Editing Appendix/Mathematics/ScaleFactors (section)

===2D Oblique Coordinate System Example===

Consider a vector, <math>~\vec{v}</math>, which in Cartesian coordinates is described by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\vec{v}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{e}_1 v_x + \hat{e}_2 v_y \, .
</math>
  </td>
</tr>
</table>
Referencing Figure 1.16.4 of [http://homepages.engineering.auckland.ac.nz/~pkel015/SolidMechanicsBooks/Part_III/ Kelly's Part III], we appreciate that in a two-dimensional (2D) '''oblique''' coordinate system where <math>~\alpha</math> is the (less than 90&deg;) angle between the two basis vectors, the same vector will be represented by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\vec{v}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{g}_1 v^1 + \hat{g}_2 v^2 \, .
</math>
  </td>
</tr>
</table>
The angle between <math>~\hat{g}_2</math> and <math>~\hat{e}_2</math> is, (&pi;/2 - &alpha;), so we appreciate that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~v_y</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~v^2\cos\biggl(\frac{\pi}{2} - \alpha \biggr) = v^2 \sin\alpha</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~v^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{v_y}{\sin\alpha} \, .</math>
  </td>
</tr>
</table>
Next, from a visual inspection of the figure, we appreciate that <math>~v_x</math> is longer than <math>~v^1</math> by the amount, <math>~v^2\cos\alpha</math>; that is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~v_x</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~v^1 + v^2\cos\alpha = v_1 + \frac{v_y}{\tan\alpha}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ v^1</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~v_x -  \frac{v_y}{\tan\alpha} \, .</math>
  </td>
</tr>
</table>
(These are the same pair of transformation relations that appear as Eq. (1.16.3) of [http://homepages.engineering.auckland.ac.nz/~pkel015/SolidMechanicsBooks/Part_III/ Kelly's Part III].)

<table border="1" align="center" cellpadding="8" width="80%"><tr><td align="left">
'''<font color="red">Covarient:'''</font> &nbsp; The set of basis vectors, <math>~\hat{g}_1</math> and <math>~\hat{g}_2</math> (note the subscript indices), that are aligned with the coordinate directions, <math>~\Theta_1</math> and <math>~\Theta_2</math>, are generically referred to as '''covariant''' base vectors.

'''<font color="red">Contravarient:'''</font> &nbsp; A second set of vectors, which will be termed '''contravariant''' base vectors, <math>~\hat{g}^1</math> and <math>~\hat{g}^2</math> (denoted by superscript indices), will be aligned with a new set of coordinate directions, <math>~\Theta^1</math> and <math>~\Theta^2</math>.

This new set of base vectors is defined as follows (see Fig. 1.15.5 of [http://homepages.engineering.auckland.ac.nz/~pkel015/SolidMechanicsBooks/Part_III/ Kelly's Part III]):  the base vector <math>~\hat{g}^1</math> is perpendicular to <math>~\hat{g}_1</math> &#8212; that is, <math>~\hat{g}^1 \cdot \hat{g}_2 = 0</math> &#8212; and the base vector <math>~\hat{g}^2</math> is perpendicular to <math>~\hat{g}_2</math> &#8212; that is, <math>~\hat{g}_1 \cdot \hat{g}^2 = 0</math>.  Further, we ensure that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{g}_1 \cdot \hat{g}^1 = 1 \, ,</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\hat{g}_2 \cdot \hat{g}^2 = 1 \, .</math>
  </td>
</tr>
</table>

(''Verbatim'' from p. 138 of [http://homepages.engineering.auckland.ac.nz/~pkel015/SolidMechanicsBooks/Part_III/ Kelly's Part III]) <font color="orange">A good trick for remembering which are the covariant and which are the contravariant is that the third letter of the word tells us whether the word is associated with subscripts or with superscripts.  In "covariant," the "v" is pointing down, so we use subscripts; for "contravariant," the "n" is (with a bit of imagination) pointing up, so we use superscripts.</font>
</td></tr></table>

Continuing with our 2D '''oblique''' coordinate system example and appreciating that Kelly has chosen to align the <math>~\hat{g}_1</math> basis vector with the <math>~\hat{e}_1</math> (Cartesian) basis vector, we see that the transformation between the two sets of '''covariant''' basis vectors is given by the relations,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{g}_1 = \hat{e}_1 \, ,</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\hat{g}_2 = \hat{e}_1\cos\alpha +  \hat{e}_2\sin\alpha \, .</math>
  </td>
</tr>
</table>
These conditions lead to the following complementary set of '''contravariant''' basis vectors:
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{g}^1 = \hat{e}_1 - \hat{e}_2 \biggl( \frac{1}{\tan\alpha}\biggr) \, ,</math>
  </td>
  <td align="center">
&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; 
  </td>
  <td align="left">
<math>~\hat{g}^2 = \hat{e}_2 \biggl( \frac{1}{\sin\alpha} \biggr) \, .</math>
  </td>
</tr>
</table>

Note that, as defined herein, the magnitude (''i.e.,'' scalar lengths) of these contravariant basis vectors is not unity; they are, instead,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~|\hat{g}^1| \equiv \biggl[ \hat{g}^1 \cdot \hat{g}^1 \biggr]^{1 / 2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
\biggl[\hat{e}_1 - \hat{e}_2 \biggl( \frac{1}{\tan\alpha}\biggr)\biggr] \cdot \biggl[\hat{e}_1 - \hat{e}_2 \biggl( \frac{1}{\tan\alpha}\biggr)\biggr]
\biggr\}^{1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{
1 + \frac{1}{\tan^2\alpha}
\biggr\}^{1 / 2}
= \frac{1}{\sin\alpha} \, ;
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~|\hat{g}^2| \equiv \biggl[ \hat{g}^2 \cdot \hat{g}^2 \biggr]^{1 / 2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\sin\alpha} \, .
</math>
  </td>
</tr>
</table>