Generate an uniform distribution on the sky between limits

Question

I have an analog question of this one : Generate an uniform distribution on the sky

I generate an uniform distribution on the sky, but just a patch of it, with $$\alpha_{min} \leq \alpha \leq \alpha_{max}$$ and $$\delta_{min} \leq \delta \leq \delta_{max}$$

for Right Ascention, I just have to take uniform distribution between $\alpha_{min}$and $ \alpha_{max}$.

For declination it works if I take $$\delta_{random} = \sin^{-1}\left( \mathrm{Uniform}(\sin\delta_{min},\, \sin\delta_{max}) \right)$$

My problem is that I don't know how to demonstrate this formula for the generation of delinations. Any idea how to make the proof of this formula ?

I saw in Generate an uniform distribution on the sky @RobJeffries post which gives an expression of $P(\delta)$ and demonstrates that $\delta = \sin^{-1}(2P-1)$ where $P$ is a random number between 0 and 1, which is compatible to my formula with $\delta_{min} = -\frac{\pi}{2}$ and $ \delta_{max} = +\frac{\pi}{2}$, but I was not able to generalize his demonstration.

Thanks in advance for your help!

Answer 1

To derive the expression for uniformly selecting a point within a solid angle bounded in right ascension, $\alpha_{min} \leq \alpha \leq \alpha_{max}$ and declination, $\delta_{min} \leq \delta \leq \delta_{max}$, we start with the definition for the differential element of solid angle: $$d\Omega = \cos\delta d\delta\ d\alpha$$

As observed in the question, the right ascension is uniformly distributed between the upper and lower bounds, $\alpha \sim U(\alpha_{min}, \alpha_{max})$. This now leaves the distribution of declination to be determined along a narrow "slice" or constant right ascension.

Borrowing from Generate an uniform distribution on the sky: $$dP = \cos\delta d\delta$$

The cumulative distribution function (CDF)of selecting a point at a specific declination is then given by: $$P(\delta) = \frac{\int_{\delta_{min}}^{\delta} \cos\delta' \,d\delta'}{\int_{\delta_{min}}^{\delta_{max}} \cos\delta' \,d\delta'}$$

$$P(\delta) = \frac{\left.\sin\delta\right|_{\delta_{min}}^{\delta}}{\left.\sin\delta\right|_{\delta_{min}}^{\delta_{max}}} = \frac{\sin\delta-\sin\delta_{min}}{\sin\delta_{max}-\sin\delta_{min}}$$

The CDF maps the declination to a quantile, $v$, between 0 and 1 : $P(\delta)=v$. We can then substitute $v$ for $P(\delta)$, solve for $\delta$ and arrive at an inverse CDF mapping $v$ to the declination.

$$v = \frac{\sin\delta-\sin\delta_{min}}{\sin\delta_{max}-\sin\delta_{min}}$$

Now solving for $\delta$ $$v\left(\sin\delta_{max}-\sin\delta_{min}\right) = \sin\delta-\sin\delta_{min}$$

$$\sin\delta = v\left(\sin\delta_{max}-\sin\delta_{min}\right)+\sin\delta_{min}$$

$$\delta = \arcsin\left(v\left(\sin\delta_{max}-\sin\delta_{min}\right)+\sin\delta_{min}\right)$$

The desired distribution in declination is then realize by letting $v \sim U(0, 1)$. This is equivalent to: $$\delta = \arcsin(u)$$ where $u \sim U(\sin\delta_{min}, sin\delta_{\max})$