, ATLAS, which earns my award for worst acronym ever (“A Toroidal LHC ApparaTus”). Anyway, these two detectors are both designed to discover new physics, and there’s a (usually) friendly rivalry between the two.

Okay, so CMS is a detector, but maybe you’d like to know what it actually looks like. Here’s a picture of (part of) the thing. (I can't show you the whole thing because it's not fully constructed yet.)

When it's done, it'll all be one giant machine, and yet it’s packed unbelievably densely with sensitive electronics. You can’t really get a great sense of scale from the picture, but this thing is 15 meters tall. That number rolled off me the first time I heard it, and maybe it isn’t impressing you much either, but when you’re standing in front of a piece of equipment that’s almost six stories high it hits you that the pictures don’t entirely do it justice.

The whole thing weighs 12,500 metric tons, which is a lot even for the volume. The reason it’s so dense is a component called the electromagnetic calorimeter, which is made up of 80,000 crystals of something called lead tungstate. Believe it or not, this stuff is 98% lead, an opaque metal, and yet it’s completely transparent. Seriously. Here’s a picture:

Yup. If you pick one up, it feels like lead, and if you bang two together they make a metallic ringing (and people will get mad at you, because they're very expensive). In total, they weigh about as much as 24 adult African elephants, but they’re supported by carbon fiber structures about 0.4 millimeters thick.

Everything about CMS is epic, but in my opinion nothing is more impressive than the rate of data flow: at the speed the protons are going, they’ll circle the LHC 40 million times per second. Each time they pass there will be about 25 collisions, so we’re looking at about a billion “events” per second. The amount of data stored for each event is about 1 MB, so the data is pouring in at a rate of a million GB per second. (I used GB in case any of you have a 100 GB hardrive on your computers; this thing could fill up ten thousand of those every second.) In fact, most of this stuff is useless and gets weeded out immediately, but about 500 Gbits/s is actually transferred through the “event builder”. Just to put this in perspective, this is equal to the total amount of data exchanged by the entire world’s telecom networks. I'm counting data from every phone conversation, every file download, every email and every internet video viewed on the entire planet.

CMS is cool, but I don’t really see it very often. For one thing, it’s 100 meters underground, and I don’t like working where there are no windows. For another, it’s in a town called Cessy (in France), and I typically work in my office in Meyrin, Switzerland. In fact, if it’s a nice day out, I’ll sometimes take my laptop to an outdoor table near the cafeteria because they’ve got good coffee.

One of the things I love about CERN is how easy it is to run into famous or important people. This week is CMS week, which mostly means lots of meetings, but it also means people from all over the world fly in to Geneva. Just sitting in the CERN cafeteria, you can see Nobel laureates, highly cited researchers, and world experts on pretty much anything that has to do with physics or hardware or computing. Yesterday a bunch of grad students and postdocs got together for a game of ultimate frisbee and the Deputy Physics Director for CMS skipped out of a meeting to play with us. He was much better than us, too, in spite of being forty years older.

I haven’t really talked much about what exactly I do, in part because I didn’t want this blog entry to be too long, but if enough people are interested I could talk a little about my research. Some of it’s a bit technical and wouldn’t be interesting to you guys, but I have to deliver a “preliminary defense” in mid-December, and I’d guess that kind of thing would resonate with anyone who has to write a senior thesis at the end of the year. Anyway, if you’re interested let me know.

Tuesday, September 18, 2007

The twin "paradox"

Okay, so I promised last week that I’d try to clear up the confusion about Alice and Bob. Alice says time is running normally on Earth, but is slowed down on Bob’s spaceship. Meanwhile, Bob is claiming that time is running normally on his spaceship, and it’s on Earth that time is slowed down. Which one is right? You’re not going to like it, but it turns out they’re both right.

What gives? Well, let’s be a little clearer about what we’re saying. To know that time is moving slow, you’d have to check the time difference between two separate events. Suppose Alice and Bob agree that right when they pass each other, they’ll both hold up big digital stopwatches, and reset them to 0 just as they go by. Then they’ll simply watch each other through high powered binoculars and compare how long it takes for each of their stopwatches to get to 1 minute. Presumably, someone’s watch is going to reach 1 minute first.

