gravity – nebusresearch

Reading the Comics, September 19, 2017: Visualization Edition

Comic Strip Master Command apparently doesn’t want me talking about the chances of Friday’s Showcase Showdown. They sent me enough of a flood of mathematically-themed strips that I don’t know when I’ll have the time to talk about the probability of that episode. (The three contestants spinning the wheel all tied, each spinning $1.00. And then in the spin-off, two of the three contestants also spun $1.00. And this after what was already a perfect show, in which the contestants won all six of the pricing games.) Well, I’ll do what comic strips I can this time, and carry on the last week of the Summer 2017 A To Z project, and we’ll see if I can say anything timely for Thursday or Saturday or so.

Jim Scancarelli’s Gasoline Alley for the 17th is a joke about the student embarrassing the teacher. It uses mathematics vocabulary for the specifics. And it does depict one of those moments that never stops, as you learn mathematics. There’s always more vocabulary. There’s good reasons to have so much vocabulary. Having names for things seems to make them easier to work with. We can bundle together ideas about what a thing is like, and what it may do, under a name. I suppose the trouble is that we’ve accepted a convention that we should define terms before we use them. It’s nice, like having the dramatis personae listed at the start of the play. But having that list isn’t the same as saying why anyone should care. I don’t know how to balance the need to make clear up front what one means and the need to not bury someone under a heap of similar-sounding names.

Mac King and Bill King’s Magic in a Minute for the 17th is another puzzle drawn from arithmetic. Look at it now if you want to have the fun of working it out, as I can’t think of anything to say about it that doesn’t spoil how the trick is done. The top commenter does have a suggestion about how to do the problem by breaking one of the unstated assumptions in the problem. This is the kind of puzzle created for people who want to motivate talking about parity or equivalence classes. It’s neat when you can say something of substance about a problem using simple information, though.

'How are you and David doing?' 'Better, with counseling.' (As Ben takes his drink bottle.) 'But sometimes he still clings to hope that Ben's autism is 'curable'. Admittedly, I've wondered that myself. Then Ben strips naked and solves a trigonometry problem.' 'Whoa.' (Ben throws his drink bottle in the air and says) 'A = (1/2)(4)(2) sin 45 deg.' — Terri Libenson’s **Pajama Diaries** for the 18th of September, 2017. When I first read this I assumed that of course the base of the triangle had length 4 and the second leg, at a 45-degree angle to that, had length 2, and I wondered if those numbers could be consistent for a triangle to exist. Of course they could, though. There is a bit of fun to be had working out whether a particular triangle could exist from knowing its side lengths, though.

Terri Libenson’s Pajama Diaries for the 18th uses trigonometry as the marker for deep thinking. It comes complete with a coherent equation, too. It gives the area of a triangle with two legs that meet at a 45 degree angle. I admit I am uncomfortable with promoting the idea that people who are autistic have some super-reasoning powers. (Also with the pop-culture idea that someone who spots things others don’t is probably at least a bit autistic.) I understand wanting to think someone’s troubles have some compensation. But people are who they are; it’s not like they need to observe some “balance”.

Lee Falk and Wilson McCoy’s The Phantom for the 10th of August, 1950 was rerun Monday. It’s a side bit of joking about between stories. And it uses knowledge of mathematics — and an interest in relativity — as signifier of civilization. I can only hope King Hano does better learning tensors on his own than I do.

Guest Woman: 'Did you know the King was having trouble controlling the young hotheads in his own tribe?' Phantom: 'Yes. He's an old friend of mine. He probably looks like an ignorant savage to you. Actually, he speaks seven languages, is an expert mathematician, and plays a fine hand of poker.' Guest Woman: 'What?' Cut to the King, in his hut, reading The Theory Of Relativity. 'Thank goodness that's over ... Now where was I?' — Lee Falk and Wilson McCoy’s **The Phantom** for the 10th of August, 1950 and rerun the 18th of September, 2017. For my money, just reading a mathematics book doesn’t take. I need to take notes, as if it were in class. I don’t quite copy the book, but it comes close.

Mike Thompson’s Grand Avenue for the 18th goes back to classrooms and stuff for clever answers that subvert the teacher. And I notice, per the title given this edition, that the teacher’s trying to make the abstractness of three minus two tangible, by giving it an example. Which pairs it with …

Will Henry’s Wallace the Brace for the 18th, wherein Wallace asserts that arithmetic is easier if you visualize real things. I agree it seems to help with stuff like basic arithmetic. I wouldn’t want to try taking the cosine of an apple, though. Separating the quantity of a thing from the kind of thing measured is one of those subtle breakthroughs. It’s one of the ways that, for example, modern calculations differ from those of the Ancient Greeks. But it does mean thinking of numbers in, we’d say, a more abstract way than they did, and in a way that seems to tax us more.

Wallace the Brave recently had a book collection published, by the way. I mention because this is one of a handful of comics with a character who likes pinball, and more, who really really loves the Williams game FunHouse. This is an utterly correct choice for favorite pinball game. It’s one of the games that made me a pinball enthusiast.

Ryan North’s Dinosaur Comics rerun for the 19th I mention on loose grounds. In it T-Rex suggests trying out an alternate model for how gravity works. The idea, of what seems to be gravity “really” being the shade cast by massive objects in a particle storm, was explored in the late 17th and early 18th century. It avoids the problem of not being able to quite say what propagates gravitational attraction. But it also doesn’t work, analytically. We would see the planets orbit differently if this were how gravity worked. And there’s the problem about mass and energy absorption, as pointed out in the comic. But it can often be interesting or productive to play with models that don’t work. You might learn something about models that do, or that could.

Why Stuff Can Orbit, Part 13: To Close A Loop

Why Stuff Can Orbit, featuring a dazed-looking coati (it's a raccoon-like creature from Latin America) and a starry background. — Art courtesy of Thomas K Dye, creator of the web comic **Newshounds**. He has a Patreon for those able to support his work. He’s also open for commissions, starting from US$10.

Previously:

And the supplemental reading:

Today’s is one of the occasional essays in the Why Stuff Can Orbit sequence that just has a lot of equations. I’ve tried not to write everything around equations because I know what they’re like to read. They’re pretty to look at and after about four of them you might as well replace them with a big grey box that reads “just let your eyes glaze over and move down to the words”. It’s even more glaze-y than that for non-mathematicians.

But we do need them. Equations are wonderfully compact, efficient ways to write about things that are true. Especially things that are only true if exacting conditions are met. They’re so good that I’ll often find myself checking a textbook for an explanation of something and looking only at the equations, letting my eyes glaze over the words. That’s a chilling thing to catch yourself doing. Especially when you’ve written some obscure textbooks and a slightly read mathematics blog.

What I had been looking at was a perturbed central-force orbit. We have something, generically called a planet, that orbits the center of the universe. It’s attracted to the center of the universe by some potential energy, which we describe as ‘U(r)’. It’s some number that changes with the distance ‘r’ the planet has from the center of the universe. It usually depends on other stuff too, like some kind of mass of the planet or some constants or stuff. The planet has some angular momentum, which we can call ‘L’ and pretend is a simple number. It’s in truth a complicated number, but we’ve set up the problem where we can ignore the complicated stuff. This angular momentum implies the potential energy allows for a circular orbit at some distance which we’ll call ‘a’ from the center of the universe.

From ‘U(r)’ and ‘L’ we can say whether this is a stable orbit. If it’s stable, a little perturbation, a nudging, from the circular orbit will stay small. If it’s unstable, a little perturbation will keep growing and never stop. If we perturb this circular orbit the planet will wobble back and forth around the circular orbit. Sometimes the radius will be a little smaller than ‘a’, and sometimes it’ll be a little larger than ‘a’. And now I want to see whether we get a stable closed orbit.

The orbit will be closed if the planet ever comes back to the same position and same momentum that it started with. ‘Started’ is a weird idea in this case. But it’s common vocabulary. By it we mean “whatever properties the thing had when we started paying attention to it”. Usually in a problem like this we suppose there’s some measure of time. It’s typically given the name ‘t’ because we don’t want to make this hard on ourselves. The start is some convenient reference time, often ‘t = 0’. That choice usually makes the equations look simplest.

The position of the planet we can describe with two variables. One is the distance from the center of the universe, ‘r’, which we know changes with time: ‘r(t)’. Another is the angle the planet makes with respect to some reference line. The angle we might call ‘θ’ and often do. This will also change in time, then, ‘θ(t)’. We can pick other variables to describe where something is. But they’re going to involve more algebra, more symbol work, than this choice does so who needs it?

Momentum, now, that’s another set of variables we need to worry about. But we don’t need to worry about them. This particular problem is set up so that if we know the position of the planet we also have the momentum. We won’t be able to get both ‘r(t)’ and ‘θ(t)’ back to their starting values without also getting the momentum there. So we don’t have to worry about that. This won’t always work, as see my future series, ‘Why Statistical Mechanics Works’.

So. We know, because it’s not that hard to work out, how long it takes for ‘r(t)’ to get back to its original, ‘r(0)’, value. It’ll take a time we worked out to be (first big equation here, although we found it a couple essays back):

$T_r = 2\pi\sqrt{ \frac{m}{ -F'(a) - \frac{3}{a} F(a) }}$

Here ‘m’ is the mass of the planet. And ‘F’ is a useful little auxiliary function. It’s the force that the planet feels when it’s a distance from the origin. It’s defined as $F(r) = -\frac{dU}{dr}$ . It’s convenient to have around. It makes equations like this one simpler, for one. And it’s weird to think of a central force problem where we never, ever see forces. The peculiar thing is we define ‘F’ for every distance the planet might be from the center of the universe. But all we care about is its value at the equilibrium, circular orbit distance of ‘a’. We also care about its first derivative, also evaluated at the distance of ‘a’, which is that $F'(a)$ talk early on in that denominator.

So in the time between time ‘0’ and time ‘T_r‘ the perturbed radius will complete a full loop. It’ll reach its biggest value and its smallest value and get back to the original. (It is so much easier to suppose the perturbation starts at its biggest value at time ‘0’ that we often assume it has. It doesn’t have to be. But if we don’t have something forcing the choice of what time to call ‘0’ on us, why not pick one that’s convenient?) The question is whether ‘θ(t)’ completes a full loop in that time. If it does then we’ve gotten back to the starting position exactly and we have a closed orbit.

Thing is that the angle will never get back to its starting value. The angle ‘θ(t)’ is always increasing at a rate we call ‘ω’, the angular velocity. This number is constant, at least approximately. Last time we found out what this number was:

$\omega = \frac{L}{ma^2}$

So the angle, over time, is going to look like:

$\theta(t) = \frac{L}{ma^2} t$

And ‘θ(T_r)’ will never equal ‘θ(0)’ again, not unless ‘ω’ is zero. And if ‘ω’ is zero then the planet is racing away from the center of the universe never to be seen again. Or it’s plummeting into the center of the universe to be gobbled up by whatever resides there. In either case, not what we traditionally think of as orbits. Even if we allow these as orbits, these would be nudges too big to call perturbations.

So here’s the resolution. Angles are right awful pains in mathematical physics. This is because increasing an angle by 2π — or decreasing it by 2π — has no visible effect. In the language of the hew-mon, adding 360 degrees to a turn leaves you back where you started. A 45 degree angle is indistinguishable from a 405 degree angle, or a 765 degree angle, or a -315 degree angle, or so on. This makes for all sorts of irritating alternate cases to consider when you try solving for where one thing meets another. But it allows us to have closed orbits.

Because we can have a closed orbit, now, if the radius ‘r(t)’ completes a full oscillation in the time it takes ‘θ(t)’ to grow by 2π. Or to grow by π. Or to grow by ½π. Or a third of π. Or so on.

So. Last time we worked out that the angular velocity had to be this number:

$\omega = \frac{L}{ma^2}$

And that looked weird because the central force doesn’t seem to be there. It’s in there. It’s just implicit. We need to know what the central force is to work out what ‘a’ is. But we can make it explicit by using that auxiliary little function ‘F(r)’. In particular, at the circular orbit radius of ‘a’ we have that:

$F(a) = -\frac{L^2}{ma^3}$

I am going to use this to work out what ‘L’ has to be, in terms of ‘F’ and ‘m’ and ‘a’. First, multiply both sides of this equation by ‘ma³‘:

$F(a) \cdot ma^3 = -L^2$

And then both sides by -1:

$-ma^3 F(a) = L^2$

Take the square root — don’t worry, that it will turn out that ‘F(a)’ is a negative number so we’re not doing anything suspicious —

$\sqrt{-ma^3 F(a)} = L$

Now, take that ‘L’ we’ve got and put it back into the equation for angular velocity:

$\omega = \frac{L}{ma^2} = \frac{\sqrt{-ma^3 F(a)}}{ma^2}$

We might look stuck and at what seems like an even worse position. It’s not. When you do enough of these problems you get used to some tricks. For example, that ‘ma²‘ in the denominator we could move under the square root if we liked. This we know because $ma^2 = \sqrt{ \left(ma^2\right)^2 }$ at least as long as ‘ma²‘ is positive. It is.

So. We fall back on the trick of squaring and square-rooting the denominator and so generate this mess:

$\omega = \sqrt{\frac{-ma^3 F(a)}{\left(ma^2\right)^2}} \\ \omega = \sqrt{\frac{-ma^3 F(a)}{m^2 a^4}} \\ \omega = \sqrt{\frac{-F(a)}{ma}}$

That’s getting nice and simple. Let me go complicate matters. I’ll want to know the angle that the planet sweeps out as the radius goes from its largest to its smallest value. Or vice-versa. This time is going to be half of ‘T_r‘, the time it takes to do a complete oscillation. The oscillation might have started at time ‘t’ of zero, maybe not. But how long it takes will be the same. I’m going to call this angle ‘ψ’, because I’ve written “the angle that the planet sweeps out as the radius goes from its largest to its smallest value” enough times this essay. If ‘ψ’ is equal to π, or one-half π, or one-third π, or some other nice rational multiple of π we’ll get a closed orbit. If it isn’t, we won’t.

So. ‘ψ’ will be one-half times the oscillation time times that angular velocity. This is easy:

$\psi = \frac{1}{2} \cdot T_r \cdot \omega$

Put in the formulas we have for ‘T_r‘ and for ‘ω’. Now it’ll be complicated.

$\psi = \frac{1}{2} 2\pi \sqrt{\frac{m}{-F'(a) - \frac{3}{a} F(a)}} \sqrt{\frac{-F(a)}{ma}}$

Now we’ll make this a little simpler again. We have two square roots of fractions multiplied by each other. That’s the same as the square root of the two fractions multiplied by each other. So we can take numerator times numerator and denominator times denominator, all underneath the square root sign. See if I don’t. Oh yeah and one-half of two π is π but you saw that coming.

$\psi = \pi \sqrt{ \frac{-m F(a)}{-\left(F'(a) + \frac{3}{m}F(a)\right)\cdot ma} }$

OK, so there’s some minus signs in the numerator and denominator worth getting rid of. There’s an ‘m’ in the numerator and the denominator that we can divide out of both sides. There’s an ‘a’ in the denominator that can multiply into a term that has a denominator inside the denominator and you know this would be easier if I could use little cross-out symbols in WordPress LaTeX. If you’re not following all this, try writing it out by hand and seeing what makes sense to cancel out.

$\psi = \pi \sqrt{ \frac{F(a)}{aF'(a) + 3F(a)} }$

This is getting not too bad. Start from a potential energy ‘U(r)’. Use an angular momentum ‘L’ to figure out the circular orbit radius ‘a’. From the potential energy find the force ‘F(r)’. And then, based on what ‘F’ and the first derivative of ‘F’ happen to be, at the radius ‘a’, we can see whether a closed orbit can be there.

I’ve gotten to some pretty abstract territory here. Next time I hope to make things simpler again.

Why Stuff Can Orbit, Part 12: How Fast Is An Orbit?

Previously:

And the supplemental reading:

On to the next piece of looking for stable, closed orbits of a central force. We start from a circular orbit of something around the sun or the mounting point or whatever. The center. I would have saved myself so much foggy writing if I had just decided to make this a sun-and-planet problem. But I had wanted to write the general problem. In this the force attracting the something towards the center has a strength that’s some constant times the distance to the center raised to a power. This is easy to describe in symbols. It’s cluttered to describe in words. This is why symbols are so nice.

