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In the derivation of Fresnel equations, I often see the first step as writing down the equations of the incident, reflected and transmitted waves as sinusoidal. I understand why we do not lose generality by assuming that the incoming wave is sinusoidal via a Fourier decomposition argument, but how do we know that the reflected and transmitted waves are sinusoidal?

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  • $\begingroup$ Please include a source where this kind of treatment is used. In many cases, we just apply boundary conditions, make use of Snell's law and the law of reflection to arrive at Fresnel's equations like here. $\endgroup$
    – exp ikx
    Mar 10, 2019 at 15:57
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    $\begingroup$ @exp ikx, The source would be Feynman lectures volume 2 chapter 33-1 equation 33.6. Griffith’s has a similar treatment $\endgroup$ Mar 10, 2019 at 16:49

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Why do we assume the reflected and transmitted waves are sinusoidal

This has no a priori justification - it is simply a useful Ansatz, and it gets all of its validity a posteriori when we use this Ansatz to build useful solutions of the Maxwell equations.

In other words, we have no guarantee that the method will work, so that at the time that the assumption is introduced, its status is little more than a shrewd guess at what we hope that the solution will look like. However, once we do the boundary matching correctly, we obtain a full solution to the Maxwell equations over all of space, with the behaviour at infinity that we were initially looking for (i.e. with three components clearly identifiable as an incident, a reflected, and a transmitted wave), and that's all that we really need.

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Any periodic function can be expressed as an (infinite) summation of (co)sinusoidal functions at integer multiples of the principle frequency. These individual functions are orthogonal.

Therefore it doesn't matter what functional form you ascribe to the incident, reflected and transmitted waves. The coefficients of each mutually orthogonal frequency component must separately satisfy the electromagnetic boundary conditions.

This means that if the incident wave is sinusoidal at a particular frequency, then so are the reflected and transmitted waves.

If the incident wave is more complex, then so are the reflected and transmitted waves, but the amplitudes of the various frequency components will still obey the same Fresnel equations.

Edit: The boundary conditions must be satisfied for all positions on the boundary and at all times. If the waves are a superposition of orthogonal sinusoidal functions at different frequencies this can only be true if each frequency component satisfies the boundary conditions separately.

A simple example; if $$ A_1 \sin \omega t + A_2 \sin 2\omega t = B_1\sin \omega t + B_2 \sin 2\omega t + B_3 \sin 3\omega t,$$ for all $t$, then $A_1 = B_1$, $A_2 =B_2$ and $B_3=0$.

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  • $\begingroup$ Why must each component separately satisfy the boundary Conditions? I agree that the sum should. $\endgroup$ Mar 10, 2019 at 17:24
  • $\begingroup$ @JacksonRudd Because an arbitrary summation can only match the boundary conditions at an instant of time. To make it work for all times then an equation has to be written for each orthogonal component. A similar argument tells you that boundaries do not change the frequency of light. $\endgroup$
    – ProfRob
    Mar 10, 2019 at 17:30
  • $\begingroup$ So the field of non linear optics does not exist? $\endgroup$
    – lalala
    Mar 10, 2019 at 18:23
  • $\begingroup$ @lala Non linear optics is the study of light where the principle of superposition (and Fresnel's relations) don't apply. $\endgroup$
    – ProfRob
    Mar 10, 2019 at 18:28
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The waves are not necessarily sinusoidal but can always be written as a linear expansion of these. The amplitude/phase version, not the squared, Fresnel equations are then applied to each component. Note that the frequency is not affected by the medium. Also momentum parallel to the interface cannot change, as this would require a grating. So each incoming plane wave gives rise again to a single plane wave of the same frequency and with the same direction (transmitted) or with its z component inverted.

Note that if there is a heating the wave will split into waves of multiple directions and if there is inelastic scattering waves of multiple frequencies may result.

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  • $\begingroup$ Yes, I see that argument and am fine with assuming that the incoming wave is sinesoidal. But your argument does not imply that the reflected and transmitted waves of a sinesoidal incedent wave will be sinesoidal. $\endgroup$ Mar 10, 2019 at 16:54
  • $\begingroup$ That is implied by Maxwell. Btw I am not assuming a shape of the incoming wave. I am expanding it in plane waves. $\endgroup$
    – my2cts
    Mar 10, 2019 at 17:40
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    $\begingroup$ This answer doesn't explain why if we start with a single incident wavevector we get also single reflected wavevector and single transmitted wavevector. Yes, we can expand the transmitted and reflected waves in plane waves, but the question was why the expansions of derived waves have only one wavevector each when the incident wave has only one too. $\endgroup$
    – Ruslan
    Mar 10, 2019 at 21:44

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