Computing Volume in General Relativity: Use of Tensor & Friedmann Eqns

[Tertius]

TL;DR Summary

Is this simple computation of volume from the timelike energy density term of a tensor correct?

When we compute the stress energy momentum tensor $T_{\mu\nu}$, it has units of energy density. If, therefore, we know the total energy $E$ of the system described by $T_{\mu\nu}$, can we compute the volume of the system from $V = E/T_{00}$?
If it holds, I would assume this would hold for any type of object used to compute $T_{\mu\nu}$, whether it be a perfect fluid or a more complex field (like the inflation field, or EM field).

One interesting consequence is that, since $G_{\mu\nu}/k= T_{\mu\nu}$, the Friedmann equations would suggest that one could compute the total volume of the universe directly from the Hubble constant, $$V = kE/G_{00} = kE/(3H^2)$$

Since we don't actually know the total energy of the universe, I am simply asking if we did, would this method be correct?

 

[PeterDonis]

When we compute the stress energy momentum tensor $T_{\mu\nu}$, it has units of energy density.

Yes.

If, therefore, we know the total energy $E$ of the system described by $T_{\mu\nu}$

In general there is no such thing as "total energy" in GR. Also, $T_{\mu\nu}$, like any tensor, is defined at a single event in spacetime; to cover a region you need a field of such tensors, one for each point. In other words, if you are thinking of a spacetime region, $T_{\mu\nu}$ is a tensor-valued function.

can we compute the volume of the system from $V = E/T_{00}$?

"Volume", meaning 3-volume, the volume of a 3-dimensional spacelike region of spacetime, is frame-dependent in relativity. So in general there is no such thing as "the" volume of a system.

 

[PeterDonis]

the Friedmann equations would suggest that one could compute the total volume of the universe directly from the Hubble constant

This obviously can't be right since the Friedmann equations can describe universes that are spatially infinite (as ours is in our best current model).

 

[Tertius]

Yes, the volume of the universe is relative to each observer. And so is the total energy.

Larger picture coming together hopefully: Does this mean that neither the volume nor the energy of the universe can actually be defined by us? (observers in the universe)

 

[PeterDonis]

the volume of the universe is relative to each observer.

Not if it's spatially infinite. Which, as I said, it is in our best current model.

And so is the total energy.

No, total energy is even more problematic, because there isn't even a single well-accepted definition for it that is frame-dependent, let alone one that is invariant. (There is one obvious invariant in GR that could be interpreted as the "total energy", but unfortunately this invariant is identically zero for every possible system, so it's useless.)

 

[Tertius]

No, total energy is even more problematic, because there isn't even a single well-accepted definition for it that is frame-dependent, let alone one that is invariant. (There is one obvious invariant in GR that could be interpreted as the "total energy", but unfortunately this invariant is identically zero for every possible system, so it's useless

Appreciate the answers.

Are you referring to a timelike Killing vector? And yeah, i am getting what you are saying that neither volume nor energy are defined for the universe.

 

[PeterDonis]

Are you referring to a timelike Killing vector?

No, to the Hamiltonian constraint.

 

[pervect]

Summary:: Is this simple computation of volume from the timelike energy density term of a tensor correct?

When we compute the stress energy momentum tensor $T_{\mu\nu}$, it has units of energy density. If, therefore, we know the total energy $E$ of the system described by $T_{\mu\nu}$, can we compute the volume of the system from $V = E/T_{00}$?

The short answer, as others have mentioned, is no.

If you have access to MTW's textbook, "Gravitation", chapter 5 covers the issue of how volumes are defined and their relation to the stress-energy tensor.

A volume element can be defined by three space-like vectors, forming a parallelepiped. This description of a 3 volume is called the 3-volume. And it exists not on a manifold, but in a tangent space at some point on the manifold.

Hopefully the term "tangent space" is familiar, if not, a bit of review and/or asking in another separate post might be in order.

Meanwhile, "total energy" in GR is a tricky thing. To stick with MTW's approach, though, total energy is a pseudo-tensor, not a tensor. So the argument you propose fails because the total energy is the integral of a pseudotensor, while the volume element, which exists at one point on the manifold in the tangent space of that manifold, is a true tensor.

A discussion of pseudotensors probably also would deserve a separate thread of its own.

Also of note is the fact that the volume 3-form, defined by the parallelepiped I mentioned above, can be related by a mathematical process involving the rank 4 Levi-Civita symbol $\epsilon_{abcd}$ to an alternate form, called the volume 1-form.

This process is related to the idea of the "Hodges dual", usually represented by a "*" operator.

One can also think of the volume 1-form as arising from the unit vector that is perpendicular to all three vectors of the volume 3-form. This approach intentionally glosses over the sign issue, of how one determine the postive sense of the volume.

The dual (this a a normal dual, not the Hodges dual!) of this vector is the volume 1-form. This is a bit confusing, as the word "dual" is used in two different concepts, the dual of a vector (which is the one-form), and the Hodges dual of a one-form, which is a 3-form.

The "forms" above are known as "differential forms", and represent anti-symmetric tensors. That's also a large (but interesting) topic that would be a digression to go into, but ask if it catches your fancy.

The energy momentum 4-vector of the energy and momenutum contained in a 3-volume represented by the volume one form $v_a$ is just $T^{ab} v_a$ where $T^{ab}$ is the stress-energy tensor.

However, you can't simply add together these 4-vectors to get a total vector in the curved manifolds, as every point of the manifold has its own tangent space.

 

[Tertius]

@pervect, Thanks for taking the time to explain that in a detailed way. MTW chapter 5 is indeed what I was looking for.