# Is there a e-ray refraction theory?

1. Dec 30, 2011

### enotstrebor

Snell's law can be derived by applying boundary conditions to Maxwell's equations.

Thus the o-ray's behavior has a formal theory that tells how and why it behaves as it does.

Is there any formal theory that covers the behavior of the e-ray (not obeying Snell's law)?

2. Dec 31, 2011

### Born2bwire

The e-ray still follows Snell's Law. The difference is that the index of refraction is dependent upon the polarization of the light with respect to the optical axis. A more general way of treating this problem would be to solve for the boundary conditions using a anisotropic permittivity (tensor). But this then could be reduced to a simple application of the appropriate index of refraction upon the decomposition of the light into the two polarizations that excite the ordinary and extraordinary waves. The wikipedia article on birefringence has a good demonstration of this process.

3. Jan 7, 2012

### enotstrebor

You can go to any text book (e.g. introduction to modern optics), http://physics.info/refraction/ or wiki or all of which say it does not follow Snell's Law.

But it also does not violate the laws of electrodynamics. Further research indicates that unlike Snell's law, the e-ray has a varying index of refraction rather than constant.

I have not seen any formal mathematical relationship for this variation, although it clearly varies with respect to the incident plane and angle to the media surface and to the optical axis orientation relative to these two surfaces.

But it appears that in one orientation the index does not change with angle in a second orthogonal direction you have a ?elliptical? variation in the index value with the maximum difference when the ray is vertical to the media and in the third direction you have an ellipse that is maximum at 90 degrees to the media surface.

So I guess in some respect I have found my own answer though I would prefer to have a mathematical relationship and I presently can not picture out how these ?ellipses? make a 3-D surface.

Thanks anyway

4. Jan 7, 2012

### Born2bwire

Snell's Law is just an equation that requires the continuity of the wave vector, that is phase continuity across the boundary. This still holds for both the ordinary and extraordinary wave. The difference being that the index of refraction is no longer a constant for the extraordinary wave. But if you go through the derivation in the Wikipedia article, you see that the same physics is derived here. The dispersion relationship that is derived dictates that the ordinary wave vectors lie on a sphere. So they point in any direction but must have a constant magnitude regardless of the direction that they point. This correlates to the idea that regardless of the direction of the refracted wave in the material, its wave number is a constant. Thus, we can express the boundary condition like the traditional simple Snell's Law. But for the extraordinary wave the wave vectors lie on an ellipse. So the wave number changes depending upon the direction of the wave inside the material. Thus, the derived Snell's Law is going to have an effective index of refraction that is dependent upon the incident wave and its polarization and relationship with the optical axis.

That is, we know from Snell's Law that in general, assuming that the boundary between materials lies in the x-y plane and ignoring the aspects about polarization, that $k_{1x}=k_{2x}$. For the ordinary waves we know that the eigenvectors of the k vectors in the material lie on a sphere. Thus,

$$k_{1x} = k_1\sin\theta_1 = k_{2x} = k_2\sin\theta_2$$
As for the extraordinary wave, the eigenvectors lie on the ellipse. So we would need to find an equation for the ellipse and take the projection of the k_x component from that surface and use that in the right-hand side of the above boundary condition.

In the example from Wikipedia, we are already given the surface equations from the simplified dispersion relation.

$$\left( \frac{k_x^2}{n_o^2}+\frac{k_y^2}{n_o^2}+\frac{k_z^2}{n_o^2} -\frac{\omega^2}{c^2} \right) \left(\frac{k_x^2}{n_e^2}+\frac{k_y^2}{n_e^2}+ \frac{k_z^2}{n_o^2} -\frac{\omega^2}{c^2}\right)=0$$

So the equation describing the elliptical surface is just

$$\left(\frac{k_x^2}{n_e^2}+\frac{k_y^2}{n_e^2}+ \frac{k_z^2}{n_o^2} -\frac{\omega^2}{c^2}\right)=0$$

We can again see trivially from solving the case of the first term of the previous equation for the ordinary wave modes that $\left| \mathbf{k} \right|^2 = n_o^2\omega^2/c^2$ which is the normal isotropic relationship. Substituting this back into our previous boundary condition we have,

$$k_1\sin\theta_1 = \frac{n_1\omega}{c}\sin\theta_1 = k_2\sin\theta_2 = \frac{n_o\omega}{c}\sin\theta_1$$
$$n_1\sin\theta_1 = n_o\sin\theta_2$$

So to summarize, everything you need is already shown in the derivation on Wikipedia. You just need to go through it again and try to understand it.

Last edited: Jan 7, 2012
5. Jan 7, 2012

### enotstrebor

All the text an references I referenced above above say you are wrong. I quote

"an ordinary ray (or o ray or ω [omega] ray) and an extraordinary ray (or e ray or ε [epsilon] ray). As the name implies, the o ray behaves in an "ordinary" way, following Snell's law without a problem. The ratio of the sine of the angle of incidence to the sine of the angle of ordinary refraction is a constant. The e ray gets its name because it does not obey this rule."

I could quote the other source as well.

Obviously you have your own definition of Snell's Law.

Good luck

6. Jan 8, 2012

### a-tom-ic

An a bit unkind response to someone trying to help you, don't you think? (Not judging the correctness of either of the responses)

7. Jan 8, 2012

### vanhees71

Well, obviously you discuss crystal optics, and there you need a dielectric tensor rather than a dielectric function. That's the only difference. The boundary conditions, i.e., the continuity of the various field-strengths components at the surface of the crystal remains the same since they follow from Maxwell's equations, which are generally valid.

8. Jan 8, 2012

### enotstrebor

I was not intending to be rude (Good Luck was intended as a sign off), and I will not discuss or debate this issue further with this Adviser)

After, After I gave the correct and textbook answer (Introduction to Modern Optics), the Adviser gives an answer which appear to be contrary ending with "You just need to go through it again and try to understand it." (the references I gave all say or indicate that the e-ray does not follow Snell's Law - the Adviser gives none)

In addition, (you may or may not have noticed), the Advisor's mathematical derivation was irrelevant to the e-ray as it was a derivation for a angle and orientational independent index of refraction (Snell's Law), while the e-ray has an angle and orientation dependent index of refraction.

So I do not find his responses helpful to me (fact) and I'm sure confusing to others (?yourself?).

I hope this helps.

Last edited: Jan 8, 2012
9. Jan 8, 2012

### Dickfore

yes.

10. Jan 8, 2012

### Born2bwire

The answer and derivation is straight from Wikipedia. You just aren't taking the time to understand it and apply it. The information to find the transmitted extraordinary ray is right there. By saying that Snell's Law does not apply here is showing that you are not understanding the underlying theory behind Snell's Law. The theory is still valid for a birefringent material. Take a look at the Wikipedia article for Snell's Law and you will see that their derivation is the same as the steps that I took in the bottom part of my derivation. It starts with the continuity of phase at the boundary which is still valid regardless of the extraordinary or ordinary wave. And if we go through the process for the ordinary wave, it comes out to be the simplified equation that one expects. It just now remains to go through and find an equation for the extraordinary wave. But it would just be a simpler process to find the transmitted wave given a specific incident wave.

EDIT: If you take the time to read through my post you will see that I derived the relationship for the ordinary wave, whose transmitted wave vectors lie on a sphere. For the extraordinary wave, you need to go through the process for wave vectors that lie on an ellipsoid. Slightly more complicated process, but it's still the same procedure more or less.

Last edited: Jan 8, 2012