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Maxwell Equations: Radiation requires acceleration?

  1. Dec 16, 2009 #1
    In a lecture on Maxwell's equations, I noticed that for radiation to occur there has to be acceleration. Does this have any relation to specific heat? I have many questions regarding this, actually. If radiative heat is always mediated by photons, and radiation only occurs with acceleration, does that mean that light has infinite acceleration, less it would not be radiating? Forgive any apparent stupidity in the question!

    Also, I realize this is from classical physics, but I wanted to relate this directly to QED and understand how they are related, and the same questions apply to QED as well.
    Last edited: Dec 16, 2009
  2. jcsd
  3. Dec 17, 2009 #2
    I don't understand about your infinite acceleration!.
    But the amount of radiation or amount of light is directly proportional to the radiated power. Also i remember that there is a quantity E/(m0c^2) in that relation which determined the total flux. E - energy, m0 - rest mass and c velocity of light.
    Photons are produced from charged particles traveling close to c.
    I guess you know this book. Classical electrodynamics by Jackson [sure you will get all information there]
  4. Dec 17, 2009 #3
    Right, and that radiated power is dependent on the acceleration of the charged particle. Thus, my question is, if the charged particle is at zero acceleration, it is no longer radiating, correct?

    Now consider a system of those charged particles (an ice cube for example), all at zero acceleration, but not zero velocity. They would no longer be radiating heat (light), is this correct?

    What I am saying is, I find it odd that in order for there to be thermal radiation, that charged particles must be in a state of acceleration.
  5. Dec 18, 2009 #4
    Yes acceleration are important for production of synchrotron light..It is perpendicular to the tangent to the moving particle (i mean pointing towards the center or centripetal acceleration). Therefore, if acceleration is zero then no synchrotron light.
    If i am correct, experts (who know about Maxwell relations) could give a proper explanation of this fact
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