Why does vacuum particle-antiparticle creation and annihilation result in nothing rather than photons? What is the difference between that and regular annihilation that does result in photons.
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1$\begingroup$ Hi Alex, staying on site a bit longer for various reasons, this might help you, I asked related question physics.stackexchange.com/q/16851 $\endgroup$– user81619Commented Aug 19, 2015 at 17:06
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To get an understanding on quantum field theory issues, you have to understand the difference between virtual particles and real particles.
Virtual particles, in contrast to real particles, are a mathematical construct inspired by the Feynman diagrams used to describe interactions. These diagrams start with real particles, i.e. particles that have the mass and the quantum numbers of their name. In between, the mathematical formulations carry the quantum numbers of the named particle, but not the mass, because they are off mass shell.
One particle exchange scattering diagram
In the diagram above the dark lines are input and output real particles, on mass shell. The exchanged particle is virtual.
Now in quantum field theory the vacuum has fluctuations that can be defined mathematically as virtual particles continuously appearing and and disappearing within the bounds given by the Heisenberg uncertainty.
Why does vacuum particle-antiparticle creation and annihilation result in nothing rather than photons?
There is no energy to get the particles in the virtual loop to become real.
What is the difference between that and regular annihilation that does result in photons.
The vacuum "particle antiparticle" creation are diagrams involving only virtual particles. For the virtual particle antiparticle loop of the vacuum to have one particle become real it has to take energy from an interaction with a field, as happens in hawking radiation, interacting with the field of the black hole.
Real particle annihilation happens when the particle antiparticle are on mass shell and their energy contributes to the photons ( or other particles ) generated.
electron positron annihilation
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$\begingroup$ Thank you for answering. I have a related question that I would rather not get crucified for asking so I'll ask it here. Can a quark-antiquark pair interact with the field of a black hole and effectively isolate an individual quark? $\endgroup$– AlexCommented Aug 19, 2015 at 19:29
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$\begingroup$ @anna you say that virtual particles are a mathematical construct, which I kind of agree with, but then how would you explain the resonance at mass of the Z boson for ee scattering? $\endgroup$ Commented Aug 19, 2015 at 19:57
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$\begingroup$ Mathematics is useful in physics in modelling and predicting experimental observations. Within Quantum Field Theory the Z is a virtual particle; when it becomes real , i.e. at an ouside leg of a feynman diagram, its lifetime is very short, decaying in mu+mu_, e+e- etc. $\endgroup$– anna vCommented Aug 20, 2015 at 3:15
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$\begingroup$ @Alex In principle any pair creation follows the recipe, but strong interactions have the peculiarity of getting stronger with separation , which is what keeps quarks tied up in a proton. So single quarks cannot have one quark back into the black hole and the other free, because there are no free quarks. The energy that the interaction would take would be enormous ( thus becoming very improbable) and the free quark would bind up with some other antiquark left over from another pair, and make a pion . Thus one could have pairs of pions, pairs of proton antiproton etc, in $\endgroup$– anna vCommented Aug 20, 2015 at 3:22
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$\begingroup$ complicated loop diagrams . All these processes have very low probability of happening due to the large energies required from the gravitational field of the black hole. Photons are much easier to produce, even in complicated diagrams (from two e+ e- loops for example where the partners are eaten up) $\endgroup$– anna vCommented Aug 20, 2015 at 3:25