Built on Facts

Re: Nidal Hasan's Weapons

Wed, 11 Nov 2009 23:56:19 -0500

God help me, I resisted mightily. If my fellow SB friend Greg wants to spin the Ft. Hoot shooting as a cause for gun control then frankly there's pretty much nothing further to say. You'd think a @#$% major in the @#$% army on a @#$% army base just might not have been terribly inconvenienced in procuring weaponry even if every civilian gun in the hemisphere vanished in a puff of sunshine and wishful thinking. But I was going to leave it alone, assuming that that particular point makes itself. To each his own.

But he wrote a follow-up post asserting a few points of fact, pretty much all of which rather wildly miss the mark. As a physicist, semi-pro educator (this blog!), and enthusiastic firearms owner and advocate, I simply can't help setting the record straight as to the points of fact. The political gun-control points I will grit my teeth and let slide. Let's begin:

He apparently carried two pistols, and both are designed to be effective killing weapons. The more newly designed Five-sevN that he had purchased under the noses of the FBI who was busy investigating him is specifically designed to be very effective at killing large numbers of people in close quarters, to have more controlled "follow-up shots" and to pierce body armor.

The Five-Seven (weird trademark capitalization is goofy even when Apple does it) is not designed to kill large numbers of people in close quarters, except insofar as any pistol is most effective relatively close. It's a pistol like any other, and does the same thing. With the exception of the last three words, you could replace Five-Seven with pretty much any centerfire pistol except the niche wilderness big-bore revolvers and have a statement that works just as well.

But the Five-Seven was designed to fire a rather unusual 5.7x28mm round which is itself designed to pierce body armor. That much is entirely true. But what's not true is that the armor-piercing handgun ammunition is available in the US. You cannot buy it, it is a violation of federal law. To be clear: if Hasan bought ammo for this pistol in the civilian market he bought ordinary, standard 5.7x28mm ammo. And the 5.7x28mm round sucks at pistol velocities for the purposes of incapacitating or killing. Though Greg sarcastically speculates it might be good for hunting moose, in fact in many states it would be banned for hunting even small critters on the grounds that such a small round would be cruel to the animal by virtue of the injury being too small to reliably kill quickly. For that matter the armor-piercing ammo wouldn't have been much better - the ability to penetrate armor is more or less precisely the inverse of what's needed to damage tissue. This is why police and self-defense ammo is almost exclusively hollow-point, which is good for quick incapacitation but terrible at armor penetration.

In short, the very fact that the pistol was designed for military use against armor-wearing opponents makes the pistol poor for anything else. Even if he had the armor-piercing bullets, which he didn't.

The other gun was a magnum, a.k.a., miniature cannon.

Magnum doesn't mean that. I suppose you might think so if you didn't know anything about guns, but that would make you correct. Magnum means all kinds of things depending on context. The .357 magnum Hasan possessed but apparently didn't use is a solid but not particularly unusual round. The .22 magnum is a very small bullet. The .44 magnum is ginormous, but no one uses it for actual combat because it's unwieldy. Magnum shotgun shells have more pellets but they're slower-moving. Some of the largest pistol bullets (.50 AE, .454 Casull, etc) aren't labeled magnum at all.

None of this is to say the weapons Hasan chose weren't dangerous and lethal. Certainly they were, and for his actions he deserves nothing but a short drop and a sudden stop. It is to say that a matter of empirical fact the weapons he carried were relatively ordinary, and that the weapon he actually used was in fact among the least effective weapons he could possibly have picked - certainly orders of magnitude less dangerous than a standard combat rifle. (The M-16, by the way, actually shoots a slightly narrower 5.56mm projectile, but the bullet is about twice as heavy and moving around 50% faster. This makes it much more effective, and yet it's still the target of continual controversy among military circles for its not-always-impressive terminal performance. In my state it's illegal for hunting deer for that very reason - too high risk of an escaped, injured deer rather than a quick kill. UPDATE: The previous sentence is not quite right - the Texas requirement is just that the round be centerfire, so effectively the 5.56 mm is the smallest legal hunting round. Other states vary, some do enforce a larger cutoff.)

You can make up your own mind about the politics of gun control (on an army base!) - as I said I'm not going to argue it here now - but at least now we can be clear as to what actually happened.

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Veterans Day (and the scientists who help them win)

Wed, 11 Nov 2009 13:07:36 -0500

On Veterans Day we commemorate the living veterans of the American armed forces. On Memorial day we commemorate those who lost their lives. We should also spend a moment to remember those who helped make sure more soldiers fit into the first category. Though I've made the point before on this blog, I'd like to commemorate two men in particular who between them likely saved the lives of tens or hundreds of thousands of Allied soldiers during the Second World War, simply by doing brilliant science. Their names are Alan Turing and Robert Watson-Watt.

