I have an integral at hand which has the form of

$$I = \int_{u\in \mathbb{S}^2} f(\mathbf{u}\cdot \mathbf{s}_1) f(\mathbf{u}\cdot \mathbf{s}_2) d\mathbf{u}$$ where $\mathbb{S}^2$ is the unit sphere with radius $1$ and $\mathbf{u}\cdot \mathbf{s}_1$ is the inner product of the two vectors.

Intuitively this integration should depend only on the relative position (say, the spherical angle) between $\mathbf{s}_1$ and $\mathbf{s}_2$ since $\mathbf{u}$ is integrated over the whole sphere surface. My question is how to prove $I$ only depend on the spherical angle between $\mathbf{s}_1$ and $\mathbf{s}_2$ without knowing the explicit form the $f$. I am thinking about rotating $\mathbf{u}$ but cannot write an explicit formula since the spherical angle has a complicated form in terms of the spherical coordinates.

  • 2
    $\begingroup$ If $x,y\in\mathbb{S}^2$, then $x\cdot y=\cos\theta$ where $\theta$ is the angle between them. $\endgroup$ – Joe Johnson 126 Nov 7 '12 at 17:06
  • $\begingroup$ @JoeJohnson126 If I write my $\mathbf{s}_1 = (\theta_1, \varphi_1)$ and $\mathbf{s}_2 = (\theta_2, \varphi_2)$, then their spherical angle is pretty complicated. $\endgroup$ – Patrick Li Nov 7 '12 at 17:11

Hint: Use a rotation ${\bf u}\mapsto R{\bf u}$ where $R$ is an appropriate rotation matrix. Since $\det R=1$, by the rule of substitution, we are done.

  • $\begingroup$ That's a good point. Thank you Berci. $\endgroup$ – Patrick Li Nov 8 '12 at 0:52

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