4. A Gravitational 1-Body Problem

4.1. Coordinates

This last point is important. When the two particles complete one orbit around each other, the relative vector will also complete one orbit in the non-inertial coordinate system given in fig. 13.

Now here are the two ways that physicists use `coordinates', at least as I understand it. The first way is to give the coordinates of vectors with respect to the coordinate frame in which they are defined. In that case, we have already dealt with the coordinates of the four vectors . In each case, a single vector, or equivalently the values of the components of that vector, describe the position of one particle with respect to a point in space. For example, describes the position of particle 2 with respect to the origin, and describes the position of particle 2 with respect to the point in space temporarily occupied by particle 1.

The second way to talk about coordinates is to capture the information about a whole system, and not just a single particle with respect to a point in space. We can say that we have described a two-body configuration completely when we give the information contained in the pair of vectors , or equivalently, in the values of their four components in two dimensions, or their six components in three dimensions. But we can also describe the same two-body configuration completely by giving the information contained in the pair of vectors .

So in the second way of speaking, a coordinate transformation means a change between describing the two-body system through specifying and describing the same system through specifying .

In the beginning of our session, I used the second way of speaking. But then, when we started talking about coordinate systems, I guess I slipped into the first way of speaking.

Dan: Thanks for separating those two ways of speaking. I'll have to go over it a few times more, to get fluent in this way of thinking, but I'm beginning to see the light.

Carol: Now the beautiful thing is: it is possible to use the same language for both ways of speaking. This is something else I learned in our `computer game' class as we called it, although it was really titled as a `geometrical representation' class. If a single vector lives in a -dimensional space, then a system of two such vectors can also be represented as a single vector in a -dimensional space. And then your second way of speaking with respect to the -dimensional space boils down to your first way of speaking with respect to the -dimensional space.

Mathematically speaking, each choice of a set of two vectors, whether it is or , determines a single point in the direct product of two copies of the base space in which the single vectors live. In our case, we have started from two-dimensional vectors, so the space for pairs of vectors is four-dimensional. And the coordinate transformation that Erica has introduced is really a bijective mapping between two four-dimensional spaces.

Dan: Just when I thought I understood something, Carol manages to make it sound all gobbledygook again. I'll just stick with Erica's explanation.

4.2. Equivalent coordinates

Erica: Yes, we can, or at least I'm pretty sure we can. But we'll have to scratch our heads a bit to write it down. Let us start with figure 10. We can use the same figure, but now we should consider and as given, and the question is how to derive the values for and .

When I took my classical mechanics course, many homework questions were of this type, and generally they involved a clever form of coordinate transformation. Hmmm. Let's see. Right now the origin is at an arbitrary point in space. We could shift to a coordinate frame that is centered on one of our three special points; the position of particle 1, the position of particle 2, or the position of the c.o.m.

Well, why not start with particle 1, and see what happens. Let me draw the new coordinate frame, using primed symbols: and instead of and for the coordinate axes.

Carol: Ah, I see, that is a great move, really! In this new coordinate system, the c.o.m. position is given by the vector , pointing from the position of particle 1 to the c.o.m. This means that we can reconstruct as the sum of and !

Dan: Wait a minute, not so quick. I don't see that yet. Let me try to take smaller steps. The vector must point, by definition, in the opposite direction of . So you can go from the old origin to by first following the vector from start till tip, and then following the vector , which conveniently starts at the tip of . And the tip of lands on particle 1.

So far so good. And this means that we have , or in simpler terms

Fine! But how do we compute this vector ?

Carol: Elementary, my dear Watson. In the old coordinate frame, we had Eq. (17) which gave us the position vector of the c.o.m. in that frame. Let's write it again:

This expression is valid in any coordinate frame, so we can use it for the new primed coordinate frame as well. In the new frame, because the new origin lies smack on the first particle, so that particle has distance zero to the new origin. And the position vector of the second particle is given as . Erica, your choice of coordinate frame shifting was brilliant! We have in the new frame:

I will take Eq. (22) and then use the original definitions Eqs. (17) and (18):

Erica: I agree. Okay, all three of us are happy now. Let's move on!