Turing was a mathematician and computer scientist who lead the British code-breaking efforts which broke the Nazi Enigma code and various other forms of wartime message encryption. Though it's impossible to accurately judge counterfactual history, I have read more than one historian speculate that the Allied codebreaking successes may have shortened the war in Europe by a year or more.

Watson-Watt was one of the early pioneers of radar, and the first person to develop it into a practical means of finding range and direction of enemy aircraft. The Battle of Britain might have been a very different story if his invention had not allowed the vastly outmanned and outgunned Royal Air Force to hold its ground against the Luftwaffe. (In a cute coda, many years later he was cited for speeding in Canada by a policeman with a radar gun)

We owe these two men a lot more credit than they get. May their names never be forgotten.

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Sunday Function

Mon, 09 Nov 2009 12:13:27 -0500

Again I have to apologize for the sparseness of posting lately, but I've got two research projects going full blast and time has not been something I have a lot of. I'll still be writing at least a few times a week, and you can't beat the price. ;) In any case once things cool down just a little I should be back to a more regular schedule.

Today's function isn't interesting because of the function itself, the interest comes from what we'll do with it. Let's say we have a function like this:

If we want to see where the function is equal to zero, it's clear that 0 = x^2 - 2 is solved by x equal to the square root of two, either positive or negative. That's the snappy analytic solution, but let's say we want to use this function to compute a decimal approximation to whatever accuracy we feel like. The brute force method is just to try various decimal numbers and see what gets us closest, refining our guess each time. But this is ugly, slow, and requires a lot of continual work. We'd prefer a faster method where we can just turn the crank easily and get a good answer. To do this, let's plot the function and (for reasons I'll explain in the next paragraph) its tangent line at the point x = 4.

Let's say we picked x = 4 for our initial guess as the value of the square root of two. It's an awful guess, obviously, but that's ok since we're looking for a procedure that can turn any terrible guess gradually into better and better approximations of the actual answer. So we pick our initial point and draw the tangent line. We notice that it cuts the x-axis pretty close to the square root of two (which is exactly where the parabola cuts the x-axis, since the square root of two is by definition the number that makes the function equal to zero). So why don't we take the location where the straight line cuts the x-axis, and use that as our second guess, and draw the tangent line there:

This line cuts the axis even closer to the point x = the square root of 2. If we take that point as our next guess and repeat the process, we should be closer still. So first we need to actually work this procedure into mathematical language so we can actually get some numbers out of it. The equation of a line is:

Where m is the slope and b is the y-intercept. We know the slope of a tangent line is just the derivative of the function at that point, and y is just the value of the function at that point. That means we can solve for b:

Where the prime denotes differentiation, and we're subscripting the x so we know it's the particular value of the guess we're using. Our new x for the next iteration of the guess will be the value that makes our line equation y = mx + b equal zero, i.e., 0 = m x + b. Substituting all the previous stuff into this equation and solve for x. After a little simplification, we get:

Now this is a very general expression that works to find the zeros of an arbitrary function f, whatever it happens to be. Our particular function can be plugged in (for us, f'(x) = 2x by a little bit of calculus), which gives us the complete procedure:

Let's give this a try. Plug in our initial guess of 4 and the procedure tells us our next guess is 2.25. Plug that in and the procedure gives us 1.56944. Plug that in and we get 1.42189. Repeat again and get 1.41423. So on and so forth closing in on the rounded-off real value of 1.41421, and if we didn't round off at 5 decimal places as I'm doing eventually we'd get as many digits of the square root as we wanted to arbitrary accuracy.

This procedure is called Newton's Method, and it's a fine way of calculating the zeros of a function if you know its derivative. The method is not quite perfect - if a function has more than one zero the one you get will depend on your initial guess in a not-always-predictable way. And while the method is quite general, there are certain functions that don't fulfill the relatively generous convergence conditions. Still, it's a great method with a long and continuing history of use. It's pretty likely that your pocket calculator uses a very similar method when you hit the square root button. And now if you ever find yourself without such a button, you can do it yourself if you're patient.

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Being an Absentee President

Sat, 07 Nov 2009 10:30:07 -0500

Of late president Obama has taken a little bit of heat for his frequent (and mostly male) golf outings. Before him, president Bush took the same sort of heat for his golf and vacations. If you were willing to dig a bit through the news archives, I'd bet you could find similar tut-tutting about previous presidents taking time off. It's a common theme for criticism of just about any important federal officeholder - it's no coincidence that so many congressional "fact-finding" missions are to tropical paradises or European vacation spots. In that case it's a criticism I vigorously share, as the taxpayer dime ought not be spent ferrying already rich congressweasels around the globe. What they do on their own dime I don't worry too much about.

