I've been reading Jackson's Chapter 8.10 and trying to find the Green's function for a non-homogenous Helmholtz equation. The problem is in cylindrical coordinates.

I first made a Fourier transform to frequency space to get rid of the time derivative. Then proceeded to assume a point source in order to solve a Green's function equation. I then expressed $G$ as a function of a sum of exponentials of $im(\phi-\phi')$ (from the $\delta(\phi-\phi')$), multiplied by a function $g_m(\rho)$ and an integral of $\exp(ikz)$ then put it in the differential equation and got a Bessel's equation:

$$\left(\dfrac{\partial^2}{\partial\rho^2}+\dfrac{1}{\rho}\dfrac{\partial}{\partial\rho}-\alpha^2-\dfrac{m^2}{\rho^2}\right)g_m(\rho) = 0$$ away from the source with $\alpha^2 = k^2 -\dfrac{\omega^2}{c^2}$.

My question is this: there is a physical meaning behind $\alpha^2$, $k$ is the wave vector at the end of the day and $\dfrac{\omega}{c}$ is also equal to $k$. How do we determine $\alpha$'s sign (it's not 0)?

My guess is that $\dfrac{\omega}{c}$ is the magnitude of the $k$ vector while $k$ in $\alpha$ is actually $k_z$. Therefore $\alpha^2$ should always be negative and we must take the Bessel function solution.


Jackson takes into account the case where it's positive and we therefore have the Modified Bessel's function solution.


Maybe it's the opposite of what I said, and $\dfrac{\omega}{c}=k_z$ instead, but in either case I don't see how we'd get cases. What is the hidden meaning behind this?


1 Answer 1


The equation has solutions for both cases:

  1. Positive $\alpha^2$: The radial Bessel function solution combined with a $k_z$ where $k_z^2 > \frac{\omega^2}{c^2}$.
  2. Negative $\alpha^2$: The radial modified Bessel function solution combined with a $k_z$ where $k_z^2 < \frac{\omega^2}{c^2}$. NB: this could still mean $k_z^2>0$, but also $k_z^2<0$, i.e. imaginary $k_z$.

It's of course not obvious what you would find acceptable as a "meaning behind this", but if you consider a distributed source, you could have:

  1. A source with very fast periodic variation in the $z$-direction (with shorter wavelength than the free space wavelength), leading to case 1) for the radial solution, which will then have exponential decay in the radial direction.
  2. A source with slow periodic variation in the $z$-direction (with longer wavelength than the free space wavelength or even exponentially decaying in the $z$ direction), leading to case 2) for the radial solution, which will then have the $1/\sqrt{\rho}$ behavior in the radial direction. You could in that case also combine cylindrical Bessel and Neumann functions into an outgoing cylindrical Hankel function describing outgoing cylindrical waves.

Any source, even a point source, can of course be constructed by Fourier transform from sources with periodic variation. If you do the Fourier transform only in the $z$-direction the source distributions described above would make sense. A practical case where they occur is in current (and charge) distributions in a coaxial cable, with arbitrary periodicity in the $z$-direction (not necessarily the periodicity of the freely propagating modes of the cable).

  • $\begingroup$ I've been thinking about this and it would also explain the different ways we treat the general Helmholtz equation (without boundaries) and the paraxial one right? As the first would fit into case 1 while the paraxial by definition has a slow z variation so it makes sense to use 2 (and I've seen it used). Right? $\endgroup$ Commented Apr 16 at 17:26
  • $\begingroup$ Something I have trouble understanding is this: "When the radial solution has oscillatory behaviour the solution of Z is exponentially decaying and vice versa" which refers to the choice of J,N or I,K for R in each case. While mathematically I understand it I'm having trouble fitting it into a realistic situation (such as if z is the direction of propagation shouldn't it always have exponential behaviour except for when we have certain boundary conditions?) This is the core of why I don't understand why both solutions can exist for a free wave. $\endgroup$ Commented Apr 16 at 17:29

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