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Basic Tensor Question

  1. Jun 29, 2014 #1
    So I'm looking at Schaum's outlines for Tensors and the definition of a Contravariant vector is
    [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.
     
  2. jcsd
  3. Jun 29, 2014 #2

    TSC

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    The last equation is unjustified.
     
  4. Jun 30, 2014 #3
    Can you elaborate?
     
  5. Jun 30, 2014 #4

    Matterwave

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    Science Advisor
    Gold Member

    Let's just look at a 1 dimensional example (your example), take ##\bar{x}^1=2x^1## So that ##\frac{\partial \bar{x}^1}{\partial x^1}=2##. Then your example says if ##T^1=1## then ##\bar{T}^1=2##. Let's just assume the space is flat so things are easy.

    But what does ##\bar{x}^1=2x^1## mean? At the point where the old ##x^1=1## the new ##\bar{x}^1=2##. So, if you think about it a bit, you will realize that your coordinates have actually shrank since you need a larger number to describe the same distance. In other words, it would be like if you switched from meters to half-meter measurements. Where the old coordinate says 1 meter, your new coordinate says 2 half meters. Therefore, your old vector was 1 meter long, your new vector is 2 half meters long. The length of the vector has remained invariant. In this example, the two vectors both point to a distance 1 meter from the origin!
     
  6. Jun 30, 2014 #5

    TSC

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    Good explanation!
     
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