I read that when an EM wave enters a dielectric material, it keeps the same frequency, while its wavelength is reduced, so that their product is a quantity less than c = 3 × 10^8 m/s.
Why does it keep the frequency and not the wavelength?
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2$\begingroup$ Possible duplicate of Why does the frequency of a wave remain constant? and physics.stackexchange.com/questions/22385/… $\endgroup$– ProfRobCommented Sep 26, 2019 at 8:33
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3 Answers
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Very briefly, because of boundary conditions.
At the boundary between the first and second media, we know that the tangential component of the E field will be equal on the two sides of the boundary. So if the E field on one side is oscillating at some frequency $\omega$, then the E field on the other side must also be oscillating with frequency $\omega$.
answered Sep 26, 2019 at 0:57
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$\begingroup$ A last question. I read that the wavelength in a Dielectric is reduced (with respect to that in vacuum) by a factor equal to the square root of the relative permittivity. But, in theory, should be present also the magnetic permeability? $\endgroup$ Commented Sep 28, 2019 at 17:17
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$\begingroup$ @Kinka-Byo, yes, the relative magnetic permeability will also affect the wavelength. We just usually ignore it because it's very close to 1 in a wide range of materials. $\endgroup$ Commented Sep 28, 2019 at 17:26
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This is not anything specific to electromagnetic waves. It's a general fact about how waves behave. Wave equations describe cause and effect relationships that are local. If you're "doing the wave" in a stadium, the way you know it's time for you to go is that your neighbor just went. You have no way of sensing the wavelength by looking around in your own neighborhood. You can only sense the time variation of the wave.
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$\begingroup$ Thanks for a 'physical' explanation of the phenomenon. $\endgroup$ Commented Sep 26, 2019 at 3:22
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$\begingroup$ I don't think I agree with this answer. The wave equation is local in space and in time. So you need something more to break that symmetry. In your analogy, you're implicitly breaking the symmetry by assuming that "you" get to be an extended timelike trajectiory rather than a single spacetime event. But I think it's important to point out that the OP's setup implicitly has boundary conditions (which are intrinsically nonlocal) that break the symmetry between space and time. $\endgroup$– tparkerCommented Sep 26, 2019 at 4:12
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As The Photon says, it's because of boundary conditions.
To elaborate on his/her answer: when an EM wave propagating through vacuum enters into a dielectric material occupying a different region of physical space, the boundary between the two regions is a timelike hypersurface (since its normal vector is spacelike). Therefore the "time-conjugate" part of the wave four-vector - the frequency - must stay the same in order for the EM fields to remain continuous across the boundary.
If instead you had a spacelike hypersurface boundary between the vacuum and the dielectric regions - e.g. you had an EM wave propagating through a vacuum and then all of space suddenly became filled with a dielectric - then the wavelength would stay the same and the frequency would change, because the "space-conjugate" part of the wave four-vector - the spatial wave vector - would need to stay the same in order for the EM fields to remain continuous across the boundary.
