Re: vector products

[Larry Smith <larry.smith@SNOW.EDU>]

At 5:34 PM -0400 8/26/02, John S. Denker wrote:
> Chuck Britton wrote [off list]:
> > Tina isn't the only one who is totally unfamiliar with this
> > Clifford Algebra thing.
>
> I had never heard of it until about a month ago.
> (There's no zealot like a convert :-)

I found John's evangelism intriguing so I downloaded his paper and then
some of the references, and one thing led to another...
I'd never heard of Clifford Algebra before that either, but now I have a
3-ring binder full of those papers I printed out from the web and I'm
studying them with interest.

I do recommend that others look at it too.   The claims the advocates make
for it are pretty wild, but I haven't found any place yet where they
haven't delivered.

> > Does this imply that the 'Right Hand Rule' doesn't have a place in
> > Clifford's algebra?
>
> Correct.  The RH rule is not needed.  The question just
> doesn't come up -- which is a good thing IMHO.
>
> > Surely this new math recognizes the difference between left and
> > right-handed coordinate systems.
>
> Not really.

Now, John's into this deeper than I am and smarter to start with, so I
really hesitate to contradict him, but the papers I read seem to explicitly
assume a right-handed coordinate system.

For example, page 5 of the Gull, Lasenby & Doran article "Imaginary Numbers
are not Real" says, "taking two orthonormal basis vectors \sigma_1 and
\sigma_2....we assume that \sigma_1, \sigma_2 are chosen such that this has
the conventional right-handed orientation."

And in "Handout 1 An Introduction to Geometric Algebra" it says, "...this
is the highest grade element in the algebra, which is often called a
pseudoscalar.... The pseudoscalar is defined to be right-handed, so that
e_1 sweeps onto e_2 in a right-handed sense."

And in Hestenes' Oersted lecture he says (page 14): "Unit pseudoscalar i
represents an oriented unit volume.  The volume is said to be righthanded,
because i can be generated from a righthanded vector basis by the ordered
product \sigma_1\sigma_2\sigma_3 = i."


At 9:46 AM -0400 8/25/02, John S. Denker wrote:
> Note that in the first chapters of the book, it is not
> necessary to introduce the full Clifford Algebra formalism.
> Instead, just go through and replace every instance of
> v cross r with v wedge r.  It works the same.  It is
> anti-commutative just like the cross product
>        v wedge r = - r wedge v
> and its magnitude is |v| |r| sin(theta).
>
> In these early chapters, only the diagram changes:  v wedge r
> is diagrammed as an area in the plane, not as a vector that
> sticks up perpendicular to the plane.
>
> In later chapters, we can trot out the full Geometric Product
>        A B = A dot B + A wedge B
> for applications such as rotation operators and for allowing
> an astonishing simplification of the Maxwell equations.

I'm quite interested in exploring this question of introducing Cliffor
Algebra at the earliest (freshman) levels.  Most of the papers I'm reading
are trying to convince other physicists to switch to Clifford Algebra so
the examples are junior-undergrad level to grad-school level.  John, how
would you like to write a short piece about how it should be introduced in
first-year calc-based physics--maybe even write the section to be included
in Bob Beichner's new book.

And how about in the math curriculum?  I'm teaching Calculus III
(multivariable) this semester and we hit cross products pretty soon; should
I teach it the "right" way instead?

Cheers,
Larry