Definition of Energy: Post-Relativity Questions

[maline]

Also in GR the Lagrange density is only determined up to a total four-divergence since

∇μVμ=1√−g∂μ(√−gVμ),​

Ahhh, I see it now! The point is that in curved spacetime, a constant is not, in general, a covariant total divergence! To generate a valid LD in GR from an added constant in SR, we would need to specify some particular vector function with constant divergence in flat spacetime, then generalize to GR by taking the covariant total divergence which will not be constant.

Now I can re-ask my old question: given this non-uniqueness of the Lagrangian density, how is it that taking the variation with respect to the metric yields the uniquely determined Hilbert Stress-Energy Tensor?

Is there a simple explanation of what makes this work out? If not, I'd still like to see the derivation.

 

[vanhees71]

Since the matter action is invariant when adding a total divergence to the Lagrangian, also the energy-momentum-stress tensor in Einstein's equation is invariant:
$$T^{\mu \nu}=\frac{\delta S_{\text{mat}}}{\delta g_{\mu \nu}}.$$
The Einstein-Hilbert action is uniquely defined up to a total divergence by the action of the gravitational field (using the conventions of Landau-Lifshitz vol. IV; 12th German edition)
$$S_{\text{grav}}=-\frac{c^3}{16 \pi k} \int \mathrm{d}^4 q \sqrt{-g} (R+2\Lambda),$$
where $\Lambda=\text{const}$ is the cosmological constant and $R$ the Ricci scalar, defined by the pseudometric $g_{\mu \nu}$. Variation of the toal action with respect to $g_{\mu \nu}$ leads to the Einstein field equations,
$$R_{\mu \nu}-\frac{R}{2} g_{\mu \nu}=\frac{8 \pi k}{c^4} T_{\mu \nu} + \Lambda g_{\mu \nu},$$
of the gravitational field.

 

[maline]

Since the matter action is invariant when adding a total divergence to the Lagrangian, also the energy-momentum-stress tensor in Einstein's equation is invariant:

Tμν=δSmat/δgμν.​

According to Wikipedia, (backed up by a quick google check of the literature)

The variation is of the Lagrangian density itself, not of the action. Are these two forms equivalent? How so?

 

[maline]

Oh, I think I see what you mean. Since the EFE is derived from criticality of the action, it cannot be affected by the divergence term, so uniqueness of the Einstein tensor implies uniqueness of the SET. Still, I don't really see how it is that the divergence term cancels out. Is there a simple way of demonstrating this, perhaps with an example?

 

[vanhees71]

According to Wikipedia, (backed up by a quick google check of the literature)

The variation is of the Lagrangian density itself, not of the action. Are these two forms equivalent? How so?

This expression doesn't make sense. Since $\mathcal{L}$ is local, the functional derivative of it is a Dirac-$\delta$ distribution. That's not the energy-momentum stress tensor, but it's the variation of the action!