Is hawking radiation more to do with energy than mass?

25 February 2011

Question

I was listening to the most recent episode of Naked Astronomy and in particular I noted Andrew Pontzen's response to the question from David Walker about whether there was a bias as to whether a particle or an antiparticle fell into a black hole in the particular depiction of Hawking radiation where such a pair of particles is produced around the event horizon.

Now, I am not a physicist, but it seems to me that Andrew's explanation was just plain wrong.  It seems to me that it does not matter whether the particle that escapes the black hole is a particle or an antiparticle...the black hole will lose mass equivalent to the energy of the particle emitted, so no bias in absorbing particles or antiparticles need exist.  The issue is not whether matter or antimatter falls into a black hole--in both cases, the black hole will grow by the mass of the object

I think that it's helpful to think of this in terms of the equivalency of matter and energy.  In this view, matter and antimatter are two different manifestations of energy that have the property of transforming into gamma radiation when corresponding particles of the two types (like a positron and an electron, a proton and an anti-proton, etc.) interact.  However, this interaction does not destroy the energy of the two particles.  Instead, it is emitted in the gamma radiation from the reaction.  As such, such an interaction occurring within the event horizon would have no effect as regardless of the form of the energy, it would continue on toward the singularity.

An antiparticle falling into a black hole makes the black hole more massive because antiparticles, just like their matter counterparts consist of a positive amount of energy.  So an antiparticle annihilating with a bit of matter falling into a black hole would not have any effect on the black hole's end mass--the gamma photon produced would still become a part of the black hole's mass, which is merely a manifestation of its energy.

So how does the conventional Hawking radiation explanation work?  In works on the properties of virtual particles, which come in and out of existence with no net change in energy.  In the case of Hawking radiation, a matter-antimatter pair form around the event horizon and one of the pair leaves the black hole's gravitational pull while the other falls in.  While it would seem that the one falling in would make the black hole larger, it must be remember that these particles came into being out of no energy and so the black hole ends up paying the "debt" of the particles' creation with its mass and this debt happens to be twice as large as the particle that fell in, as the other one escaped.

Ben Main
Neenah, WI, USA

Answer

This is Ben Main's complete question - for Andrew's answer, please listen to the podcast...

I was listening to the most recent episode of Naked Astronomy and in particular I noted Andrew Pontzen's response to the question from David Walker about whether there was a bias as to whether a particle or an antiparticle fell into a black hole in the particular depiction of Hawking radiation where such a pair of particles is produced around the event horizon.Now, I am not a physicist, but it seems to me that Andrew's explanation was just plain wrong. It seems to me that it does not matter whether the particle that escapes the black hole is a particle or an antiparticle...the black hole will lose mass equivalent to the energy of the particle emitted, so no bias in absorbing particles or antiparticles need exist. The issue is not whether matter or antimatter falls into a black hole--in both cases, the black hole will grow by the mass of the objectI think that it's helpful to think of this in terms of the equivalency of matter and energy. In this view, matter and antimatter are two different manifestations of energy that have the property of transforming into gamma radiation when corresponding particles of the two types (like a positron and an electron, a proton and an anti-proton, etc.) interact. However, this interaction does not destroy the energy of the two particles. Instead, it is emitted in the gamma radiation from the reaction. As such, such an interaction occurring within the event horizon would have no effect as regardless of the form of the energy, it would continue on toward the singularity.An antiparticle falling into a black hole makes the black hole more massive because antiparticles, just like their matter counterparts consist of a positive amount of energy. So an antiparticle annihilating with a bit of matter falling into a black hole would not have any effect on the black hole's end mass--the gamma photon produced would still become a part of the black hole's mass, which is merely a manifestation of its energy.So how does the conventional Hawking radiation explanation work? In works on the properties of virtual particles, which come in and out of existence with no net change in energy. In the case of Hawking radiation, a matter-antimatter pair form around the event horizon and one of the pair leaves the black hole's gravitational pull while the other falls in. While it would seem that the one falling in would make the black hole larger, it must be remember that these particles came into being out of no energy and so the black hole ends up paying the "debt" of the particles' creation with its mass and this debt happens to be twice as large as the particle that fell in, as the other one escaped.Ben MainNeenah, WI, USA

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