Let’s tell the story from Alice’s point of view. Right as Bob zooms by, she sees him set his watch to 0, and she does the same. Of course, time is moving slow on his spaceship, so for every second that ticks by on her watch, only half a second goes by on his. Finally, her watch reaches one minute exactly, and looking through the binoculars, she observes triumphantly that his watch only shows 30 seconds.*** (See "Optional note" below.) Now, how to prove to Bob that her watch got to 1 minute first? Simple! She’ll just take a picture of her stopwatch and sent it to him in a light signal. Except she quickly realizes that there’s a problem. The light races away from her at the speed c, but Bob’s spaceship is going pretty close to that speed anyway, so the light is only catching up very slowly. In fact, the ship is going so fast that the light signal ends up taking 3 whole minutes to get to him! By the time her picture gets to him, his watch (which is running at half speed), shows 2 minutes have elapsed, and he incorrectly thinks it’s her watch that’s running at half speed!

That’s not how Bob would tell the story at all. He agrees that as the Earth went shooting by they both set their watches, but since it’s the Earth that’s moving, Alice’s watch is only running at half speed. He’s looking through his binoculars, and he observes that right when his watch reads 2 minutes, hers only reads 1. In fact, just to prove it, she takes a picture and sends it to him! Of course, he’s not moving at all, so the light reaches him very quickly (since, of course, it’s traveling at the speed c while he’s completely stationary), and confirms what he already knew: that it’s Alice’s watch that’s running at half speed.

Okay, so that didn’t work out quite the way Alice planned. Note that she’s perfectly willing to admit that when her picture reached Bob, his watch showed 2 minutes, she just says it’s because the picture took so long to catch up to him. The culprit is the second postulate of relativity, which says that light travels at the speed c according to all observers. They both agreed on how fast the signal was moving; what they couldn’t agree on was how fast Bob was moving.

But Alice is quite clever, and she’s got another idea. Forget about sending messages; that’s too complicated. Instead, she gets her good friend Cassie to stand one light-minute away with a stopwatch of her own. Then when Bob goes by Alice, all three of them will start their watches at the same time, and when Bob’s spaceship gets to Cassie she can just hold up the stopwatch and show him it displays a longer time than his, proving his watch is running slow. What happens this time?

Well, no doubt about it, as Bob zooms by Cassie, her stopwatch shows 1 minute, and Bob’s only shows 30 seconds. Does that mean that Bob’s watch really is the one that’s running slow? Not at all! I claimed that Alice and Cassie started their watches at the same time. Put another way, I claimed they started their watches simultaneously. But of course, we already know that two events that are simultaneous in one frame are not in another! So Alice and Cassie may say they’ve started their watches at the same time, but according to Bob, Cassie made a rookie mistake and started her watch too early, which is the only reason her watch shows more time than his does.

Okay, well, if you’re anything like me, just these two examples won’t convince you that there’s not a legitimate time paradox here. I’ve just covered two possible ways you might show there’s a contradiction, and in both cases there’s a slippery way out. But maybe you’re not satisfied; it sure feels like there’s a problem somewhere. If you’re not yet convinced, I’d be interested to hear your take on the so-called “twin paradox”, which goes like this:

On her 21st birthday, an astronaut takes off in a spaceship at near the speed of light. After 5 years have elapsed on her watch, she turns around and heads back at the same speed to rejoin her twin brother, who stayed at home. The traveling twin has aged 10 years (5 years out, 5 years back), so she arrives at home just in time to celebrate her 31st birthday. However, as viewed from Earth, her clock has been running slow, so her twin brother will now be much older than she is!

But what happens when you try to tell this story from the point of view of the traveling twin? She sees the Earth fly off at near the speed of light, turn around after 5 years, and return. From her point of view, it would seem, she’s at rest, whereas her brother is in motion, and hence it is he who should be younger at the reunion. So which one is really younger when they meet up, and why? For the answer you’ll have to wait until next week, but in the meantime you can entertain yourselves with these questions.

*** Optional note: I’m being very slightly dishonest here, because I didn’t mention that Alice has to account for the travel time of light. What Alice would actually see, looking through the binoculars, is that Bob’s clock would read less than 30 seconds, and she’d have to figure out that at that moment his clock actually read 30 seconds, but the light hadn’t reached her yet. I left it out because I didn’t want the story to be any more confusing than it had to be. If this is too much, just forget about it; it doesn’t make any difference to the story. As long as you’re reading, though, let me remind you that Bob’s clock really is running slow (in the reference frame of the Earth), and it’s not just a trick of observation.

