Four-Vector position
In special relativity, space and time are not treated as separate entities but combined into a single mathematical object: the spacetime position four-vector $$ x^\mu = (ct, \; x, \; y, \; z), \quad \mu=0,1,2,3 $$ Here $ ct $ is the time component, where \( c \) is the speed of light.
The factor $ c $ (measured in $ m/s $) is introduced to bring all components into the same system of units.
Initially, time $ t $ is expressed in seconds, while the spatial components $ x, y, z $ are measured in meters.
$$ ( t, x,y,z ) $$
A vector mixing seconds and meters is not physically meaningful, since the components cannot be added consistently.
The product $ ct $ converts time into a length, expressed in meters:
$$ ( ct, x,y,z ) $$
This ensures that every component of the four-vector has the same physical dimension.
Note. The four-vector is often written in a more compact form, with \( x^0 = ct , \; x^1 = x, \; x^2 = y, \; x^3 = z \): $$ x^\mu = (x^0, x^1, x^2, x^3), \quad \mu=0,1,2,3 $$ Here $ x^0, x^1, x^2, x^3 $ are not powers of $ x $ but the four components of the four-vector: $ x^0 $ is the first component, $ x^1 $ the second, and so on.
Lorentz Transformation
The Lorentz transformation relates the coordinates of an event in system \( S \) to those in system \( S' \), moving with velocity \( v \) along the \( x \)-axis.
$$
\begin{cases}
t' &= \gamma\left(t - \tfrac{v}{c^2}x\right) \\
x' &= \gamma(x - vt) \\
y' &= y \\
z' &= z
\end{cases}
$$
To make the units consistent, multiply the first equation by $ c $:
$$
\begin{cases}
ct' &= c \gamma\left(t - \tfrac{v}{c^2}x\right) \\
x' &= \gamma(x - vt) \\
y' &= y \\
z' &= z
\end{cases}
$$
Which can be written as:
$$
\begin{cases}
ct' &= \gamma\left(ct - \tfrac{v}{c}x\right) \\
x' &= \gamma(x - vt) \\
y' &= y \\
z' &= z
\end{cases}
$$
Rewriting the second equation $ \gamma(x - vt) $ as $ \gamma(x - vt \cdot \tfrac{c}{c}) $ gives:
$$
\begin{cases}
ct' &= \gamma\left(ct - \tfrac{v}{c}x\right) \\
x' &= \gamma(x - \tfrac{v}{c} ct) \\
y' &= y \\
z' &= z
\end{cases}
$$
Now substitute $ x^0 = ct $ and $ x^{0'} = ct' $, with $ x^1 = x , x^2 = y , x^3 = z $:
$$
\begin{cases}
x^{0'} &= \gamma\left(x^0 - \tfrac{v}{c}x^1 \right) \\
x^{1'} &= \gamma(x^1 - \tfrac{v}{c} x^0) \\
x^{2'} &= x^2 \\
x^{3'} &= x^3
\end{cases}
$$
Defining $ \beta = \tfrac{v}{c} $ gives:
$$
\begin{cases}
x^{0'} &= \gamma\left(x^0 - \beta x^1 \right) \\
x^{1'} &= \gamma(x^1 - \beta x^0) \\
x^{2'} &= x^2 \\
x^{3'} &= x^3
\end{cases}
$$
where
$$ \gamma = \frac{1}{\sqrt{1-\beta^2}} $$
In compact notation, using Einstein’s summation convention, this becomes:
$$ x^{\mu'} = \Lambda^{\mu}{}_{\nu} x^\nu $$
where $ \Lambda^{\mu}{}_{\nu} $ is the Lorentz transformation matrix:
$$
\Lambda =
\begin{pmatrix}
\gamma & -\gamma \beta & 0 & 0 \\
-\gamma \beta & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}.
$$
This 4×4 matrix acts on the column vector \( (x^0, x^1, x^2, x^3)^T \) to produce the transformed four-vector:
$$ x^{\mu'} = \Lambda^{\mu}{}_{\nu} x^\nu $$
$$ \begin{pmatrix}
x^{0'} \\
x^{1'} \\
x^{2'} \\
x^{3'}
\end{pmatrix}
=
\begin{pmatrix}
\gamma & -\gamma\beta & 0 & 0 \\
-\gamma\beta & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}
\begin{pmatrix}
x^0 \\
x^1 \\
x^2 \\
x^3
\end{pmatrix}
$$
Carrying out the row-by-column multiplication reproduces the transformation equations:
$$
\begin{cases}
x^{0'} &= \gamma\left(x^0 - \beta x^1 \right) \\
x^{1'} &= \gamma(x^1 - \beta x^0) \\
x^{2'} &= x^2 \\
x^{3'} &= x^3
\end{cases}
$$
And so forth.
