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When we have 2 massive bodies coming close together say 2 black holes or 2 massive stars, how do their respective metrics/spacetime curvature combine in the space in between them?

Do we write

$$G_{\mu\nu}^1 + G_{\alpha\beta}^2 = \kappa (T_{\mu\nu}^1 + T_{\alpha\beta}^2)$$

With $G_{\mu\nu}^1$ being star 1 and $G_{\alpha\beta}^2$ being star 2 and same for the stress-energy tensors.

Or do we write it as something else?

I have not seen an example of this so please excuse if it sounds obvious.

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    $\begingroup$ No, you cannot simply add it in this way. Einstein's equations are highly non-linear. In fact, for the two-body problem, only numerical solutions to the Einstein equations are known. $\endgroup$
    – G. Blaickner
    Commented Nov 6 at 12:51

4 Answers 4

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A core feature of GR is that it is a nonlinear theory, meaning you cannot naively combine two solutions to get a new one. This is one of the aspects that make it particularly difficult to handle.

To give a first example, consider a Schwarzschild black hole with some mass. We can try to consider the scenario with two black holes now. However, the resulting gravitational field is not the sum of the two previous ones: in the Schwarzschild solution for a single black hole, the spacetime is time-translation invariant, but in this new setup the two black holes are going to move toward each other and eventually merge. Curiously, in both of these solutions (one black hole or two black holes) the stress tensor is the same ($T_{\mu\nu} = 0$ everywhere).

Even in what regards the Einstein tensor your expression may not be correct. Although it initially may seem like simply summing the Einstein equation for each case, notice the right-hand side of the Einstein equations has an implicit dependence on the actual metric of spacetime. When you bring two sorts of matter together, the metric changes, and hence the stress tensor itself ends up changing as well.

In short, there is no clear cut rule to combining solutions in GR. You have to start from scratch and solve the equations again whenever you change the matter content.

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    $\begingroup$ Thanks, I also read this physics.stackexchange.com/a/721414/231892 $\endgroup$
    – Tachyon
    Commented Nov 6 at 13:01
  • $\begingroup$ (+1) Drawing analogy with pirate ships (youtube.com/watch?v=vLCqSUUOmy0), suppose we have the wave metric for ship A alone in water, and for ship B alone in water. Then this knowledge is near completely useless for predicting the wave metric when both ships are interacting close to each other, right? But again, even using classical methods it is also near impossible to predict the equation of motion of either ship in turbulent waters. If the water is perfectly still, then Newton's inertia predicts the movement, but in turbulent waters, GR seems no worse than alternative methods? $\endgroup$
    – James
    Commented Nov 7 at 1:15
  • $\begingroup$ So it seems that energy is not conserved in GR, and in the context of this answer and the one I linked, I assume that spacetime curvature is not conserved either? Meaning that the combined metrics from 2 bodies may interact non linearly to produce a metric curvature that has greater curvature than any single body's curvature in general? But of course determining this would require advanced numerical methods to solve $\endgroup$
    – Tachyon
    Commented Nov 7 at 11:37
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    $\begingroup$ @Tachyon "Conserved" is a poor choice of words here, but you are correct in the sense that the resulting mass and curvature of, say, two planets together is different than the corresponding case for two planets separately $\endgroup$
    – Níckolas Alves
    Commented Nov 7 at 18:45
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    $\begingroup$ @Tachyon I would expect larger curvatures and masses, because gravity gravitates as well. For example, a black hole is made of vacuum, it is pure gravity gravitating $\endgroup$
    – Níckolas Alves
    Commented Nov 8 at 12:36
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As an addition to the other great answers: The equation you wrote doesn't make sense, because you can't add tensors with different indices and expect the result to be Lorentz-invariant. That's like adding the $x$-component of one vector to the $y$-component of another vector. Sure, you can do that but the result you get will depend on the coordinate system you chose.

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The Einstein Field Equations are not linear, in the sense that if you have two metric tensors that are solutions for a given stress-energy tensor, their linear combination is not (generally) also a solution. This is contrast to e.g. the Schrödinger equation, which is linear.

So beyond a very limited set of known analytic solutions you will have to resort to numerical methods. For the specific case you have mentioned (the 2-body problem in General Relativity) there are a few known solutions for special cases. See e.g. General Relativity 2-Body Closed Form

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    $\begingroup$ The linearity of the Schrödinger equation depends upon the nature of the potential, which can be non-linear. $\endgroup$
    – Simp
    Commented Nov 6 at 12:59
  • $\begingroup$ @Simp I don't think that's the case for the time dependent Schrödinger equation with a potential that is a function of the position: see e.g. physics.stackexchange.com/questions/313740/… $\endgroup$
    – Martin C.
    Commented Nov 6 at 13:52
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    $\begingroup$ In any case it's a linear differential equation $\endgroup$
    – Martin C.
    Commented Nov 6 at 13:53
  • $\begingroup$ Yes this is the linear Schrödinger equation but there are plenty of non-linear ones. $\endgroup$
    – Simp
    Commented Nov 6 at 14:38
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    $\begingroup$ @Simp There may be non-linear generalisations of the Schrödinger equation in certain scenarios, but the Schrödinger equation is by definition linear. It is $H(\psi(t)) = i\psi(t)$, where $H$ is a linear operator acting on vectors $\psi$! $\endgroup$
    – Jannik Pitt
    Commented Nov 8 at 16:55
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Only in the weak limit(e.g Earth+Moon system),generally no because EFE are not linear.

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