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Power radiated by several charges and the Larmor Formula

  1. Sep 22, 2014 #1
    Hello,

    The Larmor Formula tells how much power radiates a particle under the effect of acceleration if it's lonely on the space and it's often used to calculate radiative losses. However, I have seen some situations where the total radiated power cannot be obtained as the sum of Larmor formula over the charges.


    Let's consider this simple scenerio: we have only two charged particles (A and B) in the universe, both with the same charge (electrons). In some time to they have no relative velocity one another and are separated a distance D in the x axis.


    In that time both particles are suffering the same force but with opposite direction [itex]F_C = -F_D [/itex].


    Since v=0, the radiated electric field will is:

    [itex]E = (q/c)*(n x (n x a ) ) / ( [1 - n · v/c ]^3*R^2) [/itex]


    Where n is the unitary vector: n = R / R, a is the acceleration and R the distance from the charge to the point.


    Since the accelerations are the opposite, in a point P in the far distance R, where R >>> D, we can consider [itex]R_AP = R_BP[/itex] and therefore the electric field contributions from both particles will cancel one another given a null radiated power.


    Of course, if we see the field at distance R from A at time to + R/c the contribution of B will not be from to but from to - [itex]n_x[/itex]*D/c so the speed of B will not be exactly zero and the radiated field will not exactly cancel but we can neglect this little difference.

    This makes me to think that Larmor formula is not usefull to describe the radiated power in all the situations. Does anybody see any mistake in these statements?


    Best regards,

    Sergio
     
  2. jcsd
  3. Sep 26, 2014 #2

    Jano L.

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    Gold Member

    Your conclusion is correct. The Larmor formula is valid only for isolated particle with spatially extended charge distribution. If particle is a point, Larmor's formula is invalid (the assumptions of the Poynting theorem it is based on do not hold). If there are several independently moving charged particles contributing to the total field, the Larmor formula is not valid either (interference of the fields).
     
  4. Sep 26, 2014 #3
    Thanks Jano L. That was what I was suspecting.
     
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