### TENSORS 2:

A concrete explanation can be found here, and it proceeds as follows:

We have two inertial (no acceleration) coordinate systems, $$S$$ and $$S'$$, where $$S'$$ moves away from $$S$$ at constant velocity $$v$$ along the $$x$$ axis - for now we assume no rotation or movement along the other two axes, $$y$$ and $$z$$.

An event $$P$$ happens and it is recorded in $$S$$ as $$(ct, x, y, z)$$. Why $$ct$$? To express time in the same units as the space coordintes we multiply the velocity of light by the observation of time. Not important.

The same event $$P$$ will be expressed in relation to the $$S'$$ coordinates as $$P=(ct', x', y',z').$$

#### Definition: CONTRAVARIANT VECTOR:

$$x^\mu = (x^0,x^1,x^2,x^3)$$, such that

$$x^0= ct, \quad x^1=x, \quad x^2=y, \quad x^3=z$$

So $$\mu$$ is an index, $$\mu= 0,1,2,3.$$

#### Definition: COVARIANT VECTOR:

$$x_\mu=(x_0,x_1,x_2,x_3)$$, such that

$$x_0=ct, \quad x_1=\color{red}{-}x, \quad\color{red}{-}y \quad\color{red}{-}z$$

Why do we need this?

The contravariant and covariant vectors of the same event in relation to the frame $$S'$$ are:

$$x'^\mu=(x^{'0}, x^{'1},x^{'2},x^{'3})\quad \text{contravariant}$$

and

$$x'_\mu=(x'_0, x'_1,x'_2,x'_3)\quad\text{covariant}$$

Now we have to bring in the Lorentz transformations for the answer to make sense. They are the equations that change coordinates between $$S$$ and $$S'$$ taking into consideration time dilation and length contraction (along the $$x$$ axis only, since the frames are only moving with respect to each other along $$x$$):

The transformation of time from $$t$$ in frame $$S$$ to $$t'$$ in $$S'$$ is given by $$t' = \gamma \left(t - \frac{v}{c^2}x\right)$$; while the transformation of $$x$$ from $$S$$ to $$S'$$ obeys the equation $$x' =\gamma(x - vt).$$ This is explained here, and adopting the notation of contravariant and covariant vectors, and the $$ct$$ expression of $$t$$:

$$x'^0 = \gamma\left(x^0 - \frac{v}{c} x^1\right)$$

$$x'^1 = \gamma\left(x^1 - \frac{v}{c} x^0\right)$$

$$x'^2 = x^2$$

$$x'^3 = x^3$$

Here, $$\large\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$. Notice that the third and fourth dimension do not change in this simple model with two frames, one moving along the $$x$$ axis of the other at constant velocity.

We can express the change in frames in matrix form as:

$\begin{bmatrix}x'^0\\x'^1\\x'^2\\x'^3\end{bmatrix}=\begin{bmatrix}\gamma&-\gamma\beta&0&0\\-\gamma\beta&\gamma&0&0\\0&0&1&0\\0&0&0&1\end{bmatrix}\begin{bmatrix}x^0\\x^1\\x^2\\x^3\end{bmatrix}$

with $$\beta=\frac{v}{c}.$$

If there is motion of $$S'$$ in more than just the $$x$$ difrection, the matriceal formula above can be generalized as:

$\begin{bmatrix}x'^0\\x'^1\\x'^2\\x'^3\end{bmatrix}=\begin{bmatrix}\Lambda^0_{\; 0}&\Lambda^0_{\; 1}&\Lambda^0_{\; 2}&\Lambda^0_{\; 3}\\\Lambda^1_{\; 0}&\Lambda^1_{\; 1}&\Lambda^1_{\; 2}&\Lambda^1_{\; 3}\\\Lambda^2_{\; 0}&\Lambda^2_{\; 1}&\Lambda^2_{\; 2}&\Lambda^2_{\; 3}\\\Lambda^3_{\; 0}&\Lambda^3_{\; 1}&\Lambda^3_{\; 2}&\Lambda^3_{\; 3}\end{bmatrix}\begin{bmatrix}x^0\\x^1\\x^2\\x^3\end{bmatrix}$

This can be express more succintly as:

$\color{red}{x'^{\mu} = \Lambda^\mu_{\; \nu}x^\nu}$

The same can be done with the covariant vector transformation:

$\begin{bmatrix}x'_0\\x'_1\\x'_2\\x'_3\end{bmatrix}=\begin{bmatrix}\Lambda_0^{\; 0}&\Lambda_0^{\; 1}&\Lambda_0^{\; 2}&\Lambda_0^{\; 3}\\\Lambda_1^{\; 0}&\Lambda_1^{\; 1}&\Lambda_1^{\; 2}&\Lambda_1^{\; 3}\\\Lambda_2^{\; 0}&\Lambda_2^{\; 1}&\Lambda_2^{\; 2}&\Lambda_2^{\; 3}\\\Lambda_3^{\; 0}&\Lambda_3^{\; 1}&\Lambda_3^{\; 2}&\Lambda_3^{\; 3}\end{bmatrix}\begin{bmatrix}x_0\\x_1\\x_2\\x-3\end{bmatrix}$

or

$\color{blue}{x'_{\mu} = \Lambda_\mu^{\; \nu}x_\nu}$

The Lorentz transformation $$\Lambda$$ is defined so that $$x'^{\mu}x'_{\mu} = x^\mu x_mu$$ (i.e. $$\small \text{contravariant in S'} \times \text{covariant in S'}=\text{contravariant in S} \times \text{covariant in S}$$).

We can express it as:

$\Large \color{red}{x'^{\mu}}\,\color{blue}{x'_{\mu}} = x^\mu \,x_\mu= \color{red}{\Lambda^\mu_{\;\;\nu}\,x^\nu}\,\color{blue}{\Lambda_\mu^{\;\;\rho}\,x_\rho}= \Lambda^{\mu}_{\;\;\nu}\,\Lambda_\mu^{\;\;\rho}\,\,x^\nu \,x_\rho=\delta^\rho_{\;\;\nu}\,x^\nu\,x_\rho$

Why does it work? Well, when $$\rho=\nu=\mu$$ the Kronecker product is one, and the final part of the equation above will be $$x^\mu\,x_\mu,$$ fulfilling the requisite of $$x'^{\mu}x'_{\mu} = x^\mu x_\mu.$$

$\large\color{orange}{\Lambda^\mu_{\;\;\nu}\,\Lambda_\mu^{\;\;\rho}=\delta^\rho_{\;\;\nu}}$

OK. So now we have covariant and contravariant vectors expressing the event with respect to $$S$$ and $$S'$$… and Lorentz transformations… we can do this…

$\large x^\mu x_\mu= x^0 x_0 + x^1 x_1 + x^2 x_2 + x^3 x_3 = c^2t^2 - \left(x^2 + y^2 + z^2 \right)=\color{red}{c^2t^2 - \vec{x}\vec{x}}$

$$\vec{x}\vec{x}$$ is the norm! And the result is a scalar (field value).

How to express this operation generally? We use the Einstein summation convention:

$A^\mu B_\mu=A^0B_0+A^1B_1+A^2B_2+A^3B_3$

where $$A^\mu =(A^0,A^1,A^2,A^3)$$ has a covariant vector $$A_mu=(A_0,A_1,A_2,A_3) = (A_0,-A^1,-A^2,-A^3).$$ And so does $$B^\mu.$$