Deriving Einstein Equations: Questions on Linearity & Symmetry

[Decimal]

Hello,

I am currently taking a course on general relativity with the book "General Relativity: An introduction for physicists" by M.P. Hobson. I am a having a hard time understanding the derivation presented for the Einstein equations. The book states that Einstein proposed the following relation: $$ K_{\mu \nu} = \kappa T_{\mu \nu}$$ Here the tensor $K_{\mu \nu}$ is related to the curvature of spacetime. My book states that this tensor needs to fulfill two conditions. It should contain no terms higher than linear in the second order derivatives of the metric tensor, and the tensor should be symmetric. The book then shows a general form:$$ K_{\mu \nu} = aR_{\mu \nu} + bRg_{\mu \nu} + \lambda g_{\mu \nu} $$ and claims that the last constant $\lambda$ is immediately zero because of the first condition.

I have two questions. I understand that the tensor has to be symmetric, but I am not really sure what they mean with the first condition. I know the tensor is related to the laplacian of the metric tensor, so is that the reason that it should be linear in the second derivatives? Or am I reading this wrong?

As a followup question, why does this condition demand $\lambda$ to equal zero? Why is the last term not linear in the second derivatives of the metric?

Thanks!

 

[PeterDonis]

It should contain no terms higher than linear in the second order derivatives of the metric tensor

The general reason for this condition, as I understand it, is that the field equation involving this tensor (the one Einstein proposed) gives you the equation of motion for the matter, and a tensor that was nonlinear in second derivatives of the metric, or that contained higher order derivatives, would not lead to a reasonable equation of motion.

I know the tensor is related to the laplacian of the metric tensor

No, it isn't. The laplacian only includes spatial derivatives, not time derivatives, so it basically never appears in relativity by itself, since in relativity space and time are not separate.

why does this condition demand $\lambda$ to equal zero?

I'm not sure it does. The usual reason, as I understand it, for setting $\lambda = 0$ is so that the flat Minkowski metric, $g_{\mu \nu} = \eta_{\mu \nu}$, is a solution to the field equation for vacuum, i.e., for $T_{\mu \nu} = 0$. That is only the case if $\lambda = 0$. But that condition has nothing to do with having to be linear in the second derivatives of the metric; that condition is still satisfied for $\lambda \neq 0$, although in a rather trivial sense (see below).

One other possibility is that your textbook is interpreting the condition "no terms higher than linear in second derivatives of the metric" to also imply "must contain terms linear in the second derivatives of the metric". The term $\lambda g_{\mu \nu}$ contains no derivatives of the metric at all, unlike the other two terms. So the $\lambda$ term satisfies the condition "no terms higher than linear in second derivatives of the metric" in the trivial sense that it has no derivative terms at all.

 

[Decimal]

No, it isn't. The laplacian only includes spatial derivatives, not time derivatives, so it basically never appears in relativity by itself, since in relativity space and time are not separate.

Yeah I should have been clearer here. The way my book introduced Einsteins proposal uses the formula for Newtonian gravity $$ \nabla^2 \Phi = 4\pi G \rho$$ and the linearized metric for a weak gravitational field: $$g_{00} = (1+\frac{2\Phi}{c^2})$$ Combining these equations with $T_{00} = \rho c^2$ for a perfect fluid one can derive: $$ \nabla^2 g_{00} = \frac{8\pi G}{c^4} T_{00}$$ Now Einsteins proposal would suggest that $K_{\mu \nu}$ is somehow related to $\nabla^2 g_{\mu \nu}$. This is why I said the $K$ tensor is related to the laplacian of the metric. I understand the Laplacian usually doesn't appear in GR.

I think this relation might have something to do with the linear second order derivatives. The laplacian is obviously linear in second derivatives, so is that why $K_{\mu \nu}$ should be as well?