Subject: Re: Solving problems in special relativity
w/quate

From: mark@omnifest.uwm.edu (Mark
Hopkins)

Date: 1997/04/05

Message-Id:
<5i6uet$vvc@omnifest.uwm.edu>

Newsgroups:
sci.physics.research

[More Headers]

What you're doing is
essentially redeveloping the Spacetime Algebra

formalism of
Hestenes. The earliest reference for this is the
1966

"Spacetime Algebra", by David Hestenes.

A
more suitable framework for doing the algebraic manipulations is
not

the quaternions, but the 3+1 dimensional Clifford algebra --
the one which

is isomorphic to the Dirac algebra. To see how this
is related to

quaternions, note that one can write its generators in
the form:

g0 = B, g1 = Ci, g2 = Cj, g3 = Ck

in the
algebra given by the following relations:

{ B, C } commute
with { i, j, k }

{ i, j, k } satisfy the usual quaternion
relations

B C = -C B, B^2 = 1 = C^2

Defining I = C B, the
even subalgebra (that which consists of the sums of

products of
even numbers of factors) can be seen to comprise the
following

basis elements:

1, g1 g0 = Ii, g2 g0 = Ij, g3
g0 = Ik

g2 g3 = i, g3 g1 = j, g1 g2 = k, g0 g1 g2 g3 =
I

In this subalgenra, I commutes with everything else, so that
one has for

all practical purposes a copy of the "complex
quaternions". The complex

quaternions are isomorphic to
the Pauli spin algebra with the correspondences:

sigma_1
= -Ii, sigma_2 = -Ij, sigma_3 = -Ik

Finally, a Lorentz
transformation can be represented by a unit element, R,

of the
even subalgebra (i.e., one for which R' R = 1 = R R', where R'
denotes

the "reversion" operation), with:

L(v) = R' v R, where v = (sum v_i gi)

So, Lorentz
transformations correspond to unit complex quaternions.
This

correspondence provides a double covering since -R will also
yield the

same transformation as R. What we have, in fact, is a
quaternion

representation of the symmetry group SU(2).

[The reversion of an element of a Clifford algebra is obtained by
switching

the order of the factors in its products. Here, reversion
corresponds to the

quaternion conjugation, I -> I, i -> -i, j
-> -j, k -> -k]

The "Dirac" gradient
operator, d-bar = sum (gi d/dx^i) will take on the

form:

d-bar = B (1/c d/dt - I Del)

where

Del = i d/dx + j
d/dy + k d/dz

Alternatively,

I B d-bar = I/c
d/dt + Del

Note the appearance of the unit "I".

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