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While playing around with the Einstein field equations and trying to derive the Kerr metric, I came across the following derivation from Einstein's field equations:

$$R_{\mu\nu} = 8\pi \left(T_{\mu\nu} - \frac{1}{2} T g_{\mu\nu}\right)$$

With the definition $T = T_{\mu\nu}g^{\mu\nu}$.

Now, I'm wondering whether those equations are (all of a sudden) really linear? Have the field equations lost their nonlinearity while the operations that lead to this derivative? Or do I get something wrong, are they still non-linear? The right hand side contains only energy & mass, and the left hand side contains only geometry. To me, it seems that you can just add up the right side $\left(T_{\mu\nu} - \frac{1}{2} T g_{\mu\nu}\right)$ to a new $T'_{\mu\nu}$ - and then, the result is suddenly

$$R_{\mu\nu} = 8\pi T'_{\mu\nu}$$

Which, for me, looks like a linear version of the famous field equations. And, yeah, that's kind of surprising for me, because they used to be nonlinear which causes so many difficulties with their solvation. How can we (have we) to interpret that result??

And, addendum: What would be a possible solution ($R_{\mu\nu}$ and $T_{\mu\nu}$) for the equation $$R_{\mu\nu} = 8\pi T'_{\mu\nu}$$?

BarrierRemoval
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1 Answers1

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Einstein's equations are not linear in the metric tensor $g_{\mu\nu}$, which is the field we want to solve for$^\dagger$. This means that if $g^{(1)}_{\mu\nu}$ and $g^{(2)}_{\mu\nu}$ are two solutions of Einstein's equations, then in general a linear combination like $g^{(1)}_{\mu\nu}+g^{(2)}_{\mu\nu}$ will not also be a solution. This is true no matter what notation you use to write the equations.

Einstein's equations are also not linear in the connection $\Gamma^\lambda_{\mu \nu}$, which is relevant if you want to use a first-order formulation of GR.

$^\dagger$ Technically we would want to solve Einstein's equations plus the equations of motion for all matter fields simultaneously for the metric and the matter fields. If this bothers you, then to remove this small technical complication -- which is irrelevant for the question you are asking -- we can focus on the case $T_{\mu\nu}=0$ in this answer.

Andrew
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  • Ah, that is actually exactly the point which I don't get: That the field equations are not linear in the metric tensor. Thank you very much for pointing that out!! Because you uncovered a big misunderstanding in my view of the field equations. However, you only state that it is the case. Not, why. I don't understand: why are they not linear in the metric tensor? I could just add them up, the two metric tensors. Why is g1+g2 not a solution? I could just add them up, entry per entry, and at least the result is a thing which looks like a metric tensor. Why is it not a solution? – BarrierRemoval May 02 '22 at 05:30
  • @BarrierRemoval If you expand $R_{\mu\nu}$ in terms of the metric tensor, it is just obviously not linear in $g_{\mu\nu}$. It is easiest to start with the Christoffel symbols, $\Gamma \sim g^{-1} \partial g$, which involves the inverse metric. The inverse metric is not a linear function of the metric. (Hopefully it's clear that $(x+y)^{-1}\neq x+y$). The Riemann tensor $R \sim \partial \Gamma + \Gamma^2$ adds terms that are non-linear in $\Gamma$, so makes things more non-linear. Then finally the Ricci tensor and scalar involve contractions of the Riemann tensor with the inverse metric. – Andrew May 02 '22 at 05:35
  • How to expand in terms of the metric tensor (I know what an expansion is, like Taylor series, what you do with Christoffel symbols looks a bit like Taylor series...). How exactly comes the inverse metric into this derivation? Is there a book or a script where I can read through this? Also, I don't know that the Riemann tensor is $R\sim\partial\Gamma +\Gamma^2$ – BarrierRemoval May 02 '22 at 05:45
  • @BarrierRemoval See, for example, Chapters 2-3 of https://arxiv.org/abs/gr-qc/9712019. (The same material will also be in any GR textbook). – Andrew May 02 '22 at 05:46
  • Thanks for that. It's a good point, I see that the nonlinearity is mathematically sound. However, isn't it strange that, in this equation from above, $R_{\mu\nu} \sim T_{\mu\nu} $, there is something on the RHS which can be summed up linearily, but on the LHS you can't sum it up linearly??? – BarrierRemoval May 02 '22 at 18:47
  • @BarrierRemoval I think what you mean by "something on the right hand side which can be summed linearly" is: if you have $2$ fields $\phi$ and $\psi$ which are not coupled, then you can write the total stress energy as a sum of the stress energy for $\phi$ plus the stress energy for $\psi$. However... (a) Neither stress energy will be linear in $\phi$ or $\psi$, in general. (b) If $\phi$ and $\psi$ directly interact, you can't write the stress energy as a sum (since there is some interaction energy), (c) the key point is that the equations are nonlinear in the metric. – Andrew May 02 '22 at 19:00
  • Let me add though, I don't think it is strange that Einstein's equations are non-linear. There are very good physical reasons for it. Namely, that the energy and momentum of everything, including the gravitational field itself, gravitates. So there must be coupling between the gravitational field and itself, or non-linearity. – Andrew May 02 '22 at 19:04
  • Could you please expand on what you mean with "the gravitational field itself" gravitates? – BarrierRemoval May 02 '22 at 19:33
  • @BarrierRemoval https://physics.stackexchange.com/questions/293873/do-gravitons-interact-with-each-other – Andrew May 02 '22 at 19:36