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From: Pollux on 6 Jun 2010 23:26
I'm looking for examples of simple contravariant and covariant
quantities in physics. I understand that position is contravariant, and
a gradient covariant (please correct if I'm mistaken). Any other
quantities? How about mixed contravariant/covariant quantities? What
would be the simplest example?

Thanks,

Pollux

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From: eric gisse on 7 Jun 2010 04:30
Pollux wrote:

> I'm looking for examples of simple contravariant and covariant
> quantities in physics. I understand that position is contravariant, and
> a gradient covariant (please correct if I'm mistaken). Any other
> quantities? How about mixed contravariant/covariant quantities? What
> would be the simplest example?
>
> Thanks,
>
> Pollux
>
> --- news://freenews.netfront.net/ - complaints: news(a)netfront.net ---

Vector - covariant.
Vector in frequency domain [Fourier transform] - contravariant.
Riemann tensor R^a_bcd - mixed
T^a_a [trace of tensor] - mixed.


From: Pol Lux on 9 Jun 2010 15:08
On Jun 7, 1:30 am, eric gisse <jowr.pi.nos...(a)gmail.com> wrote:
> Pollux wrote:
> > I'm looking for examples of simple contravariant and covariant
> > quantities in physics. I understand that position is contravariant, and
> > a gradient covariant (please correct if I'm mistaken). Any other
> > quantities? How about mixed contravariant/covariant quantities? What
> > would be the simplest example?
>
> > Thanks,
>
> > Pollux
>
> > --- news://freenews.netfront.net/ - complaints: n...(a)netfront.net ---
>
> Vector - covariant.
> Vector in frequency domain [Fourier transform] - contravariant.
> Riemann tensor R^a_bcd - mixed
> T^a_a [trace of tensor] - mixed.

OK. So for a very trivial example, the slope of a mountain path, in
units of "meters per meters" would be covariant, because 10 m/m go to
10000 km/km in "kilometers per kilometers" (slope is also a gradient,
right?).

Can I ask why we bother with mixed tensors? You can use the metric to
change mixed tensors to all contravariant or all covariant tensors,
right? Is it easier to work with mixed tensors?

Pollux
From: BURT on 9 Jun 2010 15:57
On Jun 9, 12:08 pm, Pol Lux <luxp...(a)gmail.com> wrote:
> On Jun 7, 1:30 am, eric gisse <jowr.pi.nos...(a)gmail.com> wrote:
>
>
>
>
>
> > Pollux wrote:
> > > I'm looking for examples of simple contravariant and covariant
> > > quantities in physics. I understand that position is contravariant, and
> > > a gradient covariant (please correct if I'm mistaken). Any other
> > > quantities? How about mixed contravariant/covariant quantities? What
> > > would be the simplest example?
>
> > > Thanks,
>
> > > Pollux
>
> > > --- news://freenews.netfront.net/ - complaints: n...(a)netfront.net ---
>
> > Vector - covariant.
> > Vector in frequency domain [Fourier transform] - contravariant.
> > Riemann tensor R^a_bcd - mixed
> > T^a_a [trace of tensor] - mixed.
>
> OK. So for a very trivial example, the slope of a mountain path, in
> units of "meters per meters" would be covariant, because 10 m/m go to
> 10000 km/km in "kilometers per kilometers" (slope is also a gradient,
> right?).
>
> Can I ask why we bother with mixed tensors? You can use the metric to
> change mixed tensors to all contravariant or all covariant tensors,
> right? Is it easier to work with mixed tensors?
>
> Pollux- Hide quoted text -
>
> - Show quoted text -

Insnt rise and run put together as slope built of quantities that can
covary?

Mitch Raemsch
From: Rock Brentwood on 9 Jun 2010 20:53
Pollux wrote:
> I'm looking for examples of simple contravariant and covariant
> quantities in physics.

On Jun 7, 3:30 am, eric gisse <jowr.pi.nos...(a)gmail.com> wrote:
> Vector - covariant.
> Vector in frequency domain [Fourier transform] - contravariant.

Ignoring the untutored replies from people self-confident in their
ignorance, the archetypical cases of contravariant and covariant
vectors are, respectively,
Delta x = sum_{mu} Delta x^{mu} @/@x^{mu}
(coordinate differential) and
p = sum_{mu} p_{mu} dx^{mu}
(momentum).

