Does Energy in Curved Spacetime Depend on the Choice of Hypersurface?

[Demystifier]

I have a technical question that puzzles me.

Let $$T_{\mu\nu}$$ be a conserved energy-momentum tensor in curved spacetime

$$\nabla^{\mu}T_{\mu\nu}=0$$.

Let $$\Sigma$$ be a curved spacelike hypersurface with the unit vector $$n^{\mu}$$ normal to $$\Sigma$$.
Define energy $$H$$ on $$\Sigma$$ as

$$H \equiv \int_{\Sigma} d^3x |g^{(3)}|^{1/2} n^{\mu} n^{\nu} T_{\mu\nu}$$

where $$g^{(3)}$$ is the determinant of the induced metric on $$\Sigma$$.

The question: Does $$H$$ depend on $$\Sigma$$ ?
(It puzzles me because I have an argument that it does, and another argument that it doesn't.)

 

[Demystifier]

In the meantime, I have resolved the puzzle. Now I know which answer is the correct one and where was the mistake in one of the arguments.

But I will not yet present the solution here, because I find it very instructive for readers to try to find the solution by themselves. In particular, I want to see if someone will make the same mistake that I did.

 

[Demystifier]

Nobody?

OK, I will first present a seemingly correct argument that $$H$$ does not depend on $$\Sigma$$, and then I will explain where is the mistake.

The expression for $$H$$ in the first post above can be written as

$$H = \int_{\Sigma} dS^{\mu} J_{\mu}$$ ...(Eq.1)

where

$$dS^{\mu} = d^3x |g^{(3)}|^{1/2} n^{\mu}$$

is the covariantly defined 3-volume measure on $$\Sigma$$ and

$$J_{\mu} = n^{\nu} T_{\mu\nu}$$

In coordinates in which $$x^0=const$$ on $$\Sigma$$, we have

$$n^{\mu}=\frac{g^{\mu}_0}{\sqrt{g_{00}}}$$

Since it is a function of the metric, and since the covariant derivative of the metric vanishes, we have

$$\nabla^{\mu} n^{\nu} =0$$

We have derived it in special coordinates, but the last expression is covariant, which means that it is valid in any coordinates. Consequently, since the energy-momentum tensor is conserved, $$J_{\mu}$$ is conserved:

$$\nabla^{\mu} J_{\mu} =0$$ ...(Eq.2)

But this conservation implies that (Eq.1) does not depend on $$\Sigma$$.

----------------

Now, where is the mistake? (Eq.2) would imply that (Eq.1) does not depend on $$\Sigma$$ if $$J_{\mu}$$ was a fixed vector field which itself does not depend on $$\Sigma$$. However, $$J_{\mu}$$ does depend on $$\Sigma$$ because it depends on $$n^{\mu}$$ which, in turn, depends on $$\Sigma$$ because it is defined as a unit vector normal to $$\Sigma$$. Consequently, even though (Eq.2) and its derivation is correct, (Eq.1) still depends on $$\Sigma$$.

 

[Demystifier]

Has anyone found my analysis above useful or interesting, in any way?

 

[Phrak]

I'm going to back-up to your post #1 and go out on a limb since I haven't absorbed the other posts you have and am not sure what you're after. n^\mu is normal to \Sigma. I can choose a coordinate system for a particular point in which n^\mu = (1,0,0,0). So n^\nu must also be (1,0,0,0).

Then n^\mu=n^\nu is the product n^\mu n^\nu, so only selects the tensor T^00=T^tt term of the stress energy tensor. Is this right?

 

[Demystifier]

I'm going to back-up to your post #1 and go out on a limb since I haven't absorbed the other posts you have. n^\mu is normal to \Sigma. I can choose a coordinate system in which n^\mu = (1,0,0,0). So n^\nu must also be (1,0,0,0).

Then n^\mu=n^\nu is the product n^\mu n^\nu must select only the tensor T^tt term of the stress energy tensor. Is this right?

Yes, that's right. (That's why H can be interpreted as energy.)

 

[ApplePion]

You seem to be thinking that the energy of the material mass would be conseved, but it is not in general. The gravitational field can add or remove energy to the material mass. If the energy of the material mass were conserved then the slicing should not matter, but if it is not conserved then it would. (Indeed, even if the total energy were conserved in a situation, but some local regions lose energy and some gain, because of failure of simultaneity there would still be a difference for differebt slicing--but this is a bit tricky.)

You can probably convince yourself by brute force by considering a thin tube of mass in a gravitational field, and taking difference slices of it. Indeed, for simplicity make the gravitational field strong. Slices containing more of "the future" will be significantly affected by the transfer of energy to the mass.

 

[Demystifier]

You seem to be thinking that the energy of the material mass would be conseved, but it is not in general.

This is not necessarily what I was thinking. There are actually several inequivalent definitions of global energy in GR, but here I wanted to explore the properties of ONE PARTICULAR definition of energy, which is not necessarily the most physical one. My initial motivation for exploring this particular definition had a different origin, but this is not important here. Here I posted it as an instructive exercise in GR formalism.

 

[atyy]

Has anyone found my analysis above useful or interesting, in any way?

Yes, thanks!

 

[Demystifier]

Yes, thanks!

Good!

 

[Sam Gralla]

I don't think your explanation of the error is correct; at least, there is an earlier error than the one you point out. Your calculation, if correct, would show that any hypersurface-orthogonal timelike vector field is covariantly constant, i.e., $\nabla_\mu n_\nu=0$, which is false. The error is that if you treat $g_{\mu 0}$ as a vector, then it is not true that $\nabla_\rho (g_{\mu 0})=0$. What is true is that $\left(\nabla_\rho g_{\mu \nu} \right)_{\nu=0}=0$, sometimes written as $\nabla_\rho g_{\mu 0}=0$. The index notation is a bit confusing/ambiguous in this sense, which can lead to errors of this sort. It is often safer to introduce a vector field $t^\mu = (1,0,0,0)$ so that $\nabla_\rho (g_{\mu 0})=\nabla_\rho ( g_{\mu \nu}t^\mu ) = g_{\mu \nu} \nabla_\rho t^\mu = g_{\mu \nu} \Gamma^\mu_{\rho \sigma}t^\sigma = g_{\mu \nu} \Gamma^\mu_{\rho 0}$.

In any case defining the energy as the integral of T dotted twice into the normal is a bit misguided. Instead, define a potentially conserved quantity as the integral of T dotted once into the normal and once into an arbitrary vector field v. Then the quantity is actually conserved if v is a killing vector field, $\nabla_{(\mu}v_{\nu)}=0$.

I'll also note for the poster mentioning gravitational energy that the problem at hand has nothing to do with gravity. Instead, the question is whether there is a conserved total energy given a fixed stress-energy tensor on a fixed spacetime. The answer is yes if the spacetime has a timelike killing field.

 

[Demystifier]

It is often safer to introduce a vector field $t^\mu = (1,0,0,0)$ so that $\nabla_\rho (g_{\mu 0})=\nabla_\rho ( g_{\mu \nu}t^\mu ) = g_{\mu \nu} \nabla_\rho t^\mu = g_{\mu \nu} \Gamma^\mu_{\rho \sigma}t^\sigma = g_{\mu \nu} \Gamma^\mu_{\rho 0}$.

This is an extremely useful and illuminating point, thanks!

Your additional two points are also important, but not so new to me.