Q: If atoms are 99.99% space, what “kind” of space is it? Is it empty vacuum?

Physicist: This is a bit of a misnomer.

When we picture an atom we usually picture the “Bohr model”: a nucleus made of a bunch of particles packed together (protons and neutrons) with other particles zipping around it (electrons).  In this picture, if you make a guess about of the size of electrons and calculate how far they are from the nucleus, then you get that weird result about atoms being mostly empty.  But that guess is surprisingly hard to make.  The “classical electron radius” is an upper-limit guess based on the electron being nothing more than it’s own electric field, but it’s ultimately just a gross estimate.

The picture gives you an idea of more or less where things can be found in an atom, but does a terrible job conveying what those things are like.

The picture gives you an idea of more or less where things can be found in an atom, but does a terrible job conveying what those things are actually like.

However, electrons aren’t really particles (which is why it’s impossible to actually specify their size); they’re waves.  Instead of being in a particular place, they’re kinda “smeared out”.  If you ring a bell, you can say that there is a vibration in that bell but you can’t say where exactly that vibration is: it’s a wave that’s spread out all over the bell.  Parts of the bell will be vibrating more, and parts may not be vibrating at all.  Electrons are more like the “ringing” of the bell and less like a fly buzzing around the bell.

Just to be clear, this is a metaphor: atoms are not tiny bells.  The math that describes the “quantum wave function” of electrons in atoms and the math that describes vibrations in a bell have some things in common.

Where exactly is the ringing happening?

Where exactly is the ringing happening?

So, the space in atoms isn’t empty.  A more accurate thing to say is that the overwhelming majority of the matter in an atom is concentrated in the nucleus, which is tiny compared to the region where the electrons are found.  However, even in the nucleus the same “problem” crops up; protons and neutrons are just “the ringing of bells” and aren’t simply particles either.  The question “where exactly is this electron/proton/whatever?” isn’t merely difficult to answer, the question genuinely doesn’t have an answer.  In quantum physics things tend spread out between a lot of states (in this case those different states incl different positions).

The atom picture is from here.

Posted in -- By the Physicist, Physics, Quantum Theory | 7 Comments

Q: Is geocentrism really so wrong? Is the Sun being at the “center” (i.e. the Earth orbiting the Sun) just an arbitrary reference frame decision, and no more true than the Earth being at the center?

Physicist: When you walk around in this big crazy world, there aren’t any immediate reasons to suspect that the ground under your feet is doing anything more than sitting perfectly still (ring of fire notwithstanding).  Given that, when you look up at the sky and see things pinwheeling about; why not assume that they’re moving and that you’re sitting still?  On its face, geocentrism makes sense.

But there are a lot of physical phenomena that poke holes in it pretty quick.  For example, Foucault pendulums (more commonly known as “big pendulums“) swing as though the Earth were turning under them and in a way that exactly corresponds to the way everything in the sky turns overhead (not a coincidence).

The classic way that heliocentrism (the idea that the Sun is at the center of the solar system) is demonstrated to be better that geocentrism (the idea that the Earth is at the center of the solar system) is by looking at the motion of the other planets.  This was essentially what Copernicus did; point out that with the Earth at the center the motions of the other planets are crazy, but with the Sun at the center the motion of the planets (including Earth) are simple ellipses.  His original argument was essentially just an application of Occam’s razor: simpler is better, so the Sun must be at the center.

Occam’s razor is a great red flag for detecting ad-hoc theories, but it’s not science.  With that in mind, it’s impressive how much Copernicus got exactly right.  Fortunately, about a century and a half after Copernicus, Newton came along and squared that circle.  Newtonian physics says a lot more than “gee wiz, but ellipses are pretty”; it actually describes exactly why all of the orbits behave the way they do with a remarkably simple set of laws for gravity and movement in general.  Newtonian physics goes even farther, describing not just the motion of the planets, but also why we don’t directly notice the motion of our own.

If we still assumed that the Earth was sitting still in the universe, physicists would have spent the last couple centuries desperately trying to explain what’s hauling the Sun (and the rest of the planets) around in such huge circles. We’d need a bunch of extra, mysterious forces to explain away why the center of mass in the solar system doesn’t sit still (or move at a uniform speed), but is instead whipping by overhead daily.

What follows is a bunch of Newtonian stuff.

