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If we observe a charged particle like an electron passing us at some high speed $u$, then as $u \to c$ the field we observe looks like a superposition of plane waves normal to the trajectory of the electron. The field can be Fourier transformed, and the modes associated with virtual photons. See for example the discussion in chapter 19 of Classical Electricity and Magnetism by Panofsky and Phillips.

Is this virtual photon we talk about in classical electrodynamics the same as virtual photon that is the the force carrier in quantum electrodynamics?

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    $\begingroup$ @JohnRennie "Never say never": kirkmcd.princeton.edu/examples/virtual.pdf $\endgroup$ Commented Aug 8, 2022 at 7:15
  • $\begingroup$ @JohnRennie The book of classical electrodynamics by philips discusses about it.. $\endgroup$ Commented Aug 8, 2022 at 7:18
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    $\begingroup$ It is an interesting question. However, I suggest you add more context to it. Moreover, the meaning of the first sentence is not clear. Is it a question or a statement? $\endgroup$ Commented Aug 8, 2022 at 7:23

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It's the same photon but, in this context, it's part of a "back of the envelope" type calculation based on energy quantization ($E=hf=\hbar \omega$) and the assumption that each sufficiently small interval of the energy spectrum $U_\omega\mathrm d\omega$ consists of a number of "equivalent photons" $N_\omega$. It's probably a good approximation but it isn't the modern quantum field theory approach where the Number operator is defined in terms of the Hamiltonian/Energy operator and creation and annihilation operators.

The section on the "virtual photon" is in chapter 18. Section 3 "The Virtual Photon Concept"

Classical Electricity And Magnetism, 1929

Classical Electricity And Magnetism, 1929

The energy of the electromagnetic wave is (in terms of its Fourier components) is calculated.

$$U= \int_0^\infty U_\omega\mathrm d\omega$$

$$U_\omega =\frac{2}{\pi} \frac{e^2}{4\pi \epsilon_0 u} \ln\left(\frac{\gamma u}{\omega b_{min}}\right),$$ where $u$ is the velocity.

The number of "equivalent photons" $N_\omega$ is defined by $U_\omega\mathrm d\omega = \hbar \omega N_\omega\mathrm d\omega$ such that $$N_\omega\mathrm d\omega = \frac{2\alpha}{\pi} \ln(E/A\hbar\omega) \frac{\mathrm d\omega}{\omega}$$

and

$$N_\omega\mathrm = \frac{2\alpha}{\pi} \ln(E/A\hbar\omega) \frac{\mathrm 1}{\omega}.$$

$\alpha$ is the fine structure constant, $A$ is a numerical constant and $E$ is the energy of the particle.

Hence the spectrum of “equivalent photons” varies approximately as $l/\omega$. The number of equivalent photons per electron is small, namely, of the order of 1/137. Equation (18-38) is very useful in relating the probability of processes induced by electrons, or by other particles which act essentially only through their electromagnetic field, to the probability of processes induced by electromagnetic radiation.

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After searching lots of sources i haven't found a direct answer. But my suggestion is yes. It may not be entirely correct. But when $r→0$ We need to find a impact parameters which can't be zero because integral contain $1/x$ term that convert $\log(x)$ and $x=0$ is quite annoying. Therefore the lower limit definition required to use uncertainty principles.
$$XP=h$$ $$X=hc/(E_2-E_1)$$ The energy is corresponding to vertual photon we considering. What actually happens is that a vertual photon do exchange between them and colide with one anathor transfering momentum $dp$ or energy $dE$ and scattering result repulsion

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
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    Commented Aug 16, 2022 at 16:58

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