Dispersion relations and Plasma

  • Thread starter Moham1287
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  • #1
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Homework Statement


The dispersion relation for a plasma is given by

[tex]k^{2}=\frac{\omega^{2}}{c^{2}}(1-\frac{\omega^{2}_{p}}{\omega^{2}})[/tex]


[tex]\omega^{2}_{p}\:= \frac{Ne^{2}}{m_{e}\epsilon_{0}}[/tex]

Where N is the electron density

During re enrty of a spacecraft there was a radio blackout of all frequencies up to 10^10 Hz because it was surrounded by a plasma. Calculate the electron density in the plasma surrounding the spacecraft.

Sensitive equipment detected EM radio waves at 10^9 Hz at an amplitude of 10% of that before re entry. Calculate the thickness of the plasma.

Homework Equations



Given above

The Attempt at a Solution



Got the first part easily enough, by substituting in the expression for [tex]\omega^{2}_{p}[/tex], then solving [tex] \frac{Ne^{2}}{\omega^{2}m_{e}\epsilon_{0}}=1[/tex] for N to get N= 1.24x10^18. I don't really have any idea about how to go about the second part, I can't find anything about it in my textbook (I S Grant & W R Phillips Electromagnetism) or on the old googles

Any help would be much appreciated! Thanks
 
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Answers and Replies

  • #2
674
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Since the frequency is below the plasma frequency your 'k' will be imaginary. That means the wave will exponentially decay. So you will have something like E = E_0*exp(-kx). And you know the amplitude is decreased by 10%, so you are left with something like... 0.10 = exp(-k*x)

Now you just solve for 'x', and that is roughly how far the wave had to attenuate through before it escaped
 
  • #3
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Hi

thanks for the reply. I don't think I quite follow though. K is the wave number, given by 2pi/lambda isn't it? How can that be imaginary? k would be 2pi/0.3 recurring for EM of 10^9 Hz... Would my answer then just be 0.10=exp(x2pi/0.3) solve for x?
 
  • #4
674
2
The wave is in a plasma, so the dispersion relation w=ck doesn't apply anymore. Use the dispersion relation you listed above.
 

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