But the president comes in for special criticism no matter who pays for his downtime, in view of the fact that his job is so critical and demanding. Critical I'll grant, but contra the received opinion I'd like to argue that in fact it's easily possible to be an perfectly effective president while spending shockingly little time behind the Resolute desk. In fact for much of early American history presidents did just that - ie, very little. The presidency and the country have both changed, but in my opinion even today the president simply doesn't have to do much to do perform his job with great competence. Now I don't expect that any modern president will actually take as little time as I'm going to suggest; the very type of person who is attracted to the job and can campaign effectively is necessarily the kind of person who's willing, able, and eager to manage as much as possible. But he doesn't have to be. Let's go down the list of his constitutional responsibilities:

Sign or Veto Laws
How many of these does congress generate each week? I don't know, but given the glacial pace at which anything of consequence gets done in congress it can't be many. I believe it's a few hundred per year. The president has ten days to sign or veto each bill, so there's nothing wrong with just knocking out the previous week's bills over a Monday morning. No need to read them in their entirety - God knows congress doesn't bother. Working with official summaries, your various adviser's opinions, and the opinion of your constituents ought to be enough to decide the fate of a piece of legislation.

Take Oath of Office
Takes about a minute.

Be Military Commander-in-Chief
A huge responsibility, to be sure. But the president's job here is to set overall policy and strategic objectives at the highest levels. Getting involved in the details is both inefficient and actively a bad idea (Most famously in history, Hitler's terrible mis/micromanagement of his armed forces was a huge boon to the Allies). Given the fact that the president has a full-time civilian SecDef (with a huge staff) whose only job is to manage the military, the National Security Council, the Joint Chiefs, and a huge stack of generals and other military officials to translate strategic objectives into concrete plans, being CinC should be something the president could do each Monday after bill signings and before lunch.

Appoint Government Officials
The president has to appoint the cabinet and various other officials. This is also an important responsibility, but it's something that's pretty much done after the first week or two in office at least for the cabinet. The scores of lower officials that continually have to be appointed might be a pain, but that can easily (and in practice is) delegated to the cabinet and other advisers and then signed off on.

Make Treaties
Obviously the president doesn't actually write these. If he needs a treaty he tells the Secretary of State what he wants, and then subject to senate approval it happens. Treaties don't happen much anyway, so if there happens to be one in the pipeline give it an hour after lunch for progress reports and making any needed changes.

State of the Union
We think of it as a speech, but it is not thus mandated and in fact plenty of presidents just wrote up a SotU letter and mailed it to congress. So if you happen to have one coming up and want to actually do the speech, pencil in another after-lunch hour for practice and revisions with the speechwriters.

Ancillary Executive Duties
These amount to a few procedural matters with respect to congress, executive appointments as already dealt with above, and most importantly "receive Ambassadors and other public Ministers; he shall take Care that the Laws be faithfully executed". The ambassador stuff is easy: the State Department does all that. Faithfully executing the laws is obviously highly critical, but as chief executive this boils down to making sure your subordinates are doing their jobs with competence and fairness. So we'll say from 2-5 pm Monday grill your cabinet officials and lower agency officials (and heck, even random lower employees and the general public they affect) about the way their agencies are doing their jobs. Bring down the hammer when poor governance is spotted.

And that's that. Assuming your staff is even marginally committed to their jobs you can do your job perfectly well working one day a week, maybe less. Now sure you're neglecting the statecraft presidents like to do: visiting dignitaries, traveling the country giving speeches, campaigning for your party, wheeling and dealing with recalcitrant congresscritters, and all that. But will the wheels actually come off the country if those things don't happen? I seriously doubt it.

So go ahead Mr. Obama, golf to your heart's content.

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Cars and Crashing

Fri, 06 Nov 2009 14:02:44 -0500

Here's a question which pretty much everyone gets wrong. But the readers of this blog aren't just a random sample, so I bet most of you will get it right:

Two identical vehicles A and B, both traveling at speed V directly toward the other vehicle, collide exactly head-on. At another test track, car C collides with an indestructible concrete barrier at speed V'. All other things being equal, the crash-test dummy occupants of vehicles A, B, and C will experience the same forces (and therefore injuries) if...
1. V' = V/2
2. V' = V
3. V' = 2V
4. Something else (explain)

How might the answer change if the A and B do not have equal masses?

I'll let my astute readers answer, but I'll give a place to start for those who want to make sure they're thinking from a physics perspective. Momentum is mass times velocity. Force is the time rate of change of momentum. Go from there. Also, in this case I have to suggest you not try to find the answer experimentally!

