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Fiber Orientation Distribution Functions and Orientation Tensors for Different Material Symmetries

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Visualization and Processing of Higher Order Descriptors for Multi-Valued Data

Part of the book series: Mathematics and Visualization ((MATHVISUAL))

Abstract

In this paper we give closed-form expressions of the orientation tensors up to the order four associated with some axially-symmetric orientation distribution functions (ODF), including the well-known von Mises-Fisher, Watson, and de la Vallée Poussin ODFs. Each is characterized by a mean direction and a concentration parameter. Then, we use these elementary ODFs as building blocks to construct new ones with a specified material symmetry and derive the corresponding orientation tensors. For a general ODF we present a systematic way of calculating the corresponding orientation tensors from certain coefficients of the expansion of the ODF in spherical harmonics.

Mathematics Subject Classification (2010): 74A40, 74E10, 62H11, 92C10

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Authors and Affiliations

  1. Laboratory for Mathematical and Numerical Modeling in Engineering Science, National Engineering School at Tunis, University of Tunis El Manar, ENIT-LAMSIN, B.P. 37, 1002, Tunis-Belvédère, Tunisia

    Maher Moakher

  2. Section on Tissue Biophysics & Biomimetics, PPITS, NICHD, National Institutes of Health, 13 South Drive, Bldg. 13, Rm. 3W16, Bethesda, MD, 20892-5772, USA

    Peter J. Basser

Authors
  1. Maher Moakher
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  2. Peter J. Basser
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Corresponding author

Correspondence to Maher Moakher .

Editor information

Editors and Affiliations

  1. Linköping University, Norrköping, Sweden

    Ingrid Hotz

  2. Visualization and Medical Image Analysis, University of Bonn, Bonn, Germany

    Thomas Schultz

Appendix

Appendix

We give here the expressions for the normalized orientation-like tensors that appear in (33)–(36).

The vectors s 1, m, \(m = -1,\ldots,1\), which are obtained from \(\int _{S^{2}}S^{1,m}(\theta,\phi )\mathbf{n}\,d\sigma\) by normalization, are given by

$$\displaystyle{\mathbf{s}^{1,-1} = \left [\begin{array}{*{10}c} -1\\ 0 \\ 0\end{array} \right ],\quad \mathbf{s}^{1,0} = \left [\begin{array}{*{10}c} 0\\ 0 \\ 1 \end{array} \right ],\quad \mathbf{s}^{1,1} = \left [\begin{array}{*{10}c} 0\\ -1 \\ 0\end{array} \right ].}$$

The second-order tensors S l, m, l = 0, 2, \(m = -l,\ldots,l\), which are obtained from \(\int _{S^{2}}S^{l,m}(\theta,\phi )\mathbf{n}^{\otimes 2}\,d\sigma\) by normalization, are given by

$$\displaystyle{\mathbf{S}^{0,0} = \frac{1} {\sqrt{3}}\left [\begin{array}{*{10}c} 1&0&0\\ 0 &1 &0 \\ 0&0&1 \end{array} \right ],\quad \mathbf{S}^{2,-2} = \frac{1} {\sqrt{2}}\left [\begin{array}{*{10}c} 1& 0 &0\\ 0 &-1 &0 \\ 0& 0 &0 \end{array} \right ],\quad \mathbf{S}^{2,-1} = \frac{1} {\sqrt{2}}\left [\begin{array}{*{10}c} 0 &0&-1\\ 0 &0 & 0 \\ -1&0& 0 \end{array} \right ],}$$
$$\displaystyle{\mathbf{S}^{2,0} = \frac{1} {\sqrt{6}}\left [\begin{array}{*{10}c} -1& 0 &0\\ 0 &-1 &0 \\ 0 & 0 &2 \end{array} \right ],\quad \mathbf{S}^{2,1} = \frac{1} {\sqrt{2}}\left [\begin{array}{*{10}c} 0& 0 & 0\\ 0 & 0 &-1 \\ 0&-1& 0 \end{array} \right ],\quad \mathbf{S}^{2,2} = \frac{1} {\sqrt{2}}\left [\begin{array}{*{10}c} 0&1&0\\ 1 &0 &0 \\ 0&0&0 \end{array} \right ].}$$

Note that S l, m are traceless except for S 0, 0 which has unit trace. Furthermore, the set \(\{\mathbf{S}^{l,m},\ l = 0,2,\ m = -l,\ldots,l\}\) forms an orthonormal basis of the space of symmetric second-order tensors.

