1
$\begingroup$

I would like a simple explanation of what, in a physical problem, determines the order of the Bessel functions that describe the solutions. What are dimensional(?) parameters of the system that induce a certain order of the resulting Bessel function order?

I know that Bessel equations occur in cylindrical/spherical problems. For instance, take the example in the paper Am. J. Phys. 85, 341 (2017), about Bessel modes in cylindrical waveguides. Equation 5 reads:

$$\frac{d^2}{d\rho^2} R(\rho)+ \frac{1}{\rho} \frac{d}{d\rho} R(\rho) + \left(k_\rho^2 - \frac{m^2}{\rho^2}\right)R(\rho) = 0,$$

$\rho$ being the distance to center, $R$ radial part of the field, $m$ the parameter that describes the mode and the resulting order of the Bessel function solution, and k the propagation constant (here $k_\rho$ in radial direction).

I short circuit a few equations to give the resulting field form (eq 8) for the fundamental mode :

$$ E(\rho,z) = \left[H_0^{(1)}(k_\rho\rho)+H_0^{(2)}(k_\rho\rho)\right] e^{ik_zz} = 2 J_0(k_\rho\rho) e^{ik_zz} $$

$H_0^{(j)}$ being a Hankel function (of order zero), $J_0$ the Bessel function of first kind (and order zero). Higher modes have corresponding higher $m$.

So here the order of the function involved is related to the propagation mode, hence the number of zeros in the field. Now I know the question of zeros in Bessel functions is can get quite tricky, and I must admit i haven't done much research into that... but firstly aren't they supposed to be infinite? because this contradicts my idea that a mode is linked to the number of nodal points of the field.

Improve this question
$\endgroup$
1
  • 1
    $\begingroup$ Sine functions run out to infinity, but boundary conditions determine the modes (of a vibrating string, for example). $\endgroup$
    – user137289
    Commented Jul 13, 2018 at 14:38

1 Answer 1

Reset to default
3
$\begingroup$

The "$n$" in the $J_n(\kappa r)$ refers to the number of nodes in the angular direction. A complete set of eigenfunctions of $-\nabla^2$ in ${\mathbb R}^2$ are $$ \psi_{n,\kappa}(r,\theta)= e^{in\theta}J_n(\kappa r), \quad \kappa\in [0,\infty) $$ with eigenvalue $\kappa^2$.

Improve this answer
$\endgroup$
7
  • $\begingroup$ Interesting. But I can't make sense of the connection between the number of nodes in the angular direction and what happens in the radial direction... $\endgroup$
    – Paul Cwave
    Commented Jul 13, 2018 at 22:48
  • $\begingroup$ @Paul Cwave They are independant. If you impose a radial $\psi(r,\theta)=0$boundary condition at $r=R$, then that selects $\kappa_m = x_m/R$, $m=0,1,2,\ldots$, where $x_m$ is the $m$-th zero of $J_n(x)$, as the set of eigenvalues associated with the angular mode number $n$. $\endgroup$
    – mike stone
    Commented Jul 14, 2018 at 16:24
  • $\begingroup$ but the index $n$ in $J_n$ (radial term) is the same as the one in the exponential (angular) term? $\endgroup$
    – Paul Cwave
    Commented Jul 14, 2018 at 16:34
  • $\begingroup$ @Paul Cwave Yes. So what's the problem? Different angular momentum states have different spectra. This is hardly surprising. $\endgroup$
    – mike stone
    Commented Jul 14, 2018 at 17:57
  • 1
    $\begingroup$ @PaulCwave They're not independent - the behaviour on the angular direction has a direct impact on the ODE for the radial direction, i.e. the $m^2/\rho^2$ term on that ODE, which comes directly from the angular part of the derivative with a suitable $1/\rho^2$ geometrical factor. This is the generic behaviour on general coordinate systems, and the systems where this doesn't happen (like cartesian coordinates) are few and far between. $\endgroup$
    – Emilio Pisanty
    Commented Jul 15, 2018 at 13:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.