# Laplacian and spherical symmetry

Please can you explain me why

$$\nabla^2 G_{\omega}(R) = f(R)$$

is equivalent to

$$\frac{1}{R} \frac{d^2(RG_\omega(R))}{d R^2} = f(R)$$

Here:

• $G_\omega(R)$ is the Fourier transform of $G$
• $G$ is the Green's function
• We suppose that $G$ and $f$ are spherically symmetric
• $R = \sqrt{x^2 + y^2 + z^2}$

Thanks!

-

If $H(R)$ is spherically symmetric, we can write the Laplacian of $H$ in spherical coordinates:

$\nabla^2 H = \frac{1}{R^2}\frac{d}{dR}\bigg(R^2\frac{dH(R)}{dR}\bigg)$

(see wikipedia).

Then, $\frac{d}{dR}\bigg(R^2\frac{dH(R)}{dR}\bigg) = R^2\frac{d^2H}{dR^2} + 2R \frac{dH}{dR}$

On the other way, $\frac{d}{dR}(RH(R)) = R\frac{dH}{dR} + H$ and thus

$\frac{d^2}{dR^2}(RH(R)) = R\frac{d^2H}{dR^2} + 2\frac{dH}{dR}$

$\Rightarrow \frac{1}{R^2}\frac{d}{dR}\bigg(R\frac{dH(R)}{dR}\bigg) = \frac{1}{R}\frac{d^2}{dR^2}(RH(R))$

$\Rightarrow \nabla^2 H = \frac{1}{R^2}\frac{d}{dR}\bigg(R^2\frac{dH(R)}{dR}\bigg) = \frac{1}{R}\frac{d^2}{dR^2}(RH(R))$.

Now, apply it to $G_{\omega}$.

-
Thank you very much! – Abramo Oct 29 '10 at 13:36