I am trying to prove that the stress-energy-momentum Hilbert tensor satisfies a conservation law if one assumes diffeomorphism invariance of general relativity. I have taken the definition of the SEM Hilbert tensor field to be: $$T^{\mu \nu}=\frac{-2}{\sqrt{-det(g)}}\frac{\delta(\mathcal{L}_{M}\sqrt{-det(g)})}{\delta g_{\mu \nu}}$$ $\delta_{X}S_{M}$ is the variation originating from some difffeomorphism acting on the manifold, which corresponds to some vector field $X$ in $\Gamma(TM)$. Following the calculation performed by Uldreth in this comment: https://physics.stackexchange.com/a/394031/202198

I come upon the following expression:

$$ 0=\delta_{X}S_{M} = -\int_{M} T^{\mu \nu} (\nabla_{\partial\mu} X)_{\nu} \sqrt{-det(g)}d^{4}x $$ Now, integrating it by parts, at which I fail miserably, ought to yield something resembling (I think!):

$$ 0 = \delta_{X}S_{M} = \int_{M}\text{Div(...)} + \int_{M} (\nabla_{\partial\mu} T)^{\mu \nu} X_{\nu} \sqrt{-det(g)}d^{4}x $$

The rough idea I have is to try to use some covariant Stokes-Gauss law which I cannot really recognise in my situation - and then cross the boundary terms out due to the variation vanishing there. I am not even sure what should be in the first integral - the divergence of some tensor? Perhaps multiplied by $\sqrt{-detg}$?

Could someone help me work out in detail the steps leading to the final expression? Additionally, but not necessarily, how could one formulate all of the above in coordinate-free notation, where the Stokes theorem is much easier to operate with for me?

  • $\begingroup$ The energy momentum does not generally vanish in a diffeomorphism-invariant theory; rather, it is conserved. See e.g. the disucssion around Eq. (5.38) in arxiv.org/abs/gr-qc/9712019. $\endgroup$
    – Toffomat
    Jul 23, 2018 at 11:14
  • $\begingroup$ Yes, you're absolutely right! I must have had in my mind the idea of variation vanishing on the boundary. I have edited the question now. $\endgroup$
    – K.T.
    Jul 23, 2018 at 14:55

1 Answer 1


Your question is essentially about integral theorems. So let us see. You have the expression (up to a sign) $$ I=\int T^{\mu\nu}\nabla_\mu X_\nu\ \sqrt{-\det g}d^nx. $$

Utilizing the derivation property of $\nabla$, we can write this as $$ I=\int\nabla_\mu(T^{\mu\nu}X_\nu)\sqrt{-\det g}d^nx-\int \nabla_\mu T^{\mu\nu}X_\nu\ \sqrt{-\det g}d^nx. $$

So in the divergence term, you have the divergence of a vector field, which is $Y^\mu=T^{\mu\nu}X_\nu$.

The covariant formulation of Gauss' theorem is as follows. Let $\mathcal D$ be a regular domain in the manifold, with boundary $\partial\mathcal D$. Let us denote the coordinates in $M$ with $x$ and the coordinates in $\partial \mathcal D$ by $\xi$. The induced metric is $h_{ij}=(\phi^\ast g)_{ij}$ where $\phi:\partial\mathcal D\rightarrow M$ is the inclusion map.

The Gauss' theorem is $$ \int_{\mathcal D}\nabla_\mu X^\mu\sqrt{-\det g}d^nx=\int_{\partial\mathcal D}n_\mu X^\mu \sqrt{|\det h|}d^{n-1}\xi. $$

Here $n$ is the (usually taken to be) outward-pointing unit normal to the boundary. The boundary's orientation is defined such that if $\{n,e_2,...,e_n\}$ is an positively oriented orthonormal basis of $M$ (along $\partial\mathcal D$), then $e_2,...,e_n$ is considered to be a positively oriented orthonormal basis of $\partial\mathcal D$.

Despite the index notation, this is essentially already coordinate free. Let $đ\mu_g$ be the $n$-form on $M$ whose local expression is given by $$ đ\mu_g=\sqrt{-\det g}dx^1\wedge...\wedge dx^n $$ and let $đ\mu_h$ be the $n-1$-form on $\partial \mathcal D$, whose local expression is given by $$ đ\mu_h=\sqrt{|\det h|}d\xi^2\wedge...\wedge d\xi^n. $$

We define for any vector field $X$ the divergence as either $\text{div}X=\text{Tr}\nabla X$ or as $\text{div}(X)đ\mu_g=\mathcal L_Xđ\mu_g$, two expressions, whose equivalence can be readily verified. Then we have $$ \int_\mathcal D \text{div}Xđ\mu_g=\int_{\partial \mathcal D}\langle n,X\rangle đ\mu_h. $$

This result actually follows from the generalized Stokes theorem for differential forms, which is $$ \int_{\mathcal \partial D}\omega=\int_\mathcal D d\omega $$ (where $\omega$ is an $n-1$-form). To see that this does follow from the generalized Stokes theorem, consult Introduction to Smooth Manifolds by John Lee or General Relativity by Straumann. The omitted proofs are also found in these books.

  • $\begingroup$ Thank you for answering in such detail! I suppose I have to familiarise myself with books you recommended. $\endgroup$
    – K.T.
    Jul 23, 2018 at 15:00
  • $\begingroup$ Does Gauss' theorem work if the covariant derivative is not the metric derivative respect to the metric in the volume form? In other words, does Gauss' theorem need a specific connection to work? $\endgroup$ Apr 4 at 18:46
  • $\begingroup$ @P.C.Spaniel If you can reduce it to an exterior derivative, then yes. If $\mu$ is a volume form that is constant with respect to $\nabla$, and $\nabla$ is torsionless, then $\mathrm{Div}_\nabla(X)\mu=d(X\rfloor\mu)$, so in this case works. When $\nabla$ has torsion, it is more complicated, and the commentspace is too short to properly work it out. $\endgroup$ Apr 4 at 19:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.