Ask Alice About Tensors

We want to show you some math that shows what is necessary if you wish to go down the rabbit hole to find Invariancy. After that, we want to be brutally honest about the consequences of the choices we made to get Invariancy.

We will even show the story from an angle that indicates the Evil Queen is helping you when she is screaming “Feed your head!”

Invariancy means “no change”. What do we mean by no change? Lets briefly glimpse the idea by looking at a short, simple story.

10 in = 25.4 cm

The number gets larger, changing from 10 to 25.4
The length of the unit gets smaller, changing from 1 in to 1 cm
The two changes are such that one counterbalances the other
After all that work–the length is the same
Every multiplication was ultimately a multiplication by 1 (this is the trick).

We want this “no change” when we do work with Tensors.

If you’re doing work with 3×3 matrices, then multiplication by one is done by the 3×3 Identity Matrix, shown below:

$\begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 1 \end{bmatrix}$

This idea of multiplication by an Identity Element will be used soon.

We have definitions for Vectors (v) and One-forms given below, with a prime symbol indicating “new” (v’, w’) and the lack of a prime symbol indicating “old” (v,w) :

$v'^\alpha = \dfrac {\partial x'^j}{\partial x^{\alpha}} v^j$

$w'_\alpha = \dfrac {\partial x^{\alpha}} {\partial x'^i} w_i$

$\begin{bmatrix} 25.4 \\ 2.54 \end{bmatrix} = \begin{bmatrix} 2.54 && 0 \\ 0 && 2.54 \end{bmatrix} \begin{bmatrix} 10 \\ 1 \end{bmatrix}$

$\begin{bmatrix} \dfrac { \partial x'^1} {\partial x^1} && \dfrac { \partial x'^2} {\partial x^1} \\ \\ \dfrac { \partial x'^1} {\partial x^2} && \dfrac { \partial x'^2} {\partial x^2} \end{bmatrix}$

The notation used for the component of the matrix is a partial derivative. Because we are working with Linear Equations, every partial derivative will be a constant.

$\overrightarrow{v'} = \overrightarrow{v}$

We want the above to be true. We will start with the left side and make changes, hoping to eventually have the right side.

$\overrightarrow{v'}$

$v'^{\alpha}w'_{\alpha}$

$\dfrac {\partial y^\alpha}{\partial x^i} v^i \dfrac {\partial x^j}{\partial y^{\alpha}} w_j$

Because a factor like $\dfrac {\partial y^\alpha}{\partial x^i}$ is just a scalar, it can be moved to a new place in a term without changing the value of the term. Observe what happens when we put the two pieces together:

$\dfrac {\partial y^{\alpha}}{\partial x^i} \dfrac {\partial x^j}{\partial y^{\alpha}} v^i w_j$

We almost have what we want in those last two factors. We want $v^i$ to change to $v^j$ . That would happen with

$\delta^j_i v^i = v^j$

To get this the following must be true:

$\dfrac {\partial y^\alpha}{\partial x^i} \dfrac {\partial x^j}{\partial y^{\alpha}} = \delta^j_i$

We have the equality. We know the truth of this equality limits us to calculations where a new component is a linear equation.

$\dfrac { \partial x^i} { \partial x^j} = \delta^i_j$

Appendix A

The core idea of the trick might be made more obvious if you see it being done with very simple numbers in a story about going from inches to centimeters, one change at a time:

10 in

(10) (in)

(10) (1 in)

(1) (10) (1 in)

(1/2.54) (2.54) (10) (1 in)

(2.54) (10) (1/2.54) (1 in)

(2.54) (10) 1/2.54 1 in

(2.54) (10) 1/2.54 in

(2.54) (10) 1 cm

(2.54) (10) cm

2.54 10 cm

25.4 cm

The purpose of the above “trip” was to show the journey. I assume you already knew that 10 inches would be 25.4 centimeters.

Notice that we could add “1” as a factor to the multiplication anytime we wanted to do so, and we did this several times.

Notice that we also made use of a * (1/a) = 1.

Appendix B

One pill makes you larger…
And, one pill, makes you small…
And the ones that mother gives you, don’t do anything at all…
Ask Alice… when’s she’s ten feet tall!

Appendix C

We made a choice when we decided to use a matrix to get from the old vector to the new vector. This choice means that each component of the new vector is calculated as a linear equation in which all the components of the old vector are present. This is the point where the evil Queen screams:

“Don’t come crying to me, my pretty, when you need a curve but all you can do is draw that dreadful straight line! Feed Your Head!”

Appendix D

The math didn’t produce an Identity Matrix until we multiplied two matrices together. Each of those two matrices, by itself, was not the Identity Matrix.

If we were doing a change from inches to centimeters we would see the following two matrices:

$\begin{bmatrix} 2.54 && 0 && 0 \\ 0 && 2.54 && 0 \\ 0 && 0 && 2.54 \end{bmatrix}$

$\begin{bmatrix} \dfrac {1} {2.54} && 0 && 0 \\ 0 && \dfrac {1} {2.54} && 0 \\ 0 && 0 && \dfrac {1} {2.54} \end{bmatrix}$

Hopefully you an readily see that we could multiply these two together and get the Identity Matrix.

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