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I am trying to calculate the solid angle subtended by arbitrary-shaped loops on a sphere's surface.

First, I parametrize circular loops by:

$$\theta(t,k_{x0},r) = k_{x_0} + r \cos(t);$$ $$\phi(t,k_{y0},r) = k_{y_0} + r \sin(t);$$

where $0\leq t\leq2\pi$, and $k_{x_0}$, $k_{y_0}$ define the loop's center. So, we can say that this step draws out a circular loop on the $\theta/\phi$ plane.

Then I project these onto the sphere's surface using spherical coordinates, as follows:

$$x(\theta,\phi)=r \cos{\theta}\sin{\phi}, y(\theta,\phi)=r \sin{\theta}\sin{\phi}, z(\theta,\phi)=r \cos{\phi}$$

How do I go about calculating the surface area within these $(x,y,z)$ loops on the surface? This will allow me to calculate the solid angle I need.

The solid angle is given by: $$\Omega= \iint_S \frac{\hat{r}\cdot\hat{n}}{r^2} \, \mathrm{d}\Sigma = \iint_{\mathcal{R}}\sin \theta \, \mathrm{d} \theta \, \mathrm{d} \phi=\frac{\textrm{spherical surface area}}{r^2}$$

I tried using various types of RegionMeasures to calculate this area (such as defining the area within the loop on the sphere as a Region, and by varying the radius from 0 to r, calculating the length of each loop in between and summing it all up), but I feel like I am missing a simple answer to my problem. Maybe what I am missing is a way to somehow map my arbitrary loops into a appropriate integration bounds for $\Omega$, but I tried to avoid this by resorting to Mathematica.

So far, I found the following posts most useful:

Integrate to calculate enclosed area

https://math.stackexchange.com/questions/1832110/area-of-a-circle-on-sphere

Thanks in advance for your time!

Note: I am parametrizing these loops in a peculiar way because I am trying to investigate a physics problem where the functions $x(\theta,\phi),y(\theta,\phi),z(\theta,\phi)$ will be different, and make my loops twist and turn. The ultimate goal is to find the solid angle in these cases, but I wanted to start with the sphere.

Note: I posted a similar question before, but it was misinterpreted by answerers because I did a poor job phrasing it at first, and I am afraid that it lost attention and I still am clueless as to how to proceed (I did not want to delete it because others already put time into it).

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    $\begingroup$ One function, I don't know if you've seen it, that could perhaps compute the solid angle with your parametrization is Area together with spherical coordinates, which can be specified as the third argument. $\endgroup$
    – C. E.
    Feb 27, 2019 at 21:14
  • $\begingroup$ @C.E. thank you, I'll test that soon! I remember the page saying that the Area[] function has an argument for the metric. Just out of curiosity do you know whether the metric can be custom, i.e. for an arbitrary shape that is not a sphere? Regardless, if this is the solution, please consider posting it as an answer so that I can accept it as correct. $\endgroup$ Feb 27, 2019 at 22:10
  • $\begingroup$ @C.E., my attempt didn't work out because ultimately, my loop is controlled by one variable t (as it is a line integral). Area[] requires at least 2, and I thought about using Stokes' theorem to convert my loop integral into an area integral, but the way forward is not too clear. $\endgroup$ Feb 28, 2019 at 0:52
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    $\begingroup$ If I'm not mistaken, you could make use of the Dirac monopole vector potential,$$\vec{A} = \frac{1-\cos \theta}{r \sin \theta} \hat{\phi}.$$This potential has the property that $\vec{\nabla} \times \vec{A} = \hat{r}/r^2$, and so you could apply Stokes' theorem to turn the area integral for $\Omega$ into a path integral. $\endgroup$ Sep 9, 2020 at 19:32

1 Answer 1

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Your integral is rather simple for your parametrization. Let us introduce some notations:

  • a - is the $\theta$-coordinate of the center; the integral does not depend on the $\phi$-coordinate of the center.
  • s in the angular radius of the circular loop.

If loop is on the flat surface you simply write

Integrate[r, {r, 0, s}, {t, 0, 2 π}]
(* π s^2 *)

Notice this integral is computed in a polar coordinate system, which is good for integrating over disks. Notice that the Jacobian $r$ needs to be introduced.

For the area of a circular loop on the unit sphere we need to add one more Jacobian $\sin\theta$ due to spherical integration.

Assuming[π > a > 0 && π/2 > s > 0, 
 Integrate[r Sin[a + r Cos[t]], {r, 0, s}, {t, 0, 2 π}]]

 (* 2 π s BesselJ[1, s] Sin[a] *)

We can also visualize the loops on the sphere.

Clear[a,b,s];
θ=a+s Cos[t];
ϕ=b+s Sin[t];
a=π/5;
b=0;
g[1]=ParametricPlot3D[Table[{Sin[θ] Cos[ϕ],Sin[θ] Sin[ϕ],Cos[θ]},{s,0.1,0.5,0.1}],{t,0,2 π},PlotStyle->Red];
eqr=ParametricPlot3D[{ Cos[ϕ], Sin[ϕ],0},{ϕ,0,2 π},PlotStyle->Black];
mer[1]=ParametricPlot3D[{ Sin[θ] Cos[b],Sin[θ] Sin[b],Cos[θ]},{θ,0,2 π},PlotStyle->Directive[Black]];
ct[1]={Sin[a] Cos[b],Sin[a] Sin[b],Cos[a]};
a=π/3;
b=-π/2;
ct[2]={Sin[a] Cos[b],Sin[a] Sin[b],Cos[a]};
g[2]=ParametricPlot3D[Table[{Sin[θ] Cos[ϕ],Sin[θ] Sin[ϕ],Cos[θ]},{s,0.1,0.5,0.1}],{t,0,2 π},PlotStyle->Blue];
mer[2]=ParametricPlot3D[{ Sin[θ] Cos[b],Sin[θ] Sin[b],Cos[θ]},{θ,0,2 π},PlotStyle->Black];
Show[g[1],g[2],eqr,mer[1],mer[2],Graphics3D[{Sphere[],Gray,Sphere[{0,0,1},.025],Red,Sphere[ct[1],.025],Blue,Sphere[ct[2],.025]}]
 ,PlotRange->1.1,Axes->False,Boxed->False
]

enter image description here

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