# Invariant Mass-Energy in FRW Spacetime

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• Tertius
In summary, GR has very limited situations in which a total mass-energy can be defined. The Komar mass, for example, requires the presence of a timelike killing vector field and an asymptotically flat spacetime. The spatial curvature changes are accounted for in the proper volume element and the integration bounds are defined at infinity where it is asymptotically flat. However, the FRW metric does not have a timelike killing vector field, so the Komar mass cannot be defined. Instead, the trace of the energy momentum tensor can be computed, but it is not equivalent to the invariant mass energy density. To find the total energy contained in a spacelike slice for FRW spacetime, one could integrate ##T_{ab
Tertius
GR has very limited situations in which a total mass-energy can be defined. The Komar mass, for example, requires the presence of a timelike killing vector field and an asymptotically flat spacetime. Basically, if the metric change with time or it's spacelike curvature does not flatten out asymptotically, you cannot define the Komar mass integral. From the derivation, if you can find a timelike killing vector you can define a Noether conserved quantity in the timelike direction, because it is continuously symmetric. Then, the spatial curvature changes are accounted for in the proper volume element, and the integration bounds are defined at infinity where it is asymptotically flat.
Obviously the FRW metric does not have a timelike killing vector field, so this fails immediately.
We can, however, compute the trace of the energy momentum tensor. This is commonly interpreted as the invariant mass energy density. If we then take the trace of the energy momentum tensor and integrate it over a certain spacelike region, with the proper volume element correction (from the spacelike surface metric ##\gamma##), we have
$$\int T^{\mu\nu} g_{\mu\nu} \sqrt{\gamma} dx^3$$
This will be a function of time. I would expect the result to be the invariant mass-energy of that region of space as a function of time. Is this actually observer (defined as a timelike worldline) independent? If we take the spacelike portion to be flat, does this integration actually work? (it doesn't work at infinity because the answer would be infinite mass)

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Tertius said:
The Komar mass, for example, requires the presence of a timelike killing vector field
Yes.

Tertius said:
and an asymptotically flat spacetime.
No. That is a requirement for the ADM mass and the Bondi mass (which do not require a timelike Killing vector field), but not for the Komar mass.

Tertius said:
the spatial curvature changes
There won't be any if the spacetime has a timelike KVF. That's the point of the requirement: in a spacetime with a timelike KVF, the metric (and hence the curvature) is constant along every integral curve of the KVF.

Tertius said:
We can, however, compute the trace of the energy momentum tensor. This is commonly interpreted as the invariant mass energy density.
No, it isn't. The energy density observed by an observer with 4-velocity ##u^a## is ##T_{ab} u^a u^b##. That's not the same as ##T^a{}_a##, which is the trace.

If your goal is to find an integral for "total energy contained in a spacelike slice" for FRW spacetime, the most obvious thing to do would be to integrate ##T_{ab} u^a u^b## using the 4-velocity of comoving observers as ##u^a##, over a spacelike slice of constant comoving coordinate time, i.e., a spacelike slice that is everywhere orthogonal to ##u^a##. This will be infinite for flat and open FRW spacetimes, but finite for closed ones.

I should have specified 'the spatial curvature changes' I was referring to a Schwarzschild metric, which has a timelike KVF, but has spatial curvature.

From my understanding, the trace is a coordinate independent quantity, and should be observer independent because if any timelike observers were to sum up the invariant masses of all of the particles in a spacelike region, they would calculate the same total invariant mass. What am I missing here? Is it because the trace ##T## is not an observable quantity?

Tertius said:
I should have specified 'the spatial curvature changes' I was referring to a Schwarzschild metric, which has a timelike KVF, but has spatial curvature.
Yes, but along any integral curve of the timelike KVF, the spatial curvature does not change. Also, the spatial curvature of the surfaces orthogonal to the timelike KVF does not change from surface to surface. So there is no useful notion of "spatial curvature changes".

Tertius said:
From my understanding, the trace is a coordinate independent quantity
Yes.

Tertius said:
and should be observer independent
Yes.

Tertius said:
if any timelike observers were to sum up the invariant masses of all of the particles in a spacelike region, they would calculate the same total invariant mass.
This seemingly innocuous statement hides a number of important complexities. But we don't even need to delve into those complexities to give the simple response that the trace of the SET has nothing to do with the invariant mass of anything.

Another simple response is that, if we are modeling matter using an SET, we are assuming the matter is continuous, not made of "particles". (Technically, there are ways of constructing a sort of "stress-energy tensor" using delta functions to represent particles, but such methods quickly become very cumbersome and nobody uses them for any real problem solving.)

Tertius said:
What am I missing here? Is it because the trace ##T## is not an observable quantity?
No, it's because the trace is the wrong quantity to look at if you are interested in the things you say you are interested in.