The perturbed orbit, the one I want to see close up, looks like an oscillation around that circle. The fact it is a perturbation, a small nudge away from the equilibrium, means how big the perturbation is will oscillate in time. How far the planet (whatever) is from the center will make a sine wave in time. Whether it closes depends on what it does in space.

Part of what it does in space is easy. I just said what the distance from the planet to the center does. But to say where the planet is we need to know how far it is from the center and what angle it makes with respect to some reference direction. That’s a little harder. We also need to know where it is in the third dimension, but that’s so easy. An orbit like this is always in one plane, so we picked that plane to be that of our paper or whiteboard or tablet or whatever we’re using to sketch this out. That’s so easy to answer we don’t even count it as solved.

The angle, though. Here, I mean the angle made by looking at the planet, the center, and some reference direction. This angle can be any real number, although a lot of those angles are going to point to the same direction in space. We’re coming at this from a mathematical view, or a physics view. Or a mathematical physics view. It means we measure this angle as radians instead of degrees. That is, a right angle is $\frac{\pi}{2}$ , not 90 degrees, thank you. A full circle is $2\pi$ and not 360 degrees. We aren’t doing this to be difficult. There are good reasons to use radians. They make the mathematics simpler. What else could matter?

We use $\theta$ as the symbol for this angle. It’s a popular choice. $\theta$ is going to change in time. We’ll want to know how fast it changes over time. This concept we call the angular velocity. For this there are a bunch of different possible notations. The one that I snuck in here two essays ago was ω.

We came at the physics of this orbiting planet from a weird direction. Well, I came at it, and you followed along, and thank you for that. But I never did something like set the planet at a particular distance from the center of the universe and give it a set speed so it would have a circular enough orbit. I set up that we should have some potential energy. That energy implies a central force. It attracts things to the center of the universe. And that there should be some angular momentum that the planet has in its movement. And from that, that there would be some circular orbit. That circular orbit is one with just the right radius and just the right change in angle over time.

From the potential energy and the angular momentum we can work out the radius of the circular orbit. Suppose your potential energy obeys a rule like $V(r) = Cr^n$ for some number ‘C’ and some power, another number, ‘n’. Suppose your planet has the mass ‘m’. Then you’ll get a circular orbit when the planet’s a distance ‘a’ from the center, if $a^{n + 2} = \frac{L^2}{n C m}$ . And it turns out we can also work out the angular velocity of this circular orbit. It’s all implicit in the amount of angular momentum that the planet has. This is part of why a mathematical physicist looks for concepts like angular momentum. They’re easy to work with, and they yield all sorts of interesting information, given the chance.

I first introduced angular momentum as this number that was how much of something that our something had. It’s got physical meaning, though, reflecting how much … uh … our something would like to keep rotating around the way it has. And this can be written as a formula. The angular momentum ‘L’ is equal to the moment of inertia ‘I’ times the angular velocity ‘ω’. ‘L’ and ‘ω’ are really vectors, and ‘I’ is really a tensor. But we don’t have to worry about this because this kind of problem is easy. We can pretend these are all real numbers and nothing more.

The moment of inertia depends on how the mass of the thing rotating is distributed in space. And it depends on how far the mass is from whatever axis it’s rotating around. For real bodies this can be challenging to work out. It’s almost always a multidimensional integral, haunting students in Calculus III. For a mass in a central force problem, though, it’s easy once again. Please tell me you’re not surprised. If it weren’t easy I’d have some more supplemental reading pieces here first.

For a planet of mass ‘m’ that’s a distance ‘r’ from the axis of rotation, the moment of inertia ‘I’ is equal to ‘mr²‘. I’m fibbing. Slightly. This is for a point mass, that is, something that doesn’t occupy volume. We always look at point masses in this sort of physics. At least when we start. It’s easier, for one thing. And it’s not far off. The Earth’s orbit has a radius just under 150,000,000 kilometers. The difference between the Earth’s actual radius of just over 6,000 kilometers and a point-mass radius of 0 kilometers is a minor correction.

So since we know $L = I\omega$ , and we know $I = mr^2$ , we have $L = mr^2\omega$ and from this:

$\omega = \frac{L}{mr^2}$

We know that ‘r’ changes in time. It oscillates from a maximum to a minimum value like any decent sine wave. So ‘r²‘ is going to oscillate too, like a … sine-squared wave. And then dividing the constant ‘L’ by something oscillating like a sine-squared wave … this implies ω changes in time. So it does. In a possibly complicated and annoying way. So it does. I don’t want to deal with that. So I don’t.

Instead, I am going to summon the great powers of approximation. This perturbed orbit is a tiny change from a circular orbit with radius ‘a’. Tiny. The difference between the actual radius ‘r’ and the circular-orbit radius ‘a’ should be small enough we don’t notice it at first glance. So therefore:

$\omega = \frac{L}{ma^2}$

And this is going to be close enough. You may protest: what if it isn’t? Why can’t the perturbation be so big that ‘a’ is a lousy approximation to ‘r’? To this I say: if the perturbation is that big it’s not a perturbation anymore. It might be an interesting problem. But it’s a different problem from what I’m doing here. It needs different techniques. The Earth’s orbit is different from Halley’s Comet’s orbit in ways we can’t ignore. I hope this answers your complaint. Maybe it doesn’t. I’m on your side there. A lot of mathematical physics, and of analysis, is about making approximations. We need to find perturbations big enough to give interesting results. But not so big they need harder mathematics than you can do. It’s a strange art. I’m not sure I know how to describe how to do it. What I know I’ve learned from doing a lot of problems. You start to learn what kinds of approaches usually pan out.

But what we’re relying on is the same trick we use in analysis. We suppose there is some error margin in the orbit’s radius and angle that’s tolerable. Then if the perturbation means we’d fall outside that error margin, we just look instead at a smaller perturbation. If there is no perturbation small enough to stay within our error margin then the orbit isn’t stable. And we already know it is. Here, we’re looking for closed orbits. People could in good faith argue about whether some particular observed orbit is a small enough perturbation from the circular equilibrium. But they can’t argue about whether there exist some small enough perturbations.

Let me suppose that you’re all right with my answer about big perturbations. There’s at least one more good objection to have here. It’s this: where is the central force? The mass of the planet (or whatever) is there. The angular momentum is there. The equilibrium orbit is there. But where’s the force? Where’s the potential energy we started with? Shouldn’t that appear somewhere in the description of how fast this planet moves around the center?

It should. And it is there, in an implicit form. We get the radius of the circular, equilibrium orbit, ‘a’, from knowing the potential energy. But we’ll do well to tease it out more explicitly. I hope to get there next time.

Why Stuff Can Orbit, Part 11: In Search Of Closure

Previously:

And the supplemental reading:

I’m not ready to finish the series off yet. But I am getting closer to wrapping up perturbed orbits. So I want to say something about what I’m looking for.

In some ways I’m done already. I showed how to set up a central force problem, where some mass gets pulled towards the center of the universe. It can be pulled by a force that follows any rule you like. The rule has to follow some rules. The strength of the pull changes with how far the mass is from the center. It can’t depend on what angle the mass makes with respect to some reference meridian. Once we know how much angular momentum the mass has we can find whether it can have a circular orbit. And we can work out whether that orbit is stable. If the orbit is stable, then for a small nudge, the mass wobbles around that equilibrium circle. It spends some time closer to the center of the universe and some time farther away from it.

I want something a little more, else I can’t carry on this series. I mean, we can make central force problems with more things in them. What we have now is a two-body problem. A three-body problem is more interesting. It’s pretty near impossible to give exact, generally true answers about. We can save things by only looking at very specific cases. Fortunately one is a sun, planet, and moon, where each object is much more massive than the next one. We see a lot of things like that. Four bodies is even more impossible. Things start to clear up if we look at, like, a million bodies, because our idea of what “clear” is changes. I don’t want to do that right now.

Instead I’m going to look for closed orbits. Closed orbits are what normal people would call “orbits”. We’re used to thinking of orbits as, like, satellites going around and around the Earth. We know those go in circles, or ellipses, over and over again. They don’t, but the difference between a closed orbit and what they do is small enough we don’t need to care.

Here, “orbit” means something very close to but not exactly what normal people mean by orbits. Maybe I should have said something about that before. But the difference hasn’t counted for much before.

Start off by thinking of what we need to completely describe what a particular mass is doing. You need to know the central force law that the mass obeys. You need to know, for some reference time, where it is. You also need to know, for that same reference time, what its momentum is. Once you have that, you can predict where it should go for all time to come. You can also work out where it must have been before that reference time. (This we call “retrodicting”. Or “predicting the past”. With this kind of physics problem time has an unnerving symmetry. The tools which forecast what the mass will do in the future are exactly the same as those which tell us what the mass has done in the past.)

Now imagine knowing all the sets of positions and momentums that the mass has had. Don’t look just at the reference time. Look at all the time before the reference time, and look at all the time after the reference time. Imagine highlighting all the sets of positions and momentums the mass ever took on or ever takes on. We highlight them against the universe of all the positions and momentums that the mass could have had if this were a different problem.

What we get is this ribbon-y thread that passes through the universe of every possible setup. This universe of every possible setup we call a “phase space”. It’s easy to explain the “space” part of that name. The phase space obeys the rules we’d expect from a vector space. It also acts in a lot of ways like the regular old space that we live in. The “phase” part I’m less sure how to justify. I suspect we get it because this way of looking at physics problems comes from statistical mechanics. And in that field we’re looking, often, at the different ways a system can behave. This mathematics looks a lot like that of different phases of matter. The changes between solids and liquids and gases are some of what we developed this kind of mathematics to understand, in fact. But this is speculation on my part. I’m not sure why “phase” has attached to this name. I can think of other, harder-to-popularize reasons why the name would make sense too. Maybe it’s the convergence of several reasons. I’d love to hear if someone has a good etymology. If one exists; remember that we still haven’t got the story straight about why ‘m’ stands for the slope of a line.

Anyway, this ribbon of all the arrangements of position and momentum that the mass does ever at any point have we call a “trajectory”. We call it a trajectory because it looks like a trajectory. Sometimes mathematics terms aren’t so complicated. We also call it an “orbit” since very often the problems we like involve trajectories that loop around some interesting area. It looks like a planet orbiting a sun.

A “closed orbit” is an orbit that gets back to where it started. This means you can take some reference time, and wait. Eventually the mass comes back to the same position and the same momentum that you saw at that reference time. This might seem unavoidable. Wouldn’t it have to get back there? And it turns out, no, it doesn’t. A trajectory might wander all over phase space. This doesn’t take much imagination. But even if it doesn’t, if it stays within a bounded region, it could still wander forever without repeating itself. If you’re not sure about that, please consider an old sequence I wrote inspired by the Aardman Animation film Arthur Christmas. Also please consider seeing the Aardman Animation film Arthur Christmas. It is one of the best things this decade has offered us. The short version is, though, that there is a lot of room even in the smallest bit of space. A trajectory is, in a way, a one-dimensional thing that might get all coiled up. But phase space has got plenty of room for that.

And sometimes we will get a closed orbit. The mass can wander around the center of the universe and come back to wherever we first noticed it with the same momentum it first had. A that point it’s locked into doing that same thing again, forever. If it could ever break out of the closed orbit it would have had to the first time around, after all.

Closed orbits, I admit, don’t exist in the real world. Well, the real world is complicated. It has more than a single mass and a single force at work. Energy and momentum are conserved. But we effectively lose both to friction. We call the shortage “entropy”. Never mind. No person has ever seen a circle, and no person ever will. They are still useful things to study. So it is with closed orbits.

An equilibrium orbit, the circular orbit of a mass that’s at exactly the right radius for its angular momentum, is closed. A perturbed orbit, wobbling around the equilibrium, might be closed. It might not. I mean next time to discuss what has to be true to close an orbit.

Why Stuff Can Orbit, Part 10: Where Time Comes From And How It Changes Things

Previously:

And the supplemental reading:

And again my thanks to Thomas K Dye, creator of the web comic Newshounds, for the banner art. He has a Patreon to support his creative habit.

In the last installment I introduced perturbations. These are orbits that are a little off from the circles that make equilibriums. And they introduce something that’s been lurking, unnoticed, in all the work done before. That’s time.

See, how do we know time exists? … Well, we feel it, so, it’s hard for us not to notice time exists. Let me rephrase it then, and put it in contemporary technology terms. Suppose you’re looking at an animated GIF. How do you know it’s started animating? Or that it hasn’t stalled out on some frame?

If the picture changes, then you know. It has to be going. But if it doesn’t change? … Maybe it’s stalled out. Maybe it hasn’t. You don’t know. You know there’s time when you can see change. And that’s one of the little practical insights of physics. You can build an understanding of special relativity by thinking hard about that. Also think about the observation that the speed of light (in vacuum) doesn’t change.

When something physical’s in equilibrium, it isn’t changing. That’s how we found equilibriums to start with. And that means we stop keeping track of time. It’s one more thing to keep track of that doesn’t tell us anything new. Who needs it?

For the planet orbiting a sun, in a perfect circle, or its other little variations, we do still need time. At least some. How far the planet is from the sun doesn’t change, no, but where it is on the orbit will change. We can track where it is by setting some reference point. Where the planet is at the start of our problem. How big is the angle between where the planet is now, the sun (the center of our problem’s universe), and that origin point? That will change over time.

But it’ll change in a boring way. The angle will keep increasing in magnitude at a constant speed. Suppose it takes five time units for the angle to grow from zero degrees to ten degrees. Then it’ll take ten time units for the angle to grow from zero to twenty degrees. It’ll take twenty time units for the angle to grow from zero to forty degrees. Nice to know if you want to know when the planet is going to be at a particular spot, and how long it’ll take to get back to the same spot. At this rate it’ll be eighteen time units before the angle grows to 360 degrees, which looks the same as zero degrees. But it’s not anything interesting happening.

We’ll label this sort of change, where time passes, yeah, but it’s too dull to notice as a “dynamic equilibrium”. There’s change, but it’s so steady and predictable it’s not all that exciting. And I’d set up the circular orbits so that we didn’t even have to notice it. If the radius of the planet’s orbit doesn’t change, then the rate at which its apsidal angle changes, its “angular velocity”, also doesn’t change.

Now, with perturbations, the distance between the planet and the center of the universe will change in time. That was the stuff at the end of the last installment. But also the apsidal angle is going to change. I’ve used ‘r(t)’ to represent the radial distance between the planet and the sun before, and to note that what value it is depends on the time. I need some more symbols.

There’s two popular symbols to use for angles. Both are Greek letters because, I dunno, they’ve always been. (Florian Cajori’s A History of Mathematical Notation doesn’t seem to have anything. And when my default go-to for explaining mathematician’s choices tells me nothing, what can I do? Look at Wikipedia? Sure, but that doesn’t enlighten me either.) One is to use theta, θ. The other is to use phi, φ. Both are good, popular choices, and in three-dimensional problems we’ll often need both. We don’t need both. The orbit of something moving under a central force might be complicated, but it’s going to be in a single plane of movement. The conservation of angular momentum gives us that. It’s not the last thing angular momentum will give us. The orbit might happen not to be in a horizontal plane. But that’s all right. We can tilt our heads until it is.

So I’ll reach deep into the universe of symbols for angles and call on θ for the apsidal angle. θ will change with time, so, ‘θ(t)’ is the angular counterpart to ‘r(t)’.

I’d said before the apsidal angle is the angle made between the planet, the center of the universe, and some reference point. What is my reference point? I dunno. It’s wherever θ(0) is, that is, where the planet is when my time ‘t’ is zero. There’s probably a bootstrapping fallacy here. I’ll cover it up by saying, you know, the reference point doesn’t matter. It’s like the choice of prime meridian. We have to have one, but we can pick whatever one is convenient. So why not pick one that gives us the nice little identity that ‘θ(0) = 0’? If you don’t buy that and insist I pick a reference point first, fine, go ahead. But you know what? The labels on my time axis are arbitrary. There’s no difference in the way physics works whether ‘t’ is ‘0’ or ‘2017’ or ‘21350’. (At least as long as I adjust any time-dependent forces, which there aren’t here.) So we get back to ‘θ(0) = 0’.