Position and velocity are both entirely subjective, but acceleration is objective.  What that physically means that there is absolutely no way, whatsoever, to determine where you are or how fast you’re moving by doing tests of any kind.  Sure, you can look around and see other things passing by, but even then you’re only measuring your relative velocity (your velocity relative to whatever you’re looking at).  So, hypothetically, if you’re on a big ball of stuff flying through space, you’d never be able to tell.  Acceleration on the other hand is easy to measure.

Whether you're juggling in a place (upper left), a different place (upper right), or even a different speed (lower left), the laws of physics are indistinguishable.

The laws of physics are exactly the same regardless of where you are or how you’re moving.  Therefore there is no experiment that can tell you your “true” position or velocity.  Acceleration however does change things.  That’s why you can juggle in exactly the same way in a place (upper left), a different place (upper right), or even a different speed (lower left), but you can’t juggle, or you have to juggle differently, when accelerating (lower right).

At first blush it would seem as though there’s no way, from here on Earth, to tell the difference between the Earth moving or sitting still.  If the Earth is sitting still, we wouldn’t be able to tell.  If the Earth is moving, we also wouldn’t be able to tell.  But we’re doing more than just moving; we’re moving in circles and as it happens traveling in a circle requires acceleration.  The push you feel when you speed up or slow down comes from the exact same source as the push you feel when you turn a corner or spin around: acceleration.

The Moon orbits the Earth, and the Earth sorta orbits the Moon by wobbling.

The Moon orbits the Earth, and the Earth sorta orbits the Moon.  By wobbling.

If the Earth were “nailed to space” and never accelerated, then we‎’d only have one each of the lunar and solar tides.  If the Earth never moved, then the Moon’s gravity would pull the oceans toward it and that’s it.  But the Earth does move.  The Moon is heavy so, even though the Earth doesn’t move nearly as much, the Earth does execute little circles to balance the Moon’s big circles.  The Moon’s big circles generate enough centrifugal force to balance the Earth’s pull on it (that’s what an orbit is), and at the same time the Earth’s little circles balance the Moon’s pull on us.

If the Earth were nailed to the sky, then the Moon's gravity would cause only one tide a day.  We experience two because the Earth and Moon both orbit the same point (red dot).

If the Earth were nailed to the sky, then the Moon’s gravity would cause only one tide a day as the seas are pulled toward it. We experience two because the Earth and Moon orbit each other around the same point (red dot).  The swinging of the backside of the Earth means that the water on the far side is “flung outward”.

The same basic thing happens between the Earth and the Sun.  Things closer to the Sun orbit faster and things farther away orbit slower.  But the Earth has to travel as one big block.  The side facing the Sun is about 4,000 miles closer, and traveling slower than it would if it were orbiting at that slightly lower level.  As a result, the Sun’s gravity “wins” a little in the “noon region” of the Earth and we get a high tide (pulled toward the Sun).  The side facing away is moving a little bit faster than something at that distance from the Sun should, so it’s flung outward a little more than it should be and we get another high tide at midnight.  These are called the “solar tide” and they’re harder to notice because they’re about half as strong as the lunar (regular) tides.  That said, the solar tides are important and they exist because the Earth is traveling in a circle around the Sun.

Long story short: If the Earth were stationary (geocentrism) then we’d have to come up with lots of bizarre excuses to explain why Newton’s laws work perfectly here on the ground, but not at all in space, and we’d only have one solar and lunar tide a day.  If the Earth is moving (specifically: around the Sun), then Newton’s simple laws can be applied universally without buckets of caveats and asterisks*, and we get two lunar and solar tides a day.

*or even †’s.

Posted in -- By the Physicist, Astronomy, Experiments, Physics, Relativity | 7 Comments

Q: Is there such a thing as half a derivative?

The original question was: Another one of those questions of the type “does this make sense”.  You have first derivatives and second derivatives.  f'(x), f”(x) or sometimes dy/dx and d^2y/dx^2. Is there any sensible definition of a something like a “half” derivative, or more generally an nth derivative for a non-integer n?


Physicist: There is!  For readers not already familiar with first year calculus, this post will be a lot of non-sense.

Strictly speaking, the derivative only makes sense in integer increments.  But that’s never stopped mathematicians from generalizing.  Heck, non-integer exponentiation doesn’t make much sense (I mean, 23.5 is “2 times itself three and a half times”.  What is that?), but with a little effort we can move past that.