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Horses, Lasers, and Interferometric Autocorrelation

Wed, 04 Nov 2009 12:31:33 -0500

For most of human history, our ability to perceive and understand very fast optical events has been limited by the temporal resolution of the human eye. Things that happened too fast were a blur, and all we had to go on was guesswork. Take, for instance, this 1821 painting of a horse race:

If it looks a little bizarre, it's because it guesses wrong on the character of one of those very short events - in this case, the position of a horse's feet during a gallop. Not until the invention of high-speed photography later on in the 19th century could the gallop be better resolved and understood. As it turns out, there is a point in the stride where the horse is entirely out of contact with the ground, but it's the point where all the hooves are brought inward. During the extension of the hooves, one is always in contact with the ground:

(Both images from the Wikipedia horse gait article)

Time went on and science discovered and used faster and faster events, and the technology has had to keep pace. The overall procedure is to find an even faster event than the one you're trying to measure, and use it as part of your examination. In the case of the horse, the faster event is the flash and shutter speed of the camera. This sort of thing can get you down to events as small as a fraction of a millisecond with standard cameras, and specifically designed high-speed cameras can do a bit better. Much below the millisecond range though, and shutters simply can't move fast enough to provide that faster event. Photodiodes can respond much more quickly than film and shutters, and using those along with careful and sensitive measurement it's possible to measure flashes of light with nanosecond resolution. Using streak cameras which trade off spatial resolution for temporal resolution can improve this to resolution of a couple of picoseconds. (The one in our lab can get to about the 1ps range. State-of-the-art can improve on this by perhaps a factor of 10, though at the cost of massive expense for not so much experimental value).

A picosecond is a trillionth of a second, by the way. It's very fast - light only moves around a third of a millimeter in a picosecond - but lots of atomic processes happen much faster. We can both use and study these processes with pulsed lasers, which today can easily reach pulse lengths of a few femtoseconds. A femtosecond is a thousandth of a picosecond, so it's pretty much an unbelievably small amount of time. A sort of typical way of thinking about the orders of magnitude is to note that the speed of light is 300 nanometers per femtosecond, and an average atom might be a few tenths of a nanometer in diameter. These femtosecond pulses are close to the shortest optical events we can currently produce. We'd like to measure them, but remember how we need that shorter (or at least comparably short) event? We just haven't got any laying around, and if we did we'd be studying them which would put us right back at square one.

But we can cheat. What if we could use the pulse itself as that short event? What we do is split the pulse into two copies of itself with a beam splitter. Then we make one pulse travel a slightly longer path to the detector, and by varying that path length the crests and troughs of the two pulses will interfere with each other in a way that depends on the path length difference. This is all detected on a photodiode or photomultiplier or equivalent with only nanosecond resolution and so all we see is a total integrated intensity as a function of path length difference, but we can still extract information from that:

This image is also from Wikipedia. M1 is the movable mirror, and SHG is a nonlinear crystal that responds to the square of the intensity, which allows us a bit more information.

In a way it's like examining the rainfall in a region by looking at a rain gauge once per month. You have no idea if it's been sprinkling all month or if you just had a short but intense cloudburst. In meteorology you'd not be able to say all that much about the day-by-day rainfall by looking at the bucket, but in optics our job is easier because the variable path length gives us a way to tease more information out of the intensity.

What does an autocorrelation look like? Here's one fresh from our lab. It's just a calibration test with standard equipment, so I'm afraid you won't be able to use it to scoop us on our hopefully interesting upcoming results. ;)

Y-axis intensity (au), x-axis femtoseconds. From it, we can say the pulse is about 9 fs long and a little bit messy - the wings indicate the pulse isn't quite a perfect sech^2 shape as we might like. But it's pretty neat that we can so easily measure such a short event with a procedure that's relatively simple and fits in a box not much bigger than one you get a Big Mac in.

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Mass, Energy, and the Sun

Mon, 02 Nov 2009 12:35:57 -0500

These days pretty much everyone knows that mass and energy are two sides of the same coin, as discovered by Einstein. In fact this is so well known that the average man on the street - knowing nothing at all of physics - would still recognize the expression even if he didn't know what it meant or how to use it:

E is energy, m is mass, c is the speed of light. As with all such formulas, you need to have properly matching units. You can't just use mass in pounds, e in British thermal units, and speed in furlongs per fortnight. But any properly consistent set of units will work; for macroscopic purposes we usually use joules, kilograms, and meters per second.

