- #1
spookyfw
- 24
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Hello,
I'm currently going through Agrawal's book 'Nonlinear Fiber Optics' and got stuck with some mathematical cosmetics (pp. 40). It is the substition of:
\vec{P_L}(\vec{r},t) = \frac{1}{2} \hat{x} \left(P_L \exp{(-i \omega_0 t)} + c.c.\right)
into
\vec{P_L}(\vec{r},t) = \epsilon_0 \int_{-\infty}^{\infty} \chi^{(1)} (t-t') \cdot E(\vec{r},t') dt'
According to the book this should result in:
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'
under the assumption that \chi was diagonal, lumping \hat{x} and \chi^{(1)} (t-t') 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,
spookyfw
I'm currently going through Agrawal's book 'Nonlinear Fiber Optics' and got stuck with some mathematical cosmetics (pp. 40). It is the substition of:
\vec{P_L}(\vec{r},t) = \frac{1}{2} \hat{x} \left(P_L \exp{(-i \omega_0 t)} + c.c.\right)
into
\vec{P_L}(\vec{r},t) = \epsilon_0 \int_{-\infty}^{\infty} \chi^{(1)} (t-t') \cdot E(\vec{r},t') dt'
According to the book this should result in:
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'
under the assumption that \chi was diagonal, lumping \hat{x} and \chi^{(1)} (t-t') 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,
spookyfw
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