The third-order tensors S l, m, l = 1, 3, \(m = -l,\ldots,l\), which are obtained from \(\int _{S^{2}}S^{l,m}(\theta,\phi )\mathbf{n}^{\otimes 3}\,d\sigma\) by normalization, are given by

$$\displaystyle{\mathsf{S}^{1,-1} = \frac{1} {\sqrt{15}}\left [\begin{array}{*{10}c} -3&-1&-1&0& 0 & 0 \\ 0 & 0 & 0 &0& 0 &-\sqrt{2} \\ 0 & 0 & 0 &0&-\sqrt{2}& 0 \end{array} \right ],\quad \mathsf{S}^{1,0} = \frac{1} {\sqrt{15}}\left [\begin{array}{*{10}c} 0&0&0& 0 &\sqrt{2}&0 \\ 0&0&0&\sqrt{2}& 0 &0 \\ 1&1&3& 0 & 0 &0 \end{array} \right ],\quad \mathsf{S}^{1,1} = \frac{1} {\sqrt{15}}\left [\begin{array}{*{10}c} 0 & 0 & 0 & 0 &0&-\sqrt{2} \\ -1&-3&-1& 0 &0& 0 \\ 0 & 0 & 0 &-\sqrt{2}&0& 0 \end{array} \right ],}$$
$$\displaystyle{\mathsf{S}^{3,-3} = \frac{1} {2}\left [\begin{array}{*{10}c} -1&1&0&0&0& 0 \\ 0 &0&0&0&0&\sqrt{2} \\ 0 &0&0&0&0& 0 \end{array} \right ],\quad \mathsf{S}^{3,-2} = \frac{1} {\sqrt{6}}\left [\begin{array}{*{10}c} 0& 0 &0& 0 &\sqrt{2}&0 \\ 0& 0 &0&-\sqrt{2}& 0 &0 \\ 1&-1&0& 0 & 0 &0 \end{array} \right ],\quad \mathsf{S}^{3,-1} = \frac{1} {2\sqrt{15}}\left [\begin{array}{*{10}c} 3&3&-4&0& 0 & 0 \\ 0&0& 0 &0& 0 &\sqrt{2} \\ 0&0& 0 &0&-4\sqrt{2}& 0 \end{array} \right ],}$$
$$\displaystyle{\mathsf{S}^{3,0} = \frac{1} {\sqrt{10}}\left [\begin{array}{*{10}c} 0 & 0 &0& 0 &-\sqrt{2}&0 \\ 0 & 0 &0&-\sqrt{2}& 0 &0 \\ -1&-1&2& 0 & 0 &0 \end{array} \right ],\quad \mathsf{S}^{3,1} = \frac{1} {2\sqrt{15}}\left [\begin{array}{*{10}c} 0&0& 0 & 0 &0&\sqrt{2} \\ 1&1&-4& 0 &0& 0 \\ 0&0& 0 &-4\sqrt{2}&0& 0 \end{array} \right ],\quad \mathsf{S}^{3,2} = \frac{1} {\sqrt{6}}\left [\begin{array}{*{10}c} 0&0&0&\sqrt{2}& 0 & 0 \\ 0&0&0& 0 &\sqrt{2}& 0 \\ 0&0&0& 0 & 0 &\sqrt{2} \end{array} \right ],}$$
$$\displaystyle{\mathsf{S}^{3,3} = \frac{1} {2}\left [\begin{array}{*{10}c} 0 &0&0&0&0&-\sqrt{2} \\ -1&1&0&0&0& 0\\ 0 &0 &0 &0 &0 & 0 \end{array} \right ].}$$

We remark that the set \(\{\mathsf{S}^{l,m},\ l = 1,3,\ m = -l,\ldots,l\}\) forms an orthonormal basis of the space of totally symmetric third-order tensors.

