Trying to study Tensors (but )

[Curl]

I'm trying to figure this out but its confusing. I'm going by some notes someone put up online:
samizdat.mines.edu/tensors/ShR6b.pdf

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. Can anyone give me some pointers? If I just read without trying to do the exercises I don't get anything out of it.

Thanks

 

[Fredrik]

I can't make sense of it either. Were you able to make sense of Exercise 8.2? It seems to rely only on the fact that S and T are defined as each other's inverses. I don't see how it has anything to do with "the result of exercise 5.8". And S and T are still going to be each other's inverses even if we swap their meanings.

I also can't resist mentioning that I really dislike presentations on tensors based on the "something that transforms as blah-blah" idea. These guys are trying to avoid "difficult" mathematics, but they're just making it harder. I think it's easier to learn tensors by studying differential geometry, at least if you ignore the stuff about topology. This post would be a good place to start. (Ignore the first two paragraphs and start at "A manifold...").

 

[Curl]

Okay do you know any other good place to start on tensors? I tried Simmonds and I think it is garbage.

I'm still searching online for notes, and will try Schaums but it seems like most of them aren't good.

 

[Fredrik]

I don't know a text that's really great. I picked up some stuff about tensors from the SR section of Schutz's GR book, and then learned the rest from books on differential geometry. I used Spivak at the time, along with Wald (that was a GR book), but now I think a much better choice would be Lee, "Introduction to smooth manifolds". However, I'm reluctant to recommend that you try it this way, because you might get really confused by all the stuff about topology and then blame me for wasting your time.

If you want to try, I think the post I linked to is actually a good place to start. Instead of reading a book from page 1, you could read my posts and use the books only to look up details that you want to know more about. (Get the big picture first, and fill in the details later).

 

[Stephen Tashi]

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 [itex] {\bf e} = \{e_1,e_2,e_3 \}, {\bf f} = \{f_1,f_2,f_3\} [/itex] and [itex] {\bf g} = \{g_1, g_2, g_3\} [/itex].

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:
[tex] {\bf f} = {\bf e }{\bf S} [/tex]
[tex] {\bf e} = {\bf f }{\bf T } [/tex]

[tex] {\bf g} = {\bf f }{\bf P} [/tex]
[tex]{\bf g} = {\bf e }{\bf R} [/tex]

where each of [itex] {\bf S}, {\bf T}, {\bf P},{\bf R} [/itex] is a 3 by 3 matrix.
and [itex] { \bf T } = { \bf S}^{-1} [/itex]

The convention for writing the matrix elements is illustrated by:
[itex] {\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} [/itex]

So my version of eq 5.7 is:

[tex] \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} [/tex]

In order to be consistent the result of changing from basis [itex] {\bf e} [/itex] to basis [itex] {\bf g} [/itex] directly via the matrix [itex] {\bf R} [/itex] must produce the same result as changing from basis [itex] {\bf e } [/itex] to the basis [itex] {\bf f } [/itex] and then changing from the basis [itex] {\bf f }[/itex] to the basis [itex] {\bf g } [/itex]. This amounts to the equality:

[tex] {\bf e} {\bf R} = ( {\bf e}{\bf S} ){\bf P} [/tex]. For this to hold it is sufficient that [itex] { \bf R } = {\bf S} {\bf P} [/itex] which we will assume.

The section that develops eq. 6.2 demonstrates the interesting fact that the coordinates [itex] {\bf x } = \{x_1,x_2,x_3\} [/itex] of a vector in basis [itex] {\bf e} [/itex] transform to its coordinates [itex] {\bf y} [/itex] in basis [itex] {\bf f} [/itex] by the rule:
[tex] {\bf y} = {\bf T} {\bf x} [/tex]
where the coordinates are written as column vectors. (This is contrary what we might naively expect - namely some equation involving the matrix [itex] {\bf S} [/itex] ).

The section with eq 8.1 hypothesizes that there is a quantity [itex] {\bf A } [/itex] that has a representation as 3 coordinates [itex] \{a_1, a_2, a_3\} [/itex] in basis [itex]{\bf e}[/itex]. Visualizing the coordinates as a row vector [itex] {\bf a } [/itex] , the rules for relating the coordinates for this type of quantity to its representation [itex] {\bf b} [/itex] in basis [itex] {\bf f} [/itex] are:

[tex] {\bf b } = {\bf a} {\bf S} [/tex]
[tex] {\bf a } = {\bf b} {\bf T} [/tex]

This representation is consistent with respect two ways of changing from basis [itex] {\bf e} [/itex] to basis [itex] g [/itex].

The first way is change from basis [itex] {\bf e } [/itex] to basis [itex] {\bf g}[/itex] directly by using matrix [itex]{\bf R}[/itex]. The second way is to change from basis [itex] {\bf e } [/itex] to basis [itex] {\bf f } [/itex] and then to change from basis [itex] {\bf f } [/itex] to basis [itex] {\bf g} [/itex]. This amounts to the equality:

[tex] {\bf a} {\bf R } = ({\bf a}{\bf S}){\bf P} [/tex]

Since we assumed above that [itex] {\bf R} = {\bf S}{\bf P} [/itex] , this equality holds.

My interpretation of exercise 8.3:

Let [itex] {\bf M }[/itex] be the matrix that transforms from basis [itex] {\bf v}[/itex] to basis [itex] {\bf w} [/itex]. Suppose there is a quantity [itex] {\bf A} [/itex] who coordinates ( as a row vector) [itex] {\bf b} [/itex] in basis [itex] {\bf w} [/itex] are given as a function of its coordinates [itex] {\bf a} [/itex] in basis [itex] {\bf v} [/itex] by:

[tex] {\bf b } = {\bf a} {\bf M}^{-1} [/tex]

Show the test of consistency may fail.

The test of consistency would be that this equality holds:

[tex] {\bf a}{\bf R}^{-1} = ( {\bf a {\bf S}^{-1}}) {\bf P}^{-1} [/tex]

Although we have assumed [itex] {\bf R} = {\bf S}{\bf P} [/itex], this does not imply what would be needed , which is [itex] {\bf R}^{-1} = {\bf S}^{-1}{\bf P}^{-1} [/itex], so we should be able to find a numerical example where the equality fails.