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Since the energy of the electron at rest can be calculated by:

$$ E_e=\frac{h c}{ \lambda_e} $$

where $\lambda_e$ is the Compton wavelength value of the electron at rest, $h$ the Planck constant and $c$ the speed of light and we know that the effective Compton wavelength of the electron reduces with acceleration of the particle from its rest value then we see that the energy of the electron increases with acceleration from the above equation.

However, my problem now is that according to the equation below for an accelerated electron:

$$ \lambda=\frac{h}{m_{(eff)} c}, $$

since the effective mass of the electron $m_{(eff)}$ increases with acceleration and the Compton wavelength decreases accordingly as shown in the above equation then the intrinsic spin magnetic dipole moment of the electron $μ_{S}$ cannot remain invariant from its rest value of one Bohr magneton $μ_{Β}$ but must also decrease with acceleration as shown by the equation:

$$ \mu_S=\frac{e \hbar}{2 m_{(eff)}} $$

How is that possible? I thought that the intrinsic spin magnetic dipole moment of the electron to be an invariant? Does this apply only for the electron at rest and not for an accelerated electron?

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Acceleration is an invariant. So something that is a function of acceleration only is also an invariant.

I don’t know if your calculation is correct, but the invariance is a non-issue as described

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  • $\begingroup$ do you mean it is an invariant as long as the particle is at rest? $\endgroup$
    – Markoul11
    Commented Aug 7, 2023 at 11:30
  • $\begingroup$ It does not even have to be under acceleration. Special relativity says that the effective mass of the electron increases with an increase in the constant speed value: $\mathrm{m}_{\mathrm{e}}=\frac{\mathrm{m}_{\mathrm{e}}(\text { in rest })}{\sqrt{1-(\mathrm{u} / \mathrm{c})^2}}$ $\endgroup$
    – Markoul11
    Commented Aug 7, 2023 at 13:07
  • $\begingroup$ @Markoul11 it sounds like your question then is not about acceleration but about velocity. So it is a duplicate of your previous questions on the velocity topic physics.stackexchange.com/questions/734052/… and physics.stackexchange.com/questions/733827/… I will leave this as an answer to the question you asked. It does not answer the question you intended, but the intended question is already answered elsewhere $\endgroup$
    – Dale
    Commented Aug 7, 2023 at 19:35

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