Electrooptic Tensor, Relating Tensors to Orientations

Question

So I'm trying to gain a better understanding of electrooptic tensors:
An example of a quartz electrooptic tensor is given. I know in order to best implement this crystal, in order to get the highest birefringence or highest opto,piezoelectric output, I want to use the value that is largest in this tensor.

However, I don't understand how to look at the tensor and say "Ah, I want to put pressure vertically on the material" as opposed to horizontally if I wanted to use the r11 value (in the case of a piezoelectric). Whereas for the birefringent case, you're looking to know the direction of the input/output of the light.
I think there's a simple explanation to this, I'm just not aware of it and can't find any valuable resources. Feel free to use a value listed in the following tensor to describe orientations:
For wavelengths of 409-605nm r11=-.47 and r41=.20 under constant strain and r11=.174 under constant stress.
These values were found in Marvin J. Webers text "Handbook of Optical Materials" 2003.

Answer 1

You wrote "vertical pressure" but I think you meant to write "vertical electric field", since this is for electro-optic modulation. I assume you got the $r$-tensor from the following excellent page:

http://www.sjsu.edu/faculty/beyersdorf/Archive/Phys208F07/ch%207-electrooptics.pdf

To recap what that PDF says. the impermeability of a material may be tensor-expanded to first-order as a function of the ambient electric field as $$\eta=\eta_0+\mbox{InverseVoigt}[\mathbf{r}\cdot\mathbf{E}]$$ where $$\eta_0=\left(\begin{array}{cc} \eta_{11}&0&0\\ 0&\eta_{22}&0\\ 0&0&\eta_{33} \end{array}\right)$$ is the zero-field representation of the index ellipsoid coefficients in the principle-axis coordinates, $$\mathbf{r}=\left(\begin{array}{cc} r_{11}&0&0\\ -r_{11}&0&0\\ 0&0&0\\ r_{41}&0&0\\ 0&-r_{41}&0\\ 0&-r_{11}&0 \end{array}\right) $$ is the Voigt representation of the third-order electro-optic tensor, $$\mathbf{E}=\left(\begin{array}{cc} E_x\\ E_y\\ E_z \end{array} \right)$$ is the applied electric field, and $$\mbox{InverseVoigt}\left[\left( \begin{array}{cc} a\\ b\\ c\\ d\\ e\\ f \end{array} \right) \right]=\left(\begin{array}{cc} a&e&f\\ e&b&d\\ f&d&c \end{array}\right)$$ is just shorthand for the operator which converts between matrix and Voigt notation.

For the quartz tensor you provided, this becomes

$$\eta=\left(\begin{array}{cc} \eta_{11}+r_{11}E_x&-r_{41}E_y&-r_{11}E_y\\ -r_{41}E_y&\eta_{22}-r_{11}E_x&r_{41}E_x\\ -r_{11}E_y&r_{41}E_x&\eta_{33} \end{array}\right).$$ As you can see, there is no z-dependence.

Now, obviously this matrix is not a diagonal matrix, but by the Spectral Theorem we can rotate our coordinate system to give new principal axes for the non-zero field crystal. The 3 eigenvalues of $\eta$ are the impermittivities of the field-stressed crystal. Likewise, since there is no $z$-dependence, we can parametrize our field as $$\mathbf{E}(\theta)=|\mathbf{E}|\left(\begin{array}{cc} \mbox{Cos}(\theta)\\ \mbox{Sin}(\theta)\\ 0 \end{array} \right).$$

Now, you said that you wanted to maximize the amount of birefringence, correct? For that, you want to find $\theta$ which maximizes the difference between the minimum and maximum eigenvalue; the relevant physical birefringence plane is given by the space spanned by the two eigenvectors corresponding to those eigenvalues.

I was too lazy to actually look up the refractive indices and electro-optical activities of quartz, but you can do that yourself. I made a Mathematica notebook which computes the impermittivities and plots them as a function of field angle and inserted random, wildly-unphysical numbers for the crystal. The plot looks something like this for zero-field:

And something like this for a huge field:

As you can see, there is a splitting from uniaxial to biaxial behavior when the field is applied, although obviously the numbers I used are fictitious, so you'll need to use the real numbers from reference tables to get physically useful info on actual quartz.

The link to the code I used to generate these images is here: https://www.dropbox.com/s/fn35y4ifbdb0ztz/Electro-Optic%20Tensor%20of%20Quartz.nb

Anyways, I hope you find the code useful. It's currently configured for quartz, but you can just change the electro-optic tensor entries to whatever you need, so it's completely general for all types of crystals. I also included an example plot for KDP using its actual physical parameters.