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Mathematical Reformulation of Polarization Equation

  1. Nov 10, 2014 #1

    I hope I got the section right ;). I orginally posted this in the physics section, but as the problem is more mathematical. It would be nice if someone knows the right direction.

    I've stumbled upon a math problem while going through some physics and got stuck with some mathematical cosmetics (pp. 40). It is the substition of:
    [tex] \vec{P_L}(\vec{r},t) = \frac{1}{2} \hat{x} \left(P_L \exp{(-i \omega_0 t)} + c.c.\right) [/tex]
    [tex] \vec{P_L}(\vec{r},t) = \epsilon_0 \int_{-\infty}^{\infty} \chi^{(1)} (t-t') \cdot E(\vec{r},t') dt' [/tex]
    According to the author of the book (Agrawal, 'Nonlinear Fiber Optics') this should result in:
    [tex] P_L(\vec{r},t) = \epsilon_0 \int_{-\infty}^{\infty}\chi^{(1)}_{xx} (t-t') \cdot E(\vec{r},t') \exp{(i \omega_0 (t-t'))} dt' [/tex]
    under the assumption that the tensor [itex] \chi [/itex] was diagonal, lumping [itex] \hat{x} [/itex] and [itex] \chi^{(1)} (t-t') [/itex] together makes sense. But what I don't get is how to integrate the exponentials into the integral. It looks like the shift theorem, but the sum of the two exponentials leaves me puzzled. Can anyone give me a hint?

    Thank you very much in advance,
  2. jcsd
  3. Nov 15, 2014 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
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