# Landau-Lifschitz pseudotensor from Einstein equations (part I)

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

$$T^{ik}=\frac{1}{8\pi\kappa}\left(R^{ik}-\frac{1}{2}g^{ik}R\right)$$

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.

$$R^{ik}-\frac{1}{2}g^{ik}R=\frac{1}{2}g^{im}g^{kp}g^{nl}\left(g_{lp},_{mn}+g_{mn},_{lp}-g_{mp},_{ln}-g_{ln},_{mp}\right)-$$

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

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?

• Suggestion: Replace stress-energy tensor with Einstein tensor since this seems purely about calculating curvature. – Qmechanic Mar 26 at 11:08
• Following landau Lifschitz trying to obtain stress-energy pseudo tensor. – 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.
• How to simplify this? $$g^{nl}g_{nl,mp}$$ – Constantin Mar 27 at 5:10
• You have to simplify the entire term: $g^{im}g^{kp}g^{nl}g_{nl,mp} = (g^{im}g^{kp}g^{nl} g_{nl,m})_{,p} = (g^{im}g^{kp}g^{nl} g_{nl,p})_{,m}$ – Void Mar 27 at 6:48
• Using the identities I have this $$\frac{1}{2}\left((g^{ik}g^{mn}-g^{im}g^{nk})_{,mn}+g^{nl}g^{ik}_{,nl}+g^{im}g^{kp}g_{ln}g^{ln}_{,mp}-g^{ik}g^{ml}g_{np}g^{np}_{,lm}-g^{ik}g^{np}g_{lm}g^{lm}_{,np}\right)$$ – Constantin Mar 27 at 6:59