Find $J: S^2\rightarrow \mathbb{R}$, if given $I:S^2\rightarrow \mathbb{R}$, s.t. $I(\vec{a})=\int_{S^2}\vec{n}\cdot\vec{a} J(\vec{n}) ds$.

Question

I thought of the following problem, when we were discussing radiation intensity in an astrophysics lecture. Suppose $\mathbb{R}^3$ is filled with uniform radiation, i.e. there is a function $J:S^2\rightarrow \mathbb{R}$, so that at any point in $\mathbb{R}$ the amount of radiation in direction $\vec{n}$ is $J(\vec{n})$.
If we put a unit area with normal vector $\vec{a}$ into space, we can calculate the flux $I$ through it by $I(\vec{a})=\int_{S^2}\vec{n}\cdot\vec{a} J(\vec{n}) ds = \int_{S^2}\cos(\sphericalangle(\vec{n},\vec{a})) J(\vec{n}) ds $.

Under what conditions and how is it possible to do the inverse, i.e. calculate $J$ given $I$?

Of course there will be several $J$'s for an $I$, since many $J$'s lead to the same $I$. This leads to the next question: 'How big' are the classes of $J$'s that correspond to an $I$?

For example, when calculating the flux, the symmetric part of $J$ does not contribute anything. I think, this problem is analogous to calculating the deconvolution of a function convoluted around the cosine, just on a sphere instead of the real line, and the distance between two points on the sphere is the angle of their position vector.

Answer 1

My text is too long for a comment so I put it here: I'm not sure if I completely understand your notations. What I understand is that if I know a certain function $J:S^2\rightarrow \Bbb{R}$ like $(\phi,\theta)\mapsto \theta^2$, in spherical coordinates (with $r=1$), and a "unit area" (I think you mean something like a square or a disk having surface $=1$ and perpendicular to a vector of length $1$ named $\vec{a}$) then I can calulate $I(\vec{a})$. The location of the center of this surface object not being important (I think this is why the radiation is called uniform), lets take a square that is perpendicular to the $y$-axis then what would be the value of $I(\vec{a})$?

Answer 2

Following Nimda's hint, $J$ can be decomposed into spherical harmonics $Y_l^m$. Therefore, it is useful to look at the flux function $I_{Y_l^m}$ corresponding to the $J=Y_l^m$ for all possible $l$,$m$. It is useful to use the addition theorem for spherical harmonics: $P_l(cos(\gamma)=\frac{4\pi}{2l+1}\sum_{m=-l}^{l}Y_l^m(\theta,\phi)Y_l^{m*}(\theta',\phi')$, where $\gamma$ is the angle between the points $(\theta,\phi)$ and $(\theta',\phi')$. Because, $P_1(x)=x$, this means $\vec{n}\cdot\vec{a}=\frac{4\pi}{3}\sum_{m=-1}^{1}Y_1^m(\theta,\phi)Y_1^{m}(\theta',\phi')^*$. Hence, we get $$ I_{Y_l^m} = \int_{S^2} \frac{4\pi}{3}\sum_{m'=-1}^{1}Y_1^{m'}(\theta,\phi)Y_1^{m'}(\theta',\phi')^*Y_l^{m}(\theta',\phi') $$ This is equal to $\frac{4\pi}{3} Y_l^m(\theta,\phi)$ if $l=1$ and $0$ for other $l$.\ This shows, that only linear combinations of $Y_1^{-1}$, $Y_1^0$ and $Y_1^1$ are possible functions for $I$'s, that have a corresponding $J$. Furthermore, if we are given $I$ as a decomposition into spherical harmonics, $I(\theta,\phi)=\sum_{m=-1}^1 b_m Y_1^m(\theta,\phi)$, then $$J(\theta,\phi)=\sum_{m=-1}^1 b_m\frac{3}{4\pi} Y_1^m(\theta,\phi) + \sum_{1\neq l=0}^{\infty}\sum_{m=-1}^1 c_{l,m}Y_l^m(\theta,\phi)$$ , where $c_{l,m}$ are arbitrary coefficients.