MSA-A Gravitational 1-Body Problem

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Table of Contents = A Gravitational 1-Body Problem =

Coordinates
Dan: But I'm still confused. You started talking about a change from absolute to relative coordinates. But in fact we seem to switch from two vectors in an absolute inertial coordinate system to two other vectors, one in the old coordinate system and one in a new, non-inertial coordinate system. What is going on?

Erica: I guess the notation and terminology in physics is rather sloppy. We generally don't try to be logically precise, the way mathematicians are. But I'm glad you're forcing us to be clear about our terms. It helps me, too, to build things up again from scratch.

It seems that there are two ways to use the notion of coordinates. But let me first summarize what we have learned.

We have, so far, dealt with two coordinate frames. We have the inertial one, centered on an arbitrary point, a point that is either at rest or in uniform motion with respect to absolute space. The coordinate axes point in fixed directions with respect to absolute space. And we also have the non-inertial frame, centered on particle 1. And although this frame is non-inertial, because particle 1 feels the force of particle 2 and therefore does not move in a straight line at constant speed, the coordinate axes of the non-inertial coordinate system remain parallel to the coordinate axes of the inertial coordinate system.

This last point is important. When the two particles complete one orbit around each other, the relative vector r 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 r$1$, r$2$, R, r. 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, r$2$ describes the position of particle 2 with respect to the origin, and r 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 r$1$, r$2$, 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 R, r.

So in the second way of speaking, a coordinate transformation means a change between describing the two-body system through specifying r$1$, r$2$ and describing the same system through specifying R, r.

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 d-dimensional space, then a system of two such vectors can also be represented as a single vector in a 2d-dimensional space. And then your second way of speaking with respect to the d-dimensional space boils down to your first way of speaking with respect to the 2d-dimensional space.

Mathematically speaking, each choice of a set of two vectors, whether it is {r$1$, r$2$} or {R,r}, 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.

Equivalent coordinates
Carol: Well, I have one question left. I mentioned that the transformation between the {r$1$, r$2$} coordinate system and the {R,r} was bijective. What that means is that any pair {r$1$, r$2$} corresponds to a unique pair {R,r}, and vice versa.

I think that is true, but I would like to prove it, to make sure, and to see explicitly which pair of {r$1$, r$2$} corresponds to which pair of {R,r} vectors.

Dan: Good questions! Erica has shown how to derive R and r from r$1$ and r$2$, with the definitions above. But when we are given only R and r, can we then really recover the original r$1$ and r$2$?

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 R and r as given, and the question is how to derive the values for r$1$ and r$2$.

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: xʹ and yʹ instead of x and y 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 Rʹ, pointing from the position of particle 1 to the c.o.m.  This means that we can reconstruct r$1$ as the sum of 'R and -Rʹ!

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

So far so good. And this means that we have r$1$=R+(-Rʹ), or in simpler terms

$$

Fine! But how do we compute this vector -Rʹ?

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, rʹ$1$}=0 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 rʹ$2$=r$2$-r$1$-r. Erica, your choice of coordinate frame shifting was brilliant! We have in the new frame:

$$

If we now use Eq. 19 that Dan just derived, and substitute the value of Rʹ that we found in Eq. 21, we get:

$$

Closing the Circle
Dan: And the second particle's position is obtained simply by interchanging the subscripts 1 and 2 everywhere, right?

Erica: Wrong, but almost right: there is an additional sign change. We can show that in the same way as Carol just did, but putting the origin now in the position of particle 2. Or, even simpler, we can look at the picture, which tells us that:

$$

So you see, this boils down to an expression with a plus sign, rather than the minus sign in Eq. 22:

$$

Dan: Okay, okay, I grant you your positive sign. But I'm still not fully satisfied.

Carol: First, starting with {r$1$,r{sub|2}}}, we have derived expressions for {R,r}. Then, starting with {R,r} we have derived expressions for {r$1$,r$2$}. What more could you possibly want?!?

