💼 Case: Consider the space-time interval between two space-time points on the world line of a particle
Where the 2 events happen simultaneously in the rest frame of the particle $S'$
In $S'$ we thus can write
$$ \begin{aligned} \Delta x'^i&=0 \\ \Delta x'^\mu \Delta x_\mu ' &=c^2 \Delta t'^2-0 \equiv c^2 \Delta \tau^2 \end{aligned} $$
where $\Delta \tau^2$ is the proper time interval and is a Lorentz scalar ie can be measured by a single clock
Applying the Lorentz transformations in the $S$ frame we get
$$ \Delta x^0 = c\Delta t = \gamma(c\Delta \tau +0) \qquad \Rightarrow \qquad \Delta t= \gamma \Delta \tau $$
🗒️ Note: The time interval is always $\Delta t \ge \Delta \tau$
Now if we consider the velocity is usually represented by
$$ \vec u = \frac{\text d \vec r} {\text dt} $$
Now taking this to 4 vectors we write
$$ u^\mu= \frac{\text d x^\mu}{\text d \tau} $$
where $x^\mu$ is a 4-vector and $\text d \tau$ is Lorentz invariant
🗒️ Note: this is a bit of weird object because $\text dx^\mu$ is in the $S$ frame where particle is moving and $\text d \tau$ is in $S'$ where particle is stationary
Analysis of the components where $i=1,2,3$
$$ \begin{aligned} \text{general: }\quad u^i &= \frac{\text d x^i}{\text d \tau} = \gamma \frac{\text dx^i}{\text d t} \\ \text{specific: } \quad u^0 &= \frac{\text d x^0}{\text d \tau} = \gamma c \frac{\text d t}{\text dt} = \gamma c \qquad \text{since $ \gamma \text d \tau = \text d t $} \\ \Rightarrow \qquad u^\mu&= (\gamma c, \gamma \vec u) \end{aligned} $$
⚙️ Properties
- Lorentz transformation: $u'^\mu = \Lambda ^\mu _\nu u^\nu$
- Magnitude: $u_\mu u^\mu = \gamma^2 c^2 -\gamma^2 u^2 = c^2$
$$ p^\mu = m u^\mu = m \frac{\text dx^\mu}{\text d \tau} = \gamma m(c,\vec u) $$
where $m$ is the mass of the particle in its rest frame (rest mass, invariant mass or just mass)
🗒️ Note: $p^\mu$ is a 4-vector because $u^\mu$ is a 4-vector and $m$ is invariant
Analysis of components where $i=1,2,3$
$$ \begin{aligned} \text{general: }\quad p^i &= \gamma m u^i\\ \text{specific: } \quad p^0 &= \gamma mc = \frac{\gamma mc^2}{c} = \frac{E}{c} \quad \text{where $E=\gamma mc^2$ is total relative energy}\\ \Rightarrow \qquad p^\mu&= (\frac{E}{c}, \vec p) \end{aligned} $$
⚙️ Properties:
- Lorentz transformation: $p'^\mu = \Lambda ^\mu_\nu p ^\nu$
- Magnitude: $p^\mu p_\mu =\left ( \frac{E}{c} \right )^2-p^2= (mc)^2$
This allows us to write the general result
$$ \boxed{E^2 - (pc)^2= (mc^2)^2} $$
$$ j^\mu =\rho_0 u^\mu $$
where $\rho_0$ is the charge density in the rest frame, which is Lorentz invariant
🗒️ Note: $j^\mu$ is a 4-vector because $u^\mu$ is a 4-vector and $\rho_0$ is invariant
Analysis of components where $i=1,2,3$
$$ j^i=\gamma \rho_0 u^i $$