Struggling With Plasma Physics Question

[fizicksiscool]

Homework Statement

The Equation of motion is: $m\frac{d\textbf{v}}{dt} = -e(\textbf{E}(\textbf{r} ,t) + \textbf{v} \times \textbf{B}_{ext})$
Where $\textbf{B}_{ext}$ is a constant in the +z direction.
$E = \begin{bmatrix} E_x \\ E_ y \\ 0 \\ \end{bmatrix} = \begin{bmatrix} cos(k_{\pm}z - \omega t)\\ \pm sin(k_{\pm}z - \omega t) \\ 0 \\ \end{bmatrix}$
The number density of mobile electrons is $n(r, t) = n_0 + \delta n(r, t)$
There is also an immobile background charge density $\rho = +en_0$

Assuming that such a wave exits, Show that a possible solution of the equation of motion results in E creating a current j obeying:
$\frac{d\textbf{j}}{dt} = \alpha_{\pm}\textbf{E}$
Find the constants $\alpha_{\pm}$ in terms of $\omega$, $\omega_c = \frac{eB_{ext}}{m}$, and $\omega_p^2 = \frac{n_0e^2}{m\epsilon_0}$

Relevant Equations

The Maxwell Equations:
$\nabla \times B = \mu_0 j +\frac{1}{c^2}\frac{\partial E}{\partial t}$
$\nabla \times E = -\frac{\partial B}{\partial t}$
$\nabla \cdot E = \frac{\rho}{\epsilon_0}$
$\nabla \cdot B = 0$
and
$\textbf{j} = -ne\textbf{v}$

First, assuming, $v \alpha e^{i(k{\pm}z - \omega t)}$ I worked with the equation of motion to get:
$-i\omega v_x = -\frac{e}{m}E_x - \omega_c v_y$ and $-i\omega v_y = -\frac{e}{m}E_y + \omega_c v_x$
Solving this system of equations, I end up with:
$v_x = \frac{-e}{m(\omega^2 - \omega_c^2)}(i\omega E_x +\omega_c E_y)$ and $v_x = \frac{-e}{m(\omega^2 - \omega_c^2)}(i\omega E_y -\omega_c E_x)$
Then I plug this into $j = -nev$ and get $j = \frac{e^2(n_0 + \delta n(r, t))}{m(\omega^2 - \omega_c^2)}\begin{bmatrix} i\omega E_x + \omega_c E_y\\ i\omega E_y - \omega_c E_x\\ 0 \end{bmatrix}$
And this is where I get stuck. I do not know how to deal with $\delta n$, and I can't figure out how $\epsilon_0$ gets involved so that I can have $\omega_p$. I also can't see how the time derivative will end up being just some constant times E, as $\frac{\partial E_x}{\partial t} \alpha E_y$ and $\frac{\partial E_y}{\partial t} \alpha E_x$ so it seems like $\alpha_{\pm}$ should be a tensor and not a constant.
Pushing forward with the time derivative despite this, I get:
$\frac{\partial j}{\partial t} = \frac{e^2(\frac{\partial \delta n(r, t)}{\partial t}}{m(\omega^2 - \omega_c^2)}\begin{bmatrix} i\omega E_x + \omega_c E_y\\ i\omega E_y - \omega_c E_x\\ 0 \end{bmatrix} + \frac{e^2(n_0 + \delta n(r, t))}{m(\omega^2 - \omega_c^2)}\begin{bmatrix} \pm i\omega^2 E_y \mp \omega_c \omega E_x\\ \mp i\omega^2 E_x \pm \omega_c \omega E_y\\ 0 \end{bmatrix}$
Which isn't proportional to E and has no $\epsilon_0$

 

[billtodd]

Multiply Lorentz's EOM BY -ne to get:
$mdj/dt=ne^2E+ej\times B_{ext}$.
Take: $j=w e^{\alpha t}$
So you get the equation: $m\alpha w=ne^2 E+ew\times B_{ext}$.
Now, just match the coordinates of $w=(w_1,w_2,w_3)$ with with the rhs of the last above equation. don;t forget: $B_{ext}=B\hat{z}$. where $B$ is a constant both in time and space.

You need to solve a Linear equations system

 

[billtodd]

wait a minute in the last equation E should be multiplied by $e^{\alpha t}$, in which case you should solve the equaiton whilst t=0.

 

[fizicksiscool]

I appreciate the reply, but I still have a few issues. $\epsilon_0$ is not involved, so I can't express my answer in terms of $\omega_p$. Also, your solution would still result in $\alpha_{\pm}$ being tensors, while the problem requires them to be constants. Also, would your $\alpha$ be the same as the $\omega$ in the expression for $\textbf{E}(\textbf{r}, t)$ ? If not, then how would I derive $\alpha$?

