## The Lemniscate and Spacetime

```The advance of the quantum phase of a system over a timelike interval
of spacetime equals the Lorentz-invariant magnitude of that interval.
In the xt plane the square of the magnitude of the interval from
the origin to the event (x,t) is s^2 = t^2 - x^2.  Similarly we can
define the squared magnitude of a spacelike interval from the origin
to the event (x,t) as s^2 = x^2 - t^2.

Also, for any given inertial coordinate system x,t, we can associate
an angle q with each event, defined by tan(q)=t/x.  Thus we have
t = x tan(q), so we can express the squared magnitude of a spacelike
interval as
cos(2q)
s^2  =  x^2 - t^2   =   x^2 [1 - tan(q)^2]  =   x^2 ---------
cos(q)^2

Multiplying through by cos(q)^2 gives

s^2 cos(q)^2  =  x^2 cos(2q)

Substituting  t^2 / tan(q)^2  for x^2 gives the analogous expression

s^2 sin(q)^2  =  t^2 cos(2q)

Adding these two expressions gives the result

s^2  =  (x^2 + t^2) cos(2q)

This means that if we select an event whose coordinates are (x,t)
with respect to a given inertial system, the interval from the origin
to this event makes an angle q relative to the positive x axis in
the xt plane, and the squared Lorentz-invariant magnitude of the
interval is given by (x^2 + t^2) cos(2q).

Consequently the "circular" locus of events such that x^2 + t^2 = r^2
for any fixed r can be represented in polar coordinates (s,q) by the
equation
s^2  =  r^2 cos(2q)

This is the equation of a lemniscate, illustrated in the figure below
for several values of r.

Actually the original form of the lemniscate, first discussed by
Jakob Bernoilli in 1694, was the locus of points satisfying the
equation
__________
x^2 + y^2  =  k^2 / x^2 - y^2

which is, in Bernoulli's words, "like a lying eight-like figure,
folded in a knot of a bundle, or of a lemniscus, a knot of a French
ribbon", from which the curve gets its name.  Solving for y as a
function of x gives the quartic

y^4  +  (2x^2 + k^2) y^2  +  (x^4 - k^2 x^2)  =  0

For any given real value of x this has four roots, but they are purely
imaginary except in the range x=-k to +k, where there are two real
roots and two imaginary roots.  The figure below shows the real y
roots in red and the imaginary y roots in purple.

Of course if we force y to be real but allow x to be imaginary, we
rotate the figure 90 degrees, and if both x and y are imaginary we
have just the rotated version of the interior region of the green
circle.

It's also worth noting that the lemniscate is the inverse (in the
sense of inversive geometry) of the hyperbola relative to the circle
of radius k.  In other words, if we draw a line emanating from the
origin and it strikes the lemniscate at the radius s, then it strikes
the hyperbola at the radius R where sR = k^2.  This is clear from the
fact that the equation for a hyperbola in polar coordinates is
R^2 = k^2/[E^2 cos(q)^2 - 1] where E is the eccentricity.  For an
equilateral hyperbola we have E = sqrt(2), so the denominator is
2cos(q)^2 - 1 = cos(2q), and hence the equation of the hyperbola is
R^2 = k^2/cos(2q).  Since the polar equation for the lemniscate is
s^2 = k^2 cos(2q) we have sR = k^2.

Not surprisingly, the study of this curve led Fagnano, Euler, Legendre,
Gauss, and others to the discovery of addition theorems for integrals
and more generally to the theory of elliptic integrals.  This is a
direct consequence of the velocity addition law in special relativity,
which amounts to an algebraic relation between arc lengths along a
hyperbola.  For example, Fagnano showed that if
________
/ 1 + v
u =  / -------
/   1 - v

then the variables u and v satisfy the integral equation

dv                              du
---------------  =    - sqrt(2) ---------------
sqrt(1 - v^4)                   sqrt(1 + u^4)

Similarly he showed that if

__________
/ 1 - v^2
u = - / ---------
/   1 + v^2

then u and v satisfy

/      dv              /      du
| -------------   =    | -------------
/ sqrt(1 - v^4)        / sqrt(1 - u^4)

```