Polarizations of plane waves propagating in anisotropic media


by loehre
Tags: anisotropy, eigenvectors, wave propagation
loehre
loehre is offline
#1
Mar26-13, 08:53 AM
P: 5
Hey guys! (I am not sure if I should post this thread in Physics or Mathematics)

I have had some issues with developing expressions for the polarizations (material displacement) of waves propagating in anisotropic media. To bring you guys up to speed I have to start a few steps before the problem appears. We start with Christoffel's equation (http://en.wikipedia.org/wiki/Linear_...offel_equation):

\begin{equation}
(C_{ijkl}n_jn_l - \rho c^2\delta_{ik})u_k = 0,
\end{equation}

where Cijkl is the stiffness tensor, ni is the normal to the plane wave, ρ is the density, c is the phase velocity, δik is the Kroenecker delta and uk is the polarization of the wave. This is obviously and eigenvalue problem where ρc2 is the eigenvalue. Introducing reduced notation: 11, 22, 33, 23, 13 and 12 is replaced by 1, 2, 3, 4, 5 and 6, respectively. Now, implementing a transverse isotropic material (http://en.wikipedia.org/wiki/Linear_...ogeneous_media) and looking in a direction in the x1-x3 plane (ni = [sin(θ), 0 , cos(θ)]T), the eigenvalue problem becomes

\begin{equation}
\begin{vmatrix}
C_{11}n_1^2+C_{44}n_3^2-\rho c^2 & 0 & (C_{13}+C_{44})n_1n_3 \\
0 & \frac{1}{2}(C_{11}-C_{12})n_1^2+C_{44}n_3^2 -\rho c^2 & 0 \\
(C_{13}+C_{44})n_1n_3 & 0 & C_{44}n_1^2+C_{33}n_3^2 -\rho c^2
\end{vmatrix} = 0.
\end{equation}

To avoid working with many symbols, we can insert values for the stiffness coefficients: C11 = 5, C33 = 6, C44 = 2, C12 = 4 and C13 = 3. The eigenvalues then become:

\begin{eqnarray}
\rho c^2_1 &=& cos(\theta)^2+1 \\
\rho c^2_2 &=& 0.5cos(\theta)^2+3.5+0.5\sqrt{-51cos(\theta)^4+58cos(\theta)^2+9} \\
\rho c^2_3 &=& 0.5cos(\theta)^2+3.5-0.5\sqrt{-51cos(\theta)^4+58cos(\theta)^2+9}
\end{eqnarray}

and the corresponding eigenvectors are:

\begin{eqnarray}
\mathbf{u}_1 &=& \begin{Bmatrix}
0\\
1 \\
0
\end{Bmatrix}, \\
\mathbf{u}_2 &=& \begin{Bmatrix}
10sin(\theta)cos(\theta)/(7cos(\theta)^2-3+\sqrt{-51cos(\theta)^4+58cos(\theta)^2+9})\\
0 \\
1
\end{Bmatrix},\\
\mathbf{u}_3 &=& \begin{Bmatrix}
10sin(\theta)cos(\theta)/(7cos(\theta)^2-3-\sqrt{-51cos(\theta)^4+58cos(\theta)^2+9}) \\
0 \\
1
\end{Bmatrix}.
\end{eqnarray}

For those interested: the eigensolutions tell us that we can have three different types of plane waves propagating in direction n, each with its characteristic wave speed. One is termed quasi-longitudinal wave and the two other are called quasi-transverse waves. The eigenvectors tells us what the polarization (material displacement direction) of the wave is.

Here is the issue: u2 is not defined for θ = ∏/2 and u3 is not defined for θ=0 (zero divided by zero). If I insert these values for θ before I solve the eigenvalue problem, I obtain the correct eigenvectors. Why does this happen? If I use Maple to take the limit of θ goes to ∏/2 on u2, I get the answer 'undefined' (I guess this means 'infinity'). Is it possible to avoid this division by zero for certain θs through some mathematical trick? I have tried scaling up the eigenvectors etc. Also, can I say that the vector ['infinity',0,1] is equivalent with [1, 0, 0]?
Phys.Org News Partner Physics news on Phys.org
The hemihelix: Scientists discover a new shape using rubber bands (w/ video)
Mapping the road to quantum gravity
Chameleon crystals could enable active camouflage (w/ video)
Darwin123
Darwin123 is offline
#2
Mar26-13, 09:31 AM
P: 741
Quote Quote by loehre View Post
For those interested: the eigensolutions tell us that we can have three different types of plane waves propagating in direction n, each with its characteristic wave speed. One is termed quasi-longitudinal wave and the two other are called quasi-transverse waves. The eigenvectors tells us what the polarization (material displacement direction) of the wave is.