For a circular orbit, the dynamic equilibrium case, these are pretty boring, but at least they’re easy to write. They’re:

$r(t) = a \\ \theta(t) = \omega t$

Here ‘a’ is the radius of the circular orbit. And ω is a constant number, the angular velocity. It’s how much a bit of time changes the apsidal angle. And this set of equations is pretty dull. You can see why it barely rates a mention.

The perturbed case gets more interesting. We know how ‘r(t)’ looks. We worked that out last time. It’s some function like:

$r(t) = a + A cos\left(\sqrt{\frac{k}{m}} t\right) + B sin\left(\sqrt{\frac{k}{m}} t\right)$

Here ‘A’ and ‘B’ are some numbers telling us how big the perturbation is, and ‘m’ is the mass of the planet, and ‘k’ is something related to how strong the central force is. And ‘a’ is that radius of the circular orbit, the thing we’re perturbed around.

What about ‘θ(t)’? How’s that look? … We don’t seem to have a lot to go on. We could go back to Newton and all that force equalling the change in momentum over time stuff. We can always do that. It’s tedious, though. We have something better. It’s another gift from the conservation of angular momentum. When we can turn a forces-over-time problem into a conservation-of-something problem we’re usually doing the right thing. The conservation-of-something is typically a lot easier to set up and to track. We’ve used it in the conservation of energy, before, and we’ll use it again. The conservation of ordinary, ‘linear’, momentum helps other problems, though not I’ll grant this one. The conservation of angular momentum will help us here.

So what is angular momentum? … It’s something about ice skaters twirling around and your high school physics teacher sitting on a bar stool spinning a bike wheel. All right. But it’s also a quantity. We can get some idea of it by looking at the formula for calculating linear momentum:

$\vec{p} = m\vec{v}$

The linear momentum of a thing is its inertia times its velocity. This is if the thing isn’t moving fast enough we have to notice relativity. Also if it isn’t, like, an electric or a magnetic field so we have to notice it’s not precisely a thing. Also if it isn’t a massless particle like a photon because see previous sentence. I’m talking about ordinary things like planets and blocks of wood on springs and stuff. The inertia, ‘m’, is rather happily the same thing as its mass. The velocity is how fast something is travelling and which direction it’s going in.

Angular momentum, meanwhile, we calculate with this radically different-looking formula:

$\vec{L} = I\vec{\omega}$

Here, again, talking about stuff that isn’t moving so fast we have to notice relativity. That isn’t electric or magnetic fields. That isn’t massless particles. And so on. Here ‘I’ is the “moment of inertia” and $\vec{w}$ is the angular velocity. The angular velocity is a vector that describes for us how fast the spinning is and what direction the axis around which the thing spins is. The moment of inertia describes how easy or hard it is to make the thing spin around each axis. It’s a tensor because real stuff can be easier to spin in some directions than in others. If you’re not sure that’s actually so, try tossing some stuff in the air so it spins in each of the three major directions. You’ll see.

We’re fortunate. For central force problems the moment of inertia is easy to calculate. We don’t need the tensor stuff. And we don’t even need to notice that the angular velocity is a vector. We know what axis the planet’s rotating around; it’s the one pointing out of the plane of motion. We can focus on the size of the angular velocity, the number ‘ω’. See how they’re different, what with one not having an arrow over the symbol. The arrow-less version is easier. For a planet, or other object, with mass ‘m’ that’s orbiting a distance ‘r’ from the sun, the moment of inertia is:

$I = mr^2$

So we know this number is going to be constant:

$L = mr^2\omega$

The mass ‘m’ doesn’t change. We’re not doing those kinds of problem. So however ‘r’ changes in time, the angular velocity ‘ω’ has to change with it, so that this product stays constant. The angular velocity is how the apsidal angle ‘θ’ changes over time. So since we know ‘L’ doesn’t change, and ‘m’ doesn’t change, then the way ‘r’ changes must tell us something about how ‘θ’ changes. We’ll get into that next time.

Why Stuff Can Orbit, Part 9: How The Spring In The Cosmos Behaves

Previously:

And the supplemental reading:

First, I thank Thomas K Dye for the banner art I have for this feature! Thomas is the creator of the longrunning web comic Newshounds. He’s hoping soon to finish up special editions of some of the strip’s stories and to publish a definitive edition of the comic’s history. He’s also got a Patreon account to support his art habit. Please give his creations some of your time and attention.

Now back to central forces. I’ve run out of obvious fun stuff to say about a mass that’s in a circular orbit around the center of the universe. Before you question my sense of fun, remember that I own multiple pop histories about the containerized cargo industry and last month I read another one that’s changed my mind about some things. These sorts of problems cover a lot of stuff. They cover planets orbiting a sun and blocks of wood connected to springs. That’s about all we do in high school physics anyway. Well, there’s spheres colliding, but there’s no making a central force problem out of those. You can also make some things that look like bad quantum mechanics models out of that. The mathematics is interesting even if the results don’t match anything in the real world.

But I’m sticking with central forces that look like powers. These have potential energy functions with rules that look like V(r) = C rⁿ. So far, ‘n’ can be any real number. It turns out ‘n’ has to be larger than -2 for a circular orbit to be stable, but that’s all right. There are lots of numbers larger than -2. ‘n’ carries the connotation of being an integer, a whole (positive or negative) number. But if we want to let it be any old real number like 0.1 or π or 18 and three-sevenths that’s fine. We make a note of that fact and remember it right up to the point we stop pretending to care about non-integer powers. I estimate that’s like two entries off.

We get a circular orbit by setting the thing that orbits in … a circle. This sounded smarter before I wrote it out like that. Well. We set it moving perpendicular to the “radial direction”, which is the line going from wherever it is straight to the center of the universe. This perpendicular motion means there’s a non-zero angular momentum, which we write as ‘L’ for some reason. For each angular momentum there’s a particular radius that allows for a circular orbit. Which radius? It’s whatever one is a minimum for the effective potential energy:

$V_{eff}(r) = Cr^n + \frac{L^2}{2m}r^{-2}$

This we can find by taking the first derivative of ‘V_eff‘ with respect to ‘r’ and finding where that first derivative is zero. This is standard mathematics stuff, quite routine. We can do with any function whether it represents something physics or not. So:

$\frac{dV_{eff}}{dr} = nCr^{n-1} - 2\frac{L^2}{2m}r^{-3} = 0$

And after some work, this gets us to the circular orbit’s radius:

$r = \left(\frac{L^2}{nCm}\right)^{\frac{1}{n + 2}}$

What I’d like to talk about is if we’re not quite at that radius. If we set the planet (or whatever) a little bit farther from the center of the universe. Or a little closer. Same angular momentum though, so the equilibrium, the circular orbit, should be in the same spot. It happens there isn’t a planet there.

This enters us into the world of perturbations, which is where most of the big money in mathematical physics is. A perturbation is a little nudge away from an equilibrium. What happens in response to the little nudge is interesting stuff. And here we already know, qualitatively, what’s going to happen: the planet is going to rock around the equilibrium. This is because the circular orbit is a stable equilibrium. I’d described that qualitatively last time. So now I want to talk quantitatively about how the perturbation changes given time.

Before I get there I need to introduce another bit of notation. It is so convenient to be able to talk about the radius of the circular orbit that would be the equilibrium. I’d called that ‘r’ up above. But I also need to be able to talk about how far the perturbed planet is from the center of the universe. That’s also really hard not to call ‘r’. Something has to give. Since the radius of the circular orbit is not going to change I’m going to give that a new name. I’ll call it ‘a’. There’s several reasons for this. One is that ‘a’ is commonly used for describing the size of ellipses, which turn up in actual real-world planetary orbits. That’s something we know because this is like the thirteenth part of an essay series about the mathematics of orbits. You aren’t reading this if you haven’t picked up a couple things about orbits on your own. Also we’ve used ‘a’ before, in these sorts of approximations. It was handy in the last supplemental as the point of expansion’s name. So let me make that unmistakable:

$a \equiv r = \left(\frac{L^2}{nCm}\right)^{\frac{1}{n + 2}}$

The $\equiv$ there means “defined to be equal to”. You might ask how this is different from “equals”. It seems like more emphasis to me. Also, there are other names for the circular orbit’s radius that I could have used. ‘r_e‘ would be good enough, as the subscript would suggest “radius of equilibrium”. Or ‘r₀‘ would be another popular choice, the 0 suggesting that this is something of key, central importance and also looking kind of like a circle. (That’s probably coincidence.) I like the ‘a’ better there because I know how easy it is to drop a subscript. If you’re working on a problem for yourself that’s easy to fix, with enough cursing and redoing your notes. On a board in front of class it’s even easier to fix since someone will ask about the lost subscript within three lines. In a post like this? It would be a mess.

So now I’m going to look at possible values of the radius ‘r’ that are close to ‘a’. How close? Close enough that ‘V_eff‘, the effective potential energy, looks like a parabola. If it doesn’t look much like a parabola then I look at values of ‘r’ that are even closer to ‘a’. (Do you see how the game is played? If you don’t, look closer. Yes, this is actually valid.) If ‘r’ is that close to ‘a’, then we can get away with this polynomial expansion:

$V_{eff}(r) \approx V_{eff}(a) + m\cdot(r - a) + \frac{1}{2} m_2 (r - a)^2$

where

$m = \frac{dV_{eff}}{dr}\left(a\right) \\ m_2 = \frac{d^2V_{eff}}{dr^2}\left(a\right)$

The “approximate” there is because this is an approximation. $V_{eff}(r)$ is in truth equal to the thing on the right-hand-side there plus something that isn’t (usually) zero, but that is small.

I am sorry beyond my ability to describe that I didn’t make that ‘m’ and ‘m₂‘ consistent last week. That’s all right. One of these is going to disappear right away.

Now, what $V_{eff}(a)$ is? Well, that’s whatever you get from putting in ‘a’ wherever you start out seeing ‘r’ in the expression for $V_{eff}(r)$ . I’m not going to bother with that. Call it math, fine, but that’s just a search-and-replace on the character ‘r’. Also, where I’m going next, it’s going to disappear, never to be seen again, so who cares? What’s important is that this is a constant number. If ‘r’ changes, the value of $V_{eff}(a)$ does not, because ‘r’ doesn’t appear anywhere in $V_{eff}(a)$ .

How about ‘m’? That’s the value of the first derivative of ‘V_eff‘ with respect to ‘r’, evaluated when ‘r’ is equal to ‘a’. That might be something. It’s not, because of what ‘a’ is. It’s the value of ‘r’ which would make $\frac{dV_{eff}}{dr}(r)$ equal to zero. That’s why ‘a’ has that value instead of some other, any other.

So we’ll have a constant part ‘V_eff(a)’, plus a zero part, plus a part that’s a parabola. This is normal, by the way, when we do expansions around an equilibrium. At least it’s common. Good to see it. To find ‘m₂‘ we have to take the second derivative of ‘V_eff(r)’ and then evaluate it when ‘r’ is equal to ‘a’ and ugh but here it is.

$\frac{d^2V_{eff}}{dr^2}(r) = n (n - 1) C r^{n - 2} + 3\cdot\frac{L^2}{m}r^{-4}$

And at the point of approximation, where ‘r’ is equal to ‘a’, it’ll be:

$m_2 = \frac{d^2V_{eff}}{dr^2}(a) = n (n - 1) C a^{n - 2} + 3\cdot\frac{L^2}{m}a^{-4}$

We know exactly what ‘a’ is so we could write that out in a nice big expression. You don’t want to. I don’t want to. It’s a bit of a mess. I mean, it’s not hard, but it has a lot of symbols in it and oh all right. Here. Look fast because I’m going to get rid of that as soon as I can.

$m_2 = \frac{d^2V_{eff}}{dr^2}(a) = n (n - 1) C \left(\frac{L^2}{n C m}\right)^{n - 2} + 3\cdot\frac{L^2}{m}\left(\frac{L^2}{n C m}\right)^{-4}$

For the values of ‘n’ that we actually care about because they turn up in real actual physics problems this expression simplifies some. Enough, anyway. If we pretend we know nothing about ‘n’ besides that it is a number bigger than -2 then … ugh. We don’t have a lot that can clean it up.

Here’s how. I’m going to define an auxiliary little function. Its role is to contain our symbolic sprawl. It has a legitimate role too, though. At least it represents something that it makes sense to give a name. It will be a new function, named ‘F’ and that depends on the radius ‘r’:

$F(r) \equiv -\frac{dV}{dr}$

Notice that’s the derivative of the original ‘V’, not the angular-momentum-equipped ‘V_eff‘. This is the secret of its power. It doesn’t do anything to make $V_{eff}(r)$ easier to work with. It starts being good when we take its derivatives, though:

$\frac{dV_{eff}}{dr} = -F(r) - \frac{L^2}{m}r^{-3}$

That already looks nicer, doesn’t it? It’s going to be really slick when you think about what ‘F(a)’ is. Remember that ‘a’ is the value for ‘r’ which makes the derivative of ‘V_eff‘ equal to zero. So … I may not know much, but I know this:

$0 = \frac{dV_{eff}}{dr}(a) = -F(a) - \frac{L^2}{m}a^{-3} \\ F(a) = -\frac{L}{ma^3}$

I’m not going to say what value F(r) has for other values of ‘r’ because I don’t care. But now look at what it does for the second derivative of ‘V_eff‘:

$\frac{d^2 V_{eff}}{dr^2}(r) = -F'(r) + 3\frac{L^2}{mr^4}$

Here the ‘F'(r)’ is a shorthand way of writing ‘the derivative of F with respect to r’. You can do when there’s only the one free variable to consider. And now something magic that happens when we look at the second derivative of ‘V_eff‘ when ‘r’ is equal to ‘a’ …

$\frac{d^2 V_{eff}}{dr^2}(a) = -F'(a) - \frac{3}{a} F(a)$

We get away with this because we happen to know that ‘F(a)’ is equal to $-\frac{L}{ma^3}$ and doesn’t that work out great? We’ve turned a symbolic mess into a … less symbolic mess.

Now why do I say it’s legitimate to introduce ‘F(r)’ here? It’s because minus the derivative of the potential energy with respect to the position of something can be something of actual physical interest. It’s the amount of force exerted on the particle by that potential energy at that point. The amount of force on a thing is something that we could imagine being interested in. Indeed, we’d have used that except potential energy is usually so much easier to work with. I’ve avoided it up to this point because it wasn’t giving me anything I needed. Here, I embrace it because it will save me from some awful lines of symbols.

Because with this expression in place I can write the approximation to the potential energy of:

$V_{eff}(r) \approx V_{eff}(a) + \frac{1}{2} \left( -F'(a) - \frac{3}{a}F(a) \right) (r - a)^2$

So if ‘r’ is close to ‘a’, then the polynomial on the right is a good enough approximation to the effective potential energy. And that potential energy has the shape of a spring’s potential energy. We can use what we know about springs to describe its motion. Particularly, we’ll have this be true:

$\frac{dp}{dt} = -\frac{dv_{eff}}{dr}(r) = -\left( F'(a) + \frac{3}{a} F(a)\right) r$

Here, ‘p’ is the (linear) momentum of whatever’s orbiting, which we can treat as equal to ‘mr’, the mass of the orbiting thing times how far it is from the center. You may sense in me some reluctance about doing this, what with that ‘we can treat as equal to’ talk. There’s reasons for this and I’d have to get deep into geometry to explain why. I can get away with specifically this use because the problem allows it. If you’re trying to do your own original physics problem inspired by this thread, and it’s not orbits like this, be warned. This is a spot that could open up to a gigantic danger pit, lined at the bottom with sharp spikes and angry poison-clawed mathematical tigers and I bet it’s raining down there too.

