So I'm looking at Schaum's outlines for Tensors and the definition of a Contravariant vector is(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

\bar{T}^i=T^r\frac{\partial\bar{x}^i}{\partial x^r}

[/tex]

Where [itex]\bar{x}^i[/itex] and [itex]x^r[/itex] denote components of 2 different coordinates (the superscript does not mean 'to the power of') and [itex]T^i[/itex] and [itex]T^r[/itex] are contravariant tensors of order 1 (aka, a vector).

Lets say you have some 2-D vector [itex]{\bf v}[/itex]. It can be described as

[tex]

{\bf v}=\bar{T}^1\hat{\bar{e}}_1+\bar{T}^2\hat{\bar{e}}_2=T^1\hat{e}_1+T^2 \hat{e}_2

[/tex]

The vector [itex]{\bf v}[/itex] is the same length, but the basis for each vector may be different. If the operation from [itex](\hat{e}_1,\hat{e}_2)\rightarrow(\hat{\bar{e}}_1,\hat{\bar{e}}_2)[/itex] performs elongation, then [itex](T^1,T^2)\rightarrow(\hat{T}^1,\hat{T}^2)[/itex] will shrink (and vice versa) to preserve the shape of [itex]{\bf v}[/itex]. In this case, [itex](T^1,T^2)[/itex] are said to be contravariant vectors because they grow contrary to the direction that the basis grows in. However, the definition I found in Schaum's outlines seem to say the opposite.

For example, if

[tex]

\bar{T}^i=2,\,T^r=1,\,\frac{\partial\bar{x}^i}{\partial x^r}=2

[/tex]

Does that not say that going from unbarred to barred coordinates, the vector components are growing and so is the coordinate system? I must be confusing myself.

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# Basic Tensor Question

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