Here is the issue: u2 is not defined for θ = ∏/2 and u3 is not defined for θ=0 (zero divided by zero). If I insert these values for θ before I solve the eigenvalue problem, I obtain the correct eigenvectors. Why does this happen? If I use Maple to take the limit of θ goes to ∏/2 on u2, I get the answer 'undefined' (I guess this means 'infinity'). Is it possible to avoid this division by zero for certain θs through some mathematical trick? I have tried scaling up the eigenvectors etc. Also, can I say that the vector ['infinity',0,1] is equivalent with [1, 0, 0]?
"Undefined" does not always mean "infinity". "Not unique" is also "undefined". For example, 1/0→∞. However, 0/0→z where n can be any value at all including ∞. I think you have a case of 0/0. This would be "indeterminable" rather than infinite.

I suspect the latter case in your example. To take the limit, you should use L'Hospitals Law.

Never trust symbolic manipulation software. If the software can't do it, then try it on your own.

Several things can happen that make an eigenvector solution indeterminable. I'll mention two possibilities that don't involve ∞.

1) One of your eigenvalues may be zero. If an eigenvalue is zero, then all vectors are eigenvectors.

2) Two of your eigenvectors may be equal in value but not zero. Then, any linear combination of those two eigenvectors will be a solution to the eigenvector problem.


There are corresponding physical implications to an indeterminate solution. However, you can check the mathematics before you think about the physics.
DrDu
DrDu is offline
#3
Mar26-13, 09:56 AM
Sci Advisor
P: 3,378
For either ##\theta=0 ## and ##\pi/2## your matrix whose eigenvalues you are looking for becomes diagonal because either n_1 or n_3 is zero, so that both the eigenvalues and the eigenvectors can be read off directly.

loehre
loehre is offline
#4
Mar26-13, 10:45 AM
P: 5

Polarizations of plane waves propagating in anisotropic media


"Undefined" does not always mean "infinity". "Not unique" is also "undefined". For example, 1/0→∞. However, 0/0→z where n can be any value at all including ∞. I think you have a case of 0/0. This would be "indeterminable" rather than infinite.
If I first insert θ=0.1 and then insert θ=0.001, I see that the norm of the eigenvector u3 increases rapidly. So I have all reason to suspect that 'undefined' in this case means 'infinity'.

There are corresponding physical implications to an indeterminate solution. However, you can check the mathematics before you think about the physics.
I know that for θ=0, for example, the eigenvectors should be u1=[0, 1, 0], u2=[0, 0, 1] and u3=[1, 0, 0]. I know this from physical reasoning, and I also get these vector if I insert θ=0 into my eigenvalue problem (second equation).

For either θ=0 and π/2 your matrix whose eigenvalues you are looking for becomes diagonal because either n_1 or n_3 is zero, so that both the eigenvalues and the eigenvectors can be read off directly.
Yes, I know. However, I would like expressions for the eigenvectors with θ that are valid for all θs. So the question is: can I say that the eigenvector [∞, 0 , 1] is equivalent with [1, 0, 0], for example? Or if there is some way to work around this problem with infinite values.
DrDu
DrDu is offline
#5
Mar26-13, 01:46 PM
Sci Advisor
P: 3,378
Quote Quote by loehre View Post
Ican I say that the eigenvector [∞, 0 , 1] is equivalent with [1, 0, 0], for example? Or if there is some way to work around this problem with infinite values.
Yes, as 1/infinity=0. However, I don't see why you need maple to solve a simple 3x3 eigenvalue equation.
loehre
loehre is offline
#6
Mar26-13, 06:07 PM
P: 5
However, I don't see why you need maple to solve a simple 3x3 eigenvalue equation.
You are right; I do not need Maple to solve the problem. Nevertheless, it is convenient if I want to change the stiffness coefficient values or if I want to insert different values for θ. Also, taking the limit of the mentioned eigenvectors is not straight forward by hand (L'Hopital's rule etc.).

Anyway, thank you guys for the replies. I guess I now have to just convince my supervisor that the expressions for the eigenvectors are satisfactory.
Darwin123
Darwin123 is offline
#7
Mar26-13, 07:21 PM
P: 741
Quote Quote by loehre View Post
If I first insert θ=0.1 and then insert θ=0.001, I see that the norm of the eigenvector u3 increases rapidly. So I have all reason to suspect that 'undefined' in this case means 'infinity'.
I don't think that the divergence of the "norm" guarantees the uniqueness of the eigenvalues.

I think that I figured out in terms of both physics and mathematics what your undefined eigenvectors mean.

The mathematics is basically my first possibility. The values of θ where the eigenvector is undefined corresponds to a pair of eigenvalues are equal. These values correspond to a uniaxial crystal.

Physically, the eigenvalues that you are describing are dielectric-constants of the material. The eigenvectors correspond to directions in which electromagnetic radiation (EMR) can propagate in a straight line. However, electromagnetic radiation can move in a curved path at certain polarizations when the material is not birefringent.