So we can rewrite all this as

$m\frac{d^2r}{dt^2} = -\frac{dv_{eff}}{dr}(r) = -\left( F'(a) + \frac{3}{a} F(a)\right) r$

And when we learned everything interesting there was to know about springs we learned what the solutions to this look like. Oh, in that essay the variable that changed over time was called ‘x’ and here it’s called ‘r’, but that’s not an actual difference. ‘r’ will be some sinusoidal curve:

$r(t) = A cos\left(\sqrt{\frac{k}{m}} t\right) + B sin\left(\sqrt{\frac{k}{m}} t\right)$

where, here, ‘k’ is equal to that whole mass of constants on the right-hand side:

$k = -\left( F'(a) + \frac{3}{a} F(a)\right)$

I don’t know what ‘A’ and ‘B’ are. It’ll depend on just what the perturbation is like, how far the planet is from the circular orbit. But I can tell you what the behavior is like. The planet will wobble back and forth around the circular orbit, sometimes closer to the center, sometimes farther away. It’ll spend as much time closer to the center than the circular orbit as it does farther away. And the period of that oscillation will be

$T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{m}{-\left(F'(a) + \frac{3}{a}F(a)\right)}}$

This tells us something about what the orbit of a thing not in a circular orbit will be like. Yes, I see you in the back there, quivering with excitement about how we’ve got to elliptical orbits. You’re moving too fast. We haven’t got that. There will be elliptical orbits, yes, but only for a very particular power ‘n’ for the potential energy. Not for most of them. We’ll see.

It might strike you there’s something in that square root. We need to take the square root of a positive number, so maybe this will tell us something about what kinds of powers we’re allowed. It’s a good thought. It turns out not to tell us anything useful, though. Suppose we started with $V(r) = Cr^n$ . Then $F(r) = -nCr^{n - 1}$ , and $F'(r) = -n(n - 1)C^{n - 2}$ . Sad to say, this leads us to a journey which reveals that we need ‘n’ to be larger than -2 or else we don’t get oscillations around a circular orbit. We already knew that, though. We already found we needed it to have a stable equilibrium before. We can see there not being a period for these oscillations around the circular orbit as another expression of the circular orbit not being stable. Sad to say, we haven’t got something new out of this.

We will get to new stuff, though. Maybe even ellipses.

Why Stuff Can Orbit, Part 8: Introducing Stability

Previously:

And the supplemental reading:

I bet you imagined I’d forgot this series, or that I’d quietly dropped it. Not so. I’ve just been finding the energy for this again. 2017 has been an exhausting year.

With the last essay I finished the basic goal of “Why Stuff Can Orbit”. I’d described some of the basic stuff for central forces. These involve something — a planet, a mass on a spring, whatever — being pulled by the … center. Well, you can call anything the origin, the center of your coordinate system. Why put that anywhere but the place everything’s pulled towards? The key thing about a central force is it’s always in the direction of the center. It can be towards the center or away from the center, but it’s always going to be towards the center because the “away from” case is boring. (The thing gets pushed away from the center and goes far off, never to be seen again.) How strongly it’s pulled toward the center changes only with the distance from the center.

Since the force only changes with the distance between the thing and the center it’s easy to think this is a one-dimensional sort of problem. You only need the coordinate describing this distance. We call that ‘r’, because we end up finding orbits that are circles. Since the distance between the center of a circle and its edge is the radius, it would be a shame to use any other letter.

Forces are hard to work with. At least for a lot of stuff. We can represent central forces instead as potential energy. This is easier because potential energy doesn’t have any direction. It’s a lone number. When we can shift something complicated into one number chances are we’re doing well.

But we are describing something in space. Something in three-dimensional space, although it turns out we’ll only need two. We don’t care about stuff that plunges right into the center; that’s boring. We like stuff that loops around and around the center. Circular orbits. We’ve seen that second dimension in the angular momentum, which we represent as ‘L’ for reasons I dunno. I don’t think I’ve ever met anyone who did. Maybe it was the first letter that came to mind when someone influential wrote a good textbook. Angular momentum is a vector, but for these problems we don’t need to care about that. We can use an ordinary number to carry all the information we need about it.

We get that information from the potential energy plus a term that’s based on the square of the angular momentum divided by the square of the radius. This “effective potential energy” lets us find whether there can be a circular orbit at all, and where it’ll be. And it lets us get some other nice stuff like how the size of the orbit and the time it takes to complete an orbit relate to each other. See the earlier stuff for details. In short, though, we get an equilibrium, a circular orbit, whenever the effective potential energy is flat, neither rising nor falling. That happens when the effective potential energy changes from rising to falling, or changes from falling to rising. Well, if it isn’t rising and if it isn’t falling, what else can it be doing? It only does this for an infinitesimal moment, but that’s all we need. It also happens when the effective potential energy is flat for a while, but that like never happens.

Where I want to go next is into closed orbits. That is, as the planet orbits a sun (or whatever it is goes around whatever it’s going around), does it come back around to exactly where it started? Moving with the same speed in the same direction? That is, does the thing orbit like a planet does?

(Planets don’t orbit like this. When you have three, or more, things in the universe the mathematics of orbits gets way too complicated to do exactly. But this is the thing they’re approximating, we hope, well.)

To get there I’ll have to put back a second dimension. Sorry. Won’t need a third, though. That’ll get named θ because that’s our first choice for an angle. And it makes too much sense to describe a planet’s position as its distance from the center and the angle it makes with respect to some reference line. Which reference line? Whatever works for you. It’s like measuring longitude. We could measure degrees east and west of some point other than Greenwich as well, and as correctly, as we do. We use the one we use because it was convenient.

Along the way to closed orbits I have to talk about stability. There are many kinds of mathematical stability. My favorite is called Lyapunov Stability, because it’s such a mellifluous sound. They all circle around the same concept. It’s what you’d imagine from how we use the word in English. Start with an equilibrium, a system that isn’t changing. Give it a nudge. This disrupts it in some way. Does the disruption stay bounded? That is, does the thing still look somewhat like it did before? Or does the disruption grow so crazy big we have no idea what it’ll ever look like again? (A small nudge, by the way. You can break anything with a big enough nudge; that’s not interesting. It’s whether you can break it with a small nudge that we’d like to know.)

One of the ways we can study this is by looking at the effective potential energy. By its shape we can say whether a central-force equilibrium is stable or not. It’s easy, too, as we’ve got this set up. (Warning before you go passing yourself off as a mathematical physicist: it is not always easy!) Look at the effective potential energy versus the radius. If it has a part that looks like a bowl, cupped upward, it’s got a stable equilibrium. If it doesn’t, it doesn’t have a stable equilibrium. If you aren’t sure, imagine the potential energy was a track, like for a toy car. And imagine you dropped a marble on it. If you give the marble a nudge, does it roll to a stop? If it does, stable. If it doesn’t, unstable.

The sort of wiggly shape that serves as every mathematical physicist's generic potential energy curve to show off the different kinds of equilibrium. — A phony effective potential energy. Most are a lot less exciting than this; see some of the earlier pieces in this series. But some weird-shaped functions like this were toyed with by physicists in the 19th century who were hoping to understand chemistry. Why should gases behave differently at different temperatures? Why should some combinations of elements make new compounds while others don’t? We needed statistical mechanics and quantum mechanics to explain those, but we couldn’t get there without a lot of attempts and failures at explaining it with potential energies and classical mechanics.

Stable is more interesting. We look at cases where there is this little bowl cupped upward. If we have a tiny nudge we only have to look at a small part of that cup. And that cup is going to look an awful lot like a parabola. If you don’t remember what a parabola is, think back to algebra class. Remember that curvey shape that was the only thing drawn on the board when you were dealing with the quadratic formula? That shape is a parabola.

Who cares about parabolas? We care because we know something good about them. In this context, anyway. The potential energy for a mass on a spring is also a parabola. And we know everything there is to know about masses on springs. Seriously. You’d think it was all physics was about from like 1678 through 1859. That’s because it’s something calculus lets us solve exactly. We don’t need books of complicated integrals or computers to do the work for us.

So here’s what we do. It’s something I did not get clearly when I was first introduced to these concepts. This left me badly confused and feeling lost in my first physics and differential equations courses. We are taking our original physics problem and building a new problem based on it. This new problem looks at how big our nudge away from the equilibrium is. How big the nudge is, how fast it grows, how it changes in time will follow rules. Those rules will look a lot like those for a mass on a spring. We started out with a radius that gives us a perfectly circular orbit. Now we get a secondary problem about how the difference between the nudged and the circular orbit changes in time.

That secondary problem has the same shape, the same equations, as a mass on a spring does. A mass on a spring is a central force problem. All the tools we had for studying central-force problems are still available. There is a new central-force problem, hidden within our original one. Here the “center” is the equilibrium we’re nudged around. It will let us answer a new set of questions.

Why Stuff Can Orbit, Part 7: ALL the Circles

Previously:

And some supplemental reading:

Last time around I showed how to do a central-force problem for normal gravity. That’s one where a planet, or moon, or satellite, or whatever is drawn towards the center of space. It’s drawn by a potential energy that equals some constant times the inverse of the distance from the origin. That is, V(r) = C r^-1. With a little bit of fussing around we could find out what distance from the center lets a circular orbit happen. And even Kepler’s Third Law, connecting how long an orbit takes to how big it must be.

There are two natural follow-up essays. One is to work out elliptical orbits. We know there are such things; all real planets and moons have them, and nearly all satellites do. The other is to work out circular orbits for another easy-to-understand example, like a mass on a spring. That’s something with a potential energy that looks like V(r) = C r².

I want to do the elliptical orbits later on. The mass-on-a-spring I could do now. So could you, if you look follow last week’s essay and just change the numbers a little. But, you know, why bother working out one problem? Why not work out a lot of them? Why not work out every central-force problem, all at once?

Because we can’t. I mean, I can describe how to do that, but it isn’t going to save us much time. Like, the quadratic formula is great because it’ll give you the roots of a quadratic polynomial in one step. You don’t have to do anything but a little arithmetic. We can’t get a formula that easy if we try to solve for every possible potential energy.

But we can work out a lot of central-force potential energies all at once. That is, we can solve for a big set of similar problems, a “family” as we call them. The obvious family is potential energies that are powers of the planet’s distance from the center. That is, they’re potential energies that follow the rule

$V(r) = C r^n$

Here ‘C’ is some number. It might depend on the planet’s mass, or the sun’s mass. Doesn’t matter. All that’s important is that it not change over the course of the problem. So, ‘C’ for Constant. And ‘n’ is another constant number. Some numbers turn up a lot in useful problems. If ‘n’ is -1 then this can describe gravitational attraction. If ‘n’ is 2 then this can describe a mass on a spring. This ‘n’ can be any real number. That’s not an ideal choice of letter. ‘n’ usually designates a whole number. By using that letter I’m biasing people to think of numbers like ‘2’ at the expense of perfectly legitimate alternatives such as ‘2.1’. But now that I’ve made that explicit maybe we won’t make a casual mistake.

So what I want is to find where there are stable circular orbits for an arbitrary radius-to-a-power force. I don’t know what ‘C’ and ‘n’ are, but they’re some numbers. To find where a planet can have a circular orbit I need to suppose the planet has some mass, ‘m’. And that its orbit has some angular momentum, a number called ‘L’. From this we get the effective potential energy. That’s what the potential energy looks like when we remember that angular momentum has to be conserved.

$V_{eff}(r) = C r^n + \frac{L^2}{2m} r^{-2}$

To find where a circular orbit can be we have to take the first derivative of V_eff with respect to ‘r’. The circular orbit can happen at a radius for which this first derivative equals zero. So we need to solve this:

$\frac{dV_{eff}}{dr} = n C r^{n-1} - 2\frac{L^2}{2m} r^{-3} = 0$

That derivative we know from the rules of how to take derivatives. And from this point on we have to do arithmetic. We want to get something which looks like ‘r = (some mathematics stuff here)’. Hopefully it’ll be something not too complicated. And hey, in the second term there, the one with L² in it, we have a 2 in the numerator and a 2 in the denominator. So those cancel out and that’s simpler. That’s hopeful, isn’t it?

$n C r^{n-1} - \frac{L^2}{m}r^{-3} = 0$

OK. Add $\frac{L^2}{m}r^{-3}$ to both sides of the equation; we’re used to doing that. At least in high school algebra we are.

$n C r^{n-1} = \frac{L^2}{m}r^{-3}$

Not looking much better? Try multiplying both left and right sides by ‘r³‘. This gets rid of all the ‘r’ terms on the right-hand side of the equation.

$n C r^{n+2} = \frac{L^2}{m}$

Now we’re getting close to the ideal of ‘r = (some mathematics stuff)’. Divide both sides by the constant number ‘n times C’.

$r^{n+2} = \frac{L^2}{n C m}$

I know how much everybody likes taking (n+2)-nd roots of a quantity. I’m sure you occasionally just pick an object at random — your age, your telephone number, a potato, a wooden block — and find its (n+2)-nd root. I know. I’ll spoil some of the upcoming paragraphs to say that it’s going to be more useful knowing ‘r^{n + 2}‘ than it is knowing ‘r’. But I’d like to have the radius of a circular orbit on the record. Here it is.

$r = \left(\frac{L^2}{n C m}\right)^{\frac{1}{n + 2}}$

Can we check that this is right? Well, we can at least check that things aren’t wrong. We can check against the example we already know. That’s the gravitational potential energy problem. For that one, ‘C’ is the number ‘G M m’. That’s the gravitational constant of the universe times the mass of the sun times the mass of the planet. And for gravitational potential energy, ‘n’ is equal to -1. This implies that, for a gravitational potential energy problem, we get a circular orbit when

$r_{grav} = \left(\frac{L^2}{n G M m^2}\right)^{\frac{1}{1}}$

I’m labelling it ‘r_grav‘ to point out it’s the radius of a circular orbit for gravitational problems. Might or might not need that in the future, but the label won’t hurt anything.

Go ahead and guess whether that agrees with last week’s work. I’m feeling confident.

OK, so, we know where a circular orbit might turn up for an arbitrary power function potential energy. Is it stable? We know from the third “Why Stuff Can Orbit” essay that it’s not a sure thing. We can have potential energies that don’t have any circular orbits. So it must be possible there are unstable orbits.

Whether our circular orbit is stable demands we do the same work we did last time. It will look a little harder to start, because there’s one more variable in it. What had been ‘-1’ last time is now an ‘n’, and stuff like ‘-2’ becomes ‘n-1’. Is that actually harder? Really?

So here’s the second derivative of the effective potential:

$\frac{d^2V_{eff}}{dr^2} = (n-1)nCr^{n - 2} + 3\frac{L^2}{m}r^{-4}$

My first impulse when I worked this out was to take the ‘r’ for a circular orbit, the thing worked out five paragraphs above, and plug it in to that expression. This is madness. Don’t do it. Or, you know, go ahead and start doing it and see how long it takes before you regret the errors of your ways.

The non-madness-inducing way to work out if this is a positive number? It involves noticing $r^{n-2}$ is the same number as $r^{n+2}\cdot r^{-4}$ . So we have this bit of distribution-law magic:

$\frac{d^2V_{eff}}{dr^2} = (n-1)nCr^{n + 2}r^{-4} + 3\frac{L^2}{m}r^{-4}$

$\frac{d^2V_{eff}}{dr^2} = \left((n-1)nCr^{n + 2} + 3\frac{L^2}{m}\right) \cdot r^{-4}$

I’m sure we all agree that’s better, right? No, honestly, let me tell you why this is better. When will this expression be true?

$\left((n-1)nCr^{n + 2} + 3\frac{L^2}{m}\right) \cdot r^{-4} > 0$

That’s the product of two expressions. One of them is ‘r^-4‘. ‘r’ is the radius of the planet’s orbit. That has to be a positive number. It’s how far the planet is from the origin. The number can’t be anything but positive. So we don’t have to worry about that.

SPOILER: I just palmed a card there. Did you see me palm a card there? Because I totally did. Watch for where that card turns up. It’ll be after this next bit.

So let’s look at the non-card-palmed part of this. We’re going to have a stable equilibrium when the other factor of that mess up above is positive. We need to know when this is true:

$(n-1)nCr^{n + 2} + 3\frac{L^2}{m} > 0$

OK. Well. We do know what ‘rⁿ⁺²‘ is. Worked that out … uhm … twelve(?) paragraphs ago. I’ll say twelve and hope I don’t mess that up in editing. Anyway, what’s important is $r^{n+2} = \frac{L^2}{n C m}$ . So we put that in where ‘rⁿ⁺²‘ appeared in that above expression.

$(n-1)nC\frac{L^2}{n C m} + 3 \frac{L^2}{m} > 0$

This is going to simplify down some. Look at that first term, with an ‘n C’ in the numerator and again in the denominator. We’re going to be happier soon as we cancel those out.

