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Years ago as an undergraduate, I studied new-physics contributions to the reaction $ cb \rightarrow tb $ in the case that the $t$ subsequently decays as $t\rightarrow be^+\nu_e$. I considered the observable $\theta_e$, defined as the angle between the center of mass 3-momentum of the top quark and the 3-momentum of the positron as seen from the top quark's rest frame. I calculated $\dfrac{d\sigma}{d\cos\theta_e}$, and at some point of the calculation I used a Lorentz boost into the $t$ rest frame.

As far as I know, Lorentz transformations only work when the transformation is done from one inertial reference frame to another. But as I have been considering the subject recently, it occurs to me that the $t$ rest frame in the example I described is actually non-inertial (possessing centripetal acceleration) if the reaction occurs in a magnetic field, which is exactly the case for interactions in the ATLAS detector and in many other particle detectors. My question is, how does one justify a Lorentz boost into the rest frame of a charged particle when that particle is in a magnetic field? My suspicion is that particle phenomenologists commonly perform Lorentz boosts to the rest frames of charged particles in magnetic fields, so it seems like a question worth asking.

Here's a guess for the example I gave above: the top quark has a sufficiently short lifetime ($\sim 10^{-25}$ s) that the arclength of its circular trajectory is small and its path looks approximately linear; maybe this means that we can think of its rest frame as approximately inertial. Is this guess correct? If it is, what about interactions involving charged particles with longer lifetimes? Alternatively, in the hypothetical situation where a particle interaction occurs in an extraordinarily intense magnetic field (for the sake of comparison with my top quark example, why not even consider $10^{25}$ Tesla?) can these kinds of Lorentz boosts become absolutely inappropriate? (I understand that magnetic fields on the order of $10^{25}$ are completely untenable; however, it illustrates the point to consider it. If the B-field is that strong, then perhaps even the top quark could make something like a complete circle before it decays in $10^{-25}$ seconds!)

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    $\begingroup$ I think the precise statement is that you didn't use the frame of the top quark; instead, you boosted into the inertial frame that happened to be moving with the same speed as the top quark at some moment in time. If the top quark happens to decay at that moment, you can calculate the result and then boost back to the original frame. $\endgroup$
    – knzhou
    Feb 14, 2022 at 4:47
  • $\begingroup$ But if you're interested in the whole history of a particle rather than just its moment of decay, then indeed, for cases where the magnetic field can seriously bend a particle's trajectory, you wouldn't treat it by going to the particle's frame. Instead you would solve for the trajectory in the lab frame. That's implicitly what's done by the trackers, to infer which particles passed through, and all the Monte Carlo programs. $\endgroup$
    – knzhou
    Feb 14, 2022 at 4:47
  • $\begingroup$ @knzhou The measured trajectories are fitted with classical charge/magnetic-field interactions. If one wants to consider the magnetic field effect at the point (small volume) where the QFT models the interaction, one would have to use higher order QED vertices containing the magnetic field . BUT if you put in the numbers, the effect of the dimensions of the macroscopic magnetic field, the strength of those diagrams would be infinitesimally small , imo. $\endgroup$
    – anna v
    Feb 14, 2022 at 5:17
  • $\begingroup$ @annav Yes, I agree, QFT is the best tool for the decay and classical physics is good for the trajectories! Indeed it should be possible to use QED to calculate the trajectories as well, though quite difficult... $\endgroup$
    – knzhou
    Feb 14, 2022 at 5:30
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    $\begingroup$ Almost anything can be estimated analytically... but numerics is more practical if you want the detailed result, 99% of the time. $\endgroup$
    – knzhou
    Feb 14, 2022 at 17:31

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