## Thursday, March 17, 2016

### On Guessing

This is a follow-up to my Lagrange point post. At the end, I briefly mentioned the L4/L5 Lagrange points, which are stable and form equilateral triangles with the masses of a three-body system. I'd like to delve into the physics of these points a bit to illustrate something about how physicists solve problems.

That is, physicists (in general) do not like doing calculations. They don't want to sit around all day crunching numbers to arrive at an answer. When you solve a physics problem, the goal is to build as simple a model as possible that captures the essential features of what you're studying. (This is where the spherical cow jokes come in.) That way, if you're lucky, you can avoid having to do a lot of math. Instead you can arrive at the answer you want by symmetry, or dimensional analysis, or guessing.

Guessing is an important part of the physicist's toolkit and some of what makes doing these problems fun (for me, at least). It's easy to stare at a problem for hours and feel overwhelmed by the complexity of it. I liken this to how it feels when you've just begun to write something. You have a blank screen and a blinking cursor in front of you and there's nothing more terrifying or paralyzing.

In writing, sometimes the solution is to just start writing and see where the story takes you. And so it follows with physics. If you have a complex problem, at times the best strategy is to just guess at the answer and see where the physics takes you. In this way, doing physics can be a lot like playing a game or solving a puzzle. It's fun, and I seriously wouldn't still be in school if I thought otherwise.

So let's return to the L4/L5 Lagrange points. In class, when discussing the three-body problem, our professor performed enough derivation to get us to believe that stable orbits can exist. He went through the same argument I used about rotating frames and centrifugal force. So a test mass is in a stable orbit when gravity and centrifugal force cancel out. He then gave us the punch line, telling us where the Lagrange points are, but didn't go through the math of actually finding them. Why not? Because if you do the derivation, the equations of motion you end up having to solve are:

 I should probably credit Massimo Ricotti for this.
I'm not going to attempt to explain what all that means. It's ugly, and you wouldn't want to solve that unless you had no other choice. But there is another way. Our professor mentioned that when thinking about the 5 Lagrange points, you can guess where 2 of them (L4/L5) must be.

This intrigued me, which is why we're here today. What makes it possible to guess these locations? As we saw with the L2 point, its exact location is related to the square root of the ratio between the two big masses. This is (probably) not something you could just pull out of thin air. But that's not the case for L4 and L5. The location of one of these points is at the vertex of an equilateral triangle that has the two large masses at the other vertices. Flip this triangle over and you get the other one. How massive the objects are isn't relevant at all; distance is the only important variable (and two masses can basically orbit each other at any distance they like). So you could conceivably guess the answer just by looking at the problem.

 There are a lot more MS Paint illustrations coming. You've been warned.
But what makes equilateral triangles, as aesthetically pleasing as they are, physically appealing? Let's consider a special case and then move on to a more general scenario.

Forget the Earth-Moon system and consider two stars of equal mass in circular orbits about each other. In that case, the stars are actually orbiting their center of mass, which is halfway between the two for equal mass stars. A third body that's motionless in the rotating frame also orbits the center of mass, which means centrifugal force pushes away from that center. To make the problem even simpler, let's put the third body equidistant from the two stars.

 I'm a big fan of purple.
Then the forces of gravity to the left and right cancel out, leaving only gravity pulling down and centrifugal force pushing up. To get our Lagrange point, we just need those forces to balance. This means we have to guess how far up from the center the Lagrange point is.

First, let's consider gravity. The total strength of gravity depends on the inverse square of the distance to the stars, d. But we don't want the total force, only the vertical component. That part is a fraction of the total, and that fraction is equal to h/d. This means gravity now depends on the distance to the center of mass and the inverse cube of the distance to the stars.