The derivative of a function is the slope at every point along that function, and it tells you how fast that function is changing.  The “2nd derivative” is the derivative of the derivative, and it tells you how fast the slope is changing.

f(x) is a parabola.  f'(x) describes the fact that as you move to the right the parabola's slope increases.  Notice that a negative slope means "down hill".  f''(x) describes

f(x) is a parabola. f'(x) describes the fact that as you move to the right the parabola’s slope increases. Notice that a negative slope means “down hill”. f”(x) describes the slope of f'(x), which is constant.

When you want to generalize something like this to you basically need to “connect the dots” between those cases where the math actually makes sense.  For something like exponentiation by not-integers there’s a “correct” answer.  For not-integer derivatives there really isn’t.  One way is to use Fourier Transforms.  Another is to use Laplace Transforms.  Neither of these is ideal.  Just to be clear: non-integral derivatives are nothing more than a matter of choosing “what works” from a fairly short list of options that aren’t terrible.

It turns out (as used in both of those examples) that integrals are a great way of “connecting dots”.  When you integrate a function the result is more continuous and more smooth.  In order to get something out that’s discontinuous at a given point, the function you put in needs to be infinitely nasty at that point (technically, it has to be so nasty it’s not even a function).  So, integrals are a quick way of “connecting the dots”.

To get the idea, take a look at N!.  That excited looking N is “N factorial” and it’s defined as N!=1\cdot2\cdot3\cdots(N-1)\cdot N.  For example, 3!=1\cdot2\cdot3=6.  Clearly, it doesn’t make a lot of sense to write “3.5!” or, even worse, “π!”.  And yet there’s a cute way to smoothly connect the dots between 3! and 4!.

Gamma(x+1) is a fairly natural way of generalizing x! to non-natural numbers.

Γ(N+1) is a fairly natural way of generalizing N! to non-natural numbers.  The dotted lines correspond to 1!=1, 2!=2, and 3!=6.

The Gamma function, Γ(N),  (not to be confused with the gamma factor) is defined as: \Gamma(N+1) = \int_0^\infty t^{N} e^{-t}\,dt.  Before you ask, I don’t know why Euler decided to use “N+1″ instead of “N”.  Sometimes decent-enough folk have good reasons for doing confusing things.  If you do a quick integration by parts, a pattern emerges:

\begin{array}{ll}\Gamma(N+1)\\[2mm]= \int_0^\infty t^{N} e^{-t}\,dt \\[2mm]=\left[-t^Ne^{-t}\right]_0^\infty + N\int_0^\infty t^{N-1} e^{-t}\,dt \\[2mm]=N\int_0^\infty t^{N-1} e^{-t}\,dt \\[2mm]=N\Gamma(N)\end{array}

So, Γ(N+1) has the same defining property that N! has: \Gamma(N+1) = N\cdot \Gamma(N) and N! = N\cdot (N-1)!.  Even better, \Gamma(1) = \int_0^\infty e^{-t}\,dt=-e^{-t}\big|_0^\infty = 0-(-1)=1, which is the other defining property of N!, 0!=1.  We now have a bizarre new way of writing N!.  For all natural numbers N, N! = Γ(N+1).  Unlike N!, which only makes sense for natural numbers, Γ(N+1) works for any positive real number since you can plug in whatever positive N you like into \int_0^\infty t^{N} e^{-t}\,dt.

Even better, this formulation is “analytic” which means it not only works for any positive real number, but (using analytic continuation) works for any complex number as well (with the exception of those poles at each negative integer where it jumps to infinity).

|Γ(N)|, where N can now take values in the complex plane.  The smooth area on the right corresponds to

|Γ(N)|, where N can now take values in the complex plane.

Long story short, with that integral formulation you can connect the dots between the integer values of N (where N! makes sense) to figure out the values between (where N! doesn’t make sense).

So, here comes a pretty decent way to talk about fractional derivatives: fractional integrals.

If “f ‘(x)=f(1)(x)” is the derivative of f, “f(N)(x)” is the Nth derivative of f, and “f(-1)(x)” is the anti-derivative, then by the fundamental theorem of calculus f^{(-1)}(x) = \int_0^x f(t)\,dt.  It turns out that f^{(-N)}(x)=\frac{1}{(N-1)!}\int_0^x (x-t)^{N-1}f(t)\,dt.  x-t runs over strictly positive values, so there’s no issue with non-integer powers, and it just so happens that we already have a cute way of dealing with non-integer factorials, so we may as well deal with that factorial cutely: f^{(-N)}(x)=\frac{1}{\Gamma(N)}\int_0^x (x-t)^{N-1}f(t)\,dt.

