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Hi greysphere, you are definitely correct that one primary thing preventing velocity of the electron from exceeding than the speed of light is the presence of gamma in the relativistic force law, aka \partial_t (m_e \gamma v ) = q_e(E + v \times B), although the LHS doesn't quite equal \gamma m_e a, since \gamma also depends on v... In general I think it's fine to use Coulomb's law as an approximation in this case because the proton is much heavier than the electron and so we can just stay in the proton's reference frame and let the electron fall in from infinity (and we're ignoring QM and just doing relativistic EM here). We could also switch to a tritium nucleus and make it a bit better of an approximation, or indeed add a whole bunch more neutrons and get lucky that they don't beta decay to make it an arbitrarily good one. It is true that if the proton starts moving that you will no longer have a pure Coulomb field with respect to the original reference frame, as after a Lorentz boost the E field gets squished into the transverse direction somewhat, and you'll gain a B field swirling around the proton... Staying with the frozen proton approx, if we plug numbers in we get quite a bit of energy: set the proton radius r_p to 1E-15, and we get U = q_e^2 / ( 4 \pi \eps_0 r_p ) ~ 1.4 MeV, or a gamma of about 4, so yeah, it would be moving faster than c if we stayed with Newtonian mechanics. But there's another wrinkle: the 1.4 MeV of liberated potential energy won't all go into the electron's relativistic kinetic energy, because it is accelerating like crazy, especially in the final femtometers, and that acceleration (essentially Bremsstrahlung, although its not braking here) will generate an intense pulse of EM radiation as well - a decent fraction of the 1.4 MeV will go into that instead. You could perhaps estimate how much using the Larmor formula (in general calculating this radiation reaction force precisely becomes very complex, because the excitation of the EM wave modifies the acceleration, which modifies the excitation of the EM wave etc... And, now looking on Wikipedia, I'm not surprised to see that the first QM version of the calculation was done by Sommerfeld). So yeah, the electron will zip through the proton, with much of the potential energy converted to an EM pulse that zips off to infinity, and so the electron is now bound to the proton, and will continue to zig zag back and forth, emitting more radiation until it comes to a rest inside the proton. So yeah, we do need QM after all. |