Matt Springer is a graduate student of physics at Texas A&M; university. He is also an occasional writer and tinkerer, and he is probably too curious for his own good.
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If you happen to be in the Bryan/College Station area tomorrow, you might consider checking out the Texas A&M; Physics Festival. It's sort of an open house with a ridiculous number of top-notch physics demonstrations as well as some very interesting talks. It's free! I'll be there helping out with some of the optics demos.
While we're here, we may as well do some physics. This problem is from Halliday, Resnick, and Walker 14.62. It asks (fitting well with the space stealth theme from earlier):
One way to attack a satellite in Earth orbit is to launch a swarm of pellets in the same orbit as the satellite but in the opposite direction. Suppose a satellite in a circular orbit 500km above the Earth's surface collides with a pellet having a mass of 4g. a) What is the kinetic energy of the pellet in the reference frame of the satellite just before the collision? b) What is the ratio os this kinetic energy to the kinetic energy of a 4g rifle bullet from a modern army rifle with a muzzle speed of 950m/s?
Well, we could "cheat" and use the virial theorem, but since at the intro level the students don't know that yet we'll do it the long (but still not difficult) way.
To find the kinetic energy we need to know the velocity of the pellet. We don't know the velocity, but we do know that the pellet is in a circular orbit. We know that for any object to be in uniform circular motion, the force on that object has to be equal to m v^2 / r, where r is the radius of the circle.
But we also know the equation describing the force of gravity, as given by Newton. That force has to be providing the requisite pull to keep the object in uniform circular motion:
With M the mass of the Earth, and m the mass of the pellet. Solve for v:
Now r is not 500 km. r is the radius of the orbit, which is the radius of the Earth plus 500 km. It works out to about 7.6 km/s. But we can't just get the kinetic energy from that either - we're asked for the energy in the reference frame of the satellite, which is also moving at 7.6 km/s in the opposite direction. The total closing speed is 15.2 km/s. The satellite "sees" itself as at rest with a projectile incoming at that closing speed. Kinetic energy is 1/2 times the mass of the projectile times the square of its closing speed, for a total of some 462 kJ. This is a pretty hefty quantity of energy. The rifle bullet postulated in the problem only has 1805 J of kinetic energy, so the orbital pellet hits with something like 256 times more energy.
Fig 1: Aluminum block hit by 7g projectile at ~7km/s
This is why proposals for space warfare rarely involve the "explosives on rockets" paradigm which is so dominant in terrestrial warfare. At the speeds involved, adding an explosive is just an extra complication that doesn't add anything of consequence to the destructive energy available just by virtue of the orbital velocity.
Somehow - and I don't know exactly how, you know how the internet is - I came across this odd but cute song by the ineffable Weird Al. It's an almost seven-minute(!) ode to the roadside attraction that is the titular biggest ball of twine. The twine ball actually exists, and lives in Darwin, Minnesota.
Oh! What on earth would make a man decide to do that kind of thing!
Oh! Windin' up twenty-one thousand, one hundred forty pounds of string!
...
You know, I bet if we unravelled that sucker,
It'd roll all the way down to Fargo, North Dakota
'Cause it's the biggest ball of twine in Minnesota
I'm talkin' 'bout the biggest ball of twine in Minnesota
Hmm. Sounds like a Fermi problem to me. Eyeballing the twine ball, it looks like it's about 10 feet in diameter (I could look it up, but that's no fun). We'll call it 3 meters for good measure. This is just over 7 cubic meters of twine. Converting units and dividing out, the twine ball has a density of about 1.4 grams per cubic centimeter. This is not wildly unreasonable - it's a little denser than water - but I'm not sure loosely wound twine is that heavy. I think Al's probably adjusted the actual mass a bit upward to fit the rhythm of the song.
