# EM Wave equation and dispersion in a plasma

I am currently studying for qualifiers and I am really stuck on this one practice problem and I want to understand why I am having such a difficult time.

The idea is that we have a plasma with free electron density $n$, and this induces a current density $J$. From this I want to find:

1) a wave equation

2) an expression for $\frac{dJ}{dt}$ and

3) assuming a solution $\exp{(i kz-wt)} \hat{ x},$ find a dispersion relation.

My approach: Maxwell's Equations. Specifically taking the curl of $\nabla \times E$ $$\nabla \times(\nabla\times E=-\partial B/\partial t)$$ $$\nabla(\nabla\cdot E)-\nabla^2E=-\partial/\partial t (\mu_0J+\mu_0\epsilon_0\partial E/\partial t)$$ $$\nabla^2E-\nabla(\rho/\epsilon_0)=\mu_0 (\partial J/\partial t+\partial^2E/\partial t^2)$$ So that is what I have solved as my wave equation. However, I believe that $\rho$ is simply $n$ (only charge available, so my intuition tells me that $\nabla \rho=0$. This would reduce it slightly to something that looks more like a wave equation to me.

The second part asks for an expression $\frac{dJ}{dt}$ using Newton's second law with a coulomb force. So I go about:

$$F=ma=m(dv/dt)=qE$$ $$qm\dot v=q^2E\implies q\dot v=\frac{q^2}{m}E=AdJ/dt$$ $$dJ/dt = \frac{q^2}{mA}E$$ This is where I feel I may have made a mistake. I know Ohms law tells us that $J=\sigma E$ so I believe that $\dot J$~$\dot E$. Have I thought about this correctly, in that $q\dot v= AJ$ , where A is the cross sectional area.

I am not really sure if I am approaching this properly. I think once I have these two down, then the dispersion question should be simple, so I would appreciate any help.

The first part is solved correctly. The second part, I guess is more complicated. Assuming motion of electrons under an applied alternating field, by Newton's laws,

$$m\dfrac{d^2x}{dt^2} + m\gamma \dfrac{dx}{dt} + {\omega_0}^2 x= qE_0 \cos\omega t$$

EDIT: The question requires $\gamma$, $\omega$ both going to zero.

from which you can find out that the time dependent solution for $x$ is proportional to $E_0\exp(-i\omega t)\;.$

Now by definition, $$J = Nqv$$

So differentiating $v = \dfrac{dx}{dt}$, you get

$$J = \sigma E$$

Notice that $\sigma$ here is a complex quantity unlike the case of conductors.

• Am I correct in that del p=0 – yankeefan11 Jan 14 '16 at 1:39
• @yankeefan11 it has to be as charge cannot accumulate inside plasma. – Bruce Lee Jan 14 '16 at 1:42
• I get that. So for the second part, we are assumed to ignore the $\gamma$ effects., so it is just m(dv/dt)=qE(r,t). And because J=Nqv, dJ/dt=Nq(dv/dt), which means that $dJ/dt=\frac{NE(r,t)q^2}{m}$ if I am correct? – yankeefan11 Jan 14 '16 at 1:49
• @yankeefan11 i got that. I got an incorrect term in my answer which I edited just now. Please see it. The answer will be clear. – Bruce Lee Jan 14 '16 at 1:59
• So will the $\omega_0^2$ impact that at all? – yankeefan11 Jan 14 '16 at 2:01