How is the zero point of energy defined in general relativity?

[maline]

Which Lagrangian to choose for a given situation must thus follow other principles, like symmetry principles and finally is subject to empirical justification as any postulated physical law used to derive the equations of motion.

So it's true that GR forces a choice of a particular Lagrangian density, where SR would have allowed an equivalence class of Lagrangian densities? Could you say something about the "symmetry principles" that can be used to make this selection?

 

[vanhees71]

GR doesn't force anything. As I said you need to employ (heuristic) principles to write down the Lagrangian for a given physical situation.

A first guess for generalizing physical laws from special to general relativity is to substitute covariant derivatives, wherever you have a partial derivative in the special theory (using Galileian coordinates). That's however not unique, because covariant derivatives do not commute, while partial derivatives do. That's already a problem for naively translating Maxwell's equations (which are 2nd order equations when written in terms of the four-potential). The Lagrangian, however only involves first-order derivatives, and particularly simple only four-curls, i.e., you have
$$\mathcal{L}=-\frac{1}{4} F_{\mu \nu} F^{\mu \nu}, \quad \text{with} \quad F_{\mu \nu} = \nabla_{\mu} A_{\nu}-\nabla_{\nu} A_{\mu} = \partial_{\mu} A_{\nu} -\partial_{\nu} A_{\mu}, \quad F^{\mu \nu} = g^{\mu \rho} g^{\nu \sigma} F_{\rho \sigma}.$$
This you plug into the action
$$S=\int \mathrm{d}^4 q \sqrt{-g} \mathcal{L}$$
leading uniquely to the correct (free) Maxwell equations using Hamilton's principle for the variation of the vector field $A_{\mu}$. The variation with respect to $g_{\mu \nu}$ gives you uniquely the correct symmetric (and gauge invariant!) energy-momentum tensor of the electromagnetic field, and this must (within General Relativity) provide the correct local energy and momentum density of the electromagnetic field, while within Special Relativity the energy-momentum tensor is not uniquely defined from Noether's theorem (applied to space-time translation invariance of the Minkowski space), but only total energy and momentum are uniquely defined.

 

[PeterDonis]

So it's true that GR forces a choice of a particular Lagrangian density

No. The point is that, in GR, adding a constant to the Lagrangian density is not just a "change of zero point" with no physical meaning. It is an actual physical change to the model. Physically, it corresponds to adding a cosmological constant (or more generally changing the magnitude of the cosmological constant), which is a physical change that leads to different predictions for physical observables. Mathematically, GR let's you add any constant to the Lagrangian density that you want; but physically, only one value for that constant will give you a model that matches observations.

 

[maline]

GR doesn't force anything. As I said you need to employ (heuristic) principles to write down the Lagrangian for a given physical situation.

A first guess for generalizing physical laws from special to general relativity is to substitute covariant derivatives, wherever you have a partial derivative in the special theory (using Galileian coordinates)

Mathematically, GR let's you add any constant to the Lagrangian density that you want; but physically, only one value for that constant will give you a model that matches observations.

In SR, whenever we write down a Lagrangian density, we are actually referring to the equivalence class of all LD's that differ from this one by a total divergence (or a total time derivative for actual Lagrangians). When generalizing to GR, this equivalence will no longer obtain, so before even beginning to deal with covariant derivatives etc., we must first select one representative of the class as the "correct" one to work with. Is there any general way to make this selection?
It seems that in many cases, the "correct" LD is also already the most familiar one, such as L=−1/4 FμνFμν for classical EM. Perhaps this is because the familiar forms are usually the most simple or symmetric?

 

[vanhees71]

Also in GR the Lagrange density is only determined up to a total four-divergence since
$$\nabla_{\mu} V^{\mu}=\frac{1}{\sqrt{-g}} \partial_{\mu} (\sqrt{-g}V^{\mu}),$$
and thus
$$S'=\int \mathrm{d}^4 q \sqrt{-g} [\mathcal{L}+\nabla_{\mu} V^{\mu}] = \int \mathrm{d}^4 q [\sqrt{-g} \mathcal{L} + \partial_{\mu} (\sqrt{-g}V^{\mu})]= \int \mathrm{d}^4 q \sqrt{-g} \mathcal{L}=S,$$
i.e., the action doesn't change when you change the Lagrangian by a total divergence.

 

[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!