# Photon gas and relativity

1. Jun 12, 2007

### bernhard.rothenstein

Consider a gas of photons in a vessel. The total momentum of the radiation is equal to zero. Its total (rest) mass is m=E(rad)/cc. Consider an unidirectional motion of the vessel. Question: Will the (rest) mass of the radiation increase due to the fact that the total momentum is no longer equal to zero. Do you know a transparent treatment of the problem in the limits of special relativity? Thanks for help.

2. Jun 12, 2007

### jostpuur

A quick answer is no, the rest mass of the system will not change.

You can also calculate the assumed change in mass, and see that it does not change. So the situation is this. You have photons, indexed with $$k$$, and each of them has four-momentum $$(E_k/c,\boldsymbol{p}_k)$$. Total energy of the system is $$E=\sum_k E_k$$, and total momentum $$\boldsymbol{p}=\sum_k\boldsymbol{p}_k=0$$.

You then boost the system with $$\Lambda^{\mu\nu}$$. The new four-momentum of each photon is then $$p'^\mu = \Lambda^\mu{}_\nu p^\nu$$. This means that the new total energy of the system is

$$E'=\sum_k E'_k = \sum_k (\Lambda^0{}_0 E_k + \Lambda^0{}_j p^j_k c) = \Lambda^0{}_0 \sum_k E_k + \Lambda^0{}_j \sum_k p^j_k c = \Lambda^0{}_0 E$$

where the $$E$$ is the old total energy. Notice that the sum of old momentums vanishes. The new total momentum can be calculated similarly to be

$$p'^i = \sum_k p'^i_k = \Lambda^i{}_0 E/c$$

The new total mass is now

$$m'=\sqrt{(E')^2/c^4 - |\boldsymbol{p}'|^2/c^2} = (E/c^2)\sqrt{(\Lambda^0{}_0)^2-|\Lambda^i{}_0|^2} = m\sqrt{\Lambda_{\mu 0}\Lambda^{\mu 0}}$$

In order to conclude that the mass has not changed, it suffices to show that $$\Lambda_{\mu 0}\Lambda^{\mu 0}=1$$. You can calculate this explicitly by substituting some boost in some particular direction, or you can also note that this follows from $$g_{\alpha\beta}\Lambda^\beta{}_\mu\Lambda^\alpha{}_\nu =g_{\mu\nu}$$, which is always true for all lorentz transformations.

(I hope there wasn't many mistakes)

Last edited: Jun 12, 2007
3. Jun 12, 2007

### smallphi

The mass of a system is an invariant i.e. can be calculated in any inertial frame you want. If the vessel is boosted, you can always go back to the rest frame of the vessel where the total photon 3-momentum is zero and the total photon energy is still the same as before the boost. Calculating the mass in the rest frame shows it doesn't change which is confirmed by the jostpuur calculation in the lab frame.

Last edited: Jun 12, 2007
4. Jun 12, 2007

### pervect

Staff Emeritus
Unfortunately, the mass of the photon gas is NOT a relativistic invariant, for the following reason.

A system must have either zero volume, or be isolated, for it to have an invariant invariant mass. The system being described above is not isolated (because of the container walls) and it has a nonzero volume, thus its invariant mass is not invariant.

See for example http://arxiv.org/abs/physics/0505004

The mass of the photon gas plus its enclosing pressure vessel is a relativistic invariant, however, because that system is isolated.

To correctly calculate the energy and momentum of the photon gas excluding the pressure vessel, the best route is to write down the stress-energy tensor of the photon gas, which in a cartesian (t,x,y,z) coordinate system in geometric units would be just

[corrected]
diag(rho, rho/3, rho/3, rho/3)

boost it, and find the energy by the intergal of T^00 over the (lorentz-contracted) volume, and the momentum by the integral of T^01.

which works out the energy and momentum of a swarm of particles bouncing around inside a container.

You might also check out the reference in that thread to Rindler, who uses the stress-energy approach to work out the energy and momentum in a stressed rod. This would be "Rindler, Introduction to SR", 2nd edition, pg 130-132. Tollman might also have something along the same lines.

