Spheres as Objects - May 1 Special

 We humans are poor creatures. We are bound by our genetics, imprisoned within three spatial dimensions, residing on a beautiful but otherwise insignificant planet in the vast Cosmos—a planet periodically bombarded by comets that wipe out most of its life.

And yet, we strive to understand as much of the world—outside and inside us—as it is possible for us to grasp. What we perceive with our senses at any given moment is just one side of a multifaceted reality. Plato's allegory of the cave reveals only a small part of the whole truth.

Since today is May 1st, it is more than appropriate to quote from Mario Bunge’s 1950 paper, “The Inexhaustible Electron” (Science & Society, Vol. 14, No. 2, Spring 1950, pp. 115–121):

“The electron is as inexhaustible as the atom; nature is infinite, but it exists infinitely.”
—Lenin, Materialism and Empirio-Criticism (1908)

Lenin wrote this during the crisis of modern physics, at a time when the classical theory of the electron had been given a consistent form, mainly through the work of H. A. Lorentz (1853–1928). The central problem of this theory—which was then expected to explain all the properties of matter—was the structure of the electron.

What is the nature of the electron’s mass: mechanical, electromagnetic, or both? What is the nature of the field within the electron? What is its radius? These were some of the questions that intrigued physicists forty years earlier.

The electron is as inexhaustible

Physicist and philosopher Mario Bunge wrote those words 75 years ago. Today, the situation is not much better. In 1995, J. Keller and Z. Oziewicz organized an international conference, Theory of the Electron, in Mexico City. A group photo shows 37 participants, and the conference proceedings span 500 pages. In 2002, J. Keller published a 275-page monograph, The Theory of the Electron (Kluwer, 2002), summarizing just one such theory.

Why would the electron be considered inexhaustible? Clearly, it is more than just a point. But if it is not a point, then what is it? What “shape” does it have?

I think a better approximation would be to imagine the electron as a sphere. But what lies inside this sphere? Perhaps it contains a window. A window to where? To other dimensions—what else could it be?

And how many of these other dimensions are there? Well, that depends on how you choose to look at them. Which face of the multidimensional universe do you want to observe? One face might have a finite number of dimensions; another might well be infinite-dimensional.

This should not be surprising. We need to get used to it. One face of a quantum object may appear as a “particle,” another as a “wave.” We can see only one face at a time. Sometimes it will be sharp, sometimes it will be blurry.

Anyway, we are discussing spheres—in fact, we are discussing their geometry. We treat oriented spheres as “objects.” We are learning how to navigate in the space of these objects.

Object Oriented Programming

Lie sphere geometry reminds me of "object-oriented programming," which caught my attention in the past. We will return to the math in the next post.

Lie Sphere Geometry Part 11: Reprojection of Spheres

 

This post is a continuation of Part 9. Working on this post gave me a real headache. All was going fine until it came to deciding the value of epsilon (it appears in (9a) and (9b) below) that takes care of the sign of the radius of the sphere in R³. I went to bed last night thinking about how to solve the contradiction I have arrived at.

It took me half the day today to figure out the solution of the problem. It was my simple algebra error, and I had to pay for it with Sisyphus  efforts.

It was my simple algebra error, 

and I had to pay for it with Sisyphus  efforts.

Let us first recall the definition from Lie Sphere Geometry: Part 9: Spheres of negative radius:

Definition 1. The oriented sphere with center c in R³ and signed radius ρ, 0≠ρ∈R, is

S_ρ(c) = {y∈R³: (y-c)² = ρ²}                (0)

with unit normal vector field

n'(y) = (c-y)/ρ.             (1)

It is an image, by the stereographic projection of some oriented sphere S_t(m) on S³.

Our aim now is to find (t,m) and to express them in terms of c and ρ. We recall that, with 0<t<π, m∈S³,  S_t(m) denotes the set

S_t(m) = {x∈S³: x·m = cos(t)},        (2)

together with the normal vector field

n(x) = (m - cos(t)x)/sin(t).         (3)

The pair (t,m) should be such that the image of n(x(y)) has the same direction as n'(y), where x(y) is the inverse stereographic projection of y.

We start with recalling the stereographic projection formula from x = (x⁰,...,x³) in S³, with the removed "South Pole" (-1,0,0,0), to y=(y¹,y²,y³) in R³:

yⁱ  = xⁱ/(1+x⁰),      ( i=1,2,3).        (4)

It will be convenient to introduce x'=(x¹, x², x³), so that (4) can be written as a vector formula:

y = x'/(1+x⁰).                (4a)

Let us now use (4a) to write (0) as

((x'/(1+x⁰) - c)² = ρ²,

or

(x' - (1+x⁰)c)² = (1+x⁰)²ρ².         (5)

Expanding the left-hand side we obtain

x'² - 2(1+x⁰)x'·c + (1+x⁰)²c² = (1+x⁰)²ρ².        (6)

Eq. (6) does not look like (2) at all. Eq. (2) is linear in x, while (6) is a quadratic equation. However there is the following algebra magic that does the job: we have that x²=1, therefore (x⁰)²+x'²=1. Or x'²=1-(x⁰)² =(1+x⁰)(1-x⁰). Using this, and dividing both sides of (6) by (1+x⁰)≠0, we obtain

(1-x⁰) - 2x'·c + (1+x⁰)c² = (1+x⁰)ρ².        (6a)

We now collect the coefficients in terms linear in x, and move the rest to the right hand side, we also multiply both sides with (-1):

x⁰(1+ρ²-c²)+x'·(2c) = 1+c²-ρ².        (7)

Eq. (7) already has the form of Eq. (2), with m⁰=(1+ρ²-c²) and mⁱ=2cⁱ (i=1,2,3), but m should satisfy m²=1. Therefore we need to normalize. To this end we define

D =((1+ρ²-c²)²+4c²)^1/2,            (8)

and define

m=ε((1+ρ²-c²),2c)/D        (9a)

t= cos^-1(ε(1+c²-ρ²)/D),        (9b)

where ε=±1.

We have to decide now on the sign of  ε so that we have the correct direction of the normal. We know from Proposition 2 of Part 9 that for the normals to be correct we should have, in particular, the identity:

ρ = sin(t)/(m⁰+cos(t)).

Now, here, 

m⁰ = ε(1+ρ²-c²)/D,

t = cos^-1(ε(1+c²-ρ²)/D),

thus

m⁰+cos(t) = 2ε/D.

On the other hand,  we have the trigonometric identity sin(cos^-1(x))=(1-x²)^1/2. With x=ε(1+c²-ρ²)/D, we easily find that

sin(t)=(4ρ²)^1/2 /D=2|ρ|/D.

It follows that ρ=|ρ|/ε, and thus ε = sgn(ρ).

In this way we have arrived at the following Proposition:

Proposition 1. Let S_ρ(c) be a sphere in R³ of signed radius ρ∈R,  ρ≠0, and center c∈R³. The image of S_ρ(c) by the inverse stereographic projection is the oriented sphere S_t(m) in S³, with

m=ε((1+ρ²-c²),2c)/D

t = cos^-1(ε(1+c²-ρ²)/D),

where

D =((1+ρ²-c²)²+4c²)^1/2,

 

ε = sgn(ρ).

In the next post we will calculate the inverse stereographic image of an oriented plane in R³.