Relativistic Lagrangian for a Free Particle

Take a stopwatch from here to there — same start event, same finish event — and the reading depends on the route. Coast in a straight line and the watch runs long. Dart out at a rifle bullet's speed and back, and it runs short: for a one-second trip at $1000\ \mathrm{m/s}$ , short by $5.56\times10^{-12}\ \mathrm{s}$ . Nature, left alone, always takes the long route. That single fact is a complete dynamical theory. Ask what a free particle extremises and there is exactly one Lorentz scalar attached to a bare worldline: its length, which is $c$ times the proper time. So $S = -mc^2\!\int\!\mathrm{d}\tau$ — and the minus sign is not decoration. It is what converts most proper time into least action. What comes out is a Lagrangian nobody would have guessed. $L = -mc^2/\gamma$ is not kinetic minus potential; for a free particle it is not even positive. Turn the Euler–Lagrange crank on it and out drops $\mathbf{p} = \gamma m\mathbf{v}$ , conserved, and a straight worldline.

Key takeaways

Signature $(+,-,-,-)$ : $\eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)$ , $ds^2 = c^2dt^2 - d\mathbf{x}^2$ , $u\!\cdot\!u = +c^2$ , $p\!\cdot\!p = +m^2c^2$ . Timelike $\Rightarrow$ positive norm. Every metric expression below is re-derived from this, never transcribed.
The only scalar a bare worldline supplies is its length, so the action is forced: $S = -mc^2\!\int_A^B\!\mathrm{d}\tau$ , the constant fixed by the Newtonian limit. In lab time, $L = -mc^2\sqrt{1 - v^2/c^2} = -mc^2/\gamma$ : not $T - V$ , not $\tfrac12\gamma mv^2$ , and negative. Only $L + mc^2 \to \tfrac12 mv^2$ as $v \ll c$ .
Euler–Lagrange gives $\partial L/\partial\mathbf{v} = \gamma m\mathbf{v}$ — the canonical momentum is the relativistic momentum — and $\mathrm{d}(\gamma m\mathbf{v})/\mathrm{d}t = 0$ : the worldline is straight.
The covariant form $S = -mc\!\int\!\sqrt{\eta_{\mu\nu}\dot x^\mu \dot x^\nu}\,\mathrm{d}\lambda$ carries the prefactor $mc$ — not $mc^2$ — and there is no minus under the root in this signature.
Owned result — "least action" is literal here, and provably so. For a family departing and returning at speed $w$ , $\tau(w) = T\sqrt{1-w^2/c^2}$ , so $\boxed{\tau''(0) = -T/c^2 < 0}$ — a strict maximum of proper time — and hence $S''(0) = +mT > 0$ , a strict minimum of the action. And $\tau \le T$ for every timelike rival, so the minimum is global.
Key idea: the free relativistic particle has no equation of motion to postulate. It has a stopwatch, and it takes the route that maximises the reading. The momentum, the energy and the straight line are all bookkeeping on that one sentence.

The core insight

Newtonian mechanics builds $L$ out of energies and then checks that the action is stationary. Relativity runs the logic backwards: it builds $L$ out of geometry — the invariant length of the worldline — and lets the energies fall out. There is precisely one scalar available, $\int\!\mathrm{d}\tau$ , so the action is not chosen, it is forced; and the minus sign in $S = -mc^2\!\int\!\mathrm{d}\tau$ is forced too, by the demand that the low-speed limit give $+\tfrac12 mv^2$ . Once both signs are fixed, "least action" and "maximal ageing" are the same statement.

Where this lives today

Every relativistic theory in existence is written as an action, for the reason at the top of this article: an action can be a Lorentz scalar, and only a scalar's extremisation means the same thing to every observer. Field theory does not generalise $F = ma$ ; it generalises $S = -mc^2\!\int\!\mathrm{d}\tau$ . Add $-q\!\int\!A_\mu\,\mathrm{d}x^\mu$ and you have the electrodynamics of a charged particle (`aq-sr-charged-particle-lagrangian-em-field`). Add nothing at all, but let the metric vary, and $S = -mc\!\int\!\sqrt{g_{\mu\nu}\dot x^\mu\dot x^\nu}\,\mathrm{d}\lambda$ is the geodesic action of general relativity — the same action, $\eta$ promoted to $g$ . Free fall is free motion, and a falling apple obeys the principle of maximal ageing. And Feynman's path integral says the classical worldline is not chosen but dominant: every path contributes $e^{iS/\hbar}$ , and the stationary-phase point is the one computed above. The worked example gives that a hard edge — a 1 kg mass over one second has $|S| = 8.99\times10^{16}\ \mathrm{J\,s} \approx 8.5\times10^{50}\,\hbar$ . The phase turns over $10^{50}$ times, everything cancels except the stationary point, and classical mechanics is what is left. The variational principle is not a principle at all — it is what quantum mechanics looks like when $\hbar$ is small.