$(n-1)\frac{L^2}{m} + 3\frac{L^2}{m} > 0$

And now we get to some fine distributive-law action, the kind everyone likes:

$\left( (n-1) + 3 \right)\frac{L^2}{m} > 0$

Well, we know $\frac{L^2}{m}$ has to be positive. The angular momentum ‘L’ might be positive or might be negative but its square is certainly positive. The mass ‘m’ has to be a positive number. So we’ll get a stable equilibrium whenever $(n - 1) + 3$ is greater than 0. That is, whenever $n > -2$ . Done.

No we’re not done. That’s nonsense. We knew that going in. We saw that a couple essays ago. If your potential energy were something like, say, $V(r) = -2 r^3$ you wouldn’t have any orbits at all, never mind stable orbits. But 3 is certainly greater than -2. So what’s gone wrong here?

Let’s go back to that palmed card. Remember I mentioned how the radius of our circular orbit was a positive number. This has to be true, if there is a circular orbit. What if there isn’t one? Do we know there is a radius ‘r’ that the planet can orbit the origin? Here’s the formula giving us that circular orbit’s radius once again:

$r = \left(\frac{L^2}{n C m}\right)^{\frac{1}{n + 2}}$

Do we know that’s going to exist? … Well, sure. That’s going to be some meaningful number as long as we avoid obvious problems. Like, we can’t have the power ‘n’ be equal to zero, because dividing by zero is all sorts of bad. Also we can’t have the constant ‘C’ be zero, again because dividing by zero is bad.

Not a problem, though. If either ‘C’ or ‘n’ were zero, or if both were, then the original potential energy would be a constant number. V(r) would be equal to ‘C’ (if ‘n’ were zero), or ‘0’ (if ‘C’ were zero). It wouldn’t change with the radius ‘r’. This is a case called the ‘free particle’. There’s no force pushing the planet in one direction or another. So if the planet were not moving it would never start. If the planet were already moving, it would keep moving in the same direction in a straight line. No circular orbits.

Similarly if ‘n’ were equal to ‘-2’ there’d be problems because the power we raise that parenthetical expression to would be equal to one divided by zero, which is bad. Is there anything else that could be trouble there?

What if the thing inside parentheses is a negative number? I may not know what ‘n’ is. I don’t. We started off by supposing we didn’t know beyond that it was a number. But I do know that the n-th root of a negative number is going to be trouble. It might be negative. It might be complex-valued. But it won’t be a positive number. And we need a radius that’s a positive number. So that’s the palmed card. To have a circular orbit at all, positive or negative, we have to have:

$\frac{L^2}{n C m} > 0$

‘L’ is a regular old number, maybe positive, maybe negative. So ‘L²‘ is a positive number. And the mass ‘m’ is a positive number. We don’t know what ‘n’ and C’ are. But as long as their product is positive we’re good. The whole equation will be true. So ‘n’ and ‘C’ can both be negative numbers. We saw that with gravity: $V(r) = -\frac{GMm}{r}$ . ‘G’ is the gravitational constant of the universe, a positive number. ‘M’ and ‘m’ are masses, also positive.

Or ‘n’ and ‘C’ can both be positive numbers. That turns up with spring problems: $V(r) = K r^2$ , where ‘K’ is the ‘spring constant’. That’s some positive number again.

That time we found potential energies that didn’t have orbits? They were ones that had a positive ‘C’ and negative ‘n’, or a negative ‘C’ and positive ‘n’. The case we just worked out doesn’t have circular orbits. It’s nice to have that sorted out at least.

So what does it mean that we can’t have a stable orbit if ‘n’ is less than or equal to -2? Even if ‘C’ is negative? It turns out that if you have a negative ‘C’ and big negative ‘n’, like say -5, the potential energy drops way down to something infinitely large and negative at smaller and smaller radiuses. If you have a positive ‘C’, the potential energy goes way up at smaller and smaller radiuses. For large radiuses the potential drops to zero. But there’s never the little U-shaped hill in the middle, the way you get for gravity-like potentials or spring potentials or normal stuff like that. Yeah, who would have guessed?

What if we do have a stable orbit? How long does an orbit take? How does that relate to the radius of the orbit? We used this radius expression to work out Kepler’s Third Law for the gravity problem last week. We can do that again here.

Last week we worked out what the angular momentum ‘L’ had to be in terms of the radius of the orbit and the time it takes to complete one orbit. The radius of the orbit we called ‘r’. The time an orbit takes we call ‘T’. The formula for angular momentum doesn’t depend on what problem we’re doing. It just depends on the mass ‘m’ of what’s spinning around and how it’s spinning. So:

$L = 2\pi m \frac{r^2}{T}$

And from this we know what ‘L²‘ is.

$L^2 = 4\pi^2 m^2 \frac{r^4}{T^2}$

That’s convenient because we have an ‘L²‘ term in the formula for what the radius is. I’m going to stick with the formula we got for ‘rⁿ⁺²‘ because that is so, so much easier to work with than ‘r’ by itself. So we go back to that starting point and then substitute what we know ‘L²‘ to be in there.

$r^{n + 2} = \frac{L^2}{n C m}$

This we rewrite as:

$r^{n + 2} = \frac{4 \pi^2 m^2}{n C m}\frac{r^4}{T^2}$

Some stuff starts cancelling out again. One ‘m’ in the numerator and one in the denominator. Small thing but it makes our lives a bit better. We can multiply the left side and the right side by T². That’s more obviously an improvement. We can divide the left side and the right side by ‘r^{n + 2}‘. And yes that is too an improvement. Watch all this:

$r^{n + 2} = \frac{4 \pi^2 m}{n C}\frac{r^4}{T^2}$

$T^2 \cdot r^{n + 2} = \frac{4 \pi^2 m}{n C}r^4$

$T^2 = \frac{4 \pi^2 m}{n C}r^{2 - n}$

And that last bit is the equivalent of Kepler’s Third Law for our arbitrary power-law style force.

Are we right? Hard to say offhand. We can check that we aren’t wrong, at least. We can check against the gravitational potential energy. For this ‘n’ is equal to -1. ‘C’ is equal to ‘-G M m’. Make those substitutions; what do we get?

$T^2 = \frac{4 \pi^2 m}{(-1) (-G M m)}r^{2 - (-1)}$

$T^2 = \frac{4 \pi^2}{G M}r^{3}$

Well, that is what we expected for this case. So the work looks good, this far. Comforting.

Why Stuff Can Orbit, Part 6: Circles and Where To Find Them

Previously:

And some supplemental reading:

So now we can work out orbits. At least orbits for a central force problem. Those are ones where a particle — it’s easy to think of it as a planet — is pulled towards the center of the universe. How strong that pull is depends on some constants. But it only changes as the distance the planet is from the center changes.

What we’d like to know is whether there are circular orbits. By “we” I mean “mathematical physicists”. And I’m including you in that “we”. If you’re reading this far you’re at least interested in knowing how mathematical physicists think about stuff like this.

It’s easiest describing when these circular orbits exist if we start with the potential energy. That’s a function named ‘V’. We write it as ‘V(r)’ to show it’s an energy that changes as ‘r’ changes. By ‘r’ we mean the distance from the center of the universe. We’d use ‘d’ for that except we’re so used to thinking of distance from the center as ‘radius’. So ‘r’ seems more compelling. Sorry.

Besides the potential energy we need to know the angular momentum of the planet (or whatever it is) moving around the center. The amount of angular momentum is a number we call ‘L’. It might be positive, it might be negative. Also we need the planet’s mass, which we call ‘m’. The angular momentum and mass let us write a function called the effective potential energy, ‘V_eff(r)’.

And we’ll need to take derivatives of ‘V_eff(r)’. Fortunately that “How Differential Calculus Works” essay explains all the symbol-manipulation we need to get started. That part is calculus, but the easy part. We can just follow the rules already there. So here’s what we do:

The planet (or whatever) can have a circular orbit around the center at any radius which makes the equation $\frac{dV_{eff}}{dr} = 0$ true.
The circular orbit will be stable if the radius of its orbit makes the second derivative of the effective potential, $\frac{d^2V_{eff}}{dr^2}$ , some number greater than zero.

We’re interested in stable orbits because usually unstable orbits are boring. They might exist but any little perturbation breaks them down. The mathematician, ordinarily, sees this as a useless solution except in how it describes different kinds of orbits. The physicist might point out that sometimes it can take a long time, possibly millions of years, before the perturbation becomes big enough to stand out. Indeed, it’s an open question whether our solar system is stable. While it seems to have gone millions of years without any planet changing its orbit very much we haven’t got the evidence to say it’s impossible that, say, Saturn will be kicked out of the solar system anytime soon. Or worse, that Earth might be. “Soon” here means geologically soon, like, in the next million years.

(If it takes so long for the instability to matter then the mathematician might allow that as “metastable”. There are a lot of interesting metastable systems. But right now, I don’t care.)

I realize now I didn’t explain the notation for the second derivative before. It looks funny because that’s just the best we can work out. In that fraction $\frac{d^2V_{eff}}{dr^2}$ the ‘d’ isn’t a number so we can’t cancel it out. And the superscript ‘2’ doesn’t mean squaring, at least not the way we square numbers. There’s a functional analysis essay in there somewhere. Again I’m sorry about this but there’s a lot of things mathematicians want to write out and sometimes we can’t find a way that avoids all confusion. Roll with it.

So that explains the whole thing clearly and easily and now nobody could be confused and yeah I know. If my Classical Mechanics professor left it at that we’d have open rebellion. Let’s do an example.

There are two and a half good examples. That is, they’re central force problems with answers we know. One is gravitation: we have a planet orbiting a star that’s at the origin. Another is springs: we have a mass that’s connected by a spring to the origin. And the half is electric: put a positive electric charge at the center and have a negative charge orbit that. The electric case is only half a problem because it’s the same as the gravitation problem except for what the constants involved are. Electric charges attract each other crazy way stronger than gravitational masses do. But that doesn’t change the work we do.

This is a lie. Electric charges accelerating, and just orbiting counts as accelerating, cause electromagnetic effects to happen. They give off light. That’s important, but it’s also complicated. I’m not going to deal with that.

I’m going to do the gravitation problem. After all, we know the answer! By Kepler’s something law, something something radius cubed something G M … something … squared … After all, we can look up the answer!

The potential energy for a planet orbiting a sun looks like this:

$V(r) = - G M m \frac{1}{r}$

Here ‘G’ is a constant, called the Gravitational Constant. It’s how strong gravity in the universe is. It’s not very strong. ‘M’ is the mass of the sun. ‘m’ is the mass of the planet. To make sense ‘M’ should be a lot bigger than ‘m’. ‘r’ is how far the planet is from the sun. And yes, that’s one-over-r, not one-over-r-squared. This is the potential energy of the planet being at a given distance from the sun. One-over-r-squared gives us how strong the force attracting the planet towards the sun is. Different thing. Related thing, but different thing. Just listing all these quantities one after the other means ‘multiply them together’, because mathematicians multiply things together a lot and get bored writing multiplication symbols all the time.

Now for the effective potential we need to toss in the angular momentum. That’s ‘L’. The effective potential energy will be:

$V_{eff}(r) = - G M m \frac{1}{r} + \frac{L^2}{2 m r^2}$

I’m going to rewrite this in a way that means the same thing, but that makes it easier to take derivatives. At least easier to me. You’re on your own. But here’s what looks easier to me:

$V_{eff}(r) = - G M m r^{-1} + \frac{L^2}{2 m} r^{-2}$

I like this because it makes every term here look like “some constant number times r to a power”. That’s easy to take the derivative of. Check back on that “How Differential Calculus Works” essay. The first derivative of this ‘V_eff(r)’, taken with respect to ‘r’, looks like this:

$\frac{dV_{eff}}{dr} = -(-1) G M m r^{-2} -2\frac{L^2}{2m} r^{-3}$

We can tidy that up a little bit: -(-1) is another way of writing 1. The second term has two times something divided by 2. We don’t need to be that complicated. In fact, when I worked out my notes I went directly to this simpler form, because I wasn’t going to be thrown by that. I imagine I’ve got people reading along here who are watching these equations warily, if at all. They’re ready to bolt at the first sign of something terrible-looking. There’s nothing terrible-looking coming up. All we’re doing from this point on is really arithmetic. It’s multiplying or adding or otherwise moving around numbers to make the equation prettier. It happens we only know those numbers by cryptic names like ‘G’ or ‘L’ or ‘M’. You can go ahead and pretend they’re ‘4’ or ‘5’ or ‘7’ if you like. You know how to do the steps coming up.

So! We allegedly can have a circular orbit when this first derivative is equal to zero. What values of ‘r’ make true this equation?

$G M m r^{-2} - \frac{L^2}{m} r^{-3} = 0$

Not so helpful there. What we want is to have something like ‘r = (mathematics stuff here)’. We have to do some high school algebra moving-stuff-around to get that. So one thing we can do to get closer is add the quantity $\frac{L^2}{m} r^{-3}$ to both sides of this equation. This gets us:

$G M m r^{-2} = \frac{L^2}{m} r^{-3}$

Things are getting better. Now multiply both sides by the same number. Which number? r³. That’s because ‘r^-3‘ times ‘r³‘ is going to equal 1, while ‘r^-2‘ times ‘r³‘ will equal ‘r¹‘, which normal people call ‘r’. I kid; normal people don’t think of such a thing at all, much less call it anything. But if they did, they’d call it ‘r’. We’ve got:

$G M m r = \frac{L^2}{m}$

And now we’re getting there! Divide both sides by whatever number ‘G M’ is, as long as it isn’t zero. And then we have our circular orbit! It’s at the radius

$r = \frac{L^2}{G M m^2}$

Very good. I’d even say pretty. It’s got all those capital letters and one little lowercase. Something squared in the numerator and the denominator. Aesthetically pleasant. Stinks a little that it doesn’t look like anything we remember from Kepler’s Laws once we’ve looked them up. We can fix that, though.

The key is the angular momentum ‘L’ there. I haven’t said anything about how that number relates to anything. It’s just been some constant of the universe. In a sense that’s fair enough. Angular momentum is conserved, exactly the same way energy is conserved, or the way linear momentum is conserved. Why not just let it be whatever number it happens to be?

(A note for people who skipped earlier essays: Angular momentum is not a number. It’s really a three-dimensional vector. But in a central force problem with just one planet moving around we aren’t doing any harm by pretending it’s just a number. We set it up so that the angular momentum is pointing directly out of, or directly into, the sheet of paper we pretend the planet’s orbiting in. Since we know the direction before we even start work, all we have to care about is the size. That’s the number I’m talking about.)

The angular momentum of a thing is its moment of inertia times its angular velocity. I’m glad to have cleared that up for you. The moment of inertia of a thing describes how easy it is to start it spinning, or stop it spinning, or change its spin. It’s a lot like inertia. What it is depends on the mass of the thing spinning, and how that mass is distributed, and what it’s spinning around. It’s the first part of physics that makes the student really have to know volume integrals.

We don’t have to know volume integrals. A single point mass spinning at a constant speed at a constant distance from the origin is the easy angular momentum to figure out. A mass ‘m’ at a fixed distance ‘r’ from the center of rotation moving at constant speed ‘v’ has an angular momentum of ‘m’ times ‘r’ times ‘v’.

So great; we’ve turned ‘L’ which we didn’t know into ‘m r v’, where we know ‘m’ and ‘r’ but don’t know ‘v’. We’re making progress, I promise. The planet’s tracing out a circle in some amount of time. It’s a circle with radius ‘r’. So it traces out a circle with perimeter ‘2 π r’. And it takes some amount of time to do that. Call that time ‘T’. So its speed will be the distance travelled divided by the time it takes to travel. That’s $\frac{2 \pi r}{T}$ . Again we’ve changed one unknown number ‘L’ for another unknown number ‘T’. But at least ‘T’ is an easy familiar thing: it’s how long the orbit takes.

Let me show you how this helps. Start off with what ‘L’ is:

$L = m r v = m r \frac{2\pi r}{T} = 2\pi m \frac{r^2}{T}$

Now let’s put that into the equation I got eight paragraphs ago:

$r = \frac{L^2}{G M m^2}$

Remember that one? Now put what I just said ‘L’ was, in where ‘L’ shows up in that equation.

$r = \frac{\left(2\pi m \frac{r^2}{T}\right)^2}{G M m^2}$

I agree, this looks like a mess and possibly a disaster. It’s not so bad. Do some cleaning up on that numerator.

