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The Landau-Lifshitz energy-momentum pseudotensor is defined as follows:

$t^{\mu\nu}_{LL}=-\dfrac{c^4}{8\pi G}\left(G^{\mu\nu}+\Lambda g_{\mu\nu}\right)+\dfrac{c^4}{16\pi G\left(-g\right)}\left(\left(-g\right)\left(g^{\mu\nu}g^{\alpha\beta}-g^{\mu\alpha}g^{\nu\beta}\right)\right)_{,\alpha\beta}$

We have used the comma notation to denote the partial derivative: $\left(A^x\right)_{,\lambda}=\frac{\partial{A^x}}{\partial{\lambda}}$. If this is the case, for the Minkowski metric we end up having:

$\left(-g\right)\left(T^{22}+t^{22}_{LL}\right)=c^6\left(\dfrac{\Lambda r^2}{8\pi G}+\dfrac{1}{8\pi G}\right)\sin^2{\theta}$

And:

$\left(-g\right)\left(T^{33}+t^{33}_{LL}\right)=c^6\left(\dfrac{\Lambda r^2}{8\pi G}+\dfrac{1}{8\pi G}\right)$

Where $T^{\mu\nu}=G^{\mu\nu}=R^{\mu\nu}=R=0$. These clearly not vanish when $r\rightarrow 0$, as they become constants for $\theta \neq 0$ in $t^{22}_{LL}$ and for every value of $\theta$ in $t^{33}_{LL}$. Shouldn't it have to vanish locally, when $r$ is small enough? Or does it only vanish when applying the derivative with respect to every coordinate $x^\lambda$? That is:

$\frac{\partial{}}{\partial{x^\nu}}\left[\left(-g\right)\left(T^{\mu\nu}+t^{\mu\nu}_{LL}\right)\right]=0$

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  • $\begingroup$ How do you end up with a sin²θ in your equation? The pseudotensor only works with cartesian coordinates, see physics.stackexchange.com/a/676035/24093 $\endgroup$
    – Yukterez
    Commented Jul 19, 2023 at 22:22
  • $\begingroup$ Yeh, it seems so. Nonetheless, then it always goes to zero anyway doesn't it? At least when summed with $T^{\mu\nu}$. $\endgroup$
    – Antoniou
    Commented Jul 19, 2023 at 23:43
  • $\begingroup$ They don't always sum to 0, in case of Schwarzschild with locally stationary observers the regular energy tensor is 0 but the pseudotensor isn't, see here at output 9 and 10. In the homogenous FLRW both energies, real and pseudo, add to 0 though, see here, so it depends on the metric and the energy of the local observers defined by your coordinates. $\endgroup$
    – Yukterez
    Commented Jul 20, 2023 at 0:16
  • $\begingroup$ In their derivarion they assumed that we picked a coordinate system such that $\partial_{\sigma}g_{\mu\nu}=0$. So that kind of sentence it to be zero, since only scaled Minkowski spacetimes or similar can be valid solutions. See elegio.it/mc2/… page 316 of the PDF. $\endgroup$
    – Antoniou
    Commented Jul 20, 2023 at 6:09

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