Last edited: Jun 12, 2007
5. Jun 12, 2007

### smallphi

From the links you gave I formed the impression that the mass of extended object is not a relativistic invariant in GR because there is no unique way to add 4 vectors at different points.

This problem though is in the context of SR where adding 4 vectors at two different points is perfectly well defined. Note that any particle collision in SR can be thought of as conservation of the invariant mass of the extended system consisting of the colliding particles since you can consider the particles long before they collide and they are far away from each other.

It's true that the only the total mass of the photons+container should be conserved in SR. One can cosider though the limit of zero mass of container and obtain again m_photons ' = m_photons. Since the mass of the photon gas doesn't seem to depend on the container mass, the container is there only to reflect the photons, then it must be invariant for any container mass.

If you insist the result m'=m is not valid in SR, you will have to point out the mistake in the jostpuur's derivation. There will be of course issue of simultaneity, if you take all the particles in the rest frame at a given time t, in the lab frame at time t' some of them wold have collided with the walls already and reflected 3 momentum, while others won't. While it is not clear how that would change the calculation, the reasoning in the previous paragraph says it won't.

Last edited: Jun 12, 2007
6. Jun 12, 2007

### jostpuur

So the thing I ignored in the calculation was, that the momentum of each photon changes in time, and when I calculate the total momentum at some instant in one reference frame, in other frame these momentums were not existing at the same instant?

Would an assumption, that the momentum distribution remains constant, suffice to fix this?

7. Jun 12, 2007

### pervect

Staff Emeritus
[corrected]
The unboosted stress-energy tensor for the photon gas is:
\left[ \begin {array}{cccc} \rho&0&0&0\\\noalign{\medskip}0&1/3\,\rho &0&0\\\noalign{\medskip}0&0&1/3\,\rho&0\\\noalign{\medskip}0&0&0&1/3\, \rho\end {array} \right]

Boosted by a velocity v, it becomes:

\left[ \begin {array}{cccc} -1/3\,{\frac {\rho\, \left( 3+{v}^{2} \right) }{-1+{v}^{2}}}&4/3\,{\frac {\rho\,v}{-1+{v}^{2}}}&0&0 \\\noalign{\medskip}4/3\,{\frac {\rho\,v}{-1+{v}^{2}}}&-1/3\,{\frac { \rho\, \left( 3\,{v}^{2}+1 \right) }{-1+{v}^{2}}}&0&0 \\\noalign{\medskip}0&0&1/3\,\rho&0\\\noalign{\medskip}0&0&0&1/3\,\rho \end {array} \right]

Because the volume should decrease by a factor of gamma, the total energy should be

$$E = E_0 \frac{\ 1+v^2/3 }{\sqrt{1-v^2}}$$
and the total momentum should be
$$P = 4/3\,{\frac {{\it E_0}\,v}{\sqrt {1-{v}^{2}}}}$$

giving an invariant mass m^2 E^2 - P^2 (which is not actually invariant!) of

$$m^2 = \left( 1 - \frac{v^2}{9} \right) E_0^2$$

Last edited: Jun 12, 2007
8. Jun 12, 2007

### jostpuur

Why those diagonal terms? Shouldn't photon gas at rest have $$T^{00}$$ as the only non-zero component?

9. Jun 12, 2007

### smallphi

Off diagonal terms because you are not in the rest frame anymore.

Pervect did you take into account that p = rho/3 for photon gas?

10. Jun 12, 2007

### jostpuur

No, I mean those diagonal terms. Don't they describe 3-momentum currents? I just got a feeling that I'm lost on something fundamental

11. Jun 12, 2007

### pervect

Staff Emeritus
While this is true in general, it's not what I meant, and as you point out it's not relevant to the problem at hand. The problem you point out arises in GR, not in SR, when attempting to form a more general GR definition of mass. But at the moment we don't have to worry about this, it's a red herring.

As the reference I gave states, the problem is the relativity of simultaneity. Consider a moving rod, where both ends are under pressure. The rod is not isolated - there is a force on the rod doing work that pushes on one end of the rod, there is another force on the rod doing negative work that pushes on the other end.

