A Energy-Momentum Tensor for 2-Body Problem: Approach

captainbleak
Messages
1
Reaction score
0
How do you go about writing down the energy momentum tensor for the 2-body problem. Just looking for the approach.
 
Physics news on Phys.org
That's too vague; what's the precise setup? Are the bodies interacting only gravitationally, or also e.g. electromagnetically (in which case you have to also account for the energy-momentum of the field?). In the simple case of two gravitationally interacting particles, you can write the components in terms of delta functions e.g. ##T^{00}(\mathbf{r}, t) = m_1 \delta(\mathbf{r} - \mathbf{r}_1(t)) + m_2 \delta(\mathbf{r} - \mathbf{r}_2(t))## etc.
 
captainbleak said:
How do you go about writing down the energy momentum tensor for the 2-body problem. Just looking for the approach.
Well, a large part of the problem is defining what you mean by the two body problem. I'd say you're probably looking for a vacuum solution of Einstein's field equations, henceforth EFE, in asymptotically flat space-time with appropriate boundary conditions. But that's my interpretation, it might not be what you're interested in at all. You should be able to reformulate the boundary conditions in question in terms of the "contained mass" and "contained angular momentum", given the metric at infinity. That's actually the way I think of the boundary conditions, but I haven't justified how I'm able to think of it in this way. I believe I've seen that discussed for a more exact meaning for these somewhat ambiguous phrases in MTW's textbook "Gravitation", for what it's worth.

The actually solution, which would be the metric, would have to be done numerically and approximately, there's no exact solution. And it'd be much easier if you were only interested in the solution in the "far field". But by definition, the proposed problem is a vacuum solution, so the stress energy tensor ##T_{ab}## would be zero everywhere, which defines the meaning of the problem. By the EFE, this means that the Einstein tensor ##G_{ab}## would also be zero. Since it's a vacuum soultion, you can equivalently say that the Ricci tensor, ##R_{ab}## is zero.

So, in that simplified case, the stress energy tensor would be zero, as it's a vacuum solution. But that may or may not be the problem you actually want to solve.

Some old related ramblings, https://www.physicsforums.com/threads/linearized-metric-for-gw-emitting-orbiting-bodies.880101/

and

https://www.physicsforums.com/threa...-a-pair-of-bodies-in-a-circular-orbit.881740/

In the later case you can see that I decided to abandon the transverse traceless gauge that is usually used in lineraized theory. This made the linearized equations considerably more complicate due to the lack of the TT gauge conditions,; but essentially allowed me to use spherical coordinates for the far field metric, which I found physically meaningful.

Note that the proposed solution a very rough approximation, zero to order (1/r^2). And because of this, it's only valid for large r, it's hopeless at small r. I'm reasonably sure there are better solutions in the literature, I believe some fairly high order approximations have been done to study binary black hole inspirals. However, I don't have a reference, though I recall reading some back in the day when I was fiddling with this. I also recall they were pretty hard to read :).

Note that there are no singularities in the solution. This is because it's just a far field solution for large r, a much easier problem. I'd expect singularities in the near field, though.

Also note that I I didn't actually present my stumbling attempts to deal with the projection operators I used, though they are briefly referenced in the Lanadau-Lifschitz textbook I referenced in the first link. Instead, I worked something out, and got a result that I was happy with, though I doubt anyone has ever checked it for correctness.

To summarize, I presented a solution for a metric that had the Ricci tensor R (and hence the Einstein tensor G) zero to order 1/r^2 "far away" (i.e. large r) from the pair of orbiting bodies. There's one free parameter in the solution, which I think of as related to the angular momentum contained in the solution by the pair of bodies, though I haven't justified why I think of it that way. Various other parameters (such as the oribital period and the enclosed mass) have been normalized away.

Note that the angular momentum wouldn't actually be constant with time for a more accurate solution, but the basic approximation used ignores this issue completely. From what I remember from my reading, it takes a fairly high order solution to correctly deal with the issue of angular momentum decay, though I don't recall where I read this anymore.
 
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Does the speed of light change in a gravitational field depending on whether the direction of travel is parallel to the field, or perpendicular to the field? And is it the same in both directions at each orientation? This question could be answered experimentally to some degree of accuracy. Experiment design: Place two identical clocks A and B on the circumference of a wheel at opposite ends of the diameter of length L. The wheel is positioned upright, i.e., perpendicular to the ground...
According to the General Theory of Relativity, time does not pass on a black hole, which means that processes they don't work either. As the object becomes heavier, the speed of matter falling on it for an observer on Earth will first increase, and then slow down, due to the effect of time dilation. And then it will stop altogether. As a result, we will not get a black hole, since the critical mass will not be reached. Although the object will continue to attract matter, it will not be a...
Back
Top