$\sqrt{\frac{\omega ^2}{c^2}-k_z^2}$ in cylindrical harmonics The radial component of the solution of the wave equation in cylindrical coordinates is $$J_\nu \bigg(\rho\sqrt{\frac{\omega ^2}{c^2}-k_z^2}\,\,\bigg).$$
But I always thought that $\frac \omega c$ was the same thing as $k$. What am I missing? What changed?
 A: You're not missing anything. You are right, $k=\omega/c$. The argument $\sqrt{\frac{\omega ^2}{c^2}-k_z^2}$ in the Bessel function is the projection of the wavevector onto the radial direction.
The use of Bessel functions beclouds what's going on a bit. Recall that a plane wave with wavevector $\vec{k}$ has the functional variation $\psi(\vec{r}) = e^{i\,\vec{k}\cdot\vec{r}} = \exp(i\,(k_x\,x + k_y\,y+k_z\,z))$ with $k^2 = k_x^2 + k_y^2 + k_z^2$. 
Now, the cylindrical modes are superpositions of plane waves with common $k_z$, and thus common $k_x^2+k_y^2 = k^2 - k_z^2 = \left(\frac{\omega}{c}\right)^2 - k_z^2$. Parameterizing  such plane waves by $k_x = -k_r \,\cos\theta,\,k_y = -k_r \,\sin\theta$, where $k_r = \sqrt{k^2-k_z^2}$, writing $x = r\,\cos\phi$ and $y = r\,\sin\phi$ (to convert Cartesians to polars) and then summing all such waves with weights $e^{i\,\nu\,\theta}$ in the superposition (which is still a solution of the linear Helmholtz equation), we get:
$$\begin{array}{lcl}\psi_S(r,\,\phi,\,z) &=& \int_{\theta=0}^{2\,\pi} \exp\left(i\,\left(\nu\,\theta -k_r\,r\,\cos(\theta-\phi)\right)\right)\,e^{i\,k_z\,z}\,\mathrm{d}\theta\\&=&e^{i\,\nu\,\phi}\,e^{i\,k_z\,z}\,\int_{\theta=0}^{2\,\pi} \exp\left(i\,\left(\nu\,\theta -k_r\,r\,\cos\theta\right)\right)\mathrm{d}\theta\\&=&\frac{2\,\pi}{i^n}\,e^{i\,\nu\,\phi}\,e^{i\,k_z\,z}\,J_\nu(\sqrt{k^2-k_z^2}\,r)\end{array}$$
where we've used the definition $J_n(z)= \frac{i^n}{2 \pi} \int_0^{2\,\pi} e^{i(n \theta- z \cos\theta)} \,\mathrm{d}\theta$. Thus you see that the wavenumber is still $k$ and we've recovered the function you cite, with radial wavenumber $\sqrt{k^2-k_z^2}$ and we recognize the full expression for a cylindrical mode.