The total net rate of energy being imparted to the rod is zero.

Suppose we shut off both forces - what happens to the energy of the rod?

If we shut off both forces at the same time in the lab frame, we don't change the energy or momentum in the lab frame.

But when we shut off both forces at the same time in the lab frame, we shut off both forces at different times in the comoving frame, implying that we do change the energy and momentum in the comvoing frame.

There's no way to shut off both forces "at the same time" in all frames, unless the rod has zero extent.

The final result is that the stress-energy tensor is always covariant, however the total energy-momentum of an object of finite volume is not covariant, unless the object of finite volume is an isolated system.

This does not have anything to do with a non-Minkowskian metric, it's a purely SR effect.

It's a rather tricky point. For a textbook treatment, see Rindler (and I think Tollman discusses it as well).

12. Jun 12, 2007

### smallphi

When I said that total mass is invarian in SR I meant isolated system like container + photons. If you take the limit of the mass of container to be zero (which practically means m_container << E_photons), you obtain again m_photons ' = m_photons.

Your result predicts significant decrease of the total mass of photons so one of those two lines of reasoning will be wrong. The question is where is the error ...

13. Jun 12, 2007

### pervect

Staff Emeritus
You're right, I'm missing a factor of 1/3.

14. Jun 12, 2007

### jostpuur

Why do you have, for each fixed i=1,2,3, a momentum current $$(0,\delta_{1i},\delta_{2i},\delta_{3i})$$ with some constant, instead of plain $$(0,0,0,0)$$? Is there momentum drifting in those directions, or does the stress-energy tensor mean something else than what I think it means?

15. Jun 12, 2007

### pervect

Staff Emeritus
The diagonal terms are the pressure terms.

As you pointed out, for a photon gas, rho = 3P, or equivalently, P = rho/3.

For an ideal fluid, the stress energy tensor is diag(rho, P, P, P). Since for a photon gas, rho = 3P, we can write that as diag(rho, rho/3, rho/3, rho/3).

16. Jun 12, 2007

### pervect

Staff Emeritus
If you work the problem out including the walls of the container, you will find that the pressure terms do cancel out (as they must), and that the mass of the total system of gas + container is invariant as it should be.

Note that if you assume that the container has a tension that exceeds its mass, you'll violate the weak energy condition.

17. Jun 12, 2007

### pervect

Staff Emeritus
For a reference for the stress-energy tensor of an ideal fluid, look at Sean Carroll's GR notes:

http://xxx.lanl.gov/PS_cache/gr-qc/pdf/9712/9712019v1.pdf

pg 29, 1.107 for the fluid in its rest frame, 1.110 for the fluid in an arbitrary frame. Instead of boosting the stress-energy tensor as I did, you can use the results from 1.110 to calculate it directly, given a 4-velocity
$\vec{v} = [\gamma, \gamma v, 0, 0 ]$

18. Jun 12, 2007

### smallphi

I forgot that even if the container has zero rest mass without the photon gas inside, when you put the photons inside, the container is under pressure. Iimagine it's made of massless ellastic bands, they stretch till they balance the pressure of the photon gas so the container's rest mass (it's internal potential energy) becomes nonzero and cannot be neglected.

Last edited: Jun 12, 2007
19. Jun 12, 2007

### smallphi

I redid the calculation suggested by pervect for arbitrary ideal gas, not only photons. The result is

$$M^2 = M^2_0(1-\frac{p^2}{\rho^2} v^2)$$

where rho and p are the energy density and pressure of the gas in the rest frame of the container, M_0 is the rest mass of gas in the rest frame of the container, M is the rest mass calculated in inertial frame in which the container moves at constant speed v. The formula shows that indeed the nonzero pressure of the gas on the container leads to the non-invariance of the mass M.