 

[billtodd]

I appreciate the reply, but I still have a few issues. $\epsilon_0$ is not involved, so I can't express my answer in terms of $\omega_p$. Also, your solution would still result in $\alpha_{\pm}$ being tensors, while the problem requires them to be constants. Also, would your $\alpha$ be the same as the $\omega$ in the expression for $\textbf{E}(\textbf{r}, t)$ ? If not, then how would I derive $\alpha$?

w is a vector alpha is a scalar, Obviously you need to use here vector analysis identities with the Dell operator; you should tinker with the identity. I would first take divergence on the equation and after that another equation with curl, and then equating the vector identities with their componenets while t=0.

 

[renormalize]

And this is where I get stuck.

Your Homework Statement looks like it relates to so-called "extraordinary" electromagnetic L,R waves in a plasma. I suggest you look at pp. 10-19 of these notes:
http://sun.stanford.edu/~sasha/PHYS312/2007/L11/phys312_2007_l11.pdf
for a discussion of these waves, which should help you solve your problem.

 

[fizicksiscool]

Your Homework Statement looks like it relates to so-called "extraordinary" electromagnetic L,R waves in a plasma. I suggest you look at pp. 10-19 of these notes:
http://sun.stanford.edu/~sasha/PHYS312/2007/L11/phys312_2007_l11.pdf
for a discussion of these waves, which should help you solve your problem.

Thanks, It definitely is. I actually have already looked at those notes, although now that I'm looking at them again, I realized I was making some errors earlier, but even after correcting those and ignoring the $\delta n$ part of n, I still have a few problems.
Using the relations between $E_x$ and $E_y$ for R and L waves, I am able to get: $\frac{dj}{dt} = \frac{\epsilon_0 \omega \omega_p^2}{1 \mp \frac{\omega_c}{\omega}}E$. My problem is that I have an extra factor of $\epsilon_0$ that I'm not supposed to have, as I'm required to express $\alpha_{\pm}$ solely in terms of $\omega$, $\omega_c$, and $\omega_p$. With $\omega_p = \frac{n_0^2e}{m\epsilon_0}$. I'm not sure how to resolve this. I also have a similar problem with k, where I'm stuck with an extra factor of 1/c.

 

[renormalize]

I still have a few problems.

You don't show your work so I have to ask if you're verifying your derivation against the reference (which uses gaussian units) that I cited in post #6?
The first equation there on top of pg. 12 is easily rewritten to read:$$\left(\frac{\omega^{2}}{c^{2}}-k^{2}\right)\vec{E}=\frac{4\pi}{c^{2}}\frac{d\vec{j}}{dt}\tag{1}$$whereas the two dispersion relations appearing on pg. 13 for L,R waves can be combined into the one relation:$$k^{2}c^{2}=\frac{\omega^{2}-\omega_{p}^{2}\pm\omega\omega_{c}}{1\pm\frac{\omega_{c}}{\omega}}\tag{2}$$Inserting (2) into (1) and solving for $d\vec{j}/dt$ gives:$$\frac{d\vec{j}}{dt}=\frac{1}{4\pi}\left(\frac{\omega_{p}^{2}}{1\pm\frac{\omega_{c}}{\omega}}\right)\vec{E}\tag{3}$$There's no evidence of anomalous factors of $c^{-1}$ in (2) or $\epsilon_0\omega$ in (3).

 

[fizicksiscool]

Note in my problem: $\omega_p^2 = \frac{ne^2}{m\epsilon_0}$, unlike the lecture slides where $\omega_p^2 = \frac{4\pi ne^2}{m}$.
Starting from my earlier solution for j:
$j = \frac{e^2n}{m(\omega^2 - \omega_c^2)}\begin{bmatrix} i\omega E_x + \omega_c E_y\\ i\omega E_y - \omega_c E_x\\ 0 \end{bmatrix}$, and then using the relations for R waves that: $E_x = -iE_y$ and for L waves that: $E_x = iE_y$ and then combining these two as $E_x = \mp iE_y$(The R wave corresponds to the + solution for the wave and the L wave corresponds to the - solution as defined in the beginning of the problem) I can rewrite this expression as:
$j = \frac{e^2n}{m(\omega^2 - \omega_c^2)}\begin{bmatrix} i\omega E_x \pm i\omega_c E_x\\ i\omega E_y \pm i\omega_c E_y\\ 0 \end{bmatrix}$
which can be rewritten as:
$j = \frac{ie^2n}{m(\omega \mp \omega_c)}\textbf{E}$
Which after taking a time derivative, and using $E \propto e^{-i\omega t}$ becomes:
$\frac{dj}{dt} = \frac{\omega e^2n}{m(\omega \mp \omega_c)}\textbf{E} = \frac{\epsilon_0 \omega \omega_p^2}{(\omega \mp \omega_c)}\textbf{E}$ a solution which has an extra $\epsilon_0$ I am not allowed to have.

