Homogeneity of space implies that the laws of physics are form invariant under translations $(x, y, z)\rightarrow (x+a, y+b, z+c)$. This makes sense for Newtonian mechanics and special relativity.

However, it's not clear to me what spatial homogeneity could even mean in general relativity. If there is mass-energy, then the spacetime curvature varies from place to place. There is no longer a choice of coordinates that correspond to a global inertial reference frame, so how can we talk about translations? Perhaps we can consider geodesic flows in place of translations in the usual sense, but I'm not sure if homogeneity is preserved in such a case. In light of this, how does Noether's theorem imply that momentum is conserved in GR? (Or rather, is momentum conserved in GR?)

  • $\begingroup$ Are you claiming that the Friedmann equations are invalid? $\endgroup$
    – D. Halsey
    Jan 27, 2023 at 0:24
  • $\begingroup$ @D.Halsey My understanding is that the FLRW metric (from which the Friedmann equations are derived) assumes a homogeneous mass-energy distribution in the universe. This is a valid model when you are taking an averaged coarse-grained view of the universe, but on a granular level we know planets, galaxies, filaments all exist. Friedmann equations are a useful approximation on cosmological scales, but matter is not completely homogeneous (otherwise we wouldn't exist). I am asking how do we think about Noether's theorem when we take that inhomogeneity into account. $\endgroup$ Jan 27, 2023 at 0:43
  • $\begingroup$ Even if my understanding is wrong (if it is you can correct me), I would still ask, how does spatial homogeneity and momentum conservation work in the Schwarzschild metric? $\endgroup$ Jan 27, 2023 at 0:44
  • $\begingroup$ I've had an impression from somewhere (unfortunately, I--a layperson--can't remember where) that the Schwarzschild metric is rather artificial or hypothetical, in view of the fact that the collapsed stars comprising most black holes (except such huge ones as the one that resulted from dust collapse in Sagitarrius A) appear to have resulted from the collapse of stars, which generally have at least a faint residual rotation: If I remember correctly, Schawarzschild BH's--if they'd really exist at all--don't rotate, unlike Kerr & Kerr-Newman BH's. $\endgroup$
    – Edouard
    Jan 27, 2023 at 8:54

1 Answer 1


As you note, in general space times, there is no globally conserved quantity momentum.

What we do have in general relativity, is a kind of local conservation law for the stress-energy tensor: $$ {T^{\mu\nu}}_{;\nu} = 0. \tag{1} $$ Wikipedia has a few more details: https://en.wikipedia.org/wiki/Stress%E2%80%93energy_tensor#In_general_relativity_2. (This follows directly from the Einstein field equations and the properties of the Riemann tensor.)

Note the covariant derivative here, in a coordinate system we can decompose this into two pseudo-tensors (that is, coordinate dependent quantities), that represent the "gravitational" and "matter" parts of the stress energy: $$ {T^{\mu\nu}}_{,\nu} + {t^{\mu\nu}}_{,\nu} = 0$$ (the gravitational part $t$ is totally under-specified, as we can add any divergence free part to it.). But we can connect this with our classical understanding that energy is exchanged between the gravitational field and the matter, and we now have a continuity equation for $T + t$, so by the Gauss law we get that "all momentum and energy that flows into a bounded piece of space time must flow out of it".

But this also means, that momentum is not conserved globally, unless the space admits a Killing vector field $\xi^\mu$ that can be understood as linear translation.

A Killing vector field is a vector field whose associated flow leaves the space-time invariant, that is a Killing vector field is a generator of a continuous space-time symmetry. Such Killing vector fields obey the Killing equation $$ \xi_{\mu;\nu} + \xi_{\nu;\mu} = 0. \tag{2} $$

If such a symmetry exists, then we know that an associated conserved quantity must exist by the Noether theorem. We can then contract this quantity with the locally conserved stress-energy tensor as follow: $$ J^\mu = T^{\mu\nu} \xi_\nu $$ And this $J^\mu$ will then correspond to a momentum density and momentum current in the direction of $\xi$. $$ {J^\mu}_{;\mu} = {T^{\mu\nu}}_{;\mu} \xi_\nu + T^{\mu\nu} \xi_{\nu;\mu} = 0 + T^{\mu\nu} \xi_{\nu;\mu} = 0. \tag{3}$$ (Here we used eq. (1) to show the first term is zero, then according to eq. (2) the $\xi_{\mu;\nu}$ is anti-symmetric when swapping indices $\mu \leftrightarrow \nu$, and contracting that with the symmetric tensor $T^{\mu\nu}$ gives zero.)

You can then work out a conservation law for a global quantity under time evolution if the space-time can be foliated by space-like slices by integrating this quantity over a such a slice.

(All this is fleshed out in more detail in about any introduction to general relativity, e.g. Ryder: Introduction to General Relativity or more old-school Landau/Lifshitz Volume 2)


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