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I have an integral on the unit sphere as follows.

$$I(\mathbf{s}_1, \mathbf{s}_2) = \int_{\mathbb{S}^2} f(\mathbf{x} \cdot \mathbf{s}_1)f(\mathbf{x}\cdot\mathbf{s}_2)d\mathbf{x} $$ where the integral is on the whole unit sphere $\mathbb{S}^2$. $\mathbf{s}_1$ and $\mathbf{s}_2$ are also unit-length vectors on the unit sphere. For example, $\mathbf{s}_1$ is the vector from the origin to a point on the sphere $\vec{O s_1}$.

Intuitively, since the integral is over the whole sphere, $I(\mathbf{s}_1, \mathbf{s}_2)$ should only depend on the relative angle between $\mathbf{s}_1$ and $\mathbf{s}_2$. This is actually done here.

Now I want to go further and do a Taylor expansion in terms of the relative spherical angle between $\mathbf{s}_1$ and $\mathbf{s}_2$, say $\theta$, so that $\theta = \angle (\mathbf{s}_1, \mathbf{s}_2)$. When $\mathbf{s}_1$ is close to $\mathbf{s}_2$, $\theta$ will be small and I want to do an expansion in terms of $\theta$. The problem is that I don't know how to explicit write $I(\mathbf{s}_1, \mathbf{s}_2)$ as a function of $\theta$, so I am stuck.

I need help on write this $I(\mathbf{s}_1, \mathbf{s}_2)$ as an Taylor expansion in terms of $\theta$, the relative angle between vectors $\mathbf{s}_1$ and $\mathbf{s}_2$. Any comments will be greatly appreciated.

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There's no reason for this integral to depend only on the angle. In the other question this was ensured by the dot products, but here the dependence on $s_1$ and $s_2$ is completely arbitrary; they essentially act as parameters only. –  joriki Jan 10 '13 at 18:06
    
@joriki You are right. I edited the post so $f$ only depends on the dot product as the other post. Thank you. –  Patrick Li Jan 10 '13 at 18:10

1 Answer 1

up vote 1 down vote accepted

Here is a way how to explicitly write $I$ in terms of the inner product.

For such integrals it is usually most convenient to use the addition theorem of the spherical harmonics $$P_l( \mathbf{x}\cdot\mathbf{y} ) = \frac{4\pi}{2l+1}\sum_{m=-l}^l Y_{l m}^*(\theta_\mathbf{x},\varphi_\mathbf{x}) \, Y_{l m}(\theta_\mathbf{y},\varphi_\mathbf{y})$$ valid for two vectors $\mathbf{x}, \mathbf{y}$ on the unit sphere; here $(\theta_\mathbf{x},\phi_\mathbf{x})$ are the spherical coordinates of $\mathbf{x}$, respectively.

Due to the completeness of the Legendre polynomials, we can introduce $$f_l =\frac{2}{2l+1}\int_{-1}^1 f(x) P_l(x) dx$$ such that $$f(x) = \sum_{l=0}^\infty f_l P_l(x).$$

Due to the integral over the unit sphere, we can choose $\mathbf{s}_1$ to be along the $z$ axis and $\mathbf{s}_2$ in the $xz$ plane. Then $$\begin{align}I(\mathbf{s}_1, \mathbf{s}_2) &= \int f(\cos \theta_\mathbf{x})f(\mathbf{x}\cdot\mathbf{s}_2)d\Omega_{\mathbf{x}}= \sum_{l l'} f_l f_{l'} \int P_l(\cos \theta_\mathbf{x}) P_{l'}(\mathbf{x}\cdot\mathbf{s}_2)d\Omega_{\mathbf{x}}\\ &=\sum_{l l'} f_l f_{l'} \sqrt{\frac{4\pi}{2l'+1}}\sum_{m'=-l'}^{l'}Y_{l'm'}(\theta_{\mathbf{s}_2},0)\underbrace{\int Y_{lm}(\theta_\mathbf{x},0) Y^*_{l'm'}(\theta_\mathbf{x},\varphi_\mathbf{x})d\Omega_{\mathbf{x}}}_{\delta_{ll'}\delta_{m'0}}\\ &=\sum_l f_l^2 P_l(\cos \theta_{\mathbf{s}_2})\\ &=\sum_l f_l^2 P_l(\mathbf{s}_1\cdot\mathbf{s}_2),\end{align}$$ where in the last step we have reintroduced a coordinate independent way of writing the expression.

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