# The wavelength of the virtual photons of the Coulomb force

Is there a way to calculate the wavelength, frequency or energy (per photon) of the virtual photons that charges exchange to account for the Coulomb force? This is assuming that it's always the same wavelength, of course.

## Answers and Replies

They have wavelength just of the length between interacting bodies. If an electron is half angstrem away of a proton, the wavelength of the virtual photons will be half an angstrem.

I think this is wrong. It would mean they carried energy and they don't. The frequency is zero and the wavelength is unbounded.

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Vanadium 50
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I don't see how that can be correct.

Virtual photons don't propagate; they don't satisfy the wave equation, so how can they have a wavelength? What does a wavelength even mean in this case?

What does a wavelength even mean in this case?
Virtual particles break conservation of energy in the classical sense. The maximum amount of extra energy can be $$\Delta E = \hbar / d$$, where $$d$$ is the distance between interacting particles. If you now count the wavelength of a photon of such energy, you will get exactly $$d$$.

Vanadium 50
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I don't follow. The fact that you have a length scale d is not the same as saying it's a wavelength. Also, 4-momentum (one term of which is energy) is conserved at every vertex, so I don't see where this non-conservation of energy is coming from.

Thanks everyone. Didn't realize it was ambiguous. I thought that virtual photons were more like regular photons than they actually are I suppose. Thanks for the insight.