# Why is Hawking Radiation not probabilistically symmetric?

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The properties Strange mentioned don't have an uncertainty relation. You don't get to violate them in combination with some other property or variable.

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Spooky action was his description of entanglement, not AFAIK tunneling.

We're discussing Hawking radiation which uses entanglement. You can see its debate in the no hair theorem under the Hawking radiation section on this link

Here it's also noted under firewall

http://en.m.wikipedia.org/wiki/Firewall_(physics)

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But entanglement is not relevant to the question of the OP.

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The two virtual particles are entangled, this is the aspect of the process that led to the quantum information paradox

The virtual particle that escapes becomes a real particle

Edited by Mordred
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The two virtual particles are entangled, this is the aspect of the process that led to the quantum information paradox

The virtual particle that escapes becomes a real particle

Yes, it does. And AFAICT the mass loss of the BH is not dependent on the information but rather the energy, so it is not inherently tied up in the information paradox.

Yeah true enough

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So what type particle is in a virtual particle then? Is it...a photon? So how does it collide with itself again if it doesn't have electric charge? Is it...an electron and positron? Why don't they produce a photon anyway?

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Virtual particles can be any type of particle: photon, electron, muon, quark .... Lower mass (energy) particles will exist for longer.

They can't produce a photon because that would violate conservation.

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So what type particle is in a virtual particle then?

You should think of virtual particles as the particles that only appear internally in Feynman diagrams. For now it is probabily best you just think of these diagrams as a nice way to keep track of the interactions that are allowed. For example when an electron interacts with another electron the photon they exchange is virtual.

Importantly, as these virtual particles are not directly observed they do not need to obey the standard laws of physics. One calls these things 'off mass-shell', they do not obey the equations of motion for classical particles.

Edited by ajb
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Well I'm aware of bosonic exchanges in the standard model between two real particles, but virtual particles in the vacuum of space near a black hole don't seem like typical bosons.

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The similarities is closer to Parker and Unruh radiation. Which is similar enough to the instaton. Or quasi particles as opposed to the SM particles.

virtual particles can be particle like.

However in the case of Hawking radiation as its thermodynamic, it's virtual photons

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So, the reason the black hole mass change due to virtual particles isn't symmetric is because black holes don't gain mass whenever a member of a virtual pair doesn't escape, they only lose mass when one does?

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Right one particle has to escape, keep in mind in order for Hawking radiation to work in terms of mass loss the blackbody temperature of the universe must be less than the blackbody temperature of the BH. If it's higher then the BH absorbs the surrounding temperature.

One key note on Hawking radiation, is it works via the particles frequency, the particle that escapes gains the frequency (flux,energy)of the infalling particle.

$\lambda_{H}=8\pi^2R_H$

$T=\frac{h}{2\pi}k$

$k=\frac{1}{2}f(R_H)=\frac{h}{8\pi M}$

The familiar form being

$T_H=\frac{hc^3}{8\pi GMk_b}$

As you can see from this the loss of mass isn't due to the anti particle as per se but due to that particle having a lower wavelength(energy) from the blackbody vacuum outside the BH, as opposed to the vacuum inside the BH. (Though there are other effects due to the entanglement, and anti particle,particle interactions inside the BH

Edited by Mordred

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