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Nonlinear optics waveform

  1. Aug 30, 2010 #1
    From a textbook - The reason why the polarization plays a key role in the description of nonlinear optical phenomena is that a time-varying polarization can act as the source of new components of the electromagnetic field....... the wave equation in nonlinear optical media often has the form:

    [tex]\nabla[/tex] 2 E - [tex]\frac{n2}{c2}[/tex] [tex]\frac{d2E}{dt2}[/tex] = [tex]\frac{1}{\epsilon c2}[/tex][tex]\frac{d2PNL}{dt2}[/tex]

    This equation is given with no derivation or justification. Can someone explain where this comes from?

    EDIT: I'm having a really hard time getting the equation to come out correctly on the website. Its nabla to the second power operating on the electric field E minus the second time derivative of E times n squared over c squared (where n is the linear refractive index and c is the speed of light) equal to 1 over epsilon c squared times the second time derivative of the polarization. I'll try to uplaod a photo of the equation.
     
  2. jcsd
  3. Sep 3, 2010 #2
    See Boyd's book, section 2.
     
  4. Sep 4, 2010 #3

    Redbelly98

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    Can't offer help, but I think this is what the equation in the OP is supposed to be:

    [tex]\nabla^2E - \frac{n^2}{c^2} \ \frac{d^2 E} {dt^2}
    = \frac{1}{\epsilon c^2} \ \frac{d^2P^{NL}}{dt^2}[/tex]
     
  5. Sep 9, 2010 #4
    thats exactly it - i think maybe cause im on a mac? or maybe cause i'm using firefox? I'll see if my linux machine does a better job.
     
  6. Sep 9, 2010 #5

    Redbelly98

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    You can click on the equation I wrote to see the correct LaTex code. For example, superscripts in LaTex are made using the "^" character, not the [noparse][/noparse] tags.

    Other users with macs have been able to write LaTex equations.
     
  7. Sep 12, 2010 #6
    It's pretty much the usual derivation of the wave equation, except with a nonlinear polarization term kept along for the ride. That is, take the cross product of Faraday's law, substitute in Ampere's Law, and simplify. You have to also assume that the E field is divergenceless (which is not strictly true here, but is what people do nonetheless).
     
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