Do all virtual particle travel at light speed in a vacuum? else wouldn't that imply they should have rest mass however tiny? When they pop back out of existence do their mass disappear instantly? BTW what is the heaviest virtual particle ever found?


Do all virtual particle travel at light speed in a vacuum?

Light speed is the limit for any transfer of energy/momentum and information.

else wouldn't that imply they should have rest mass however tiny?

As virtual particles are described by a four vector, it will have a length value which by definition is the invariant mass of a particle. Real particles have positive and fixed invariant mass. Virtual particles can have any value of invariant mass allowed within the limits of integration, where they are defined.

Here is the definition of a virtual particle, in this pictorial representation of the integration that must be carried out to get the crossection of e-e- scattering.


Virtual particles live only within integration limits, they have the quantum numbers of the named particle but their mass is off shell, within the limits of the implied integration.

When they pop back out of existence do their mass disappear instantly?

They do not exist outside integration limits, which supply the energy for the interaction. If you are thinking of vacuum loops of pair produced particle antiparticle, they can only exist in corrections to real particle interactions. If no real particles supply four vectors for the interaction, there are no observable virtual particles.

BTW what is the heaviest virtual particle ever found?

Virtual particles cannot be observed. They can be stated as a mathematical hypothesis, but their mass has to be within the limits of the integration.

In e+e- annihilation , the closer to the mass of the Z the incoming energy is, the closer the virtual Z is to the on shell mass of 90+ GeV of the Z.


  • $\begingroup$ thanks, may I know what is another term for limits of integration because google results only show up math? $\endgroup$ – user6760 Mar 2 '18 at 8:09
  • $\begingroup$ The feynman diagrams are a recipe for computing interactions with prescribed integrals. The limits go usually over the square root of s as above in the last plot, or over four momentum tranfer, Q^2 , depending on the observables to be fitted/predicted. see here physics.stackexchange.com/questions/389068/… for the feynman diagrams for above e+e- plot $\endgroup$ – anna v Mar 2 '18 at 8:13
  • $\begingroup$ why are limits of integration of invariant masses of the "virtual particles" defined only on real values? shouldn't they be allowed to have values in all the complex plane? $\endgroup$ – lurscher Dec 17 '18 at 13:59
  • $\begingroup$ @lurscher afaik in feynman diagrams q^2, energy, momentum are all defined as real numbers. The end result of the calculation has to be real numbers otherwise there is no prediction to compare with measurement. We measure real numbers. $\endgroup$ – anna v Dec 17 '18 at 14:14
  • $\begingroup$ if $q^2$ is restricted to the real line, then $q$ must belong on the union of the real and the imaginary line, right? $\endgroup$ – lurscher Dec 17 '18 at 14:18

Massive fields can give rise to virtual particles (e.g. virtual electrons). They can have any momentum since they aren't required to satisfy the mass shell condition $p^2=-m^2$.

Remember that rest mass isn't necessarily conserved during relativistic scattering events. What is conserved is four-momentum. Virtual particles aren't an exception to this rule. As far as I know, the top quark is the heaviest fundamental particle (virtual or otherwise) ever found.

  • $\begingroup$ Just curious is virtual electron the same excitation of the electron field as with electron? If so how come one is extremely short lived while the latter seems to outlive almost everything in the universe? (Sorry I'm easily side tracked) $\endgroup$ – user6760 Mar 2 '18 at 4:33
  • $\begingroup$ there was something wrong with your formula, if p is the momentum. have replaced it $\endgroup$ – anna v Mar 2 '18 at 7:38
  • $\begingroup$ There wasn’t a problem with it, $p$ was the four momentum. Was that not obvious from the form of the equation $p^2=-m^2$? $\endgroup$ – BRT Mar 2 '18 at 11:53

It is my understanding that, at least from my basic familiarity with the Casimir effect, virtual particles can achieve significant duration. They may well have a rest mass. How the binaries that pop out related to Newtonian physics, I can't say. But I think that when they pop out, they rather expel a measurable amount of radiation - Hawking radiation, if I'm not mistaken (which is likely). But this radiation is how they detect black holes.

  • $\begingroup$ if they can achieve significant duration can they fuse into a virtual atom? my understanding is that they are immediately annihilated by their counterpart after popping into existence. $\endgroup$ – user6760 Mar 2 '18 at 3:41
  • $\begingroup$ In the Casimir context, they exert a force, like mass, but negative. They pull rather than repel. I'm not sure what differentiates them from being atomic at that point, other than having the complete opposite effect of matter. $\endgroup$ – Austin Alexander Mar 2 '18 at 3:51
  • $\begingroup$ In the context of Hawking Radiation, it is required that they go their separate ways... just one of them has to go into the black hole's singularity. $\endgroup$ – Austin Alexander Mar 2 '18 at 3:52
  • $\begingroup$ I'm asking this is because I'm thinking about double slit experiment, an electron actually appears everywhere along all possible paths as virtual electron each with their respective quantum states until one of them are forced into existence. Is this thinking dangerous? $\endgroup$ – user6760 Mar 2 '18 at 4:13
  • $\begingroup$ @user6760 look at this answer of mine physics.stackexchange.com/questions/389279/… . Do not confuse the mathematics with the experimental results . You are confusing the formalism of paths, i.e. how one can calculate an interaction, with reality. $\endgroup$ – anna v Mar 2 '18 at 8:04

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