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Maxwells Equations and Time Invariance

  1. Nov 26, 2011 #1

    In my book it says that if the dielectric function ε is time invariant, we can write a solution to Maxwells equations of the form E(r, t) = E(r)exp(jωt). I agree that the ME are separable, but I don't see how they know that the time-dependence is harmonic? What is so special about exp(jωt) in this respect?

    Last edited: Nov 26, 2011
  2. jcsd
  3. Nov 26, 2011 #2
    I think I understand it actually.. I just had to refresh my QM. Thanks.
  4. Nov 26, 2011 #3
    [itex]\epsilon\;[/itex] is not time invariant in the sense because it is frequency dependent which means there is a [itex]\frac{dV}{dt}\;[/itex] dependent. There is no pure lossless dielectric, there is always a loss tangent for all dielectrics.

    [tex]\vec E_{(\vec r,t)}=Re[\tilde E_{(\vec r)}e^{j\omega t}] \;\hbox { not }\;\vec E_{(\vec r,t)}=\tilde E_{(\vec r)}e^{j\omega t}[/tex]

    And The phasor [itex]\tilde E_{\vec r}\;[/itex] do have the attenuation constant that has frequency dependent part. The question on your statement is whether frequency depend implies time dependent.
    Last edited: Nov 26, 2011
  5. Nov 26, 2011 #4
    Well, there is two things here:

    1) ε might vary because of some external perturbation, but that will generally be very slow compared to optical frequencies, so in this sense can say that it is time invariant, so it is quasi stationary

    2) ε might be influenced by the field itself assuming the field is strong enough -> nonlinear optics

    I guess (1) is relevant for our discussion. That ε = ε(ω) means we have dispersion. Only if ε is complex do we have absorption.
  6. Nov 26, 2011 #5
    I am going to stop here, I don't know optics, I based on RF microwave tx line which is still EM, but the environment is completely different and I am not qualified to talk on this. Sorry.
  7. Nov 26, 2011 #6
    No need to apologize. Thanks for helping. By the way, the answer to my original question was (in principle) energy conservation.
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