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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 $\frac{\omega}{k}\frac{d\omega}{dk}=c^2$

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  • $\begingroup$ 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. $\endgroup$ – honeste_vivere May 17 '16 at 14:40
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When the phase velocity is a constant (with respect to wavelength), the group velocity will indeed be equal to it, as you yourself have shown.

What you've got wrong here is the assumption for this case that the product of the phase velocity and group velocity equal the square of the speed of light, which can be true in other cases but not for plasma electromagnetic waves.

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  • $\begingroup$ I have added a proof that $v_p v_g=c^2$ $\endgroup$ – binaryfunt May 17 '16 at 8:13
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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 $k$.

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