Notes on Relativity
18 Oct 2026Throughout the note, we use the following tensor:
\[\Lambda^\mu_{\;\nu} = \begin{pmatrix} \gamma & -\beta\gamma & 0 & 0\\ -\beta\gamma & \gamma & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix}\]Definition. A transformation $M$ is Lorentz invariant if it leaves the Minkowski metric (inner product) between two events invariant, i.e.
\[\eta(Mu, Mv) = \eta(u,v)\]Note that this condition is equivalent to $M^T\eta M = \eta$.
Corollary. $d\tau$ is invariant under Lorentz invariant transformation.
See “Notes on Tensor”. Due to Lorentz invariance, we are free to use index raising/lowering & index contraction to construct tensor equations.
Clock
A clock should measure a Lorentz invariant quantity.
Consider a light clock, i.e. light traversing between two mirrors:
- Trajectory of light is Lorentz invariant
- The distance betwee mirrors is Lorentz invariant under smooth impulse, if the mirrors are connected by a rigid rod (⇒ rigidity: tendency to maintain equilibrium)
- Hence, a rigid rod with mirrors attached at each end is an ideal clock
Lagrangian of a free relativistic particle
Criteria for Lagrangian:
- Local (Up to first derivative only)
- Lorentz invariant
- (Gauge invariant => Action must be invariant under Gauge transform)
Since $d\tau$ is Lorentz invariant, we will use this to construct the Lagrangian:
\[S = -mc\int d\tau\]Note that $c$ is simply a constant, added to match the dimensions. From $d\tau^2 = dt^2 - \frac{dx^2}{c^2}$, we have
\[d\tau = \sqrt{\dot{t}^2 - \frac{\dot{x}^2}{c^2}} d\lambda\]where $\lambda$ is a formal variable and dot stands for $d/d\lambda$. We then have:
\[S = \int \underbrace{-mc\sqrt{\dot{t}^2 - \frac{\dot{x}^2}{c^2}}}_{L(x, t, \dot{x}, \dot{t})}d\lambda\]To minimize $S$, we use Euler-Lagrange equation:
\[\frac{d}{d\lambda}\frac{\partial L}{\partial \dot{X^\mu}} = \frac{\partial L}{\partial X^\mu}\]4-momentum
The canonical momentum for $L$ are:
\[P_t =-mc\frac{\dot t} {\sqrt{\dot t^{2}-\dot{x}^{2}/c^2}}\]If we let $\lambda = \tau$,
\[\dot{t} = \gamma, \; \dot{x} = \gamma v, \; \sqrt{\dot t^{2}-\dot{x}^{2}/c^2} = 1\]Hence,
\[P_t = -mc \gamma\]or alternatively, $P^t = mc\gamma$. Indeed, we have:
\[P^\mu = mU^\mu\]Conservation Laws
Since $P^\mu$ is the canonical momentum obtained from the Lagrangian, we can use Noether’s law:
\[\sum P^\mu f(\mathbf{x}) \text{ is conserved}\]Hence we obtain that $P^t$ is conserved under time translation. Hence, $P^0$ is the energy of the system. Matching the dimension, we have $cP^t = E$.
Mass
Consider a free particle moving at velocity $v$. Since 4-momentum is a 4-vector, we can Lorentz transform it to have zero velocity, i.e.
\[P^\mu = \begin{pmatrix} E/c \\ 0 \\ 0 \\ 0 \end{pmatrix}\]Note that $E/c$ is a time-invariant quantity that does not depend on $v$ (since we can always Lorentz transform to an appropriate frame). Thus, this quantity is intrinsic to the system. We thus define this quantity to be the mass. Matching the dimension, we have $m = E/c^2$.
Translation between relativistic unit and standard unit
General rule:
\[t \mapsto ct, \quad v \mapsto v/c\]
Hence, with $\gamma = \frac{1}{\sqrt{1 - v^2}}$ and $\gamma_s = \frac{1}{\sqrt{1 - (v/c)^2}}$,
| Relativistic unit | Standard unit |
|---|---|
| $x’ = \gamma(x - vt)$ | $x’ = \gamma_s(x - vt)$ |
| $t’ = \gamma(t - vx)$ | $\displaystyle t’ = \gamma_s\left(t - \frac{vx}{c^2}\right)$ |
| $X^\mu = (t, x, y, z)$ | $X^\mu = (ct, x, y, z)$ |
| $d\tau^2 = dt^2 - dx^2 - dy^2 - dz^2$ | $c^2 d\tau^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2$ |
| $\displaystyle U^\mu = \frac{dX^\mu}{d\tau} = (\gamma, \gamma\dot{x})$ | $U^\mu = (c\gamma_s, \gamma_s\dot{x})$ |