Since the speed of light c = 299,792,458 m/s, the speed of light squared is about 8.99x10¹⁶ m^2/s^2. As such one kilogram of mass is the equivalent of a ridiculously huge amount of energy, as we see in nuclear reactors and weapons. It's a little less commonly understood that this isn't just true for nuclear reactions, it's true for everything. Burn some logs in your fireplace and the total mass of the wood and oxygen will have been just microscopically higher than the final mass of the combustion products. The remainder has gone into the energy released to warm your house. It's the very same equation on a much smaller scale.

To get a specific number for this, we need to know how much energy is released by wood. Some googling seems to indicate a figure around 15 million joules per kilogram, so if we assume the burning of 10 kilograms of wood, we get 150 million joules total. Divide that by the speed of light squared and you'll get a mass of about 1.7 micrograms. Very tiny, much smaller than your average sand grain.

In fact the total solar power output absorbed by the earth is about 1.8 ¹⁷ watts. Over the course of a year, this comes out to a staggering 5.68²⁴ joules. It's a huge amount of energy, which you might expect since it's responsible for warming the earth, driving the weather, and powering essentially the entire biosphere. Dividing that by c^2 and you'll see that a total of about 63 million kilograms of solar mass is converted into sunlight each year in order to illuminate the earth. It sounds like a lot, but the same mass of water would fit in a cube 40 meters on a side.

The sun is quite a bit bigger than that, and though it doesn't come close to achieving perfect conversion efficiency, it probably won't be running out of nuclear fuel any time soon.

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Physics and insomina

Fri, 30 Oct 2009 13:41:23 -0500

There's an interesting article in New Scientist that purports to describe "seven questions that keep physicists up at night". The list is very heavy on the "deep questions" that tend to percolate around the more esoteric quarters of the high-energy physics world, and not so much on the vast bulk of physicists who (ad Chad and I like to harp on) do physics that's much more directly connected to the real (i.e., observable) world. For instance, it's the question of high-temperature superconductivity probably dominates the dreams of lots of solid state physicists, but it's not on the list. So let's look at the questions and rate them on Matt's AMO Physicist Insomnia Index:

Why this universe?
In other words, "why these laws of physics and not others?". I can't say I worry about this much. The laws are what they are. I hold out zero hope for the idea that only one set of laws will turn out to be mathematically possible.

What is everything made of?
Electrons and photons - wait, you mean there's more? Seriously, it would be nice to know what's what in the world of fundamental particles and fields. But it doesn't keep AMO physicists up at night, since as far as we're concerned atoms are pretty much as small as we need to worry about.

How does complexity happen?
Ah, now this is an interesting question and has lots of bearing everywhere in physics. The trivial answer is "because the mathematics puts it there", but nonlinearity and all the rest of that difficult math is not completely understood despite its vast importance. This one actually registers at a modest level on the insomnia scale for me. (Not literally of course.)

Will string theory ever be proved correct?
Nope. But I wouldn't be at all shocked if it's proved incorrect. In any case I worry about it exactly none, as it affects experimental physics exactly none.

What is the singularity?
What is a singularity, not what is The Singularity. I don't really worry about this, but it's an interesting and important question. We have great mathematical descriptions of gravity and quantum mechanics, but they don't mesh well in the domains where both effects are of similar scale. How to resolve this? Beats me, but I'd say this one at least registers at a small level on the insomnia scale.

What is reality really?
Ok, this is just ridiculous.

How far can physics take us?
The article is a little vague and airily philosophical about this, but I think it's asking how close physics is to being "finished". This isn't a purely hypothetical question. As in chemistry, large pieces of the subject are - if not finished - then at least only in need of some work around the periphery. But I'd say even if the Theory of Everything were found tomorrow, we'd have many decades of work left both exploring its consequences and continuing work in non-fundamental physics. In any case, not worth worrying about.

So of the seven questions, I think two are actually relevant to most physics. Well, it's a start.

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Book-throwing and physics.

Wed, 28 Oct 2009 12:57:35 -0500

Grab a book, or an empty DVD case, or anything else that's a uniform rectangular solid. If it's a book, you might want to secure the book closed with tape or a very lightweight clip, because we'll be throwing it in the air. We want to test a theory.

In classical mechanics we know that each solid object has three special rotational axes, called "principal axes". Intuitively, imagine the object is shaped from styrofoam. The principal axes are the axes along which you can spear the object with a wooden dowel, and the object doesn't try to "wobble" when you rotate the dowel. Aligning a car tire is simply the process of making sure the axle and the principal axis of the tire are collinear. There's a somewhat involved mathematical procedure you can use to calculate the principal axes (there's always three of them*), but frequently we can find them a priori without any math just by looking for axes of symmetry. Pulling a picture from Wikipedia, here's one principal axis of our book:

There are two more principal axes, and we can find them in the same way. One runs through the center, parallel to the spine. One runs through the center, parallel to the lines of text. If we throw the book in the air such that it's rotating exactly about one of those axes, it won't wobble.

