## 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

## Wednesday, March 9, 2016

### Lagrange Point 2: Newton's Redemption

This past November, I had the opportunity to tour Goddard Space Flight Center. Although we saw many cool operations (including a gigantic cryogenic chamber!), the most interesting was the under construction James Webb Space Telescope. I had intended to write about the visit at the time, but I spent much of my fall semester trying not to hyperventilate instead. However, we just covered some relevant material in my theoretical astrophysics course, so let's take a look now.

 A full-scale model. Credit: NASA
JWST gets called the successor to Hubble, but calling it the sequel would probably be more appropriate. It promises to explore material untouched by the first one, it's going to have even more spectacular visuals, and it's way over budget and behind schedule. The two features that most distinguish it from Hubble are its size (bigger) and its wavelengths of interest (longer).

Longer means infrared. Being an infrared telescope, JWST will see through dust, directly image planets, and peer further back in time at objects redshifted out of the visible range. But infrared telescopes come with some complications. On Earth, we don't do a lot of infrared astronomy, partly because the atmosphere absorbs too much of it, but also because stuff too cold to emit visible light (basically everything on Earth) is usually spilling out lots of infrared instead. We can't do IR astronomy on Earth for the same reason we can't do visible astronomy during the day: it's too bright.

That's why JWST will be in space. But even in space, the Earth and sun loom large. Keep the telescope too near the Earth, and the Earth warms it up, generating noise in the cameras. JWST must be kept cold, much colder than the objects it wants to look at. The only way to accomplish that is to put it far away from the Earth and hold up a shield to block the Earth and sun. The trick is that you want to be able to block both bodies at the same time, which wouldn't work if you just flung the satellite into any old orbit. The farther you get from the sun, the longer your year (Kepler's third law says the cube of your semi-major axis is proportional to the square of your year), so the sun and Earth will change relative positions in the sky.

You need to find an orbit that's far away, stable, and lets you block two objects at once--tricky. Arranging three objects in space is known as the three-body problem in celestial mechanics , and it has a long history. When Newton first formulated his laws of motion and gravity, he was able to solve the one- and two-body problems. That is, he could tell you how a tiny, insignificant planet would orbit a gigantic star (the one-body problem) or how two comparable objects would orbit each other (the two-body problem), but he was not able to count any higher than 2. Newton reasoned that miniscule interactions from nearby planets would build up over time and slowly destabilize orbits, and he assumed the only solution was divine intervention.

Astronomers, physicists, and mathematicians spent a long time looking for more precise answers. It turns out there is no generic solution to the three-body problem, no simple orbit that works for any configuration of three or more masses. Using perturbation theory, you can account for the infinitesimal, cumulative influences of many bodies over time, but in the long run (millions of years), orbits become chaotic. Chaotic doesn't necessarily mean that a planet will be flung from the solar system, but that we eventually can't say with any precision where in an orbit a planet will be at any given time.

A couple mathematicians were able to work out very specific periodic solutions to what gets called the restricted three-body problem, or the 2+1 body problem: two large gravitating masses, one tiny mass that is virtually insignificant. In just the right location relative to the big ones, the small one can be stable. Nowadays these are known as the Lagrange points, in honor of one of the mathematicians who worked them out (Euler already had enough named after him).

This seems perfect for JWST. If there's a line between the sun and the Earth, we want JWST to be on that line out past Earth. Can we find a Lagrange point there?

 In space, lines are purple.
Well first let's backtrack just a second. There isn't really a line connecting the sun and the Earth, because the Earth is constantly in motion about the sun at ~30 km/s. The only way to draw such a line is if we imagine ourselves moving along at the same angular speed as the Earth so that it appears stationary.

Notice I said angular speed, which is how long it takes to move a given angle rather than a given distance. If you think about a spinning tire, the outer bits are moving faster than the inner bits, because the bigger the radius, the larger the circumference covered in the same amount of time. But they are both covering the same fraction of a circle in the same time, and thus both have the same angular speed. If different bits moved at different angular speeds, they wouldn’t keep the same relative positions and the tire would spin apart.