$r = \frac{4 \pi^2 m^2}{G M m^2} \frac{r^4}{T^2}$

That’s looking a lot better, isn’t it? We even have something we can divide out: the mass of the planet is just about to disappear. This sounds bizarre, but remember Kepler’s laws: the mass of the planet never figures into things. We may be on the right path yet.

$r = \frac{4 \pi^2}{G M} \frac{r^4}{T^2}$

OK. Now I’m going to multiply both sides by ‘T²‘ because that’ll get that out of the denominator. And I’ll divide both sides by ‘r’ so that I only have the radius of the circular orbit on one side of the equation. Here’s what we’ve got now:

$T^2 = \frac{4 \pi^2}{G M} r^3$

And hey! That looks really familiar. A circular orbit’s radius cubed is some multiple of the square of the orbit’s time. Yes. This looks right. At least it looks reasonable. Someone else can check if it’s right. I like the look of it.

So this is the process you’d use to start understanding orbits for your own arbitrary potential energy. You can find the equivalent of Kepler’s Third Law, the one connecting orbit times and orbit radiuses. And it isn’t really hard. You need to know enough calculus to differentiate one function, and then you need to be willing to do a pile of arithmetic on letters. It’s not actually hard. Next time I hope to talk about more and different … um …

I’d like to talk about the different … oh, dear. Yes. You’re going to ask about that, aren’t you?

Ugh. All right. I’ll do it.

How do we know this is a stable orbit? Well, it just is. If it weren’t the Earth wouldn’t have a Moon after all this. Heck, the Sun wouldn’t have an Earth. At least it wouldn’t have a Jupiter. If the solar system is unstable, Jupiter is probably the most stable part. But that isn’t convincing. I’ll do this right, though, and show what the second derivative tells us. It tells us this is too a stable orbit.

So. The thing we have to do is find the second derivative of the effective potential. This we do by taking the derivative of the first derivative. Then we have to evaluate this second derivative and see what value it has for the radius of our circular orbit. If that’s a positive number, then the orbit’s stable. If that’s a negative number, then the orbit’s not stable. This isn’t hard to do, but it isn’t going to look pretty.

First the pretty part, though. Here’s the first derivative of the effective potential:

$\frac{dV_{eff}}{dr} = G M m r^{-2} - \frac{L^2}{m} r^{-3}$

OK. So the derivative of this with respect to ‘r’ isn’t hard to evaluate again. This is again a function with a bunch of terms that are all a constant times r to a power. That’s the easiest sort of thing to differentiate that isn’t just something that never changes.

$\frac{d^2 V_{eff}}{dr^2} = -2 G M m r^{-3} - (-3)\frac{L^2}{m} r^{-4}$

Now the messy part. We need to work out what that line above is when our planet’s in our circular orbit. That circular orbit happens when $r = \frac{L^2}{G M m^2}$ . So we have to substitute that mess in for ‘r’ wherever it appears in that above equation and you’re going to love this. Are you ready? It’s:

$-2 G M m \left(\frac{L^2}{G M m^2}\right)^{-3} + 3\frac{L^2}{m}\left(\frac{L^2}{G M m^2}\right)^{-4}$

This will get a bit easier promptly. That’s because something raised to a negative power is the same as its reciprocal raised to the positive of that power. So that terrible, terrible expression is the same as this terrible, terrible expression:

$-2 G M m \left(\frac{G M m^2}{L^2}\right)^3 + 3 \frac{L^2}{m}\left(\frac{G M m^2}{L^2}\right)^4$

Yes, yes, I know. Only thing to do is start hacking through all this because I promise it’s going to get better. Putting all those third- and fourth-powers into their parentheses turns this mess into:

$-2 G M m \frac{G^3 M^3 m^6}{L^6} + 3 \frac{L^2}{m} \frac{G^4 M^4 m^8}{L^8}$

Yes, my gut reaction when I see multiple things raised to the eighth power is to say I don’t want any part of this either. Hold on another line, though. Things are going to start cancelling out and getting shorter. Group all those things-to-powers together:

$-2 \frac{G^4 M^4 m^7}{L^6} + 3 \frac{G^4 M^4 m^7}{L^6}$

Oh. Well, now this is different. The second derivative of the effective potential, at this point, is the number

$\frac{G^4 M^4 m^7}{L^6}$

And I admit I don’t know what number that is. But here’s what I do know: ‘G’ is a positive number. ‘M’ is a positive number. ‘m’ is a positive number. ‘L’ might be positive or might be negative, but ‘L⁶‘ is a positive number either way. So this is a bunch of positive numbers multiplied and divided together.

So this second derivative what ever it is must be a positive number. And so this circular orbit is stable. Give the planet a little nudge and that’s all right. It’ll stay near its orbit. I’m sorry to put you through that but some people raised the, honestly, fair question.

Why Stuff Can Orbit, Part 5: Why Physics Doesn’t Work And What To Do About It

Less way previously:

My title’s hyperbole, to the extent it isn’t clickbait. Of course physics works. By “work” I mean “model the physical world in useful ways”. If it didn’t work then we would call it “pure” mathematics instead. Mathematicians would study it for its beauty. Physicists would be left to fend for themselves. “Useful” I’ll say means “gives us something interesting to know”. “Interesting” I’ll say if you want to ask what that means then I think you’re stalling.

But what I mean is that Newtonian physics, the physics learned in high school, doesn’t work. Well, it works, in that if you set up a problem right and calculate right you get answers that are right. It’s just not efficient, for a lot of interesting problems. Don’t ask me about interesting again. I’ll just say the central-force problems from this series are interesting.

Newtonian, high school type, physics works fine. It shines when you have only a few things to keep track of. In this central force problem we have one object, a planet-or-something, that moves. And only one force, one that attracts the planet to or repels the planet from the center, the Origin. This is where we’d put the sun, in a planet-and-sun system. So that seems all right as far as things go.

It’s less good, though, if there’s constraints. If it’s not possible for the particle to move in any old direction, say. That doesn’t turn up here; we can imagine a planet heading in any direction relative to the sun. But it’s also less good if there’s a symmetry in what we’re studying. And in this case there is. The strength of the central force only changes based on how far the planet is from the origin. The direction only changes based on what direction the planet is relative to the origin. It’s a bit daft to bother with x’s and y’s and maybe even z’s when all we care about is the distance from the origin. That’s a number we’ve called ‘r’.

So this brings us to Lagrangian mechanics. This was developed in the 18th century by Joseph-Louis Lagrange. He’s another of those 18th century mathematicians-and-physicists with his name all over everything. Lagrangian mechanics are really, really good when there’s a couple variables that describe both what we’d like to observe about the system and its energy. That’s exactly what we have with central forces. Give me a central force, one that’s pointing directly toward or away from the origin, and that grows or shrinks as the radius changes. I can give you a potential energy function, V(r), that matches that force. Give me an angular momentum L for the planet to have, and I can give you an effective potential energy function, V_eff(r). And that effective potential energy lets us describe how the coordinates change in time.

The method looks roundabout. It depends on two things. One is the coordinate you’re interested in, in this case, r. The other is how fast that coordinate changes in time. This we have a couple of ways of denoting. When working stuff out on paper that’s often done by putting a little dot above the letter. If you’re typing, dots-above-the-symbol are hard. So we mark it as a prime instead: r’. This works well until the web browser or the word processor assumes we want smart quotes and we already had the r’ in quote marks. At that point all hope of meaning is lost and we return to communicating by beating rocks with sticks. We live in an imperfect world.

What we get out of this is a setup that tells us how fast r’, how fast the coordinate we’re interested in changes in time, itself changes in time. If the coordinate we’re interested in is the ordinary old position of something, then this describes the rate of change of the velocity. In ordinary English we call that the acceleration. What makes this worthwhile is that the coordinate doesn’t have to be the position. It also doesn’t have to be all the information we need to describe the position. For the central force problem r here is just how far the planet is from the center. That tells us something about its position, but not everything. We don’t care about anything except how far the planet is from the center, not yet. So it’s fine we have a setup that doesn’t tell us about the stuff we don’t care about.

How fast r’ changes in time will be proportional to how fast the effective potential energy, V_eff(r), changes with its coordinate. I so want to write “changes with position”, since these coordinates are usually the position. But they can be proxies for the position, or things only loosely related to the position. For an example that isn’t a central force, think about a spinning top. It spins, it wobbles, it might even dance across the table because don’t they all do that? The coordinates that most sensibly describe how it moves are about its rotation, though. What axes is it rotating around? How do those change in time? Those don’t have anything particular to do with where the top is. That’s all right. The mathematics works just fine.

A circular orbit is one where the radius doesn’t change in time. (I’ll look at non-circular orbits later on.) That is, the radius is not increasing and is not decreasing. If it isn’t getting bigger and it isn’t getting smaller, then it’s got to be staying the same. Not all higher mathematics is tricky. The radius of the orbit is the thing I’ve been calling r all this time. So this means that r’, how fast r is changing with time, has to be zero. Now a slightly tricky part.

How fast is r’, the rate at which r changes, changing? Well, r’ never changes. It’s always the same value. Anytime something is always the same value the rate of its change is zero. This sounds tricky. The tricky part is that it isn’t tricky. It’s coincidental that r’ is zero and the rate of change of r’ is zero, though. If r’ were any fixed, never-changing number, then the rate of change of r’ would be zero. It happens that we’re interested in times when r’ is zero.

So we’ll find circular orbits where the change in the effective potential energy, as r changes, is zero. There’s an easy-to-understand intuitive idea of where to find these points. Look at a plot of Veff and imagine this is a smooth track or the cross-section of a bowl or the landscaping of a hill. Imagine dropping a ball or a marble or a bearing or something small enough to roll in it. Where does it roll to a stop? That’s where the change is zero.

It’s too much bother to make a bowl or landscape a hill or whatnot for every problem we’re interested in. We might do it anyway. Mathematicians used to, to study problems that were too complicated to do by useful estimates. These were “analog computers”. They were big in the days before digital computers made it no big deal to simulate even complicated systems. We still need “analog computers” or models sometimes. That’s usually for problems that involve chaotic stuff like turbulent fluids. We call this stuff “wind tunnels” and the like. It’s all a matter of solving equations by building stuff.

We’re not working with problems that complicated. There isn’t the sort of chaos lurking in this problem that drives us to real-world stuff. We can find these equilibriums by working just with symbols instead.

Why Stuff Can Orbit, Part 4: On The L

Less way previously:

We were chatting about central forces. In these a small object — a satellite, a planet, a weight on a spring — is attracted to the center of the universe, called the origin. We’ve been studying this by looking at potential energy, a function that in this case depends only on how far the object is from the origin. But to find circular orbits, we can’t just look at the potential energy. We have to modify this potential energy to account for angular momentum. This essay I mean to discuss that angular momentum some.

Let me talk first about the potential energy. Mathematical physicists usually write this as a function named U or V. I’m using V. That’s what my professor used teaching this, back when I was an undergraduate several hundred thousand years ago. A central force, by definition, changes only with how far you are from the center. I’ve put the center at the origin, because I am not a madman. This lets me write the potential energy as V = V(r).

V(r) could, in principle, be anything. In practice, though, I am going to want it to be r raised to a power. That is, V(r) is equal to C rⁿ. The ‘C’ here is a constant. It’s a scaling constant. The bigger a number it is the stronger the central force. The closer the number is to zero the weaker the force is. In standard units, gravity has a constant incredibly close to zero. This makes orbits very big things, which generally works out well for planets. In the mathematics of masses on springs, the constant is closer to middling little numbers like 1.

The ‘n’ here is a deceiver. It’s a constant number, yes, and it can be anything we want. But the use of ‘n’ as a symbol has connotations. Usually when a mathematician or a physicist writes ‘n’ it’s because she needs a whole number. Usually a positive whole number. Sometimes it’s negative. But we have a legitimate central force if ‘n’ is any real number: 2, -1, one-half, the square root of π, any of that is good. If you just write ‘n’ without explanation, the reader will probably think “integers”, possibly “counting numbers”. So it’s worth making explicit when this isn’t so. It’s bad form to surprise the reader with what kind of number you’re even talking about.

(Some number of essays on we’ll find out that the only values ‘n’ can have that are worth anything are -1, 2, and 7. And 7 isn’t all that good. But we aren’t supposed to know that yet.)

C rⁿ isn’t the only kind of central force that could exist. Any function rule would do. But it’s enough. If we wanted a more complicated rule we could just add two, or three, or more potential energies together. This would give us $V(r) = C_1 r^{n_1} + C_2 r^{n_2}$ , with C₁ and C₂ two possibly different numbers, and n₁ and n₂ two definitely different numbers. (If n₁ and n₂ were the same number then we should just add C₁ and C₂ together and stop using a more complicated expression than we need.) Remember that Newton’s Law of Motion about the sum of multiple forces being something vector something something direction? When we look at forces as potential energy functions, that law turns into just adding potential energies together. They’re well-behaved that way.

And if we can add these r-to-a-power potential energies together then we’ve got everything we need. Why? Polynomials. We can approximate most any potential energy that would actually happen with a big enough polynomial. Or at least a polynomial-like function. These r-to-a-power forces are a basis set for all the potential energies we’re likely to care about. Understand how to work with one and you understand how to work with them all.

Well, one exception. The logarithmic potential, V(r) = C log(r), is really interesting. And it has real-world applicability. It describes how strongly two vortices, two whirlpools, attract each other. You can write the logarithm as a polynomial. But logarithms are pretty well-behaved functions. You might be better off just doing that as a special case.

Still, at least to start with, we’ll stick with V(r) = C rⁿ and you know what I mean by all those letters now. So I’m free to talk about angular momentum.

You’ve probably heard of momentum. It’s got something to do with movement, only sports teams and political campaigns are always gaining or losing it somehow. When we talk of that we’re talking of linear momentum. It describes how much mass is moving how fast in what direction. So it’s a vector, in three-dimensional space. Or two-dimensional space if you’re making the calculations easier. To find what the vector is, we make a list of every object that’s moving. We take its velocity — how fast it’s moving and in what direction — and multiply that by its mass. Mass is a single number, a scalar, and we’re always allowed to multiply a vector by a scalar. This gets us another vector. Once we’ve done that for everything that’s moving, we add all those product vectors together. We can always add vectors together. And this gives us a grand total vector, the linear momentum of the system.

And that’s conserved. If one part of the system starts moving slower it’s because other parts are moving faster, and vice-versa. In the real world momentum seems to evaporate. That’s because some of the stuff moving faster turns out to be air objects bumped into, or particles of the floor that get dragged along by friction, or other stuff we don’t care about. That momentum can seem to evaporate is what makes its use in talking about ports teams or political campaigns make sense. It also annoys people who want you to know they understand science words better than you. So please consider this my authorization to use “gaining” and “losing” momentum in this sense. Ignore complainers. They’re the people who complain the word “decimate” gets used to mean “destroy way more than ten percent of something”, even though that’s the least bad mutation of an English word’s meaning in three centuries.

Angular momentum is also a vector. It’s also conserved. We can calculate what that vector is by the same sort of process, that of calculating something on each object that’s spinning and adding it all up. In real applications it can seem to evaporate. But that’s also because the angular momentum is going into particles of air. Or it rubs off grease on the axle. Or it does other stuff we wish we didn’t have to deal with.

The calculation is a little harder to deal with. There’s three parts to a spinning thing. There’s the thing, and there’s how far it is from the axis it’s spinning around, and there’s how fast it’s spinning. So you need to know how fast it’s travelling in the direction perpendicular to the shortest line between the thing and the axis it’s spinning around. Its angular momentum is going to be as big as the mass times the distance from the axis times the perpendicular speed. It’s going to be pointing in whichever axis direction makes its movement counterclockwise. (Because that’s how physicists started working this out and it would be too much bother to change now.)

You might ask: wait, what about stuff like a wheel that’s spinning around its center? Or a ball being spun? That can’t be an angular momentum of zero? How do we work that out? The answer is: calculus. Also, we don’t need that. This central force problem I’ve framed so that we barely even need algebra for it.