To solve for $k_{\pm}$ I start with $\nabla \times B = \mu_0 j + \mu_0\epsilon_0 \frac{\partial E}{\partial t}$ and use $\nabla \times B = ik\times B$ with $B = \frac{k}{\omega} \times E$ from $\nabla \times E = -\frac{\partial B}{\partial t}$ to get $\frac{i}{\omega}k \times k \times E = \mu_0 j + \mu_0\epsilon_0 \frac{\partial E}{\partial t}$ then using the triple vector product identity along with $E \propto e^{-i\omega t}$ and $k \dot E = 0$ I get: $-\frac{i}{\omega}k^2 E = \mu_0 j - i\omega\mu_0\epsilon_0 E$. Then, after some manipulation, I can get: $c^2k^2E = \frac{i\omega j}{\epsilon_0)} + \omega^2 E$ which can be rewritten as $\frac{i \epsilon_0}{\omega}(\omega^2 - c^2k^2)E = j$, which using my previously found expression for j can be written as: $i(\omega - \frac{c^2k^2}{\omega})E = \frac{\omega_p^2}{\omega^2 - \omega_c^2}\begin{bmatrix} i\omega E_x + \omega_c E_y\\ i\omega E_y - \omega_c E_x\\ 0 \end{bmatrix}$, which, defining $d = \frac{\omega_p^2}{\omega^2 -\omega_c^2}$, gives the system of equations:
$(\omega - \frac{c^2k^2}{\omega})E_x = d(\omega E_x - i\omega_c E_y)$ and $(\omega - \frac{c^2k^2}{\omega})E_y = d(\omega E_y + i\omega_c E_x)$, which can be manipulated to give: $(1 - \frac{c^2k^2}{\omega^2} - d)E_x = -d i\frac{\omega_c}{\omega} E_y)$ and $(1 - \frac{c^2k^2}{\omega^2} - d)Ey = d i\frac{\omega_c}{\omega} E_x)$, which can be multiplied together to give:
$(1 - \frac{c^2k^2}{\omega^2} - d)^2 = \frac{\omega_c^2}{\omega^2}d$, which gives $1 - \frac{c^2k^2}{\omega^2} - d = \pm\frac{\omega_c}{\omega}d$, which gives $k^2 = \frac{\omega^2}{c^2}(1 - d(1 \pm \frac{\omega_c}{\omega}))$ which is $k^2 = \frac{\omega^2}{c^2}(1 - \frac{omega_p^2}{\omega^2 - \omega_c^2}(\frac{\omega \pm \omega_c}{\omega}))$ which leads to $k_{\pm} = \frac{\omega}{c}\sqrt{1 - \frac{\omega_p^2}{\omega^2 \mp \omega\omega_c}}$(choosing the positive solution as $k > 0$ since $k \propto \frac{1}{\lambda}$ and $\lambda > 0$) and this solution is clearly proportional to $\frac{1}{c}$. Then, just as confirmation, I can plug this value for k back into my earlier expression: $\frac{i \epsilon_0}{\omega}(\omega^2 - c^2k^2)E = j$ to get $\frac{i \epsilon_0}{\omega}(\omega^2 - \omega^2( 1 - \frac{\omega_p^2}{\omega^2 \mp \omega\omega_c})) = j$, which gives: $i\epsilon_0 \frac{\omega_p^2}{\omega \mp \omega_c} = j$, which using $E \propto -e^{-i\omega t}$, gives $\epsilon_0\omega\frac{\omega_p^2}{\omega \mp \omega_c} = \frac{dj}{dt}$, the same as I got earlier.

 

[renormalize]

...a solution which has an extra $\epsilon_0$ I am not allowed to have.

Mea culpa! And my apologies for leading you astray by trying to compare your SI results directly to those in the gaussian class notes I referenced. I have independently verified your derivation and agree that the proper solution in SI units is $\frac{d\vec{j}}{dt}=\left(\frac{\epsilon_{0}\omega_{p}^{2}}{1\mp\frac{\omega_{c}}{\omega}}\right)\vec{E}$. Note that the SI units on the left side of that equation are $\frac{C}{m^{2}s^{2}}$ and to get those same units on the right side, the permittivity $\epsilon_{0}$ must appear on the right. How certain are you that the requested solution to the problem cannot involve dimensional constants like $\epsilon_0,\mu_0,\text{or }c$ in addition to $\omega,\omega_p,\omega_c$?