4.4. Newton's Equations of Motion

Erica: Yes, time to rewrite Newton's equations of motion into the new system. For the one-dimensional case above, we used Eqs. (27) in scalar form. Let me write it in vector form. So this is the equation for the acceleration of the first particle, due to the gravitational force that the second particle exerts on it:

Here I have used the abbreviation

for the absolute value of the vector , which in our two-dimensional case can be written as

while in general, in 3D, it will be

I'm glad you both have at least some familiarity with differential equations. It may not be a bad idea to brush up your knowledge, if you want to know more about the background of Newtonian gravity. There are certainly plenty of good introductory books. At this point it is not necessary, though, to go deeply into all that. I can just provide the few equations we need to get started, and for quite a while our main challenge will be to figure out how to solve these equations.

Dan: Glad to hear that! But I'm puzzled about one thing. Why is there a third power in the denominator? I thought that gravity shows an inverse square attraction, not an inverse cube! And after all, that was what we wrote in Eq. (27).

Erica: Yes, the magnitude of the acceleration is indeed proportional to the minus second power of the separation. However, we also need to indicate the direction of the acceleration. We can define a unit vector pointing in the direction of the second particle, as seen from the position of the first particle:

Carol: I guess that Erica wrote it in the form of Eq. (27) because it will be easier to program that way.

Erica: Indeed. Once you get used to this way of writing the equations of motion, there is no need to introduce the new quantity , since it is not used anywhere separately.

Let me also write the acceleration for the second particle:

Okay, onward with the equations of motion. Given Eqs.(27) and (33), we can calculate the accelerations for the alternative coordinates by using the defining equations (17) and (18), as follows.

Dan: That does make life simple. In this coordinate frame, all the information about the motions in the two-body problem is now bundled in Eq. (35). It almost looks as if we are dealing with a 1-body problem, instead of a 2-body problem!

Erica: Yes, this is what I meant when I announced that we could map the two-body problem into an equivalent one-body problem, for any choice of the masses. The original equations (27) and (33) are coupled: both and occur in both equations, indirectly through the fact that . In contrast, the equations (34) and (35) are decoupled.

A clear way to show this is to draw two separate figures, Figs. 12 and 13. The c.o.m. vector in Fig. 12 moves in a way that is totally independent of the way the relative vector in 13 moves. The c.o.m. vector moves at a constant speed, while the relative vector moves as if it follows an abstract particle in a gravitational field.

To see this, notice that Eq. (35) is exactly Newton's equation for the gravitational acceleration of a small body, a test particle, that feels the attraction of a hypothetical body of mass . To check this, look at Eq. (22), and replace by and replace by .

Dan: Indeed. So the relative motion between two bodies can be described as if it was the motion of just one body under the gravitational attraction of another body, that happens to have a mass equal to the sum of the masses of the two original bodies.

Erica: Exactly. And to complete this particular picture, we have to make sure that the other body stays in the origin, at the place of the center of mass. We can do this by giving our alternative body a mass zero. In other words, we consider the motion of a massless test particle under the influence of the gravitational field of a body with mass , that is located in the center of the coordinate system.

Carol: I prefer to give it a non-zero mass. No material body can have really zero mass. Instead, we can consider it to have just a very very small mass. We could call it , as mathematicians do when they talk about something so small as to be almost negligible.

Dan: If you like. I'm happy with the physical limit that Erica mentioned, rather than the type of mathematical nicety that you introduced. Zero I can understand. Because the test particle has zero mass, it exerts zero gravitational pull on the central body. Therefore, the central body does not move at all, and the only task we are left with is to determine the motion of the test particle around the center.

And that is a one-body problem. Okay, I now see the whole picture.

4.6. Wrapping Up