Talk:Birefringence

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The refractive index is referred to several times as a tensor. This is incorrect, because the property does not transform by the laws of tensors (although permittivity does). See for example "Modern Crystallography IV" Springer series in solid state science v.37, L. A. Shuvalov (Ed.) --FPD.

Could someone also include a discussion on how the ionosphere is birefringent for HF radio waves?

Contents

[edit] Errors in formal definition

The "formal" definition section is wrong, I think, because it fails to take into account that the divergence constraint on the electric field is changed by the anisotropy: now we have \nabla \cdot \varepsilon \mathbf{E} = 0. This means, among other things, that the electromagnetic wave equation has to be written in its full form - c^2 \nabla \times \nabla \times \mathbf{E} = \varepsilon \frac{\partial^2\mathbf{E}}{\partial t^2}, which does not simplify to c^2 \nabla ^2 \mathbf{E} = \varepsilon \frac{\partial^2\mathbf{E}}{\partial t^2} if the electric field is not divergence-free.

The easy way to see the error is that the conclusion is absurd: if \mathbf{E} simply lies along one of the principal axes of ε, then for most directions of \mathbf{k} there will be no eigensolutions that satisfy the divergence constraint \nabla \cdot \varepsilon \mathbf{E} = i\mathbf{k} \cdot \varepsilon \mathbf{E} = 0.

I don't have time to fix it now, but I tagged the article as disputed.



(By the way, the subsection on fibers is goofy too, because it mixes up nonlinearity and birefringence when the two have nothing to do with one another. It's also kind of obscurely written, because the semi-vectorial Schrodinger(-like, it's not quantum mechanics) equation is simply an approximation that can be used in low-contrast waveguides.)

—Steven G. Johnson 20:29, 16 December 2005 (UTC)

Sorry for the error in the definition - and thanks for pointing it out. Using the wave equation properly as you quote, I get k^2 \mathbf{E_0} -\mathbf{(k.E_0) k} = \frac{\omega^2}{c^2}\mathbf{\varepsilon . E_0}, which then satisfies the divergence constraint. I will try to correct it, unless you do first.

User:Salthebad

The "new" version comes to the same conclusion, and is still wrong. The solution electric fields for propagating waves are not simply eigenvectors of the dielectric tensor, nor are they restricted to propagate in only three directions in the crystal. There are propagating solutions in every direction, but the dispersion surface depends upon polarization. Again, you made the same mistake: equatino (4a) does not follow from equation (3a) because the divergence of E is not zero. I suspect that you are doing something like commuting the ∇ and ε operations, which you can't do if ε is not isotropic. —Steven G. Johnson 18:57, 29 December 2005 (UTC)


Note that this problem is, I think, much easier to describe in terms of the magnetic field, since to use the electric field you must inherently solve a generalized eigenproblem (matrices on both sides) or a non-Hermitian problem if you move both matrices to one side. For the magnetic field, you get an ordinary eigenproblem:

-\mathbf{k} \times \left( \frac{1}{\varepsilon} \mathbf{k} \times \mathbf{H}_0 \right) = \frac{\omega^2}{c^2} \mathbf{H}_0

with the constraint \mathbf{k}\cdot\mathbf{H}_0=0, where we have written the magnetic field as:

\mathbf{H} = \mathbf{H}_0 \exp(i\mathbf{k}\cdot\mathbf{x} - i\omega t)

This is an ordinary 3×3 eigenproblem for \mathbf{H}_0, which is Hermitian if we neglect material absorption (i.e. if ε is Hermitian) and thus has three orthogonal solution vectors and corresponding eigen-frequencies ω. One of these solutions is ω=0 but violates the divergence constraint, so it is discarded. The other two solutions have magnetic fields \mathbf{H}_0 orthogonal to k, and correspond to the propagating transverse waves. The corresponding electric field amplitude is given by:

\mathbf{E}_0 = -\frac{c}{\omega \varepsilon} \mathbf{k} \times \mathbf{H}_0

The electric field amplitude is not in general orthogonal to \mathbf{k}, nor does it generally lie along one of the principal axes of ε. Nor are the two electric field polarizations orthogonal; rather, they satisfy \mathbf{E}_0 \cdot \varepsilon \mathbf{E}_0' = 0.

Note also that k is the direction of the phase velocity, but is not generally the direction of the group velocity for anisotropic media, and it is the group velocity that determines the direction in which the refracted beams propagate. Note also that the "effective index" c|\mathbf{k}|/\omega is not in general one of the three eigenvalues of \sqrt{\varepsilon}, but rather lies in between the eigenvalues. (I seem to recall that it forms an ellipsoid, or at least an ellipse for each possible orientation of \mathbf{E}_0.

My own research involves media more complicated than simple homogeneous dielectrics such as this, so I'm probably forgetting some interesting details here, and there might be simpler analysis tricks. e.g. there seem to be all sorts of graphical methods for analyzing these birefringent systems that I've never looked into. But I think I can still spot erroneous approaches.

