Vectors Question

[A5...@cunyvm.bitnet]

I'm reading Principles of Electrodynamics by Schwartz. In it he discusses
two types of vectors, polar (where the components negate under pure
reflective transformation) and axial (where the components are invariant
for pure reflective transformations).

Either my thinking is wrong or he contradicts himself :

Let us set c = a cross b. In my mind, if a and b are both polar or they are
both axial then c is axial. I do this by picturing reflections performed on
the vectors, and I am fairly certain of this result. If we try to cross two
vectors of different types, then the c vectors before and after the
reflection (c and c') differ in magnitude as well as direction and so
I feel that it's impossible to cross different type vectors. Be that as it
it may you won't get a polar vector by crossing a polar with and axial
(Unless my mental pictures are totally wrong.)

From Coulomb's law he gets that force is a polar vector. Fine.

Now from magnetism we know that force F = qv cross B. Velocity v is polar.
He's already assumed that charge q is scalar (totally invariant). From all
this he thinks this tells us that B is axial!

So my question is, how is he crossing a polar with an axial and getting a
polar vector??
-------
CHARLES HOPE A54SI@CUNYVM A5...@CUNYVM.CUNY.EDU
GOVERNMENT BY REPORTERS...MEDIA-OCRACY.

[Daniel E. Platt]

In article <93025.22...@CUNYVM.BITNET>, <A5...@CUNYVM.BITNET> writes:
|> I'm reading Principles of Electrodynamics by Schwartz. In it he discusses
|> two types of vectors, polar (where the components negate under pure
|> reflective transformation) and axial (where the components are invariant
|> for pure reflective transformations).
|>
|> Either my thinking is wrong or he contradicts himself :
|>
|> Let us set c = a cross b. In my mind, if a and b are both polar or they are
|> both axial then c is axial. I do this by picturing reflections performed on
|> the vectors, and I am fairly certain of this result. If we try to cross two
|> vectors of different types, then the c vectors before and after the
|> reflection (c and c') differ in magnitude as well as direction and so
|> I feel that it's impossible to cross different type vectors. Be that as it
|> it may you won't get a polar vector by crossing a polar with and axial
|> (Unless my mental pictures are totally wrong.)
|>
|> From Coulomb's law he gets that force is a polar vector. Fine.
|>
|> Now from magnetism we know that force F = qv cross B. Velocity v is polar.
|> He's already assumed that charge q is scalar (totally invariant). From all
|> this he thinks this tells us that B is axial!

Jackson calls polar vectors 'vectors' and axial vectors 'pseudovectors.' He
identifies location, velocity, momentum, force, electric field, etc as vectors
(polar), and angular momentum and magnetic field as pseudovectors (axial).

You should not think of the transformation as a reflection about some plane.
It is a transformation x -> x' = -x for all components. For functions of
location, it goes like B -> B'(x, t) = +B(-x, t), E -> E'(x, t) = -E(-x, t).

Cross products of even with even and odd with odd are even (Angular momentum
is axial), while odd with even are odd (polar). F = v X B with F and v odd
implies B must be even.

Dan

--
-------------------------------------------------------------------------------
Daniel E. Platt pl...@watson.ibm.com
The views expressed here do not necessarily reflect those of my employer.
-------------------------------------------------------------------------------

[Matt McIrvin]

<A5...@CUNYVM.BITNET> writes:

> Let us set c = a cross b. In my mind, if a and b are both polar or they are
> both axial then c is axial. I do this by picturing reflections performed on
> the vectors, and I am fairly certain of this result. If we try to cross two
> vectors of different types, then the c vectors before and after the
> reflection (c and c') differ in magnitude as well as direction and so
> I feel that it's impossible to cross different type vectors. Be that as it
> it may you won't get a polar vector by crossing a polar with and axial
> (Unless my mental pictures are totally wrong.)

I think you will, actually. The terms axial and polar don't refer to
mirror reflections, exactly; they refer to how the vector transforms under
the transformation x -> -x, where x is a position vector. In general
that's equivalent to a mirror reflection combined with some rotation; you
can think of it as mirror reflection through *all three* coordinate planes.
Under this space inversion a polar vector reverses direction and an
axial vector doesn't.

Now suppose a is axial and b is polar. The product a x b points in
a direction given by the right hand rule, of course, and its magnitude
is given by |a||b| sin theta where theta is the angle between the two
vectors. a is unchanged and b goes to -b. This leaves theta unchanged,
as well as |a| and |b|; all that changes is the direction given by the
right-hand rule. Therefore the product reverses direction and is a
polar vector.
--
Matt McIrvin

[A5...@cunyvm.bitnet]

Now I see where I was wrong. (-a) X b = -(a X b). I was thinking they were
actually differing in magnitude but no. I am grateful for all the responses.

[John C. Baez]

Matt McIrvin writes:
><A5...@CUNYVM.BITNET> writes:

>The terms axial and polar don't refer to
>mirror reflections, exactly; they refer to how the vector transforms under
>the transformation x -> -x, where x is a position vector. In general
>that's equivalent to a mirror reflection combined with some rotation; you
>can think of it as mirror reflection through *all three* coordinate planes.
>Under this space inversion a polar vector reverses direction and an
>axial vector doesn't.

While we're at it, we might as well say that this transformation x -> -x
is known in physics as "parity." Also that the fact that the cross
product of two polar vectors is axial is related to the fact that the
wedge product of two one-forms is a two-form. And also that a fairly
general sort of tensor field (built up from vector fields and covector
fields, aka one-forms) is describe by a pair of nonnegative integers
(p,q) saying how many co- and contravariant indices there are (if one
condescends to write indices), *and* a sign saying how the tensor
transforms under parity, *and* a real number w saying how it transforms
under scaling. w is called the conformal weight, and if it is different
than one would naively expect one says that one has a "tensor density."
So axial and polar vectors also come with any conformal weight you want.
There are more flavors of vector fields than they teach you in basic
vector calculus!!

[Michael Weiss]

John Baez brings up the subject of parity in his post on axial and polar
vectors. As nearly everyone reading this newgroup doubtless knows, parity
(P) is violated by weak interactions; this was discovered in 1957,
following suggestions put forward the year before by T.D. Lee and C.N.
Yang. Yang and Lee promptly became the youngest recipients of the Nobel
Prize in physics for their work. Failure of time-reversal (T) invariance
was discovered in 1964.

Some may find it interesting that both failures were anticipated by
Dirac, as I discovered in reading Pais' "Inward Bound". Pais quotes
himself from a paper published in 1968:

I asked Dirac why he had not introduced parity in his famous book on
quantum mechanics. With characteristic simplicity Dirac answered:
"Because I did not believe in it", and he showed me a paper in the
Reviews of Modern Physics [vol 21, p. 393, 1949] in which he said so.
In the same paper, Dirac expressed doubts about time reversal
invariance as well, but, four years ago I did not pay much
attention to that. Unjustifiedly...