Pair creation and annihilation

meopemuk

Thank you, strangerep, your answer is more than sufficient.

Eugene.

strangerep

Yes, I was just about to say the same thing. So let me try to answer the
original question...

It depends on what one means by "vacuum". If one has a naive picture of
the vacuum as empty space with no particles of matter in it, but allows
the possibility of static fields, then the previous answers are relevant.
However, in QFT the notions of particle and field become hard to
separate. The "vacuum" is merely the state of lowest energy and this
state is common to all types of particles and fields. So a region of
otherwise empty space upon which an intense Coulomb field has been
imposed certainly does not qualify as "vacuum" in the QFT sense.
So my answer to the original question is that spontaneous creation of
real (on-shell) particles from the physical QFT vacuum does not occur
(neither theoretically nor experimentally). If it did, the vacuum would
be unstable, and none of us would exist.

Demystifier

My point is, if we deal with collisions, then the field is far from being static. On the other hand, the Schwinger effect talks specifically about static fields. I am not surprised that a collision may lead to a pair creation, but I am very skeptical about the claim that a STATIC field may produce a pair. See, e.g.,
http://xxx.lanl.gov/abs/hep-th/0103251
http://xxx.lanl.gov/abs/hep-ph/0105176

passingthru

I have more elementary questions about pair production. Why does it have to be near a large nucleus. I've read that it's because of relativity and some observer may see a reduced frequency of the photon because of the relativistic doppler effect. Why does a large nulceus fix that problem? Is it because the mass of the nucleus is dilated and provides the extra energy/mass for the pair production? Also, I understand that if one of the pair is charged, the other particle has to have the opposite charge, but why do the particles have to have any charge at all? Is it because the only particles that we know of with small enough mass are charged?

malawi_glenn

The large nucleus makes the calculations simpler, since it's kinetic energy gain in the "collision" will be neglected. There is possibility for pair production "on" electrons as well (but then there will be triple production if I remember my class in radiation physics 3years ago LOL)

The particle has to be charged, the photon couples to electric charge.

Bob S

Virtual pair production, and its polarization in a Coulomb field, leads to the renormalization of "bare" charge. The charge that we all use in our calculations is the renormalized charge, corresponding to 1.6 x 10^-19 Coulombs per electron (or proton). The present renormalization calculations are based on an article published by Uehling in 1935 (google Uehling integral). When two charges are closer than an electron Compton wavelength, the charges begin to "see" some un-renormalized charge, and this correction (called vacuum polarization) must be included in energy calculations. Perhaps the best example of this are the energy level calculations of muonic atoms (because the muons are very close to nuclei), where the measured atomic transition energies agree very well with calculations using a muon mass based on muonic g/m and g-2 experiments. The comparison of electron and muon g-2 measurements also provide a very accurate estimete of the virtual electron-positron pair production in a Coulomb field.
Pair production and annihilation is not limited to electrons. In muonic g-2, for example, there are corrections for both electronic and muonic pair production. In principle, any charged particle-antiparticle pairs could be included, but the higher mass severely limits the rate.

Abbas Sherif

"the energy exceeding the rest mass of the two particles which in turn means exceeding the rest energy of the two particles". Are you implying translation from virtual particles to real particles violates conservation of energy? I find this hard to believe. Please make yourself clearer.