Chapter 11 exercise 15

An alternate pure geometric approach...

Start with a source at the origin, its streamlines and equipotentials are consistent with a sink at infinity.  To see this clearly project them onto the Riemann sphere.  The streamlines of the source are the spheres' great circles of longitude and the equipotentials become circles of latitude. (In other words you can think of a source as part of a dipole of infinite separation)

Now  apply a mobius transformation bringing the sink at infinity to, say 2a, from the source.  As a mobius transformation maps circles to circles, the great circle streamlines between the source and sink at infinity become arcs of circles joining the poles of the dipole, centred, by symmetry, on the perpendicular bisector of the dipole axis.

Such a transformation can be seen graphically in [29] p163. 

Therefore streamlines between the poles of a balanced doublet are the arcs of circles whose centres are on the perpendicular bisector of the line joining the poles.

Consider the sources and sinks of the question to be a pairs of doublets one above the other. one to the right of the other. Let the doublets be 2a apart, centred on the origin (purely for convenience of notation).

Consider the bottom doublet, there will be a streamline for the bottom doublet whose radius of curvature will be sqrt(2) * a. The centre of curvature of this streamline will be on the perpendicular bisector of the line joining the poles of the doublet.  In this case, the centre of curvature will therefore be at the origin.

By symmetry, this is also true for the top doublet. This will also be true for the left and right hand doublets.

Therefore all the sources and sinks lie on a streamline forming a complete circle centred on the origin of radius sqrt (2) * a.

Done

telemeter