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In classical relativistic Electrodynamics, we are often told that any accelerating point charge inherently radiates (Bremstrallung). (This is the basis for the need for a QM conception of electrons.)

We are also told that the innate (not due to the fields) energy of a moving particle is based on relativity, and is written as $E = \gamma m c^2$. Thus, the only dynamic dependence that the relativistic energy has is speed (not velocity ... no direction to $\gamma$).

This puzzles me. If both of these are true, then the speed (and thus the energy) of a point charge accelerating in a direction perpendicular to its velocity (i.e. turning without speeding up/slowing down) would not change during the acceleration. How then can it radiate away energy?

Note: I do realize that this radiation is experimentally confirmed. It is the generally accepted theory behind such observations that is leaving me confused.

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The earth turning around the sun has a kinetic energy given by 1/2Mv**2. It also has an angular acceleration en.wikipedia.org/wiki/… . It is the angular acceleration that enters the electromagnetic argument. The function of E does not enter in the argument. –  anna v Oct 4 '13 at 3:35
    
I don't see how this answers my question. The explicit assumption is that the particle moves around a curved path without speeding up/slowing down. In this case there would be no angular acceleration. Did I miss something? Please clarify. –  happyt Oct 5 '13 at 6:48
    
You are missing the definition of angular acceleration the change in the direction of motion is also an acceleration= dv(vector)/dt –  anna v Oct 5 '13 at 7:06
    
    
@anna v No. I got the fact that a change in direction is an acceleration. The angular acceleration is proportional to tangential acceleration (describing speed changes, not direction changes). For motion around a curve at constant speed, there is no angular acceleration, but there is radial (centripetal) acceleration. –  happyt Oct 5 '13 at 7:47
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The uniform circular motion is only a first approximation to the path of a classical electron orbiting the proton. When you add the effect of the radiation it's path rapidly turns into a inward spiral.

The energy $\gamma m c^2$ is the mass plus kinetic energy, but does not (as you note) include the potential energy of the system. The only part of that that is "intrinsic" in the sense that it can not be taken away from the electron is the energy associated with it's actual (i.e. rest) mass.

Not that this matters, because the electron actually gains kinetic energy (i.e. goes faster) as it gets closer to the proton, but the kinetic energy gained is more than offset by the potential energy lost (as it must be because electromagnetic energy is being carries out of the system).

BTW--it makes no difference to this argument that the electron has relativistic velocity.

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You say that "uniform circular motion is only a first approximation" to the electron path. Is this because the electromagnetic force is "time-delayed" due to the finite speed of field propagation? –  happyt Oct 5 '13 at 6:26
    
I said that the relativistic energy was "intrinsic" in the sense that it was not due to any external interaction. What word should I have used? If, for clarification, an electron could somehow be constrained to go in a uniform circular path, its intrinsic energy wound not be able to change. If so, then how could it radiate, despite its acceleration? Are you saying that constraining the electron to such a path somehow causally influences the E-M field causing it to spontaneously radiate? –  happyt Oct 5 '13 at 6:46
    
In the classical model the fact that the electron radiates is the main perturbation: it is the radiation which causes the path to spiral in. There is no way to reconcile fixed paths with Maxwell's equations classically. Bohr just didn't worry about it. Quantum Mechanically the electron does not follow a curved path it exists in a bound state that does not accelerate (has a zero derivative with respect to time). –  dmckee Oct 5 '13 at 14:26
    
You said that "There is no way to reconcile fixed paths with Maxwell's equations classically". By this do you mean that integrating Maxwell's equations for a charge/current density (to find functions for the fields) of a point particle (using a delta function) on a circular path at constant speed, would lead to some mathematical contradiction? –  happyt Oct 6 '13 at 7:00
    
I know that electron's radiate when they are accelerated (that is an experimental fact). Yet it is also known that electrons are not continuously radiating when they are "orbiting" the nucleus in a classical sense. Are you saying that Maxwell's equations cannot explain why the electron doesn't radiate when "orbiting" the nucleus, and thus a new type of mechanics (quantum) was needed? –  happyt Oct 6 '13 at 7:09
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