# Why doesn't the group velocity of a plasma EM wave equal the phase velocity here?

For plasma EM waves we have the dispersion relation $$\omega^2=\omega_p^2+c^2k^2$$ where the plasma frequency $$\omega_p^2=\frac{n_e e^2}{\epsilon_0 m_e}$$ One can show that $v_p v_g=c^2$, i.e., the product of phase and group velocities is the speed of light squared (see edit at bottom).

The critical density is when $$\omega^2=\frac{n_{crit} e^2}{\epsilon_0 m_e}$$ $$\Rightarrow \frac{\omega_p}{\omega}=\sqrt{\frac{n_e}{n_{crit}}}$$ Then subbing into the dispersion relation $$\omega^2=\frac{n_e}{n_{crit}}\omega^2+c^2k^2$$ $$\Rightarrow \omega^2(1-\frac{n_e}{n_{crit}})=c^2k^2$$ $$\Rightarrow v_p=\frac{\omega}{k}=c \left(1-\frac{n_e}{n_{crit}}\right)^{-1/2}$$ is the phase velocity. The group velocity is $\frac{d\omega}{dk}$, so one might expect $v_g=v_p$ here as $\omega$ seems linear in $k$. But using $v_g=c^2/v_p$, we get $$v_g=c\left(1-\frac{n_e}{n_{crit}}\right)^{1/2}$$ instead. So why does the group velocity not equal the phase velocity?

Edit

Just to show that $v_p v_g=c^2$ $$\omega=\left(\omega_p^2+c^2 k^2\right)^{1/2}$$ $$\Rightarrow \frac{\omega}{k}=\left(\frac{\omega_p^2}{k^2}+c^2\right)^{1/2}$$ $$\frac{d\omega}{dk}=\frac{1}{2}\left(\omega_p^2+c^2 k^2\right)^{-1/2}2c^2 k$$ $$=c^2\left(\frac{\omega_p^2}{k^2}+c^2\right)^{-1/2}$$ Thus $\frac{\omega}{k}\frac{d\omega}{dk}=c^2$

• Because $\nabla_{\mathbf{k}} \omega \neq \frac{\omega}{k}$... Many electromagnetic waves in plasmas are dispersive because they polarize the medium, thus there is a wavenumber-/frequency-dependent effective inertia term for wave propagation. May 17, 2016 at 14:40

• I have added a proof that $v_p v_g=c^2$ May 17, 2016 at 8:13
I think I've figured out why. It's simply because the critical density $n_{crit}$ is a function of $\omega$.
The critical density of the plasma is that which is required for the EM wave frequency to equal the plasma frequency $\omega_p$, so is dependent on the frequency of the radiation. So $$\omega=ck\left(1-\frac{n_e}{n_{crit}(\omega)}\right)^{-1/2}$$ therefore differentiating $\omega$ with respect to $k$ involves differentiating $n_{crit}(\omega)$ with respect to $k$; it is not as simple as dividing by $k$.