Tensors and Vector Invariance

You understand vector invariance, at least to some degree, if you agree with the following:

Vector $\overrightarrow{a}$ is a velocity of 10 in/min east
Vector $\overrightarrow{b}$ is a velocity of 25.4 cm/min east
$\overrightarrow{a}$ = $\overrightarrow{b}$

Vector invariance is part of a larger goal, to be able to make math changes without doing anything that changes the physics of the story. A vector is pointed at a point. The vector represents a the velocity of a charging bull. The point is you. The bull will get to you in five seconds. Rotating the coordinate system and making the appropriate changes to components won’t help you escape (you need to do something with a matrix that is NOT a tensor).

Math is show below to prove one of the things we know in Tensor Calculus. This proof should do two separate things. First, it should give you the idea, but second, and more importantly, it should show something about how the thinking is done, the more basic ideas–that probably weren’t mentioned in the other tutorial you read.

We begin with the definition of a vector:

$\overrightarrow{s} = x^{\mu} y_{\mu}$
$\overrightarrow{s}' = x'^{\mu} y'_{\mu}$

The first vector, $\overrightarrow{s}$ , has a basis vectors and components that we chose, probably to match up to some story. It’s all arbitrary.

The next choice (and the last choice) is the basis vectors for $\overrightarrow{s}'$ . The new components will depend on these choices.

A matrix can operate on a vector to make a new vector. Study the example below:

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

Some constraints get built into the story as a consequence of the above statement. If we say that our vectors are changed by matrices and only by matrices, then the only changes possible are those that can be done by matrices.

Another rule: we won’t accept changes to vectors that change the physics of the story.

$A^\mu_\alpha x^\alpha = x'^\mu$

The matrix $A^\mu_\sigma$ is whatever it takes to change x to x’.

In the example at the top of this page, the change to units (from inches to centimeters) was accompanied by a change to the xcomponent (from 10 to 25.4). Our matrix A is changing components. We need a second matrix, B, to make changes to the basis vector.

$B^\beta_\mu y_\beta = y'_\mu$

The matrix B, like A, is whatever it takes to change y to y’. The equations that have A and B can be used to make substitutions to some definitions shown earlier:

$\overrightarrow{s} = x^{\mu} y_{\mu}$

$\overrightarrow{s}' = x'^{\mu} y_{\mu}$

$\overrightarrow{s}' =A^\mu_\alpha x^\alpha B^\beta_\mu y_\beta$

The way vectors ans matrices work in Tensor Calculus, we can move pieces around as is done below. Your brain may be thinking, hey wait a minute, in general, that is not true! This tells us that we only allow vectors and matrices for which what we do below, is true. All other vectors and matrices are thrown into the trash can. Below we change the positions of a matrix in a vector so that the two matrices will be next to each other.

$\overrightarrow{s}' =A^\mu_\alpha B^\beta_\mu x^\alpha y_\beta$

Compare this to the following:

$\overrightarrow{s} = x^{\mu} y_{\mu}$

We need two things to happen: 1) we need the matrices A and B to disappear and we need either x or y to make an alpha/beta switch so both x and y will have the same symbol.

Multiplication by a Kronecker Delta will change a symbol.

$\delta^\beta_\alpha y_\beta = y_\alpha$

$\delta^\beta_\alpha x^\alpha = y^\beta$

To achieve our goal of $\overrightarrow{s}$ = $\overrightarrow{s}'$ we need the following to be true:

$A^\mu_\alpha B^\beta_\mu = \delta^\beta_\alpha$

Please take careful note of the distinction. We did not prove that the above was true. Rather, we said that if it isn’t true then the whole thing falls apart.

Appendix A

It is fair for you to ask, what transformations are legal? Do we, somewhere, have a list of all of them?

At one place we discover that for a math something to be true, we need to have it so that the inverse of Matrix A is also the transpose of Matrix A.

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