What about the ball's odds of rolling to Fargo? We need to figure out how much the string weighs per meter and then extrapolate out from the 21,140 lb mass postulated. As it happens, I have a pair of calipers and a round cross-sectional shoestring. It's hard to measure precisely without squashing the string, but it appears to be about 4.5mm in diameter. Assuming the twine is similar, each meter of twine will have a volume of about 1.6e-5 cubic meters. We could just divide the 7 cubic meter twine ball volume by that and call it a day, but the twine probably isn't so densely packed that the volume is entirely filled with the twine. There's probably some air spacing between the windings. On the other hand, we have no good way to guess a correction factor, which is not likely to be large. So in the spirit of the Fermi problem, we'll just scrap the mass factor entirely and just divide out volumes: 7 cubic meters divided by 1.6e-5 cubic meters per meter (how's that for units!) is 437,500 meters, or 272 miles. And darned if it isn't 204 miles from Darwin MN, to Fargo, ND.
Now our estimate is pretty rough, but that's not half bad for a Weird Al song.
And yes, I know this is a ridiculous thing to write about, but I'm hoping it will finally get the song out of my head...
If you put your finger down on any point on that line, and then put another finger on another point on that line, you find that the total change in the y-coordinate divided by the total change in the x-coordinate between those two positions is 1. Move two units to the right, and the line rises by two units, etc. This is the same no matter which two points you pick. Every line has this sort of property, which is called slope. Steep lines that rise greatly for each unit of x displacement have a large slope, flat horizontal lines have zero slope, and descending lines have negative slope. In each case the way to calculate the number describing the slope is just to divide the amount of vertical rise by the amount of horizontal run. If you have the equation of the line, the slope is just the number in front of the x. I.e., y = 3x has a slope of 3, because if you increase x by some amount, y increases by three times that amount.
But what if your equation isn't a line? Here's y = x^2:
Clearly this curve hasn't got a constant slope. But this doesn't stop us from at least informally describing each point on the curve as if it had a slope. Over on the left, the slope seems to be negative, since a little ball places on the curve would be sliding downward. Then it gets less and less negative, until at the bottom of the parabola the slope seems to be zero. Then the slope becomes more and more positive as the graph rises more and more for each bit of distance run.
That's pretty informal though. We might want to assign a number to that ever-changing slope, and we need a good way to define that slope. I propose we define the "slope of a curve at a particular point" as the slope of the line (which is well defined) which happens to be tangent - just touching - the curve at that point. For instance, if we wanted to find the slope at x = 1, we'd look at the slope of the line touching the graph at that point:
Which is nice, but right now we have no way of knowing what that line is. To write the equation of a line, you either need a point on that line and a slope, or two points on that line. We only know one point, and we're trying to find the slope. Can we get around this? Well, we can try to do so by putting one point where we're trying to find the slope and one point farther to the right by a distance delta x. Then we can shrink delta x and see what happens to the slope. Since the slope is the change in y over the change in x, we'll just call it Δy/Δx.
Well, it looks like the slope as x = 1 is about 2. It sure seems to be closing in on that value anyway. But we want to be sure of this. We'll calculate it by hand for a generic x, though in this case of course x = 1 which we can substitute in at the end if we want.
The change in y (which we called Δy) is easy to write down. The change between two numbers is the end value minus the start value. The start height for a given x is just x^2, and the end height will be the square of x plus however far we went to the right:
Therefore the slope between the points x and (x + Δx) is:
We can go ahead and multiply out the top:
Cancel the x^2 and divide by delta x:
But the whole point is that we want the change in x to approach zero. This means the only surviving term is 2x. And that's the slope of the tangent line to the parabola y = x^2 at any point x. It's our Sunday Function, and for its official presentation I'll switch the deltas on the left side into the d-notation favored in modern calculus:
The notation dy/dx is the standard way of expressing the statement "the slope of the tangent line to the function y at a given point x", which takes up way too much space to write down. "d" is just part of the notation, it's not a new variable or anything.
And that's one half of the basic concepts of calculus - finding the slope (or more generally the rate of change) of a function at a given point.
The eagle-eyed among you may not be happy with this. You might say something like this: "If delta x is greater than zero, you can't really be said to have truly found rate of change at exactly one point. But if delta x is actually equally to zero, then you've divided by zero, which is impossible. Either way this method doesn't quite work."
And you'd be right. But calculus did nonetheless work perfectly for solving problems (heck, we just solved a problem with it), so for a while mathematicians and scientists were more or less willing to press on and keep developing the calculus in the anticipation that a more formally careful method of finding slopes could be found, avoiding the divide-by-zero problem. Sure enough, the 19th century mathematicians Augustin-Louis Cauchy and Karl Weierstrass came up with a formally correct if conceptually recondite way to re-express this procedure without division by zero. With calculus on a firm logical footing, this process of differentiation is ubiquitous in modern science and engineering.
