Is this a valid approach to think about the $N$-Body - Problem in General Relativity?

Question

I would like to analyze the following problem: Some point masses, between and around them just empty space ($N$-Body-problem). I would like to analyze the space between and around those point masses. If there are singularities, even if they are only coordinate singularities due to my chosen coordinates (see below), I have not the objective to calculate "through" them. I just leave out the part "behind the singularities" of the manifold (as 'not defined'). Therefore, the whole construction will be a manifold with holes in it.

I'm only interested in approaching a solution for empty space far away from the singularities.

I would like to make the following construction:

1. I assume there's a solution of GR for that problem. Meaning that there exists a well-defined metric tensor $g_{\mu\nu}$ at every point. (Edit: outside the singularities)

2. I chose $(x, y, z, t)$ coordinates and write the metric tensor down in those coordinates. That results in $g_{\mu\nu}$ at every point in $(x, y, z, t)$ coordinates.

3. The determinant of $g_{\mu\nu}$ at every point is well defined (since $g_{\mu\nu}$ is well defined). Lets call it $|g_{\mu\nu}|$.

4. I want to find for every point a coordinate system which leads to a $g'_{\mu\nu}$ where the $g'_{0i}$ with $i = (1,2,3)$ are zero: $g'_{0i}$=0. Furthermore, it shall be $|g_{\mu\nu}|$ = $|g'_{\mu\nu}|$.

$$g'_{\mu\nu} = \begin{pmatrix} g_{00} & 0 & 0 & 0 \\ 0 & g_{11} & g_{12} & g_{13} \\ 0 & g_{12} & g_{22} & g_{23} \\ 0 & g_{13} & g_{23} & g_{33} \end{pmatrix},$$

That means, the coordinate system in general changes from point to point - but the determinant stays the same as in the starting $(x, y, z, t)$ coordinate system.

The reason why I use $(x, y, z, t)$ - coordinates in the beginning is that the Jacobian determinant of this coordinate system is 1.

Is this construction possible or not? (If not, why not?)

Answer 1

The $g'_{0i}=0$ components of the metric tensor are zero in a static gravitational field (constant gravitational fields).

Strictly speaking, only the field created by a single body (immobile=>$g'_{0i}=0$ ) can be constant.

In a system of several bodies, their mutual gravitational attraction generates a movement, which means that the field of these bodies cannot be constant.

Field Theory (Landau, Lifchitz Theoretical Physics Vol 02),§88 :The constant gravitational field.

Answer 2

You seem VERY confused about all of that, as i already said in one of your old posts

N-body problem has nothing to do with the question: once you have a solution to the Einsten equation (i.e. the tensor $g_{\mu\nu}$) it doesn't matter for your question whether it comes from an N-body problem, a rotating disk or as a gift from gods. Known the metric tensor you can do the rest.

I now list all the problems contained in your question:

1 - For point masses you cannot have a metric tensor which is regular everywhere in cartesian coordinates, point masses are singular and have singular behaviour at horizons.

3 - Since the metric is singular, so does the determinant.

4.a - You cannot find a coordinate system for every point, this has absolutely no meaning. A coordinate system is a description of your manifold. It means nothing to have a different choice of coordinates at every point. The answer could stop there since all the rest is plain non-sense.

4.b - As i already told you in your other post, you CANNOT put the shift vector everywhere to zero for every metric, they need to satisfy some conditions. If they are satisfied then you are allowed, with ONE coordinate transformation.

Given 2 metrics in different coordinates systems (which cover ALL THE SPACETIME interested, again) you have:

$$g' = (J^{-1})^T g J^{-1}$$ Where $J$ is the jacobian of the transformation.

Given both metrics you can solve for the jacobian to find the desired coordinate transformation.

As i already told you in the other post, instead of trying to solve the problem nobody solved$^1$, go study GR since you are very confused about every aspect of the theory.

Good standard references are the books by Carroll, Wald, Schutz of Weinberg.

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$^1$ i want to state that to solve the N body problem the difficult is not to have a coordinate transformation to remove the shift vector but to find the metric tensor from the Einstein differential equations: your metric for the N body problem here is god-given, and you don't know its form, so you still did not solve the N body problem.

Answer 3

$x,y,z$ are usually use for cartesian coordinates, if so the metric tensor should be diagonal (with the metric locally flat). Or, if by x,y,z did you meant arbitrary coordinates, then you can always have a metric tensor of that shape by choosing the right coordinates. However the values of the $g_{ij}$'s will depend on position, will not be constant in a N body problem. And not even in a 1 body problem (the schwarzschild solution), unless you are at infinity (in which case you can use cartesian coordinates for the spatial part).