Electromagnetic field in a medium

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An electromagnetic field in a medium can be expressed as E = E0cos(ωt-kx), with refraction indices affecting its behavior in transparent media. When analyzing the field in a moving reference frame with velocity v = c/n, the transformation of the wave equation must consider Lorentz invariance. The new frequency ω' cannot simply be zero; instead, it is determined by the relativistic Doppler effect, which relates ω' to the velocity of the medium. The discussion highlights the complexities of wave behavior in different reference frames, particularly regarding frequency and wave vector transformations. Understanding these principles is essential for accurately describing electromagnetic waves in varying conditions.
Bob_for_short
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An electromagnetic field, as a plane wave, has a known space-time dependence:

E = E0cos(ωt-kx).

In a transparent medium it is the same except for involving the refraction indices n.

Now, let us look at the field in a moving reference frame - that with v = c/n. What solution is for the wave in such a frame? Isn't it

E' = E'0cos(ω't') ?
 
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Isn't it a challenging exercise?
 
The combination (ωt-kx) is a Lorentz invariant and so will transform to (ω't'-k'x').
 
Meir Achuz said:
The combination (ωt-kx) is a Lorentz invariant and so will transform to (ω't'-k'x').

Sorry, I meant cos(k'x'). In this moving reference frame the new frequency ω'=0, isn't it?
 
Last edited:
Bob_for_short said:
Sorry, I meant cos(k'x'). I this moving reference frame the new frequency ω'=0, isn't it?
No. Omega' is given by the usual relativistic Doppler in terms of v/c, which in your case would be 1/n.
 
Meir Achuz said:
No. Omega' is given by the usual relativistic Doppler in terms of v/c, which in your case would be 1/n.

Strange. As soon as ω and k are connected via c and n>1, there should be the reference frame with the zero frequency ω'.
 

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