Re: contravariant and covariant objects

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

        Without going into boring detail, and referring again to Morse and
Feshbach (and a plethora of similar texts, no doubt), it's not the
topology (whether the space is curved or not) that makes the difference,
but the coordinate system that counts.  In curvilinear coordinates there
are different options for defining the metric tensor (I'm talking about
ordinary 3-d Euclidean space) some lead to metrics where the covariant
vectors and tensors differ from the contravariant ones.
        The difference can be important in inohomogeneous media where the
stress and strain tensors cannot be reduced to scalars.
                                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