There are three dielectric constants corresponding to the three principle axes of the material. If the material is totally isotropic, then the three dielectric constants are equal. If the material is birefingent with only one optical axis, only two dielectric-constants are equal while the third dielectric-constant is different. If the material is biaxial, all three dielectric-constants are unequal.


You are subconsciously assuming that the material is biaxial. If the material is biaxial, then there really only three directions in which light can travel in a straight line. These three directions are orthogonal. They are the principle axes of the material. Therefore, the eigenvector problem uniquely identifies these three directions.


In an isotropic material, like vacuum, light of any polarization can travel in a straight line in any direction. Therefore, all three eigenvectors are undefined. This is because all three dielectric-constants are equal.

In a uniaxial crystal, two dielectric-constants corresponding to different axes are equal. The third axis is the optical axis. The plane perpendicular to this is the isotropic plane. Two of the eigenvectors corresponding to the isotropic plane are undefined. EMR of any polarization can travel in a straight line if the direction of propagation is either in the direction of the optical axis or in the isotropic plane. That angle θ=0 and θ=90° correspond to the two directions in the isotropic plane.

I suspect that you are looking at a model for a uniaxial crystal. That is why only two eigenvectors are undefined. The eigenvector that is always defined is the optical axis.

In a biaxial crystal, there are are two optical axes. There is no isotropic plane. The only way an EMR ray can travel in a straight line is to travel in the direction of a principle axis. There are only three of them.

I note that you didn't check the case of an limiting isotropic material. Go ahead. Look at the problem for a matrix where all three eigenvalues are equal. I conjecture that all three eigenvectors, not just two, are "undefined" in the case of an isotropic material.
loehre
loehre is offline
#8
Mar27-13, 06:42 AM
P: 5
Thank you Darwin123 for taking the time to write such a thorough answer.

I should maybe have been more precise in that I am working with elastic waves (stress waves) in solid media. I know many of the same principles apply to optics as to elasticity, but there are some differences in the phyical interpretation. For example: eigenvectors represent direction of material displacement. Typical transverse isotropic material in elasticity are fibrous materials, e.g. skeletal muscles. This type of material exhibits hexagonal symmetry. Also, in the derivation of Christoffel's equation (the equation I start with), plane waves are assumed. Plane waves are equi-phase planes of infinite size propagating in direction n. It therefore not relevant to talk about EMR moving in curved paths (that would be relevant for finite-sized elastic waves). The terminology with EMR and crystals is unfamiliar to me, and I am sorry if I misunderstand some of your points.

The mathematics is basically my first possibility. The values of θ where the eigenvector is undefined corresponds to a pair of eigenvalues are equal. These values correspond to a uniaxial crystal.
Well, two of the eigenvalues are only equal at θ=0, not θ=∏/2.

I note that you didn't check the case of an limiting isotropic material. Go ahead. Look at the problem for a matrix where all three eigenvalues are equal. I conjecture that all three eigenvectors, not just two, are "undefined" in the case of an isotropic material.
An isotropic material has a stiffness tensor on the form

\begin{equation*}
\mathbf{C}= \begin{bmatrix}
\lambda+2\mu & \lambda & \lambda & 0 & 0 & 0 \\
\lambda & \lambda+2\mu & \lambda & 0 & 0 & 0 \\
\lambda & \lambda & \lambda+2\mu & 0 & 0 & 0 \\
0 & 0 & 0 & \mu & 0 & 0 \\
0 & 0 & 0 & 0 & \mu & 0 \\
0 & 0 & 0 & 0 & 0 & \mu
\end{bmatrix},
\end{equation*}

where λ and μ are the Lamè constants. If I set λ=2 and μ=1, I obtain the following eigenvalues:

\begin{eqnarray}
\rho c_1^2 &=& 4, \\
\rho c_2^2 &=& 1, \\
\rho c_3^2 &=& 1.
\end{eqnarray}

As expected from theory, the eigenvalues are independent of θ, and two of them are always equal. The first wave speed is related to longitudinal wave, while the two other are related to transverse (shear) waves. The eigenvectors in the x1-x3 plane are:

\begin{equation*}
\mathbf{u}_1 = \begin{Bmatrix}
\frac{sin(\theta)}{cos(\theta)} \\
0 \\
1
\end{Bmatrix},\quad
\mathbf{u}_2 = \begin{Bmatrix}
-\frac{cos(\theta)}{sin(\theta)} \\
0 \\
1
\end{Bmatrix},\quad
\mathbf{u}_3 = \begin{Bmatrix}
0\\
1 \\
0
\end{Bmatrix}.
\end{equation*}