On the other hand, centrifugal force depends on the distance to the center of mass, h, and the inverse cube of the distance between the stars, a. Our gravity and centrifugal terms are nearly the same, except one uses a and the other d. But we're trying to find d, so let's just guess that d=a. Then all the lengths of our triangle are equal and we've found a point where all the forces cancel out--a Lagrange point. (This guess works because the constants in each equation are the same. Otherwise, d might just be proportional to a.)

So there we have it. Using a few reasonable assumptions, a simple model, and nothing more than geometry, we've found the Lagrange points. Where do we go from here? How about back to the Sun-Earth system, where one of the two masses is much, much bigger than the other. If that's the case, then the center of mass moves to the sun, and centrifugal force points directly away from it.

 It's a trap!
If we maintain our equilateral triangle guess, where does that leave us? With a problem. The problem is that if you rotate the above picture so that the sun's gravity vector and the centrifugal vector are horizontal, you're left with the Earth's gravity vector at an angle of 60° away from horizontal. This is bad because the "vertical" component of the Earth's gravity isn't balanced by anything else, which means that no matter what values you insert into your equation, there is no equilibrium point. Uh, oh.

But our graph has fooled us here. You see, by moving the center of mass directly on top of the sun, we are implicitly saying that the Earth has no mass whatsoever. And if that's the case, then it has no gravitational force, which means it doesn't need to be counteracted at all. In the limit where the Earth has no mass, the three-body problem reduces to the one-body problem. So there is a point of stability at the equilateral triangle, but also at any point along the same circular orbit.

This wasn't a totally useless exercise, however. It shows us that it's reasonable to expect L4/L5 to be stable from one extreme of equal masses to the other extreme of just one big mass. But we haven't yet proven that the L4/L5 points exist where they do for any arbitrary masses. How do we do that? First, let's make a generic diagram describing the situation.

Let's say that Star A has a mass of m and Star B has a mass of km, where k is some fraction between 0 and 1. This means we can vary between the two extremes of equal mass (k=1) and one dominant mass (k=0). The smaller k is, the farther to the left the center of mass moves, the smaller Star B's gravity vector is, and the more horizontal the centrifugal vector gets. This should mean that the forces pointing to the right stay balanced. Additionally, as k gets smaller, there is less overall gravity pointing down, but because the centrifugal force is getting more horizontal, that gravity has less it needs to counteract. So our equilateral triangle still looks good.

To prove the general validity of our guess, let's see what happens if the interior angles are some arbitrary angle, rather than the 60° they must be. We have to compare the combined vertical force of gravity to the vertical centrifugal force. Using trig, we can find the distance from the test mass to a star in terms of a and θ. Because of the inverse square law of gravity, a is going to be squared. Trig also gets us the vertical component of that force in terms of θ.

On the other hand, centrifugal force depends on the distance to the center of mass, l. But because we only want the vertical component, the actual location of the center of mass is irrelevant and all we need is h, which again can be found in terms of a and θ. As before, centrifugal force also depends on the inverse cube of a, so some canceling of exponents means it's the inverse square of a that shows up.

Because both expressions depend on the square of a, we can get rid of it. Both forces are also equally dependent on the sum of the masses of the two stars, so we can cancel the mass terms, too. This means our equation is now defined entirely in terms of θ. After a little algebra, we can arrive at the following equality:

sin(θ) = 1/2

Everything else in our equation is gone. All that matters is the angle between h and d. Now, I just happen to know that the sine of 30° is 1/2. This means the full interior angle is 60°. With our guess that the test mass is halfway between the two stars, the only possibility is an equilateral triangle with interior angles of 60° and lengths of a. (A similar argument can be made for the horizontal components of the forces.)

I should note that this doesn't prove that there aren't other Lagrange points forming different triangles when the test mass is not half way between. To see that there can't be other points of stability (except on the line joining the two stars), you need to solve for the effective potential of the force fields at work in this system. That can't be done by guessing, but it can be done by drawing! Unfortunately, drawing equipotential surfaces would strain my artistic talents past their breaking point. Here's some computer art instead.

 Credit: NASA / WMAP Science Team