Can something be perpendicular to itself? At first glance, this seems impossible. A "thing," in the ordinary sense, cannot stand at right angles to itself. But reality is richer than just "things." Perhaps even the solidity of things is an illusion, a condensation of what might be called "nothing." Take light, for example: it is not a "thing" in the usual sense. In Minkowski space, the four-momentum of a photon has zero length—it is, quite literally, perpendicular to itself.
In this note, we explore a related geometric curiosity. We will look at
the set of all points perpendicular to any point lying on the null cone
of R2,2. The surprising outcome will be a
pair of intersecting circles. In the following post, we will uncover
their deeper meaning: these circles turn out to be nothing less than two
representations of light rays. That, however, lies ahead. For now, let
us enjoy the game of pure geometry—elementary trigonometry, sines and
cosines leading us into unexpected structures.
Let us first recall some basic definitions pertaining to our subject. We start with more "geometric" coordinate independent way.
Let V be a real 4-dimensional vector space endowed with a quadratic form Q of signature (2,2). Let B: V⨉V → V be the associated bilinear form, so that Q(x) = B(x,x).
Let O(V) denote the orthogonal group of (V,B), that is the set of all linear transformations L of V which leave B invariant: B(Lx,Ly) = B(x,y), for all x,y in V.
Note. Later on we will also consider two subgroups of O(V): special orthogonal group SO(V), and its connected component of the identity SO0(V).
We define N to be the null cone of V:
N = {x∈V: Q(x) =0}
and PN to be the projective null cone:
PN = N⟍{0} / ℝ×,
where Rx is the multiplicative group of non-zero real numbers. But we will be mostly interested in PN+ defined as
PN+ = N⟍{0} / ℝ×>0,
where ℝ×>0 is the multiplicative group of positive (i.e. >0) real numbers.
PN can be obtained from PN+ by identifying pairs of opposite elements: if we denote by [x] the equivalence class of x, representing a point in PN+, then we obtain a point in PN by identifying [x] with [-x]. The null cone N⟍{0} is 3-dimensional, therefore PN+ and PN are 2-dimensional. The group O(V) acts on PN+ and on PN
in a natural way, by L[x] = [Lx], its action being transitive. So, we
have a geometry in the sense of Felix Klein. We will identify the
corresponding subgroups later on. Studying geometry is nothing else but
studying invariants of this action, or better: studying constructions invariant under this action. Our first construction will be the construction of the space perpendicular to a point.
We first define the relation of orthogonality "⟘" in PN+:
Definition 1. For any two points p,q in PN+, we write p⟘q if p=[x], q=[y], and B(x,y) = 0.
Notice
that the definition makes sense, since the condition B(x,y)=0 does not
depend on the choice of representatives of equivalence classes. The
definition is also invariant in the sense that if p⟘q, the Lp⟘Lq for any L in O(V).
Exercise 1. Verify that, for all p, we have p⟘p, and p⟘(-p).
Note. Notice that -p is well defined. If p = [x] then -p is defined as [-x]. Moreover L(-p) = -L(p) for any L in O(V).
We now define:
p⟘ = {q ∈ PN+ : q⟘p } (1)
Now, while PN+ is 2-dimensional, p⟘ has one condition more, thus it is 1-dimensional. So it is a curve in PN+. It contains at least these two points: p and -p. The construction is O(V) invariant: L(p⟘)=(Lp)⟘. Thus through each point of PN+
we have two special curves. They meet again at opposite point -p. What
are these curves? How to represent them graphically? We know already
that PN+ can be faithfully represented by a torus. How these
two lines will look like on the torus? And what about PN, where p and -p
are identified?
Graphical representation of p⟘.
So far we didn't use coordinates. But to arrive at a graphical representation coordinates are unavoidable. Thus we introduce an orthonormal basis adapted to our point p. So, let e1,...,e4 be mutually perpendicular vectors in V, with Q(e1) = Q(e2) =1, Q(e3) = Q(e4) = -1, and such that p can be represented by e2 + e4, which is in N.
