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Special Relativity can be used to show that the magnetic force on a charge in parallel motion next to an infinite wire can be understood as an electrostatic force (when viewed from the rest frame of the charge). There are several posts and nice YouTube videos that explains this.

But how does this work in the case of light waves? Here magnetism cannot be an electrostatic effect in the rest frame, since no charges are present to produce the static field. So, from the perspective of special relativity, what causes the magnetic force of light in the rest frame of the charge?

(As a sidenote, I am also aware that the magnetic force is not an electrostatic effect in general. For non-parallel motion dynamic equations must be used, as this textbook shows)

Qmechanic
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Mads Vestergaard Schmidt
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  • You said that you are aware that the magnetic field is not an electrostatic effect in general. But this is one of those cases. Did you understand that or not? – naturallyInconsistent Apr 19 '23 at 18:11
  • @naturallyInconsistent - When we consider nonparallel motion next to a wire dynamic equations must be used to obtain the correct field of the charges. In case om EMR there are no charges, so that does not help us. – Mads Vestergaard Schmidt Apr 19 '23 at 18:36
  • No, I am saying that if you understood that you cannot just assume that all magnetic fields are due to Lorentz transforming an electrostatic field, then you should see that the magnetic field part of a light wave cannot be explained that way in the first place. – naturallyInconsistent Apr 19 '23 at 18:39
  • I see what you are saying. I am asking the question in a slightly naive way on pourpose. Since infinite wires does in exist, the magnetic force (not field) can never be understood as an electro static effect. It is always an electrodynamic effect. But a lot of readers are not aware of this detail, so I don't want it to take focus from the question. – Mads Vestergaard Schmidt Apr 19 '23 at 19:09
  • Infinite wire carrying a current or charged infinite wire? Or none of these two? – Mauricio Apr 19 '23 at 19:25
  • Electrostatics and magnetostatics are not even Lorentz invariant. All such scenarios have an often hidden special rest frame. Why are you expecting to derive a relativistic phenomenon from a non-relativistic border case? Static fields are the result of a poor mental model. They don't exist. It's just a pain to treat the quasi-static fields that are easily demonstrated on the lab bench with the relativistic theory. It's also pedantic to attempt that. That should not deflect from the fact that one approximation is not like the other. – FlatterMann Apr 19 '23 at 19:30
  • No, you are just completely wrong. $E^2 - B^2$ is a Lorentz invariant, so if you start with a purely static E field, you may generate some B field, but it can never produce a pure B field. Similarly, no amount of boosting can start from pure E field and get light waves, where |E| = |B|. – naturallyInconsistent Apr 20 '23 at 01:58
  • @Mauricio Infinite wire carrying a current. There is a couple of links where you can see what I have in mind. – Mads Vestergaard Schmidt Apr 21 '23 at 19:26

1 Answers1

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In short, what is experienced as a magnetic force in the moving frame of the charge, is experienced as an electric force in the rest frame, due to the Doppler shift of the amplitude of the electric wave. Einstein published the following amplitude transformation equation in 1905.

\begin{equation} A’ = \gamma\left( 1 - \beta \cos \theta \right) A \end{equation}

If we evaluate the combined electric and magnetic force in the moving frame, and the purely electric force in the rest frame, we can link them with the relativistic force transformations. When we do this, the amplitude transformation above can be derived. This shows us that the magnetic force of the wave accounts for the change in electric force the moving charge experiences, due to the Doppler amplitude shift of the electric wave it interacts with, in the rest frame perspective. You can find the derivation and full explanation in this preprint.

S

Electric wave with amplitude $A_E$ and wavelength $\lambda_0$ in the moving frame. The charge $q$ moves to the left with $v$ while the wave propagates to the right along $\hat{K}$.

S'

Same situation viewed in the rest frame of the charge. The Amplitude is now altered by the factor $\Gamma=\gamma\left( 1 - \beta \cos \theta \right) A_E$, and the wavelength by the factor $ \frac{1}{\Gamma} $.

Mads Vestergaard Schmidt
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    Why are you giving a weird tangent on your own question? – naturallyInconsistent Apr 19 '23 at 18:40
  • Because both question and answer was missing in here. Now it is here so people can find it. I see other users who make these Q&A style posts. For the record the explanation I give has been through peer review in two journals. It was rejected because they did not find it interesting enough, but aside from that all reviewers confirmed that the analysis is correct. You are welcome to provide a better answer to the question. – Mads Vestergaard Schmidt Apr 20 '23 at 07:23