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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?)

MartyMcFly
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  • Related questions by OP: https://physics.stackexchange.com/q/788780/
    https://physics.stackexchange.com/q/788632/ . https://physics.stackexchange.com/q/788282/     – MartyMcFly Nov 16 '23 at 19:25
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    Why mention the N-body problem if you're just asking whether metrics can be diagonalized? – Connor Behan Nov 16 '23 at 19:32
  • It's not diagonalizing. With diagonalizing I would change the overall coordinate system. I think that my 1. $|g'|$ =$|g|$ and 2. $|g'{0i}| =0$ leads to a different CS at each point. Furthermore, I'm interested in every possible metric (outside event horizon). And the N-Body Problem seems to me capable of producing any possible (non-singular) $g{\mu\nu} $. Last but not least, the N-Body-problem is unsolved so far, therefore, it cannot be that simple. So, either the approach is not valid, or the idea that one gets new insights to the n-body problem with that approach is not true. – MartyMcFly Nov 16 '23 at 19:46
  • Even the two body problem is unsolved in GR. And point masses are by definition singular. Finding all possible metrics is only meaningful when you specify energy conditions. – Connor Behan Nov 16 '23 at 20:09
  • As the determinants shall be the same, $|g|$ =$|g'|$, the Jacobian determinant of the change of coordinates must be 1. – MartyMcFly Nov 16 '23 at 20:29
  • Energy condition: The same as in our solar system. – MartyMcFly Nov 16 '23 at 20:30
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    A post-Newtonian approach to the $N$-body problem, based on General Relativity, leads to the Einstein–Infeld–Hoffmann equations. (They’re used to calculate accurate spacecraft trajectories, I believe.) I don’t think it has any similarity to what you are suggesting. – Ghoster Nov 17 '23 at 04:46
  • Just a side comment that not all initial configurations of mass points are physically possible. For example, if the total mass of all mass points is contained within the corresponding Schwarzschild circumference, then this configuration cannot evolve from realistic initial conditions as you observe it from outside. – safesphere Nov 22 '23 at 01:22
  • I'm not sure: Wouldn't that mean that you just "debunked" the big bang? – MartyMcFly Nov 22 '23 at 15:34
  • Another question: safesphere says "from the outside", and I know this interpretation of (x, y, z, t) as an observer in the far-away-limit. However, I visualize/imagine the (x, y, z, t) more as 'from behind', unaffected from $g_{\mu\nu}$, but nevertheless 'inside'; unaffected but capable to go+measure everywhere. Is this wrong? Does the observer need a certain position to be???? – MartyMcFly Nov 22 '23 at 15:43

3 Answers3

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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.

The Tiler
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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.

$^1$ 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.

LolloBoldo
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    Your answer is not friendly and doesn't explain anything. I said I want to leave out the singularities. I know that there are problems. I try to understand curvature of spacetime without thinking about singularities. Singularities are very obviously something that cannot be understood using GR. Curvature of spacetime, however, I'm convinced, can be understood. And I simply try. There is curvature without singularities, for example all around us on earth. Why is everyone so keen on the singularities? We cannot get a grip on them anyway using GR. – MartyMcFly Nov 27 '23 at 18:29
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    I'm looking for an answer from a person who has read my question. – MartyMcFly Nov 27 '23 at 18:32
  • I've read your question. Did you skip all the "coordinates change at every point mean nothing" part of my answer? – LolloBoldo Nov 27 '23 at 21:11
  • If you want singularities just read from the 4.a point ahead – LolloBoldo Nov 27 '23 at 21:12
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    I'm NOT interested in singularities. – MartyMcFly Nov 28 '23 at 05:34
  • I'm looking for, for example, the combination of two Schwarzschild-metrics. One singularity at point x1, second singularity at point x2, far away from x1. I know that one cannot combine simply g1 and g2. It's an unsolved problem. However, nature obviously knows how to combine the curvature of more than one object. And it's not that complicated (i mean, there are no "special effects" in their combination) We only don't understand how. – MartyMcFly Nov 28 '23 at 05:58
  • The truth is, (now I got a question ban anyway, I can tell the truth), I am actually looking for an answer to the following question (but did not dare ro ask because I was afraid of a question ban): How about the equation div (det $g_{ij}$) = 4 $\pi$ m? Probably some work left with the constants. i, j are in [1,3]. $g_{00}$ is dependent on $g_{ij}$. That leads in the spherical symmetric case to A=1+1/r, which is very, very similar to the Schwarzschild solution. The wording of the formula is "mass is the source of volume of space" (not curvature-source as EFE) - Could you answer this, please? – MartyMcFly Nov 28 '23 at 06:46
  • Of course it must be div(grad (det ($g_{ij}$))) = 4 $\pi \rho$. $\rho$ is mass density. div (grad(|$g_{ij}$|)) = 4 $\pi \rho$. i, j in [1,3]. – MartyMcFly Nov 28 '23 at 11:35
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$x,y,z$ 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).

Pato Galmarini
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