See, we only have a single object that’s moving. That’s the planet or satellite or weight or whatever it is. It’s got some mass, the value of which we call ‘m’ because why make it any harder on ourselves. And it’s spinning around the origin. We’ve been using ‘r’ to mean the number describing how far it is from the origin. That’s the distance to the axis it’s spinning around. Its velocity — well, we don’t have any symbols to describe what that is yet. But you can imagine working that out. Or you trust that I have some clever mathematical-physics tool ready to introduce to work it out. I have, kind of. I’m going to ignore it altogether. For now.

The symbol we use for the total angular momentum in a system is $\vec{L}$ . The little arrow above the symbol is one way to denote “this is a vector”. It’s a good scheme, what with arrows making people think of vectors and it being easy to write on a whiteboard. In books, sometimes, we make do just by putting the letter in boldface, L, which is easier for old-fashioned word processors to do. If we’re sure that the reader isn’t going to forget that L is this vector then we might stop highlighting the fact altogether. That’s even less work to do.

It’s going to be less work yet. Central force problems like this mean the object can move only in a two-dimensional plane. (If it didn’t, it wouldn’t conserve angular momentum: the direction of $\vec{L}$ would have to change. Sounds like magic, but trust me.) The angular momentum’s direction has to be perpendicular to that plane. If the object is spinning around on a sheet of paper, the angular momentum is pointing straight outward from the sheet of paper. It’s pointing toward you if the object is moving counterclockwise. It’s pointing away from you if the object is moving clockwise. What direction it’s pointing is locked in.

All we need to know is how big this angular momentum vector is, and whether it’s positive or negative. So we just care about this number. We can call it ‘L’, no arrow, no boldface, no nothing. It’s just a number, the same as is the mass ‘m’ or distance from the origin ‘r’ or any of our other variables.

If ‘L’ is zero, this means there’s no total angular momentum. This means the object can be moving directly out from the origin, or directly in. This is the only way that something can crash into the center. So if setting L to be zero doesn’t allow that then we know we did something wrong, somewhere. If ‘L’ isn’t zero, then the object can’t crash into the center. If it did we’d be losing angular momentum. The object’s mass times its distance from the center times its perpendicular speed would have to be some non-zero number, even when the distance was zero. We know better than to look for that.

You maybe wonder why we use ‘L’ of all letters for the angular momentum. I do. I don’t know. I haven’t found any sources that say why this letter. Linear momentum, which we represent with $\vec{p}$ , I know. Or, well, I know the story every physicist says about it. p is the designated letter for linear momentum because we used to use the word “impetus”, as in “impulse”, to mean what we mean by momentum these days. And “p” is the first letter in “impetus” that isn’t needed for some more urgent purpose. (“m” is too good a fit for mass. “i” has to work both as an index and as that number which, squared, gives us -1. And for that matter, “e” we need for that exponentials stuff, and “t” is too good a fit for time.) That said, while everybody, everybody, repeats this, I don’t know the source. Perhaps it is true. I can imagine, say, Euler or Lagrange in their writing settling on “p” for momentum and everybody copying them. I just haven’t seen a primary citation showing this is so.

(I don’t mean to sound too unnecessarily suspicious. But just because everyone agrees on the impetus-thus-p story doesn’t mean it’s so. I mean, every Star Trek fan or space historian will tell you that the first space shuttle would have been named Constitution until the Trekkies wrote in and got it renamed Enterprise. But the actual primary documentation that the shuttle would have been named Constitution is weak to nonexistent. I’ve come to the conclusion NASA had no plan in mind to name space shuttles until the Trekkies wrote in and got one named. I’ve done less poking around the impetus-thus-p story, in that I’ve really done none, but I do want it on record that I would like more proof.)

Anyway, “p” for momentum is well-established. So I would guess that when mathematical physicists needed a symbol for angular momentum they looked for letters close to “p”. When you get into more advanced corners of physics “q” gets called on to be position a lot. (Momentum and position, it turns out, are nearly-identical-twins mathematically. So making their symbols p and q offers aesthetic charm. Also great danger if you make one little slip with the pen.) “r” is called on for “radius” a lot. Looking on, “t” is going to be time.

On the other side of the alphabet, well, “o” is just inviting danger. “n” we need to count stuff. “m” is mass or we’re crazy. “l” might have just been the nearest we could get to “p” without intruding on a more urgently-needed symbol. (“s” we use a lot for parameters like length of an arc that work kind of like time but aren’t time.) And then shift to the capital letter, I expect, because a lowercase l looks like a “1”, to everybody’s certain doom.

The modified potential energy, then, is going to include the angular momentum L. At least, the amount of angular momentum. It’s also going to include the mass of the object moving, and the radius r that says how far the object is from the center. It will be:

$V_{eff}(r) = V(r) + \frac{L^2}{2 m r^2}$

V(r) was the original potential, whatever that was. The modifying term, with this square of the angular momentum and all that, I kind of hope you’ll just accept on my word. The L² means that whether the angular momentum is positive or negative, the potential will grow very large as the radius gets small. If it didn’t, there might not be orbits at all. And if the angular momentum is zero, then the effective potential is the same original potential that let stuff crash into the center.

For the sort of r-to-a-power potentials I’ve been looking at, I get an effective potential of:

$V_{eff}(r) = C r^n + \frac{L^2}{2 m r^2}$

where n might be an integer. I’m going to pretend a while longer that it might not be, though. C is certainly some number, maybe positive, maybe negative.

If you pick some values for C, n, L, and m you can sketch this out. If you just want a feel for how this V_eff looks it doesn’t much matter what values you pick. Changing values just changes the scale, that is, where a circular orbit might happen. It doesn’t change whether it happens. Picking some arbitrary numbers is a good way to get a feel for how this sort of problem works. It’s good practice.

Sketching will convince you there are energy minimums, where we can get circular orbits. It won’t say where to find them without some trial-and-error or building a model of this energy and seeing where a ball bearing dropped into it rolls to a stop. We can do this more efficiently.

Why Stuff Can Orbit, Part 3: It Turns Out Spinning Matters

Way previously:

Before the big distractions of Theorem Thursdays and competitive pinball events and all that I was writing up the mathematics of orbits. Last time I’d got to establishing that there can’t be such a thing as an orbit. This seems to disagree with what a lot of people say we can observe. So I want to resolve that problem. Yes, I’m aware I’m posting this on a Thursday, which I said I wasn’t going to do because it’s too hard on me to write. I don’t know how it worked out like that.

Let me get folks who didn’t read the previous stuff up to speed. I’m using as model two things orbiting each other. I’m going to call it a sun and a planet because it’s way too confusing not to give things names. But they don’t have to be a sun and a planet. They can be a planet and moon. They can be a proton and an electron if you want to pretend quantum mechanics isn’t a thing. They can be a wood joist and a block of rubber connected to it by a spring. That’s a legitimate central force. They can even be stuff with completely made-up names representing made-up forces. So far I’m supposing the things are attracted or repelled by a force with a strength that depends on how far they are from each other but on nothing else.

Also I’m supposing there are only two things in the universe. This is because the mathematics of two things with this kind of force is easy to do. An undergraduate mathematics or physics major can do it. The mathematics of three things is too complicated to do. I suppose somewhere around two-and-a-third things the mathematics hard enough you need an expert but the expert can do it.

Mathematicians and physicists will call this sort of problem a “central force” problem. We can make it easier by supposing the sun is at the center of the universe, or at least our coordinate system. So we don’t have to worry about it moving. It’s just there at the center, the “origin”, and it’s only the planet that moves.

Forces are tedious things to deal with. They’re vectors. In this context that makes them bundles of three quantities each related to the other two. We can avoid a lot of hassle by looking at potential energy instead. Potential energy is a scalar, a single number. Numbers are nice and easy. Calculus tells us how to go from potential energy to forces, in case we need the forces. It also tells us how to go from forces to potential energy, so we can do the easier problem instead. So we do.

To write about potential energy mathematical physicists use exactly the letter you would guess they’d use if every other letter were unavailable for some reason: V. Or U, if they prefer. I’ll stick with V. Right now I don’t want to say anything about what rule determines the values of V. I just want to allow that its value changes as the planet’s distance from the star — the radius ‘r’ of its orbit — changes. So we make that clear by writing the potential energy is V = V(r). (The potential energy might change with the mass of the planet or sun, or the strength of gravity in the universe, or whatever. But we’re going to pretend those don’t change, not for the problem we’re doing, so we don’t have to write them out.)

If you draw V(r) versus r you can discover right away circular orbits. They’re ones that are local maximums or local minimums of V(r). Physical intuition will help us here. Imagine the graph of the potential energy as if it were a smooth bowl. Drop a marble into it. Where would the marble come to rest? That’s a local minimum. The radius of that minimum is a circular orbit. (Oh, a local maximum, where the marble is at the top of a hill and doesn’t fall to either side, could be a circular orbit. But it isn’t going to be stable. The marble will roll one way or another given the slightest chance.)

The potential energy for a force like gravity or electric attraction looks like the distance, r, raised to a power. And then multiplied by some number, which is where we hide gravitational constants and masses and all that stuff. Generally, it looks like V(r) = C rⁿ where C is some number and n is some other number. For gravity and electricity that number is -1. For two particles connected by a spring that number n is +2. Could be anything.

The trouble is if you draw these curves you realize that a marble dropped in would never come to a stop. It would roll down to the center, the planet falling into the sun. Or it would roll away forever, the planet racing into deep space. Either way it doesn’t orbit or do anything near orbiting. This seems wrong.

It’s not, though. Suppose the force is repelling, that is, the potential energy gets to be smaller and smaller numbers as the distance increases. Then the two things do race away from each other. Physics students are asked to imagine two positive charges let loose next to each other. Physics students understand they’ll go racing away from each other, even though we don’t see stuff in the real world that does that very often. We suppose the students understand, though. These days I guess you can make an animation of it and people will accept that as if it’s proof of anything.

Suppose the force is attracting. Imagine just dropping a planet out somewhere by a sun. Set it carefully just in place and let it go and get out of the way before happens. This is what we do in physics and mathematics classes, so that’s the kind of fun stuff you skipped if you majored in something else. But then we go on to make calculations about it. But that’ll orbit, right? It won’t just drop down into the sun and get melted or something?

Not so, the way I worded it. If we set the planet into space so it was holding still, not moving at all, then it will fall. Plummet, really. The planet’s attracted to the sun, and it moves in that direction, and it’s just going to keep moving that way. If it were as far from the center as the Earth is from the Sun it’ll take its time, yes, but it’ll fall into the sun and not do anything remotely like orbiting. And yet there’s still orbits. What’s wrong?

What’s wrong is a planet isn’t just sitting still there waiting to fall into the sun. Duh, you say. But why isn’t it just sitting still? That’s because it’s moving. Might be moving in any direction. We can divide that movement up into two pieces. One is the radial movement, how fast it’s moving towards or away from the center, that is, along the radius between sun and planet. If it’s a circular orbit this speed is zero; the planet isn’t moving any closer or farther away. If this speed isn’t zero it might affect how fast the planet falls into the sun, but it won’t affect the fact of whether it does or not. No more than how fast you toss a ball up inside a room changes whether it’ll eventually hit the floor. </p.

It’s the other part, the transverse velocity, that matters. This is the speed the thing is moving perpendicular to the radius. It’s possible that this is exactly zero and then the planet does drop into the sun. It’s probably not. And what that means is that the planet-and-sun system has an angular momentum. Angular momentum is like regular old momentum, only for spinning. And as with regular momentum, the total is conserved. It won’t change over time. When I was growing up this was always illustrated by thinking of ice skaters doing a spin. They pull their arms in, they spin faster. They put their arms out, they spin slower.

(Ice skaters eventually slow down, yes. That’s for the same reasons they slow down if they skate in a straight line even though regular old momentum, called “linear momentum” if you want to be perfectly clear, is also conserved. It’s because they have to get on to the rest of their routine.)

The same thing has to happen with planets orbiting a sun. If the planet moves closer to the sun, it speeds up; if it moves farther away, it slows down. To fall into the exact center while conserving angular momentum demands the planet get infinitely fast. This they don’t typically do.

There was a tipoff to this. It’s from knowing the potential energy V(r) only depends on the distance between sun and planet. If you imagine taking the system and rotating it all by any angle, you wouldn’t get any change in the forces or the way things move. It would just change the values of the coordinates you used to describe this. Mathematical physicists describe this as being “invariant”, which means what you’d imagine, under a “continuous symmetry”, which means a change that isn’t … you know, discontinuous. Rotating thing as if they were on a pivot, that is, instead of (like) reflecting them through a mirror.

And invariance under a continuous symmetry like this leads to a conservation law. This is known from Noether’s Theorem. You can find explained quite well on every pop-mathematics and pop-physics blog ever. It’s a great subject for pop-mathematics/physics writing. The idea, that the geometry of a problem tells us something about its physics and vice-versa, is important. It’s a heady thought without being so exotic as to seem counter-intuitive. And its discoverer was Dr Amalie Emmy Noether. She’s an early-20th-century demonstration of the first-class work that one can expect women to do when they’re not driven out of mathematics. You see why the topic is so near irresistible.

So we have to respect the conservation of angular momentum. This might sound like we have to give up on treating circular orbits as one-variable problems. We don’t have to just yet. We will, eventually, want to look at not just how far the planet is from the origin but also in what direction it is. We don’t need to do that yet. We have a brilliant hack.

We can represent the conservation of angular momentum as a slight repulsive force. It’s not very big if the angular momentum is small. It’s not going to be a very big force unless the planet gets close to the origin, that is, until r gets close to zero. But it does grow large and acts as if the planet is being pushed away. We consider that a pseudoforce. It appears because our choice of coordinates would otherwise miss some important physics. And that’s fine. It’s not wrong any more than, say, a hacksaw is the wrong tool to cut through PVC pipe just because you also need a vise.

This pseudoforce can be paired with a pseduo-potential energy. One of the great things about the potential-energy view of physics is that adding two forces together is as easy as adding their potential energies together. We call the sum of the original potential energy and the angular-momentum-created pseudopotential the “effective potential energy”. Far from the origin, for large radiuses r, this will be almost identical to the original potential energy. Close to the origin, this will be a function that rises up steeply. And as a result there can suddenly be a local minimum. There can be a circular orbit.

Spring potential, which is a parabola growing with the distance r from the origin. And the effective potential, which grows to a vertical asymptote where the radius is zero. — Figure 1. The potential energy of a spring — the red line — and the effective potential energy — the blue line — when the angular momentum is added as a pseudoforce. Without angular momentum in consideration the only equilibrium is at the origin. With angular momentum there’s some circular orbit, somewhere. Don’t pay attention to the numbers on the axes. They don’t mean anything.

Gravitational potential, with a vertical asymptote at the radius equalling zero going down to negative infinitely great numbers and a horizontal asymptote at the radius going off to infinity. And the effective potential, with the vertical asymptote at radius of zero going to positive infinitely great numbers, forcing there to be some minimum: a circular orbit. — Figure 2. The potential energy of a gravitational attraction — the red line — and the effective potential energy — the blue line — when the angular momentum is added as a pseudoforce. Without angular momentum in consideration there’s no equilibrium. The thing, a planet, falls into the center, the sun. With angular momentum there’s some circular orbit. As before the values of the numbers don’t matter and you should just ignore them.

The location of the minimum — the radius of the circular orbit — will depend on the original potential, of course. It’ll also depend on the angular momentum. The smaller the angular momentum the closer to the origin will be the circular orbit. If the angular momentum is zero we have the original potential and the planet dropping into the center again. If the angular momentum is large enough there might not even be a minimum anymore. That matches systems where the planet has escape velocity and can go plunging off into deep space. And we can see this by looking at the plot of the effective velocity even before we calculate things.

This only goes so far as demonstrating a circular orbit should exist. Or giving some conditions for which a circular orbit wouldn’t. We might want to know something more, like where that circular orbit is. Or if it’s possible for there to be an elliptic orbit. Or other shapes. I imagine it’s possible to work this out with careful enough drawings. But at some point it gets easier to just calculate things. We’ll get to that point soon.

