My understanding is that Hawking radiation isn't really radiated from a black hole, but rather occurs when a particle anti-particle pair spontaneously pop into existence, and before they can annihilate each other, the antiparticle gets sucked into the black hole while the particle escapes. In this way it appears that matter has escaped the black hole because it has lost some mass and that amount of mass is now zipping away from it.

Is this correct? If so, wouldn't it be equally likely that the particle be trapped in the black hole and the antiparticle go zipping away, appearing as if the black hole is spontaneously growing and emitting antimatter?

How is it that this process can become unbalanced and cause a black hole to eventually emerge from its event horizon and evaporate into cosmic soup over eons?

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    $\begingroup$ The way you ask suggests that you confuse anti-matter and negative matter. Both matter and anti-matter is affected by gravity as we know it. When it's anihilated, it yields energy worth its mass. Negative matter, which is purely theoretical, would be pushed away from normal matter, reacting inversely to gravity. It would, as well, curve spacetime "up" instead of "down". Some have suggested that this could allow to construct structures which would let you exceed the speed of light. $\endgroup$ – Tomáš Zato Jun 17 '14 at 0:14

To add to Rory's answer-

The ability to radiate particles in a random, statistical way, is in a deep sense identical to an object having the property we know as "temperature." So, black holes have a temperature. It has a particular formula that is inversely proportional to the mass of the black hole. If you set that temperature equal to the current temperature of the Cosmic Microwave Background (CMB) that is 2.725 K, then you get a mass of about 4.503 X 10^22 kg, or a little over half the mass of the Moon. Black holes above this mass will be cooler than the CMB incident upon them, so will gather mass-energy from it. Black holes below it will lose energy due to Hawking radiation faster than they gain it from the CMB, so will head towards a catastrophic, runaway "pop." Note that the CMB is also getting cooler as time goes on, so the equilibrium mass shifts upwards. No one that I know of has bothered to do any detailed "race" calculations between a black hole's Hawking radiation and the changing temperature of the CMB.

Another important mass related to Hawking radiation is the mass at which the black hole is so cool that it would have emitted negligible radiation even if had been around since the beginning of the universe. This is about 2 X 10^11 kg, roughly comparable to the total mass of all humans.

The second mass is less than the first, so if a whole range of black holes had been created at the beginning of the universe, the upshot is that some would be popping right now! Astronomers are on the lookout for these events.


You sort of have the answer in your question - but you are assuming mass is positive, as opposed to viewing it as an amount of energy.

Since the particle that is emitted has positive energy, the particle that gets absorbed by the black hole has a negative energy relative to the outside universe. This results in the black hole losing energy, and thus mass.

Smaller primordial black holes can emit more energy than they absorb, which results in them losing net mass. Larger black holes, such as those that are one solar mass, absorb more cosmic radiation than they emit through Hawking radiation.

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    $\begingroup$ Is this kind of negative energy a real thing we can experimentally prove exists? $\endgroup$ – trampster Apr 10 '17 at 9:27
  • $\begingroup$ It's relative, not absolute. $\endgroup$ – Rory Alsop Apr 10 '17 at 9:57
  • $\begingroup$ In the 1st (1997) ed. of his popular science book titled "The Inflationary Universe", Guth makes a fairly convincing case (on its p.289-293) for ordinary gravity being "negative energy", but it's definitely not standard terminology. Curiel has an online "Primer on energy conditions", but it's a little too abstruse for me. $\endgroup$ – Edouard Jul 24 at 3:22

The virtual particle/antiparticle explanation is common, but (from what I understand) not very accurate; see e.g. this explanation by John Baez. To summarize it in less technical terms, spacetime near the black hole's event horizon is so strongly curved that what a nearby observer would call "absolute zero" (i.e. zero emission of radiation) looks like a greater-than-zero temperature to someone far away. That means the black hole is emitting energy, and by conservation of mass/energy the hole must be getting smaller as a result.


Simply put, the particles that pop in and out of existence are not matter/anti-matter pairs but rather virtual particles (particle/anti particle pairs) that both have net mass, therefore contributing to the net mass of the universe when one of them is swallowed up by a black hole at the event horizon and the other escapes.

It seems illogical to assume that the escape of one particle and the entrapment of the other could lead to a net gain in mass/energy of our universe, but the newly escaped particle is in fact a particle that we otherwise wouldn't have without the this effect at the event horizon.


First of all, you should know that anti particle and anti matter are two different names of the same thing. When matter and anti matter pop up near an event horizon, the theory days that sometimes one of them may get absorbed by the black hole and the other one runs free. Since both the particles are not eliminated because they didn't collide with each other, this violates the law of thermodynamics as it says that energy can neither be created nor destroyed. Note all this comes down to perspective. To an outside observer watching this happen from s distance, it will appear as if black hole is glowing and emitting some energy which is actually one of the particles escaping. Now black hole is losing mass or energy by absorbing one of the created particles to avoid violating law of thermodynamics.

  • $\begingroup$ How does a particle/anti-particle pair not colliding violate laws of thermodynamics? $\endgroup$ – Kyle Kanos Nov 21 '15 at 18:44

It seems logical to me that a black hole has no temperature, as temperature is the moving around of molecules, and there is no movement in a black hole, as it's all compressed so densely, in fact it's meant to be infinitely dense.

It also seems logical that a black hole with the mass of 100,000 stars would never allow anything to escape from it, ever.

If a particle of any type popped out, it would have to travel at faster than the speed of light until it reached the event horizon, which in a black hole of that size would be many kilometers away.

This theory has never been proven, it's impossible to test, and just basic logic suggests that all energy has gravity, and with the immense gravity of one of those black holes in the center of galaxies, nothing could escape ever, or if it did, it would get sucked straight back in.

  • $\begingroup$ This is just a bunch of speculation not based on any significant knowledge of physics. It would to look up the answers on this site to some of the points you raise: physics.stackexchange.com/questions/23099/…. $\endgroup$ – rghome Sep 12 '16 at 11:59
  • $\begingroup$ @rghome Note, currently it is far impossible to measure the Hawking-radiation, and also the QG is only an open problem, although the majority of the physicists think it exists. But, if it is isn't, then this answer may answer the original question (although some currently unknown QG effect is estimated instead of a singularity). $\endgroup$ – user259412 Sep 12 '16 at 12:18
  • $\begingroup$ @peterh I think the OP's question assumes Hawking Radiation and the OP already understands this is created outside the event horizon, so talking about how radiation can't get out from inside a black hole doesn't really answer the question since that is not what the OP asked and the OP already knows it can't. Obviously, we are talking about the current state of theory here. $\endgroup$ – rghome Sep 12 '16 at 14:12
  • $\begingroup$ @rghome Right, I had to read also the question, not only the answer. Thanks! $\endgroup$ – user259412 Sep 12 '16 at 14:41

protected by Qmechanic Nov 21 '15 at 18:17

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