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Energy carried by an EM wave

  1. Apr 2, 2005 #1

    Given that the electric field vector of an EM wave is described as:
    [tex] E=E_0[1+\cos(\Omega t)]\cos(\omega t) [/tex]
    How would one go about finding the energy carried by this wave? On another note, what's the actual frequency of the wave that can be used in [tex] E=hf [/tex]? Graphed in a CAS, this wave actually looks like that of circularly polarised radiation ("packets" are discernible).

  2. jcsd
  3. Apr 2, 2005 #2


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    Use the formula for the energy density for a plane wave [itex]u = \epsilon_0 E^2 [/tex]. This is energy per unit volume, which can be multiplied by c to get the power per unit area, or intensity. You can get the average of this by integrating over one period, assuming the ratio of the frequencies is rational, and divding by the period. The E=hf formula only works for a photon with a well defined wavelength, and can't be applied here.
    Last edited: Apr 2, 2005
  4. Apr 2, 2005 #3
    Thanks a ton for the reply StatusX!

    Unfortunately, your suggestion only makes things more complicated for me. The original point of finding the energy carried by the wave was to find the kinetic energy of an electron scattered by this wave, with a work function W (I did not mention this earlier). Hence, I need simply energy, not power per unit area. With the latter, I'm forced to introduce arbitrary constants to get rid of little barriers like the "unit area" part of things. Constants which I cannot justify. Even with power, I must integrate over some arbitrary time to find the total energy transmitted to the electron, even though the question assumes that the collision is instantaneous. So I'm stuck.
  5. Apr 2, 2005 #4


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    Well, you could decompose this into a superposition of monochromatic waves. Use the identity cos(a)cos(b) = (cos(a+b) + cos(a-b))/2. Then I'm pretty sure this means there will be a mix of photons of three different energies, in a proportion found by the coefficients of the corresponding cos term.
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