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Real photons do have frequencies, which is directly related to its energy. So, can virtual photons that take part in EM interactions have frequencies too?

When my hand is pressed up against a glass window, do the virtual photons taking part in the EM interaction keeping my hand from falling through the window have a frequency compared to the photons that passing through the window (visible spectrum) or those reflecting off it (frequencies the glass is opaque to)?

Also, since virtual photons may be massive and definitely have four-momentum, they definitely do have some energy - so is there any notion of frequency?

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  • $\begingroup$ What makes you think that they have an energy but no frequency? $\endgroup$ – user17574 Nov 16 '14 at 7:11
  • $\begingroup$ That's why I think they should have one. But I have never seen it discussed in physics courses. $\endgroup$ – XYZT Nov 17 '14 at 2:39
  • $\begingroup$ True, I was actually seriously wondering what your reasons were :) When we hand-wavingly introduced them in my particle physics course we simply treated them as normal photons with the exception that they cannot exist as asymptotic states, so I would've thought the same as you. But now you got some solid answers down there! $\endgroup$ – user17574 Nov 17 '14 at 8:38
  • $\begingroup$ Related: physics.stackexchange.com/q/19015/17609 $\endgroup$ – Keep these mind Apr 12 '15 at 15:32
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They have both frequency and wavelength (or wave vector $\mathbf{k}$), but the frequency and the wavelength are not related as in a real photon. Instead, they are independent integration variables of a Fourier integral representing a near field. For example, you can represent a "moving" Coulomb field $V(\mathbf{r}(t))$ as a Fourier integral over $\omega$ and $\mathbf{k}$.

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Virtual photons are not on the mass shell, that means that their mass can be different than zero in the calculations using the Feynman diagrams. They are not observable in any sense other than as the result of the calculations that need them. They appear because in an interaction quantum numbers are conserved, and these photons are used to carry the quantum number balance from the input particles to the output particles.

On mass shell photons have E=h*nu , and mass 0.

What does the frequency nu mean? It is the classical electromagnetic wave frequency that is built up by zillions of photons carrying a quantum of energy , and it was found experimentally that that quantum is equal to h*nu where h is the Planck constant and nu the frequency.

edit second time around:

The four vectors of the virtual photons in the Feynman diagrams are under an integration sign. That means that they will have variable four momentum vectors to the limits of integration

As there is no way to measure the energy of a virtual particle per se, or for an ensemble of them to compose a light beam , it has no meaning to ask whether it has a frequency.

The energy transferred by the individual molecules in your hand is very low , whereas the energies of visual light to which the glass is transparent are higher, so in any case there is no comparison. The real photons that will come from the pressure of your hand will be infrared photons.

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  • $\begingroup$ "As there is no way to measure the energy of a virtual particle per se, or for an ensemble of them to compose a light beam , it has no meaning to ask whether it has a frequency". Mass of virtual photons cannot be measured as well, and yet it is meaningful to talk about it? $\endgroup$ – bright magus Nov 16 '14 at 10:00
  • $\begingroup$ @brightmagus it is the length of the four vector that carries the same quantum numbers as the on mass shell particle. It is a known number $\endgroup$ – anna v Nov 16 '14 at 11:40
  • $\begingroup$ But if you can give a number for mass without measuring, and assume it is meaningful, why can't you extend this logic to energy? Mass is still equivalent to energy, isn't it? $\endgroup$ – bright magus Nov 16 '14 at 13:38
  • $\begingroup$ @brightmagus for off mass shell the "mass squared " (four dot product of fourvector) may be even negative, it has no meaning as "mass" except by position. $\endgroup$ – anna v Nov 16 '14 at 15:25
  • $\begingroup$ And it is assumed to be "meaningful"? $\endgroup$ – bright magus Nov 16 '14 at 16:44

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