Analyzing the Green function is through its frequency components

Question

In the book Introduction to Wave Scattering, Localization and Mesoscopic Phenomena (Springer Series in Materials Science) one reads the following

The specific initial condition for the Green funcion is that at $t=0$, the system is excited by a pulse localized at $\mathbf{r}=\mathbf{r}'$; i.e., the right-hand sides of $(2.4)$ and $(2.7)$ are replaced by $\delta(t)\delta(\mathbf{r}-\mathbf{r}')$, where $\mathbf{r}'$ denotes the source position. The reason for such a source is that it can excite the natural resonances (or eigenfunctions) of all temporal and spatial frequency in the system (since the source contains all temporal and spatial frequencies), so that the subsequent development may contain information about them. A useful approach for analyzing the Green function is hrough its frequency components. The left-hand side of the wave equation for the frequency components $\omega$ is given by $(2.8)$, and the same frequency component on the right-hand side is simply $\delta(\mathbf{r}-\mathbf{r}')$ because $\delta(t)$ gives $1$ as the amplitude for every frequency component. Equation both sides yields $$(\nabla^2+\kappa^2)G(\omega,\mathbf{r},\mathbf{r}')=\delta(\mathbf{r}-\mathbf{r}')\tag{2.16}$$

where

For waves of a given energy or frequency, the quantum and classical wave equations can be put in the same form $$\nabla^2\phi+\kappa^2\phi=0,\tag{2.8}$$ where $$\kappa^2=\dfrac{2m(E-V)}{\hbar^2}\tag{2.9}$$ for the quantum case, and $$\kappa^2=\frac{\omega^2}{v^2}\tag{2.10}$$

I don't see how to deduce eq $(2.16)$.

1. what does it mean to say that the left-hand side of the wave equation for the frequency component $\omega$?

2. why is it given by $(2.8)$?

Answer 1

You can think about the time dependent Green's function $G(\boldsymbol{r},t, \boldsymbol{r}^\prime,t^\prime)$ (remember primed variables are source variables, while unprimed are observation variables), as a pulse oscillating in time (for example in one-dimension, the pulse may be your hand waving a rope). Of course, such oscillating function can be Fourier transformed:

$$G(\boldsymbol{r},t, \boldsymbol{r}^\prime,t^\prime)=\int d{\omega}G(\omega,\boldsymbol{r},\boldsymbol{r}^\prime)e^{-i\omega t} \qquad (1)$$

so, if you introduce that into a wave equation, you obtain:

$$(\nabla^{2}-\frac{1}{c}\frac{\partial^2}{\partial t^2})G(\boldsymbol{r},t, \boldsymbol{r}^\prime,t^\prime)=-4\pi\delta{(\boldsymbol{r}-\boldsymbol{r}^\prime)}\delta(t-t^\prime)$$

$$\Longrightarrow (\nabla^2+k^2)G(\boldsymbol{r},t, \boldsymbol{r}^\prime,t^\prime)=-4\pi\delta{(\boldsymbol{r}-\boldsymbol{r}^\prime)}\delta(t-t^\prime) \qquad (2)$$

where $k=\omega/c$. Notice that, the delta function in the right hand side, tells you about the localization in space and time of the pulse in consideration.

Now, instead of working with all of the frequencies of the Green's function, you can work with one single frequency $\omega$ (you waving the rope with only one frequency). To do that mathematically, you should take the Fourier Transform of $\delta(t-t^\prime)$ and set an equality for each term corresponding to a signle $\omega$ of equation (2) above. That would yield your equation (2.16).

I think that the same idea is applied in your equation (2.8). It is equal to zero, maybe because they're considering a region where the source point $\boldsymbol{r}^\prime$ is not defined, so in that case the delta function would give zero.

If you want to investigate the mathematics more deeply, you may want to read Jackson's Classical Electrodynamics 2nd edition book, chapter 6 (section 6.6 more concretely, which is about Green's functions for the wave equation). Also, you can read this.