GrrlScientist sends a link to this rather wild stunt from India:
How is it possible? What kind of friction is necessary, and is it any more difficult for the cars to do the stunt than it is for the motorcycles?
Before we do any math, I want to think about the problem qualitatively. Let's tally up the forces acting on the vehicles. First, there's gravity. It points vertically downward, straight toward the center of the earth. Second, there's the normal force. "Normal force" means the force normal (in this context a physics technical term meaning perpendicular) to the surface. It's the force the track exerts on the vehicle. It points, as the name indicates, perpendicular to the track surface. Third, there's friction. It points in the opposite direction of wherever the vehicle would be sliding if there were no friction.
You may be a little suspicious of that phrasing, and you should be. How do we know what direction that is? Well, we could calculate it but we don't have to. Let's take a look at the diagram of the situation, conveniently found on the Wikipedia article on the inclined plane:
The gravitational force downward is mg, and the normal force is N. The terms with sine and cosine are just the gravitational force expressed in components - those two forces are exactly equivalent to the downward vertical mg. We do know from Newton's laws that F = ma, i.e,, the sum of all the forces F is equal to the mass times the acceleration. In the situation in the video, the total acceleration parallel to the track is zero. The motorcycles neither ascend nor descend once they reach their cruising altitude, so to speak. But that doesn't mean there's no acceleration. If there were no acceleration, the vehicles would continue in straight lines, sinking into the track like ghosts. Instead, the riders are essentially orbiting; their acceleration is directed toward the center of the circle. Their speed doesn't change, but their direction does.
From freshman physics, we know the force required to keep an object in uniform circular motion is F = v^2/r, toward the center of the circle of radius r. But looking at the diagram, it's easy to see that N points toward the center of the circle, and so does the parallel component of mg. The frictional force f doesn't, and so even if there were no friction and the track were made of ice, N and mg could produce the required acceleration, keeping the riders in place.
Now I'm going to depart from the diagram just a bit to make the math easier. Instead of decomposing mg into parts with respect to the plane, I'm going to decompose N into vertical and horizontal components with respect to the ground.
Since the cars neither rise nor fall, the vertical component of N has to be equal to the downward force mg:
Therefore:
The only horizontal force is the horizontal component of N, and it has to be equal to the force needed for circular motion:
But we already figured out what N was, so we can substitute:
Now with we can solve for v, the speed necessary to stay orbiting the wall even in the absence of friction:
Because sin/cos = tan, and the m's cancel out. Which is nice - it means that it's equally easy (or hard) for the heavy cars and light motorcycles to do the trick, provided they can actually get up to that speed and provided they're both not too big with respect to the turning radius r.
So how fast do they have to go? It's hard to estimate what r is, but we can guess perhaps 10 meters. The angle is also tough, but let's guess 70 degrees just as an estimate. That gives us a value of about 16.4 m/s, or 36 miles per hour. To my unpracticed eye this seems a bit high, but then I've probably overestimated the angle. A 60 degree incline would mean 29mph, for instance. Either way though, it's pretty clearly doable for motorcycles and cars alike. I don't know what the Indian equivalent of OSHA might think, but the laws of physics are fine with this bit of daring.
Sine, cosine, and tangent are of course the workhorse functions of trigonometry. You learn 'em in high school, and if you go on in math and science you never stop using them. Now on many occasions you might have the sine or cosine or tangent of some angle, and you want a way to invert those functions to recover the angle from those values. Let's take a look at the inverse tangent function:
Now we'll do one of those things were we skip any motivation and just do some stuff for reasons we'll get to at the end. Let's find the derivative of the arctangent function. To start, take the tangent of both sides of the above definition. The tangent undoes the arctangent, so we get:
Differentiate both sides. The derivative of tangent is secant squared, which we could prove if we felt like it, and since f is a function we have to use the chain rule. Doing thus, and recalling that the derivative of x is 1:
Solve for the derivative:
Which isn't all that helpful except for the fact that there's a basic trig identity which connects the secant and tangent functions: sec(x)^2 - tan(x)^2 = 1. Substituting:
Ah! But tangent(f) is x, so we can swap that in:
Ok, ok. What's that good for? Well, if just write that down as our Sunday Function and graph it, we'll have something to celebrate:
That is the graph of a function, 1/(1+x^2), which has nothing at all to do with circles or geometry in any obvious way. But we know it's the derivative of the inverse tangent, and conversely the inverse tangent is the integral of that function, so we can easily plug in and see that the area under the curve is pi. Which is nice, because to day is Pi Day. Happy Pi Day!