Tertius said:
I originally thought about the trace as the invariant mass density of the system because of this paper: https://projecteuclid.org/journals/...ntum-tensor-and-the-tensor/cmp/1103860087.pdf
But I should have read further, as it maybe does not apply to curved spacetimes.
More precisely, it does not apply to GR. It is a paper about quantum field theory, not GR. You should not be reading this paper to learn about GR. You should be reading papers or textbooks about GR. Those will give you a much better understanding of the physical meaning of various possible invariants that can be formed from the stress-energy tensor.

Tertius said:
If the trace of the SEM does not represent the invariant mass, what does it represent, precisely?
That depends on the specific kinds of stress-energy that are present. The only way to figure it out is to find an expression for the SET in terms of the kinds of stress-energy that are present, and take its trace and see what you get.

For example, the stress-energy tensor for a perfect fluid with energy density ##\rho## and pressure ##p## (both quantities being measured by an observer at rest in the fluid) is

$$T_{ab} = \left( \rho + p \right) u_a u_b + p g_{ab}$$

where ##u_a## is the 4-velocity of the fluid and ##g_{ab}## is the metric tensor. Taking the trace of this is straightforward:

$$T^a{}_a = \left( \rho + p \right) u^a u_a + 4 p = - \left( \rho + p \right) + 4 p = - \rho + 3 p$$

where we have used the fact that the 4-velocity is a timelike unit vector.

Tertius said:
How would we observe the trace?
Generally you wouldn't observe it directly. You would observe other quantities and compute it from those. For example, for the perfect fluid above, you would measure its energy density and pressure and calculate the trace from that using the above formula.

vanhees71
Right, so the trace with a perfect fluid is the expected $$-\rho + 3p$$
For dust, we use ##p=0##, and for photons (traceless) we use ##\rho=3p##
So a timelike KVF, if it exists, can give us the total energy.
A spacelike KVF, which does exist in FRW spacetimes, can give us momentum.

So in FRW, we can rigorously define momentum density because of continuous spatial symmetry but we cannot get an energy density.

In flat space, we can get invariant mass squared by taking the square of a momentum vector. $$p^\mu p_\mu = -m^2$$
This effectively subtracts the squared momentum from the squared total energy.

Even though we can get momentum density from spatial symmetry, we still cannot get energy density from temporal symmetry, so is there no hope of defining an invariant mass quantity in FRW?

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Tertius said:
Even though we can get momentum density from spatial symmetry, we still cannot get energy density from temporal symmetry, so is there no hope of defining an invariant mass quantity in FRW?
An invariant mass quantity of what?

vanhees71
Tertius said:
A spacelike KVF, which does exist in FRW spacetimes, can give us momentum.
No, it can't. See below.

Tertius said:
in FRW, we can rigorously define momentum density
No, we can't. See below.

Tertius said:
In flat space, we can get invariant mass squared by taking the square of a momentum vector.
Yes, but first you have to find the 4-momentum vector. It isn't just handed to you. And there isn't always a well-defined 4-momentum vector at all.

For an isolated system in an asymptotically flat spacetime, you can find a well-defined 4-momentum vector for the system by a suitable integral of the stress-energy tensor. The norm of this 4-momentum then gives you an invariant mass, which allows you, for many purposes, to just treat the system as a point particle with that invariant mass and 4-momentum vector. The particle's "energy" and "momentum" are then just the appropriate components of the 4-momentum vector.

Note that doing the above does not, in general, require the spacetime to have any KVFs. Basically what you're doing is approximating the finite-sized "world tube" in spacetime that is occupied by the object, as a one-dimensional worldline followed by a point particle.

The key limitation is that the system must be isolated (its "world tube" must be of finite "width") in an asymptotically flat spacetime. Obviously this does not apply to any FRW spacetime since none of them are asymptotically flat and none of them describe isolated systems.

Killing vector fields give different ways of defining "energy" and "momentum" by making use of Noether's theorem. While there are special cases (for example, an asymptotically flat spacetime containing an isolated system that is also stationary) where the definitions give the same or compatible answers, there is nothing that ensures that that will always be the case, or that the Noether's theorem answers derived from KVFs will have the same physical interpretations as the "components of 4-vector" that you are used to.

vanhees71
I hadn't thought of it in terms of needing isolation before. But I see what you are saying.

I should have specified my flat space 4-momentum comment was just an example (a mostly not precise relationship) of what I am trying to understand about curved spacetime.

To be precise, I want to find the invariant mass in an FRW universe. Which I can only get a finite answer if it is closed. I understand I cannot get a total energy, but I don't understand why a spacelike KVF in FRW spacetimes does not give me a conserved momentum density. It seems from Noether's theorem that would be the case.

Thanks for the help, btw.

Tertius said:
I want to find the invariant mass in an FRW universe.
There isn't one.

Tertius said:
Which I can only get a finite answer if it is closed.
No, even in that case the thing you get is not an invariant mass, because it's not the norm of a 4-momentum vector of an isolated object. It's a property of the spacetime. Spacetimes don't have 4-momentum vectors.

Tertius said:
I don't understand why a spacelike KVF in FRW spacetimes does not give me a conserved momentum density. It seems from Noether's theorem that would be the case.
I think you are waving your hands and using your untrained intuition instead of actually doing the math. Try doing the math. Be sure you are very careful and rigorous about formulating Noether's theorem. What quantity actually appears in a rigorous statement of that theorem? (Hint: it's not the stress-energy tensor.)

