# I Gravitational mass of a ball of photons

#### PeterDonis

Mentor
Summary
What is the gravitational mass of a ball of photons?
This is based on "Concept Question 10.4" in Andrew Hamilton's General Relativity, Black Holes, and Cosmology. I have modified the question somewhat in order to focus on what seem to me to be the key issues.

Suppose we have a spherically symmetric ball of stress-energy surrounded by vacuum. More precisely, the ball contains a perfect fluid with an equation of state given by $p = w \rho$, where $w$ is a constant that can vary according to the type of stress-energy. We will consider three commonly used values for $w$ taken from cosmology: $w = \frac{1}{3}$, which is usually called "radiation" (the original question in Hamilton's book used "photons" to refer to this, hence the title of this thread); $w = 0$, which is usually called "matter"; and $w = -1$, which is usually called "dark energy" (Hamilton calls it "vacuum").

Since the spacetime is spherically symmetric, the geometry of the vacuum region is known; by Birkhoff's Theorem, it must be the Schwarzschild geometry. This geometry has one free parameter, $M$, which is the externally measured mass of whatever is inside the non-vacuum region. This obviously gives rise to the question: how does $M$ relate to the stress-energy tensor inside the non-vacuum region? In particular, how does it relate to the equation of state parameter $w$ (since that is the key thing that characterizes the stress-energy inside the non-vacuum region)?

We can sharpen this question by considering two possible cases:

(1) The ball of stress-energy has no restriction at its boundary: it is free to expand or contract, as long as it remains spherically symmetric. For this case, does the mass $M$ depend on $w$? If so, how?

(2) The ball of stress-energy is contained by a stationary wall with a fixed surface area $A$. The wall itself must contain stress-energy, so there will be some externally measured mass $M_{\text{wall}}$ associated with it. The question for this case is, for a given value of $w$, is the "net" mass $M - M_{\text{wall}}$, which can be interpreted as the externally measured mass of the stress-energy inside the wall (i.e., the same stress-energy as in #1 above) the same as the mass $M$ in #1 above? If not, how does it change?

Now, this question occurs in a part of Hamilton's book which is discussing FRW spacetime, which seems to suggest (at least to me) that the intended answers to the above are, for case #1, that $M$ should vary with $w$ (since the dynamics of an FRW spacetime certainly does), and for case #2, that the presence of the wall should make a difference (since the wall prevents the stress-energy inside from exhibiting the same dynamics as it would in the free case).

However, I can think of two simple heuristic arguments that seem to lead to the opposite conclusion:

For case #1, we can extend the obvious Schwarzschild coordinates in the external vacuum region inside the region containing the stress-energy, and then consider a spacelike slice of constant time $t$ in these coordinates. In such a slice, from the known properties of the Einstein Field Equation for the case of spherical symmetry, we should have the following:

$$M = \int_0^R 4 \pi r^2 \rho (r) dr$$

where $\rho (r)$ is the energy density and $R$ is the areal radius of the surface of the ball. But this integral only contains $\rho$; it does not contain $p$. So the equation of state does not affect this integral at all. So $M$ should be independent of $w$.

For case #2, we can do the same integral as above, but now just split up into two pieces:

$$M = M_{\text{net}} + M_{\text{wall}} = \int_0^R 4 \pi r^2 \rho (r) dr + \int_R^{R_{\text{wall}}} 4 \pi r^2 \rho (r) dr$$

where $R$ is the areal radius of the inner surface of the wall and $R_{\text{wall}}$ is the areal radius of its outer surface. The first term in this integral is the same as above, so we should have $M_{\text{net}}$ in the above be the same as $M$ for case #1 (with the obvious proviso that we have to compare them for the same value of $R$, i.e., at an instant where the areal radius in case #1 is the same as that of the inner surface of the wall in case #2).

So what do others think? Is the answer the one that I think Hamilton seems to expect? Or are my heuristic arguments correct?

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#### martinbn

However, I can think of two simple heuristic arguments that seem to lead to the opposite conclusion:

For case #1, we can extend the obvious Schwarzschild coordinates in the external vacuum region inside the region containing the stress-energy, and then consider a spacelike slice of constant time $t$ in these coordinates. In such a slice, from the known properties of the Einstein Field Equation for the case of spherical symmetry, we should have the following:

$$M = \int_0^R 4 \pi r^2 \rho (r) dr$$

where $\rho (r)$ is the energy density and $R$ is the areal radius of the surface of the ball. But this integral only contains $\rho$; it does not contain $p$. So the equation of state does not affect this integral at all. So $M$ should be independent of $w$.
Intuitively $R$ should depend on $t$, hence on the dynamics of the space-time, hence on the equation of state.

#### vanhees71

Gold Member
Isn't this just the usual TOV problem for different equations of state? Of course you can always give the solution in terms of FLRW metrics (also for the exterior, where you just use the usual Schwarzschild coordinates and transform them to Gaussian normal coordinates to enable the matchting condition to the solution for the interior). It's all worked out in Landau Lifshitz vol. II or Weinberg, Gravitation and Cosmology.

#### pervect

Staff Emeritus
For the second case, assume that everything is static. Then for a small ball, with isotroic pressures, the Komarr mass of the ball is basically just the integral of $\rho + 3P$.

By small ball, I mean that we use linearized gravity, linearized about a flat Minkowskii space-time.

I believe that in this case, the virial theorem, as presented by Carlip in"Kinetic energy and the equivalence principle", the pressure term averages out to have zero net contribution. Therefore you just need to integrate the energy density $\rho$ of both the wall and the fluid inside the wall to get the total mass. I.e. you just have to integrate $T^{00}$ over the volume.

