quick, simple question about contraviant and covariant components


by enfield
Tags: components, contraviant, covariant, simple
enfield
enfield is offline
#1
Apr9-12, 04:37 PM
P: 22
Can the covariant components of a vector, v, be thought of as v multiplied by a matrix of linearly independent vectors that span the vector space, and the contravariant components of the same vector, v, the vector v multiplied by the *inverse* of that same matrix?

thinking about it like that makes it easy to see why the covariant and contravariant components are equal when the basis is the normalized mutually orthogonal one, for example, because then the matrix is just the identity one, which is its own inverse.

that's what the definitions i read seem to imply.

Thanks!
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Mentz114
Mentz114 is offline
#2
Apr9-12, 08:40 PM
PF Gold
P: 4,081
The contravariant and covariant components of a vector are linear combinations of each other, but the transformation is performed by the metric and the inverse metric. It is a change of basis between the tangent and cotangent spaces.

The zero'th component of Vμ is given by

V0= g0aVa = g00V0+g01V1+g02V2+g03V3

If the metric is the identity matrix then the components are the same.


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