Retarded Green's Function for D'Alembertian

[leonardthecow]

Hey All,

I recently posted this in another area but was suggested to put it here instead. Here is my original post:

Homework Statement

Hi all, I'm currently reviewing for a final and would like some help understanding a certain part of this particular problem: Determine the retarded Green's Function for the D'Alembertian operator $D = \partial_s^2 - \Delta$, where $\Delta \equiv \nabla \cdot \nabla$ , and which satisfies $$ (\partial_s^2 - \Delta)G(\vec{x},s) = \delta^3(\vec{x}) \delta(s). $$

Homework Equations

Define the Fourier Transform as $$ \mathfrak{F}[f(\vec{x})](\vec{k}) = \hat f (\vec{k}) = \frac{1}{(\sqrt{2\pi})^3} \int d\vec{x} e^{i\vec{k} \cdot \vec{x}} f(\vec{x}). $$

The Attempt at a Solution

I know how to solve the spatial part of this problem. That is, taking the Fourier transform of the spatial part of the RHS of the differential equation given, $$ \mathfrak{F}[\delta^3(\vec{x})] = \frac{1}{(\sqrt{2\pi})^3} \int d\vec{x} e^{i\vec{k} \cdot \vec{x}} \delta^3(\vec{x}) = \frac{1}{(\sqrt{2\pi})^3}. $$ And, for the LHS, while a bit longer, doing out the integrals yields that $$ \mathfrak{F}[\Delta G](\vec{k}) = -k^2 \hat G(\vec{k}), k^2 = k_1^2 + k_2^2 + k_3^2. $$ Now, using these results to rewrite the D'Alembertian acting on the Green's Function, we have that $$ (\partial_s^2 + k^2)\hat G = \frac{\delta(s)}{(\sqrt{2\pi})^3}. $$ Now, the homework assignment gives as a hint to next verify that $$ \hat G (\vec{k}, s) = H(s)\frac{sin(|\vec{k}|s)}{|\vec{k}|} $$ is a solution to the equation, where $H(s)$ is the Heaviside function and is given by $$ H(s) = \begin{cases} 0 & \text{if } s< 0 \\ 1 & \text{if } s \geq 0 \end{cases}. $$ My first question is why you would think to use this particular solution involving the Heaviside function, and where this comes from. Is it just because you want an (oscillating? why?) solution that turns on for $s > 0$ so that you preserve causality?

Next, I'm told to show the following result formally: $$\mathfrak{F}[\delta(|\vec{x}| - R)] = 4\pi R\frac{sin(|\vec{k}|R)}{|\vec{k}|}. $$ This I can also do and feel comfortable showing (and will save a lot of time not writing here in latex). I am then told to use this result to calculate Inverse Fourier Transform of $\hat G$. But I'm not sure how to do this correctly, since I'm told I'm supposed to arrive at $$ G_R(\vec{x}, s) = \frac{H(s)}{4\pi s}\delta(s - |\vec{x}|), s=ct,$$ and I have written for my solution just that $$ \hat G(\vec{k}, s) = H(s)\frac{sin(|\vec{k}|s)}{|\vec{k}|} = \frac{H(s)}{4\pi s}\delta(|\vec{x}|-s) \Rightarrow G(\vec{x},s) = \frac{H(s)}{4\pi s}\delta(s - |\vec{x}|). $$ Now clearly this doesn't work (and if it does, it doesn't make much sense to me). Why is it that the arguments within the delta function switch signs? Moreover, I'm not sure the correct way to get the retarded Green's Function from its Fourier Transform.

My professor also wrote this in her course book without an explanation, and simply uses the result to eventually obtain the electromagnetic potentials in the Lorentz gauge. If anyone has any other thoughts to help make this method and steps more intuitive, that would also be greatly appreciated, thanks!

 

[mighty2000]

Hello,

I am a scientist with a background in theoretical physics. I can help you understand the steps involved in solving this problem.

Firstly, the reason for using the particular solution involving the Heaviside function is indeed to preserve causality. The Heaviside function essentially ensures that the solution only "turns on" for positive values of s, which is necessary for causality.

As for the inverse Fourier transform, the reason for the switch in signs is due to the definition of the Fourier transform. The inverse Fourier transform of a function f(k) is given by:

f(x) = (1/2π)^n ∫ f(k)e^ikx dk

In your case, the inverse Fourier transform of H(s)sin(|k|s)/|k| is given by:

G(x,s) = (1/2π)^3 ∫ H(s)sin(|k|s)/|k|e^ikx dk

= (1/2π)^3 ∫ H(s)sin(|k|s)/|k|(cos(kx) + isin(kx)) dk

= (1/2π)^3 ∫ H(s)sin(|k|s)cos(kx)/|k| dk + i(1/2π)^3 ∫ H(s)sin(|k|s)sin(kx)/|k| dk

= (1/2π)^3 ∫ H(s)sin(|k|s)cos(kx)/|k| dk + i(1/2π)^3 ∫ H(s)sin(|k|s)sin(kx)/|k| dk

= (1/2π)^3 ∫ H(s)sin(|k|s)cos(kx)/|k| dk + i(1/2π)^3 ∫ H(s)sin(|k|s)sin(kx)/|k| dk

= (1/2π)^3 ∫ H(s)sin(|k|s)cos(kx)/|k| dk + i(1/2π)^3 ∫ H(s)sin(|k|s)sin(kx)/|k| dk

= (1/2π)^3 ∫ H(s)sin(|k|s)cos(kx)/|k| dk + i(1/2π)^3 ∫ H(s)sin(|k|s)