Maximum power (luminosity) limit in General Relativity

Question

Is the following thought experiment - to determine the maximum limit for power $P_m$ in General Relativity - correct?

It is by Craig J Hogan, a member of the team that discovered dark energy, so pretty mainstream I would think.

Consider an (almost) Kugelblitz sphere of radius $R$ $\geqslant$$2Gp \over c^3$ filled with photons with a total mass-energy of $E=pc$ that leave after a time $t=R/c$. Average power (luminosity) $P=E/t$. Therefore:

$P_m$ $\leqslant$$c^5 \over2G$

Original paper

The term $c^5 \over G$ is Planck power

Answer 1

There is a discussion of this in Misner, Thorne, and Wheeler, p. 980. Their treatment is more careful than Hogan's and more clear about the underlying assumptions. They derive this estimate by considering violent processes of astrophysical collapse in which the virial theorem is valid. They don't claim that it's an upper limit on the luminosity of any object.

Hogan's discussion seems to have a lot of logical problems. He starts by assuming a sphere "filled with light...and released [in] an instant." This is a very specific system, so there is no logical reason to conclude, as he does, that therefore, "An absolute luminosity limit for anything is imposed by General Relativity."

That would be pretty illogical, since, for example, if you had two such sources side by side, they would release twice the power, and you could consider them as one big object. I suppose MTW evade this counterexample because the two objects would not be equilibrated with one another, and therefore the virial theorem wouldn't apply to them if they were considered as a single system. I think this is also the answer to mmeent's counterexample of a thin shell of energy being released. It is indeed a counterexample to Hogan's claim, but not to MTW's, since the thin shell probably doesn't obey the virial theorem.

It's also weird and unfortunate that Hogan wants to connect this to the Planck mass. This would suggest, incorrectly, that there is something quantum-mechanical about this, when in fact (as he admits) there isn't. As he explains, the h-bars go away when you form units of power in Planck units.

Answer 2

Via dimensional analysis, you might think the 'Planck Luminosity' $L_p=c^5/G$ would be the maximum luminosity (power). This is also the result from MWT (page 980 btw).

However: As Cardoso pointed out in 2018, the existence of black hole horizons implies a maximum power (luminosity) limit in General Relativity, leading to related conjectures first posed around the time of Hogan’s 1999 paper. Not even gravitational waves can escape a black hole.

Thus, consider a process with finite duration $t$, which produced some radiation with total energy $E = p \ c$ contained in a shell of thickness $R=c \ t$. Imagine evolving this system back into the past, so that the shell is focused near the origin where its self-gravity is large. In order to have escaped its self-gravity we must impose that a BH horizon was not present, so require \begin{equation} \notag R\geq \frac{2Gp}{c^3} \end{equation} The power (luminosity) is therefore bounded by:  \begin{equation} \notag P = \frac{E}{t}=\frac{c^5}{2G} \end{equation} This is maximum power in GR, regardless of the nature of the system. You might be tempted to call this half a 'Planck Power' (or half a 'Planck Luminosity') but there is no $\hslash$ in this expression, it is purely classical. This is why you won't see an equation for 'Planck Power' in wiki.