Exquisite Geometry
 

As a physicist, I have found many beautiful
applications of the circle of Apollonius:
the locus of a point whose distances to two fixed
points are in a constant ratio.

Anyone care to start the ball rolling?

David

First of all prove it is a circle/sphere and find
its centre and radius.

Done that some years ago. I have been so fascinated that I wrote a text about it and illustrated it with some figures. Investigating this subject further leads to beautiful families of circles and lines. Somewhere I took up the notion "curves with constant curvature" (ccc) for those things.

Proving that it is a circle has been easiest with algebra (cartesian coordinates). However, a purely geometric proof would be nice.

Let's call the two fixed points "poles". I'll drop just one note here: They are inverses of each other with respect to the circle.

Here is a plot showing a family of ccc, each one featuring a specific distance ratio [tex]\rho[/tex]. The plot sets [tex]\rho = \tan\alpha[/tex] with [tex]\alpha[/tex] stepping from 0 to 180 degrees, stepsize 5 degrees. The vertical line corresponds to [tex]\rho = 1[/tex] of course.

It is tempting to regard the curves in the previous plot as equipotential lines, which is why I speak of "poles". Exercise: find the field lines. As an intermediate step, name a (mostly) analytic complex function [tex]f(z)[/tex] such that the curves [tex]\Re f(z)=\text{const.}[/tex] can be associated with the equipotential lines. (Herein, [tex]\Re[/tex] means "real part".)

Hint: The distance ratio can be defined as a function on the (punctuated) plane.

Hint 2: No answers [tex]\Rightarrow[/tex] no more graphics (from me, that is).

[QUOTE=ccorn;217368]The plot sets [tex]\rho = \tan\alpha[/tex] with [tex]\alpha[/tex] stepping from 0 to 180 degrees, stepsize 5 degrees.[/QUOTE]

Of course, [tex]\alpha[/tex] runs from 0 to 90 degrees only (We don't need negative [tex]\rho[/tex]).

[quote=ccorn;217348]Done that some years ago. I have been so fascinated that I wrote a text about it and illustrated it with some figures. Investigating this subject further leads to beautiful families of circles and lines. Somewhere I took up the notion "curves with constant curvature" (ccc) for those things.

Proving that it is a circle has been easiest with algebra (cartesian coordinates). However, a purely geometric proof would be nice.

Let's call the two fixed points "poles". I'll drop just one note here: They are inverses of each other with respect to the circle.[/quote]

Many thanks for the beautiful pictures: they justify my OTT
title of the thread!
I would describe "ccc" as a bit pedantic though, since "circle" will
do adequately since we are not concerned with things like the helix here.

[quote=ccorn;217368]Here is a plot showing a family of ccc, each one featuring a specific distance ratio [tex]\rho[/tex]. The plot sets [tex]\rho = \tan\alpha[/tex] with [tex]\alpha[/tex] stepping from 0 to 90 degrees, stepsize 5 degrees. The vertical line corresponds to [tex]\rho = 1[/tex] of course.[/quote]

[quote=ccorn;217385]It is tempting to regard the curves in the previous plot as equipotential lines, which is why I speak of "poles". Exercise: find the field lines. As an intermediate step, name a (mostly) analytic complex function [tex]f(z)[/tex] such that the curves [tex]\Re f(z)=\text{const.}[/tex] can be associated with the equipotential lines. (Herein, [tex]\Re[/tex] means "real part".)

Hint: The distance ratio can be defined as a function on the (punctuated) plane.

Hint 2: No answers [tex]\Rightarrow[/tex] no more graphics (from me, that is).[/quote]

When you refer to "poles" and "equipotentials" you are clearly thinking
about 2D problems, for example two infinite parallel lines, one charged
positively the other negatively: V = ln(r1/r2).*
The field lines are perpendicular to the equipotentials and are circles
passing through both poles.
Note that this is not the "dipole field" of a bar magnet or 2 opposite
point charges in 3D.
There was a class experiment passing a current between two copper
discs on blotting paper soaked in copper sulpate solution.
Using a probe connected via a galvanometer to the point
between two resistors, they had to plot equipotentials.
Nice idea. Shame that of all things to draw on, soggy blotting
paper is the pits:smile:

More anon

David

* Or the magnetic field lines of two infinite straight
wires carrying currents in opposite directions (see definition
of the ampere).