Tuesday, September 11, 2007

Time dilation

By the way, as a quick note, I should probably mention that simultaneity, and all the other consequences of special relativity, are not a matter of one observer or another “getting it wrong”. When I say that event B comes before A according to one observer, but another observer says that A comes before B, it’s not that one of them has made a mistake. In the first observer’s frame, B really does come before A, and in the other observer’s frame, it really doesn’t. It’s weird.

I mention this because sooner or later one of you is going to read a bad relativity book, and it’s going to say something about one observer not accounting for the travel time of light, or something. But that has nothing to do with relativity. If I take a measurement of something, I have to be clever enough to subtract out any effect due to the signal taking time to reach me. Actually, if I’m really clever, I’ll get a bunch of low-ranking grad students and space them at intervals with a stopwatch and a ruler so they can just write down measurements and bring them back to me. It’s like when you see lightning and then later you hear the thunder, you might incorrectly believe that they didn’t come from the same source, because you didn’t account for their different travel times. But this isn’t relativity, this is just making a mistake.

Anyway, the weirdness this week is about time dilation. Simply put, time dilation says this:

In a moving frame, time runs slow.

How much slower? That’s the subject of this week’s question. (By the way, if you happen to get stuck, you can always write me a comment; I’m happy to help.)

Let’s say Bob hops on a spaceship that goes rocketing by the Earth at near the speed of light. Just as he passes by, Alice peeks in the window. If right at that moment Bob is, I dunno, say playing a game of pool, then Alice sees it as though it were happening in slow motion. If Bob jumps the cueball, it seems to float slowly through the air; if he sinks a ball, it drifts gently downwards to the bottom of the pocket; and a blast break doesn’t look nearly as impressive when Bob’s traveling at 90% of the speed of light.

Of course, if you ask Bob about all this, he’ll say he’s just moving at normal speed. As far as he’s concerned, his spaceship is completely stationary, and it’s the Earth that’s rocketing by in the other direction. Hmm…

With any luck, at least a couple of you are scratching your heads at this point. I just told you that time runs slow in a moving frame, so Bob is moving slow according to Alice because Bob is moving with respect to her. But by the same argument, Alice must be moving slow according to Bob, because in Bob’s reference frame (the spaceship) it’s Alice who’s doing the moving.

So what’s the deal? Common sense says that if Bob looks slow to Alice, then Alice must look fast to Bob, right? If they both pull out stop watches and start them just as they pass by, whose watch reaches one minute first? Is it possible that one of them is wrong? If so, which one? I’ll try to unstick this one for you next week, but in the meantime, if you have any opinions or guesses, leave me a comment. (Christos, if you’re reading this blog, you’re not allowed to give away the answer.)

Problem on time dilation:

Monday, September 3, 2007

Special Relativity

The plan for the next few weeks is to start with Einstein’s special theory of relativity. (It’s “special” because it only deals with things that aren’t accelerating; if you want acceleration, you’re looking for the general theory of relativity, but I would advise you not to go looking for it anytime soon.) Special relativity is a good place to start because you already have all the math you need, namely algebra and trig. Plus it was figured out before all the other stuff I’ll talk about in this blog, and some of the other stuff is based on it.

The whole theory comes from just two postulates:

1. Principle of relativity: The laws of physics are the same in all inertial (non-accelerating) reference frames.
2. Universal speed of light c: The speed of light in vacuum c (about 300 000 m/s) is the same for all inertial observers, regardless of the motion of the source.

The first postulate has actually been around for hundreds of years, but it probably deserves a little explanation. As far as we’re concerned, it means this: if you’re moving along without speeding up or slowing down, you can never really prove that you’re moving at all. After all, how would you prove it? You’d have to do some sort of physics experiment, and according to postulate 1 you’ll get the same result as if you weren’t moving. If you see me walking down the street you could say I’m moving while the ground stays still, but it’s just as fair to say I’m staying in one place while the ground moves underneath me. Like what happens on a treadmill, for example.

The upshot is that you can pick any (non-accelerating) reference frame you like and decide it’s stationary. If someone else picks a reference frame that’s moving relative to yours and claims that’s the one that’s stationary, that’s cool too. Neither one of you is wrong; you’re just describing the same thing two different ways. If I walk into a wall, it doesn’t matter whether you think I’m moving forward or the wall is moving backward, the physics when I hit it stays the same.

