The Relativistic Free-Particle Lagrangian

This note derives the Lagrangian of a free relativistic particle,

from two independent directions. Route 1 consumes the relativistic energy and momentum derived in the kinetic-energy note and recovers $L$ by a Legendre transform — a consistency check showing that those quantities are compatible with a unique Lagrangian. Route 2 ignores the collision results and derives $L$ straight from Lorentz symmetry (the action must be a Lorentz scalar), then re-derives $p$ and $E$ from $L$ as its own check.

Both routes require adjoining one postulate that the collision analysis never used — the action principle — and with it the relativistic free-particle Lagrangian is fixed with no free constants.

The action principle: the one new postulate¶

The collision derivation obtained $E$ and $p$ from symmetries alone — no equation of motion, no action. To speak of a Lagrangian we adjoin Hamilton’s principle:

• A particle’s trajectory extremizes the action

  $$S = \int_{t_1}^{t_2} L(\mathbf x, \mathbf v, t)\,dt, \qquad \mathbf v = \dot{\mathbf x}.$$

  (3)

• The canonical momentum and Hamiltonian are, in one spatial dimension,

  $$p = \frac{\partial L}{\partial v}, \qquad H = p\,v - L.$$

  (4)

  (In three dimensions $p_i = \partial L/\partial v_i$ and $H = \mathbf p\cdot\mathbf v - L$.)

For a free particle $L$ has no explicit time dependence, so $H$ is conserved and equals the total energy $E$; and $L$ depends on $\mathbf v$ only through $v^2$ (isotropy, the relativistic analog of (P1)), so the canonical momentum coincides with the mechanical momentum $\mathbf p=\gamma m\mathbf v$. Under these identifications, (4) is the bridge between the Lagrangian and the dynamical quantities ($p$, $E$) already derived. Two routes now pin down $L$.

Route 1: the Legendre transform (from the derived $p$ and $E$)¶

This is the direct route: take the collision-derived $p(v)$ and $E(v)$ and invert (4).

From energy. Since $E = H = p v - L$, we have $L = p v - E$. Substituting $p = \gamma m v$ and $E = \gamma m c^2$,

Now use

so $L = -m\cdot c^2/\gamma$, which is (5).

From momentum. Equivalently, integrate $p = \partial L/\partial v$:

With the substitution $u = 1 - v^2/c^2$, $du = -2v\,dv/c^2$, the integral is

Hence

for an integration constant $C$. It is fixed by the derived rest energy: at $v=0$, $p(0)=0$ and $L(0) = -mc^2 + C$, so

The kinetic-energy note derived $E(0) = mc^2$ (the rest energy, output of the non-relativistic limit), so $C = 0$. Thus $L = -mc^2/\gamma$.

Route 2: Lorentz invariance of the action (symmetry-first)¶

The deeper derivation uses only Lorentz symmetry, not the collision results. It has three steps.

Step 1 — the action is a Lorentz scalar¶

The equations of motion are Lorentz-covariant, and the action generates them via $\delta S = 0$. For the variational principle to be frame-independent, $S$ itself must be a Lorentz scalar — a number unchanged by boosts. (Any non-scalar part of $S$ would single out a preferred frame and break the symmetry.)

Step 2 — the only worldline invariant is the proper time¶

Along a particle’s worldline the Lorentz-invariant line element is

so the invariant proper time is

Any Lorentz-scalar functional of the worldline alone is built from $\int d\tau$. For a free particle — no external fields, no internal structure — the action must be linear in this invariant, with a coefficient proportional to the particle’s only attribute, its rest mass $m$:

for a constant $\alpha$ with dimensions $[\text{energy}]/[\text{mass}]=(\text{velocity})^2$. The only relativistic velocity scale is $c$, so $\alpha = k\,c^2$ for a dimensionless $k$.

