I read that solving the Einstein Field Equations can sometimes lead to the problem of non-conservation of energy and that the Landau-Lifschitz Pseudotensor resolves this problem. I can't however find literature on a Schwarzschild solution with a Energy-Stress tensor that also takes into account the Landau-Lifschitz Pseudotensor. Hasn't this been worked out yet? And what happens to the singularity? Is the curvature at the center of black hole still infinite? Wouldn't this imply a divergent energy?

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    $\begingroup$ As the pseudotensor does not represent the physical energy density, this depends on the coordinates used. ui.adsabs.harvard.edu/abs/1991PhLA..157..195V/abstract $\endgroup$ Commented Sep 1, 2021 at 11:56
  • $\begingroup$ @DaddyKropotkin but then how does the pseudotensor resolve the energy conservation when it doesn't represent the physical energy density? $\endgroup$
    – eeqesri
    Commented Sep 1, 2021 at 12:05
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    $\begingroup$ The stress-energy tensor is zero everywhere in the Schwarzschild spacetime. So your question is moot. Also, energy is conserved in the Schwarzschild spacetime, so there is no problem. $\endgroup$
    – safesphere
    Commented Sep 1, 2021 at 19:03
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    $\begingroup$ Ive read the section in L&L and I think you've misunderstood the passage. They demonstrate that conservation of energy can hold for the specific example that they consider, but they conclude: "Thus, in any case, it has no meaning to speak of a definite localization of the energy of the gravitational field in space... On the other hand, from the vanishing of a pseudotensor at some point in one reference system it does not at all follow that this is so for another reference system, so that it is meaningless to talk of whether or not there is gravitational energy at a given place..... $\endgroup$ Commented Sep 1, 2021 at 21:05
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    $\begingroup$ Schwarzschild is a vacuum solution so $G_{\mu \nu}=0$, so the only calculation is just to plug in the metric into the LL pseudotensor, but it doesn't provide any physical insight. $\endgroup$
    – Eletie
    Commented Sep 2, 2021 at 15:25

2 Answers 2


As stated in the comments, for the Schwarzschild vacuum spacetime there is no physical insight to be gained from examining the Landau–Lifshitz pseudotensor. It is of course a coordinate dependent object, and in the standard Schwarzschild coordinates it's nonzero components are $$t^{11}_{LL}= \frac{2 (5r - 6r_s + r \cot^2\theta )}{r^2(r-r_s)} $$ $$ t^{22}_{LL}= -2 \frac{(r-r_s) (\cot^2 \theta -1)}{r^3} $$ $$ t^{23}_{LL}= t^{32}_{LL}= 2 \frac{(r-r_s) \cot \theta }{r^4} $$ $$ t^{33}_{LL}= -\frac{2}{r^4} \quad \quad t^{44}_{LL}= -\frac{2 \csc^2 \theta}{r^4} $$ Change to some other coordinates and you'll have different expressions. This doesn't represent the energy of the gravitational field, and at a point you can always find coordinates where $t_{LL}$ is zero. (I'm pretty sure in isotropic Cartesian coordinates at $R=0$ this object is zero, but you can check).

  • $\begingroup$ The fact that there are θ dependencies all over the place in a spherical symmetric solution is weird, the tt- and rr-components being dependend on θ makes no sense. The rθ-crossterm is also weird since there is no rφ-crossterm. If I take the formula from Wikipedia I get the same result though (with the difference that you set 16πG=1 while I only set G=1), i.ibb.co/K67D8gm/Unbenannt-1.png but on arxiv.org/pdf/1308.0394.pdf#page=8 they say: "The LL formulation is defined to work in Cartesian-like coordinates, in which ηαβ=diag(−1,1,1,1)" $\endgroup$
    – Yukterez
    Commented Nov 6, 2021 at 0:44
  • $\begingroup$ @Yukterez this shouldn't seem weird at all, because the LL is coordinate dependent and this $\theta$ dependence means nothing physical here - it just comes from the fact that $\theta$ explicitly appears in the metric quite a lot. In the paper they're basically saying which coordinates 'look nice' for the LL formulation. $\endgroup$
    – Eletie
    Commented Nov 6, 2021 at 11:58
  • $\begingroup$ If you mean the paper Daddy Kropotkin mentioned in the comments, that is behind a paywall, I tried to get it with Sci Hub but they didn't have it either. Anyways, the paper I found also mentions something like that, so this basically means that you would have to transform the Schwarzschild coordinates from spherical to cartesian like at f.yukterez.net/einstein.equations/files/8.html before you can apply the Pseudotensor and then get more meaningful results than you have with spherical coordinates. In cartesian coordinates everything except the t⁰⁰ component is 0, which looks much better $\endgroup$
    – Yukterez
    Commented Nov 7, 2021 at 4:31
  • $\begingroup$ @Yukterez no, I'm talking about the paper you linked (which probably would also be relevant to the OP as it has some explicit calculations). $\endgroup$
    – Eletie
    Commented Nov 7, 2021 at 11:05

The result Eletie gave is correct if you plug in spherical Schwarzschild coordinates, but since the Landau Lifschitz formulation is designed to work in cartesian coordinates only:

https://arxiv.org/pdf/1308.0394.pdf#page=8 Fromholz, Poisson & Will wrote: "The LL formulation is defined to work in Cartesian-like coordinates, in which ηₐᵦ=diag(−1,1,1,1)"

http://www.elegio.it/mc2/LandauLifshitz_TheClassicalTheoryOfFields_text.pdf Landau & Lifschitz wrote: "On the other hand, we can get values of the tᶦᵏ different from zero in flat space, i.e. in the absence of a gravitational field, if we simply use curvilinear coordinates instead of cartesian."

the results make more sense if you transform your metric into the cartesian form before feeding it into the Pseudotensor, otherwise you get nonsensical $\theta$-dependencies all over the place, infinite energy density at the poles and weird crossterms even if the metric you plug in is spherically symmetric.

In cartesian Raindrop and Finkelstein coordinates the density of the Pseudoenergy is 0 everywhere, but in cartesian Droste coordinates we get

$$t_{LL}^{ \ tt} = -\frac{r_s \ M}{4 \ \pi \ r^2 \ (r-r_s)^2} \ \ , \ \ \ \frac{t_{LL}^{ \ tt}}{g^{tt}} = -\frac{r_s \ M}{4 \ \pi \ r^3 \ (r-r_s)}$$

so in the frame of the free falling raindrop there is no Pseudoenergy, while in the frame of a stationary observer there is. All the other components except the ${t}_{LL}^{tt}$ component are 0 in these coordinates, although it is the same metric Eletie used, just in the cartesian form.

As in the case of the electromagnetic field energy or the Newtonian gravitational field energy the Pseudoenergy density also falls off approximately proportional to $r^{-4}$.

References on the subject can be found at 42(4):261-264.2005 and arxiv:1308.0394, with more examples here.


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