The fourth-order tensors \(\mathbb{S}^{l,m}\), l = 0, 2, 4, \(m = -l,\ldots,l\), which are obtained from \(\int _{S^{2}}S^{l,m}(\theta,\phi )\mathbf{n}^{\otimes 4}\,d\sigma\) by normalization, are given by

$$\displaystyle{\mathbb{S}^{0,0} = \frac{1} {3\sqrt{5}}\left [\begin{array}{*{10}c} 3&1&1&0&0&0\\ 1 &3 &1 &0 &0 &0 \\ 1&1&3&0&0&0\\ 0 &0 &0 &2 &0 &0 \\ 0&0&0&0&2&0\\ 0 &0 &0 &0 &0 &2 \end{array} \right ],\quad \mathbb{S}^{2,-2} = \frac{1} {2\sqrt{21}}\left [\begin{array}{*{10}c} 6& 0 & 1 & 0 &0&0\\ 0 &-6 &-1 & 0 &0 &0 \\ 1&-1& 0 & 0 &0&0\\ 0 & 0 & 0 &-2 &0 &0 \\ 0& 0 & 0 & 0 &2&0\\ 0 & 0 & 0 & 0 &0 &0 \end{array} \right ],\quad \mathbb{S}^{2,-1} = \frac{1} {2\sqrt{21}}\left [\begin{array}{*{10}c} 0 & 0 & 0 & 0 &-3\,\sqrt{2}& 0 \\ 0 & 0 & 0 & 0 & -\sqrt{2} & 0 \\ 0 & 0 & 0 & 0 &-3\,\sqrt{2}& 0 \\ 0 & 0 & 0 & 0 & 0 &-2 \\ -3\,\sqrt{2}&-\sqrt{2}&-3\,\sqrt{2}& 0 & 0 & 0 \\ 0 & 0 & 0 &-2& 0 & 0 \end{array} \right ],}$$
$$\displaystyle{\mathbb{S}^{2,0} = \frac{1} {2\sqrt{63}}\left [\begin{array}{*{10}c} -6&-2& 1 &0&0& 0\\ -2 &-6 & 1 &0 &0 & 0 \\ 1 & 1 &12&0&0& 0\\ 0 & 0 & 0 &2 &0 & 0 \\ 0 & 0 & 0 &0&2& 0\\ 0 & 0 & 0 &0 &0 &-4 \end{array} \right ],\quad \mathbb{S}^{2,1} = \frac{1} {2\sqrt{21}}\left [\begin{array}{*{10}c} 0 & 0 & 0 & -\sqrt{2} & 0 & 0 \\ 0 & 0 & 0 &-3\,\sqrt{2}& 0 & 0 \\ 0 & 0 & 0 &-3\,\sqrt{2}& 0 & 0 \\ -\sqrt{2}&-3\,\sqrt{2}&-3\,\sqrt{2}& 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 &-2\\ 0 & 0 & 0 & 0 &-2 & 0 \end{array} \right ],\quad \mathbb{S}^{2,2} = \frac{1} {2\sqrt{21}}\left [\begin{array}{*{10}c} 0 & 0 & 0 &0&0&3\,\sqrt{2} \\ 0 & 0 & 0 &0&0&3\,\sqrt{2} \\ 0 & 0 & 0 &0&0& \sqrt{2} \\ 0 & 0 & 0 &0&2& 0\\ 0 & 0 & 0 &2 &0 & 0 \\ 3\,\sqrt{2}&3\,\sqrt{2}&\sqrt{2}&0&0& 0 \end{array} \right ],}$$
$$\displaystyle{\mathbb{S}^{4,-4} = \frac{1} {2\sqrt{2}}\left [\begin{array}{*{10}c} 1 &-1&0&0&0& 0\\ -1 & 1 &0 &0 &0 & 0 \\ 0 & 0 &0&0&0& 0\\ 0 & 0 &0 &0 &0 & 0 \\ 0 & 0 &0&0&0& 0\\ 0 & 0 &0 &0 &0 &-2 \end{array} \right ],\quad \mathbb{S}^{4,-3} = \frac{1} {4}\left [\begin{array}{*{10}c} 0 & 0 &0&0&-\sqrt{2}&0 \\ 0 & 0 &0&0& \sqrt{2} &0 \\ 0 & 0 &0&0& 0 &0\\ 0 & 0 &0 &0 & 0 &2 \\ -\sqrt{2}&\sqrt{2}&0&0& 0 &0 \\ 0 & 0 &0&2& 0 &0 \end{array} \right ],\quad \mathbb{S}^{4,-2} = \frac{1} {\sqrt{14}}\left [\begin{array}{*{10}c} -1& 0 & 1 & 0 &0&0\\ 0 & 1 &-1 & 0 &0 &0 \\ 1 &-1& 0 & 0 &0&0\\ 0 & 0 & 0 &-2 &0 &0 \\ 0 & 0 & 0 & 0 &2&0\\ 0 & 0\! & 0 & 0 &0 &0 \end{array} \right ],}$$
