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I have a $6\times6$ matrix having its elements being functions of Euler's angles (ZXZ rotation scheme) representing a tensor physical property. To find the average of the tensor property, I need to integrate this matrix over space in a spherical coordinate system. I found at many places that two of the Euler's angles are same as the spherical coordinates $(\theta, \phi)$, but I am not able to imagine how to deal with the third Euler's angle in this integration. Would appreciate any help.

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    $\begingroup$ I don't think you can convert one into the other. Euler angles can define any rotation having the radius $r$ fixed. If you have this fixed, which is one of the spherical coordinates you won't be able to get it by any transformation. $\endgroup$
    – Mr. Nobody
    Commented Oct 24, 2018 at 10:24

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I believe that the orientational average of a function $f(\phi, \theta, \chi)$ of the three Euler angles $\phi$, $\theta$, $\chi$ may be written $$ \langle f\rangle = \frac{1}{8\pi^2} \int_0^{2\pi} d\phi \, \int_0^{\pi} \sin\theta d\theta \, \int_0^{2\pi} d\chi \, f(\phi, \theta, \chi) $$ Here $\theta$ is the angle of rotation about the $x$ axis, which is applied in between the two rotations about the $z$ axes. You can see an example of this being used in this paper (in that paper the intermediate rotation is about the $y$ axis, but I don't believe that affects the volume elements appearing in the integral).

[EDIT following OP comments]

I'm just going to make a few more remarks here, following the comments made on my original answer. (Generally, extended discussion in the comments is discouraged, and I can see the danger of going far beyond the original question).

From your comments, and your other question on Math SE, it seems that you wish to orientationally average a $6\times 6$ stiffness tensor for a crystal with some symmetry.

Certainly, you can use symmetry to restrict the range of integration. You need to identify symmetry elements (rotations) of the crystal, let's say $\mathcal{R}$, which satisfy $f(\Omega)=f(\mathcal{R}\Omega)$ where $\Omega$ is short for $(\phi,\theta,\chi)$. Then you need to identify the corresponding angle ranges, i.e. the sub-regions of angular integration that are mapped into each other by this same operation $\mathcal{R}$. It may be easiest to represent $\mathcal{R}$ as a rotation about an axis by a specified angle, but this is easily converted into other forms. If you are doing this numerically, it would make sense to check (numerically) once or twice that you get the same answer from the full integration as from the symmetry-reduced one.

However, for a symmetric crystal, surely the angular dependence of a fourth-rank tensor is known exactly? For example, for cubic crystals, this open source paper by KM Knowles and PR Howie, J Elast, 120, 87 (2015) describes in eqn (5) how the stiffness tensor $C_{ijk\ell}$ (written in its full form, not in the reduced Voigt $6\times6$ notation) depends on orientation, which is represented as a $3\times3$ rotation matrix with components $a_{ij}$. (These can be written, if you like, in terms of the Euler angles). Once you know the elements of $C_{ijk\ell}$ in a convenient set of axes based on the crystal unit cell, computing the orientational average of these rotation matrix elements is relatively straightforward, and for an isotropic orientational average (which seems to be what you want, i.e. you haven't mentioned an orientational distribution function) it can surely be evaluated analytically. An example is given in section 7 of that paper.

Of course, your crystal may not be cubic. But for any crystal symmetry, it should be possible to apply the same principles, and it may have already been done for your crystal, if you search in the literature.

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    $\begingroup$ These notes on the Haar measure back up your equation; they also contain a reference to Naimark's Linear Representations of the Lorentz Group if you want more details. (The Haar measure is basically a "volume element" for a group which is preserved under the action of that group, in this case $SO(3)$.) $\endgroup$ Commented Oct 24, 2018 at 12:09
  • $\begingroup$ Many thanks @MichaelSeifert that's helpful. I was looking around for something to point the OP to, about SO(3), but settled for a practical example of use! $\endgroup$
    – user197851
    Commented Oct 24, 2018 at 12:16
  • $\begingroup$ Thanks for the response and reference. One more small point, which I did not mentioned in my last post. To take under the consideration of symmetry of crystal, I need to integrate this tensor within the given range of (theta, phi). Say theta (0, Pi/4) and phi (0, ArcCot[theta]). How should I approach. Are phi and theta in above formula same as in spherical coordinate system ? Thanks once again. $\endgroup$
    – user49535
    Commented Oct 24, 2018 at 12:40
  • $\begingroup$ Only you can answer that question! You need to think about the definition of the angles you are using. And $\text{ArcCot}(\theta)$ looks a little odd if $\theta$ is an angle. I usually interpret $\theta$ as a rotation about the intermediate $y$ axis, not about $x$. Have a look at the Wikipedia page and decide for yourself if this is the same $\theta$ as yours. $\endgroup$
    – user197851
    Commented Oct 24, 2018 at 13:16
  • $\begingroup$ Thanks for the link. Considering the range of angles used in formula given by you, it seems I would need to integrate phi/chi in the range of 0-Pi/4 and theta 0-ArcCot[Cos phi/chi]. Sorry for typo in last message. Still have some confusion between phi and chi. Since spherical coordinates are easy to imagine- I am dead sure that phi (angle having full range of 0 - 2Pi) needs to be integrated 0 - Pi/4 and theta (full range of 0 - Pi) 0 - ArcCot[Cos theta]. $\endgroup$
    – user49535
    Commented Oct 25, 2018 at 4:44

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