Re: contravariant and covariant objects

[Jack Uretsky <jlu@HEP.ANL.GOV>]

        It's not the perpendicularity that counts.  It's the
curvilinearity.  I think that all authors assume that their axes are
perpendicular at the points of intersection.
        A good example is my very clumsy description of acceleration in
spherical coordinates in my posting of the Kepler problem (I'll soon clean
up the Gibbsian version of that demo to avoid dealing with coordinates).
        It's too clumsy to explain in a text posting (there are others
who, I am sure will try), but working out a few simple problems, as at the
end of Morse & Feshbach Ch. 1 should clarify the issue.
                                        Regards,
                                                Jack

On Sun, 22 Sep 2002, Joe Heafner wrote:

> > From: "John S. Denker" <jsd@MONMOUTH.COM>
> >
> > For vectors in ordinary non-curved Euclidean space, there
> > is no distinction between covariant and contravariant
> > vectors, so the issue isn't usually mentioned when
> > vectors are first introduced.
> This is one thing that confuses me. In a non-curved non-orthogonal Euclidean coordinate system (I have in mind a two dimensional coordinate system in which the x and y axes are not perpendicular), the term *covariant* components refers to the components that are perpendicular to the axes (the covariant x component is perpendicular to the y axis and the covariant y component is perpendicular to the x axis). The term *contravariant* components refers to the components that are parallel to the axes (the contravariant x component is parallel to the x axis and the contravariant y component is parallel to the y axis). I quote from the web site <http://www.mathpages.com/rr/s5-02/5-02.htm>:
>
> "Figure 1 shows an arbitrary coordinate system, with the axes X1 and X2, as well as the contravariant and covariant components of the position vector P with respect to these coordinates. As can be seen, the jth component of the "contravariant path" from O to P consists of a segment parallel to jth coordinate axis, whereas the jth component of the "covariant path" consists of a segment perpendicular to all the axes other than the jth. This is the essential distinction between the contravariant and covariant ways of expressing a vector (or, more generally, a tensor)."
>
> This is completely understandable to me, and I totally see how the contravariant and covariant components become equivalent if the coordinate axes become orthogonal. However, Weinreich (p. 7) emphasizes that there is also a contravariant/covariant distinction in a change of scale of the units along the axes. To me, this implies that even in an orthogonal coordinate system, there is still a visual distinction between the two. Since Weinreich uses electric field as a prototype, I'm lead to assume that {\vec E} is always a contravariant vector even in an orthogonal system and this is a point of confusion for me. Weinreich (p. 8) also points out that there are other transformation properties that could be considered.
>
> > (Note the reference cites the E-field as an example of
> > a 1-form, but I'm not 100% happy with that;  I suppose
> > it's OK for electrostatics but it's not relativistically
> > correct.)
> Aha! See above.
>
>
> Cheers,
> Joe Heafner - Instructional Astronomy and Physics
> Home Page http://users.vnet.net/heafnerj/index.html
> I don't have a Lexus, but I do have a Mac. Same thing.

--
        "What did Barrow's lectures contain?  Bourbaki writes with some
scorn that in his book in a hundred pages of the text there are about 180
drawings.  (Concerning Bourbaki's books it can be said that in a thousand
pages there is not one drawing, and it is not at all clear which is
worse.)"
                                        V. I. Arnol'd in
                                Huygens & Barrow, Newton & Hooke