Step 3 — fix the prefactor from the non-relativistic limit¶

Since $d\tau = dt/\gamma$, the Lagrangian is

Expanding at small speed,

For this to reproduce the Newtonian free-particle Lagrangian $L_{\text{Newt}}=\tfrac12 m v^2$, the coefficient of $m v^2$ must be $+\tfrac12$, i.e. $-k/2 = 1/2$, so

Both the factor $c^2$ and the minus sign are forced; nothing is put in by hand. Therefore

in agreement with Route 1, (5). Equivalently, the action is the proper time weighted by rest energy,

Closing the loop: $L \;\longrightarrow\; p,\,E$¶

Route 2 never used the collision results, so it must re-derive them. From $L = -m c^2(1-v^2/c^2)^{1/2}$,

and

These are exactly the energy and momentum derived from collisions in the kinetic-energy note. The two routes — symmetry-first (Route 2) and collision-first (Route 1) — meet here, each confirming the other.

Checks¶

Non-relativistic limit. Expanding (5),

The leading dynamical term $\tfrac12 m v^2$ is the Newtonian free-particle Lagrangian; the additive constant $-m c^2$ is a total time derivative and drops out of the Euler–Lagrange equations; and $\tfrac18\,m v^4/c^2$ is the first post-Newtonian correction. The expansion matches that of $E_{\text{kin}}=(\gamma-1)mc^2$, as it must, since both come from the same $\gamma$.

Equation of motion. For a free particle $\partial L/\partial \mathbf x = 0$, so the Euler–Lagrange equation gives $d(\gamma m\mathbf v)/dt = 0$: momentum is conserved and the particle moves inertially.

Hamiltonian. $H = \mathbf p\cdot\mathbf v - L = \gamma m v^2 + m c^2/\gamma = \gamma m c^2 = E$, the total energy (including rest energy), conserved because $L$ has no explicit time dependence.

Energy–momentum relation. From the recovered $p$ and $E$,

The additive constant and covariance¶

In non-relativistic mechanics one may add a constant to $L$ freely: it changes the Lagrangian by a total time derivative and leaves the equations of motion untouched. Why, then, can we not replace $L \to L + C$ here?

Because of covariance. The Hamiltonian shifts as $H \to H - C$ when $L \to L + C$, and $H = E$ is the time component of the energy–momentum 4-vector, which mixes with momentum under boosts. A constant shift $-C$ would not transform as a component of a 4-vector (it would be a frame-independent additive constant in $E$), breaking Lorentz symmetry. Hence $C$ is fixed, not free: requiring $E$ to be the time component of a 4-vector — equivalently, requiring $S$ to be the Lorentz scalar $-mc^2\int d\tau$ — sets $C=0$ and pins the rest energy to exactly $mc^2$.

This is the Lagrangian-side echo of the point made in the kinetic-energy note: the rest energy $\mathcal E(0)=c^2$ is a genuine physical scale, not a convention, precisely because Lorentz covariance forbids shifting it. The collision derivation derives that fact from the functional equation; the Lagrangian derivation respects it through the invariance of the action.

Remarks: dimensions, interactions, massless particles¶

Three dimensions. Replace $v^2$ by $\lVert\mathbf v\rVert^2$ throughout:

Isotropy guarantees $L$ depends on $\mathbf v$ only through its magnitude, so the one-dimensional derivation carries over unchanged.

Interactions. The free Lagrangian is the building block: external fields couple by adding terms rather than modifying $-mc^2/\gamma$. For electromagnetism (minimal coupling),

with $(\phi,\mathbf A)$ the scalar and vector potentials. This separability — free part plus interaction — is a feature of the Lagrangian formulation that the pure collision/symmetry analysis does not by itself provide; it is an additional principle (local coupling to fields).

Massless particles. For $m=0$ the action $-mc^2\int d\tau$ vanishes, and a massless particle’s worldline has $d\tau=0$ (it travels at $v=c$), so this Lagrangian does not apply. Massless particles are described by an action extremized over an arbitrary affine parameter, e.g. $S\propto\int g_{\mu\nu}\dot x^\mu\dot x^\nu\,d\lambda$, a separate construction outside the scope of this note.

Summary¶

Both routes give

and each confirms the other. Route 1 shows that the collision-derived $p$ and $E$ are compatible with a unique Lagrangian (no free constant); Route 2 shows that same Lagrangian is forced by Lorentz invariance of the action, and it reproduces $p$ and $E$ without using the collision. The single postulate added beyond the kinetic-energy note is the action principle; given it, the relativistic free-particle Lagrangian is determined with no freedom.