In Landau-Lifschitz (Volume II):

Actually, it is not difficult to bring $T^{ik}$ to this form. To do this we start from the field equation


and for $R^{ik}$ we have $$R^{ik}=\frac{1}{2}g^{im}g^{kp}g^{ln}\Big\{\frac{\partial^2g_{lp} }{\partial x^m\partial x^n}+\frac{\partial^2g_{mn} }{\partial x^l\partial x^p}-\frac{\partial^2g_{ln} }{\partial x^m\partial x^p}-\frac{\partial^2g_{mp} }{\partial x^l\partial x^n}\Big\}$$

(we recall that at the point under consideration, all the $\Gamma^i_{kl}=0$). After simple transformations, the tensor $T^{ik}$ can be put in the form

$$T^{ik}=\frac{\partial}{\partial x^l}\Big\{\frac{1}{16\pi\kappa}\frac{1}{-g}\frac{\partial}{\partial x^m}[(-g)(g^{ik}g^{lm}-g^{il}g^{km})]\Big\}$$

My question is: what is the hint to obtain the expression for $T^{ik}$? After a straightforward calculation, I have a long-expression different from this one.


$$ -\frac{1}{2}g^{ik}g^{np}g^{ml}\left(g_{lp},_{mn}+g_{mn},_{lp}-g_{np},_{lm}-g_{lm},_{np}\right) $$

Using the identities in the comments obtained this: Using the identities I have this $$\frac{1}{2}\left((g^{ik}g^{mn}-g^{im}g^{nk})_{,mn}+g^{nl}g^{ik}_{,nl}+g^{im}g^{kp}g_{ln}g^{ln}_{,mp}-g^{ik}g^{ml}g_{np}g^{np}_{,lm}-g^{ik}g^{np}g_{lm}g^{lm}_{,np}\right)$$

How to proceed then?

  • 1
    $\begingroup$ Suggestion: Replace stress-energy tensor with Einstein tensor since this seems purely about calculating curvature. $\endgroup$ – Qmechanic Mar 26 at 11:08
  • $\begingroup$ Following landau Lifschitz trying to obtain stress-energy pseudo tensor. $\endgroup$ – Constantin Mar 26 at 11:14

The derivation you refer to is done locally in a set of coordinates in which $g_{ik,l} = 0$, which can be also characterized as related to a set of Riemann normal coordinates by a constant linear transform. This means, that you can make operations such as this one $$g^{ij} (g_{kl,mn}) = (g^{ij}g_{kl,m})_{,n}$$ Another useful identity is $g^{ij}g_{jk,l} = -g^{ij}_{\;\,,l}g_{jk}$ since $g^{ij}g_{jk} = \delta^i_k\,,\; (\delta^i_k)_{,l} = 0$. One last formulas that you need is $g^{ij}g_{ij,k} = - g^{ij}_{\;\,,k}g_{ij} = g_{,k}/g$, where $g$ is the metric determinant (deriving this formula is a nice little exercise on its own).

It is a little bit laborous to get to the Landau-Lifschitz pseudotensor from the bottom up, but with the formulas above it is really just regrouping the terms in the right way. Start with regrouping the Einstein tensor as a divergence of a 3-index tensor using the first set of identities. Then identify terms that can be written as a $1/g \partial_m$ divergence within this tensor using the $g_{,k}/g$ identity and that should do the trick.

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  • $\begingroup$ How to simplify this? $$g^{nl}g_{nl,mp}$$ $\endgroup$ – Constantin Mar 27 at 5:10
  • $\begingroup$ You have to simplify the entire term: $g^{im}g^{kp}g^{nl}g_{nl,mp} = (g^{im}g^{kp}g^{nl} g_{nl,m})_{,p} = (g^{im}g^{kp}g^{nl} g_{nl,p})_{,m}$ $\endgroup$ – Void Mar 27 at 6:48
  • $\begingroup$ Using the identities I have this $$\frac{1}{2}\left((g^{ik}g^{mn}-g^{im}g^{nk})_{,mn}+g^{nl}g^{ik}_{,nl}+g^{im}g^{kp}g_{ln}g^{ln}_{,mp}-g^{ik}g^{ml}g_{np}g^{np}_{,lm}-g^{ik}g^{np}g_{lm}g^{lm}_{,np}\right)$$ $\endgroup$ – Constantin Mar 27 at 6:59
  • $\begingroup$ How to proceed then? $\endgroup$ – Constantin Mar 27 at 7:02
  • $\begingroup$ As said in the post, in the first step you want to make the Einstein tensor look like a coordinate divergence of a three-index (pseudo)tensor. Then you have to use the identities involving the metric determinant. $\endgroup$ – Void Mar 27 at 7:41

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