Last edited: Jun 12, 2007
20. Jun 12, 2007

### jostpuur

I quickly grasped the idea behind those pressure terms, and cought up with you guys! Although I still don't understand what went wrong in my original calculation. So the stress-energy tensor approach gives

$$E'=\frac{1+u^2 p/\rho}{\sqrt{1-u^2}}E$$

and the simpler approach that I tried first gives

$$E'=\frac{1}{\sqrt{1-u^2}}E$$

That doesn't make any sense. This is what is happening: When I start calculating the new energy directly from old energies, the momentums do not matter at all. If I instead start calculating the energy density, then the momentum density currents matter, and the pressure terms have an effect on the energy density. Then, magically, the effect from pressure terms survives when I compute the energy.

(Now I remember again the old good times, when I was a kid and believed that the relativity was contradictory...)

Last edited: Jun 12, 2007
21. Jun 12, 2007

### smallphi

The total 3-momentum in the container rest frame is zero if the momenta of the photons are taken at the SAME time t in the rest frame. When you transform the individual photon energies and momenta to the lab frame, you have to get them at the same fixed time t' in lab frame. The same t' in lab frame demands different t in rest frame for photons at different positions so the sum of 3-momenta of those photons at DIFFERENT times t in rest frame is no longer zero. The zero of that sum is the assumption that breaks down.

Last edited: Jun 12, 2007
22. Jun 12, 2007

### jostpuur

I see that is worth attention, but I'm not fully conviced that the solution lies in that remark. For example, suppose that the photon density was so small, that none of the photons happens to collide anywhere along their world lines, in between the events, that are simultaneous in the first and the second frame. In such special case my original calculation should have been valid.

Also, let us first fix some constant $$\lambda$$, and let energy density and pressure be $$\lambda\rho$$ and $$\lambda p$$, and the volume of the box be $$\lambda^{-1}V$$. Now the energy does not depend on $$\lambda$$. Letting $$\lambda\to\infty$$ shrinks the box to a small point, and puts an infite energy density into it. Shouldn't the contradiction between the two approaches somehow dissappear on this limit?

23. Jun 12, 2007

### smallphi

Maybe the explanation lies in the fact that energy density and pressure of the gas are time averaged quantities so the energy-momentum description is averaged description. With each reflection of a photon from the walls of the container, there will be transfer of momentum between the gas and the container so the relativistic mass of the gas will fluctuate. At certain moments t' in lab, there won't be any collisions with walls in the rest frame and your result will hold true. At other moments you will have 1,2,3 ... collisions so other results will hold true. Averaging over time, the result derived through the energy-momentum tensor will hold true.

Experimentally, if your aparatus for measuring the total energy and momentum of the gas in the lab frame is ultra-sensitive, it will show fluctuations due to collisions of photons with the container walls. After you time-average your measurements, the result must coinside with the one preducted by the energy-momentum tensor.

Last edited: Jun 12, 2007
24. Jun 12, 2007

### bernhard.rothenstein

giving up the vessel!

Thanks to all participants. Consider a scenario proposed by Gron
[O.Gron, "A modification of Einstein's first deduction of the inertia-energy relationship," Eur.J.Phys. 8, 24-26 1987] involving a body at rest in I' which emits a short pulse of light isotraopically. We can consider that the pulse consits in pairs of photons having equal and opposite momentums.
I use the notations g for gamma, k(+)=sqrt[(1+b)/1-b)], b=V/c k(-)=sqrt[(1-b)/1+b)], K(+)=g(1+bcostheta') and K(-)=g(1-bcostheta').
Let N be the number of photons emitted by the source. The energy of the system of photons is in I' Nhf' the momentum of the system being equal to zero.
In I the energy of a pair of photons is
E=hf'[K(+)+K(-)]=2ghf' (1)
having there a momentum
p=hf'(k(+)+k(-))=2ghbf' (2)
both being theta' independent.
The (rest) mass of the system of photons emitted by the body is in I the same as in I' Nhf'/cc.
Would that suffice for illustrating the invariance of the (rest) mass and the fact that it is not additive for beginners avoiding the complications introduced by the vessel?

25. Jun 12, 2007

### smallphi

You don't need to use photons to show the invariance of the total mass. Any collision of particles, reaction or decomposition of particles will do as long as you have isolated system (no vessel).

The simplest reaction imaginable that illustrates what you want is annihilation of electron and positron at rest into two photons.