In the Stealth in Space post earlier this week, we discussed the problem of detecting the thermal emission from a spacecraft. If the interior isn't generating a lot of power, there's not much thermal radiation being emitted, making it a tough job to detect.
But it was pointed out in the comments that the heat from the sun would itself warm the spacecraft exterior, increasing the thermal signature by potentially a large amount. Let's verify this. If you take a perfectly absorbing sphere and set it in orbit around the sun (say, at the distance of the Earth), it'll absorb all the light from the sun that intersects its cross-sectional area. But assuming it conducts heat well, it'll be radiating equally in all directions - the area of emission will be the whole surface of the sphere. Thus at equilibrium, the power in from the sun equals the power out via radiant heat.
The power in is the power-per-area from the sun (about 1300 W/m^2 at the distance of the earth) time the area of cross-section, and the power out will be equal to the thermal-power-per-area given by the Stefan-Boltzmann law times the whole area:
Where Ac is the area of cross section, At is the total surface area, capital phi is the flux from the sun, and sigma is the Stefan-Boltzmann constant. For a uniform sphere, we can plug in the area:
Solve for T:
Notice the r's have canceled; the final temperature is independent of the size of the sphere. With our numbers, the temperature is a toasty 275 K. Well, toasty with respect to space. You'd still want to wear a jacket.
But that's a pretty hefty temperature if you want to stay stealthy. In the comments I suggested making the spacecraft out of reflective or transparent materials, both of which have potentially serious problems. It would probably be a better bet to just tweak our formula above by simply giving our spacecraft a convenient shape. If for instance we shaped our spaceship like a pencil and kept either the point or the eraser aimed at the sun, there would only be a small fraction of our total radiating area available to absorb sunlight. I'll leave the equation-modification aside and quote some results: if there's 100 times as much radiating area as there is absorbing area, the temperature is down to 123 K. If there's 1000 times, it's down to 69 K. Returns are diminishing (it scales as the inverse fourth power of the ratio), but still significant. That combined with some judicious reflectors and I think you can get the hull temperature pretty low.
Fig 1: Pencil-shaped spacecraft design, from 2001
The problems the sun's heat poses are also strongly influenced by where you're trying to be stealthy. Around Mercury there's a lot more solar light to deal with. Around Saturn, much less. All other things being equal, your hull temperature scales with the inverse square root of your distance from the sun. At Saturn, for instance, our 100x area blackbody spacecraft would have a hull temp of a frigid 39 K.
So it's something to think about if people ever come to blows over mining Ceres or something.
UPDATE: In the previous post, commenter Anthony brings up something well worth discussing here:
...if a ship is being hit by X watts of sunlight, it's pretty much going to emit X watts worth of photons (unless it has somewhere to store the heat, and it usually only takes hours to days to overwhelm any practical heat sink), and tweaks to albedo and heat distribution across the surface just modify the spectrum and direction of the emissions.
Which is pretty important, lest we fall into the same trap I criticized in my previous post. In the end, total incoming = total outgoing. Lowering the temperature is good for reducing that part of the total emissions which are promiscuously broadcast in all directions. Hopefully you can catch most of the rest with a mirror and reflect it in a narrow, specific direction away from enemy sensors.
While doing some poking around online, I came across a website called Project Rho, which tries to provide some science background for science fiction writers who want some degree of technical accuracy in their imaginative work. Generally it looks like they're on the right track.
