Measuring distances with quaternions

Subject: measuring distances w/quaternions
From: sweetser@alum.mit.edu (Doug B Sweetser)
Date: 1997/05/09
Message-Id: <E9x4A6.HsM@world.std.com>
Newsgroups: sci.physics.research

Hello:

In this post, I want to calculate 4 distances: the spatial separation in
a plane, in a volume, the interval in flat spacetime and the
interval in curved spacetime. What I hope to show is that by using
quaternions, my method of calculation remains the same, only the input
changes. The first three may be obvious, but the fourth is at the heart
of general relativity.

Let's start with the simple case of measuring distance in a plane. To
make this trivial case tricky, I will use quaternions. Take the
difference between two quaternions A and B in the xy plane. Square
it and examine the result:

s^2 = (0 + (Ax - Bx)i + (Ay - By)j + 0 k)^2

= - (delta x)^2 - (delta y)^2 + 0 i + 0 j + 0 k

The magnitude of the scalar is the square of the distance. The sign
indicates that the distance is "spacelike", an issue that never
concerned the Pythagoreans! This minus sign does not indicate that
the answer is wrong; just that it has a link to an idea that arises in
special relativity.

The case of measuring distance in a volume is no different, except
that the third spatial measure is not zero.

s^2 = (0 + (Ax - Bx)i + (Ay - By)j + (Az - Bz)k)^2

= - (delta x)^2 - (delta y)^2 - (delta z)^2
+ 0 i + 0 j + 0 k

Again, the expected result with the identical technique. This is
fortunate because the A and B in the plane with delta z = 0 could
have been rotated to A' and B' with delta z not = 0, and the form of
the calculation and its result should not change.

Since the transition from a plane to a volume was simple, let's hope
the same works for the move up to events in spacetime, where delta
t is no longer zero.

s^2 = ((At - Bt) + (Ax - Bx)i + (Ay - By)j + (Az - Bz)k)^2

= (delta t)^2 - (delta x)^2 - (delta y)^2 - (delta z)^2
+ 2 delta t delta x i + 2 delta t delta y j
+ 2 delta t delta z k

The scalar part is the square of the invariant interval seen in special
relativity. The form of the calculation has not changed from the
previous example, which is fortunate because a boost in the
reference frame would make delta t not zero.

The vector portion is a valid measure, but I do not think it currently
plays a role in physics. Its presence does not make the calculation
wrong, just different. I suspect that is may someday play a role in
doing physics when the interval is zero (lightlike), but I'm not sure.

What happens in curved spacetime? In the process of parallel
transport, the origin of event A is different from event B. Perform
the calculation just as before: take the difference between event A
and B - including the difference in their origins - square it, then look
at the resulting scalar. (OAt is the time at the origin of A)

s^2 = (((At + OAt) - (Bt + OBt)) + ((Ax + OAx) - (Bx + OBx))i
+ ((Ay + OAy) - (By + OBy))j + ((Az + OAz) - (Bz + OBz))k)^2

= (delta t)^2 - (delta x)^2 - (delta y)^2 - (delta z)^2
+ (delta Ot)^2 - (delta Ox)^2 - (delta Oy)^2 - (delta Oz)^2
+ 2 delta t delta Ot - 2 delta x delta Ox
- 2 delta y delta Oy - 2 delta z delta Oz

+ 2 (delta t + delta Ot)(delta x + delta Ox) i
+ 2 (delta t + delta Ot)(delta y + delta Oy) j
+ 2 (delta t + delta Ot)(delta z + delta Oz) k

Curved spacetime is more complex that flat! This should look
somewhat familiar though. In general relativity, the affine
parameter is defined as

lambda = a tau + b

or squared

lambda^2 = a^2 tau^2 + 2 a b tau + b^2

The a's and b's are scalars, so the square of the affine parameter
would be invariant under a Lorentz boost. Compare this with the s^2
calculated above. a^2 is one, and b^2 is the interval between the
origins and also an invariant. The cross terms are different however,
because they would change under a boost for the quaternion
calculation. This may make working with quaternion measures in
curved spacetime ever so slightly different from general relativity in
different reference frames!

With perfect hindsight, could this result have been expected?
General relativity is based on Riemann geometry. Riemann geometry
in turn has both a topological structure and a metric structure.
Quaternions too have both of these structures in their own way. A
theorem by Pontryagin states that the only arc-wise connected,
topological number fields are the real, complex and quaternion
numbers. Quaternions do not have a metric in the usual sense of the
word. Yet as the first three distances discussed here have shown, the
method of squaring the difference of two quaternions generates a
scalar which is identical to the Minkowski metric. This method
actually creates another quaternion, so it is mathematically distinct.

Not having taken a graduate level general relativity class (just a
survey course taught by Edwin F. Taylor), I am not sure how to proceed.
This connection between Riemann geometry and quaternions appears
to me to be a critical step in doing general relativity with
quaternions. Since I was able to get as far as the affine parameter
using only ASCII text, quaternions must be simper pedagogically :-)
There is hope that the two approaches may be ever so slightly
different (just like me :-)


Doug Sweetser
http://world.std.com/~sweetser

Riemann throws a perfect quaternion curve ball.
"Strike 8 pi T!" Shouts the ump with frizzy white hair.