Expert: Philip A. Stahl - 2/13/2008

Question
What is the difference between the X- and O-modes of plasma waves? Btw, I never did receive a notice of the previous answered question (for cutoff and resonance frequencies) but had to find it in the answer library. What gives??

Answer
Hello,

Using the graph below, the “O-mode” occurs for the specific case in which the perturbed E –field is parallel to the B-field B(o) along the z-axis.  

^ z
!
!
!
^ B_o
!
!
^ E1
!
(*)----------------------------------->y

x (k)


Thus:

E1 ll  B_0

and k =  k x^  (e.g the wave number vector k is in the same direction as x. (Note that in the preceding graph, the symbol (*) denotes the x –direction out of the paper or screen.

The dispersion relation for the O-mode is:

w^2 =  w_e^2  + k^2 c^2

where the symbols have the same meaning as in the previous question I addressed (to do with cut-off, resonant frequency)

This follows exactly the same form as an EM wave propagating in an unmagnetized plasma

X-mode, or extraordinary mode waves presume:  k X E1  =  wB1


Here, both the uniform (B_0) and perturbed (B1) B-fields are parallel to the z-axis. While E1 (perturbed electric, or E-field) is perpendicular to B_0, B1 and in the x-y plane as shown in the graphic below (imagine a plane coming out of the screen bounded by x and y and the E1 vector forming a diagonal to either):



^ z
!
!
!
^ B_o
!
!
^ B1
!
(*)----------------------------------->y
x (k)


The dispersion relationship is:

w^4 – w^2(w_uH^2  + w_e^2  + k^2 c^2) + w_e^4 + k^2 c^2 w_uH^2 = 0

where the symbols mean as they did earlier, and now (w_uH) is the upper hybrid frequency.

Note: the upper hybrid frequency is equal to the *square root* of the SUM of the squares of the electron plasma frequency and the electron cyclotron frequency, e.g.

w_uH =  [ w_e^2  +   W_e^2]^ ˝

Hope this helps.

Regarding your not receiving a notification for the previous answer, this may sometimes happen – though hopefully not often! Call it a glitch! (It's amazing with the tens of thousands of exchanges each day on 'allexperts' it has so few glitches!)

The important thing is you were still able to access the answer on account of your resourcefulness, as opposed to just waiting for it to materialize in e-mail.