Relativistic mechanics — 4-vectors and the free-particle Lagrangian

Make proper time the action and the free-particle Lagrangian falls out in one line — that single line then yields both $E^2 = p^2 + m^2$ and Newtonian mechanics in one stroke.

Opening

The Lagrangian $L = \tfrac{1}{2} m v^2 - V$ that has carried us so far is clean under Galilean transformations but breaks under Lorentz transformations. This chapter rebuilds the skeleton so it respects the principle of relativity. Throwing a single Lorentz invariant — the proper time — into the action fixes the free-particle Lagrangian automatically, and from that one line we read off the 4-momentum, the mass–energy relation, and the Newtonian equations of motion in the low-velocity limit. When chapter 8 moves on to field theory, the same prescription — “build the action out of a Lorentz invariant” — lifts straight up to infinitely many degrees of freedom. Throughout this chapter we work in natural units $c = 1$ and restore SI units only for the final numerical table.

Main 1 — Proper time and the 4-velocity

Write a particle’s worldline in Minkowski spacetime as a 4-tuple $x^\mu = (t, \vec x)$. Here $\mu$ (mu) is the spacetime index running over $0, 1, 2, 3$. The invariant interval between two neighbouring events is

$$d\tau^2 = dt^2 - |d\vec x|^2,$$

and this $\tau$ (tau) is called the proper time — the time read by a clock the particle itself carries. The crucial property is that every inertial observer assigns the same numerical value to it.

Reparametrising in terms of the lab time $t$ brings in the 3-velocity $\vec v = d\vec x/dt$, and

$$d\tau = dt\, \sqrt{1 - v^2},$$

which forces the familiar factor

$$\gamma \equiv \frac{dt}{d\tau} = \frac{1}{\sqrt{1 - v^2}}.$$

$\gamma$ (gamma) is the rate at which lab time accumulates relative to proper time; it diverges as $v \to 1$.

Define the 4-velocity as the derivative with respect to proper time,

$$u^\mu = \frac{dx^\mu}{d\tau} = (\gamma, \, \gamma \vec v).$$

Multiplying by the mass $m$ gives the 4-momentum $p^\mu = m u^\mu = (\gamma m, \gamma m \vec v)$. The zero-component is the quantity we will call energy, $E = \gamma m$, and the spatial three components form the relativistic momentum $\vec p = \gamma m \vec v$.

Main 2 — Free-particle Lagrangian and $E^2 = p^2 + m^2$

The power of the Lagrangian recipe lies in the demand that the action be invariant under coordinate transformations. The simplest quantity automatically invariant under Lorentz transformations is the proper time itself, so we postulate

$$S = -m \int d\tau$$

for a free particle. The sign and the overall factor $m$ will be justified shortly. Rewriting the action in terms of the lab time via $d\tau = dt/\gamma = dt\sqrt{1-v^2}$ gives

$$S = -m \int dt\, \sqrt{1 - v^2}, \qquad L = -m\sqrt{1 - v^2}.$$

This is the relativistic free-particle Lagrangian. It depends only on the velocity, not on the coordinates, so the conjugate momentum is read off by direct differentiation:

$$p_i = \frac{\partial L}{\partial v^i} = \frac{m v^i}{\sqrt{1 - v^2}} = m \gamma v^i.$$

As expected, this is exactly the spatial part of the 4-momentum. The Legendre transform produces the Hamiltonian

$$H = p_i v^i - L = \frac{m v^2}{\sqrt{1 - v^2}} + m\sqrt{1 - v^2} = \frac{m}{\sqrt{1 - v^2}} = m\gamma = E,$$

confirming that the energy is precisely the zero-component of the 4-momentum. Finally, using $E = \gamma m$ and $\vec p = \gamma m \vec v$ together with $\gamma^2(1 - v^2) = 1$ to eliminate $\gamma$ yields the one-line identity

$$E^2 = |\vec p|^2 + m^2.$$

(Restoring units, $E^2 = (pc)^2 + (mc^2)^2$.) At rest, $\vec p = 0$, so $E = m$, i.e. $E = mc^2$ — Einstein’s relation drops out of the Lagrangian formalism as a by-product.

Main 3 — Newtonian limit and rest energy

Expanding $L = -m\sqrt{1 - v^2}$ for $v^2 \ll 1$ gives

$$L = -m + \tfrac{1}{2} m v^2 + \tfrac{1}{8} m v^4 + O(v^6).$$

The first term, $-m$, is a velocity-independent constant; it drops out of the Euler–Lagrange equations $\frac{d}{dt}\frac{\partial L}{\partial v^i} - \frac{\partial L}{\partial x^i} = 0$ entirely. From the standpoint of equations of motion the term is invisible. It does, however, survive in the energy and shows up there as the rest energy $E_0 = m$ (in SI, $mc^2$).

The second term, $\tfrac{1}{2} m v^2$, is exactly the Newtonian kinetic energy. The third term, $\tfrac{1}{8} m v^4$, is the first relativistic correction: at $v/c \approx 0.1$ it already produces a $\sim 0.4\%$ deviation. So the relativistic Lagrangian connects smoothly to Newtonian mechanics while carrying along the new piece of information called rest energy. When chapter 8 writes down a Lagrangian for a field, exactly the same principle — integrate an invariant — does the work.

In Python

# Tabulate gamma, pc, E, KE for an electron (m c^2 = 0.511 MeV)
# and verify E^2 - (pc)^2 = (mc^2)^2 on the last row.
import numpy as np

mc2 = 0.511  # electron rest energy [MeV]
beta = np.array([0.1, 0.3, 0.5, 0.7, 0.9, 0.99, 0.999])  # v/c

gamma = 1.0 / np.sqrt(1.0 - beta**2)
E = gamma * mc2                # total energy [MeV]
pc = gamma * beta * mc2        # momentum * c [MeV]
KE = E - mc2                   # kinetic energy [MeV]

print(f"{'v/c':>7} {'gamma':>10} {'pc [MeV]':>12} {'E [MeV]':>12} {'KE [MeV]':>12}")
for b, g, p, e, k in zip(beta, gamma, pc, E, KE):
    print(f"{b:>7.3f} {g:>10.4f} {p:>12.5f} {e:>12.5f} {k:>12.5f}")

# Check the dispersion relation on the last row
inv = E[-1]**2 - pc[-1]**2     # should equal (mc^2)^2
print(f"\nE^2 - (pc)^2 = {inv:.8f}  vs  (mc^2)^2 = {mc2**2:.8f}")
print(f"relative error = {abs(inv - mc2**2) / mc2**2:.2e}")

The last two printed lines should agree to floating-point precision. At $v/c = 0.999$ one finds $\gamma \approx 22.4$ and $E \approx 11.4$ MeV, so the kinetic energy is roughly twenty-two times the rest energy — directly visible in the table.

To the next chapter

Chapter 8: Noether’s theorem in field theory lifts the recipe of this chapter — “integrate a Lorentz invariant to obtain an action” — to infinitely many degrees of freedom. Just as conservation of 4-momentum for a single particle was the shadow of spacetime translation symmetry, in field theory the energy–momentum tensor will drop out of the same place automatically. That bridge is the field-theoretic generalisation of Noether’s theorem.