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To find the solid angle subtended at a point O by an arbitrary surface element $d{\vec S}=dS\hat{{n}}$, one joins the peripheral points of $d{\vec S}$ to O by straight lines which generates a cone at O. Then, if we draw a sphere of radius $r$ centered at O, the cone intercepts a surface element of a sphere given by $dS\cos\alpha$ where $\hat{{r}}\cdot \hat{{n}}=\cos\alpha$. The corresponding solid angle is given by $$ d\Omega=\frac{dS\cos\alpha}{r^2} $$ How can we integrate this expression over a closed surface $S$ of arbitrary shape to obtain that the total solid angle subtended at O is $4\pi$? I want to use spherical polar coordinates. Can we say that $dS\cos\alpha=r^2\sin\theta d\theta d\phi$ and integrate over $\theta,\phi$? Thiis does give $4\pi$ indeed.

Solidification
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1 Answers1

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I am not quite sure what $$ \int_S\frac{dS\cos\alpha}{r^2} $$ means because with $dS=r^2\sin\theta\,d\theta\,d\phi$ this will depend on $\alpha$ and not equal $4\pi\,.$

Let's start from scratch. In polar coordinates a point $\boldsymbol{r}$ on the sphere is $$ \boldsymbol{r}=\left(\begin{matrix}r\cos\theta\\r\sin\theta\cos\phi\\r\sin\theta\sin\phi \end{matrix}\right)\,,~~\theta\in[0,\pi]\,,~~\phi\in[0,2\pi)\,. $$ If the angles change by small amounts this point changes by $$\tag{1} \left(\begin{matrix}-r\sin\theta\\r\cos\theta\cos\phi\\r\cos\theta\sin\phi \end{matrix}\right)\,d\theta\,,\text{ resp. by }\left(\begin{matrix}0\\-r\sin\theta\sin\phi\\r\sin\theta\cos\phi \end{matrix}\right)\,d\phi\,. $$ The infinitesimal oriented surface element is the cross product of the two vectors from (1): $$ r^2\left(\begin{matrix}\cos\theta\sin\theta\\\sin^2\theta\cos\phi~\\-\sin^2\theta\sin\phi\end{matrix}\right)\,d\theta\,d\phi\,. $$ The length of this cross product is the infinitesimal surface area $$ r^2\sqrt{\sin^4\theta+\cos^2\theta\sin^2\theta}\,d\theta\,d\phi=r^2\sin\theta\,d\theta\,d\phi\,. $$ If the cone $O$ that has the angle $\alpha$ at the center of the sphere traces out a surface around the north pole that surface is $$ r^2\int_0^{\alpha/2}\int_0^{2\pi}\sin\theta\,d\theta\,d\phi=2\pi r^2(1-\cos(\alpha/2))\,. $$ For $\alpha=\pi$ we get the surface of the northern hemisphere which is $2\pi r^2\,.$

Kurt G.
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