This addendum serves as supplementary material for a recently published piece in Cantor's Paradise discussing Green's functions.

In the final part of the article, I elaborate on the Poisson equation within electrostatics and clarify the significance of the Green's function, which represents the potential arising from a point charge. This is often denoted as 1 / r, along with additional factors.

Originally, I planned to conclude the article by deriving the Green's function. However, with the text already exceeding 3000 words, such an inclusion seemed more tedious than enlightening. Instead, I opted to share this information on my personal profile for those interested.

The approach to deriving the Green's function is quite straightforward:

1. Perform the Fourier Transform.
2. Determine the Green's function in Fourier space using simple algebra.
3. Apply the inverse transform to revert to real space.

Steps one and two are relatively uncomplicated, while the third poses more challenges. Given that this is a three-dimensional equation, the inverse Fourier Transform involves three separate integrals. Two of these are manageable, but the last requires more effort; however, it is a well-known integral that can be looked up.

This same methodology applies to the Green's function for the Helmholtz equation.

In the case of the Helmholtz equation, the last integral is significantly more complex. The only solution I am aware of involves contour integration within the complex plane, which I will address in a separate article. Enjoy.

The solution for the Poisson Green's function is as follows:

To recap, we aim to find the Green's function corresponding to the Poisson equation. In the realm of electrostatics, this equation describes the electric potential, or voltage, V(r), based on a given charge density ?(r).

By identifying the Green's function, we can compute the potential for any specified charge density by integrating the charge density with the Green's function.

First, we observe that the Laplacian operator is self-adjoint. The equation for our Green's function is:

Note that for the delta function, the input is a vector r rather than a single variable x, which we can interpret as a product of delta functions for each spatial dimension.

Next, we perform the Fourier Transform. For clarity, I will exclude the limits. Here, empty limits imply from negative infinity to positive infinity. The Fourier Transform is:

The differential d³r is a convention in physics for a volume element since we are dealing with a three-dimensional Fourier Transform.

It is important to note that while I am using a single integral notation, this actually represents a triple integral. I will clarify this further when we analyze the integrals in detail, but this notation serves to keep the expressions organized.

On the right side of the Fourier Transform, we have the delta function, which we can evaluate using the sifting property.

The left side appears more complex, but this is where the Fourier Transform proves advantageous. Differentiating a function f(x) corresponds to multiplying its Fourier transform by ik across each dimension.

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

Here, Gˆ denotes the Fourier Transform of G or the unprimed r. The Laplacian operator only acts on r, while the coordinates k or r’ are disregarded.

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

Where:

This leads us to the outcome for the left side of the Green's function equation:

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

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

This process can be particularly challenging in general, but a few favorable factors emerge.

Firstly, the Poisson Green's function exhibits spherical symmetry because the delta function is only non-zero at the source point. The only relevant factor is the distance from the delta source.

This encourages us to perform the integral using spherical coordinates, allowing us to orient the axes freely. Specifically, we align the r - r' vectors along the k_z axis, or the z-axis in Fourier space.

We can evaluate the dot product in terms of the angle between the vectors (refer to equation 14 of the main article):

Where ? will be one of the angles integrated in the inverse Fourier Transform.

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

? is once again the angle in the dot product, while ? represents the usual azimuthal angle, ranging from 0 to 360 degrees, or 0 to 2?.

Substituting this into the integral, the k² terms cancel out, resulting in a greatly simplified expression:

Though this may still appear daunting, it can now be tackled using standard calculus II techniques, which we will execute step-by-step.

First, observe that there is no r present, so that component of the integral simplifies to multiplying by 2?. For the ? integral, we can employ a simple u-substitution.

This leads to:

The last expression should be familiar. Recall that the sine function can be defined using complex exponentials:

Thus, our final result for the ? integral is:

By combining the results of the r and ? integrals in equation 10, we focus on the final integral concerning the radial coordinate k:

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

To evaluate it, we can utilize Feynman’s technique, as illustrated in a post by Kensei S.

While this process is not overly complex, Kensei presents it exceptionally well, making it unnecessary for me to rehash the details. I will simply present the result:

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

This concludes the derivation.

One last point to consider:

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

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

How might you adjust the Poisson Green's function to yield the same 1 / r form as k approaches zero?

Recently, I created a “buy me a coffee” link. If you enjoyed this article and wish to see more, your support would be greatly appreciated. Thank you for reading. -Tom