Integrals of isotropic tensors, for expansion over spherical harmonics

[ergospherical]

Homework Statement

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Relevant Equations

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Consider an expansion for the density $\rho(t,\mathbf{x})$ of the form$$\rho(t,\mathbf{x}) = \sum_{l=0}^{\infty} a_{i_1 i_2 \dots i_{\mathscr{l}}}(t,r) \hat{x}_{i_1} \hat{x}_{i_2} \dots \hat{x}_{i_{\mathscr{l}}}$$where $r = |\mathbf{x}|$ and $\hat{x}_i = x_i/r$. Also, $a_{i_1 i_2 \dots i_{\mathscr{l}}}(t,r)$ is totally symmetric and traceless on any pair of indices. We would like to compute\begin{align*}
I_{ij}(t) &= \int d^3 x \ \rho(t,\mathbf{x}) x_i x_j \\
&= \sum_{l=0} \int d^3 x \ a_{i_1 i_2 \dots i_{\mathscr{l}}}(t,r) \hat{x}_{i_1} \hat{x}_{i_2} \dots \hat{x}_{i_l} \ x_i x_j
\end{align*}For $l=0$, I assume that we can take the coefficient to be unity (I could be wrong about this too...). In this case the contribution to the sum (which I'll denote by $I^{(0)}$) will be:$$I^{(0)}_{ij} = \int d^3 x \ x_i x_j$$and by considering the effect of rotations, I can figure out that:$${\tilde{I}^{(0)}_{ij}} = R_{ip} R_{jq} I^{(0)}_{pq} = \int d^3 x \ x_{i}' x_{j}' = I^{(0)}_{ij}$$therefore $I^{(0)}_{ij} = \alpha \delta_{ij}$ for some constant $\alpha$. From this you easily find $\alpha$ by taking the trace, i.e. $I^{(0)} = \int d^3 x \ r^2 = 3\alpha$.

Now the more tricky (?) case of the higher poles. Say $l=1$, for example:$$I^{(1)}_{ij} = \int d^3 x \ \frac{a_k}{r} x_k \ x_i x_j $$In this case, I'm not sure if the answer is also $\propto \delta_{ij}$? I don't think so, because I expect the $x_k$ needs to be rotated as well. However, obviously $k$ does not appear in the index structure of $I$. How do I treat these cases?

 

[Orodruin]

What happens with the integrand if you let $x_i \to -x_i$?

 

[ergospherical]

That makes sense - if you have an odd number of x’s, then under inversion the integral flips sign. But since the integral is extended over all space, it must be the same as before, so zero.

What about all the even cases, e.g. $I^{(2)}$?

 

[Orodruin]

Use that the result of the angular integral must be isotropic. An isotropic tensor of rank 2n can be written as a multiple of the fully symmetrized products of n Kronecker deltas.

Edit: I think I spoke a bit fast there. Of course, with the Kronecker delta being isotropic, so will any product of Kronecker deltas, but the argumentation remains the same.