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Without considering the radiation of the charge, the charge will move in circle in an uniform external magnetic field.

But when the radiation of the charge is considered, the charge will lose energy and momentum, result in a trajectory that is not a circle exactly. It could be a spiral trajectory that finally converge to a single point.

(1) Is there a numerically exact formular of the trajectory, if all radiation and relativity effect are considered but no quantum mechanics is considered (i.e. classical)?

(2) Will the trajectory of a charge with a higher initial speed (e.g. closer to light speed) shrink (relative to its initial circle trajectory which depends on the initial speed) faster than a charge with lower speed?

Answers doesn't need to address both subquestions.

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  • $\begingroup$ In case that the speed is reduced by radiation, the cycling radius willl get bigger, not "converge to a single point". $\endgroup$
    – ytlu
    Jan 22, 2021 at 4:16
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    $\begingroup$ @ytlu The paper I have cited in my answer does not agree with you. Figure 9 shows an inward spiral. $\endgroup$
    – G. Smith
    Jan 22, 2021 at 4:38
  • $\begingroup$ @ytlu If the paper I linked to is behind a paywall, read this instead. “The orbital radius, given as above by $r=mvc/qB$, decreases as the speed decreases: the electron spirals inward...”. $\endgroup$
    – G. Smith
    Jan 22, 2021 at 5:05
  • $\begingroup$ My mistake. $r = mv / qB$, I mixed the denominator with numerator. $\endgroup$
    – ytlu
    Jan 22, 2021 at 5:10
  • $\begingroup$ A similar question What is the equation of the spiral path of an electron under the influence of a magnetic field? $\endgroup$ Jan 22, 2021 at 19:55

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A 1961 paper by Plass in Reviews of Modern Physics, “Classical Electrodynamic Equations of Motion with Radiation Reaction,” considers the case of a point charge in a uniform magnetic field in section III 3(b), taking into account the Abraham-Lorentz-Dirac force. It states

An exact, physical solution can be obtained when the velocities are nonrelativistic.

and, in this solution,

the coefficient $\alpha$, which determines the exponential decay of the motion, increases in first approximation as the square of the Larmor frequency $\omega$.

However

It is not possible to solve the relativistic equations of motion exactly ...

So apparently the answer to your question (1) is No and the answer to (2) is probably Yes.

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