Okay, so hopefully you believe me about the principle of relativity. It’s the universal speed of light which ought to really upset you. You certainly wouldn’t believe me if I’d said that the speed of a bowling ball is the same for all observers, regardless of their relative motion. If I’m on a train going 60 mph and I roll a bowling ball at 5 mph down the corridor, then the ball’s total speed relative to the ground is “obviously” 65 mph—the speed of the bowling ball (A) with respect to the ground (C) is equal to the speed of the ball (A) relative to the train (B) plus the speed of the train (B) relative to the ground (C). Our common sense tells us that this should work for any three objects A, B, and C. But according to postulate 2, if instead I send a light signal down the corridor, its speed is c relative to the train and c relative to the ground!

Impossible? When I was a kid I sure thought so. In fact, I was absolutely convinced there was a logical paradox there, and I’d spend days coming up with thought experiments that would give contradictory results. But, infuriatingly, each time there was some clever way out so that you could just barely avoid a contradiction. As it turns out, there really isn’t any paradox at all, but special relativity does require you to change dramatically how you look at space and time.

There are three major consequences to special relativity. We’ll work out the first one this week and I’ll give you a little problem, and we’ll save the last two for next week. If I can direct your attention here, we’ll get started:

Friday, August 24, 2007

The Large Hadron Collider

There are several experiments going on at CERN, but the big one, the one everyone’s excited about, is the LHC. It’s still being built, but when it’s done it will be by far the largest accelerator ever made. We’re talking an order of magnitude more powerful than the runner-up, the Tevatron at Fermilab (just outside of Chicago).

But what is a collider? I’m so glad you asked. A collider is a big ring (actually a pair of rings, one right inside the other). Its job is to accelerate particles (in this case protons) in both directions. It gets them up to very close to the speed of light (about 99.9999997% of the speed of light) and then smashes them together in a head-on collision. This isn’t very healthy for the proton (the parts of it that aren’t annihilated instantly are shredded into several hundred pieces) but it turns out it’s a pretty good way to find new physics. We build huge detectors at the points on the ring where we’re colliding the particles and see what comes out.

Physically, the LHC is an enormous underground ring 27 km in circumference, which comes to about 5.3 miles in diameter. Seriously. It's 5.3 miles in diameter. That’s the distance from Townview to the Café Brazil on Central, and it’s all one giant machine. You could fit several small villages on the inside of this thing. And I know, because when I drive out to Point 5 on the circumference of the ring, I have to go through several small villages. So the thing is pretty big.


CERN is the European Organization for Nuclear Research, but I guess when you say it in French the acronym makes more sense. Actually, the acronym doesn’t make a whole lot of sense anyway, since nobody really does any nuclear research there. CERN is the world’s largest particle physics lab, and the particles we’re talking about are much, much smaller than nuclei. It turns out that in physics going down in size means going up in energy, so what I do is sometimes called high energy physics.
Aerial view of CERN. Off to the right is where I live. Just
out of the picture to the left is the only McDonalds in miles.

CERN itself lies on the border between Switzerland and France, which you would think makes for a lot of paperwork, but I guess not. I usually work at one of the sites in France, but sometimes I go over the main office in Meyrin, Switzerland. The transition is so smooth that sometimes you forget you’re passing between two countries. In total, about 8,000 physicists of all ranks work on CERN experiments, which is about half of all particle physicists in the world.

About Me

I’m a former TAG student, and I graduated in 2001. During my senior year I got Rebecca to let me take mechanics and E&M concurrently. (At the time she only offered mechanics, so you guys are lucky.) Anyway, after I left TAG I did my undergrad at Rice, where I majored in physics and math, which was a fairly equal mix of entertaining and agonizing. On graduating from Rice, I hopped clear across the country to the University of Wisconsin, because I wanted to work with a guy there named Wesley Smith. That seems to have worked out pretty well, because I spent the next couple of years finishing up my grad classes, and on June 4th he moved me out to CERN in Geneva, Switzerland, which is where I am at the moment. I’m currently doing research on something called supersymmetry, and in the next couple of years I’ll write and defend a dissertation, at which point I’ll get to be “Dr. Weinberg”.


Hello physics people! If you’re currently enrolled in Dr. Jensen’s physics class, then this blog is for you. My goal in writing this is to give you an impression of what physics is like out at the frontiers, and what it means to actually “work” in physics. So I’ll try to cover lots of stuff, but I’ll also post some physics puzzles here from time to time that relate to areas you wouldn’t normally run into, like relativity, quantum mechanics, quantum field theory, and particle phenomena. Maybe, secretly, you’re thinking about majoring in physics and you’d like to know what it’s like. If that’s not you, that’s cool; maybe you’ll find something in here to impress your friends and confound your enemies.
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