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Re: model of EM wave



Hi all-
The trouble with John's interpretation is that there are 15
independent elements in the Dirac Algebra (I will use, in his notation,
g_5=g_0g_1g_2g_3, note that (g_5)^2=-1). There are the 4 g_'s, 3
sigma_i's (that I will abbreviate s_i's), 3 g_5s_i's, 4 g_5g_i's and g_5.
The s_i's are given by is_1=g_2g_3 1,2,3 cyclic.
The s_i's exactly obey the Pauli algebra, as do the (1+ig_5)s_i's
and their counterparts with +i<-> -i. The respective members of the
latter 2 Pauli algebras commute.
Note that any ordinary product of anti-commuting elements may be
written as 1/2 of a wedge product without changing anything.
John's solution can be written in terms of the independent
elements of the Dirac algebra as:
F = i(g_5s_1+s_2)sin(kz-wt+ph), which again exhibits the B-field
at right angles to the E-field. Note, now that E and B are both elements
of the even subalgebra (product of an even number of Dirac matrices) of
the Dirac algebra, putting them on similar footing.
There is a mechanical equivalent to ExB drift that perhaps
emphasizes the importance of the pseudovector nature of the B-field.
Spin a quarter on a tilted surface that is sufficiently rough so that the
coin does not slide down the surface. The quarter will "drift"
perpendicular to the direction of tilt, the direction of drift will depend
upon the direction of rotation. If you analyze the motion you will see
that the gravitational force down the slope plays the role of an Electric
force and the angular momentum plays the role of the B-field (This is a
familiar effect with gyroscope toys). The motion is at right angles to
the two directions: spin axis and gravity component.


On Tue, 3 Sep 2002, John S. Denker wrote:

Earlier I gave a solution to the Maxwell equations and said:

It's hard to imagine anything much simpler than that:
it has an electric bivector and a magnetic bivector and
an overall sinusoidal running-wave factor.

But I was wrong. There _is_ something simpler!
Simpler and more elegant.

I wrote:
F = (g_0 g_1 + g_3 g_1) sin(k z - w t + ph)

We can simplify this either by adding the two bivectors
edge-to-edge (with due regard to sign) using the method
diagrammed at
http://www.monmouth.com/~jsd/physics/gif48/add-bivectors.gif
or we can perform the equivalent mathematics, applying
the distributive law in reverse:
F = (g_0 + g_3) g_1 psi(z - t)
or, since all the vectors involved are orthogonal,
F = (g_0 + g_3) /\ g_1 psi(z - t)

That is to say: F need not be visualized as the sum of
two bivectors (electric plus magnetic) but can perfectly
well be visualized as a single bivector. One edge is in
the direction of the wave's world-line (the t+z direction,
i.e. g_0+g_3) while the other edge is in the direction of
the polarization.

It might be edificational to make a model of this by gluing
bits of cardboard (representing snapshots of F) to a meter
stick. Let the vertical axis of the room represent t. Let,
say, the north-south axis represent the direction of
polarization, and the east-west axis represent the direction
of propagation. The meter stick will be inclined 45 degrees.
Let the snapshots be equally spaced along the stick. Each
snapshot will have one edge in the t+z direction (colinear
with the stick). I suggest making them so that this edge has
constant length; the magnitude of the bivector is then set
by choosing the length of the other (x) edge appropriately,
according to the wavefunction psi(). Also draw an arc-with-
arrowhead on each card to represent the _sign_ of the bivector.

Note that the t+z direction is a null vector, lying on the
light-cone, neither timelike nor spacelike. All observers
agree that null vectors are null.

Note that this object is not chiral. Linearly-polarized
EM waves are not chiral.

Finally note that the wavefunction psi() can be any
differentiable function of (z-t) ... certainly not restricted
to simple sine waves. Homework: verify this by plugging
into the free-space Maxwell equation (del F = 0) and turning
the crank. (The calculation fits on a post-it note.)


--
"What did Barrow's lectures contain? Bourbaki writes with some
scorn that in his book in a hundred pages of the text there are about 180
drawings. (Concerning Bourbaki's books it can be said that in a thousand
pages there is not one drawing, and it is not at all clear which is
worse.)"
V. I. Arnol'd in
Huygens & Barrow, Newton & Hooke