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4-vectors
Using index notation, the time and space coordinates can be combined into a single "4-vector" [math]x^{\mu},\ \mu=0,\ 1,\ 2,\ 3[/math], that has units of length, i.e. ct is a distance.
[math]\begin{bmatrix}
x^0 \\
x^1 \\
x^2 \\
x^3
\end{bmatrix}=
\begin{bmatrix}
ct \\
x \\
y \\
z
\end{bmatrix}[/math]
We can express the space time interval using the index notation
[math](ds)^2\equiv c^2 dt^{'2}-dx^{'2}-dy^{'2}-dz^{'2}= c^2 dt^{2}-dx^2-dy^2-dz^2[/math]
[math](ds)^2\equiv (dx^0)^{'2}-(dx^1)^{'2}-(dx^2)^{'2}-(dx^3)^{'2}= (dx^0)^{2}-(dx^1)^2-(dx^2)^2-(dx^3)^2[/math]
Since [math]ds^2 [/math] is nothing more than a dot product of a vector with itself, we should expect the components of the indices to follow a similar relationship.
[math](ds)^2\equiv x_{\nu} x^{\mu}[/math]
[math](ds)^2\equiv
\begin{bmatrix}
dx_0 & -dx_1 & -dx_2 & -dx_3
\end{bmatrix} \cdot
\begin{bmatrix}
dx^0 \\
dx^1 \\
dx^2 \\
dx^3
\end{bmatrix}= (dx^0)^{2}-(dx^1)^2-(dx^2)^2-(dx^3)^2=(dx^0)^{'2}-(dx^1)^{'2}-(dx^2)^{'2}-(dx^3)^{'2}[/math]
The change in signs in the covariant term,
[math]x_{\nu}= \begin{bmatrix}
dx_0 & -dx_1 & -dx_2 & -dx_3
\end{bmatrix}[/math]
To the contravarient term
[math]x^{\mu}=
\begin{bmatrix}
dx^0 \\
dx^1 \\
dx^2 \\
dx^3
\end{bmatrix}
[/math]
Comes from the Minkowski metric
[math]
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 &-1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{bmatrix}[/math]
Using the Lorentz transformations and the index notation,
[math]
\begin{cases}
t'=\gamma (t-vz/c^2) \\
x'=x' \\
y'=y' \\
z'=\gamma (z-vt)
\end{cases}
[/math]
[math]\begin{bmatrix}
x'^0 \\
x'^1 \\
x'^2\\
x'^3
\end{bmatrix}=
\begin{bmatrix}
\gamma (x^0-vx^3/c) \\
x^1 \\
x^2 \\
\gamma (x^3-vx^0)
\end{bmatrix}
=
\begin{bmatrix}
\gamma (x^0-\beta x^3) \\
x^1 \\
x^2 \\
\gamma (x^3-vx^0)
\end{bmatrix}[/math]
Where [math]\beta \equiv \frac{v}{c}[/math]
This can be expressed in matrix form as
[math]\begin{bmatrix}
x'^0 \\
x'^1 \\
x'^2\\
x'^3
\end{bmatrix}=
\begin{bmatrix}
\gamma & 0 & 0 & -\gamma \beta \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-\gamma \beta & 0 & 0 & \gamma
\end{bmatrix}
\cdot
\begin{bmatrix}
x^0 \\
x^1 \\
x^2 \\
x^3
\end{bmatrix}[/math]
Letting the indices run from 0 to 3, we can write
[math]x'^{\mu}=\sum_{\nu=0}^3 (\Lambda_{\nu}^{\mu})x^{\nu}[/math]
Where [math]\Lambda[/math] is the Lorentz transformation matrix for motion in the z direction.
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