I face some trouble solving Maxwell's equations inside a cylinder with perfect conductor boundaries (in 3D) ? We work with cylindrical coordinates $(r, \phi, z)$ and we make the assumption that fields have a sinusoidal "$e^{i\omega t}$" time dependence. Note that we have a $\phi$ symmetry. First, and in any coordinates system, by taking the rotational and injecting one equation in the other we reduce Maxwell's equations to the following, $$ \nabla\times\nabla\times E = -\partial_t^2 E = \omega^2 E $$ In vacuum, from the $curl curl$ identity, it leads, $$ \nabla\times\nabla\times E = \nabla(\nabla . E) - \nabla^2 E = - \nabla^2 E $$ Where $- \nabla^2 E$ is the laplacian operator applied to each coordinate.

Now, in cylindrical coordinates, we can only compute the $z-$coordinate since, in this case we get the wave equation, $$ \nabla^2 E_z = \omega^2 E_z $$ For the other coordinates, the change of coordinates introduce other terms such that (for the $\phi-$ coordin. ate)$\frac{E_r}{r^2} - \frac{2}{r^2}\frac{\partial E_\phi}{\partial\phi}$.

Then, a fastidious step consists in performing a separation of variable which leads us quite easily to the solution for every separated variable and also to the Bessel differential equation which brings its solution, the Bessel function.

Together with boundary conditions we can get the solution according to $z$ but what about the other coordinates ?

  • $\begingroup$ Be careful of the identity $\nabla \times (\nabla\times\mathbf{E}) = \nabla (\nabla\cdot\mathbf{E}) - \nabla^2 \mathbf{E}$: it really is only a definition of the vector Laplacian and the wonted expression that involves the scalar Laplacian $\nabla^2 \mathbf{E} = \nabla^2(E_x) \hat{\mathbf{x}} + \nabla^2(E_y) \hat{\mathbf{y}} + \nabla^2(E_z) \hat{\mathbf{z}}$ only holds for Cartesian components of a vector field (of course the scalar Laplacians in this one can be expressed in any co-ordinates as long as they operate on Cartesian components) ... $\endgroup$ Sep 3, 2013 at 0:45
  • $\begingroup$ ...The [wiki page "Del in cylindrical and spherical co-ordinates"] (en.wikipedia.org/wiki/…) gives the right expression for the vector Laplacian in cylindrical co-ordinates, which is the one you will need to use. $\endgroup$ Sep 3, 2013 at 0:47

1 Answer 1


For an infinite perfectly conducting cylinder, any solution of the Maxwell equations can be represented as a superposition of cylindrical waves of TM type (for which $H_z=0$) and cylindrical waves of TE type (for which $E_z=0$). For cylindrical waves of TM type, you can find all field components if you know $E_z$, and for cylindrical waves of TE types, you can find all field components if you know $H_z$. You may wish to look at Eqs. (78), (79) of my article http://arxiv.org/abs/physics/0405091 . For example, if you know $E_z$ in Eq. (78), you get all the other field components in that equation by replacing the coefficient and replacing the cylindrical function $Z_n$ with its derivative or $Z_n/\rho$. You choose $Z_n$ based on your boundary conditions. Don't forget that there are also solutions of TE type!

  • $\begingroup$ Thanks but it still unclear, how did equations (78)-(79) were found ? $\endgroup$
    – Amzocks
    Sep 3, 2013 at 7:21
  • $\begingroup$ @Amzocks: These equations are no invention of mine - this is a quite standard material. See, e.g., the book by Stratton, Electromagnetic theory, Section 6.1 (books.google.com/… ). Then you need to apply his formulas for cylindrical coordinates. Maybe you can find the final formulas in an explicit form elsewhere. $\endgroup$
    – akhmeteli
    Sep 3, 2013 at 8:04
  • $\begingroup$ @Amzocks: Actually, it looks like you should be able to find formulas similar to Eqs. 78-79 of my article in Section 6.4 of the book by Stratton, but they seem to be missing from the Google preview. $\endgroup$
    – akhmeteli
    Sep 3, 2013 at 8:11

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