$$\displaystyle{\mathbb{S}^{4,-1} = \frac{1} {\sqrt{28}}\left [\begin{array}{*{10}c} 0 & 0 & 0 &0& 3\sqrt{2} &0 \\ 0 & 0 & 0 &0& \sqrt{2} &0 \\ 0 & 0 & 0 &0&-4\sqrt{2}&0 \\ 0 & 0 & 0 &0& 0 &2 \\ 3\sqrt{2}&\sqrt{2}&-4\sqrt{2}&0& 0 &0 \\ 0 & 0 & 0 &2& 0 &0 \end{array} \right ],\quad \mathbb{S}^{4,0} = \frac{1} {2\sqrt{70}}\left [\begin{array}{*{10}c} 3 & 1 &-4& 0 & 0 &0\\ 1 & 3 &-4 & 0 & 0 &0 \\ -4&-4& 8 & 0 & 0 &0\\ 0 & 0 & 0 &-8 & 0 &0 \\ 0 & 0 & 0 & 0 &-8&0\\ 0 & 0 & 0 & 0 & 0 &2 \end{array} \right ],\quad \mathbb{S}^{4,1} = \frac{1} {2\sqrt{28}}\left [\begin{array}{*{10}c} 0 & 0 & 0 & \sqrt{2} &0&0 \\ 0 & 0 & 0 & 3\sqrt{2} &0&0 \\ 0 & 0 & 0 &-4\sqrt{2}&0&0 \\ \sqrt{2}&3\sqrt{2}&-4\sqrt{2}& 0 &0&0 \\ 0 & 0 & 0 & 0 &0&2\\ 0 & 0 & 0 & 0 &2 &0 \end{array} \right ],}$$
$$\displaystyle{\mathbb{S}^{4,2} = \frac{1} {2\sqrt{14}}\left [\begin{array}{*{10}c} 0 & 0 & 0 &0&0&-\sqrt{2} \\ 0 & 0 & 0 &0&0&-\sqrt{2} \\ 0 & 0 & 0 &0&0& 2\sqrt{2} \\ 0 & 0 & 0 &0&4& 0\\ 0 & 0 & 0 &4 &0 & 0 \\ -\sqrt{2}&-\sqrt{2}&2\sqrt{2}&0&0& 0 \end{array} \right ],\quad \mathbb{S}^{4,3} = \frac{1} {4}\left [\begin{array}{*{10}c} 0 & 0 &0&-\sqrt{2}& 0 & 0 \\ 0 & 0 &0& \sqrt{2} & 0 & 0 \\ 0 & 0 &0& 0 & 0 & 0 \\ -\sqrt{2}&\sqrt{2}&0& 0 & 0 & 0 \\ 0 & 0 &0& 0 & 0 &-2\\ 0 & 0 &0 & 0 &-2 & 0 \end{array} \right ],\quad \mathbb{S}^{4,4} = \frac{1} {2\sqrt{2}}\left [\begin{array}{*{10}c} 0 & 0 &0&0&0& \sqrt{2} \\ 0 & 0 &0&0&0&-\sqrt{2} \\ 0 & 0 &0&0&0& 0\\ 0 & 0 &0 &0 &0 & 0 \\ 0 & 0 &0&0&0& 0 \\ \sqrt{2}&-\sqrt{2}&0&0&0& 0 \end{array} \right ].}$$

It should be noted that \(\mathbb{S}^{l,m}\) are traceless except for \(\mathbb{S}^{0,0}\) which has unit trace. The set \(\{\mathbb{S}^{l,m},\ l = 0,2,4,\ m = -l,\ldots,l\}\) forms an orthonormal basis of the space of totally symmetric fourth-order tensors.

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Moakher, M., Basser, P.J. (2015). Fiber Orientation Distribution Functions and Orientation Tensors for Different Material Symmetries. In: Hotz, I., Schultz, T. (eds) Visualization and Processing of Higher Order Descriptors for Multi-Valued Data. Mathematics and Visualization. Springer, Cham. https://doi.org/10.1007/978-3-319-15090-1_3

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