📖 Definition:
- Four vectors: They have one time components and 3 spatial components we assume same units for all four components
- Lorentz scalars: Scalar products of four vectors are Lorentz invariant
We can write a general 4 vector as
$$ \begin{aligned} \text{contravariant:} \quad a^\mu &= (a^0, a^1,a^2,a^3)^T &\quad &\text{where} \quad \mu =\{0,1,2,3\} \\ \text{covariant:} \quad a_\mu &= (a^0, -a^1,-a^2,-a^3)^T &\quad &\text{where} \quad \mu =\{0,1,2,3\} \end{aligned} $$
🗒️ Note: $0$ is time $123$ are $xyz$
We can define the four-position vector and the Lorentz transformation matrix
$$ \begin{aligned} x^\mu &=(ct,\vec r)^T=(ct,x,y,z)^T \\ \Lambda &= \begin{bmatrix} \gamma & - \gamma \beta & 0 & 0 \\ -\gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \end{aligned} $$
where $\beta=v/c$ and $\gamma=(1-\beta^2)^{-1/2}$
💃 Example: to introduce index notation lets look at 2 ways to write the the transformation of $x^\mu$
$$ \begin{aligned} \begin{pmatrix} x'^0 \\ x'^1 \\ x'^2 \\ x'^3 \end{pmatrix}& = \begin{bmatrix} \gamma & - \gamma \beta & 0 & 0 \\ -\gamma \beta & \gamma & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{pmatrix} x^0 \\ x^1 \\ x^2 \\ x^3 \end{pmatrix} \\[2.6em] x'^\mu &=\Lambda ^\mu_\nu x^\nu \qquad \text{where} \quad \Lambda ^ \mu_\nu = \frac{\partial x'^\mu}{\partial x^\nu} \end{aligned} $$
We can thus write that a general vector must satisfy
$$ \boxed{ \begin{aligned} \text{Contravariant:} \quad a'^\mu&=\sum_{\nu=0}^3\frac{\partial x'^\mu}{\partial x^\nu} a^\nu =\sum^3_{\nu=0}\Lambda ^\mu_\nu a^\nu = \Lambda ^\mu_\nu a^\nu \\[1.5em] \text{Covariant:} \quad a'\mu&=\sum{\nu=0}^3\frac{\partial x^\nu}{\partial x'^\mu} a_\nu =\sum^3_{\nu=0}\Lambda ^\nu_\mu a_\nu = \Lambda ^\nu_\mu a_\nu \end{aligned}} $$
We define the four-vector derivative
$$ \partial \mu = \frac{\partial}{\partial x^\mu} = \left ( \frac{1}{c}\frac{\partial}{\partial t}, \, \vec \nabla \right )= \left ( \frac{\partial}{\partial x^0}, \, \frac{\partial}{\partial x^1}, \, \frac{\partial}{\partial x^2}, \, \frac{\partial}{\partial x^3}\right ) \\ \partial ^\mu = \frac{\partial}{\partial x\mu} = \left ( \frac{1}{c}\frac{\partial}{\partial t}, \, -\vec \nabla \right )= \left ( \frac{\partial}{\partial x_0}, \, \frac{\partial}{\partial x_1}, \, \frac{\partial}{\partial x_2}, \, \frac{\partial}{\partial x_3}\right ) $$
And it has the following transformations
$$ \begin{aligned} \partial '_\mu = \Lambda _\mu ^ \nu \partial _\nu \qquad \partial '^\mu = \Lambda _\nu ^ \mu \partial ^\nu
\end{aligned} $$
We define the Minkowski metric tensor
$$ \eta = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \end{bmatrix} = \operatorname{diag} (1,-1,-1,-1) $$
which is used in the following way
$$
a_\mu = \eta_{\mu \nu} a^\nu \qquad a^\mu = \eta^{\mu \nu} a_\nu \qquad \eta_{\mu \nu} = \eta ^{\mu \nu} \qquad \eta=\eta^{-1}
$$
We define the four-vector dot product as
$$ \begin{aligned} \tilde a \cdot \tilde b &= a^0 b^0 - (a^1 b^1 +a^2 b^2 + a^3 b^3) \\ \tilde a \cdot \tilde b &=a_\mu b^\mu = a^\mu b_\mu = \eta^{\mu \nu} a_\nu b_\mu = \eta_{\mu \nu} a^\nu b^\mu \end{aligned} $$
Which we can show is Lorentz invariant
$$ \tilde a' \cdot \tilde b'=\Lambda \mu ^\nu A^\mu\rho a_\nu b^\rho = \delta ^\nu_\rho a_\nu b^\rho = a_\nu b^\nu = \tilde a \cdot \tilde b $$
Now looking at our Klein Gordon equation:
$$ (\square + \mu^2) \psi(\vec r,t)=0 $$