Holy crap!  We now have a way to describe fractional integrals that works pretty generally.  Finally, and this is very round-about, but it turns out that a really good way to do half a derivative is to do half an integral and then do a full derivative of the result:

f^{\left(\frac{1}{2}\right)}(x)=\frac{d}{dx}f^{\left(-\frac{1}{2}\right)}(x)=\frac{d}{dx}\left[\frac{1}{\Gamma\left(\frac{1}{2}\right)}\int_0^x (x-t)^{-\frac{1}{2}}f(t)\,dt\right]=\frac{d}{dx}\left[\frac{1}{\sqrt{\pi}}\int_0^x \frac{1}{\sqrt{x-t}}f(t)\,dt\right]

That “root pi” is just another math thing.  If you want to do, say, a third of a derivative, then you can first find f(-2/3)(x) and then differentiate that.  This isn’t the “correct” way to do fractional derivatives, just something that works while satisfying a short wishlist of properties and re-creating regular derivatives without making a big deal about it.


Answer Gravy: You can show that f^{(-N)}(x)=\frac{1}{(N-1)!}\int_0^x (x-t)^{N-1}f(t)\,dt (or even better, f^{(-N)}(x)=\frac{1}{\Gamma(N)}\int_0^x (x-t)^{N-1}f(t)\,dt) through induction.  The base case is f^{(-1)}(x)=\frac{1}{(1-1)!}\int_0^x (x-t)^{1-1}f(t)\,dt=\int_0^x f(t)\,dt.  This is true by the fundamental theorem of calculus, which says that the anti-derivative (the “-1″ derivative) is just the integral.  So… check.

To show the equation in general, you demonstrate the (N+1)th case using the Nth case.

\begin{array}{ll}  f^{(-N-1)}(x)\\[2mm]    =\int_0^x f^{(-N)}(t)\,dt \\[2mm]  = \int_0^x \frac{1}{\Gamma(N)}\int_0^t (t-u)^{N-1}f(u) \,du\,dt \\[2mm]    = \frac{1}{\Gamma(N)}\int_0^x \int_0^t (t-u)^{N-1}f(u) \,du\,dt \\[2mm]  = \frac{1}{\Gamma(N)}\int_0^x \int_u^x (t-u)^{N-1}f(u) \,dt\,du \\[2mm]    = \frac{1}{\Gamma(N)}\int_0^x f(u)\int_u^x (t-u)^{N-1} \,dt\,du \\[2mm]    = \frac{1}{\Gamma(N)}\int_0^x f(u)\left[\frac{1}{N}(t-u)^{N}\right]_u^x\,du \\[2mm]    = \frac{1}{\Gamma(N)}\int_0^x f(u)\frac{1}{N}(x-u)^{N}\,du \\[2mm]    = \frac{1}{\Gamma(N+1)}\int_0^x f(u) (x-u)^{N}\,du \\[2mm]  \end{array}

Huzzah!  Using the formula for f(-N)(x) we get the formula for f(-N-1)(x).

There’s a subtlety that goes by really quick between the fourth and fifth lines.  When you switch the order of integration (dudt to dtdu) it messes up the limits.  Far and away the best way to deal with this is to draw a picture.  At first, for a given value of t, we integrate u from zero to t, and then integrating t from zero to x.  When switching the order we need to make sure we’re looking at the same region.  So for a given value of u, we integrate t from u to x and then integrate u from zero to x.

You can either

Integrating over the same region in two different orders.

So that’s what happened there.

Posted in -- By the Physicist, Conventions, Equations, Math | 13 Comments

Q: Why is our Moon drifting away while Mars’ moons are falling?

The original question was: I know the Moon is getting further away because tides/friction/conservation of angular momentum.  This video claims Phobos is getting closer to Mars because of tidal forces, what gives?  Obviously no oceans to drag around but what else?


Physicist: A moon causes the material of the planet under it to distend toward it (and away, which is why there are two tides).  This is especially obvious on Earth where the water is free to move a lot more than the ground.  However that bump takes a little while to relax and, because planets turn and moons orbit, that bump is never exactly under the moon.

The tidal bulge created by a moon doesn't stay directly under that moon, either because the planet is turning, because the moon is orbiting or both.

The tidal bulge created by a moon doesn’t stay directly under that moon, either because the planet is turning, because the moon is orbiting, or both.  This is really, really not to scale.

Because the Earth spins in the same direction that the Moon orbits, our bump leads the Moon a little.  If the Moon orbited in the opposite direction, or even orbited so fast that its orbit were faster than the Earth’s spin, then the bump would trail the Moon.

