sci.physics.research (moderated) #6351 (10 more)
[1]

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

[1] The
Equivalence of the Dirac and Maxwell Equations

Date: Thu Feb 13
12:51:57 EST 1997

Organization: Omnifest

Lines:
138

Originator: anderson@bohr

In this treatment, which
employs Hestenes spacetime algebra approach

and explains the
correspondence between that approach and Hestenes'

notation to
the standard spinor notation used in Quantum Physics, we'll
show

the intriguing relation between the Dirac and (classical)
Maxwell equations.

This has been demonstrated a couple
times before in the recent literature,

but not in a way that is (in
my opinion) as accessible as the treatment

given below.

---
-----------------

A spinor, according to the treatment given
by mathematicians, is an

element of the even subalgebra of a
Clifford algebra. In the context of the

Dirac algebra, spinors take
on the form:

psi_H = b + (B1 g01 + B2 g02 + B3 g03) + (E1 g23
+ E2 g31 + E3 g12) + e g0123

where { g_0, g_1, g_2, g_3 } are
the generators of the Dirac algebra, and

g_ij.. = g_i g_j ... This is
the representation of spinors that Hestenes

adopts, thus the
subscript. (The actual notation with the B's, E's, etc.,

however, is
mine, used to expedite the discussion below. Hestenes writes
his

psi_H in the form R exp(g0123 S)).

The relation
between this spinor and the one normally used in
Quantum

Physics (psi_D, with D for Dirac) can be stated as so:

psi_M |0> = psi_D

where

|0> = / 1 \ and
psi_M = psi_H in matrix form

| 0 |

| 0 |

\ 0 /

So what Hestenes did, in effect, is
unify the notation used for spinors and

operators by
"factoring out the trailing |0>".

Given the
restriction of psi_H to the even subalgebra, this relation

defines a
one-one correspondence between psi_H and psi_D. The Dirac
matrix

representation (stated in block form) is:

g0 = 1
0 g1 = 0 X g2 = 0 Y g3 = 0 Z

0 -1 X 0 Y 0
Z 0

where X = 0 1, Y = 0 -i, Z = 1 0,

1 0
i 0 0 -1

where I'm using scalars k in the blocks to denote
scalar multiples kI of the

identity matrix I. In this
representation, psi_H is:

psi_M = b - i E B - i e

B - i e b - i E

where B = (B1 X + B2 Y + B3 Z), and E = (E1 X + E2 Y
+ E3 Z). Finally, this

yields:

psi_D = / b - i E3
\

| E2 - i E1 |

| B3 - i
e |

\ B1 + i B2 /

The Dirac equation
in the standard notation is:

p-bar psi_D = m c
psi_D

where p-bar = h-bar D, D = g0/c d/dt + g1 d/dx + g2
d/dy + g3 d/dz, and the

d's are supposed to denote partial
derivatives. This can be written in the

form:

h-
bar D psi_M |0> = m c psi_M |0>

or

D psi_M
|0> = (mc/h-bar) psi_M |0>

This implies

D psi_M = (mc/h-bar) psi_M F

where F is some factor for which F
|0> = |0>. Since D psi_M is in the odd

subalgebra, then F
must be too. This pins F down to F = g0. Therefore,

D psi_M = (mc/h-bar) psi_M g0

Since the matrix
representation is faithful, the same equation holds in

abstract
form, as well:

D psi_H = (mc/h-bar) psi_H
g0,

where

D = g0/c d/dt + g1 d/dx + g2 d/dy + g3
d/dz.

So the effect of converting to the Hestenes notation is to
stick an

extra factor of g0 on the right. This means that in this
form Dirac's

equation has to be stated with a timelike direction
explicitly selected.

I think this is directly related to the
"problem of time" often mentioned

by people in
Quantum Gravity.

Now express this equation in the
following representation:

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

where

QC = -CQ, Q^2 = 1 = C^2

Q,
C commute with the quaternion units i, j and k.

and define I = QC.
This also provides a faithful representation of the Dirac

algebra.
In this form

g01 = I i, g02 = I j, g03 = I k

g23 = i, g31 = j, g12 = k

g0123 = -I

and

psi_H = (b + E) + I (B - e)

where, now, we write B = B1 i + B2 j + B3
k, E = E1 i + E2 j + E3 k. Dirac's

equation takes on the form:

(Q/c d/dt + C Del) psi_H = mc/h-bar psi_H Q

where Del = i d/dx + j
d/dy + k d/dz. Multiplying on the left by Q, this

yields:

(1/c d/dt + I Del) psi_H = mc/h-bar (Q psi_H Q)

and since Q I Q = Q
Q C Q = C Q = -Q C = -I we can write:

Q psi_H Q = (b + E) - I
(B - e) -- the conjugate of psi_H

Taking the real and imaginary
parts, we can then decompose Dirac's equation

into the pair of
quaternion equations:

1/c d/dt (b + E) - Del (B - e) =
(mc/h-bar) (b + E)

1/c d/dt (B - e) + Del (b + E) = - (mc/h-
bar) (B - e)

The quaternions Del B and Del E decompose further
into:

Del B = -(div B) + (curl B); Del E = -(div E) + (curl
E)

where

div (Ui + Vj + Wk) = dU/dx + dV/dy + dW/dz

curl (Ui + Vj + Wk) = i(dW/dy-dV/dz) + j(dU/dz-dW/dx) + k(dV/dx-
dU/dy)

grad S = Del S, for scalars S

corresponding to the
original quaternion definitions of these operations made

by
Hamilton in the 19th Century. Therefore, taking scalar and vector
parts of

each of the two equations above results in the following
system of 4

equations:

div E = -1/c (d/dt + w) e

div B = -1/c (d/dt - w) b

curl E + 1/c (d/dt + w) B =
-grad b

curl B - 1/c (d/dt - w) E = grad e

where w =
mc^2/h-bar is the angular frequency associated with the mass m.

Look familiar? When m = 0 and when psi_H consist only of
bivector

components (i.e., e = b = 0), this reduces to the Maxwell
equations for the

free field (apart from a factor of c here and
there). Therefore, Maxwell's

equations for the free field can be
written in the form:

p-bar psi_H =
0

where

psi_H = c (B1 g01 + B2 g02 + B3 g03) + (E1 g23 +
E2 g31 + E3 g12)

When m = 0 and when there are sources, e
and b are not necessarily both 0 and

one has the following
relations:

b = 0, in the absence of magnetic monopole
sources

e = - time integral (rho c/epsilon_0)

where
rho is the charge density. Therefore, one may identify the
occurrence

of e and b in psi_H in the general case (m > 0) as a
kind of potential

respectively for "electric" and
"magnetic" monopole sources.

The appearance of the
non-zero angular frequency is all that really

distinguishes the
fermion field from the combined classical (electromagnetic

field +
sources).

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