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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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