The bump itself has mass and therefore a little extra gravity.  If it leads the moon, then the moon speeds up because it’s getting a tiny, tiny extra pull in the direction its orbiting.  Speeding up causes things in orbit to assume higher orbits, which we often and not-quite-accurately describe as “drifting away”.  Our Moon gets a couple cm farther away every year.

On the other hand, if the tidal bump trails behind a moon, then that moon is slowed down and drops lower as a result.  Phobos’ orbital period is about 8 hours (it’s already very low), and Mars’ day (a “sol“) is about as long as ours, so the bump Phobos creates necessarily trails behind it.  As a result Phobos is slowly dropping and will eventually impact Mars.  Mars is going to have a really bad sol in about 50 million years.

But raising moons consumes a lot of energy and that energy has to come from somewhere.  The same tiny pull that the Earth applies to the Moon to speed up its orbit is applied to the Earth to slow down our day.  When the Moon formed around 4.5 billion years ago, it about 15 times closer to the Earth (give or take) and a day was only about 6 hours long.  Back then a full moon would have provided about 200 times as much light and solar eclipses would have blacked out swaths of the Earth’s surface nearly the size of Australia.

Our Moon has more than 7 million times the mass of Phobos, so Phobos doesn’t have nearly as pronounced an impact on the spin of Mars.

The Earth and Moon as they are now and the Earth and Moon as they were when the Moon formed.

The Earth and Moon as they are now and the Earth and Moon as they were when the Moon formed.

We live in a remarkably unlikely time, when the size of the Moon in the sky perfectly matches the size of the Sun.  In fact, since the Moon’s orbit around Earth is a little elliptical, the Moon is sometimes a little smaller and sometimes a little bigger.  We live on the only planet that gets to see both annular and total solar eclipses.  But see a total eclipse while you can; in a few million years we’ll be stuck with only annular eclipses.  Sucks to be you, unforeseeable future generations!

Posted in -- By the Physicist, Astronomy, Physics | 10 Comments

Q: Why do we (people) wave our arms when we fall? Is it for attention?

The original question was: When I am about to fall backwards I spin my arms up over my head then down . This seems to help in preventing my fall somewhat.

Is this some form of conservation of angular momentum?


Physicist: Actually, it’s some form of conservation of angular momentum.

Assuming you still have at least one foot on the ground, falling over is just a rotation that stops either comically or sadly or both.  You start upright and 90° later you’re prone.  Everything that moves is subject to the conservation of momentum: if you push something one way, then you’ll be pushed the other way.  Similarly, everything that rotates is subject to the conservation of angular momentum: if you rotate something one way, then you’ll be rotated the other way.

If you're falling to the left (counterclockwise), then you want your body to rotate clockwise.  To create

Mr. Anderson wants his body to rotate clockwise (to stand back up), so he rotates his arms counter-clockwise.  This sort of insight is why it’s so unpleasant to watch movies with physicists.

If you start to fall in some direction, then you want your body to rotate in the opposite direction.  Rotating your arms in the direction of the fall causes the rest of you to rotate back to upright.

Someone with gigantic hands and stick-thin-but-long arms would practically never fall over.  The more mass you have far from the pivot point (shoulders) the greater the “moment of inertia“.  Something with a lot of mass is hard to move or stop (it has a lot of “inertia“).  Something with a high moment of inertia is hard to turn and to stop turning.  In fact, that’s why flywheels are designed the way they are.  Most of their mass is far from their pivot so that they have a high moment of inertia without being ridiculously heavy.

She couldn't fall if she wanted to; her moment of inertia is way to high.

She couldn’t fall if she wanted to: her moment of inertia is way to high.

Until we can find the coveted giant-hands-long-arms gene, tightrope walkers will be forced to continue using poles to keep their balance.  They use their pole to keep balanced in exactly the same way the rest of us ground-dwellers use our arms.  If you can believe it, the search for this gene isn’t a priority for most geneticists.

Secretly sad that his arms are so short.

Such tiny arms.

The same trick we (humans with arms) use to keep our balance is the same trick that space craft use to rotate in space.  Here on the ground we can push off of things.  If you want to face a new direction, use your feet to push on the ground and voilà: a new vista lays before you (or another wall if you’re inside).

However, space is notable for its remarkable dearth of stuff.  There’s nothing in space to push on, so spacecraft engineers (notable for their remarkable plenitude of cleverness) literally provide stuff for their spacecraft to push: flywheels.  By turning a tiny flywheel clockwise a lot, the rest of the spacecraft turns counter-clockwise a little.

