# Retarded Green's Function for D'Alembertian

• leonardthecow
In summary: H(s)sin(|k|s)(cos(kx) + isin(kx)) dk= (1/2π)^3 ∫ H(s)sin(|k|s)e^ikx dk= (1/2π)^3 ∫ H(s)e^ikx dk= (1/2π)^3 ∫ H(s)e^ikx dk= (1/2π)^3 ∫ H(s)e^ikx dk= (1/2π)^3 ∫ H(s)e^ikx dk= (1/2π)^3 ∫ H(s)e^
leonardthecow
Hey All,

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

leonardthecow said:

## 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!

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)

## 1. What is the Retarded Green's Function for D'Alembertian?

The Retarded Green's Function for D'Alembertian is a mathematical function used in the study of wave equations, specifically the D'Alembertian operator. It describes the response of a system to a point source at a specific time, taking into account the time it takes for the wave to travel from the source to a given point in the system.

## 2. How is the Retarded Green's Function for D'Alembertian derived?

The Retarded Green's Function for D'Alembertian is derived using the method of images, which involves creating a mirrored version of the system with a source at a negative distance from the original source. By superposing the two systems, the solution to the wave equation can be found.

## 3. What are the properties of the Retarded Green's Function for D'Alembertian?

The Retarded Green's Function for D'Alembertian has several important properties, including causality, which means that the response of the system only depends on past events. It also satisfies the wave equation, is symmetric under time reversal, and approaches zero as the distance from the source increases.

## 4. How is the Retarded Green's Function for D'Alembertian used in practical applications?

The Retarded Green's Function for D'Alembertian is commonly used in physics and engineering to solve problems involving wave propagation, such as in acoustics, electromagnetics, and fluid dynamics. It can also be used in the study of general relativity and quantum field theory.

## 5. Are there any limitations or drawbacks to using the Retarded Green's Function for D'Alembertian?

One limitation of the Retarded Green's Function for D'Alembertian is that it assumes a linear and isotropic system, which may not always be the case in practical applications. Additionally, it can be difficult to calculate in systems with complex geometries, and may not provide an accurate solution in all cases.

• Calculus and Beyond Homework Help
Replies
20
Views
743
• Calculus and Beyond Homework Help
Replies
5
Views
958
• Calculus and Beyond Homework Help
Replies
6
Views
2K
• Cosmology
Replies
2
Views
761
• Classical Physics
Replies
1
Views
510
• Calculus and Beyond Homework Help
Replies
5
Views
680
• Calculus and Beyond Homework Help
Replies
6
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
3
Views
935
• Calculus and Beyond Homework Help
Replies
1
Views
864