## Everything Interesting There Is To Say About Springs

I need another supplemental essay to get to the next part in Why Stuff Can Orbit. (Here’s the last part.) You probably guessed it’s about springs. They’re useful to know about. Why? That one killer Mystery Science Theater 3000 short, yes. But also because they turn up everywhere.

Not because there are literally springs in everything. Not with the rise in anti-spring political forces. But what makes a spring is a force that pushes something back where it came from. It pushes with a force that grows just as fast as the distance from where it came grows. Most anything that’s stable, that has some normal state which it tends to look like, acts like this. A small nudging away from the normal state gets met with some resistance. A bigger nudge meets bigger resistance. And most stuff that we see is stable. If it weren’t stable it would have broken before we got there.

(There are exceptions. Stable is, sometimes, about perspective. It can be that something is unstable but it takes so long to break that we don’t have to worry about it. Uranium, for example, is dying, turning slowly into stable elements like lead and helium. There will come a day there’s none left in the Earth. But it takes so long to break down that, barring surprises, the Earth will have broken down into something else first. And it may be that something is unstable, but it’s created by something that’s always going on. Oxygen in the atmosphere is always busy combining with other chemicals. But oxygen stays in the atmosphere because life keeps breaking it out of other chemicals.)

Now I need to put in some terms. Start with your thing. It’s on a spring, literally or metaphorically. Don’t care. If it isn’t being pushed in any direction then it’s at rest. Or it’s at an equilibrium. I don’t want to call this the ideal or natural state. That suggests some moral superiority to one way of existing over another, and how do I know what’s right for your thing? I can tell you what it acts like. It’s your business whether it should. Anyway, your thing has an equilibrium.

Next term is the displacement. It’s how far your thing is from the equilibrium. If it’s really a block of wood on a spring, like it is in high school physics, this displacement is how far the spring is stretched out. In equations I’ll represent this as ‘x’ because I’m not going to go looking deep for letters for something like this. What value ‘x’ has will change with time. This is what makes it a physics problem. If we want to make clear that ‘x’ does depend on time we might write ‘x(t)’. We might go all the way and start at the top of the page with ‘x = x(t)’, just in case.

If ‘x’ is a positive number it means your thing is displaced in one direction. If ‘x’ is a negative number it was displaced in the opposite direction. By ‘one direction’ I mean ‘to the right, or else up’. By ‘the opposite direction’ I mean ‘to the left, or else down’. Yes, you can pick any direction you like but why are you making life harder for everyone? Unless there’s something compelling about the setup of your thing that makes another choice make sense just go along with what everyone else is doing. Apply your creativity and iconoclasm where it’ll make your life better instead.

Also, we only have to worry about one direction. This might surprise you. If you’ve played much with springs you might have noticed how they’re three-dimensional objects. You can set stuff swinging back and forth in two directions at once. That’s all right. We can describe a two-dimensional displacement as a displacement in one direction plus a displacement perpendicular to that. And if there’s no such thing as friction, they won’t interact. We can pretend they’re two problems that happen to be running on the same spring at the same time. So here I declare: we can ignore friction and pretend it doesn’t matter. We don’t have to deal with more than one direction at a time.

(It’s not only friction. There’s problems about how energy gets transmitted between ways the thing can oscillate. This is what causes what starts out as a big whack in one direction to turn into a middling little circular wobbling. That’s a higher level physics than I want to do right now. So here I declare: we can ignore that and pretend it doesn’t matter.)

Whether your thing is displaced or not it’s got some potential energy. This can be as large or as small as you like, down to some minimum when your thing is at equilibrium. The potential energy we represent as a number named ‘U’ because of good reasons that somebody surely had. The potential energy of a spring depends on the square of the displacement. We can write its value as ‘U = ½ k x2‘. Here ‘k’ is a number known as the spring constant. It describes how strongly the spring reacts; the bigger ‘k’ is, the more any displacement’s met with a contrary force. It’ll be a positive number. ½ is that same old one-half that you know from ideas being half-baked or going-off being half-cocked.

Potential energy is great. If you can describe a physics problem with its energy you’re in good shape. It lets us bring physical intuition into understanding things. Imagine a bowl or a Habitrail-type ramp that’s got the cross-section of your potential energy. Drop a little marble into it. How the marble rolls? That’s what your thingy does in that potential energy.

Also we have mathematics. Calculus, particularly differential equations, lets us work out how the position of your thing will change. We need one more piece for this. That’s the momentum of your thing. Momentum is traditionally represented with the letter ‘p’. And now here’s how stuff moves when you know the potential energy ‘U’:

$\frac{dp}{dt} = - \frac{\partial U}{\partial x}$

Let me unpack that. $\frac{dp}{dt}$ — also known as $\frac{d}{dt}p$ if that looks better — is “the derivative of p with respect to t”. It means “how the value of the momentum changes as the time changes”. And that is equal to minus one times …

You might guess that $\frac{\partial U}{\partial x}$ — also written as $\frac{\partial}{\partial x} U$ — is some kind of derivative. The $\partial$ looks kind of like a cursive d, after all. It’s known as the partial derivative, because it means we look at how ‘U’ changes as ‘x’ and nothing else at all changes. With the normal, ‘d’ style full derivative, we have to track how all the variables change as the ‘t’ we’re interested in changes. In this particular problem the difference doesn’t matter. But there are problems where it does matter and that’s why I’m careful about the symbols.

