Due to Hawking Radiation, after a very long time, a black hole will eventually disappear.

But where does this "evaporation" go? What is this made of?


1 Answer 1


Although black holes are widely believed to emit Hawking radiation, it should be stressed that it has not actually been observed (yet?). The Hawking radiation should consist of electromagnetic radiation/waves that have a near-perfect black body spectrum, which is at a temperature that is inversely proportional to the mass of the black hole - the smaller the black hole, the higher the temperature.

The radiation is caused because "vacuum" is not empty when considered in quantum mechanics. Particle/anti=particle pairs are created for brief moments of time, before they annihilate again. Close to the event horizon of a black hole, it is possible that one particles from a virtual particle pair travels within the event horizon of the black hole and cannot escape to annihilate with its anti-particle. In principle, these particle/anti-particle pairs could be any types of particle, but in practice they are more likely to be the lightest particles. The lightest charged particles are electron/positron pairs, since these need to "borrow" less energy from the vacuum to be created. I do not think it is essential for one of the pair to disappear, since "Unruh radiation", a close relation of Hawking radiation should also be seen whenever there is acceleration with respect to the vacuum.

To stay out of the black hole event horizon the particles must be accelerating. Accelerating charged particles locally emit electromagnetic radiation, which is then gravitationally redshifted when seen by an observer a long way from the black hole. It turns out that to be in a thermal equilibrium, the radiation must have a blackbody spectrum form.

The temperatures and amount of radiation emitted are very small for stellar-sized black holes. The temperature of a non-rotating Schwarzschild black hole is given by $$ T = \frac{6.2\times 10^{-8}}{M}\ K,$$ where the mass $M$ is in solar maases.

The power emitted by a black hole as Hawking radiation is $$ P = \frac{9\times 10^{-29}}{M^2}\ W$$

  • 1
    $\begingroup$ +1, but why single out electron/positron pairs when neutrino are much lighter leptons and therefore theoretically more relevant to a much wider mass range? That should start to happen roughly when the temperature (surface gravity) is comparable to the particle mass in natural units. There's more species of them too, which also important, since power emitted scales as $\propto(\eta/2)$, where $\eta$ is the number of thermodynamically relevant degrees of freedom. $\endgroup$
    – Stan Liou
    Aug 10, 2015 at 15:31
  • $\begingroup$ @StanLiou I'm sure you are right. Perhaps I should have said charged particles. I guess the neutrinos would have their own separate Planck distribution? $\endgroup$
    – ProfRob
    Aug 10, 2015 at 15:40
  • $\begingroup$ As an analogy, yes, but one probably shouldn't call it 'Planck'... while a Planck distribution has the maximum entropy for a given well-defined temperature for the electromagnetic field/photon gas (and therefore also a limit of a Bose-Einstein distribution), the leptons are massive fermions rather than massless photons, so they should follow Fermi-Dirac statistics. But yes, specifying charged particles would avoid these issues, and the power formula would be correct when the black hole temperature is low compared to the mass of the lightest elementary massive particle. $\endgroup$
    – Stan Liou
    Aug 10, 2015 at 16:14

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .