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##?