Using complex quaternions to do special relativity

Subject: Re: Solving problems in special relativity w/quate
From: (Mark Hopkins)
Date: 1997/04/05
Message-Id: <5i6uet$>
Newsgroups: sci.physics.research
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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
d-bar = B (1/c d/dt - I Del)
Del = i d/dx + j d/dy + k d/dz

I B d-bar = I/c d/dt + Del
Note the appearance of the unit "I".

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