A Leap Day 2016 Mathematics A To Z: Yukawa Potential

Yeah, ‘Y’ is a lousy letter in the Mathematics Glossary. I have a half-dozen mathematics books on the shelf by my computer. Some is semi-popular stuff like Richard Courant and Herbert Robbins’s What Is Mathematics? (the Ian Stewart revision). Some is fairly technical stuff, by which I mean Hidetoshi Nishimori’s Statistical Physics of Spin Glasses and Information Processing. There’s just no ‘Y’ terms in any of them worth anything. But I can rope something into the field. For example …

Yukawa Potential

When you as a physics undergraduate first take mechanics it’s mostly about very simple objects doing things according to one rule. The objects are usually these indivisible chunks. They’re either perfectly solid or they’re points, too tiny to have a surface area or volume that might mess things up. We draw them as circles or as blocks because they’re too hard to see on the paper or board otherwise. We spend a little time describing how they fall in a room. This lends itself to demonstrations in which the instructor drops a rubber ball. Then we go on to a mass on a spring hanging from the ceiling. Then to a mass on a spring hanging to another mass.

Then we go onto two things sliding on a surface and colliding, which would really lend itself to bouncing pool balls against one another. Instead we use smaller solid balls. Sometimes those “Newton’s Cradle” things with the five balls that dangle from wires and just barely touch each other. They give a good reason to start talking about vectors. I mean positional vectors, the ones that say “stuff moving this much in this direction”. Normal vectors, that is. Then we get into stars and planets and moons attracting each other by gravity. And then we get into the stuff that really needs calculus. The earlier stuff is helped by it, yes. It’s just by this point we can’t do without.

The “things colliding” and “balls dropped in a room” are the odd cases in this. Most of the interesting stuff in an introduction to mechanics course is about things attracting, or repelling, other things. And, particularly, they’re particles that interact by “central forces”. Their attraction or repulsion is along the line that connects the two particles. (Impossible for a force to do otherwise? Just wait until Intro to Mechanics II, when magnetism gets in the game. After that, somewhere in a fluid dynamics course, you’ll see how a vortex interacts with another vortex.) The potential energies for these all vary with distance between the points.

Yeah, they also depend on the mass, or charge, or some kind of strength-constant for the points. They also depend on some universal constant for the strength of the interacting force. But those are, well, constant. If you move the particles closer together or farther apart the potential changes just by how much you moved them, nothing else.

Particles hooked together by a spring have a potential that looks like $\frac{1}{2}k r^2$ . Here ‘r’ is how far the particles are from each other. ‘k’ is the spring constant; it’s just how strong the spring is. The one-half makes some other stuff neater. It doesn’t do anything much for us here. A particle attracted by another gravitationally has a potential that looks like $-G M \frac{1}{r}$ . Again ‘r’ is how far the particles are from each other. ‘G’ is the gravitational constant of the universe. ‘M’ is the mass of the other particle. (The particle’s own mass doesn’t enter into it.) The electric potential looks like the gravitational potential but we have different symbols for stuff besides the $\frac{1}{r}$ bit.

The spring potential and the gravitational/electric potential have an interesting property. You can have “closed orbits” with a pair of them. You can set a particle orbiting another and, with time, get back to exactly the original positions and velocities. (Three or more particles you’re not guaranteed of anything.) The curious thing is this doesn’t always happen for potentials that look like “something or other times r to a power”. In fact, it never happens, except for the spring potential, the gravitational/electric potential, and — peculiarly — for the potential $k r^7$ . ‘k’ doesn’t mean anything there, and we don’t put a one-seventh or anything out front for convenience, because nobody knows anything that needs anything like that, ever. We can have stable orbits, ones that stay within a minimum and a maximum radius, for a potential $k r^n$ whenever n is larger than -2, at least. And that’s it, for potentials that are nothing but r-to-a-power.

Ah, but does the potential have to be r-to-a-power? And here we see Dr Hideki Yukawa’s potential energy. Like these springs and gravitational/electric potentials, it varies only with the distance between particles. Its strength isn’t just the radius to a power, though. It uses a more complicated expression:

$-K \frac{e^{-br}}{r}$

Here ‘K’ is a scaling constant for the strength of the whole force. It’s the kind of thing we have ‘G M’ for in the gravitational potential, or ‘k’ in the spring potential. The ‘b’ is a second kind of scaling. And that a kind of range. A range of what? It’ll help to look at this potential rewritten a little. It’s the same as $-\left(K \frac{1}{r}\right) \cdot \left(e^{-br}\right)$ . That’s the gravitational/electric potential, times e^-br. That’s a number that will be very large as r is small, but will drop to zero surprisingly quickly as r gets larger. How quickly will depend on b. The larger a number b is, the faster this drops to zero. The smaller a number b is, the slower this drops to zero. And if b is equal to zero, then e^-br is equal to 1, and we have the gravitational/electric potential all over again.

Yukawa introduced this potential to physics in the 1930s. He was trying to model the forces which keep an atom’s nucleus together. It represents the potential we expect from particles that attract one another by exchanging some particles with a rest mass. This rest mass is hidden within that number ‘b’ there. If the rest mass is zero, the particles are exchanging something like light, and that’s just what we expect for the electric potential. For the gravitational potential … um. It’s complicated. It’s one of the reasons why we expect that gravitons, if they exist, have zero rest mass. But we don’t know that gravitons exist. We have a lot of trouble making theoretical gravitons and quantum mechanics work together. I’d rather be skeptical of the things until we need them.

Still, the Yukawa potential is an interesting mathematical creature even if we ignore its important role in modern physics. When I took my Introduction to Mechanics final one of the exam problems was deriving the equivalent of Kepler’s Laws of Motion for the Yukawa Potential. I thought then it was a brilliant problem. I still do. It struck me while writing this that I don’t remember whether it allows for closed orbits, except when b is zero. I’m a bit afraid to try to work out whether it does, lest I learn that I can’t follow the reasoning for that anymore. That would be a terrible thing to learn.

Reading the Comics, January 27, 2015: Rabbit In A Trapezoid Edition

So the reason I fell behind on this Reading the Comics post is that I spent more time than I should have dithering about which ones to include. I hope it’s not disillusioning to learn that I have no clearly defined rules about what comics to include and what to leave out. It depends on how clearly mathematical in content the comic strip is; but it also depends on how much stuff I have gathered. If there’s a slow week, I start getting more generous about what I might include. And last week gave me a string of comics that I could argue my way into including, but few that obviously belonged. So I had a lot of time dithering.

To make it up to you, at the end of the post I should have our pet rabbit tucked within a trapezoid of his own construction. If that doesn’t make everything better I don’t know what will.

Mark Pett’s Mr Lowe for the 22nd of January (a rerun from the 19th of January, 2001) is really a standardized-test-question joke. But it brings up a debate about cultural biases in standardized tests that I don’t remember hearing lately. I may just be moving in the wrong circles. I remember self-assured rich white males explaining how it’s absurd to think cultural bias could affect test results since, after all, they’re standardized tests. I’ve sulked some around these parts about how I don’t buy mathematics’ self-promoted image of being culturally neutral either. A mathematical truth may be universal, but that we care about this truth is not. Anyway, Pett uses a mathematics word problem to tell the joke. That was probably the easiest way to put a cultural bias into a panel that

T Lewis and Michael Fry’s Over The Hedge for the 25th of January uses a bit of calculus to represent “a lot of hard thinking”. Hammy the Squirrel particularly is thinking of the Fundamental Theorem of Calculus. This particular part is the one that says the derivative of the integral of a function is the original function. It’s part of how integration and differentiation link together. And it shows part of calculus’s great appeal. It has those beautiful long-s integral signs that make this part of mathematics look like artwork.

Leigh Rubin’s Rubes for the 25th of January is a panel showing “Schrödinger’s Job Application”. It’s referring to Schrödinger’s famous thought experiment, meant to show there are things we don’t understand about quantum mechanics. It sets up a way that a quantum phenomenon can be set up to have distinct results in the everyday environment. The mathematics suggests that a cat, poisoned or not by toxic gas released or not by the decay of one atom, would be both alive and dead until some outside observer checks and settles the matter. How can this be? For that matter, how can the cat not be a qualified judge to whether it’s alive? Well, there are things we don’t understand about quantum mechanics.

Roy Schneider’s The Humble Stumble for the 26th of January (a rerun from the 30th of January, 2007) uses a bit of mathematics to mark Tommy, there, as a frighteningly brilliant weirdo. The equation is right, although trivial. The force it takes to keep something with a mass m moving in a circle of radius R at the linear speed v is $\frac{m v^2}{R}$ . The radius of the Moon’s orbit around the Earth is strikingly close to sixty times the Earth’s radius. The Ancient Greeks were able to argue that from some brilliantly considered geometry. Here, R_E gets used as a name for “the radius of the Earth”. So the force holding the Moon in its orbit has to be approximately $\frac{m v^2}{60 R_e}$ . That’s if we say m is the mass of the Moon, and v is its linear speed, and if we suppose the Moon’s orbit is a circle. It nearly is, and this would give us a good approximate answer to how much force holds the Moon in its orbit. It would be only a start, though; the precise movements of the Moon are surprisingly complicated. Newton himself could not fully explain them, even with the calculus and physics tools he invented for the task.

Dave Whamond’s Reality Check for the 26th of January isn’t quite the anthropomorphic-numerals joke for this essay. But we do get personified geometric constructs, which is close, and some silly wordplay. Much as I like the art for Over The Hedge showcasing a squirrel so burdened with thoughts that his head flops over, this might be my favorite of this bunch.

Dave Blazek’s Loose Parts for the 27th of January is a runner-up for the silly jokes trophy this time around.

Our pet rabbit flopped out inside of a cardboard box. The box was set up, upside-down, so he could go inside and chew on the contents. He's pulled the side flaps inward, so that the base is a trapezoidal prism. — Cardboard boxes are normally pretty good environments for rabbits, given that they’re places the rabbits can do in and not be seen. We set the box up, but he did all the chewing.

Now I know what you’re thinking: isn’t that actually a trapezoidal prism, underneath a rectangular prism? Yes, I suppose so. The only people who’re going to say so are trying to impress people by saying so, though. And those people won’t be impressed by it. I’m sorry. We gave him the box because rabbits generally like having cardboard boxes to go in and chew apart. He did on his own the pulling-in of the side flaps to make it stand so trapezoidal.

Reblog: Multivariable Calculus

I believe that I used this textbook, or at least one similar to it, in learning Multivariable Calculus. (Admittedly there are certain similarities among introductory textbooks on this subject which can’t be avoided.) Still, it should be useful for people not sure how you get from coordinate pairs to proving swiftly that the force of gravity on the inside of a solid-shell style Dyson sphere is zero.

Are You Stronger Than Jupiter?

A comment on my earlier piece comparing the acceleration due to gravity that we feel from the Moon compared to what we feel from someone else in the room challenged me: how strong is the gravitational pull from Jupiter, compared to that of someone else in the room? Jupiter has a great edge in mass: someone else in the room weighs in at somewhere around 75 kilograms, while the planet comes to around 1,898,600,000,000,000,000,000,000,000 kilograms. On the other hand, your neighbor is somewhere around one meter away, while Jupiter will be something like 816,520,800,000 meters away. Maybe farther: that’s the farthest Jupiter gets from the Sun, and the Earth can be on the opposite side of the Sun from Jupiter, so add up to another 152,098,232,000 meters on top of that.

That distance is going to be a little bit of a nuisance. The acceleration we feel towards any planet will be stronger the nearer it gets, and while, say, Neptune is always about the same distance from the Earth, there are times that Venus and Mars are surprisingly close. Usually these are announced by clouds of forwarded e-mails announcing that Mars will be closer to the Earth than it’s been in 33,000 years, and will appear to be as large as the Empire State Building. Before you have even had the chance to delete these e-mails unread the spoilsport in your group of e-mail friends will pass along the snopes.com report that none of this is true and the e-mail has been going around unaltered since 1997 anyway. But there is still a smallest distance and a greatest distance any planet gets from the Earth.

If we want to give planets the best shot, let’s look at the smallest distance any planet gets from the Earth. For Mercury and Venus, this happens when the planet is at aphelion, the farthest it gets from the Sun, and the Earth at perihelion, the nearest it gets. For the outer planets, it happens with the Earth at aphelion and the other planet at perihelion. (Some might say ‘apogee’ and ‘perigee’, although these are properly speaking only the words to use when something orbits the Earth. Some might say ‘apoapsis’ and ‘periapsis’, which talk about the nearest and farthest points of an orbit without being particular about what is being orbited, but no one actually does.) Here I’m making the assumption that there’s no weird orbital locks where, say, the Earth can’t be at perihelion while Venus is at aphelion, which might even be true. It’s probably close enough.

Continue reading “Are You Stronger Than Jupiter?”

A Planet Is Not A Dot

A lot of what I said in describing how we might fall into the Moon, if you and I were in the same room and suddenly the rest of the world stopped existing, was incorrect. That isn’t to say it was wrong or even bad to consider; it just means that the equations that I produced and the numbers that came out from them aren’t exactly what would happen if the sudden-failure-of-planet-Earth case were to happen. I knew the wouldn’t be exactly right going in, which leaves us the question of what I thought I was doing and why I bothered doing it.

The first reason, and the reason why it wasn’t a waste of time to consider these simple approximations of how strongly the Moon is attracting us — how fast we are falling into it, and how fast we would be falling if the Earth weren’t falling into the Moon along with us — is thanks to something which Isaac Asimov perfectly described. In an essay called “The Relativity Of Wrong”, he wrote about — well, the title says it. Ideas are not just right or wrong; they can be wrong by differing amounts, and can be wrong by such a tiny amount that it isn’t worth the complications to get it exactly right. Probably the most familiar example is the flatness of the Earth. To model the globe, or a large nation, the idea that the Earth is nearly flat is sufficiently wrong as to produce measurable, important errors where plots of land are justifiably claimed by multiple owners, maybe from multiple governments, or aren’t claimed at all and form the basis for nowhere towns in which mild fantasy or comic stories can be set. But if one wants to draw a map of the town, or of one’s own property, the curvature of the Earth is not worth considering. We can pretend the Earth is flat and get our work done a lot sooner. Other sources of error will mess up the precise result before that does.

Continue reading “A Planet Is Not A Dot”

In Case Of Sudden Failure Of Planet Earth

Have you ever figured out just exactly what you would do if the Earth were to suddenly disappear from the universe, leaving just you and whatever’s around to fall towards whatever the nearest heavenly bodies are? No, me neither. Asked to improvise one, I suppose I’d suffocate within minutes and then everything else becomes not so interesting to me, although possibly my heirs might be interested, if they’re somewhere.

My mother accidentally got me thinking about this fate. She’s taking a course about the science and world view of the Ancient Egyptians. (In talking about it she asked if I knew anything about the science of the Ancient Egyptians, and I tried to say I didn’t really know more than the average lay reader might, although when I got to mentioning that I knew of their awareness of the Sothic Cycle and where we get the name “Sothic Cycle” I realized that I can’t really call myself ignorant about the science of the Ancient Egyptians.) But the class had reached the point where astrological beliefs of the ancients were under discussion, and apparently some of the students insisted that astrology really and truly works, and my mother wanted me to figure out how strong the force of gravity of the Moon is, compared to the force of gravity of another person in the same room. This would allow her to go into class armed with numbers which have never dissuaded anyone from believing in astrology, but, it’s fun for someone.

I did double-check, though, that she meant the gravitational pull of the Moon, rather than its tidal pull. The shorthand reason for this is that arguments for astrology having some physical basis tend to run along the lines of, the Moon creates the tides (the Sun does too, but smaller ones), tides are made of water (rock moves, too, although much less), human bodies are mostly water (I don’t know what the fluid properties of cytoplasm are, but I’m almost curious enough to look them up), so there must be something tide-like in human bodies too (so there). The gravitational pull of the Moon, meanwhile, doesn’t really mean much: the Moon is going to accelerate the Earth and the people standing on it by just about the same amount. The force of gravity between two objects grows with the two objects’ masses, and the Earth is more massive than any person on it. But this means the Earth feels a greater force pulling it towards the Moon, and the acceleration works out tobe just the same. The force of gravity between two objects falls off as the square of the distance between them, and the people on the surface of the Earth are a little bit closer or a little bit farther away from the Moon than the center of the Earth is, but that’s not very different considering just how far away the Moon is. We spend all our lives falling into the Moon, as fast as we possibly can, and we are falling into the Moon as fast as the Earth is.

Continue reading “In Case Of Sudden Failure Of Planet Earth”

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Yukawa Potential

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