Dan: Let me play the devil's advocate. We are doing science, so I want to have hard evidence! As you already saw, it is all too easy to replace a plus sign with a minus sign and stuff like that, so we'd better make sure we really get things right. Let me try to prove it my way.

Carol: Prove what?

Dan: Prove that we are consistent, and that we can close the circle of transformations, from {r$1$,r$2$} to {R,r} and then back again to {r$1$,r$2$}.

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

$$

Yes, now I am convinced. And I can already see more or less how it works for r$2$. But, to be really sure, I'd like to finish the job:

$$

Carol: That is nice, I must admit, to see the truth in front of us so clearly.

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

Newton's Equations of Motion
Carol: Well, we got a new system of relative coordinates. I presume we're going to use it for something, right?

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 r, 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 $$\mathbf{\hat{r}}$$ pointing in the direction of the second particle, as seen from the position of the first particle:

$$

Using this unit vector, we can rewrite Eq. 27 as:

$$

Dan: I see. So the magnitude is indeed inverse square, but the direction is given by the unit vector, which has length one, and therefor does not influence the length of the acceleration vector. I like this way of writing better, since it brings out the physics more clearly.

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 $$\mathbf{\hat{r}}$$, since it is not used anywhere separately.

Let me also write the acceleration for the second particle:

$$

Carol: Look Dan, another sign change: the force of attraction exerted by the first particle on the second points toward the first particle.

Dan: Hmm, I'm still a bit confused about these signs. When the force points to the first particle, why does that imply a minus sign?

Carol: The easiest way to see this is to take a particular case. Imagine that the second particle is positioned in the origin of the coordinates. Since gravity pulls particle 2 in the direction of particle 1, the acceleration that particle 2 experiences points in the direction of r$1$. Notice that in this particular case r = r$2$-r$1$ ~=~ -r$1$. Therefore, the direction r$1$ is the opposite of the direction of r, hence the minus sign.

Dan: Ah, that is neat. Instead of trying to figure things out in full generality, you take a particular limiting case, and check the sign. Sort of like what Erica did, in constructing here a primed coordinate system. Since you already know the magnitude and the line along which the acceleration is directed, once you know the sign in one case, you know the sign in all cases.

Carol: Yes. Technically we call this invariance under continuous deformation. If you bring the second particle a little bit out of the origin, by small continuous changes, the acceleration between the particles must change continuously as well; it cannot suddenly flip to the opposite direction.

Erica: Neat indeed: this means that understanding the sign in one place let you know the sign in all places. I'll remember using that trick.

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.

$$

and

$$

An Equivalent 1-Body Problem
Dan: The second equation looks like a form of Newton's law of gravity, but what does it mean that the first equation gives just zero as an answer?

Carol: Well, it seems that there is zero acceleration for the position of the center of mass.

Dan: Ah, so the c.o.m. moves with constant velocity? But of course, that is what we found in the one-dimensional case too.

Erica: Yes, and this means that we can choose an inertial coordinate frame that moves along with the c.o.m., and in that coordinate frame the center of mass does not move at all. And to make it really simple, we can choose a coordinate frame where the c.o.m. is located at the origin of the coordinates.

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 r$1$ and r$2$ occur in both equations, indirectly through the fact that r=r$2$-r$1$. 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 M$1$+M$2$. To check this, look at Eq. 22, and replace r$1$ by r and replace M$2$ by M$1$+M$2$.

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 M$1$+M$2$, 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 ε, 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.

Wrapping Up
Carol: Let us gather the formulas we have obtained so far, for our coordinate transformation, from absolute to relative coordinates.

$$

$$

$$

$$

Dan: So from all these equations, what we are going to solve with our computer program is the last one above, right?

Erica: Yes, and let me write the equation once again here:

$$

And there is one more thing: let's make life as simple as we can, by choosing a system of physical units in which the gravitational constant and the total mass of the 2-body system are both unity:

$$

Our original equation of motion now becomes simply:

$$

Dan: I can't imagine anything simpler than that! Let's start coding.