I had some thoughts from a previous question of mine. If I have a metric $g^{\mu \nu}$

$$g^{\mu \nu} \to \lambda g^{\mu \nu}$$

Then does it automatically follow for the stress energy tensor $T^{\mu \nu}$:

$$T^{\mu \nu} \to \lambda^2 T^{\mu \nu}~?$$

If not, a counter-example will suffice. If yes, a proof will do.


Consider the stress energy tensor for a perfect fluid:

$$T^{\mu \nu} = \left(\rho + \frac{p}{c^2} \right) U^{\mu} U^\nu + p g^{\mu \nu}, $$

Now keeping our notation ambiguous:

$$g^{\mu \nu} \to \lambda^2 g^{\mu \nu}$$

But $$g^{\mu \nu} g_{\mu \nu} = 4$$


$$ g_{\mu \nu} \to \frac{1}{\lambda^2}g_{\mu \nu} $$

We also know:

$$ g_{\mu \nu} U^{\mu} U^\nu = c^2 $$


$$ U^{\mu} U^\nu \to \lambda^2 U^{\mu} U^\nu $$

Thus we have effectively done the following:

$$ T^{\mu \nu} \to \lambda^2 T^{\mu \nu} $$


I seem to have come to the opposite conclusion from the answers posted below. I was hoping someone could address this? Consider the Einstein tensor:

$$G^{\alpha \beta} = (g^{\alpha \gamma} g^{\beta \zeta} -\frac{1}{2} g^{\alpha \beta} g^{\gamma \zeta})( \Gamma^{\epsilon}_{\gamma \zeta , \epsilon} - \Gamma^{\epsilon}_{\gamma \epsilon, \zeta} + \Gamma^{\epsilon}_{\epsilon \tau} \Gamma^{\tau}_{\gamma \zeta} - \Gamma^{\epsilon}_{\zeta \tau} \Gamma^{\tau}_{\epsilon \gamma} )$$

(The Christoffel symbols are scale invariant). Thus

$$ G^{\mu \nu} \to \lambda^2 G^{\mu \nu}$$



2 Answers 2


What you're asking about is a specific case of Weyl transformations of the form $$ g_{ab} \rightarrow e^{-2\phi(x)}g_{ab} $$ where $\phi$ is a constant. Under these, the Ricci scalar is not invariant (even for a constant rescaling with $\phi(x)=c$). To answer your question about $T^{\mu \nu}$, you need to have the precise form of your matter action. In general, you cannot say one way or the other whether the energy-momentum tensor is scale-invariant, or what factors of $\lambda$ it'll pick up. But the relation $T^{\mu \nu} \rightarrow \lambda T^{\mu \nu}$ won't hold in general.

You can find details about this sort've thing when looking at paper at papers about scale-invariance in GR and Weyl gravity (which is constructed to be scale invariant), e.g. https://arxiv.org/abs/gr-qc/0312024. The issues of matter coupling (which is more in line with your question) is a little more complicated and usually taken on a case-by-case basis.


To supplement Eletie's answer, note that under the transformation $g\mapsto \lambda g$ with $\lambda$ an $\mathbb R$-valued constant, the Christoffel symbols remain unchanged because $\Gamma \sim g^{-1}(\partial g + \partial g - \partial g)$, which means that the Riemann tensor $\mathrm {Riem} = \mathrm d\Gamma - \Gamma\wedge \Gamma$ and the Ricci tensor remain unchanged while the Ricci scalar transforms as $R \sim g^{-1} \mathrm{Ric} \mapsto \lambda^{-1} R$. The Einstein tensor therefore remains invariant, since $\mathrm{Ein}\sim \mathrm{Ric} - \frac{1}{2}R g.$

In other words, if $g$ is a solution to Einstein's equations for a given stress-energy tensor $T$, then $\lambda g$ is a solution for the same $T$. If $T\mapsto T'=\lambda T$, then the corresponding solution $g'$ is not related to $g$ by a simple scaling.

  • $\begingroup$ These are good points, thanks for adding. $\endgroup$
    – Eletie
    Commented May 12, 2022 at 19:56
  • $\begingroup$ Umm I think I managed to confused myself. $G^{\alpha \beta} = (g^{\alpha \gamma} g^{\beta \zeta} -\frac{1}{2} g^{\alpha \beta} g^{\gamma \zeta})( \Gamma^{\epsilon}_{\gamma \zeta , \epsilon} - \Gamma^{\epsilon}_{\gamma \epsilon, \zeta} + \Gamma^{\epsilon}_{\epsilon \tau} \Gamma^{\tau}_{\gamma \zeta} - \Gamma^{\epsilon}_{\zeta \tau} \Gamma^{\tau}_{\epsilon \gamma} )$. While the Christoffel symbol remains invariant. Clearly the Einstein Tensor scales by $\lambda^2$? $\endgroup$ Commented May 13, 2022 at 3:50
  • 1
    $\begingroup$ @MoreAnonymous The Einstein tensor is naturally $(0,2)$-tensor, and in that form it is invariant under $g\mapsto \lambda g$. If you choose to work with its $(2,0)$-tensor counterpart - whose components are $g^{\mu\alpha}g^{\nu\beta}G_{\alpha\beta}$ - then that scales like $\lambda^{-2}$. $\endgroup$
    – J. Murray
    Commented May 13, 2022 at 4:26

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