Solution for the Poisson Green's Function

To summarize, we are aiming to find the Green's function for the Poisson equation, which in electrostatics relates to the electric potential, or voltage, V (r), given a charge density ?(r).

By determining the Green's function, we can solve for any specified charge density by calculating the integral, or inner product, of the charge density with the Green's function.

It's crucial to note that the Laplacian operator is self-adjoint. The equation for our Green's function is:

For the delta function, the argument is represented as a vector r instead of a single variable x, which can be interpreted as a product of delta functions across each dimension.

Next, we apply the Fourier Transform. For clarity, I will omit the limits; from here on, empty limits will imply -? to +?. The Fourier Transform is given by:

The differential d³r is a convention in physics for a volume element, as we are performing the Fourier Transform in three dimensions.

It's important to note that although I am using a single integral sign, it actually represents a triple integral. I will clarify this further when we work through the integrals in detail, but this notation serves to keep everything organized.

On the right side of the Fourier Transform equation, we encounter the delta function. By invoking the sifting property, we can evaluate it.

The left side may appear more complex, but this is where the Fourier transform proves its power. Differentiating a function f(x) corresponds to multiplying its Fourier transform by ik for each dimension.

To validate this, we express G(r, r’) in terms of the Fourier Transform of the unprimed r:

where G? denotes the Fourier Transform of G or the unprimed r. The Laplacian operates solely on r, while the coordinates k or r’ are disregarded.

Since only the exponential term is dependent on r, it is the only term we need to differentiate, which is straightforward:

where:

This implies that the left side of the Green’s function equation simplifies to:

Our initially complex differential equation is now transformed into an algebraic equation. Dividing both sides by ?k² yields the solution for the Green's function in "Fourier space."

To derive the solution in real space, G(r, r’), we must apply the inverse transform.

While this can be quite challenging in general, there are several factors that work to our advantage.

Firstly, the equation for the Poisson Green’s function exhibits spherical symmetry, as the delta function is zero everywhere except at its location. The only relevant factor is the distance from the delta source.

This prompts us to attempt the integral in spherical coordinates and allows us the flexibility to orient the coordinate axes. Specifically, we align the vectors r and r’ along the kz axis, or the z-axis in Fourier space.

We can express the dot product in terms of the angle between the vectors (as shown in equation 14 of the main article):

Here, ? will be one of the angles we integrate in the inverse Fourier Transform.

Secondly, since we are working in three dimensions, the volume element is:

Where ? is the angle in the dot product, and ? is the usual azimuthal angle, ranging from 0 to 360 degrees, or 0 to 2?.

Substituting this into the integral allows the k² terms to cancel, resulting in a significantly simplified expression:

This may still appear daunting, but rest assured, it can now be solved using standard calculus II techniques, which we will break down step by step.

Firstly, observe that there are no k terms present, so that portion of the integral simply multiplies by 2?. For the theta integral, we employ a straightforward u-substitution.

This leads us to:

The last line should look familiar. Recall the definition of the sine function in terms of complex exponentials:

Thus, our final outcome for the ? integral is:

Incorporating the results of the ? and ? integrals back into equation 10, we are left with the last integral concerning the radial coordinate k:

This integral is the most challenging, but it is indeed a well-known integral.

To solve it, we can employ Feynman’s technique, as outlined in a post by Kensei S.

While it’s not overly complicated, Kensei provides a thorough explanation of all the details, so it would be redundant to replicate it here. I will simply share the result:

Finally, we compute the inner product with the charge density (as indicated in equation 2):

This completes the derivation.

A Final Thought:

You might have observed that the Helmholtz equation can be transformed into the Poisson equation by setting k = 0.

This suggests that a similar modification should apply to the Green’s function for the Helmholtz equation.

How could one adjust the Poisson Green’s function to retrieve the familiar 1 / r form as k approaches 0?

I have recently created a "buy me a coffee" link. If you appreciate this article and desire more content, I would greatly value your support. Thank you for reading. -Tom