The physical picture

Suppose you knew nothing about relativistic dynamics and had to invent it. One constraint: the action must be a Lorentz scalar, or different observers would disagree about which worldline is physical, and the relativity principle is dead on arrival. What scalars can a worldline supply? Not $\mathbf{v}$ — frame-dependent. Not $t$ — a coordinate. A bare curve in Minkowski space, with no fields on it and no other structure, has exactly one invariant attached: its length, $\int\!\mathrm{d}s = c\!\int\!\mathrm{d}\tau$ . So the action must be $S = \kappa\!\int\!\mathrm{d}\tau$ , and all of free-particle relativistic dynamics is the job of fixing the constant $\kappa$ . The next section does it in two lines and gets $\kappa = -mc^2$ — magnitude and sign, out of the Newtonian limit alone. Now look at what that minus sign has done. Proper time is what is extremised, and a straight timelike worldline is the longest curve between two timelike-separated events — the opposite of the Euclidean intuition, and the whole content of the twin paradox. The minus sign turns that maximum into a minimum of $S$ . "Least action" survives as a literal statement, and it survives because ageing is maximal. Get the sign wrong and you predict that a free particle takes the route that ages it least — the moving twin's route — a particle that spontaneously accelerates for no reason. The Lagrangian this produces is genuinely strange: $L = -mc^2/\gamma$ is negative, is not kinetic minus potential, and does not vanish when the particle stops — it goes to $-mc^2$ . It has no business working. It works.

The core equations

Proper time is owned by `aq-sr-proper-time` and the interval by `aq-sr-invariant-spacetime-interval`; we recall them and build.

c^2\,\mathrm{d}\tau^2 \;=\; \eta_{\mu\nu}\,\mathrm{d}x^\mu \mathrm{d}x^\nu \;=\; c^2\mathrm{d}t^2 - \mathrm{d}\mathbf{x}^2 \qquad\Longrightarrow\qquad \mathrm{d}\tau = \sqrt{1 - \frac{v^2}{c^2}}\;\mathrm{d}t = \frac{\mathrm{d}t}{\gamma} \qquad (1)

where: $\eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)$ . On a timelike worldline $\eta_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu$ is positive here, so $\mathrm{d}\tau$ is real with no minus sign under the root. A source writing $\sqrt{-\eta_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu}$ is in $(-,+,+,+)$ ; transcribed here it asks for the root of a negative number — which this article's own base does, one card away from getting it right.

\boxed{\;S = -mc^2\!\int_A^B\!\mathrm{d}\tau\;} \qquad\Longrightarrow\qquad S = \int\! L\,\mathrm{d}t, \qquad \boxed{\;L = -mc^2\sqrt{1-\frac{v^2}{c^2}} = -\frac{mc^2}{\gamma}\;} \qquad (2)

where: $L$ is negative, bounded above by $0$ , below by $-mc^2$ , and depends on $|\mathbf{v}|$ only.

\mathbf{p} \;=\; \frac{\partial L}{\partial\mathbf{v}} \;=\; \gamma m\mathbf{v}, \qquad\qquad \frac{\mathrm{d}}{\mathrm{d}t}\big(\gamma m\mathbf{v}\big) = 0 \qquad (3)

where: for a free particle the canonical and mechanical momenta coincide — a coincidence destroyed the instant a field is switched on (`aq-sr-charged-particle-lagrangian-em-field`).

\boxed{\;S = -mc\!\int\!\sqrt{\eta_{\mu\nu}\,\dot x^\mu \dot x^\nu}\;\mathrm{d}\lambda\;}, \qquad \dot x^\mu \equiv \frac{\mathrm{d}x^\mu}{\mathrm{d}\lambda} \qquad (4)

where: the prefactor is $\boldsymbol{mc}$ , not $mc^2$ — a dimensional lock, not a taste. $\sqrt{\eta_{\mu\nu}\dot x^\mu\dot x^\nu}\,\mathrm{d}\lambda$ is a length, so $mc\times$ length $=\mathrm{J\,s}$ ✓, while $mc^2\times$ length is J·m, which is not an action. And (4) makes no reference to $\lambda$ : reparametrise and the $\mathrm{d}\lambda$ cancels the Jacobian in $\dot x$ . That invariance is what makes the mass shell automatic.

