Basically what the question says -- there is reason to expect that, if allowed to continue long enough for the radiation to reach future null infinity, the fact that the radiation will fall off at $\frac{1}{r}$ will break the set of conditions required for the enveloping metric to stay asymptotically flat, which will, in turn, combined with the fact that the spacetime is inherently neither static nor stationary, will subsequently remove most of the nice features that enabled us to derive the Hawking radiation in the first place.

If this is true, why is it common for people to make claims like "The black hole will eventually completely evaporate?", and other statements relating to conditions where an appreciable proportion of the black hole's initial mass has left the horizon?

  • $\begingroup$ Does it really have to be strictly stationary/static? According to Visser, (section 8) it's OK if the evolution is just slow compared to the mode frequencies. $\endgroup$
    – twistor59
    Sep 12, 2013 at 7:23
  • $\begingroup$ I thought that the big surprise was that even after the collapse had settle and the spacetime outside the black hole is stationary it still continues to emit. Hawking in "Particle creation by black holes" has a section on back reaction, but I don't know if you addresses your question. $\endgroup$
    – MBN
    Sep 12, 2013 at 9:34
  • $\begingroup$ Could the metrics be asympotically flat ?. For instance (for t>0) : $ds^2=-(1-\frac{r_S}{r+ct})dt^2 + \frac{1}{1-\frac{r_S}{r+ct}}dr^2$ $\endgroup$
    – Trimok
    Sep 12, 2013 at 11:01
  • $\begingroup$ @Trimok: that's the point. That metric can't possibly describe a spacetime where Hawking radiation is present--where is the matter? $\endgroup$ Sep 12, 2013 at 14:37
  • $\begingroup$ @MBN: sure. The "continuing to emit" breaks the stationarity. The actual radiation is derived, originally, in such a way that the back-reaction on the spacetime is neglected. $\endgroup$ Sep 12, 2013 at 14:37

1 Answer 1


the fact that the radiation will fall off at $\frac{1}{r}$ will break the set of conditions required for the enveloping metric to stay asymptotically flat

I'm not sure this is right. There are various definitions of asymptotic flatness. Older definitions were written in terms of coordinates, newer ones in terms of conformal transformations. The original motivation, as described in ch. 11 of Wald, was to accomplish for GR what had already been done for E&M. In E&M in SR, the coordinate-based requirements given by Wald are that the fields fall off like $1/r^2$ at $i^0$, but only like $1/r$ at $\mathscr{I}^+$. This is clearly designed to allow radiation.

The definition of asymptotic flatness in Wald is actually framed in a pretty restrictive context. He first gives a definition that's purely for a vacuum spacetime (not electrovac), and then remarks that the definition carries over automatically to a spacetime in which there is a vacuum in some open neighborhood of the boundary. Obviously it should be possible to extend this to a case in which the matter fields fall off fast enough, but it looks like he just wants to avoid making the already technical discussion even more technical. But the definition of asymptotic flatness for vacuum spacetimes definitely allows for spacetimes with gravitational radiation, since the ADM energy, which is only defined in asymptotically flat spacetimes, includes the energy of gravitational radiation at null infinity. (This could probably be checked explicitly by power-counting. For an asymptotically flat spacetime, the metric differs from Minkowski by $O(1/v)$, where $v$ is an affine parameter defined in the lightlike direction.)

As further confirmation that these spacetimes with Hawking radiation are asymptotically flat, you can find Penrose diagrams for them. For example, there's one in figure 2.41 in Penrose, Cycles of Time.

  • $\begingroup$ I think the issue is that if the matter density falls off like $\frac{1}{r}$, then the potential will fall off more slowly, since Gauss's law, and by extension, Birchoff's theorem tells us that $V \propto \frac{\int \rho}{r}$, which means that we'd expect $g_{tt}$ to scale as $O(r)$ $\endgroup$ Sep 12, 2013 at 21:23
  • $\begingroup$ @JerrySchirmer: We could work out the explicit power counting, but I don't think the sketch in your comment works. Birkhoff's theorem doesn't apply because this isn't a vacuum spacetime. Also, on a surface of simultaneity, the density isn't $\propto 1/r$; the Hawking radiation is a spherical shell with a finite thickness. $\endgroup$
    – user4552
    Sep 12, 2013 at 21:44
  • $\begingroup$ Also, if a radiation field falls off like $1/v$ along some null geodesic, where $v$ is the kind of null coordinate described in my answer, the mass-energy density falls off like $1/v^2$, not $1/v$. $\endgroup$
    – user4552
    Sep 12, 2013 at 22:00
  • $\begingroup$ yes, I was being kind of dumb in my response in retrospect--I was imagining something like a permanent stream of hawking radiation leaving the horizon in perpetuity. But in reality, it would start radiating only when the collapse formed a horizon, and would radiate away any finite fraction of the hole's mass in a finite time, so my image of a stream of matter extending out to null infinity is wrong. $\endgroup$ Sep 12, 2013 at 22:14
  • $\begingroup$ Or I would say that a stream of matter does extend to certain points at null infinity, but not to other points at null infinity. $\endgroup$
    – user4552
    Sep 12, 2013 at 22:31

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