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 \(P=(ct', x', y',z').\)


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

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

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


\(x_\mu=(x_0,x_1,x_2,x_3)\), 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:

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


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

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:


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:

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


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

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 \(x'^{\mu}x'_{\mu} = x^\mu x_\mu.\)

Leading to the important result:


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 \(B^\mu.\)