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When you study the spin-orbit interaction in quantum mechanics, even for a simple hydrogen atom, you find only the electric field in the nucleus reference system, while in the electron reference system you find the magnetic field due to the relative orbit of the nucleus around it (through Lorentz Transformations). Why is that? Shouldn't there be a magnetic field in the nucleus reference system due to the orbit of the electron as well? Is it just a convention, since electric and magnetic field are actually just one field?

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    $\begingroup$ Related: Spin-orbit coupling from the rest frame of the proton? $\endgroup$ Sep 2, 2014 at 13:56
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    $\begingroup$ I believe that indeed, in the rest frame of the nucleus, there is a magnetic field generated by the motion of the electron. But this magnetic field, being generated by the electron itself, doesn't interact with that electron. It's the fields generated by the nucleus that interact with the electron spin. $\endgroup$
    – gj255
    Sep 2, 2014 at 14:20
  • $\begingroup$ Using Lorentz Transformations, you put B=0, where B is the magnetic field in the nucleus frame reference. That's what I can't understand. $\endgroup$ Sep 2, 2014 at 14:46
  • $\begingroup$ But you only need to Lorentz transform the fields generated by the nucleus, since they're the only fields that are going to affect the electron. In the rest frame of the nucleus, there is no magnetic field generated by the nucleus. Hence we set B = 0. $\endgroup$
    – gj255
    Sep 2, 2014 at 15:12

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You do this mainly because we want to calculate the spin-orbit interaction precisely for the electron. And we do this because we observe that movement of the electrons (within the validity of this view) is decoupled from the movement of the nucleon.

This makes sense because the nucleus mass is about N*2000 times larger than that of the electron, where N is the number of nucleons in the nucleus. This is, for hydrogen, larger than the ratio of mass between the Sun and Jupiter.

So, while indeed what you are saying is correct, whatever the reference frame it is, the action of the electron on the nucleus must be considerably small and negligible for most purposes.

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