[math]\displaystyle{ \textbf{\underline{Navigation}} }[/math]
[math]\displaystyle{ \vartriangleleft }[/math]
[math]\displaystyle{ \triangle }[/math]
[math]\displaystyle{ \vartriangleright }[/math]
4-vectors
Using index notation, the time and space coordinates can be combined into a single "4-vector" [math]\displaystyle{ x^{\mu},\ \mu=0,\ 1,\ 2,\ 3 }[/math], that has units of length, i.e. ct is a distance.
[math]\displaystyle{ \begin{bmatrix}
x^0 \\
x^1 \\
x^2 \\
x^3
\end{bmatrix}=
\begin{bmatrix}
ct \\
x \\
y \\
z
\end{bmatrix} }[/math]
Using the Lorentz transformations and the index notation,
[math]\displaystyle{
\begin{cases}
t'=\gamma (t-vz/c^2) \\
x'=x' \\
y'=y' \\
z'=\gamma (z-vt)
\end{cases}
}[/math]
[math]\displaystyle{ \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]\displaystyle{ \beta \equiv \frac{v}{c} }[/math]
We can express the space time interval using the index notation
[math]\displaystyle{ (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]\displaystyle{ (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]
[math]\displaystyle{ \textbf{\underline{Navigation}} }[/math]
[math]\displaystyle{ \vartriangleleft }[/math]
[math]\displaystyle{ \triangle }[/math]
[math]\displaystyle{ \vartriangleright }[/math]