Why don't E&M fields change orientation after hitting a surface? In essentially every derivation of the Fresnel equations, the general problem of radiation hitting a surface at a certain angle is broken into two parts (out of which we hope the solution any general situation can be constructed):


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*Incident electric field s-polarized, Incident magnetic field p-polarized

*Incident electric field p-polarized, Incident magnetic field s-polarized


Here is a diagram showing these two general cases, taken from this paper.

My question is, how do we know in either case (1) or (2), individually, that the reflected and transmitted fields are polarized in the same direction as the incident field? For example, how do we know that with an incident $E$-field that is initially s-polarized, both the transmitted and reflected $E$-fields will also be entirely s-polarized?
The boundary conditions at the interface don't seem to give the necessary rationale for why this is so. As an example, consider again case (1), where the incident $E$-field is s-polarized. In this case, the boundary conditions for the $E$-field are, evaluated at the interface,
$$(\mathbf{E}_i)_{\parallel}+(\mathbf{E}_r)_{\parallel}=(\mathbf{E}_t)_{\parallel} \tag{1}$$
$$\epsilon_1(\mathbf{E}_i)_{\bot}+\epsilon_1(\mathbf{E}_r)_{\bot}=\epsilon_2(\mathbf{E}_t)_{\bot} \tag{2}$$
where in both of these equations the direction indicated is relative to an arbitrary vector lying on the surface (i.e. in the picture, up is perpendicular to the surface). In the case an incident s-polarized E-field (i.e. one where the incident E-field is parallel to the surface), equations (1) and (2) reduce to
$$\mathbf{E}_i+(\mathbf{E}_r)_{\parallel}=(\mathbf{E}_t)_{\parallel} $$
$$\epsilon_1(\mathbf{E}_r)_{\bot}=\epsilon_2(\mathbf{E}_t)_{\bot} $$
By what reasoning can we further deduce the directions of the reflected and transmitted fields?
 A: There is a simple answer: Symmetry.
Suppose the material is isotropic, and consider the initial condition of the p-polarized case, with a p-polarized light wave about to hit the surface. In this case, reflecting in the plane that contains the incident and scattered wave vectors leaves both the (vector) electric field and (pseudovector) magnetic field unchanged. You might think that reflecting in this plane would reverse the direction of the magnetic field, but because the magnetic field is a pseudovector with its direction set arbitrarily by the right-hand rule, there's an extra minus sign involved in a mirror-reflection transformation. The result is a physically equivalent situation that still satisfies Maxwell's equations, with the correct direction of motion in time and the right-hand rule fixed properly.
So the initial conditions have mirror symmetry in this plane, and the time evolution equations (Maxwell's equations) have mirror symmetry in this plane, so the outcome has to have the same symmetry. So the reflected wave has to be p-polarized.
Here's another way of thinking about it: If you sent a p-polarized wave in and got an s-polarized component, how does the physics decide whether the electric field should initially go left or right? More precisely: Choose a point in time and space where the incident wave hits the surface with the z component of the electric field maximized. Either the electric field for the s-polarized component of the scattered wave is nonzero at this point, and thus must be in either the +y or the -y direction, or the field is zero and its time derivative is nonzero, so the time derivative is in either the +y or the -y direction. But there's nothing in the physics to decide which of those two cases (+y or -y) will come out. And if we deduce that it must be one of them, and we redo the analysis of the mirror-image of the same experiment (which, we've already seen, is identical to the original experiment), then we get the opposite answer. So either Maxwell's equations allow two distinct solutions (which we know from various uniqueness relations shouldn't be possible), or the s-polarized component has a magnitude of exactly zero.
Similarly, suppose we start with s polarization and again we figure out what the field would look like in a mirror. The electric field now reverses sign under the mirror reflection, as does the magnetic field (again, because of the extra minus sign arising from the magnetic field's pseudovector nature), and we have exactly the same situation but with an extra minus sign. And a similar argument tells us that the s-polarized incident wave can't produce a p-polarized reflected wave.
In summary, and using slightly more technical language:
Reflecting in the x-z plane has no effect on the p-polarized light. The initial conditions therefore have what's called "even parity" under this symmetry operation.
Reflecting in the x-z plane has no effect on the s-polarized light except for multiplication by a minus sign. The initial conditions in this case have "odd parity."
The equations of motion, Maxwell's equations, are symmetric under mirror reflections (again, so long as we understand about that extra minus sign for the magnetic field). So, by Noether's theorem, parity is conserved. Even-parity input states can only give rise to even-parity output states, and vice versa.