But it's not possible to be perfect. We will inevitably be a tiny bit off. In that case, what happens? To answer, we look at Euler's equations which describe this sort of scenario:

The various "I" represent the moment of inertia about each principal axis, and the little omega represent the angular velocity about those axes, which is just the rate of spin. Notice the equations are coupled: the rates of spin about one axis affect the change of the rate of spin in the others. However, we see that if the rotation is initially about just one axis (say, ω1 ) the rotation about the other two will stay zero because their initial rotation rates are zero.

But if the initial rotation isn't perfectly aligned with a principal axis, all three omegas may be nonzero. What happens then? Well, we can solve these equations approximately and we'll see that in fact the rotation stays nearly aligned with the principal axis, with a slight precession with a frequency given by Ω. I'll skip the math. It isn't hard, but it's not very instructive either. In any case, the frequency of precession is:

Here we've numbered the axes such that 3 is the one about which we were mostly initially rotating modulo the very slight rotations about the other two. ω with the 0 subscripted is just the initial rotational speed.

Now here's the tricky part. Ω is square, so it must be a positive number. This will be true if I3 is the largest moment of inertia, as both terms in the denominator will be positive. This will also be true if I3 is the smallest moment of inertia, as both terms will be negative, and negative times negative is positive. But if I3 is the middle-sized moment of inertia, clearly we get a negative, and a squared real number can't be negative. Our assumption that the tiny imperfection in our throw doesn't matter must therefore fail if we're initially rotating about that middle-inertia axis.

So throw that book such that it's spinning about the axis in the picture. It'll work great. Throw it such that it's spinning about the axis parallel to the spine. It'll work great. Throw it so that it's spinning about the axis parallel to the text. It will wobble crazily, no matter how hard you work on the careful precision of your throw. Bet your friend a dollar that he can't do it, and physics will make you a dollar richer.

*With especially symmetrical objects like spheres and car tires the axes are not necessarily uniquely specified. There are still three of them, just with sort of a "phase freedom" in their orientation.

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Football and Trick Questions

Mon, 26 Oct 2009 11:33:25 -0500

There's a stereotype that it's declasse for us intellectual aesthetes to enjoy football, but I don't care. I enjoy it anyway. Whether you spent any time this weekend watching football or not, I'd like to pose a quick and (maybe!) easy vaguely football-related problem to exercise your brain to make up for all the boob tube time:

You're standing on the goal line of a standard football field and walk the 100 yards to the opposite goal line at a uniform speed of 2 miles per hour. Upon reaching the other side, you immediately turn around and jog back. How fast do you have to jog so that your average speed over the whole round trip is 4 miles per hour?

Remember: average speed is total distance divided by total time.

Bonus: Rework the problem by developing a nonstandard definition of "average speed" that better fits this particular question.

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Deep Physics

Fri, 23 Oct 2009 10:00:59 -0500

The deserts of New Mexico can get blazingly hot and bone-chillingly cold, extremes of temperature familiar to the outdoorsman in that kind of terrain. A few hundred feet below the ground in Carlsbad Caverns, the temperature is essentially stable in the mid-50s no matter how scorching or frigid the air a few feet above happens to be.

Fig 1: Experimental physicists also tend to live in caves.

You can think of the earth's surface in New Mexico as subject to two superimposed sinusoidal periodic heat pulses. One has a period of 24 hours and corresponds to the heat rising and falling over the day/night cycle. The other has a period of 1 year and corresponds to the average heat changing through the seasons. Clearly this approximation is quite rough, but we can take it as a starting point for some mathematical spelunking of our own. There's some math, but you can skim over it if you wish without missing the result.

We'll need the heat conduction equation as our first piece of caving equipment. It says that the rate of change of temperature at a given point is proportional to how different it is from the average temperature of its surroundings. We also have the proportionality constant k to measure the ease with which temperature flows:

To get the temperature below ground level we also need to give the boundary condition - ie, the temperature at ground level. We assume it's just the temperature T0 multiplied by an oscillating function. (T0 is the size of the oscillation.) We're going to take the real part of an oscillating exponential for mathematical convenience:

Now this is a differential equation and boundary condition, and we'd like to solve it as painlessly as possible. One of the best ways is to start with an educated guess - try it out, and if it works you're done. If it doesn't, try something else. Let's assume that the solution giving the temperature as a function of time and depth is an oscillating function of time multiplied by some function of depth z.