The operators, themselves, @/@x^{mu} (using @ to denote partial
differential operator) and dx^{mu} are respectively contravariant and
covariant vectors. The "co-" is taken with respect to the coordinate
differential, which by custom are given indexes in the upper position.
So the components of the co-variant vectors have indexes in the
opposite position -- lower. They transform as coordinate differentials
and linear combinations thereof. The contravariant vectors transform
as the differential operators do. The opposing nature "contra-" is
reference to the fact that coordinate transformations respect the
invariant:
sum dx^{mu} @/@x^{mu}.
So transformations on the differentials dx^{mu} are opposite to that
on the operators @/@x^{mu}, so that the two cancel and leave the
invariant fixed.

In electromagnetism, the E and H fields were introduced by Maxwell in
invariant form (up to change in names) via the following differential
forms
E_x dx + E_y dy + E_z dz, H_x dx + H_y dy + H_z dz
These are therefore covariant 3-vectors. In contrast, the B and D
"induction" fields were introduced as 2-forms
D^x dydz + D^y dzdx + D^z dxdy, B^x dydz + B^y dzdx + B^z dxdy.
Therefore, they are contravariant, when regarded as 3-vectors.

Similarly, the potential A and current density J were introduced as
the following
A_x dx + A_y dy + A_z dz
J^x dydz + J^y dzdx + J^z dxdy
(up to change of names). So, as 3-vectors they are, respectively,
covariant and contravariant.

(Though it's not widely known, Maxwell did use Grassmann algebra for
differential forms, making actual use of the anti-commutativity rule
(e.g. dxdy = -dydx) in his 1873 treatise).

Another example comes out of the underlying geometry. In addition to
the invariant
dr.del + dt @/@t
(where I'm using vector notation dr = (dx,dy,dz) and del = (@/@x, @/
@y, @/@z)), you have the following quadratic invariants:
dt^2 - alpha dr^2, del^2 - alpha (@/@t)^2
(where, here, product denotes tensor product, not Grassmann product).

The value of alpha determines the nature of the geometry. For non-
relativistic theory, alpha = 0, and for relativistic theory, one has
alpha > 0 (i.e. alpha = (1/c)^2).

Closely linked to these invariants are the covariant metric
dt^2 - alpha dr^2 = g_{mu nu} dx^{mu} dx^{nu}
and the contravariant metric
del^2 - alpha (@/@t)^2 = g^{mu nu} @/@x^{mu} @/@x^{nu}.

In non-relativistic theory, the two metrics are necessarily separate
and independent. In contrast, in relativistic theory, both metrics
give us non-singular quadratic forms can are equal to one another's
inverse up to a factor. So, by convention, one of the metrics is
normally rescaled to the other's inverse and the two are treated as
one. The two conventions in force, roughly speaking, are the Quantum
Theorists', which takes the covariant metric and rescales
contravariant metric as
(@/@t)^2 - (1/alpha) del^2 = (@/@t)^2 - c^2 del^2
while the other is more commonly found with General Relativists:
taking the contravariant metric and rescaling the covariant metric as
dr^2 - (1/alpha) dt^2 = dr^2 - c^2 dt^2.
(Both metrics may, in turn, be subject to further rescaling and one
normally also adopts the convention c = 1).

Another example is the stress tensor. In the defining context -- that
associated with the action of diffeomorphism on a field -- the stress
tensor comprises the coefficients of the integral expression
T_{Delta x} = integral sum_{mu nu} Delta x^{mu} T_{mu}^{nu}
(dS)_{nu}
suitable for integration over 3 dimensional regions of space-time. (In
3-D statics, the corresponding expression would be a 2-D integral).
This gives the effect of a transformation on the field by a vector
Delta x = sum_{mu} Delta x^{mu} @/@x^{mu}.

Thus, the stress tensor is mixed covariant and contravariant T_{mu}
^{nu}. In contrast, in gravitational models based on the metric
tensor, one sees a stress tensor defined by the derivative of the
field Lagrangian with respect to the metric. The metric taken is the
contravariant metric g^{mu nu}, and the derivatives are thus covariant
T_{mu nu}.
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