The speed of the guided waves
The wave velocity we have used above is the phase velocity, which is the speed of a node of the wave; it is a function of frequency. If we combine Eqs. (24.17) and (24.18), we can write \begin{equation} \label{Eq:II:24:25} v_{\text{phase}}=\frac{c}{\sqrt{1-(\omega_c/\omega)^2}}. \end{equation}  For frequencies above cutoff—where travelling waves exist—$\omega_c/\omega$ is less than one, and $v_{\text{phase}}$ is real and greater than the speed of light. We have already seen in Chapter 48 of Vol. I that phase velocities greater than light are possible, because it is just the nodes of the wave which are moving and not energy or information. In order to know how fast signals will travel, we have to calculate the speed of pulses or modulations made by the interference of a wave of one frequency with one or more waves of slightly different frequencies (see Chapter 48, Vol. I). We have called the speed of the envelope of such a group of waves the group velocity; it is not $\omega/k$ but $d\omega/dk$: \begin{equation} \label{Eq:II:24:26} v_{\text{group}}=\ddt{\omega}{k}. \end{equation}  Taking the derivative of Eq. (24.17) with respect to $\omega$ and inverting to get $d\omega/dk$, we find that \begin{equation} \label{Eq:II:24:27} v_{\text{group}}=c\sqrt{1-(\omega_c/\omega)^2}, \end{equation}  which is less than the speed of light. The geometric mean of $v_{\text{phase}}$ and $v_{\text{group}}$ is just $c$, the speed of light: \begin{equation} \label{Eq:II:24:28} v_{\text{phase}}v_{\text{group}}=c^2. \end{equation}  This is curious, because we have seen a similar relation in quantum mechanics. For a particle with any velocity—even relativistic—the momentum $p$ and energy $U$ are related by \begin{equation} \label{Eq:II:24:29} U^2=p^2c^2+m^2c^4. \end{equation}  But in quantum mechanics the energy is $\hbar\omega$, and the momentum is $\hbar/\lambdabar$, which is equal to $\hbar k$; so Eq. (24.29) can be written \begin{equation} \label{Eq:II:24:30} \frac{\omega^2}{c^2}=k^2+\frac{m^2c^2}{\hbar^2}, \end{equation}  or \begin{equation} \label{Eq:II:24:31} k=\sqrt{(\omega^2/c^2)-(m^2c^2/\hbar^2)}, \end{equation}  which looks very much like Eq. (24.17) … Interesting! The group velocity of the waves is also the speed at which energy is transported along the guide. If we want to find the energy flow down the guide, we can get it from the energy density times the group velocity. If the root mean square electric field is $E_0$, then the average density of electric energy is $\epsO E_0^2/2$. There is also some energy associated with the magnetic field. We will not prove it here, but in any cavity or guide the magnetic and electric energies are equal, so the total electromagnetic energy density is $\epsO E_0^2$. The power $dU/dt$ transmitted by the guide is then \begin{equation} \label{Eq:II:24:32} \ddt{U}{t}=\epsO E_0^2abv_{\text{group}}. \end{equation}  (We will see later another, more general way of getting the energy flow.)