We want our frame and JWST to be moving at the same angular speed as the Earth. But in establishing this frame of reference, we have invalidated Newton's laws of motion. We are no longer in an inertial frame, which is one moving at a constant velocity. Circular motion is not constant, because velocity includes direction.

What does it mean for Newton's laws to be invalidated? It means that an object not experiencing any net force will seem to accelerate away. For our rotating frame, maintaining circular motion requires constant force toward the center of the circle. Tie a ball to the end of a string and spin the ball in a circle. The tension along the string is the radial force that maintains circular motion. If the ball comes loose, it will fly off in a straight line. But from the frame of the spinning string, which can continue spinning as long as you supply a force, the ball will appear to curve away. This tendency to accelerate away from a spinning frame can be accounted for if we invent a fictitious force--centrifugal force--that acts in opposition to whatever force maintains circular motion--centripetal force.

So if we look at the Earth from a rotating frame, JWST will seem to experience a centrifugal force pushing it away from the Earth. In order to have the telescope remain stationary in our rotating frame, the force from gravity must balance the centrifugal force.

 Doing physics really involves making diagrams like this.
So here's our three-body problem. JWST is pulled inward by the gravity of the sun at a distance of a+d and by the gravity of the Earth at a distance of just d. That sum is:

Fg = Gmsunmjwst/(a+d)2 + Gmearthmjwst/d2

And it's pulled outward by the centrifugal force which results from the angular motion of the system. How do we characterize the centrifugal force? It's the square of the angular speed times the distance from the center of mass (the sun, in this case) times the mass of the accelerating object. The angular speed is inversely proportional to the period, the Earth’s year. So centrifugal force involves the square of the period. Using Kepler’s relation between period and semi-major axis, we can substitute in that quantity (a in our diagram). Doing some algebra, that gives us a centrifugal force of:

Fc = G(msun+mearth)mjwst(a+d)/a3

And we want Fg to equal Fc. If we cancel some stuff out, we arrive at the following expression, which is defined purely in terms of the masses and the distances between them:

msun/(a+d)2 + mearth/d2 = (msun+mearth)(a+d)/a3

We're trying to solve for d, the point at which all these forces cancel out. But there's a problem. If we were to multiply all these terms out (FOIL!), we'd find this was a quintic function, which means there'd be a d5. And there is no equivalent of the quadratic formula for quintic equations. So we have to make some approximations. We have to assume that the sun is so much bigger than the Earth (true, in this case) that the Earth can be ignored whenever the two terms are added together. And we also assume that d is much smaller than a, which lets us do some mathematical tricks. If you make those approximations, and then do some more algebra, you eventually find that:

d = a(mearth/3msun)1/3

That is the location of the second Lagrange point (and the first one, but on the other side). Plugging in the relevant numbers, d = 1.5 million km, which is curiously 1/100 Earth’s distance from the sun. The sun is a little more than a hundred times wider than the Earth, which means that from L2, the Earth and sun appear just about the same size--more or less the moon. And that means JWST can easily block both of them with the same shield. (The similarity in angular size really is a happy coincidence that has to do with an accidental congruence of densities, radii, and that factor of 3 up there. Try it with any other planet and it doesn't work.)

So there you have it. When the combined gravitational pull of the Earth and sun cancel out the centrifugal force pushing JWST away, the telescope remains stationary with respect to Earth’s motion about the sun. It sits 1.5 million km behind the Earth and completes an orbit in a year despite being farther away from the sun.

But that's not quite the end of the story. It turns out that L1, L2, and L3 (on the other side of the sun from the Earth) are only metastable, which means a slight push sends an object flying off into a new orbit. So we can put satellites there, but they require station keeping to prevent them from falling away. L4 and L5, which form equilateral triangles with the two big masses of the 2+1 problem, are stable. Consequently, we actually find families of asteroids called the Trojans at the Sun-Jupiter L4 and L5 points. Also, I’ve totally neglected the Coriolis effect here, which is another fictitious force that pops up when… oh dear, look at that word count.