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I've been given the question:

"Consider the tensor: $$ C_{ij}=\int_{V}{x_ix_j|\mathbf {x}|^2 + x_ix_j(\mathbf {x.n})^2} dV $$ where V is the volume of a sphere radius R centred on the origin. What are the eigenvectors and corresponding eigenvalues of this tensor?"

I assume that the vector n is one of them since it breaks the symmetry, however, I don't know how to show it. Any help would be greatly appreciated as I am completely stuck.

This is actually part of a longer question:

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If it is at all useful, the answers I got for the previous part of the question are: $$ A_{ij}=\frac {4\pi R^7}{21}\delta_{ij} $$ and $$ B_{ijkl}= \frac {4\pi R^7}{189}(\delta_{ij}\delta_{kl}+\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}) $$

Franck Dernoncourt
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Hans Schmuber
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1 Answers1

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The expression for $B_{ijk\ell}$ is off by a factor $\frac{9}{5}$

Using the same technique as in this post (i.e. assume it's isotropic, contract it with identity matrices, calculate the scalar factor) the correct value is $$\eqalign{ B_{ijk\ell} &= \frac{4\pi R^7}{3\cdot 5\cdot 7}\bigg( \delta_{ij}\delta_{k\ell} + \delta_{ik}\delta_{j\ell} + \delta_{i\ell}\delta_{jk} \bigg) }$$ Contraction with $\delta_{k\ell}$ yields $A_{ij}$ -- which is the first half of $C_{ij}$ $$\eqalign{ A_{ij} &= B_{ijk\ell}\,\delta_{k\ell} = \frac{4\pi R^7}{3\cdot 5\cdot 7}\bigg(5\,\delta_{ij}\bigg) = \frac{4\pi R^7}{3\cdot 7}\,\delta_{ij} \\ }$$ The other half is the contraction with the constant dyadic tensor $\,N_{k\ell}=n_kn_\ell$ $$\eqalign{ F_{ij} &= B_{ijk\ell}\,n_kn_\ell = \frac{4\pi R^7}{3\cdot 5\cdot 7}\bigg(n_kn_k\,\delta_{ij}+n_in_j+n_jn_i\bigg) = \frac{4\pi R^7}{105}\bigg(\delta_{ij}+2N_{ij}\bigg) \\ }$$ Putting the two halves together
$$C_{ij} = A_{ij} + F_{ij} = \frac{8\pi R^7}{105}\bigg(3\delta_{ij}+N_{ij}\bigg) \\$$ NB:   Your calculation of $A_{ij}$ was correct.

greg
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