So you now know on general principles that the result has to hold. But what if you wanted to verify it in detail? Well, there's an easy enough way to do that as well. You can add to the analysis an additional reflected wave of the other polarization. So for example you can set up a problem where you have an incident p-polarized wave, a reflected p-polarized wave, a refracted p-polarized wave, and two extra waves not in the original problem: a reflected s-polarized wave and a refracted s-polarized wave. Then you set about solving the problem just as before. If you crank through the algebra, you'll find that the amplitudes of the s-polarized components have to be zero. There just will be no way to have them simultaneously match up the boundary conditions at all points in space and time without adding an extra incident s-polarized wave, which is cheating because you're changing the initial conditions. In fact, you might want to go ahead and go through this exercise and see exactly why it doesn't work. At one point, you'll find that if you have the s-wave match in amplitude at all points in space for a given point in time, then the time derivatives all have the wrong signs to match in nearby points in time. And vice versa: If you match the time derivatives, the amplitudes will have the wrong signs. And there's just no way to get the phase factors to match up; there will always be a minus sign that doesn't go away. This is how a parity violation shows up in the algebra. Once you've gone through that exercise, you might better understand why the symmetry has to work that way in general.
Finally, we come to birefringence. Why can birefringence mix s and p polarizations? Simple: Because a birefringent material breaks the mirror-plane symmetry in general. Sure, there will be ways to cut the surface and define the incident direction where the physical situation will still have either pure-even or pure-odd parity, but for a random cut and a random incident direction, it generally won't. The physical situation will have what's called "mixed parity," and this is what allows you to mix the s and p components without violating Noether's theorem.
This is an extremely important principle in physics. When you get used to symmetry laws, you can skip huge parts of complicated derivations, just crossing out terms that are "obviously" zero because they have the wrong symmetry. You know how in freshman physics you can often skip a lot of details of a calculation because you know that, say, momentum has to be conserved? This is exactly the same thing, but just more general. Especially when you get into quantum mechanics, this is a really big deal.
A: The definition of s-polarised light is that the electric field is polarised so that it is perpendicular to the plane of incidence. Where there is a specular reflection, the plane of incidence contains the k-vector of the incoming wave and the reflected wave.
Since the electric field of an EM wave must be perpendicular to the k-vector. This then leaves the possibility that you have a component of the reflected E-field that is p-polarised (as in the second diagram).
As the tangential component of the electric field must be continuous across the boundary, and this relationship must apply to the components of the electric field perpendicular to the plane of the diagram (the plane of incidence) and tangential to the boundary in the plane of incidence, then if we call the p-polarised component of the reflected wave $E_{p,r}$ and the corresponding (and necessary) component in the transmitted wave $E_{p,t}$, then we know that
$$ E_{p,r} \cos \theta_r = E_{p,t} \cos \theta_t \tag{1}$$
and because the s-polarised components have no contribution perpendicular to the interface, we know that
$$ \epsilon_1 E_{p,r} \sin \theta_r = \epsilon_2 E_{p,t} \sin \theta_t. \tag{2}$$
This forms a complete separate and independent pair of equations from those that determine the relationship between the angle of incidence and angle of reflection using the s-polarised components. Therefore we already know from these and the standard derivation, that $\theta_i = \theta_r$ and that $\epsilon_1^{2}\sin\theta_r = \epsilon_2^{2}\sin\theta_t$ (Snell's law for non-magnetic media).
Using Snell's law and equation (2) we can thus derive that
$$\epsilon_2 E_{p,r} = \epsilon_1 E_{p,t} \tag{3}$$
and if $\epsilon_2 > \epsilon_1$, then $E_{p,r}<E_{p,t}$.
Substituting  Snell's law into equation (1) and using $\cos \theta = (1 - \sin^2 \theta)^{1/2}$ it can be shown that
$$ E_{p,r}^{2} = E_{p,t}^{2} \left[ \frac{1 - \sin^2 \theta_t}{1 - (\epsilon_2/\epsilon_1)^4 \sin^2 \theta_t}\right] \tag{4}$$
But in this case, if $\epsilon_2 > \epsilon_1$, we see that $E_{p,r} > E_{p,t}$ and so there is no simultaneous solution for equations (3) and (4), except $E_{p,r}=E_{p,t}=0$, for any real value of $\theta_t$.
I feel sure there must be a more elegant way to show this.
A: Here's a simple explanation based on the dipolar nature of the medium:
For most of the materials, we can assume that the source of the reflected and refracted waves are the induced tiny dipoles in the dielectric medium. In an isotropic medium, polarization vector is proportional to the (total) electric field vector with a constant (as opposed to a tensor for anisotropic media). Therefore they are always in the same direction.
Also, the far-field radiation of a dipole is polarized in the same direction as the dipole moment vector and hence the incident (exciting) wave. 
Therefore, if the material is isotropic, the polarizations of reflected and refracted waves are the same as that of the incident wave. In general, this is not true for birefringent crystals, for which the direction of the induced polarization may not be the same as the incident electric field.