Any x∈N is of the form x = xiei, with
(x1)2 + (x2)2 - (x3)2 - (x4)2 = 0, (2)
or
(x1)2 + (x2)2 = (x3)2 - (x4)2 = ρ2. (3)
Since we removed the origin from N, we may assume that ρ = 1. Then x1, x2, x3, x4 can be interpreted as cos(α), sin(α), cos(β), sin(β) resp. We will assume that α and β
are in [0,2π]:
x1 = cos(α),
x2 = sin(α),
x3 = cos(β), (4)
x4 = sin(β).
For graphical representation we draw the torus as a parametric 3D curve
tor(α,β) = { (R + r cos(β)) cos(α), (R + r cos(β)) sin(α), r sin(β)}, (5)
where R and r stand for the major radius and minor radius of the torus. For our graphic representation I choose R = 3, r = 1.
For the point p we have
x1 = x3 = 0, x2 = x4 =1, (6)
which corresponds to α = β = π/2. Point -p has coordinates x1 = x3 = 0, x2 = x4 =-1, which corresponds to α = β = 3π/2. Here is the graphics of our torus with point p depicted in color blue, while -p is depicted in red.
Let us now examine the rest of p⟘. For x to be orthogonal to (0,1,0,1) we must have x2 - x4 = 0, or x2 = x4. (Why?) But then, from (2), we must have (x1)2 = (x3)2 , or
x3 = ∓ x1. (7)
1) The case of x3 = x1
Let us consider first the case x3 = +x1. Then x = (x1,x2,x1,x2), therefore (4) implies cos(β) = cos(α) and sin(β) = sin(α), thus β = α. The path is depicted in blue on the image below.
2) The case x3 = -x1.
Now x = (x1,x2,-x1,x2), thus cos(β) = -cos(α), sin(β) = sin(α), which implies β = π - α. The path is depicted in red.
The unoriented case
In this case [x] is identified with [-x]. Thus -p and p represent just one point.
For plotting we will use formulas from "Unoriented and oriented conformal completion of 1+1 dimensional spacetime"
cos(α) = x1x3 - x2x4,
sin(α) = x2x3 + x1x4,
cos(β) = x1x3 + x2x4,
sin(β) = x2x3 - x1x4.
These formulas assume normalization (x1)2 + (x3)2 = (x2)2 + (x4)2 = 1. Otherwise we need to divide the right hand side by (x1)2 + (x3)2 = (x2)2 + (x4)2 . Assuming, as before, that p is represented by x = (0,1,0,1), we get for this point
cos(α) = - 1,
sin(α) = 0,
cos(β) = 1,
sin(β) = 0.
This corresponds to α = π, β = 0.
Let us now examine the rest of p⟘.As in the oriented case we must have x3 = x1 or x3 = -x1.
1) The case of x3 = x1
For x3 = x1 we have the path (x1,x2,x1,x2), which leads to
cos(α) = (x1)2 - (x2)2,
sin(α) = x2x1 + x1x2 = 2x1x2,
cos(β) = (x1)2 + (x2)2 = 1,
sin(β) = 0.
We get the coordinate line of α, for β = 0.
2) The case of x3 = -x1
For x3 = -x1 we have the path (x1,x2,-x1,x2), which leads to
cos(α) = -1,
sin(α) = 0,
cos(β) =-(x1)2 + (x2)2,
sin(β) = -2x1x2.
We get the coordinate line of β, for α = π.
The two circles intersect at α = π, β = 0 - the point p.
Note. Notice the important difference: in the oriented case the complement PN+⟍ p⟘ of p⟘ consists of two disconnected regions, each being path-connected and simply connected, while in the unoriented case the complement PN⟍ p⟘ of p⟘ is path-connected and simply connected.
But what is the meaning of it all?
The physical meaning of so obtained p⟘ will be discussed in the next post