So now we fall back on how to take derivatives. This gives us the equation that describes how the physics of your thing on a spring works:

$\frac{dp}{dt} = - k x$

You’re maybe underwhelmed. This is because we haven’t got any idea how the momentum ‘p’ relates to the displacement ‘x’. Well, we do, because I know and if you’re still reading at this point you know full well what momentum is. But let me make it official. Momentum is, for this kind of thing, the mass ‘m’ of your thing times how its position is changing, which is $\frac{dx}{dt}$. The mass of your thing isn’t changing. If you’re going to let it change then we’re doing some screwy rocket problem and that’s a different article. So its easy to get the momentum out of that problem. We get instead the second derivative of the displacement with respect to time:

$m\frac{d^2 x}{dt^2} = - kx$

Fine, then. Does that tell us anything about what ‘x(t)’ is? Not yet, but I will now share with you one of the top secrets that only real mathematicians know. We will take a guess to what the answer probably is. Then we’ll see in what circumstances that answer could possibly be right. Does this seem ad hoc? Fine, so it’s ad hoc. Here is the secret of mathematicians:

It’s fine if you get your answer by any stupid method you like, including guessing and getting lucky, as long as you check that your answer is right.

Oh, sure, we’d rather you get an answer systematically, since a system might give us ideas how to find answers in new problems. But if all we want is an answer then, by definition, we don’t care where it came from. Anyway, we’re making a particular guess, one that’s very good for this sort of problem. Indeed, this guess is our system. A lot of guesses at solving differential equations use exactly this guess. Are you ready for my guess about what solves this? Because here it is.

We should expect that

$x(t) = C e^{r t}$

Here ‘C’ is some constant number, not yet known. And ‘r’ is some constant number, not yet known. ‘t’ is time. ‘e’ is that number 2.71828(etc) that always turns up in these problems. Why? Because its derivative is very easy to take, and if we have to take derivatives we want them to be easy to take. The first derivative of $Ce^{rt}$ with respect to ‘t’ is $r Ce^{rt}$. The second derivative with respect to ‘t’ is $r^2 Ce^{rt}$. so here’s what we have:

$m r^2 Ce^{rt} = - k Ce^{rt}$

What we’d like to find are the values for ‘C’ and ‘r’ that make this equation true. It’s got to be true for every value of ‘t’, yes. But this is actually an easy equation to solve. Why? Because the $C e^{rt}$ on the left side has to equal the $C e^{rt}$ on the right side. As long as they’re not equal to zero and hey, what do you know? $C e^{rt}$ can’t be zero unless ‘C’ is zero. So as long as ‘C’ is any number at all in the world except zero we can divide this ugly lump of symbols out of both sides. (If ‘C’ is zero, then this equation is 0 = 0 which is true enough, I guess.) What’s left?

$m r^2 = -k$

OK, so, we have no idea what ‘C’ is and we’re not going to have any. That’s all right. We’ll get it later. What we can get is ‘r’. You’ve probably got there already. There’s two possible answers:

$r = \pm\sqrt{-\frac{k}{m}}$

You might not like that. You remember that ‘k’ has to be positive, and if mass ‘m’ isn’t positive something’s screwed up. So what are we doing with the square root of a negative number? Yes, we’re getting imaginary numbers. Two imaginary numbers, in fact:

$r = \imath \sqrt{\frac{k}{m}}, r = - \imath \sqrt{\frac{k}{m}}$

Which is right? Both. In some combination, too. It’ll be a bit with that first ‘r’ plus a bit with that second ‘r’. In the differential equations trade this is called superposition. We’ll have information that tells us how much uses the first ‘r’ and how much uses the second.

You might still be upset. Hey, we’ve got these imaginary numbers here describing how a spring moves and while you might not be one of those high-price physicists you see all over the media you know springs aren’t imaginary. I’ve got a couple responses to that. Some are semantic. We only call these numbers “imaginary” because when we first noticed they were useful things we didn’t know what to make of them. The label is an arbitrary thing that doesn’t make any demands of the numbers. If we had called them, oh, “Cardanic numbers” instead would you be upset that you didn’t see any Cardanos in your springs?

My high-class semantic response is to ask in exactly what way is the “square root of minus one” any less imaginary than “three”? Can you give me a handful of three? No? Didn’t think so.

