My objective is to calculate (or find an expression) for the solid angle of a circular loop (parametrized by $(x,y)=(r \cos{x_i}, r \cos{y_i})$, where $0\leq x, y \leq 2\pi$) on the surface defined as:

\begin{split} h_{x}(x,y) &= t \left[ 1 + \cos(k \cdot b_{1} ) + \cos(k \cdot b_{2} ) \right ], \\ h_{y}(x,y) &= t \left[\sin(k \cdot b_{1} ) - \sin(k \cdot b_{2} ) \right ], \\ h_{z}(x,y) &= M - 2 t_{2} \sin{\phi} \sum_{i=1}^{3} \sin(k \cdot b_{i}), \end{split}

where $k=(x,y)$ is a row vector, $b_i$ are column vectors, $t,t_2,M, \phi$ are constants.

I plotted this surface, as seen below, and it appears as if it has regions where its surface normal vector points towards its center whereas other regions point outwards: enter image description here

When normalized onto the unit sphere, and when the inward-pointing-normal-regions are highlighted in red, the surface looks like:

enter image description here

It appears as if the projection onto the sphere results in folding. However, according to Wikipedia (https://en.wikipedia.org/wiki/Solid_angle):

even if the projection on the unit sphere to the surface S is not isomorphic, the multiple folds are correctly considered according to the surface orientation described by the sign of the scalar product ${\hat {r}}\cdot {\hat {n}}$.

How do I go about finding an expression for the solid angle of a loop on this surface? For instance, a loop could look like:

enter image description here

According to Wolfram (http://mathworld.wolfram.com/SolidAngle.html), the integrals for a relatively simpler cube seem hard to setup. Is the easiest way to do this numerically (using Mathematica, etc)?

To reiterate, I am trying to find an expression of the form (see Wikipedia article linked above):

$$ \Omega ={\frac {A}{r^{2}}} =\iint \limits _{S}{\frac {{\hat {r}}\cdot {\hat {n}}\,d\Sigma }{r^{2}}}\ $$

It is not clear to me how I could even go about starting to tackle this physics-inspired investigation, so I would appreciate any tips/resources/feedback-on-feasibility. Thanks!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.