Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.