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At 12:37 PM -0400 8/27/02, John S. Denker wrote:
2) Angular momentum = velocity cross position
This is better, because the rotation invariance is now
manifest. The law is written without mentioning anybody's
coordinate system.
However, the physics has a left/right symmetry that is
not manifest. You can't draw the cross product without
invoking a Right-Hand rule. But I emphasize that the
physics is still left/right symmetric!!!! The law is
just written in a misleading form.
3) Angular momentum = velocity wedge position
Of course the wedge product is manifestly rotationally
invariant, so we're OK in that department, too.
======
To summarize:
++ writing a cross product or wedge product is better
than writing out the components, because the product
is manifestly rotationally invariant, whereas the
components are not.
-- the cross product doesn't exist in D=2
-- the cross product conceals the left/right invariance
of the real physics in D=3
-- the cross product is almost never what you want
in D=1+3 spacetime.
++ the wedge product works just fine in D=2, D=3, and
D=1+3. It looks the same and means the same in all cases.
++ The wedge product allows a left/right symmetric law
to be written in a way that looks left/right symmetric.
=======================
I decided to see if I once knew about these wedge products previously
(teaching at a 2-yr college means I tend to forget a lot of stuff I did in
upper division and grad school). Turns out that I found this wedge product
thing in my freshman "Vector Calculus" book by Marsden and Tromba, but it
isn't in relation to a Geometric Algebra or a Clifford Algebra (neither of
which are mentioned); it is in the context of differential forms. But it
sure looks like the differential form wedge product has a lot of the
properties of the Clifford stuff John is introducing to us now.
It looks like dx, dy, and dz are like the vectors in the basis (my book
calls them 1-forms); and dxdy, dydz, etc., are bivectors (my book calls
them 2-forms); and it says "a basic 3-form is a formal expression
dxdydz.... There seems to be little difference between a 0-form and a
3-form since both involve a single real-valued function. But we
distinguish them for a purpose that will become clear when we multiply and
differentiate them." The differential form wedge product is distributive,
associative, and anticommutative.
Is this differential form wedge product the same thing as the Geometric
Algebra wedge product?
I sure do wish I could remember all the stuff I learned in school. My
major professor told me to "never forget anything" and I only wish I'd
complied.
Larry