Jam that into our original differential equation and see what happens:

This is what we in the business call an "ordinary" differential equation, and they're easy to solve if you've done it a few times. Teaching it from scratch would take us too far afield, so let's just say we did a few more lines of work and arrived at the solution:

Where for convenience we've defined delta as:

This has to match the boundary condition given at the start, so we end up seeing that A = T0. which leads to a final answer:

Ok, let's interpret this. We have an oscillating term, which indicates that the subsurface temperature goes up and down as the temperature outside cycles. But it's damped by an exponential factor - so the further down you go, the smaller the variations about the average are. There's a "typical depth" given by delta which says how far the variations reach. At that point the variations are only about 37% of their surface magnitude. At twice that depth the variations are only 14%, and very quickly falling off from there.

Further, this depth is dependent on the frequency of the oscillations. The depth increases with decreasing frequency. In other words, the daily changes will not penetrate as far as the seasonal changes. Even more weirdly, there's a phase factor in the cosine term - there will be a lag between the maximum temperature at the surface and the maximum temperature at depth. But let's plug in some numbers to get a feel for what actually might be happening under the ground at Carlsbad. The thermal diffusivity of limestone is about 1e-6 m^2/s. Plugging in for the daily variation, delta is about... 16 centimeters. Below even that small amount of rock, the temperature is pretty constant all other things being equal. It's one of the reasons burrowing animals can survive in the inhospitable desert.

The yearly variation has a much smaller frequency and thus penetrates deeper. I calculate the penetration depth is about 3.16 meters. Below that however, as many caves are, neither the daily nor seasonal variation makes a dent in that otherwise constant average temperature. Incidentally, the phase delay at 3.16 meters is about 2 months for the yearly variation - the warmest time of the year at that depth is actually about 2 months later than the warmest time of year above ground.

(Note: This fascinating example taken mostly from the excellent textbook by Fetter and Walecka)

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Myth Confirmed.

Mon, 19 Oct 2009 13:40:34 -0500

There's a question that gets posed toward the beginning of intro physics classes to gauge the students' understanding of acceleration. If you fire a bullet horizontally while at the same instant dropping a bullet from the same height, which hits the ground first?

The point is to think clearly about the equation describing accelerated motion. The equation is this:

The bold letters represent vectors. Lower case r is position, a is acceleration, v is velocity. The vectors with 0 subscripted represent the starting values, so v0 is the initial velocity and r0 is the initial position.

But the whole point of a vector equation is that the equation compactly describes the motion in 3-d space. If we pick coordinates such that the x-axis is parallel with the ground and the y-axis is vertical, the equations remains true if we replace r with x (or y), a with the acceleration in that x or y direction, and v0 with the initial velocity in that direction.

So consider that fired bullet and that dropped bullet. Both of them have the same acceleration in the y direction (-9.8 m/s^2, vertically downward, from gravity). Both have the same initial velocity in the y-direction (0, since the fired bullet's initial velocity is all in the x direction). And both have the same initial position in the y-direction - call it h, the initial height. That means both bullets are obeying the same equation in the y-direction, so they both have to reach the ground at y = 0 at the same time.

When we replace r with x, the x-direction equations are different for the two bullets, since the dropped bullet has an initial velocity of 0 in the x-direction while the fired bullet has a large initial velocity from being fired. But that matters not at all to the vertical motion.

Incidentally we can use this to find the range of a horizontally fired bullet. Take the y-direction equation, set y = 0 (which is when the bullet hits the ground), solve it for t, and plug that t back into the x-direction. The x-distance attained during that time is just the range. Doing this, I get:

For a bullet at 1,000 feet per second fired at 5 feet, that's about 550 feet. (304 m/s and 168 m)

But until relatively recently, no one had ever actually done the experiment. It's difficult, both in terms of dropping the bullets properly and making sure the gun fires exactly horizontally. Horizontal fire is critical, because if there is an initial vertical velocity for the fired bullet, the equation will be different from the dropped bullet and they won't hit the ground at the same time. Nonetheless there is a group of experimenters who are very good at this sort of thing, and not so long ago they actually did the experiment. They are of course the Mythbusters:

They do the experiment and the bullets land within a few milliseconds of each other. Of course things aren't quite mathematically perfect in real life, as air resistance and other factors will serve to distort the basic acceleration equation we're using. Nonetheless it's a quite good approximation and in this case works out beautifully.

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Sunday Function

Sun, 18 Oct 2009 13:35:22 -0500

Edit: The previous version of this post required some fixing, as I boneheadedly mixed up the O and Ω notations. The rest of it should still be good.

