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The energy-momentum tensor is defined by

$$T^{\mu\nu}=-\frac{2}{\sqrt{-g}}\frac{\delta S_m}{\delta g_{\mu\nu}}$$

where $S_m$ is the matter action

$$S_m =\int d^4x\sqrt{-g}\mathcal{L}_m$$

and $\mathcal{L}_m$ is the matter Lagrangian-density. If the total expression is a tensor, and the determinant of the metric is not a Lorentz scalar but rather a density, then that leads me to conclude that

$$\frac{\delta S_m}{\delta g_{\mu\nu}}$$

must be some kind of tensor density, rather than a full-fledge Lorentz tensor. Is this correct? If so, how is that possible since the action integrated over $d^4x$ already contains a $\sqrt{-g}$ in the measure? How is the variation of a Lorentz scalar ($S_m$) with respect to a tensor (the metric) not itself a tensor? When people write the energy-momentum tensor as a variation of the action, do they actually mean a variation of $\sqrt{-g}\mathcal{L}$? In that case I think we're really definining the Energy-Momentum Tensor-Density.

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2 Answers 2

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OP's question touches upon the very definition of a functional/variational derivative:

  1. If we define $$\frac{\delta\phi(x)}{\delta\phi(y)}~=~\delta^d(x-y),\tag{1}$$ then we have to live with consequences that the RHS is a density.

  2. If the theory has a density $\rho$ [e.g. if there's a metric tensor $g$, we can construct $\rho=\sqrt{|\det g|}$], then it is possible to define a covariant functional derivative $$\frac{\delta\phi(x)}{\delta\phi(y)}~=~\frac{\delta^d(x-y)}{\rho(x)},\tag{2}$$ such that the RHS is a scalar.

There is an analogous ambiguity for the functional derivative wrt. other tensor fields, e.g. the metric tensor field that OP asks about: $$ \frac{\delta g_{\mu\nu}(x)}{\delta g_{\alpha\beta}(y)} ~=~\frac{1}{2}\left( \delta_{\mu}^{\alpha}\delta_{\nu}^{\beta} + \delta_{\nu}^{\alpha}\delta_{\mu}^{\beta}\right)\delta^d(x-y) \tag{1'} $$ versus $$ \frac{\delta g_{\mu\nu}(x)}{\delta g_{\alpha\beta}(y)} ~=~\frac{1}{2}\left( \delta_{\mu}^{\alpha}\delta_{\nu}^{\beta} + \delta_{\nu}^{\alpha}\delta_{\mu}^{\beta}\right)\frac{\delta^d(x-y)}{\sqrt{|\det g(x)|}}. \tag{2'} $$

See also this related Phys.SE post for a related issue for the Euler-Lagrange (EL) equations.

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  • $\begingroup$ This is definitely something I hadn't considered, but we have identities like $\frac{\delta g_{\alpha\beta}}{\delta g_{\mu\nu}}=\frac{1}{2}(\delta^\mu_\alpha\delta^\nu_\beta+\delta^\mu_\beta\delta^\nu_\alpha)$. Given the RHS, it looks like this variational derivative still is a tensor, not a density. Is there something I'm missing there? $\endgroup$
    – Adots005
    Nov 10, 2021 at 18:41
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Nov 10, 2021 at 19:13
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If you define $T^{\mu\nu}$ explicitly, by writing $$ \delta S= -\int d^dx \sqrt{g}\, T^{\mu\nu} \delta g_{\mu\nu}, $$ then the invariance of $S$ and $d^dx \sqrt{g}$ under coordinate transformations shows that $T^{\mu\nu}$ is a tensor, rather than a tensor density.

I think the definition in terms of $\delta S/\delta g_{\mu\nu}$ is ambiguous because some people include and some omit the $\sqrt{g}$ in the definition of $\delta S/\delta g_{\mu\nu}$.

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