Spacetime Double Fourier Transform to get a wave

Question

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 $1/\sqrt{2\pi}$?

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?