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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