Curl said:
Look at Exercise 8.3 on page 19. I got no idea how to do this, and actually I'm not even sure what it's asking
I scanned the notes. It's probably considered gauche to use linear algebra to talk about tensors, but by doing that, I formed an interpretation. I've changed his notation a little.
(This long post has many opportunities for errors. Check it!)
Three bases for the same vector space are {\bf e} = \{e_1,e_2,e_3 \}, {\bf f} = \{f_1,f_2,f_3\} and {\bf g} = \{g_1, g_2, g_3\}.
Considering the elements of each basis to be a row vector, the change of coordinate formulae can be visualized as matrix multiplication.
Assume the following:
{\bf f} = {\bf e }{\bf S}
{\bf e} = {\bf f }{\bf T }
{\bf g} = {\bf f }{\bf P}
{\bf g} = {\bf e }{\bf R}
where each of {\bf S}, {\bf T}, {\bf P},{\bf R} is a 3 by 3 matrix.
and { \bf T } = { \bf S}^{-1}
The convention for writing the matrix elements is illustrated by:
{\bf S } = \begin{pmatrix} S_1^1 & S_2^1 & S_3^1 \\ S_1^2 & S_2^2 & S_3^2 \\ S_1^3 & S_2^3 & S_3^3 \end{pmatrix}
So my version of eq 5.7 is:
\begin{pmatrix} f_1 & f_2 & f_3 \end{pmatrix} = \begin{pmatrix} e_1 & e_2 & e_3 \end{pmatrix}\begin{pmatrix} S_1^1 & S_2^1 & S_3^1 \\ S_1^2 & S_2^2 & S_3^2 \\ S_1^3 & S_2^3 & S_3^3 \end{pmatrix}
In order to be consistent the result of changing from basis {\bf e} to basis {\bf g} directly via the matrix {\bf R} must produce the same result as changing from basis {\bf e } to the basis {\bf f } and then changing from the basis {\bf f } to the basis {\bf g }. This amounts to the equality:
{\bf e} {\bf R} = ( {\bf e}{\bf S} ){\bf P}. For this to hold it is sufficient that { \bf R } = {\bf S} {\bf P} which we will assume.
The section that develops eq. 6.2 demonstrates the interesting fact that the coordinates {\bf x } = \{x_1,x_2,x_3\} of a vector in basis {\bf e} transform to its coordinates {\bf y} in basis {\bf f} by the rule:
{\bf y} = {\bf T} {\bf x}
where the coordinates are written as column vectors. (This is contrary what we might naively expect - namely some equation involving the matrix {\bf S} ).
The section with eq 8.1 hypothesizes that there is a quantity {\bf A } that has a representation as 3 coordinates \{a_1, a_2, a_3\} in basis {\bf e}. Visualizing the coordinates as a row vector {\bf a } , the rules for relating the coordinates for this type of quantity to its representation {\bf b} in basis {\bf f} are:
{\bf b } = {\bf a} {\bf S}
{\bf a } = {\bf b} {\bf T}
This representation is consistent with respect two ways of changing from basis {\bf e} to basis g.
The first way is change from basis {\bf e } to basis {\bf g} directly by using matrix {\bf R}. The second way is to change from basis {\bf e } to basis {\bf f } and then to change from basis {\bf f } to basis {\bf g}. This amounts to the equality:
{\bf a} {\bf R } = ({\bf a}{\bf S}){\bf P}
Since we assumed above that {\bf R} = {\bf S}{\bf P} , this equality holds.
My interpretation of exercise 8.3:
Let {\bf M } be the matrix that transforms from basis {\bf v} to basis {\bf w}. Suppose there is a quantity {\bf A} who coordinates ( as a row vector) {\bf b} in basis {\bf w} are given as a function of its coordinates {\bf a} in basis {\bf v} by:
{\bf b } = {\bf a} {\bf M}^{-1}
Show the test of consistency may fail.
The test of consistency would be that this equality holds:
{\bf a}{\bf R}^{-1} = ( {\bf a {\bf S}^{-1}}) {\bf P}^{-1}
Although we have assumed {\bf R} = {\bf S}{\bf P}, this does not imply what would be needed , which is {\bf R}^{-1} = {\bf S}^{-1}{\bf P}^{-1}, so we should be able to find a numerical example where the equality fails.