Note also that for a particular plane of incidence, the resulting solutions have a phase velocity that lies in the same plane, and I believe there are only two refracted solutions even when ε has three distinct eigenvalues. Hence the errors at trirefringence. (The reason that there are only two refracted solutions is obvious from above: for a particular k, there are only two eigenfrequencies corresponding to the two transverse polarizations. Correspondingly, in a given plane of incidence there are only two solution ellipsoids that can be coupled to.) This comes back to the basic misconception that the refracted rays do not in general lie along the principal axes nor are they polarized in those directions.

—Steven G. Johnson 20:27, 29 December 2005 (UTC)


whoops - please check eqns 3-4 again. Do they not follow now? If E0 is an eigenvector of n, then surely k is perpendicular to E0. A general E0 can be written in the basis of these eigenvectors, so as far as I can see, they can then be treated independently. However, I now understand that that does not make them different rays. My use of birefringence is a lot simpler than this stuff, and I agree that your discussion is more sensible, so please edit as you see fit. User:salthebad Salman Rogers
Your equation (4a) looks okay now (although 3a has a typo...the time derivative should be 2nd deriv). And yes, you can always write any vector in the basis of the eigenvectors of ε, but that's misleading because the eigenvectors of ε are not themselves solutions (except when k lies along certain directions), so they cannot be treated independently. (Because the polarization is not generally an eigenvector of ε, it is not in general orthogonal to k.) Because of this, your equation (7) and the subsequent discussion are wrong. The phase velocity is not given by c divided by one of the eigenvalues of √ε.
I would want to look it up in a textbook or two before writing it up myself (I can't because I'm on vacation now). Not so much because I think my analysis is wrong, but there may be a more standard approach, terminology, and perhaps some algebraic tricks that I haven't seen. —Steven G. Johnson 01:36, 30 December 2005 (UTC)

Third time lucky... I hope! Birefringence is more complicated than I thought. My rewrite follows KD Moller Optics p238ff, and it should be correct although it is still not a complete description of course. —Salman Rogers

It looks okay now, although I haven't checked the detailed algebra I assume it's okay since you took it from the textbook. It's a bit unsatisfying because I think you can't quite see the forest for the trees in this kind of approach (I think that's a common weakness of textbooks that dive into the component equations without investigating the higher-level algebraic structure of the problem), but it will do for now. Thanks for looking it up. —Steven G. Johnson 23:43, 5 January 2006 (UTC)

However, I should point out that there is one hidden assumption in the algebra as you presented it: you are assuming that the eigenvectors of ε are real and orthogonal. This is in general true only for materials in which absorption loss is negligible (or can be treated as a first-order perturbation) and in the absence of magneto-optic effects, in which case ε is a real symmetric matrix. Does your textbook comment on this? —Steven G. Johnson 23:51, 5 January 2006 (UTC)

Thanks for your approval :-) I agree that component equations are not very satisfying, but in this case, I think the fact that they neatly end up with the isosurfaces of a sphere and ellipsoid multiplied together, is satisfying enough. As for absorption - the textbooks that I've seen, as well as research papers on optical anisotropy in liquids/soft matter generally reserve birefringence for retardation only (real n), and dichroism for any absorptive component. --Salman Rogers

[edit] More errors?

Hello. According to :-\nabla \times \nabla \times \mathbf{E}=\frac{1}{c^2}\mathbf{\epsilon} \cdot \frac{\part^2 \mathbf{E}}{\partial t^2} (3a)

the ε is wrong. I just checked it out and I am sure. Any idea? with regards MSchwarz

Looks right to me (neglecting material dispersion and assuming μ=1). \nabla \times \mathbf{E} = - \mu \partial\mathbf{H}/\partial ct and \nabla \times \mathbf{H} = \varepsilon \partial\mathbf{E}/\partial ct, so combining these two one gets \nabla \times \frac{1}{\mu} \nabla \times \mathbf{E} = -\frac{\varepsilon}{c^2} \partial^2\mathbf{E}/\partial t^2. —Steven G. Johnson 03:06, 25 June 2006 (UTC)


Hi, MSchwarz again. Please have a look!

\nabla \times \mathbf{E}=-\frac{\part \mathbf{B}}{\partial t} (2.Maxwell)
\nabla \times \mathbf{E}=-{\mu}\frac{\part \mathbf{}}{\partial t} \mathbf{H}
\nabla \times \nabla \times \mathbf{E}=-{\mu}\frac{\part \mathbf{}}{\partial t} \nabla \times \mathbf{H} and with \nabla \times \mathbf{H}= \mathbf{j}+\frac{\part \mathbf{D}}{\partial t} (1.Maxwell)       (and \mathbf{j}= 0 and \varrho = 0 )
\nabla \times \nabla \times \mathbf{E}=-{\mu}\frac{\part \mathbf{}}{\partial t} \frac{\part \mathbf{D}}{\partial t} 
\nabla \times \nabla \times \mathbf{E}=-{\mu}\epsilon \frac{\part^2 \mathbf{}}{\partial t^2}  \mathbf{E}
-\nabla \times \nabla \times \mathbf{E}=\frac{1}{c^2} \cdot \frac{\part^2 \mathbf{E}}{\partial t^2} .....and there is no ε anymore......

