# Does the stress energy tensor scale with the metric tensor?

## Question

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.

## Example

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

Thus

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

We also know:

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

Thus,

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

Thus we have effectively done the following:

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

## Edit:

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

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

?

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.
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.
• 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 $\lambda^2$? Commented May 13, 2022 at 3:50
• @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 $\lambda^{-2}$. Commented May 13, 2022 at 4:26