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Re: [Phys-l] bound vectors ... or not

On 09/05/2010 09:59 PM, Stefan Jeglinski wrote:
I'm confused about this discussion too...

2) I think we all agree that in many situations when dealing
with a force, we need the idea of direction-and-magnitude of
the force and also the idea of point-of-application .....

Agreed. And to your point about viewing the angular momentum L as
(d/dt) of the torque T, I would expect what follows below to apply to
both L = r x p and T = r x F.

:-)

I [either sheepishly or sadly, not sure which] never heard of a
bound vector until this discussion.

FWIW, to the best of my recollection, I have never encountered
the term before, either. (However that's not worth much because
I have a terrible memory.)

I have encountered vague inchoate versions of the idea, but
heretofore I have just dismissed them as misconceptions. I
was quite surprised to see an attempt to formalize the idea
of bound vector.

The question is simply whether we want to express those two
things using one "vector" (i.e. a so-called bound vector)
or using two vectors (i.e. plain old vectors, aka free
vectors).

Eh? Based on the wiki article you cited, a bound vector "possesses a
definite initial point and terminal point." By this simple reading,
both r and F qualify.

I disagree. This seems to be exactly the fallacy I alluded
to a moment ago.

As I see it, a vector has, by definition, a direction and
a magnitude. A vector in 3-space can be specified by
three numbers. If some funny kind of vector actually
had a definite starting-point and a definite ending-point,
we would need six numbers to specify it.

As I see it, the vector from A to B depends on the difference
between A and B, but does not otherwise depend on A and B.
In particular, the vector from A to B is the _same vector_
as the vector from A+K to B+K for any K whatesoever.

FWIW this is an example of gauge invariance. K is the
gauge. The vector (B-A) is gauge invariant even if A
and B, individually, are not.

To say the same thing using pictures, in the following
diagram F1 and F2 portray the _same vector_ because they
have the same direction and magnitude:

--------> F1

--------> F2

As I see it, F1 and F2 acting on a body provide the same
amount of force, and if you integrate over time they impart
the same amount of momentum. They will impart a different
amount of _angular momentum_ but you don't know that based
on the force vectors alone; for each force vector you need
another vector, namely a lever arm, before you can calculate
the torque and angular momentum. Torque is a bivector.
Angular momentum is a bivector.

Let's be clear: the point-of-application is important, but
IMHO it is distinct from the force; it is not part of the
force. Exercise is important and sleep is important, but
you don't sleep while you are exercising.

I [either sheepishly or sadly, not sure which] never heard of a
bound vector until this discussion. Does it satisfy someone's need to
formalize this stuff to the nth degree, or does it serve a useful
purpose?

That's an important question.

This time of year there is much classroom discussion of F=ma
and Newton's second law. If we cannot even agree on what sort
of thing is represented by F in this equation, we are in some
kind of trouble. I'm trying to figure out what kind of trouble.

We have two choices. In the equation F=ma, either:
a) F represents direction and magnitude only, or
b) F represents direction and magnitude and point of application.

Figuring this out is IMHO the very first degree of formalization,
not the nth degree.

It seems to me that option (a) leads to small trouble, whereas
option (b) leads to big trouble.
-- The problem with (a) is that it means that the version of
the laws of motion that we first learn is incomplete. To
the extent that F=ma is an equation involving free vectors,
it says nothing about the point of application, so we need
... eventually ... to have another equation before we can
claim to have a complete set of laws.
-- The problem with (b) is that to the extent that F=ma is an
equation involve "bound vectors" I don't know how to deal with
it. The mathematical tools that I have for dealing with vectors
deal with free vectors only. I hope nobody is suggesting that
the point-of-application of the force (F) is equal to mass times
the point-of-application of the acceleration (a).

I emphasize that option (a) is OK with me. I can deal with an
idea that is incomplete but easily fixable. In contrast, I can't
deal with a situation where F is some weird chimera and the F=ma
equation applies to part of F but not all of F.

I hope everybody realizes I'm making a very weak argument here.
Just because I don't know how to do it doesn't mean it cannot
be done. So I would like to leave it as a question:
-- Does anybody know how to make sense of bound vectors
(other than by splitting them into two regular vectors,
i.e. free vectors)?
-- Or does anybody have a more convincing argument that
bound vectors are just a bad idea and deserve to be
ignored the way the physics and math literature have
been ignoring them?

My reading of the math and physics literature going back 50+
years is that vector means free vector exclusively, so that
a so-called "bound vector" is not really a vector at all, but
rather a pair of vectors, like two persons inside a horse
costume.

Are you saying that you would view r x F as one bound vector
consisting of a pair of free vectors r and F?

Yes, that's the idea. I visualize r /\ F as a bivector, i.e.
as a little patch of directed area.

I still have grave doubts about _calling_ r /\ F a "bound
vector" but I would much rather discuss ideas than fuss over
mere terminology.

===============

Also I think our friends the mathematicians should be given
a say in this discussion. Under the heading of "linear algebra"
there is a quite well-established clear-cut axiomatic definition
of vector space, and the vectors in such a space are free vectors.

Just so you know where I'm coming from, I knew about linear algebra
before I started high school physics. And when I did take HS
physics, the text was _PSSC Physics_, which was written by some
very sophisticated guys ... so for me, using a mathematically
sophisticated definition of vector was as natural as breathing.


================================
On 09/06/2010 12:17 AM, John Mallinckrodt wrote:
I don't see any advantage, much less any necessity, but I *do* see
much potential for needless confusion in trying to distinguish
between "bound" and "free" vectors. Vectors have magnitude and
direction. They do not have "location." The standard "position
vector" locates a specific point in space relative to some origin by
having a magnitude and a direction such that, *if* its tail is
positioned at the origin, its tip will be at the desired point.

That's pretty much where I'm coming from.

Am I missing something?

Not that I can see ... but let's leave the question open
for a while longer.