EM wave at boundary

[madness]

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 can't 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?

 

[Meir Achuz]

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.

 

[madness]

Thanks clem. Do you know how to show this from the boundary conditions?

 

[ksac]

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..

 

[madness]

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.

 

[Born2bwire]

Thanks clem. Do you know how to show this from the boundary conditions?

The Fresnel reflection coefficients have already done this for you.

 

[madness]

The Fresnel reflection coefficients are derived from the boundary conditions - they assume what I am trying to prove.

 

[Meir Achuz]

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'.

 

[madness]

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.

 

[Born2bwire]

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.

 

[madness]

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.