When you invest in stocks, bonds, mutual funds, or just about any other major investment vehicle, you'll hear a standard but important disclaimer. "Past performance is not a reliable indicator of future returns." And it's wholly true. The hottest mutual fund of the last five years will very frequently revert to the mean over the next five. Or, start a Harley Davidson and an F-22 on the same runway and for the first few seconds the motorcycle will leave the jet in the dust. That performance doesn't guarantee future results, and in a few more seconds the jet will pull ahead.

In mathematics and physics, we're often not interested in what happens at the low end of a function. Very frequently for some f(x) or f(n), the x or n might represent huge quantum numbers or large numbers of atoms in a gas or lengthy time intervals compared to (say) the time of some underlying interaction. If we're not interested in the small-x or small-n behavior of those functions, it's also quite frequent that these functions can become much easier to deal with.

One of the ways to simplify the large-argument behavior of functions is called Big O notation. The slightly suggestive-sounding name comes from determining the order of a function for big values of the argument of that function. To define it, let's take two functions "f" and "g". We say that f is Ω(g(x)) if eventually f(x) is larger in magnitude than c*g(x) for every x larger than some critical value. Here "c" is some arbitrary constant.

Let's do an example. Let f(x) = x and let g(x) = 1. Is f(x) = Ω(g(x))?

The equals sign is sort of an abuse of notation, we just use it as shorthand for the definition above. But let's take a look at the question. For big x, is x > 1? Well sure, obviously. But we have to remember the arbitrary constant. What if it were equal to one million? It wouldn't matter. For large enough x, x is greater than one million. The answer to our question is yes.

How about a harder one? Is this true?

Assume n is some constant natural number. In other words, we're asking if (say), there some x such that e^x is thereafter greater than (say) 1,000,000*x^65,536. Or whatever other huge constants and exponents we want to name.

Well, probably the easiest way to think of it is in terms of the power series of e^x. But let's try to figure out the answer without resorting directly to calculus. Start with the supposition that e^x > c*x^n. Take the logarithm of both sides:

Obviously (though we haven't proved it so technically really this is just a plausible (though true) argument) log(x) is much more slowly growing than x, so large enough x will satisfy the requirement. Therefore e^x = Ω(x^n).

Which is very nice. All kinds of different functions have well defined big O relationships with each other. The polynomials are Ω of exponentials, exponentials are Ω of the factorial function, the factorial function is Ω of... well, not much in mathematical physics. In most cases it's the fastest growing function we frequently encounter.

This is all very helpful. If we have (as we might in thermodynamics) a function that consists of a polynomial in the numerator and an exponential in the denominator, we know it will tend toward zero as x gets large. We know this without having to do any math, just because of the big O relationships. And I like it when math means I don't have to do math.

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Obama comes to town.

Fri, 16 Oct 2009 17:36:33 -0500

Today Texas A&M was a bit of a madhouse. Huge crowds, hundreds of police, unseasonably-suited and grim-faced men with mirrored sunglasses, unmarked helicopters circling overhead, TV cameras circling below, and completely borked traffic.

Why? Not just one but two Presidents of the United States are making speeches this afternoon. Obviously most of the fuss is about President Obama, but George H.W. Bush is there as well since he's the one putting on this particular shindig. It's is a non-partisan service-oriented civic pride sort of thing. I'd like to see them it but it's an invitation-only event and only a few hundred students and faculty were invited.

The vast majority of the crowd was of the interested bystander variety. Non-protesters outnumbered protesters at least 20:1 by my unscientific estimate, and the protesters themselves were not so unevenly split between pro and con with a few surrealist fauxtesters ("Toy Story 2 Was Ok!", "I like turtles!") thrown in for good measure. Everyone I saw was well-behaved on both sides. The uninformed reader might have suspected that A&M is overwhelmingly conservative (Hey, I'm a proud and unrepentant reactionary and I go there), but in fact it's a very typical flagship state university with a wide and fairly representative variety of student politics. There was certainly no shortage of Obama fans.

News photos show a few crazies, but in my N>5000ish sample I didn't see any. Presumably throngs of non-noteworthy people behind rope lines don't sell papers.

As for Science, well, it marched on unimpeded. I went to class, analyzed some data, and did two presentations for my group about the progress of our new experiment. Then I went home. Not as exciting as our President's day, but I like it that way.

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A Pound of Electrons?

Thu, 15 Oct 2009 10:00:59 -0500

I propose a Fermi Problem.

Over the lifetime of an average light bulb, what is the total mass of all the electrons that have flowed through?

Work on that if you have an idea how to proceed, or just take your best plausible guess. Remember it's a Fermi problem, so we're looking for estimates rather than detailed calculations. My suggested solution method under the link.

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