In their section on stealth in space, they explain with the weary air of repetition that there's no such thing. The flare of a rocket is bright enough to be seen from basically anywhere, and the thermal signature of even a spacecraft with rockets off is visible from clear across the solar system. The first I can believe (though of course it's worth checking), but the second sounds fishy. But so do lots of true things. Why not run the numbers ourselves? Before we do, let's see what they say:
"Well FINE!!", you say, "I'll turn off the engines and run silent like a submarine in a World War II movie. I'll be invisible." Unfortunately that won't work either. The life support for your crew emits enough heat to be detected at an exceedingly long range. The 285 Kelvin habitat module will stand out like a search-light against the three Kelvin background of outer space.
They go on later in the article:
The maximum range a ship running silent with engines shut down can be detected with current technology is:
Rd = 13.4 * sqrt(A) * T^2
where:
Rd = detection range (km)
A = spacecraft projected area (m^2)
T = surface temperature (Kelvin, room temperature is about 285-290 K)
If the ship is a convex shape, its projected area will be roughly one quarter of its surface area.
Example: A Russian Oscar submarine is a cylinder 154 meters long and has a beam of 18 meters, which would be a good ballpark estimate of the size of an interplanetary warship. If it was nose on to you the surface area would be 250 square meters. If it was broadside the surface area would be approximately 2770. So on average the projected area would be 1510 square meters ([250 + 2770] / 2).
If the Oscar's crew was shivering at the freezing point, the maximum detection range of the frigid submarine would be 13.4 * sqrt(1510) * 273^2 = 38,800,000 kilometers, about one hundred times the distance between the Earth and the Moon, or about 129 light-seconds. If the crew had a more comfortable room temperature, the Oscar could be seen from even farther away.
The equation given isn't derived. We have no idea where they're getting that 13.4 proportionality constant. Dimensionally it's correct, and it's pretty easy to derive the equation up to that constant which will depend on the sensitivity of the detector. That equation modulo some uncertainty with respect to that constant is accurate as far as it goes given a spacecraft of hull temperature T and cross-sectional area A.
I would take you through the steps of the derivation, but it would be pointless because the assumption that the hull temperature has anything to do with the interior temperature is simply flat wrong. We can prove this with a potato.
Switch your oven to the "Bake" setting at a temperature of 350 F. After preheating, put in the potato. The interior of the oven, and eventually the potato, are maintained at a constant temperature of 350 degrees. How hot is the exterior surface of the oven? Depends on how well insulated your oven is, but I can guarantee it's a lot less than 350 degrees.
The key is the understanding the relationship between heat and energy. Put hot coffee in a thermos - the hot coffee is hot because it contains thermal energy. If the energy can't leave, the coffee will stay hot because the energy stays inside the thermos. The outside of the thermos stays at the temperature of the surroundings. Now neither the thermos nor the oven is a perfect insulator. Some energy leaks out of the oven's interior, cooling it down. The oven thus has to pump energy into the heating elements to make up for this loss. Equilibrium is reached when the rate of energy being put into the oven equals the rate of loss through the insulation.
For a spacecraft in a vacuum, the pretty much the only way to lose energy from the interior is by radiant heat. The higher the temperature of the outside, the higher the rate of energy loss via radiation. But the temperature itself is irrelevant, since just like the oven and the thermos it's not necessarily related to the actual temperature inside the cabin at all. It is always and everywhere a function of the total power passing through the hull. If the temperature inside the cabin is constant, the power leaving the hull by radiation is exactly equal to the power being generated inside the hull.
So how far away can we detect a given amount of emitted power? According to Wikipedia, a telescope of 24" aperture can detect stars of magnitude 22 after a half-hour exposure. I think this is a pretty good realistic limit for detection with reasonable equipment in a reasonable time frame. Now we need to compare this magnitude to something of known power output. How about the Sun? The sun has magnitude -26.73 as seen from the Earth's surface (smaller magnitude is brighter), for a difference in magnitude of 48.73. The exponent used for magnitude is 2.512, so the difference in power per unit area of telescope is 2.512^48.73 = 3.1 x 1019. Since the Sun radiates about 1000 watts per square meter at the distance of the earth, the smallest radiant power we can reasonably detect in our telescope is about 3.123.1 x 10-17 watts per square meter.
