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I have a question involving surface integral on a unit sphere. Suppose $s_1$ and $s_2$ are two points on a unit sphere with spherical coordinates $(\theta_1, \psi_1)$ and $(\theta_2, \psi_2)$, respectively. I want to compute

$$\int_{{\bf x}\in \mathbb{S}^2} \exp\{s^T_1 \cdot {\bf x}\} \exp\{s^T_2 \cdot {\bf x}\}d\mathbf{x}$$ where $\mathbb{S}^2$ stands for the unit sphere. $s_1^T \cdot \mathbf{x}$ means the dot product of the vectors $\vec{os_1}$ and $\vec{o\mathbf{x}}$ where $o$ is the center of the sphere.

Here is my idea. I can write $d\mathbf{x}$ as $\sin \theta d\theta d\psi$ where $(\theta, \psi)$ is the spherical coordinates of $\mathbf{x}$. Without loss of generality, I can assume $s_1$ is the top point of the sphere (on the $z$ axis) so that $s_1^T \cdot \mathbf{x}$ becomes $\cos \theta$. Now I think if I can write $s_2^T \cdot \mathbf{x}$ as a function of $\theta, \theta_2, \psi, \psi_2$, I can try to do this double integral. So my question boils down to "What is the form of $s_2^T \cdot \mathbf{x}$ in terms of their spherical coordinates $\theta, \theta_2, \psi, \psi_2$".

Maybe I can do this surface integral in an easier way. Any comments and suggestions are welcome.

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3 Answers

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Let \begin{equation} I=\int d\Omega\, e^{({\bf s}_1+{\bf s}_2)\cdot \hat{\bf r}}, \end{equation} where $d\Omega=\sin(\theta) d\phi d\theta$ and $\hat{\bf r}$ is a unit vector in spherical coordinates.

We are free to choose our coordinate system so, we choose the $\hat{\bf z}$ to lie along the ${\bf s}_1+{\bf s}_2$ direction. Then using $({\bf s}_1+{\bf s}_2)\cdot \hat{\bf r}=|{\bf s}_1+{\bf s}_2|\cos(\theta)$, where $|{\bf s}_1+{\bf s}_2|$ is the norm of the vector, \begin{equation} I=\int\limits_0^{2\pi}d\phi\int\limits^{\pi}_{0}d\theta\, \sin(\theta)e^{|{\bf s}_1+{\bf s}_2|\cos(\theta)}. \end{equation} The $\phi$ integral is trivial, while the $\theta$ integral can be easily done with the substitution $u=\cos(\theta)$ giving \begin{equation} I=4\pi\frac{\sinh(|{\bf s}_1+{\bf s}_2|)}{|{\bf s}_1+{\bf s}_2|}. \end{equation}

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The value $(=:J)$ of the integral depends solely on the angle $2\alpha$ between the two vectors ${\bf s}_1$ and ${\bf s}_2$. Therefore we may assume $${\bf s}_1=(-\sin\alpha,0,\cos\alpha)\ ,\quad {\bf s}_2=(\sin\alpha,0,\cos\alpha)\ .$$ As ${\bf x}=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$ the integrand then becomes $$\exp\bigl(({\bf s}_1+{\bf s}_2)\cdot{\bf x}\bigr)=\exp(2\cos\alpha\cos\theta)\ .$$ Therefore we obtain $$\eqalign{J&=\int_0^{2\pi}\int_0^\pi \exp(2\cos\alpha\cos\theta)\ \sin\theta\ d\theta\ d\phi\cr &=-{2\pi\exp(2\cos\alpha\cos\theta)\over 2\cos\alpha}\biggr|_{\theta=0}^\pi\cr &={2\pi\over\cos\alpha}\ \sinh(2\cos\alpha)\ .\cr}$$

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Rotational symmetry gives you the tools to reduce a lot of the complexity in this problem. Since you have the freedom to choose your coordinate axes, one trick is to put the two points on the zx-plane, with the $z$-axis bisecting the angle between them. Then, the coordinates of the two points are $(\theta, \varphi) = (\pm \alpha, 0)$ for some fixed angle $\alpha$. The cartesian components would then be $\hat z \cos \alpha \pm \hat x \sin \alpha$ for the two vectors.

This is basically one way to arrive at Christian's insight--that the answer just depends on the angle between the points.

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