Introduce spherical coordinates with origin at P. Let A be a vector field with magnitude (1/r) tan(θ/2) pointing in the φ direction. A calculation shows that B = curl A points radially outward with magnitude 1/r^2, except along θ = π, where it is singular. The curve partitions the solid angle about the origin into two parts, one of which avoids the singularity. Consider any surface bounded by the curve. The desired solid angle is the surface integral of B.dS. Then by Stokes' theorem, it also equals the line integral of A.dl taken so that the surface lies to the left as you walk along the curve. Since div B = 0, Gauss' theorem says it doesn't matter what surface you pick, just avoid the singularity. A calculation shows that in the limit of small solid angle in the vicinity of θ = 0, this reduces to the usual line integral for the plane area enclosed by a curve. For a physical motivation, consider a semi-infinite, infinitely thin solenoid, lying along θ = π, terminating at the origin, and carrying a magnetic flux of 4π. The emerging magnetic field is the B above, and A is the vector potential. -- Gene From: Bill Gosper <billgosper@gmail.com> To: math-fun@mailman.xmission.com Sent: Thursday, January 29, 2015 1:41 PM Subject: Re: [math-fun] parametrically enclosed spherical area The better question: Given a point P and closed curve X(t) in 3Space, what is the solid angle at P of the generalized cone joining X to P? --rwg On Mon, Jan 26, 2015 at 9:31 PM, Bill Gosper <billgosper@gmail.com> wrote:
The plane area inside (x(t),y(t)) = ½∫x dy - y dx. E.g., the usual picture of the area under f(x), a<x<b becomes four line integrals--the three axis-aligned segments plus the actual curve of f, traversed anticlockwise: Integrate[f[x],{x,a,b}]==(Integrate[a - 0, {t, f[a], 0}] + Integrate[t*0 - 0, {t, a, b}] + Integrate[b - 0, {t, 0, f[b]}]+Integrate[t*D[f[t], t] - f[t]*D[t,t], {t, b, a}])/2 == (b f[b] - a f[a] + Integrate[t f'[t] - f[t], {t, b, a}])/2 , which follows from integration by parts. E.g., if f:=Exp, In[55]:= % /. f -> Exp
Out[55]= -E^a + E^b == 1/2 ((-2 + a) E^a - a E^a - (-2 + b) E^b + b E^b)
In[56]:= Simplify@%
Out[56]= True
What about on a sphere? Suppose we have a a unit vector X(t):=(x(t),y(t),z(t)) describing a closed curve on a sphere. (X(0)=X(1), say.) What's the area? There ought to be some nice thing analogous to ½∫x dy - y dx. --rwg
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