Our hypothetical spacecraft is radiating that power into space, evenly distributed over the surface of a sphere of radius r, where r is the distance to the detector. When that power-per-area is the same as the limit of our telescopic capability, that gives us the maximum detection range. Mathematically,
Where rho is the sensitivity of our detector. Solve for r:
So what's the power? Well, each human on board is going to produce about 100 W just from basic bodily metabolism. Computers, life support, sanitation, and all the rest will contribute more. We might assume 10,000 watts total for a futuristic ship that's specifically designed to emit as little power as possible. It might well be significantly lower. Plugging in, I ger r = 5.98 x 109 meters. This is pretty far, but it's only around 4% of the distance from the earth to the sun. Practically nothing in terms of solar system distances. Even a ship dumping a megawatt of power should only be visible from a third of the earth-sun distance.
The reason for this divergence in our estimate versus the Project Rho estimate is that it takes a huge amount of energy to maintain a hull exterior at cabin temperatures. But insulation means that's not necessary, all that's necessary is that the power out equals the power generated in the interior.
Or at least this is my impression. I could be wrong. Thoughts?
This Sunday I was at an event that involved a number of drawings for door prizes. There were perhaps 30 couples there - it was, not to beat around the bush, a wedding registry shindig at Bed, Bath, & Beyond. (Did I mention I'm recently engaged? I am. It's the main reason for the endemic light posting, as I'm afraid the wedding bumps the blog down the priority list a bit. ;) )
There was something in the vicinity of 8 or so drawings from separate bowls, and each couple's number was placed in each bowl. The rules were such that a couple could only win once. One number was pulled twice, and that particular prize had to be re-drawn. My friend Michael and his fiancee Christine were there, and I turned to him and said "Hey, this is the Birthday Problem!" He immediately agreed, and our very patient fiancees gave each other their patented "Oh geez, they're about to launch into some tremendously nerdy discussion again" looks. Not that they mind, but of course the middle of a wedding registration is not an opportune time. But this blog is an opportune time!
The birthday problem is a famous exercise in the theory of probability, to show students (and remind professionals) that probability can behave in very counterintuitive ways. It goes like this. Let's say you have a room containing N randomly chosen people. What's the probability that at least two people share a birthday? Or more concretely, how big does N need to be before there's an even chance of at least two people sharing a birthday?
Someone else may be partying too.
Well, let's start counting it out. If N = 1, the probability is zero because there's no one to share a birthday with. If N = 2, the first person will have some particular birthday, and the second person has a 1/365 chance of having the same birthday. Therefore the probability of sharing a birthday is 1/365 (we'll ignore leap years, which affect the numbers only slightly). If N = 3, things are more complicated because there's more possibilities. They might all have different birthdays, two might share a birthday, or all three might have the same birthday. For higher N these possibilities will grow enormously and make our calculation very difficult. So we take a shortcut. We'll only look at the probability that everyone has different birthdays. All the other cases fall into the "at least one duplication" category, so we don't have to separately worry about all the cases of multiple duplications.
So for N = 3, the first person has a birthday, the second person has a (364/365) probability of not sharing the birthday, and the third person has a (364/365)*(363/365) chance of sharing neither of those birthdays. So on and so forth for higher N. The probability of at least one duplication is just 1 minus the probability of no duplications. and our Sunday Function is just the resulting expression:
If you're not familiar with that notation, it just means write down the expression in parentheses with n = 1. Then write in down again for n = 2. Then for n = 3 and so on until you get to n = N. Multiply all those together and you have the result. It's just a compact way or writing the previous paragraph mathematically.
So for a group of N people, the probability of at least one duplication of birthdays is given by that Sunday Function. Let's graph it:
The probability turns out to be better than 50:50 for N as low as 23. Get a group of 41 together and there's a better than 90% chance two will share a birthday. Most people intuitively suspect the groups would have to be much larger, but usually this is because they're really thinking of the probability of two people having a specific pre-chosen birthday. The probability of two people sharing a birthday, whatever it happens to be, grows pretty quickly.
This generalizes to the door prize situation pretty easily. Instead of two people sharing the same birthday, we're looking at two drawings sharing the same pulled number. So replace 365 by the number of couples and repeat the calculation. Running the numbers for 30 couples and 8 drawings gives a 64% probability of at least one duplication. It's thus not exactly a surprise that one prize had to be re-drawn.
