Instead of the three points A,B,C, consider a closed loop C. What is the family of surfaces such that for all points P on a surface, the solid angle C subtends at P is constant? If C is planar and encloses an infinitesimal area dA, then the solid angle is (d Omega) = (dA cos(theta))/r^2, where r is the distance between C and P, and theta is the inclination angle of dA as seen from P. The family of surfaces is thus cos(theta)/r^2 = constant, or r = constant * sqrt(cos(theta)). Note that, in contrast, r= constant * cos(theta) is a sphere. Now suppose dA is filled with an electric dipole layer. The potential of a single charge is 1/r, and the potential of a dipole is the gradient of this along the dipole axis. But this is just cos(theta)/r^2, and the strength of the dipole is proportional to dA, so the potential at a point P is proportional to the solid angle subtended by dA as seen from P. For a finite curve bounding an area filled with a constant dipole layer, just integrate over the area, and the potential is again proportional to the total solid angle. The electric field is the gradient of the potential. Alternatively, consider the infinitesimal curve to be carrying a current. This is a magnetic dipole, and the magnetic field has the same spatial dependence as in the case of the electric dipole. And the magnetic field is the gradient of a potential (called the scalar magnetic potential) which again is proportional to the solid angle (d Omega). A finite curve C can be filled by little current loops, each carrying the same current. The internal currents cancel, leaving only the current flowing through C. The magnetic field is the gradient of a potential proportional to the solid angle subtended by C, without regard as to the choice of surface that C bounds. The electric potential is single valued, but discontinuous across the dipole layer. In the magnetic case, there is no distinguished surface, and the potential is multivalued along a path that links the curve. The surfaces of constant solid angle are equipotential surfaces, and so are orthogonal to the field lines. Up close to the curve, the curve looks like a straight line, and the magnetic field consists of circles centered on C. The orthogonal surfaces are planes through C. Thus, in general, the surfaces of constant solid angle pass through C. -- Gene From: Michael Kleber <michael.kleber@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Thursday, December 31, 2015 7:14 PM Subject: Re: [math-fun] 3D version of central angle theorem In two dimensions, another way of thinking about the central angle theorem is: "Q: Given points A and B, what is the locus of all points C such that the size of angle ACB is a given theta? A: The locus is two arcs of the two circles with centers O and O', where angles AOB and AO'B are size 2*theta, and O and O' are on the perpendicular bisector of AB." This question can certainly be generalized to three dimensions. Probably the most natural way is to ask: "Given points A,B,C, what is the locus of all points D such that the solid angle at D subtended by ABC has measure theta?" The answer is evidently not of the form "All points on some piece of some sphere", but it is certainly *some* surface... --Michael