On 7 Oct 2003 08:19:48 -0700, jim rozen wrote:
In article , Gary Coffman says...
But back to the original question, why is mechanical work
calculated with the dot product w = F dot D instead of the
cross product w = F X D?

Because it's only the component of the work in the
direction of the displacement that adds to the energy
of the particle.

The dot product is just the magnitute of each vector
times the cosine of the angle between them. For
example, if you push down really hard on a roller
skate, but it only moves slowly along the floor, it's
only the component of your hand's force *parallel*
to the floor that is actually pushing the skate to
speed it up. There the value of the dot product is
small as it should be.

Yes! In 3 space, it would be over each of the three
vector components, X, Y, and Z. But we can simplify
that by contriving the force to be orthogonal to two
of those dimensions by choosing an appropriate
reference frame, leaving us with a single cosine
factor.

OK, I managed to get a copy of halliday and resnick and
looking it over, it looks like they use the term "speed"
when calculating 1/2 mv(squared), not the vector velocity.

I don't think so, see below.

They proceed to derive an example of the "work - energy
theorem" for the case of a constant force acting on an
object. But they don't use vector velocity, but rather
the scalar speed instead. So where I put Vs below, they're
scalars, not vectors.

begin quote:
======================================

The simplest situation to consider is that of a constant
force F(vector). Such a force F(vector) will produce a
constant acceleration a(vector). Choose the x-axis to be
the common direction of F(vector) and a(vector). What is the
work done by this force on the particle in causing a
displacement x? For constant acceleration, the relations are

a= (V - Vo)/t

For 'a' to be a vector, as you correctly say it is, then you need
at least one vector on the opposite side of the equation, if you
are to obey the rules of vector math. That can't be 't', but it can
be the difference between *vectors* V and Vo. So they can't
correctly be treating V and Vo as *speeds* here.

and

x= (t) X (V + Vo)/2

where Vo is the particle's *speed* [emphasis mine] at
t = 0 and V is its speed at time t.

Ok, I don't understand this. If your 'X' is supposed to be
the cross product, that doesn't make sense, because 't'
isn't a vector. If it is meant to be '*' (simple multiplication),
then the equation makes sense, but 'x' is total displacement
*only if both Vo and V are in the same direction*. The math
can't enforce that if they are treated as scalar speeds. Only
if they are vectors, can it require that by resolving their
components.

Then the work done is:

W = (F)(x) = (ma) (x) [newton's law]

W = (m) [(V - Vo)/t] [(V + Vo)/2] (t)

W = 1/2 mV(sq) - 1/2 mVo(sq)

We call one half the product of the mass of a body and the
square of its *speed* [again, emphasis mine] the kinetic
energy of the body. If we represent kinetic energy by the
symbol K then

K = 1/2 mV(sq)

The work done by the result force acting on a particle is
equal to the change in kinetic energy of the particle.

Although this is proved for a constant force only, it holds
whether the resultant force [again, the dot product between
F(vector) and D(vector)] is constant or variable.

================================================= ============
(end quote)

They go on to give some examples, where the force is not
in the direction of accelleration (circular motion, for
example, where cos(theta) is equal to zero) and where the
force varies in magnitude as well.

So basically according to H&R I was wrong in saying that
work is proportional to the dot product of two vectors,
it's not. It's proportional to the square of the *speed*
which is not a vector.

I don't think that they've showed that at all, except in
special contrived cases. Feynman says that work is
invariant over all reference frames for all forces and
displacements, not just for the ones that conveniently
line up. The only way that can be true is if work is the
dot product of F and D.

Dot products have the vector math property that they
are invariant over all reference frames. That's not true
for cross products.

The reason Feynman says work is invariant is, if it
were not true, then you could contrive a perpetual
motion machine from which you could eternally extract
net energy by simply changing reference frames. The
conservation of energy law forbids that.

(This isn't a terribly obvious point, it takes Feynman
three chapters to work up to it. But it yields a result
which is rigorously correct in both the vector math
sense and the physics sense.)

I've been trying to get people to discover this by
using the Socratic method, but as is often the case
here, that hasn't worked very well. So now I've just
spelled it out.

Gary