furthermore you easily can get to the 'wave-equation' with the "großmann-Identität" (german-word, sorry, I dontknow):
\nabla \times \nabla \times \mathbf{E}=\nabla(\nabla \mathbf{E}) -(\nabla \nabla)\mathbf{E} and as mentioned : \nabla \mathbf{E}=0    (3.Maxwell:\nabla \mathbf{D}= \varrho ! and \varrho = 0 )
\nabla \times \nabla \times \mathbf{E}=-\nabla^2 \mathbf{E}

which is :
\nabla^2 \mathbf{E} = \frac{1}{c^2} \cdot \frac{\part^2 \mathbf{E}}{\partial t^2}
You`ll find it in any book. And that was the reason for getting into this.
Im sure. Greetings from Hamburg, MSchwarz 05:51, 25 June 2006 (UTC)
You assumed that \mu \varepsilon = 1/c^2, which is only true in vacuum. (c always denotes the speed of light in vacuum.) (If you're confused by the c factors in my explanation, above, it's because I was instinctively using cgs units. Actually, I tend to set all the dimensionful constants to unity, myself.) You also assumed that the medium was homogeneous (so that you could pull μ through the derivatives), which I did not, above. In your final equation, you assumed that the medium was not only homogeneous, but also isotropic, in order to get \nabla\cdot\mathbf{E}=0, which is definitely not the case here. —Steven G. Johnson 15:31, 27 June 2006 (UTC)

[edit] Birefringence values for ice

I suppose checking birefringence values for water ice. Now it stands: 1.309 1.313 +0.014 which are wrong, because 1.313-1.309=0.004, not 0.014. It looks like the mistake is copied from the source website. I have no idea of proper values, if someone has the sources - please check it.

[edit] Fiber Optic subsection.

On 16 December 2005, Steven Johnson wrote:

(By the way, the subsection on fibers is goofy too, because it mixes up nonlinearity and birefringence when the two have nothing to do with one another. It's also kind of obscurely written, because the semi-vectorial Schrodinger(-like, it's not quantum mechanics) equation is simply an approximation that can be used in low-contrast waveguides.)

This is Adrian Keister, and I wrote that section on fiber optics. I'm not quite sure I follow your comments.

1. I realize that nonlinearity and birefringence are not the same thing. In what way does the subsection mix them up? It is true that the coupled nonlinear Schrödinger equation takes birefringence into account and is thus a model for birefringence in a fiber optic cable, among other things. See C. Menyuk, Application of multiple-length-scale methods to the study of optical fiber transmission, Journal of Engineering Mathematics 36: 113-136, 1999, Kluwer Academic Publishers, Netherlands, for the derivation of the coupled nonlinear Schrödinger equations starting from Maxwell's equations, and applying the boundary conditions for a fiber optic cable.

Nonlinearity and birefringence are independent physical concepts. Yes, you can have both in the same fiber, but there is no reason to include both here, and it is misleading to imply that you need to include nonlinearities in order to model birefringence.

2. The coupled nonlinear Schrödinger equations are called that because in applying the Inverse Scattering Transform, the forward scattering problem uses a Lax Pair L and M, and the L operator is typically the Schrödinger operator. The eigenvalue problem, therefore, is the time-independent Schrödinger equation from quantum mechanics. There is thus quite a close link between the Inverse Scattering Transform and quantum mechanics. These equations are not called the coupled nonlinear Schrödinger-like equations.

I'm quite familiar with what the NSE is, although I think your terminology is unneccessarily obscure (the standard terminology for the derivation can be expressed in four words: slowly varying envelope approximation = SVEA). In any case, you're not seeing the forest for the trees here. First, most readers will have heard of the Schrodinger equation in the context of quantum mechanics where it has a specific physical interpretation which does not apply here (the solution is not the quantum probability amplitude, the terms are not kinetic/potential energy, etcetera). The only reason for the name is that the linear part has the same form as the quantum Schrodinger equation, but it does not have the same interpretation. This confuses people if it is not presented carefully.
Second, any mention of the NSE in this context is incomplete if it does not mention that it is only an approximation (a first-order approximation at that) for weak nonlinearities and slowly varying wave envelopes.
In any case, semi-vectorial slowly-varying envelope approximations are simply one of many ways to model birefringence in optical fibers, and it's a bit strange to single them out here. (For one thing, that approximation, as applied in the linear or nonlinear Schrodinger equation, only models the effect of birefringence; in order to actually predict birefringence for a particular geometry or perturbation you need other methods.)

3. Naturally, since actual solitons do not exist, only a soliton-effect, no one is saying that the equations model the physical situation perfectly. But saying that the equations model the physical situation perfectly is not the same thing as saying that the equations model the physical situation.

Why are you bringing up solitons? Again, a concept quite independent of birefringence. (FYI, the SVEA is only an approximation even in the linear case, too.)

4. I'm not sure why the fact that the coupled nonlinear Schrödinger equation is only an approximation renders the subsection "obscurely written."

I would appreciate any comments you might have in reply.

Regards, Adrian

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