Suppose we want to paint a parallelogram. How much paint
do we need?
The edges of the parallelogram are the vectors A and B.
We're pretty sure
-- the amount of paint is a scalar,
-- it is a symmetric function of A and B, and
-- it is positive.
In Geometric Algebra, we are told that the wedge product
A /\ B represents area. But it's not directly the answer
to the paint problem. In particular
-- A /\ B is not a scalar; it's some higher-dimensional
geometric abstraction.
-- it is antisymmetric: A /\ B = - B /\ A
-- we can't even define a ">" operator for bivectors, so
we certainly can't say A /\ B will be ">" zero.
The answer is that we should be asking about the _norm_ of
A /\ B, written ||A /\ B||.
It is easy to show that
||A /\ B|| = ||A|| ||B|| |sin(theta)|
where theta is the angle between the vectors. This is the
correct answer to the paint problem. Let's see where this
result comes from.
For any multivector M, the square of the norm is given by
||M||^2 := M M~
where M~ is the reverse of M, formed by writing the vectors
that make up M in reverse order. In particular
(A /\ B)~ = (B /\ A)
And for a simple (grade=1) vector
A~ = A
A A~ = A A = A.A = ||A||^2
We can write
A = ||A|| a'
where a' is a unit vector in the direction of A.
We can write some other vector B as
B = ||B|| (a' cos(theta) + a" sin(theta))
where a" is some unit vector perpendicular to A.
Then just take these expressions for A and B, plug
them into the definition of ||A /\ B|| and turn the
crank. The result follows immediately:
||A /\ B|| = ||A|| ||B|| |sin(theta)|
This is a good homework exercise for building confidence
in the formalism.
==========================
Let's proceed to an even more interesting problem.
Suppose we have a parallelepiped with edges A, B and C,
and we want to calculate the volume.
The old-fashioned way to do this would be to use the
"triple scalar product" A dot B cross C. But cross
products are bad news, and we would be much better off
using the Geometric Algebra formulation. The desired
expression is
volume = ||A /\ B /\ C||
Note that even though we have been using the wedge product
to get rid of cross products, it is not a one-for-one
replacement. You cannot blindly replace A dot B cross C
by A dot B wedge C. The triple scalar product is properly
written ||A /\ B /\ C|| with two wedge products and no dot
products. This has a nice geometric interpretation: A /\ B
is visualized as sweeping A in the direction of B, using
it as a brush to sweep out the area A /\ B. Similarly
A /\ B /\ C is visualized as sweeping the area A /\ B in
the direction of C, using it to sweep out the volume
A /\ B /\ C.
We can verify that the magnitude of the volume behaves as
advertised, using the same procedure as before:
A = ||A|| a'
B = ||B|| (a' cos(theta) + a" sin(theta))
C = ||C|| (ab' cos(phi) + ab" sin(phi))
where ab' is any unit vector in the AB plane, and ab" is
a unit vector perpendicular to the AB plane. The result
follows immediately:
||A /\ B /\ C|| = ||A|| ||B|| ||C|| |sin(theta)| |sin(phi)|
which makes sense.
==========
It's amusing that these results can be obtained without
establishing any basis vectors and without expanding
A, B, and C in terms of components. All you need are
the basic axiomatic properties
A B = B A if A and B are colinear
A B = - B A if A and B are perpendicular
A . B = (A B + B A)/2
A /\ B = (A B - B A)/2
assuming A and B are plain old grade=1 vectors.
This note is a continuation of the campaign to stomp out
cross products.
The traditional uses of cross products include
-- torque and angular momentum, as applied to e.g.
gyroscopic precession. This can be handled by
wedge products, as depicted at http://www.monmouth.com/~jsd/physics/gif48/add-bivectors.gif
where the green bivector is the initial angular momentum,
the small purple bivector is torque*time, and the yellow
bivector is the new angular momentum, which is in a new
orientation due to precession.
-- calculating volume using the triple scalar product,
which can be handled using wedge products, as
discussed above
-- the curl of a vector field, which can be handled
spectacularly well using Geometric Algebra, as
discussed at http://www.monmouth.com/~jsd/physics/maxwell-ga.htm
So it really does look like all the traditional applications
of the cross product can be handled better (sometimes much
better) using wedge products.