If the force felt when pushing an object is mediated by the electromagnetic interaction and hence photons, what is their frequency?
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The electrostatic force is mediated by virtual photons and one could say that they are not physical photons and they do not have a frequency, i.e. one could claim your question is invalid. However, one may also determine the frequency from the energy $E$ of the virtual photons (via $E=hf$) which actually is determined in each Feynman diagram, although the energy-momentum doesn't satisfy $E^2-p^2 c^2 = m_0^2c^4$ in general. The energy is equal, by energy conservation law, to $E_{1,{\rm final}}-E_{1,{\rm initial}}$. For the electrostatic force that only marginally changes the velocity of the charged particles, you have $$E=0$$ |
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Their frequency is determined with the velocities. For example, if you have a heavy, moving with $\vec{v}= const$ source charge that creates a time-dependent potentials $\phi(\vec{x},t)$ and $\vec{A}(\vec{x},t)$, then the Fourier expansions of those potentials will contain the following time-dependent exponentials: $\phi_{\vec{k}}, \vec{A}_{\vec{k}}\propto exp(-i\vec{k}\vec{v}t)$, i.e. the frequencies $\omega=\vec{k}\vec{v}$ are related to the wave vectors in a different way than for the true photons. Any probe charge will feel these "frequencies" in a superposition, as a unique time-dependent force $e\vec{E}(t)+\frac{e}{c}[\vec{V}_p ,\vec{B}(t)]$. Also, the number of those "photons" is uncertain, that makes the notion of virtual photons a useless concept. Finally, the transferred energy for a given scattering angle is determined solely with kinematics (the same for any kind of interaction potential), so the notion of the virtual photon is worthless. |
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