Turn a wheel in space, and turn yourself.

Turn a wheel in space, and turn yourself.  These are from the Hubble space telescope.  Hubble would be a whole lot of pointless if it couldn’t point in more than one direction, but it also can’t pollute the space around it with rocket exhaust (avoiding gases is why it’s up there in the first place).  So: flywheels.

Posted in -- By the Physicist, Physics | 2 Comments

Q: What is the state of matter in deep space?

The original question was: What might be the state of matter in [interstellar space]. Average temperature is 2.7K , so all the gases like hydrogen, helium should be in liquid or solid state?

Physicist: Matter in deep space tends to take the form of gas.  Liquids basically don’t happen, and solids are pretty rare.  The cosmic microwave background ensures that everything is at least as hot as about 2.7K, but in general the gases we see out there are quite a bit hotter.  Fortunately, there’s so little of it that you’d barely notice.

Sure it's hot, but there isn't much of it so there isn't enough heat to light things on fire.

Sure it’s hot, but there’s so little of it that there isn’t enough heat to light things on fire.  The same thing is true for gases in space: it’s hot, but there’s so little of it you’d never notice.

For example, in the core of the Orion Nebula the temperature of the gas is around 10,000K, but even if you were there your biggest problem wouldn’t be getting burned, it would be freezing to death.  Also suffocation.  Really, if you’re going to travel in space at all, bring a spacesuit.

Temperature is a measure of the average random speed of a material (technically, it’s the “variance” of the velocities).  That “random” bit is important.  For example, the air you’re (hopefully) breathing isn’t sitting still, it’s moving every-which-way at around half a km per second.  Air bounces around so much that it doesn’t get very far (about 1 ten millionth of a meter before it gets bounced), but it’s all still moving pretty fast.

By comparison, Halley’s comet is moving at a few dozen km/s (it varies between about 1 km/s and 50 km/s).  However, it’s frozen because that movement isn’t random.  Its atoms are all moving really fast, but they’re all moving together.

A comet moves faster than air molecules (on average), but it's still frozen solid.

(left) a hot air balloon and (right) an artist’s interpretation of a comet.  Temperature is the random movement of molecules.  A comet moves faster than air molecules (on average), but it’s still frozen solid because all of its molecules are moving in more-or-less the same direction.  The air molecules  in a hot air balloon move fast (relative to each other) and the bits of ice in the comet moves slowly (relative to each other).

Here comes the point.  In deep space you mostly find individual atoms cruising along at high speeds.  In order to define a temperature you look at lots of atoms passing through a region and see how random those trajectories are.  Typically: they’re very random and very fast, so the temperature of those gases is thousands of degrees Celsius (or Kelvin or Fahrenheit for that matter).

It’s not unusual for things to be randomly traveling through space really fast: space is full of stuff taking forever to go from nowhere to nowhere through nothing at break-neck speeds.  Generally, unless there’s a good reason to do otherwise, the individual atoms in space are traveling in every-which-way.  A “good reason” is often running into something, or being caught up in an accretion disk.  The majority of molecules in deep space are traveling in a straight line, very fast, without (strongly) interacting with anything else for years at a time.

A natural question to ask is: Isn’t space cold?  If this interstellar gas is so damnably hot, then why doesn’t it cool off?  The answers are: “yes it is” and “you have to think about why things cool off”.  Light is created by accelerating charges.  Traveling in a straight line involves no acceleration.  If things bounce off of each other a lot, then they change direction a lot and that involves (for lack of a better word) a lot of accelerating.

This has nothing to do with either the spirit or letter of the question, but one of the most terribly cool things ever is that entire stars obey the same rules; they travel along at high speeds without interacting with anything else for huge periods of time.  While atoms scatter by bouncing off of each other, stars interact with each other through their gravity (it’s very, very rare for stars to actually run into each other).  If you stand back far enough, you find that the stars in a galaxy act a lot like a gas.  You can even describe the “temperature” of a galaxy and talk about the movement of its stars in terms of thermodynamics.  For example, “hot galaxies” are those with lots of randomly moving stars, and these galaxies literally evaporate (eject stars into inter-galactic space) and “cool” as a result.  As they “cool” the randomness of their remaining stars’ movements decrease and the galaxy itself tends to contract.

Posted in -- By the Physicist, Astronomy, Physics | 10 Comments