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Polarizations of plane waves propagating in anisotropic media 
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#1
Mar2613, 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 C_{ijkl} is the stiffness tensor, n_{i} is the normal to the plane wave, ρ is the density, c is the phase velocity, δ_{ik} is the Kroenecker delta and u_{k} is the polarization of the wave. This is obviously and eigenvalue problem where ρc^{2} 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 x_{1}x_{3} plane (n_{i} = [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: C_{11} = 5, C_{33} = 6, C_{44} = 2, C_{12} = 4 and C_{13} = 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+3.50.5\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)^23+\sqrt{51cos(\theta)^4+58cos(\theta)^2+9})\\ 0 \\ 1 \end{Bmatrix},\\ \mathbf{u}_3 &=& \begin{Bmatrix} 10sin(\theta)cos(\theta)/(7cos(\theta)^23\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 quasilongitudinal wave and the two other are called quasitransverse waves. The eigenvectors tells us what the polarization (material displacement direction) of the wave is. Here is the issue: u_{2} is not defined for θ = ∏/2 and u_{3} 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 u_{2}, 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]? 


#2
Mar2613, 09:31 AM

P: 741

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. 


#3
Mar2613, 09:56 AM

Sci Advisor
P: 3,593

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.



#4
Mar2613, 10:45 AM

P: 5

Polarizations of plane waves propagating in anisotropic media



#5
Mar2613, 01:46 PM

Sci Advisor
P: 3,593




#6
Mar2613, 06:07 PM

P: 5

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. 


#7
Mar2613, 07:21 PM

P: 741

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 dielectricconstants 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 dielectricconstants are equal while the third dielectricconstant is different. If the material is biaxial, all three dielectricconstants 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 dielectricconstants are equal. In a uniaxial crystal, two dielectricconstants 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. 


#8
Mar2713, 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 equiphase 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 finitesized elastic waves). The terminology with EMR and crystals is unfamiliar to me, and I am sorry if I misunderstand some of your points. \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 x_{1}x_{3} 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 ) 


#9
Mar2713, 06:46 AM

Sci Advisor
P: 3,593

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##? 


#10
Mar2713, 08:00 AM

P: 5

\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)^23+\sqrt{51cos(\theta)^4+58cos(\theta)^2+9})\\ 0 \\ 1 \end{Bmatrix},\\ \mathbf{\hat{u}}_3 &=& \begin{Bmatrix} 10sin(\theta)cos(\theta)/(7cos(\theta)^23\sqrt{51cos(\theta)^4+58cos(\theta)^2+9}) \\ 0 \\ 1 \end{Bmatrix}. \end{eqnarray*} If I multiply the eigenvector u_{2} with the denominator of its first element I obtain \begin{equation*} \mathbf{\hat{u}}_2 = \begin{Bmatrix} 10sin(\theta)cos(\theta)\\ 0 \\ (7cos(\theta)^23+\sqrt{51cos(\theta)^4+58cos(\theta)^2+9}) \end{Bmatrix}. \end{equation*} Inserting θ=∏/2 gives u_{2} = [0, 0, 0]^{T}, which doens't make sense. This is my problem. 


#11
Mar2713, 08:29 AM

Sci Advisor
P: 3,593

Why don't you just normalize u1, u2 and u3 to 1 before considering any limit?



#12
Mar2713, 08:39 AM

Sci Advisor
P: 5,523

http://link.springer.com/article/10.1007%2FBF01591006 http://iopscience.iop.org/00223727/21/6/003 http://www.rci.rutgers.edu/~norris/p..._33_97108.pdf http://connection.ebscohost.com/c/ar...ansversewaves 


#13
Mar2713, 09:19 AM

Sci Advisor
P: 3,593

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. 


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