# Why doesn't Hawking radiation cancel itself out?

Hawking Radiation is formed when particle, anti particle pairs formed by the uncertainty principle are separated by the event horizon of a black hole. It seems like an equal amount of particles and anti-particles should end up on each side of the event horizon. So why don't the particles and anti-particles annihilate with a new partner once separated from their original partner by the event horizon? Thus canceling out any radiation released.

This is covered by a number of existing questions, but I think it does no harm to present a fresh summary.

Firstly the analogy of the black hole absorbing one member of a pair of virtual particles is just an analogy, and actually a rather poor one, but let's go with it for now. In this analogy you're quite correct that equal numbers of particles and particles would be produced, but they wouldn't cancel each other out. An electron and positron don't just disappear when they meet, they annihilate into two 511keV photons. So a black hole producing equal numbers of particles and antiparticles would be a net emitter of photons.

But this isn't what happens. If the above analogy were correct you'd expect the Hawking radiation from a black hole to be sharply peaked. For example you'd have a peak at 511kev due to electron-positron annihilations, but no radiation below that until you got to the peaks for the neutrino annihilations (whatever mass they are). But what Hawking actually predicted is that the radiation would have a black body spectrum with no sharp peaks at all. This is the origin of the infamous information paradox.

The effect is actually due to the curvature of spacetime near the black hole. You may have heard that the curvature causes odd effects, for example time runs slower for observers hovering near the black hole, and distances measured radially change. These changes are what cause gravitational lensing, so we have real experimental evidence that they happen - this isn't just the wild rantings of theoreticians.

Anyhow, the curvature also means that a quantum field theory vacuum measured near to the black hole looks different from a quantum field theory vacuum measured far from the black hole. If you're interested, the two are related by a Bogoliubov transformation, though the Wikipedia article I've linked will be utterly incomprehensible to non-nerds. The result is that what the observer near the black hole considers to be a vacuum looks to an observer far from the black hole to be a black body glowing at a temperature of:

$$T = \frac{\hbar c^2}{8\pi GMk_B}$$

There is a more detailed description of this in the Wikipedia article on Hawking radiation. This is fairly mathematical, but armed with my general description above you should be able to follow the main points.

Hawking's argument is quite general and doesn't rely on any specific mechanisms, like particle-antiparticle annihilation, and that's one of the reasons it predicts a featureless black body spectrum.

Ummm, not exactly. The spectrum emitted by macroscopic black holes is essentially a thermal spectrum of neutral particles, i.e. photons. If neutrinos were massless, they too, would be emitted, but with current estimates for neutrino masses, stellar size black holes are way too cold to emit even neutrinos. All known black holes are much colder than the current universe, and they are all net absorbers of cosmic microwave radiation. Hawking's prediction of black holes that could emit charged particles would require very small black holes that are assumed to be remnants of the early universe. To the best of my knowledge no such radiation has been detected, so far, and there are currently no successful models that predict the existence of a large enough density of these objects for Hawking radiation to be detectable, except, maybe, one day in an accelerator experiment, if we can accelerate particles near the Planck limit.