Cyclotron radiation?

  1. If I have a charged particle traveling into a constant B field it will bend its path and emit cyclotron
    radiation. If a magnetic field cant do work then what caused the charged particle to change its kinetic energy. I guess if I was in the electrons rest frame their would be an E field in that frame.
    Its just not clear to me whats going on.
  2. jcsd
  3. Remember the definition of work
    W = ∫F‧dl
    Applying the Lorentz force law in the absence of an electric field
    W = ∫(qv×B)‧dl
    Since dl is parallel to v [dl=vdt], it is perpendicular to v×B and thus the integrand is zero. Magnetic forces do no work because they do not change the kinetic energy. They only redirect the momentum vector, preserving kinetic energy. Cyclotron orbits are circular (absent any emitted cyclotron radiation), and this is analogous to the fact that the planets' kinetic energies are constant since the sun's gravity does no work, or the fact that a mass held in circular motion by a string does not increase in its kinetic energy due to work done by the string.

    PS: The more puzzling question is: how do those big magnets at junkyards pick up cars if magnetic fields do no work?
    Last edited: Aug 22, 2013
  4. The force comes from the self-field of the charged particle.
    The electric field of the particle and the background magnetic field combine to form a Poynting vector at every point in space around the charge.

    Consider a positive charge moving in a circle. Inside the circle, the Poynting vector points in the direction away from the motion of the charged particle, causing the particle to speed up. Outside the circle, the Poynting vector points in the direction of the motion of the charged particle, slowing it down. I'm going to make some hand-wavy arguments here. The curvature of the circle makes the outside slightly bigger than the inside, and the slowing down force exceeds the speeding up force. Or, if you take a time average over a full orbit, the inside field cancels out and the outside field remains.
  5. Khashishi is making a good point which touches on a very strange aspect of E&M. It appears to me he's trying to explain where the cyclotron radiation comes from--how can the charged particle lose energy if no work is done on it?

    The answer is indeed what Kashishi says: there is actually a force that is not included in the lorentz force law. The lorentz force law can be corrected to first order by adding to it the term [tex]\mathbf{F}_\mathrm{rad} = \frac{\mu_0 q^2}{6 \pi c} \mathbf{\dot{a}} = \frac{ q^2}{6 \pi \epsilon_0 c^3} \mathbf{\dot{a}}[/tex]This force is due to the charge accelerating--and it happens any time there is a change in the acceleration vector of the charge [I guess you could be fanciful and say that it's interacting with its own field but I don't totally buy Khashishi's poynting vector argument.]. In the case of cyclotron motion, the acceleration due to the B field always points to the center of the orbit, so the acceleration vector rotates at the cyclotron frequency. But the acceleration need not be from a magnetic field--it could be that the charge is connected to a rope or something. Thus the force that slows particles in cylcotron orbits (turning them into inward spirals) is due to the particle's acceleration, not that the B field is doing work on it directly. So I guess you could say that the magnetic field does work, albeit indirectly, if you include that term in the Lorentz force law.

    The weird thing about this is that once one includes the term in the Lorentz force law, this force causes its own acceleration and we need to correct the Lorentz force law to second order, etc. in an infinite regress. This is a problematic part of classical EM and many professors would say it only gets fixed in QED. So sometimes worrying about the radiation reaction terms can be a headache.
    Last edited: Aug 22, 2013
  6. Still waiting for someone to chime in with an answer.

    (a guess)
    Maybe the magnetic field can do work kind of indirectly, via a perpendicular force, that is the cross product of velocity and the magnetic field?
  7. Re posts #3 and #4…

    Does this mean there is now pretty much a consensus in the physics community that it is jerk and not acceleration which is the source of cyclotron radiation?

    To me that seems reassuring, since it was Feynman’s opinion, and it also is consistent with Einstein’s equivalence principle (since we don’t detect radiation from charged particles due to gravitational fields).

    Also I seem to recall from an article I read regarding this issue, that there still exists an unresolved issue related to the whether a charge is affected by its own electric field. Any new info about this?
  8. Drakkith

    Staff: Mentor

  9. Jano L.

    Jano L. 1,211
    Gold Member

    No, the radiation does not require jerk. The retarded solution of the Maxwell equations gives radiation (changes in EM field propagating at the speed of light) whenever the charged particle has non-zero acceleration.

    Feynman talked about something else - the Lorentz-Abraham formula for force of radiation reaction, which is given above.

    In gravitational field, EM field gets more complicated. I think it is better to postpone this until the concept of EM radiation in inertial frame is understood.

    Where did you read that? I would like to read that article.

    It depends on whether the charge is located at a point or has some non-zero spatial extension.

    If it has non-zero size (such as charge in the metal rods of an antenna), then the formula similar to that of Lorentz-Abraham can be derived from the EM theory, has approximate validity and is useful to estimate the forces one part of charge exerts on another part.

    If the charge is located at a point, the self-force idea always leads to inconsistency, so should be rejected.

    But since we do not know whether particles such as electrons are points or have small size (current limit is I think something like R < 10E-18 m), we do not know whether they experience self-force or not. So the article was right, at least for the electron.
  10. I have never seen a better answer to this question than that given by Griffiths (always a great explainer). Attached is the excerpt from Griffiths' "Introduction to Electrodynamics" where he addresses this question... sorry it's a bit long but the question does require some more in-depth analysis:

    Attached Files:

  11. Thanks for excellent answers everyone.

    The article is from They don't specify the author. From the third paragraph,third sentence on "The difficulty is in knowing how to correctly account for the influence of a charged particle on itself."...
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