EM wave at boundary

  • Thread starter madness
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  • #1
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I have a simple question about reflecting EM waves at dielectric boundaries. To best illustrate my question, consider normal incidence. The incident wave has the wavevector k positive, and the reflected has k negative. Since B = k x E , and k has changed sign, B must also change sign. This is my problem - why cant E change sign instead? This would satisfy the necessary equations. It also makes intuitive sense, at normal incidence E and B are both parallel to the plane, and flipping either would preserve the handedness. Why is B special?
 

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  • #2
Meir Achuz
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It depends on the dielectric constant epsilon. If epsilon is greater than 1, which is the usual case, then the reflected E will not flip. For epsilon less than 1, E would flip.
For instance, for a wave leaving a dielectric into air, the reflected E does flip.
 
  • #3
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Thanks clem. Do you know how to show this from the boundary conditions?
 
  • #4
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k is the propagation vector, it represents the direction of the wave. When the wave reflects, the direction is reversed, hence k changes sign.
If you change the sign of E instead of that of k, you have a wave traveling in the same direction as initially.
Clem, can you help me understand what role permittivity has to play in the direction of E? I don't know at all about it..
 
  • #5
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ksac I think you have misunderstood my question. Given that k changes sign, E or B (but not both) could change sign to satisfy the B = k x E condition. A further condition is required to constrain which one will flip.
 
  • #6
Born2bwire
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Thanks clem. Do you know how to show this from the boundary conditions?

The Fresnel reflection coefficients have already done this for you.
 
  • #7
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The Fresnel reflection coefficients are derived from the boundary conditions - they assume what I am trying to prove.
 
  • #8
Meir Achuz
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It is done in all textbooks.
In a dielectric, the magnitudes of E and B are related by B=sqrt{epsilon mu} E.
From Maxwell's equations, both E and B tangential are continuous at the interface.
This means (with mu =1) that E_1+E_1'=sqrt{epsilon}E_2
and E_1-E_1'=E_2.
Solve for the reflected E_1'.
 
  • #9
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This is exactly the problem! Why do you choose to put the negative sign on the second equation rather than the first? From what I can see you are choosing to flip the electric field in this case.
 
  • #10
Born2bwire
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This is exactly the problem! Why do you choose to put the negative sign on the second equation rather than the first? From what I can see you are choosing to flip the electric field in this case.

It doesn't matter, you get the same solution.
 
  • #11
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Yeah thanks I figured it out. It seems to just be a convention. You write E as the one which doesn't change, and if this turns out to be false it goes negative, which in turn makes B go positive.
 

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