I'm happy to report that one of the numbers drawn was ours. We're now the proud owners of a $100 gift certificate to Sherwin-Williams. So if we ever need paint, I guess we're set!
If you read about science at all, you've heard of Heisenberg's uncertainty principle. It's the canonical example of quantum weirdness, the strange idea that you can't simultaneously know the position and momentum of a particle. Pack a particle into a small enough box and your accurate knowledge of position will necessarily cause that particle to have a very uncertain momentum, "bouncing" around crazily inside that box.
What you may not have read is that this isn't just quantum weirdness, it happens just as often in the classical world of waves. In fact, the very fact that quantum particles have a wave nature creates the bridge between the weird quantum uncertainty principle and the classical uncertainty principle that you literally hear all the time.
Let's start off with a graph of a sine wave with a frequency of 1 Hz:
There's two equivalent ways I can specify this wave. I can say, as I did, that it's a sine wave of frequency 1 Hz. Mathematically you could say I've given the spectrum of this wave, describing it in what we in the business call the frequency domain. (And phase, but that's beyond our scope for this post.) The other way I could specify this wave is in the time domain, where I just write down the function of t that describes the time behavior of the wave.
One representation is not more fundamental than the other. Given the frequency domain, it's possible to write down the one unique waveform of that frequency spectrum - which, for the record, is itself a function of ω, the frequency. Given the description in the time domain, I can calculate the frequency spectrum.
So what if I have a wave that looks like this?:
It's almost the same 1 Hz wave as above, but it's not the same. It's contained within a envelope, and only maintains that 1 Hz for a few cycles before dying away. Clearly in the frequency domain, "1 Hz" isn't exactly the right description because it has already been taken by that pure sine wave. So what's the right frequency representation?
We can calculate it by a method called the Fourier transform, named after the French mathematician Joseph Fourier. By doing this, we can see what the right frequency representation is. It's this:
Here the units are in Hz. We see that it's sharply peaked right at 1 Hz, which makes sense because it's mostly a 1 Hz signal. But because it's not exactly a pure sine wave, the sine wave in its envelope is actually a mixture of different frequencies spread near the 1 Hz mark.
What if we plot another 1 Hz wave, but force it into an even briefer envelope, like this?
We can do the Fourier transform and see how it looks in the frequency domain:
Ah. It's even more spread out. Despite being in some sense a "1 Hz" pulse, it actually involves frequencies as low as 0.5 Hz and as high as 1.5 Hz. The general principle continues to hold as we compress the pulse more and more. Shorter pulses require a broader range of frequencies. In general, short in one domain means broad in the other. Very short pulses of sound contain such a broad range of frequencies that they can scarcely be said to constitute anything resembling a musical note at all.
Now that we've seen it, we ought to listen to it. Our 1 Hz example is far below the range of human hearing - or most speakers, for that matter - so we'll use a base frequency of 440 Hz. Here's a 440 Hz tone in an envelope that's 1/2 of a second long, measured from half-maximum on the upslope to half-maximum on the down:
It is a pulse of sound, but it's obviously a nearly pure note without a lot of frequency spread. What if we pack the same wave into a smaller envelope, one that's just 1/10 of a second long?
Shorter, but still clearly identifiable. Quite a few cycles - well, 44 of them - happen in that time, so it's not so shocking we can still pick out what the note is. But what if we make the envelope just 1/100 of a second long?
Hmm. At this point it's possible to calculate that the frequency spectrum is spread almost from 380 to 510 Hz. Compared with a piano keyboard, that's starting to spread into the frequency territories of the neighboring keys. When you listen to the note, it's possible to tell that the pitch is not very easily identifiable as one definite note. It's more of a "pop" with a hint of its original tone.
Finally at 1/500 of a second even the hint of the original tone is gone. As a musician to duplicate this note on a piano and they'll just shrug their shoulders:
The frequency content of that little pop spans several octaves. It no longer resembles a pure note at all. The large degree of certainty with respect to the time of the note means a large degree of uncertainty with respect to the frequency of the note.
It's the uncertainty principle and it's not even quantum!
Matt Springer is a graduate student of physics at Texas A&M; university. He is also an occasional writer and tinkerer, and he is probably too curious for his own good.
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