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I don't know if I understand this correctly. Let's say that we have a point source of particles and it is pointed at a circular detector. The radius of the detector is 5cm and the distance to the detector is 10cm. I therefore conclude that:

$$\Omega = S/r^2 = \pi r_{circle}^2/r^2 = \pi/4.$$

What is wrong with my reasoning?

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  • $\begingroup$ Let me present an example based on your reasoning: Suppose that the radius of the detector is some huge value like 500 km but the distance to the detector is still 10 cm. According to the approach you used above, $\Omega$ would be some huge number, whereas we know that the solid angle $\Omega$ can never be greater than $4\pi$. (And, in fact, for this example $\Omega$ should be very close to $2\pi$). What went wrong? $\endgroup$ – user93237 May 9 '17 at 2:18
  • $\begingroup$ @SamuelWeir I guess that I need to define the surface area differently? $\endgroup$ – loltospoon May 9 '17 at 2:21
  • $\begingroup$ Well, I also know that $\Omega = \int \int sin \phi d\theta d\phi$, but I have no clue how to use this $\endgroup$ – loltospoon May 9 '17 at 2:22
  • $\begingroup$ What should the limits be in that integral? $\endgroup$ – Brian Moths May 9 '17 at 2:25
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs That's what I couldn't figure out. At first I thought it was $\theta = [0,\pi /2]$ and $\phi = [0,2\pi]$, but this just gave me $2\pi$, which I don't think is right $\endgroup$ – loltospoon May 9 '17 at 2:33
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If you take an object and project it radially outward onto the surface of a sphere the area of that projection divided by the square of the sphere's radius is the solid angle. That's the general definition. In this case, because the disk is oriented so that its surface normal is parallel to the radial direction the easiest sphere to choose for measuring solid angle is one that contains the circle. The radius of that sphere will then be: $$ R_{\mathrm{sphere}} = \sqrt{r_{\mathrm{circle}}^2 + d^2},$$ where $d$ is the distance to the circle.

The area from the sphere that the circle projects onto can be found by choosing polar coordinates, with the center of the polar angle ($\theta=0$) corresponding with the circle's axis. In those coordinates the area can be found with the double integral: $$\begin{align} A &= \int_0^{2\pi} \operatorname{d}\phi \int_0^{\tan^{-1}(r_{\mathrm{circle}} / d)} \operatorname{d}\theta\ R_{\mathrm{sphere}}^2 \sin \theta \\ & = 2\pi \left(r_{\mathrm{circle}}^2 + d^2\right) \cdot \left[-\cos \theta \vphantom{\int}\right]_{\theta=0}^{\tan^{-1}(r_{\mathrm{circle}} / d)} \\ & = 2\pi \left(r_{\mathrm{circle}}^2 + d^2\right) \cdot \left(1 - \cos\left[\tan^{-1}\left(\frac{r_{\mathrm{circle}}}{ d}\right)\right]\right) \\ & = 2\pi \left(r_{\mathrm{circle}}^2 + d^2\right) \cdot \left(1 - \frac{d}{\sqrt{r_{\mathrm{circle}}^2 + d^2}}\right) \end{align}$$ This gives a solid angle of: $$\Omega = 2\pi \cdot \left(1 - \frac{d}{\sqrt{r_{\mathrm{circle}}^2 + d^2}}\right).$$

Confirm for yourself that this obeys the correct limits ($d \rightarrow \infty \Rightarrow \Omega \rightarrow 0$ and $r_{\mathrm{circle}} \rightarrow \infty \Rightarrow \Omega \rightarrow 2\pi$). Also be careful of numerical issues when actually doing calculations with this formula because it suffers from a similar numerical instability to the quadratic formula when $d\gg r_{\mathrm{circle}}$.

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