4
$\begingroup$

When a photon produces an electron-positron pair, do both these particles have mass? Why or why not?

$\endgroup$
  • $\begingroup$ 1. The title talks about virtual pairs (which cannot be "produced" since they are just lines in a diagram), the body does not. Since photons can produce real pairs in pair production, you should perhaps also add the "virtual" in the body if you really want to ask about virtual particles. 2. What would it mean for a virtual particle to "have mass" (or to not have mass)? $\endgroup$ – ACuriousMind Jun 4 at 18:22
6
$\begingroup$

Do virtual electron-positron pairs have mass?

Virtual particles are within an integral depicted by a Feynman diagram

virtual

Only lines entering or leaving the diagram represent observable particles. Here two electrons enter, exchange a photon, and then exit. The time and space axes are usually not indicated. The vertical direction indicates the progress of time upward, but the horizontal spacing does not give the distance between the particles.

The virtual lines are described by a fourvector, and the "length" of a four vector is the mass for a free particle.In an integral this mass is variable within the limits of integration.

you ask:

When a photon produces an electron-positron pair, do both these particles have mass? Why or why not?

A photon has zero mass, and conservation of energy and momentum do not allow the decay into massive particles.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

A virtual pair is by definition "off shell". That is, it does not obey the usual mass relation for that kind of particle: $$E^2 = m^2 + p^2$$ Rather, it can have any $E$ and $\vec{p}$ such that 4-momentum is conserved at vertices. That means that the meaning of the "mass" of an off-shell system is pretty nebulous. You can, of course, choose to compute a value of $m$ for the virtual pair as though the usual mass relation were accurate. If you do, you will find that it could be anything, including zero.

Computing the mass of a virtual pair serves no useful purpose, nor does contemplating whether it is nonzero.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

A free photon may not produce electron-positron loops in a correct theory. In the current (incorrect) QED those corrections are mainly subtracted from the solutions - just to remove incorrectness of the original equations.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Any reference to support these bold statements? $\endgroup$ – Ruslan Jun 4 at 6:36
  • $\begingroup$ @Ruslan: Any QFT book that states that free equations are for non-physical, non-observable bare particles. $\endgroup$ – Vladimir Kalitvianski Jun 4 at 6:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.