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This question was inspired by: How are photons "consumed"?

Imagine I have some number of photons, $N$, each of frequency $\nu$, moving randomly in a spherical "wrap-around" universe of volume $V$. To clarify what I mean by 'wrap-around', if a photon reaches a point on the surface of the sphere it is immediately mapped to the antipodal point with no changes to it's properties or velocity vector. The system is otherwise completely closed, and devoid of any other particles.

How might this system evolve over time?

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Is there any particular reason you made the spatial topology $\mathbf{PR}^3\cong\mathrm{SO}(3)$ rather than $S^3$? That's not going to affect local physics, though, so at least for uniform initial conditions on some $\mathbf{PR}^3$ hypersurface, we can simply scale a restricted round metric $$d\Omega^2 = d\psi^2 + \sin^2\psi(d\theta^2 + \sin^2\theta d\phi^2),\;\;\;\;0\leq\psi,\theta,\phi\leq\pi.$$ The Einstein equations have the same local form as for $S^3$, so we would get a spacetime that locally looks like the standard $k = +1$ Freidmann-Robertson-Walker solutions of the form $ds^2 = -dt^2 + a^2(t)d\Omega^2$. For example, a lambdavacuum (if energy density due to photons is $\rho\ll|\Lambda|$), radiation-dominated ($\rho\gg|\Lambda|$, $\rho = 3p$, scale factor $\propto t^{1/2}$), etc. These cases are extensively catalogued.

The difference being that the large-scale spatial structure is different, according to the antipodal identification you prescribe. Under the assumption of uniformity, the general answer to how this system evolves is simply: like the corresponding FRW universe. Depending on the conditions, it could even undergo transition from radiation-dominated to matter-dominated, just as our own universe did. I'm unclear as to whether you intended to exclude anything but photons in vacuum at one time or for all time.

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I wouldn't object to having $S^3$ spatial topology, and I did not intend to exclude photons for 'all time', just for the initialization state of the closed system. –  TheSheepMan Aug 14 '11 at 5:21
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Did you mean $0 \le \psi,\theta,\phi\le\pi$ in that equation? –  David Z Aug 14 '11 at 7:16
    
Yes, I did. Thanks. –  Stan Liou Aug 14 '11 at 7:30

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