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Homework Help: Electromagnetic Cross Section

  1. Apr 10, 2013 #1
    1. The problem statement, all variables and given/known data

    A plane monochromatic electromagnetic wave (polarized in x-direction and propagating in the z direction) is scattered by a free electron initially at rest.


    2. Relevant equations

    [itex]\vec{E}=E_0 \exp(i(kz-wt)) \hat{x}[/itex]
    [itex]\vec{B}=\frac{E_0}{c} \exp(i(kz-wt)) \hat{y}[/itex]

    3. The attempt at a solution

    For part (iv), the numerator of the cross section is the expression given. I'm not sure about the denominator...is it magnitude of the Poynting vector? I'm confused what "energy flux" exactly means in this context.

    I have not started (v), yet, but will once (iv) is clarified.
  2. jcsd
  3. Apr 11, 2013 #2
    The incident energy really is something like ε0E2,where E is the incident electric field for which you can take any given form.Also for calculating next one you have to take an average over initial polarization and final polarization states is calculated with angle θ given for two polarizations and then summed for getting the full cross section.
    Last edited: Apr 11, 2013
  4. Apr 11, 2013 #3
    Isn't [itex]\frac{1}{2} \epsilon_0 E^2[/itex] the energy density in J/m3?

    The Poynting vector has the same units as Energy/Time/Area, as required in the denominator. The question said electromagnetic wave, so I assumed there was a magnetic component as well. That's the reason I was thinking about the Poynting vector.
  5. Apr 11, 2013 #4
    sure,that is why there is no factor of 1/2 but 1(1/2 from electric and 1/2 from magnetic) and also multiply by c to get right units.
  6. Apr 11, 2013 #5
    Ah ok, that makes sense. But how do I get the quantity in the denominator then?
  7. Apr 13, 2013 #6
    Hey,sorry for being late but I will not be able to come here for some more days.However I will give enough hints so that you can solve this(Mentors if find any thing then delete it).Now the E electric field which contains in denominator need not be known.Just use newtons second law to find acceleration(a) in terms of E(that is very simple).use that a to put in numerator,you will see E will be canceled.Now for the next part,the cross section will contain |ε*.ε0|2 term,where ε* for outgoing wave and other ε for incoming one.you will have to take average over the incident one.which will lie in xy plane.just choose it along x axis say.Also for ε* take two polarization perpendicular to outgoing wave which should be perpendicular to each other also.For example,take first one in the plane defined by incident wave and outgoing wave and second one perpendicular to this plane.Take average over azimuth [itex]\phi[/itex] to solve it.Good luck.
    edit-Just one polarization is taken for incident one,but this one will be averaged.
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