Once again, two of the eigenvectors are not defined at θ=0 and θ=∏/2, respectively. This is basically the same issue as before. Nevertheless, I am now satisfied with my results. So you do not have to look more into it (unless you are really interested )
DrDu
DrDu is offline
#9
Mar27-13, 06:46 AM
Sci Advisor
P: 3,378
Quote Quote by loehre View Post
\begin{equation*}
\mathbf{u}_1 = \begin{Bmatrix}
\frac{sin(\theta)}{cos(\theta)} \\
0 \\
1
\end{Bmatrix},\quad
\mathbf{u}_2 = \begin{Bmatrix}
-\frac{cos(\theta)}{sin(\theta)} \\
0 \\
1
\end{Bmatrix},\quad
\mathbf{u}_3 = \begin{Bmatrix}
0\\
1 \\
0
\end{Bmatrix}.
\end{equation*}
I still don't quite understand what's your problem.
The eigenvectors you write down aren't normalized any how, so why don't you chose e.g.
##u_1=(\sin \theta, 0, \cos \theta)^T##?
loehre
loehre is offline
#10
Mar27-13, 08:00 AM
P: 5
Quote Quote by DrDu View Post
I still don't quite understand what's your problem.
The eigenvectors you write down aren't normalized any how, so why don't you chose e.g.
##u_1=(\sin \theta, 0, \cos \theta)^T##?
You are absolutely right. I was a bit general with my conclusion there. The eigenvectors for an isotropic solid are valid for all θ when normalized as you suggested. However, looking back at the original eigenvectors:

\begin{eqnarray*}
\mathbf{\hat{u}}_1 &=& \begin{Bmatrix}
0\\
1 \\
0
\end{Bmatrix}, \\
\mathbf{\hat{u}}_2 &=& \begin{Bmatrix}
10sin(\theta)cos(\theta)/(7cos(\theta)^2-3+\sqrt{-51cos(\theta)^4+58cos(\theta)^2+9})\\
0 \\
1
\end{Bmatrix},\\
\mathbf{\hat{u}}_3 &=& \begin{Bmatrix}
10sin(\theta)cos(\theta)/(7cos(\theta)^2-3-\sqrt{-51cos(\theta)^4+58cos(\theta)^2+9}) \\
0 \\
1
\end{Bmatrix}.
\end{eqnarray*}

If I multiply the eigenvector u2 with the denominator of its first element I obtain

\begin{equation*}
\mathbf{\hat{u}}_2 = \begin{Bmatrix}
10sin(\theta)cos(\theta)\\
0 \\
(7cos(\theta)^2-3+\sqrt{-51cos(\theta)^4+58cos(\theta)^2+9})
\end{Bmatrix}.
\end{equation*}

Inserting θ=∏/2 gives u2 = [0, 0, 0]T, which doens't make sense. This is my problem.
DrDu
DrDu is offline
#11
Mar27-13, 08:29 AM
Sci Advisor
P: 3,378
Why don't you just normalize u1, u2 and u3 to 1 before considering any limit?
Andy Resnick
Andy Resnick is offline
#12
Mar27-13, 08:39 AM
Sci Advisor
P: 5,468
Quote Quote by loehre View Post
Hey guys! (I am not sure if I should post this thread in Physics or Mathematics)

I have had some issues with developing expressions for the polarizations (material displacement) of waves propagating in anisotropic media. <snip>
Yikes- tough problem! All my go-to references failed me. Have you looked at:

http://link.springer.com/article/10.1007%2FBF01591006
http://iopscience.iop.org/0022-3727/21/6/003
http://www.rci.rutgers.edu/~norris/p..._33_97-108.pdf
http://connection.ebscohost.com/c/ar...ansverse-waves
DrDu
DrDu is offline
#13
Mar27-13, 09:19 AM
Sci Advisor
P: 3,378
I had a closer look at your initial expressions. Your eigenvalue problem breaks down into a 1x1 eigenvalue problem for u1 and a 2x2 problem for u2 and u3 which are eigenvalues of the matrix
[itex]\left( \begin{array}{cc}
[(C_{11}-C_{33})+(C_{11}+C_{33}-2C_{44})\cos 2\theta] & (C_{13}+C_{44}))\sin 2\theta \\
(C_{13}+C_{44})\sin 2\theta & -[(C_{11}-C_{33})+(C_{11}+C_{33}-2C_{44})\cos 2\theta]
\end{array}\right)
[/itex]
There is no problem with the normalized eigenvectors as long as not all elements of the matrix vanish simultaneously.


Register to reply

Related Discussions
waves in periodic structures - Coupling of evanescent waves to propagating waves Atomic, Solid State, Comp. Physics 5
waves in periodic structures - Coupling of evanescent waves to propagating waves Classical Physics 1
Anisotropic Dielectric Media Transition General Physics 0
Plane waves: sign of Re(ε), Re(μ) in passive media, attenuation angle Classical Physics 4
Point charge in anisotropic dielectric media Advanced Physics Homework 0