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

In Landau-Lifschitz (Volume II):

Stress-energy tensor is evaluated in a reference frame where all $$g_{ik,l}=0$$. From Einstein equations:

$$T^{ik}=\frac{1}{8\pi\kappa}\left(R^{ik}-\frac{1}{2}g^{ik}R\right)=(g^{il}g^{km}-\frac{1}{2}g^{ik}g^{lm})(\partial_n \Gamma^n_{lm}-\partial_m \Gamma^n_{ln}+\Gamma^n_{pn} \Gamma^p_{lm} - \Gamma^n_{pm} \Gamma^p_{ln})$$

Taking into account that all $$\Gamma^{l}_{ik}=0$$, $$T^{ik}$$ is derived:

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

Then $$T^{ik}$$ is denoted by partial derivative of $$h^{ikl}$$

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

So, $$(-g)T^{ik}=\frac{\partial{h^{ikl}}}{\partial x^l}$$

Further, stated that in the transformation to arbitrary reference frames $$\frac{\partial{h^{ikl}}}{\partial x^l}-(-g)T^{ik}\neq0$$ and $$(-g)t^{ik}=\frac{\partial{h^{ikl}}}{\partial x^l}-(-g)T^{ik}$$. $$t^{ik}=\frac{1}{-g}\frac{\partial{h^{ikl}}}{\partial x^l}-T^{ik}$$

Where $$t^{ik}$$ is stress-energy pseudotensor and I want to derive it explicitly.

I have calculated straightforwardly the expression for $$t^{ik}$$. After a quite long calculation, there is derived the following expression. Now I am in stuck on how to proceed with the expression. There definitely must not be second-order derivatives of metric tensor. How to proceed? Any hint is appreciated.

See my blog post where the detailed derivations and intermediate calculations are given with the help of Mathematica xAct package. Second derivatives should automatically disappear when you subtract the stress-energy-momentum tensor. More specifically, when you develop the Einstein tensor, you get $$\Gamma^{l}{}_{mn} \Gamma^{p}{}_{qr} g^{im} g^{kq} g_{lp} g^{nr} - \ \tfrac{1}{2} \Gamma^{l}{}_{mn} \Gamma^{p}{}_{qr} g^{ik} g_{lp} \ g^{mq} g^{nr} - \Gamma^{l}{}_{mn} \Gamma^{p}{}_{qr} g^{im} g^{kn} \ g_{lp} g^{qr} + \tfrac{1}{2} \Gamma^{l}{}_{mn} \Gamma^{p}{}_{qr} \ g^{ik} g_{lp} g^{mn} g^{qr} - \tfrac{1}{2} g^{il} g^{km} g^{np} \ g_{np}{}_{,l}{}_{,m} + \tfrac{1}{2} g^{il} g^{km} g^{np} \ g_{mn}{}_{,l}{}_{,p} + \tfrac{1}{2} g^{il} g^{km} g^{np} \ g_{ln}{}_{,m}{}_{,p} - \tfrac{1}{2} g^{ik} g^{lm} g^{np} \ g_{ln}{}_{,m}{}_{,p} - \tfrac{1}{2} g^{il} g^{km} g^{np} \ g_{lm}{}_{,n}{}_{,p} + \tfrac{1}{2} g^{ik} g^{lm} g^{np} \ g_{lm}{}_{,n}{}_{,p}$$ When you develop $$h_{,l}^{ikl} = \left [(-g) \left(g^{ik} g^{lm} - g^{il} g^{km} \right ) \right]_{,m,l}$$ you obtain a mixture of first and second derivatives. However, the second derivatives should be the same as in the Einstein tensor, so they cancel upon subtraction $$t^{ik} = \frac{1}{-g} h_{,l}^{ikl} - T^{ik}$$
It is difficult to check if your second derivatives are correct because some of them are on upper-index metric (4) and others are on Christoffels. However, when canonicalized, they cancel each other and leave only first derivatives $$- g^{il} g^{km} g^{np} g^{qr} g_{pr}{}_{,m} g_{ln}{}_{,q} + g^{ik} g^{lm} g^{np} g^{qr} g_{mr}{}_{,p} g_{ln}{}_{,q} - g^{il} g^{km} g^{np} g^{qr} g_{pr}{}_{,l} g_{mn}{}_{,q} + 2 g^{il} g^{km} g^{np} g^{qr} g_{ln}{}_{,q} g_{mp}{}_{,r} - \tfrac{1}{2} g^{ik} g^{lm} g^{np} g^{qr} g_{ln}{}_{,q} g_{mp}{}_{,r} - g^{il} g^{km} g^{np} g^{qr} g_{lm}{}_{,n} g_{pq}{}_{,r} + \tfrac{1}{2} g^{ik} g^{lm} g^{np} g^{qr} g_{lm}{}_{,n} g_{pq}{}_{,r}$$ which means that you may be on the right track.