2
$\begingroup$

I've been reading about the dielectric function and plasma oscillations recently and I encountered the following dispersion relation for EM waves in metals or in plasma (Is it correct to treat those the same?) $$ \omega^2 = c^2k^2 + \omega_p^2 $$ where $\omega_p^2 = 4\pi n_e e^2/m$ is the squared plasma frequency with $n_e$ the free electron density. Now I am pretty certain that both book where I found a derivation (An introduction to solid state physics - Charles Kittel ; Advanced solid state physics - Philip Phillips) of this result make a mistake. Is there a way to justify the errors I think they make? The equation they obtained seems to be stated regulary, but I find something different.

The two derivations are different so I will treat them separately. I added my doubts in bold font. I will work in CGS units.

Phillips (p122)

We start from the Maxwell equation $$ \nabla \times \vec{B} = \frac{4\pi \vec{j}}{c} +\frac{1}{c}\frac{\partial \vec{E}}{\partial t} . $$ We take the time derivative of this expression: $$ \frac{\partial}{c\partial t} \nabla \times \vec{B} = \frac{\partial}{\partial t}\frac{4\pi \vec{j}}{c^2} +\frac{1}{c^2}\frac{\partial^2 \vec{E}}{\partial t^2}. $$ Using the Maxwell–Faraday equation $\frac{\partial\vec{B}}{c\partial t} = -\nabla \times \vec{E}$ and the curl of a curl expression, the left hand side of this equation can be rewritten as $$ -\nabla\times (\nabla\times\vec{E} ) = -\nabla(\nabla\cdot\vec{E}) + \nabla^2 \vec{E}. $$ Using $\nabla \cdot \vec{E} =0$ (Is this assumption correct? Since he appears to be using the free electron model I doubt this can be used.) and $\frac{\partial \vec{j}}{\partial t} = e^2 n_e \vec{E}/m$ (This relation holds only for a free electron gas (via the Lorentz force). What makes it more legit to use this model rather than the Drude model (where the electrons experience collisions which gives rise to Ohm's law)? Using Ohm's law instead does modify the obtained result quite severely.), we find $$ (\frac{\partial^2}{\partial t^2} + c^2\nabla^2 - \omega_p^2)\vec{E} = 0. $$ From this we get $$ \omega^2 = c^2k^2 + \omega_p^2. $$

With my correction the wave equatoin becomes $(\frac{\partial^2}{\partial t^2} + c^2\nabla^2 - \omega_p^2\frac{\partial}{\partial t})\vec{E} = 0$, and hence $\omega^2 = c^2k^2 - i\omega \omega_p^2$. This result coincides with the dispersion relation for EM waves in a conductor as found in Griffiths' book on electrodynamics in section 9.4.

Kittel (p397)

Here the starting point is the wave equation in a nonmagnetic isotropic medium $$ \frac{\partial^2 \vec{D}}{\partial t^2} = c^2\nabla^2{\vec{E}} $$ If I am not mistaking, this is only correct for a perfect insulator where there can be no free currents, which I think is not a correct assumption for a plasma, a metal or an electron gas. This is then used for a linear material so that $\vec{D} = \epsilon\vec{E}$ and then $$ \epsilon(\omega) = 1-\frac{\omega_p^2}{\omega^2} $$ is used. How could one justify the steps which I think are wrong?

$\endgroup$
  • $\begingroup$ This answer uses the same relation as Phillips for the time derivative of the current, so there might be more to it. $\endgroup$ – Lagrangian Jan 2 '16 at 11:56
  • $\begingroup$ (1) For a plane electromagnetic wave, the assumption $\nabla \cdot \mathbf{E} = 0$ is valid since $\mathbf{k} \perp \mathbf{E}$ by definition. (2) Ohm's law is an approximation and should be used carefully as there are many terms which are often neglected when they should not be. (3) Your expression for $\partial \mathbf{j}/\partial t$ is a very limited approximation. Generally people drop this term because the plane wave solution involves the source free limit. $\endgroup$ – honeste_vivere May 23 '16 at 21:27
  • $\begingroup$ The wave equation and dispersion relation you give at the end of the paragraph on Phillips (p137) in the sentence "With my correction the wave equation becomes...) cannot be correct because the last term in the dispersion relation has a different dimension (time to the power of -3) as opposed to the other terms (time to the power of -2). In physical equations the dimensions of terms should be the same. $\endgroup$ – freecharly Sep 7 '16 at 21:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.