# Based on black hole thermodynamics, shouldn't empty space contain infinite energy?

In SI units, the radiation from a Schwarzschild black hole is blackbody radiation with temperature $${\displaystyle T={\frac {\hbar c^{3}}{8\pi GMk_{\text{B}}}}\;\quad \left(\approx {\frac {1.227\times 10^{23}\;{\text{kg}}}{M}}\;{\text{K}}=6.169\times 10^{-8}\;{\text{K}}\times {\frac {M_{\odot }}{M}}\right)\,,}$$ where $$\hbar$$ is the reduced Planck constant, $$c$$ is the speed of light, $$k_{\text{B}}$$ is the Boltzmann constant, $$G$$ is the gravitational constant, $$M_☉$$ is the solar mass, and $$M$$ is the mass of the black hole.

By taking the limit of $$T$$ as $$M$$ goes to zero, the following is found:

$$\lim_{M\to0^+} T=\lim_{M\to0^+}{\hbar c^3\over{8\pi G k_bM}}=+\infty$$

Wouldn't this mean that empty space would have infinite energy? As when $$M=0$$ the Schwarzschild radius is also $$0$$, so every point in space would be paradoxically hot. I know I'm probably wrong, I just don't know why I'm wrong.

• If $M$ is zero what is supposed to be radiating (nothing ?) and where does the energy for this radiation come from (nowhere ?) ? – StephenG Oct 1 '18 at 21:31
• From your link " Unlike most objects, a black hole's temperature increases as it radiates away mass. The rate of temperature increase is exponential, with the most likely endpoint being the dissolution of the black hole in a violent burst of gamma rays. A complete description of this dissolution requires a model of quantum gravity, however, as it occurs when the black hole approaches Planck mass and Planck radius." – Bruce Greetham Oct 1 '18 at 21:46
• Hawking's original computation involves quantum fields in curved spacetime, where curvatures are small compared to the Planck length. I believe that in order to take a limit all the way to $M=0$, one must know how Planck scale quantum gravity works. Maybe there's something Hawking says in his article "Black hole explosions?" but I can't open it at the moment: nature.com/articles/248030a0 – Avantgarde Oct 1 '18 at 22:16
• Why should $T=\infty,\,M=0$ imply infinite energy? It implies zero lifetime, to get rid or the energy $Mc^2=0$ by radiation. – J.G. Oct 2 '18 at 9:46
• Why don't we just mention that points in empty space just aren't blackholes, so you just can't apply black holes formulas ? Is this thought completely erroneous ? – Guiroux Oct 2 '18 at 9:57

Wouldn't this mean that empty space would have infinite energy?

So ignoring quantum issues (and in the absence of a complete theory of quantum gravity we have no choice) and staying strictly with the classical approach let's consider the problem.

Regardless of what amount of power is radiated by the black hole, that power is removed from the energy of the black hole. But the black holes you are talking about have zero energy and so there is no way for them to power hawking radiation.

The mistake you are making is ignoring that nature balances it's books and in this case the balance is that you can't reduce the mass below zero.

There's another reason why your logic is failing.

The entire idea of Hawking radiation depends on the existence of a curved space time and an event horizon. But when $$M=0$$ we just get a flat spacetime. There is no event horizon. And note that an $$R=0$$ event horizon would mean there was nothing inside the black hole - no volume, nothing.

The formula you are using for temperature is based on a model which starts out with a non-zero positive mass and then makes a first order approximation close to the event horizon (Wikipedia has a description of this). But that formula does not apply when you're using $$M\to 0$$. Again, you're using an approximation based on the assumption of a curved spacetime ($$M>0$$) and applying it outside it's "designed purpose" as an approximation.

Cute question. I don't think it's possible to give a definite answer unless we decide on a definite description of the manner in which the limit is to be taken.

If we're just considering the limit of an asymptotically flat spacetime consisting of a single black hole, as the black hole's mass goes to zero, then I think the answer is pretty straightforward. The radiated power is $$P\propto M^{-2}\propto r_s^{-2}$$, where $$r_s$$ is the Schwarzschild radius. If this power is considered to be emitted from the surface of a sphere with radius equal to $$r_s$$, then the power per unit area at $$r>r_s$$ is $$I\propto Pr^{-2} \propto r_s^{-2} r^{-2}$$. The average distance of a randomly chosen observer from the black hole is infinite, so we should consider the limit $$r\rightarrow\infty$$. So now the question arises as to whether we should compute

$$\lim_{r\rightarrow\infty}\lim_{r_s\rightarrow0} I$$

or

$$\lim_{r_s\rightarrow0}\lim_{r\rightarrow\infty} I.$$

These are both indeterminate forms, so it sort of doesn't matter which one we evaluate -- neither one gives a meaningful answer without some physics input.

I think the additional physics input comes from the fact that $$r_s$$ probably can't be any smaller than the Planck length $$\ell$$. Therefore we really shouldn't be taking the limit as $$r_s\rightarrow0$$ but as $$r_s\rightarrow\ell$$. Then the double limit is zero.

Another way of approaching the whole thing is to imagine a theory of quantum gravity in which the vacuum has virtual black holes popping in and out of existence. Then presumably one of the things you want from such a theory (which we don't yet possess) is that it doesn't predict infinite density of photons for the vacuum. I would imagine that this would happen because the density of virtual Planck-scale black holes would be small.

When the mass drops down to 228 metric tonnes, that's the signal that exactly one second remains. The event horizon size at the time will be 340 yoctometers, or 3.4 × 10^-22 meters: the size of one wavelength of a photon with an energy greater than any particle the LHC has ever produced. But in that final second, a total of 2.05 × 10^22 Joules of energy, the equivalent of five million megatons of TNT, will be released. It's as though a million nuclear fusion bombs went off all at once in a tiny region of space; that's the final stage of black hole evaporation.

• This doesn't really make sense. The quoted paragraph doesn't say that $M$ never reaches zero, and in fact $M$ does reach zero after a finite time. (This is guaranteed, because the rate of mass loss is increasing with time.) – user4552 Oct 2 '18 at 0:32