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. enter image description here

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.


1 Answer 1


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.

  • 1
    $\begingroup$ While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. - From Review $\endgroup$
    – Cross
    Feb 4, 2022 at 10:01
  • $\begingroup$ OK, I edited this with a little more complete answer. However, the intermediate calculations given in the blog post are too long and detailed to be included in the answer. $\endgroup$ Feb 4, 2022 at 18:13

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