Consider a cylindrical cavity with circular cross section of radius R. The height of the cylinder is d; the walls are considered as being a perfect conductor. I'm facing a difficulty findind the TM modes inside that cavity, I'm following Jackson's book (~ p.368, 3rd ed.).

For TM modes, we're trying to find $E_z=\psi(x,y)e^{\pm ikz}$ from which we can get the transverse field as $\vec E_t=\pm\frac{ik}{\gamma ^2} \nabla _t \psi$. Where the nabla operator is the transverse gradient, in other words there is no derivative with respect to z, the coordinate alongside the height of the cylinder. In Cartesian coordinates $\nabla _t =\frac{\partial}{\partial x} \hat x + \frac{\partial}{\partial y} \hat y$.

Now since the cavity is a conductor, the electric field must be perpendicular to its surface. So for the $z=0$ x-y plane that means that $\vec E_t$ must vanish there, as well as for the x-y plane at $z=d$.

According to Jackson, the dependence of the fields with respect to the z variable is $A \sin kz + B \cos kz$ (which makes sense). He then says that the vanishing of $\vec E_t$ at the $z=0$ and $z=d$ walls implies that $E_z=\psi(x,y) \cos \left ( \frac{p \pi z}{d} \right )$; this is where my problem lies. I can't seem to make sense out of this expression.

I get that since in the expression of $E_t$ the gradient of psi does not deal with the z variable, then $E_z$ must vanish at both $z=0$ and $z=d$ in order for $E_t$ to vanish there too. And this implies that $E_z =\psi(x,y) \sin \left ( \frac{p \pi z}{d} \right )$, $p\in \mathbb{Z}$. I really don't see how there could be a cosine there, instead of a sine. I've checked in Zangwill's book as well as on the web and they all seem to agree on the cosine term. I've also checked the Jackson's errata website and there don't seem to be any error there. I don't see what I'm missing.

  • $\begingroup$ What does TM stand for? $\endgroup$
    – Gert
    Nov 5, 2015 at 14:46
  • $\begingroup$ "transverse magnetic". For a waveguide this means $B_z=0$ everywhere and $E_z$ evaluated at the surface must vanish (that's the boundary conditions). $\endgroup$ Nov 5, 2015 at 15:00

1 Answer 1


I finally got the answer by looking at the 1st or 2nd edition of the book. The 3rd edition is wrong here.

Jackson refers to wrong equations (8.31 and 8.33), instead it should refer to eq. 8.26a : $\vec E_t = \frac{i}{\mu \varepsilon \omega ^2 -k^2} [k \nabla _t E_z - \omega \hat z \times \nabla _t B_z]$ and he should have corrected it to $\vec E_t = \frac{i}{\mu \varepsilon \omega ^2 -k^2} [k \nabla _t \frac{ \partial E_z}{\partial z} - \omega \hat z \times \nabla _t B_z]$. And a similar correction for eq. 8.26b.

Strangely this hasn't been pointed out in the errata I've found on the web: Jackson's errata


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