Re: transformation properties of vectors/covectors
- From: The TimeLord <math-n-physics-not@xxxxxxx>
- Date: Mon, 22 Sep 2008 13:11:44 -0500
Am Sun, 21 Sep 2008 07:00:36 -0700 schrieb Edward Green
<spamspamspam3@xxxxxxxxxxx> in
be7b096d-dc23-4683-83ab-43246f415d96@xxxxxxxxxxxxxxxxxxxxxxxxxxxx:
For components of vectors, v' ^ n, v ^ u
v' ^ n = (@x' ^ n / @x ^ u) v ^ u
which says roughly that in order to track the variation of a function
wrt x' ^ n, we must track its variation over all the x ^ u, multiplied
by the variation of x' ^ n wrt x ^ u, and sum.
Maybe not obviously rigorously correct, but at least plausible.
The law for covectors does not seem so clear:
w' _ n = (@x ^ u/@x' ^ n) w _ u
which says roughly that to track the number of level surfaces pierced by
a vector @/@x' ^ n, we must track the surfaces pierced by each vector
@/@x ^ u, multiplied by the rate of growth of the underlying variable
wrt the target variable, and sum.
Still plausible, but considerably _less_ intuitively obviously correct!
And why the asymmetry in the indices on the variables (all superscripts,
although the indices on the components vary from superscripts to
subscripts)?
Helpful comments?
Actually it's a bit more simple than that. Think of the chain rule in
calculus. Let v be a vector and E be a scalar then (using your notation)
dv = @v/@x1*dx1 + @v/@x2*dx2 + ...
Realizing the fact that v has components, the contravariant transformation
of v falls out as a form of @v/@x. So any tranformation that maps vectors
that are not induced from scalars to vectors is contravariant.
For scalar E we have something like this
@E/@q = @E/@x1*@x1/@q + @E/@x2*@x2/@q + ...
q here can be any generalized coordinate. We recognize @E/@q as Force
whenever a potential energy exists. The @x/@q is the transformation.
So the covariant transformation rule maps vectors that are induced from
scalars to vectors.
The overriding thing is that vectors that transform contravariantly must
have their space specified separately from the vector, whereas covariant
transformations are on vectors that don't need any special information
about the space they are in. You can see this when you multiply and
contract a contravariant vector with the metric, since the resulting
covariant vector will have the space info imbedded in it (it was a
product of the metric).
This is also the reason you can't apply F=m*a to General Relativity.
F is a covariant vector and a is a contravariant vector. m is a scalar
(which by definition does not change under transformation). Even in
Special Relativity F=m*a can be problematic even though the transformation
is affine (because the moving frame is inertial), F is parallel to a and
so the tensorial differences aren't noticed in SR. In GR F is rarely
parallel to a and since m is a scalar, the equation makes no sense in
GR.
--
// The TimeLord says:
// Pogo 2.0 = We have met the aliens, and they are us!
.
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