In lecture, my professor was trying to motivate why $$v_{group}=\frac{dw}{dk},$$ and in doing so he started by claiming that in general, we can express a wave (in its complex notation) as $$\epsilon(z,t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}dk\int_{-\infty}^{\infty}dw \ \tilde{\epsilon}(k,w) \ e^{i(kz-wt)}.$$ Then, he went on to note that in a dispersive medium, $w(k)$ and wrote $$\epsilon(z,t)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}dk \ \tilde{\epsilon}(k)\ e^{i(kz-w(k)t)}.$$

The rest involved taking Taylor series of $w$ and plugging in, which I understand. But I have two questions regarding the two integrals here.

  1. Here, we have adopted the convention that the inverse Fourier Transform gets the -$i$. So if we are doing two inverse transforms, why does the one with respect to $k$ have +$i$?
  2. How can you go from integral (1) to integral (2)? Yes, the integrand is all dependent on $k$ but how can you simply neglect the integral over $dw$?
  3. How does the $1/2\pi$ turn into a $1/\sqrt{2\pi}$?

1 Answer 1

  1. You're free to specify that Fourier transforming space and time have different signs on the $i$. You just have to be consistent when you undo the transform!

  2. He is saying that for a wave in a dispersive medium, $\epsilon(k,\omega)=\epsilon(k)\delta(\omega-\omega(k))$ for some function $\omega(k).$ This says that a wave with wavenumber $k$ necessarily has frequency $\omega(k)$ in this medium. So you can't have amplitude on ALL Fourier components $e^{i(kx-\omega t)}$. You can only have amplitude on Fourier components with $\omega=\omega(k)$. Plugging this in, you can do the integral: $$ \frac{1}{2\pi}\int dk\int d\omega\ \epsilon(k)\delta(\omega-\omega(k))e^{i(kx-\omega t)}=\frac{1}{2\pi}\int \epsilon(k)e^{i(kx-\omega(k)t)} $$

  3. Your guess is as good as mine! Perhaps he actually defined $\epsilon(k,\omega)=\sqrt{2\pi}\epsilon(k)\delta(\omega-\omega(k))$, in order to make sure each Fourier transform has a factor of $\frac{1}{\sqrt{2\pi}}$ with it. But it's not important to the argument, is it?

  • $\begingroup$ YES, you are right, what you said makes sense. Thank you $\endgroup$
    – Ptheguy
    Commented Sep 27, 2017 at 2:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.