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Note that throughout this I use the spherical mapping convention: $$(x,y,z) = (r\cos\phi\sin\theta,r\cos\theta,r\sin\phi\sin\theta)$$ I have derived that the uniform pdf for a sphere $S_1$ with radius $\rho$ and center $(0,0,0)$ is $p_A(r,\phi,\theta) = \frac{\delta(r-\rho)}{4\pi\rho^2}$. I want to translate this sphere by $(0,\rho,0)$ and find the corresponding pdf.

What I have tried is to rewrite the pdf in cartesian coordinates, that is: $$p_B(x,y,z) = \frac{p_A(r, \phi, \theta)}{|r^2\sin\theta|}$$ And translate it: $p_C(x,y,z) = p_B(x,y-\rho,z)$.

Then I believe that the pdf I am looking for is: $p_D(r',\phi',\theta') = p_C(x,y,z)|r'^2\sin\theta'|$.

I have $r^2 = x^2 + (y-\rho)^2 + z^2$, $\cos\theta = \frac{y-\rho}{r}$, $r'^2 = x^2+y^2 + z^2$, $\cos\theta' = \frac{y}{r'}$. Assuming that this is correct I get the relationship: $$r^2 = r'^2 - 2\rho r'\cos\theta' + \rho^2$$ $$\cos\theta = \cos\theta' - \frac{\rho}{r}$$

Is everything correct? Is this the correct expression for the pdf I am looking for? $$p_D(r',\phi,\theta') = p_A(r(r',\theta'),\phi,\theta(r',\theta'))\frac{|r'^2\sin\theta'|}{|r^2(r',\theta')\sin(\theta(r',\theta'))|}$$

Edit: Just for completeness I provide the derivation of the uniform pdf on the sphere centered at $(0,0,0)$ with radius $\rho$. Since I only want to have a constant probability over the surface of the sphere I can write $p_A(r,\phi,\theta) = C\delta(r-\rho)$. In order to derive the pdf normalization constant $C$ I integrate the pdf: $$\int_{0}^{2\pi}{\int_{0}^{pi}{\int_{0}^{\infty}{p_A(r,\phi,\theta)r^2\sin\theta dr}d\theta}d\phi} = $$ $$2\pi C\int_{0}^{pi}{\rho^2\sin\theta d\theta} =$$ $$4C\pi\rho^2 = 1$$ $$C = \frac{1}{4\pi\rho^2}$$

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1 Answer 1

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The denominator seems to be redundant, since it actually cancels out with the term $|r^2\sin\theta|$ in the numerator, which I forgot to add in the pdf it seems: $p_A(r,\phi,\theta) = \delta(r-\rho)\frac{|r^2\sin\theta|}{4\pi\rho^2}$.

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