Straight Dope Message Board > Main Contravariant/covariant basis question
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#1
05-06-2002, 08:26 PM
 Ring Charter Member Join Date: Dec 2000 Location: South Carolina USA Posts: 1,561
Contravariant/covariant basis question

This isn't a question that will be of interest to most science buffs here so I apologize if you wasted your time clicking on it. I'm hoping a mathematician or maybe Chronos will know the answer.

http://home.pacbell.net/bbowen/covariant.htm

The above site says that given an Euclidean orthonormally represented vector

V = < 5 , 12 > you can define two new basis sets:

A1 = < .5 , 0 >
A2 = < .75 , .5 >

And

A1 = < .2 , -3 >
A2 = < .0 , 2 >

He then says that the first set is contravariant and the second set is covariant.

How can you have a covariant or contravariant basis set for fixed vector? And if you can, then what is that that makes them covariant and contravariant?
#2
05-06-2002, 09:32 PM
 ZenBeam Charter Member Join Date: Oct 1999 Location: I'm right here! Posts: 7,725
Isn't it the using of subscripts versus superscripts all that makes the two basis sets covariant vs contravariant here? He could as easily have written
A1 = <2, -3>
A2 = <0, 2>
and called that the contravariant basis. (Well, that's kind of backwards; subscript vs superscript is determined by whether it's contravariant or covariant, not the other way around). I think maybe you're looking for something too deep for this simple example.
#3
05-06-2002, 09:35 PM
 LifeWillFall Registered User Join Date: Apr 2000
I'm not exactly sure what your question is, but it seems that the author is just saying that you can represent the vector <5,12> with any number of different coordinate systems depending on the basis you choose. if you choose your basis to be the standard basis which will yeild the 2x2 (in R^2) identity matirx. but if you choose any 2 independent vectors you will be able to refer to the same vector with a vector that looks different than <5,12> but is really just a differnet way of writing the same thing. I doubt I answered your question, but maybe I'll get the discussion started

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