[QUOTE=davieddy;217444]I would describe "ccc" as a bit pedantic though, since "circle" will do adequately since we are not concerned with things like the helix here.[/QUOTE]

For [tex]\rho=1[/tex], in 2D the circle degenerates to a line, in 3D the sphere degenerates into a plane. That is the reason for the term "constant curvature".

In n dimensions, the loci with a given distance ratio [tex]\rho[/tex] form (n-1)-dimensional hypersurfaces. Therefore, in 3D we are not looking for curves, so the helix does not count. But yes, since the first "c" of ccc means "curve", the term is appropriate only in 2D.

[QUOTE=davieddy;217444]When you refer to "poles" and "equipotentials" you are clearly thinking about 2D problems, for example two infinite parallel lines, one charged positively the other negatively: V = ln(r1/r2).*
The field lines are perpendicular to the equipotentials and are circles passing through both poles.

* Or the magnetic field lines of two infinite straight wires carrying currents in opposite directions (see definition of the ampere).[/QUOTE]

Indeed. Plot attached. (I recognize a grinning bearded face under a turban and wonder whether that is due to propaganda.)

Identifying the 2D plane as the complex plane and placing the poles at -1 and +1, your V is indeed the real part of f(z) = ln((1+z)/(1-z)) = 2 artanh z. (The real part of the logarithm depends only on the absolute value of its argument, thus only on the distance ratio.) Given that the field can be described by artanh, my first approach to plot both equipotential and field lines was to feed a rectangular grid of complex numbers to the tanh function. This makes it hard to approach the poles however, therefore the plot below has been composed of circles and lines explicitly. By the way, I have used [url=http://asymptote.sourceforge.net]Asymptote[/url] for this.

[QUOTE=davieddy;217444]Note that this is not the "dipole field" of a bar magnet or 2 opposite point charges in 3D.[/QUOTE]

Right. Let's stay in 2D and make a dipole by moving the poles together. Then you get curves like those in a [url=http://en.wikipedia.org/wiki/Smith_chart]Smith chart[/url].

[QUOTE=davieddy;217444]When you refer to "poles" and "equipotentials" you are clearly thinking about 2D problems, for example two infinite parallel lines, one charged
positively the other negatively: V = ln(r1/r2).*[/QUOTE]

Or about a "wire over ground", with the straight line representing the loci with ground potential. The wonderful thing is that the wire may have nonzero radius.

I dimly remember an old manual by Motorola on ECL circuit design, complete with formulas about transmission line properties for pairs of wires and wires over ground planes etc. I was impressed seeing such real-world things described by rather short and simple formulae.

[quote=ccorn;217460]Or about a "wire over ground", with the straight line representing the loci with ground potential. The wonderful thing is that the wire may have nonzero radius.

[/quote]

I hope I need not apologize for taking some time over my
response: this subject deserves all the time available for
contemplation!

The "nonzero" radius is no great surprize in the light of the
circularity of equipotentials.
Consider the capacitance and inductance per unit length of
two cylinders radius r distance d apart, and before you can say
"Bob's Your Uncle" you are talking about an electromagnetic wave.

David

I can't wait any longer
 

(As the actress said to the bishop)

I like to approach this problem backwards.

A sphere has centre O and radius R.
Point P is a distance r from O (r>R)
P' lies on OP a distance R^2/r fom O
(That is what ccorn was talking about when he
said "inverses wrt to the circle")
C is a point on the sphere.

Convince yourself that OPC and OCP' are similar triangles.

David

The sphere has mass M and a mass m is located at P.

Show that the mass subtended by a small solid angle Omega
centred on P' exerts a component of force GMm Omega/4 pi r^2
towards O.

David