And then the practical response is: don’t worry. Exponentials raised to imaginary numbers do something amazing. They turn into sine waves. Well, sine and cosine waves. I’ll spare you just why. You can find it by looking at the first twelve or so posts of any pop mathematics blog and its article about how amazing Euler’s Formula is. Given that Euler published, like, 2,038 books and papers through his life and the fifty years after his death it took to clear the backlog you might think, “Euler had a lot of Formulas, right? Identities too?” Yes, he did, but you’ll know this one when you see it.

What’s important is that the displacement of your thing on a spring will be described by a function which looks like this:

$x(t) = C_1 e^{\sqrt{\frac{k}{m}} t} + C_2 e^{-\sqrt{\frac{k}{m}} t}$

for two constants, ‘C1‘ and ‘C2‘. These were the things we called ‘C’ back when we thought the answer might be $Ce^{rt}$; there’s two of them because there’s two r’s. I give you my word this is equivalent to a formula like this, but you can make me show my work if you must:

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

for some (other) constants ‘A’ and ‘B’. Cosine and sine are the old things you remember from learning about cosine and sine.

OK, but what are ‘A’ and ‘B’?

Generically? We don’t care. Some numbers. Maybe zero. Maybe not. The pattern, how the displacement changes over time, will be the same whatever they are. It’ll be regular oscillation. At one time your thing will be as far from the equilibrium as it gets, and not moving toward or away from the center. At one time it’ll be back at the center and moving as fast as it can. At another time it’ll be as far away from the equilibrium as it gets, but on the other side. At another time it’ll be back at the equilibrium and moving as fast as it ever does, but the other way. How far is that maximum? What’s the fastest it travels?

The answer’s in how we started. If we start at the equilibrium without any kind of movement we’re never going to leave the equilibrium. We have to get nudged out of it. But what kind of nudge? There’s three ways you can do to nudge something out.

You can tug it out some and let it go from rest. This is the easiest: then ‘A’ is however big your tug was and ‘B’ is zero.

You can let it start from equilibrium but give it a good whack so it’s moving at some initial velocity. This is the next-easiest: ‘A’ is zero, and ‘B’ is … no, not the initial velocity. You need to look at what the velocity of your thing is at the start. That’s the first derivative:

$\frac{dx}{dt} = -\sqrt{\frac{k}{m}}A sin\left(\sqrt{\frac{k}{m}} t\right) + \sqrt{\frac{k}{m}} B sin\left(\sqrt{\frac{k}{m}} t\right)$

The start is when time is zero because we don’t need to be difficult. when ‘t’ is zero the above velocity is $\sqrt{\frac{k}{m}} B$. So that product has to be the initial velocity. That’s not much harder.

The third case is when you start with some displacement and some velocity. A combination of the two. Then, ugh. You have to figure out ‘A’ and ‘B’ that make both the position and the velocity work out. That’s the simultaneous solutions of equations, and not even hard equations. It’s more work is all. I’m interested in other stuff anyway.

Because, yeah, the spring is going to wobble back and forth. What I’d like to know is how long it takes to get back where it started. How long does a cycle take? Look back at that position function, for example. That’s all we need.

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

Sine and cosine functions are periodic. They have a period of 2π. This means if you take the thing inside the parentheses after a sine or a cosine and increase it — or decrease it — by 2π, you’ll get the same value out. What’s the first time that the displacement and the velocity will be the same as their starting values? If they started at t = 0, then, they’re going to be back there at a time ‘T’ which makes true the equation

$\sqrt{\frac{k}{m}} T = 2\pi$

And that’s going to be

$T = 2\pi\sqrt{\frac{m}{k}}$

Maybe surprising thing about this: the period doesn’t depend at all on how big the displacement is. That’s true for perfect springs, which don’t exist in the real world. You knew that. Imagine taking a Junior Slinky from the dollar store and sticking a block of something on one end. Imagine stretching it out to 500,000 times the distance between the Earth and Jupiter and letting go. Would it act like a spring or would it break? Yeah, we know. It’s sad. Think of the animated-cartoon joy a spring like that would produce.

But this period not depending on the displacement is true for small enough displacements, in the real world. Or for good enough springs. Or things that work enough like springs. By “true” I mean “close enough to true”. We can give that a precise mathematical definition, which turns out to be what you would mean by “close enough” in everyday English. The difference is it’ll have Greek letters included.

So to sum up: suppose we have something that acts like a spring. Then we know qualitatively how it behaves. It oscillates back and forth in a sine wave around the equilibrium. Suppose we know what the spring constant ‘k’ is. Suppose we also know ‘m’, which represents the inertia of the thing. If it’s a real thing on a real spring it’s mass. Then we know quantitatively how it moves. It has a period, based on this spring constant and this mass. And we can say how big the oscillations are based on how big the starting displacement and velocity are. That’s everything I care about in a spring. At least until I get into something wild like several springs wired together, which I am not doing now and might never do.

And, as we’ll see when we get back to orbits, a lot of things work close enough to springs.

## 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 Veff 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.

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.