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Let a charge $q$ enter a constant magnetic field $B$ at 90 degrees, hence radiating EM energy-momentum and spiraling inwards to some point as it slows down. Time reversing this, including the current generating the magnetic field so it now becomes $-B$, the charge still radiates outgoing EM radiation and spiral inwards. All the events are reversed compared to the original case, so why isn't the EM radiation and spiraling reversed?

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    $\begingroup$ it is reversible, by highly unlike due to the second law of thermodynamics. It is very difficult in general to find initial conditions that will result in the effect you mention $\endgroup$ – Wolphram jonny Jan 31 at 22:16
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It is time reversible, you just have to remember to time reverse the field too. Absorption is the time reverse of emission. In the fully time reversed situation, the charge absorbs coincidentally perfectly arranged incoming EM radiation, which causes it to spiral outward.

In the standard derivation of EM radiation, you find that to satisfy the equations, an accelerating charge has to either emit outgoing radiation or absorb incoming radiation, or a combination of the two. To get the standard result, we neglect incoming radiation, but in your situation you can't.

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  • $\begingroup$ Why should the charge absorb the reversed incoming EM radation? Why not re-radiate it? $\endgroup$ – Physiks lover Jan 31 at 20:59
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    $\begingroup$ @Physikslover Well, the time reverse of emission is absorption, and Maxwell's equations are perfectly time reversible. $\endgroup$ – knzhou Jan 31 at 21:03
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    $\begingroup$ The situation with incoming radiation is part of the Wheeler-Feynman time-symmetric "absorber theory" of electromagnetism. (Feynman's dissertation, IIRC.) They showed that the apparent time asymmetry you identified is in fact a consequence of boundary conditions, and that if you assumed all outgoing radiation is fully absorbed and re-radiated (by the absorber) as both advanced (time reversed) and retarded radiation, they could derive the radiation reaction force on a charge. Feynman describes this in his autobiography. $\endgroup$ – Foster Boondoggle Jan 31 at 21:06

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