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Maxwells equation and Doppler Shift.

  1. May 10, 2009 #1
    I am trying to grasp how Maxwell's equations and energy/photons are related in a doppler shift problem.
    The way I stated it seems to have attracted non-serious hecklers; so I am trying to restate the problem for people who know Maxwell's equations and are familiar with engineering.

    In Maxwell's equations, there are E & H fields which propagate at velocity=c as the the change in the E field generates a H field and vice versa.

    When radiation is incident on a perfect conductor (or appropriately designed dielectric reflector) 100% reflection of the energy occurs. For example, a perfect mirror or polished superconductor bolted down to a table -- and a laser bouncing off that mirror.

    If one allows the mirror to move -- say, in an attachment to a crooks tube radiometer -- the reflected light drops in frequency and power. The drop in power is related to the acceleration of the mirror which receives 2x the momentum of the energy striking it.

    Maxwell predicted that -- the momentum transfer being power/area * area * time / c; eg: p= E/c

    In my EE class, we worked problems of maxwell's TEM waves normally incident on a reflector by setting the boundary conditions such that the E field is zero inside the conductor/mirror, and assuming plane waves that meet the boundary condition.

    The reflected wave has the same amplitude as the incident wave -- so we know that all the energy is reflected.

    However; when the same technique is applied blindly to a moving mirror (with constant velocity non-relatavistic speed) the reflected wave frequency drops -- but the amplitude does not.

  2. jcsd
  3. May 19, 2009 #2
    I see your point. Although the photons seemingly lose Energy according to [tex]W=h\nu[/tex] the classic definition of energy for a normal wave would be [tex]W=E^2[/tex] and the amplitude does not change.

    The reason why your calculation must be too naive is this: when you only account for the transferred momentum to the mirror, you completely disregard the weight. Actually you are saying that the weight is infinite since you assume is is at constant velocity, so the momentum transfer doesn't effect it.
    For this reason and because [tex]W=\frac{p^2}{2m}[/tex] the mirror cannot receive energy.
    So your calculation seems to be correct, if you do not accelerate the mirror, because it is too heavy, then the light wave will not lose energy, even though the photons have a lower frequency.
    To kill the follow up question: So do we have more photons now? Yes we do.
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