Note though that the integral of P only vanishes if you consider the effect on the mass of the system as a whole. When you subtract out the mass of the container to weigh the energy inside it, you run into the issue that the stress in the container changes the integral of $\rho + 3P$ of the container. The center of the ball is under compression, the containing wall is in tension (negative pressure). I'll believe that you'll also find that the amount of mass you assign to the wall and the amount of mass you assign to the contents will depend on the gauge choice you make, which in the linearized gravity case is equivalent to your choice of coordinates. Carlip discusses this a bit. So there isn't any unique way to assign which part of the mass is due to the container, and which part of the mass is due to its contents, but the total mass of container+contents is well defined. And in the small ball case, we can get this total mass by integrating $T^{00}$ over the volume of the system.

I recall reading at some point a paper about a paradox involving the mass of a shell containing a bomb. The mass of the system must stay the same before and after the bomb is exploded if the shell contains all the radiation emitted by the bombs detonation. We do have to assume that no energy escapes via gravitational radiation, but with a spherically symmetric explosion this should be a good assumption. I forget the name of the "paradox" this creates, but the answer is similar - the pressure in the cetner of the bomb due to the explosion adds mass, but the tension in the walls of whatever contains the explosion subtracts mass, and the total mass of the bomb + container system stays unaffected by the explosion, as it must.

Unfortunately, I don't quite recall the name of this "paradox".

The whole approach though involves having a static space-time. I'm not aware of any really good way to define mass if you don't have either a static system, or an asymptotically flat system, and some variants of your question might not have either.

#### PeterDonis

Mentor
Intuitively $R$ should depend on $t$,
Yes, but if we are looking at a spacelike slice of constant $t$, this doesn't matter. For the spherically symmetric case, the t-t and r-r components of the Einstein Field Equation are the same in the time-dependent case as they are in the static case, and those components only include $r$ derivatives, not $t$ derivatives, so they function as equations for the $r$ dependence in a spacelike slice of constant $t$.

Also, since the ball of nonzero stress-energy is of finite extent and the geometry outside it is Schwarzschild, the overall mass $M$ cannot change, since $M$ is constant in the Schwarzschild geometry. So if we evaluate $M$ in any spacelike slice of constant $t$, we have evaluated it everywhere.

#### PeterDonis

Mentor
For the second case, assume that everything is static. Then for a small ball, with isotroic pressures, the Komar mass of the ball is basically just the integral of $\rho + 3P$.
But with correction factors in the integrand for redshift and non-Euclidean spatial geometry, that cancel the contribution of $p$. So it works out to be the same as the integral that I wrote down, which only includes $\rho$. (Note, though, that in the integral I wrote down, there are no correction factors in the integrand; it's identical to the corresponding Newtonian integral. In other words, the mass function $m(r)$ only depends on $\rho$; it does not depend on $J$ or any other functions in the problem.)

Therefore you just need to integrate the energy density ρρ\rho of both the wall and the fluid inside the wall to get the total mass. I.e. you just have to integrate $T^{00}$ over the volume.
Yes, that's what the integral I wrote down does.

Note though that the integral of P only vanishes if you consider the effect on the mass of the system as a whole.
I don't think that's correct. The integral I wrote down, which only includes $\rho$, is valid for each of the regions (ball of stress-energy and containing wall) separately. The corresponding Komar mass integral can be split up the same way.

the stress in the container changes the integral of $\rho + 3P$ of the container
It "changes" it in the sense that the spatial stresses are not isotropic, so $\rho + 3p$ is no longer correct; you have different radial and tangential stresses. (And not all of them are positive, as you note.) But the cancellation I mentioned still works; it has to, since the Komar mass has to equal the integral I wrote down, that only includes $\rho$, for each region separately.

the amount of mass you assign to the wall and the amount of mass you assign to the contents will depend on the gauge choice you make
I have already made that gauge choice by choosing Schwarzschild coordinates and integrating in a spacelike slice of constant $t$. In general I think you are right that the wall-contents split will be coordinate-dependent, although the total mass $M$ will not be (since the $M$ in the Schwarzschild geometry is not gauge dependent). However, the surfaces of constant $t$ are the unique ones that are orthogonal to the timelike Killing vector field in the exterior Schwarzschild vacuum region (and in the interior region as well for the static case), so choosing them is not a purely arbitrary choice.

The mass of the system must stay the same before and after the bomb is exploded if the shell contains all the radiation emitted by the bombs detonation.
Yes. And moreover, even if there is no shell containing the bomb, since at any finite time after the explosion the radiation emitted will occupy a finite region, the mass as measured in the vacuum region outside that finite region must be the same as before the explosion. (This is all assuming a spherically symmetric explosion.) So this is another argument that I hadn't thought of.

The whole approach though involves having a static space-time
It only has to be static in the exterior vacuum region; the size of the non-vacuum region (for case #1 where the ball is not constrained by a wall) can change with time.

#### Heikki Tuuri

https://arxiv.org/abs/gr-qc/0510041

The "paradox" of a pressurized vessel is called Tolman's paradox. If you forget the stresses in the walls of the vessel, you will get a breach of Birkhoff's theorem.

Misner and Putnam published in 1959 a resolution of this paradox. The link can be found from the above paper by Ehlers et al.

#### pervect

Staff Emeritus
https://arxiv.org/abs/gr-qc/0510041

The "paradox" of a pressurized vessel is called Tolman's paradox. If you forget the stresses in the walls of the vessel, you will get a breach of Birkhoff's theorem.

Misner and Putnam published in 1959 a resolution of this paradox. The link can be found from the above paper by Ehlers et al.
Thank you. Looking up "Tolman's paradox" directly swamps the results with other paradoxes that share Tolman name.

"Gravitational mass of a ball of photons"

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