Calculating solid angle......

Question

Here is a sketch of the problem statement : A cube of edge length $l$ is placed in three dimensional space with one vertex at the origin ${(0,0,0)}$ and all the faces parallel to the (Cartesian) coordinate planes. The cube lies entirely in the first "octant". Calculate the solid angle subtended by the cube face parallel to $YZ$ plane at the point $(0,0,l)$. There are two such faces parallel to the $YZ$ plane. The face I am referring to is one which does not have any of its vertices at the origin.

I apologize for not providing a drawing of the situation. I have a decent idea of "solid angle" concept, but I am not very familiar with tricky definite integrals for area and volume calculations. I can intuitively (and crudely) approximate solid angle by elementary methods, but I just can not use integrals for exact solid angle calculation. Can anyone help ? I can project the cube face on a spherical surface centered at $(0,0,l)$ and use the projected area to calculate solid angle. But how to calculate this projected area? I am unable to setup some suitable expression for the infinitesimal projected area $dA$. I just want a hint to start with. Thanks in advance.

Answer 1

Imagine $8$ such cubes cuddling at $(0,0,0)$. Together they have an outer surface of $6\cdot4=24$ unit squares, and each such square encompasses the same solid angle with respect to $(0,0,0)$. The solid angle per unit quare therefore is $${4\pi\over24}={\pi\over6}\ ,$$ which is the answer to your question.

In order to compute this solid angle via an integral I put $l=1$ and write the projection in the form $${\bf f}:\quad (x,y,1)\mapsto{1\over\sqrt{1+x^2+y^2}}(x,y,1)\qquad\bigl((x,y)\in Q:=[0,1]^2\bigr)\ .$$ Now use $x$ and $y$ as parameters when computing the area of the surface $S:={\bf f}(Q)$. Compute ${\bf f}_x$, ${\bf f}_y$, and then the surface element $${\rm d}\omega=|{\bf f}_x\times{\bf f}_y|={1\over(1+x^2+y^2)^{3/2}}\>{\rm d}(x,y)\ .$$ (Deriving this result "geometrically" would of course require less computation.) It follows that $${\rm area}(S)=\int_0^1\int_0^1 {1\over(1+x^2+y^2)^{3/2}}\>dx\>dy\ .$$ I didn't do this integral myself. Mathematica computed $$\int_0^1{1\over(1+x^2+y^2)^{3/2}}\>dy={1\over(1+x^2)\,\sqrt{2+x^2}}$$ ad then arrived at the final result ${\pi\over6}$, as expected.

Answer 2

You are asking for the solid angle at one vertex of a cube subtended by a face of the cube that does not include that vertex.

It does not matter which vertex you choose; the cube is symmetric with respect to the vertices and the answer will be the same. So you might as well compute the angle at the origin of a face not touching the origin; it's easier to talk about.

If you want the solid angle of one of the three faces that do not have a vertex at the origin, remember that there are three of them, that they cover the entire octant, and that the cube has a three-way symmetry around their common vertex. This gives you an exact result that is not crude at all.

For a method that actually sets up an integral, note that the solid angle is simply the area of the projection of the cube's face onto a unit sphere around the origin. Take the method described here to approximate the projection of an arbitrary small square onto that sphere. This gives you a formula, $$\frac{{\mathbf x} \cdot \hat{\mathbf n}}{\lVert \mathbf x\rVert^2}A,$$ in which $\mathbf x$ is the displacement vector from the center of the sphere to a point in a small square of area $A$ and $\hat{\mathbf n}$ is a unit vector perpendicular to that square. When the center of the unit sphere is the origin, the coordinates of $\mathbf x$ are simply the coordinates of the small square; if you must place the center of the sphere at $(0,0,l),$ you will have to subtract coordinates in order to express $\mathbf x.$

In the limit, as the size of the square goes to zero, and adding up all the projections of the individual small squares within the face of the cube, you can replace the small area $A$ with the "infinitesimal" $\mathrm dA$ and integrate the projected area over the face of the cube.