Accelerating charged particle radiation reaction

  1. It is known that if a charged particle accelerates then it emits electromagnetic wave (energy). If so then this means that the work we do on particle, W=F*s, doesn't all go to particles kinetic energy, E=0.5*m*v^2. Then this means that Newton's F=m*a doesn't hold for charged objects, particles, masses, etc.. Is that true? If yes, then what resists particle to accelerate to the speed it deserves ( F*s=0.5*m*v^2, solve for v ). I hope i could explain my point. I am sorry i ask a lot about electromagnetism but it is so damn confusing to me, i cant find peace if i dont understand it properly.
     
  2. jcsd
  3. tiny-tim

    tiny-tim 26,041
    Science Advisor
    Homework Helper

    hi roboticmehdi! :smile:

    (try using the X2 button just above the Reply box :wink:)
    that's correct :smile:

    some of the work goes into the electromagnetic field, whose energy density increases

    F = ma still holds, but you have to include the force from the field!

    however, the effect is negligible in practice … a lot less than the air resistance which we also usually ignore! :wink:
     
  4. Why negligible? i think it is not negligible, for example in antennas, which try to deliver as much work as possible to electromagnetic waves, and for example a negatively charged sphere orbiting positively charged sphere of much bigger mass, the orbiting sphere would eventually lose all of its orbit energy to electromagnetic waves. then there are particle accelerators and etc. anyway, thanks for reply. can you tell me more about this ? or give good sources? what is that force acting on a particle? how it works ? thank you.
     
  5. tiny-tim

    tiny-tim 26,041
    Science Advisor
    Homework Helper

    but we wouldn't use F = ma for an antenna …

    what would be the body with mass m in that equation? :confused:
    "orbiting"? how would that happen?
    again, we don't use F = ma, we use more complicated field equations
     
  6. ok maybe not F=ma here but definitely F=d(mv)/dt
     
  7. i dont how to do this sorry. in my upper post some of my comments are inside the quoted text. dont miss it.
     
  8. anybody has other answers to my question ?
     
  9. We don't typically use F = m a directly with electromagnetic waves, but it is still there in principle. We talk more in terms of energies than forces. The power radiated by an antenna, for instance, is not calculated as the force times velocity, but rather as the energy flow rate. The conservation of energy equation in electromagnetics accounts for both a force giving kinetic energy to a charged particle and also causing energy to radiate away.
     
  10. vanhees71

    vanhees71 4,045
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    2014 Award

    This is a problem that is unsolved in classical electrodynamics. Classical point particles interacting with their own radiation field leads to equations with weird properties, predicting among other things self-acceleration.

    The best source to learn about this difficult problem is the book

    F. Rohrlich, Classical Charged Particles, World Scientific 2007
     
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