That's fair. Just didn't want the thread to die. I'll do the maths and hopefully have a more detailed thread later on.

@PeterDonis
Ok, I've had some time to evaluate my thinking here. Using Noether's theorem results in a stress energy momentum tensor that is defined using translational symmetries. These are not only usually non-existent in curved spacetime, but there are other symmetries, such as gauge symmetry that are important. In curved spacetime the SEM tensor is defined using the Einstein-Hilbert action. This makes it symmetric and gauge invariant, unlike the one from Noether's theorem. And yes the trace is definitely not considered the invariant mass in curved spacetime, because there is no such notion.

So, my best physical estimate for the meaning of the trace here is "the density of energy that moves along timelike worldlines". Evaluating this using a perfect fluid of dust, where ##p=0## and radiation ##p=\rho/3## means the trace is 0 for radiation and ##\rho## for the dust. A combination of these two fluids and then taking the trace will still result in ##\rho## for just the dust. Dust is given by ##T^{ab}=\rho u^a u^a##. For a timelike observer, the trace of this will always be non-zero and equal to ##\rho##.

Tertius said:
my best physical estimate for the meaning of the trace here is "the density of energy that moves along timelike worldlines"
No, that's not the trace. The energy density moving along a timelike worldline with 4-velocity ##u^a## is ##T_{ab} u^a u^b##.

The trace by itself does not have any recognized physical meaning.

Tertius said:
Evaluating this using a perfect fluid of dust, where ##p=0## and radiation ##p=\rho/3## means the trace is 0 for radiation
Yes.

Tertius said:
and ##\rho## for the dust.
Not quite: we have ##T_{ab} = \rho u_a u_b##, so ##T^a{}_a = \rho u^a u_a##, which, if we use the normal convention for squared norms, gives ##- \rho## (because ##u^a u_a = - 1##).

This alone should tell you that the trace can't have any direct physical meaning, since its sign depends on which metric signature convention we adopt.

Tertius said:
For a timelike observer, the trace of this will always be non-zero and equal to ##\rho##.
No, for a timelike observer following a worldline of the fluid, i.e., whose 4-velocity is ##u^a##, the energy density measured by that observer will be, as above, ##T_{ab} u^a u^b = \rho u_a u_b u^a u^b = \rho##. Here the signature convention doesn't matter because the squared norm ##u^a u_a## appears twice. But this is not the same thing as the trace of ##T_{ab}##.

Note also that you could pick a different observer, whose 4-velocity was different from ##u^a##, and the energy density that observer would measure would be different. For an observer with 4-velocity ##v^a##, the energy density they would measure for the fluid would be ##\rho u_a u_b v^a v^b = \rho \left( u_a v^b \right)^2##.

vanhees71
The trace is by definition the scalar ##T^{ab} g_{ab}##. That it's vanishing for the electromagnetic field is due to scale invariance of the free Maxwell equations, following from Noether's theorem. A fluid consisting of massive particles of course is not scale invariant because of the mass of the particles, and thus in this case the trace does not vanish.

For the ideal fluid you have (with the (+---) signature)
$$T^{ab}=(\epsilon+P) u^{a} u^{b}-P g^{ab}.$$
in the (local rest frame) of a fluid ##(T^{ab})=\mathrm{diag}(\epsilon,P,P,P)##, i.e., ##\epsilon## is the internal-energy density as measured in the rest frame of the fluid, and ##P## is the pressure as measured in this rest frame. Both quantities are thus scalars, which can be calculated in any frame using
$$T^{ab} u_a u_b = \epsilon, \quad P=\frac{1}{3} (u_{\mu} u_{\nu}-g_{\mu \nu}).$$

@PeterDonis, @vanhees71
I see. I neglected the negative sign, if you can get different answers with different metric conventions then it can't be physical.

The quantity I am really after is the matter density that is timelike. From my understanding timelike matter should always have a non-zero trace. So really what I am trying to get is an observer-independent way to express the fraction of the SEM tensor that has a non-zero trace.

Maybe you could take a ratio of the perfect fluid tensor and make it observer independent, like this:
$$\frac{T^{dust}_{\mu\nu}u^\mu u^\nu}{T^{total}_{\mu\nu}u^\mu u^\nu}$$
I would expect that this quantity is observer independent because effects of a non-inertial observer will be the same in the denominator as the numerator.

But I already knew it was formed of dust and radiation. Is there a general way to express the non-zero trace portion of a tensor?

Tertius said:
The quantity I am really after is the matter density that is timelike.
I've already told you how to obtain that. See post #15.

Tertius said:
Maybe

Tertius said:
Is there a general way to express the non-zero trace portion of a tensor?
Um, take the trace?

Tertius said:
Yes, and you already know the dust is timelike and the radiation is null. So you already know which part is timelike and which part is not. So what's the problem?

vanhees71

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