S_e = -\frac12 \int\!\Big( e^{-1}\,\eta_{\mu\nu}\dot x^\mu\dot x^\nu \;+\; e\,m^2c^2 \Big)\,\mathrm{d}\lambda \qquad (5)

where: $e(\lambda) > 0$ is the einbein. Varying $e$ (no $\dot e$ appears, so this is algebra) gives $e_* = \sqrt{\dot x\!\cdot\!\dot x}/(mc)$ — real and positive here — and substituting back returns (4) exactly. The overall minus and the plus between the terms are fixed by our signature: the mostly-plus books write $+\tfrac12\!\int(e^{-1}\dot x^2 - em^2c^2)$ , and copying that here gives $e^2 < 0$ — an imaginary einbein. Both this base and 1975's print it.

Working it out

Derivation AFrom the only available scalar to the only available Lagrangian

The goal is to show that the free-particle action is not a modelling choice but a derivation: requiring $S$ to be a Lorentz scalar built from the worldline alone leaves exactly one candidate up to a constant, and the Newtonian limit fixes that constant — magnitude and sign — with nothing left over.

Step 1. Enumerate the invariants. A worldline $x^\mu(\lambda)$ , with no fields and no extra structure, offers the tangent $\dot x^\mu$ and the metric $\eta_{\mu\nu}$ . The only scalar is $\eta_{\mu\nu}\dot x^\mu\dot x^\nu$ , and the only reparametrisation-invariant integral over the curve is $\int\!\sqrt{\eta_{\mu\nu}\dot x^\mu\dot x^\nu}\,\mathrm{d}\lambda = \int\!\mathrm{d}s = c\!\int\!\mathrm{d}\tau$ . Why: a Lagrangian depending on the arbitrary label $\lambda$ is not a theory, and no other power of the scalar survives that test. The action is forced.

Step 2. Fix the constant. Write $S = \kappa\!\int\!\mathrm{d}\tau$ . From (1), $L = \kappa\sqrt{1-v^2/c^2} = \kappa - \kappa v^2/(2c^2) + O(v^4)$ . Newton requires the velocity-dependent part to be $+\tfrac12 mv^2$ , so $-\kappa/(2c^2) = m/2$ and $\kappa = -mc^2$ . Why: both the size and the sign fall out of one limit — and the sign is what makes the kinetic term positive and, as the next card shows, "least" true.

Step 3. Read off the Lagrangian and check what it is not. $L = -mc^2/\gamma$ : negative everywhere, equal to $-mc^2$ at rest, bounded above by $0$ . Expanding, $L = -mc^2 + \tfrac12 mv^2 + \tfrac18 m v^4/c^2 + \cdots$ . Why: only the second term resembles Newton. And the coefficient is $\tfrac18$ , not $\tfrac38$ : $\tfrac38$ belongs to the kinetic-energy series $(\gamma-1)mc^2$ , since $\gamma$ expands with plus signs and $1/\gamma$ with minus signs. This draft carried $\tfrac38$ until sympy refused it.

Step 4. Turn the crank. $L$ has no explicit $\mathbf{x}$ , so $\partial L/\partial\mathbf{x} = 0$ , and

\frac{\partial L}{\partial v_i} \;=\; -mc^2\cdot\frac{-v_i/c^2}{\sqrt{1-v^2/c^2}} \;=\; \frac{m v_i}{\sqrt{1-v^2/c^2}} \;=\; \gamma m v_i,

so Euler–Lagrange reads $\mathrm{d}(\gamma m\mathbf{v})/\mathrm{d}t = 0$ . Why: the payoff. A Lagrangian resembling nothing in Newtonian mechanics has, in one differentiation, produced the momentum that is actually conserved in collisions — without ever assuming it. Crank $\tfrac12\gamma mv^2$ , which has the correct low-speed limit, and the conjugate momentum is not $\gamma m\mathbf{v}$ .

Step 5. Deduce the motion. $|\gamma m\mathbf{v}| = mc\,\beta\gamma$ and $\beta\mapsto\beta\gamma$ is strictly increasing on $[0,1)$ , so constancy of $\gamma m\mathbf{v}$ forces constancy of $\mathbf{v}$ : the worldline is straight. Why: the monotonicity is what lets you read a velocity back off a momentum. *Verified in verify_aq-sr-relativistic-lagrangian-free-particle.py: sympy confirms $\partial L/\partial v_i = \gamma m v_i$ ; the series coefficients $-mc^2,\ \tfrac12 m,\ \tfrac18 m/c^2$ are checked term by term against the $\tfrac38$ of the kinetic-energy series; and $\tfrac12\gamma mv^2$ is shown to give a different conjugate momentum.*

Go deeper · Derivation B · 2 minThe owned result: the straight worldline is the LONGEST, so the action really is LEAST

Derivation BThe owned result: the straight worldline is the LONGEST, so the action really is LEAST

Textbooks hedge — they say "stationary action" and quietly retire the word least, because in Newtonian mechanics the stationary point genuinely can be a saddle. Here we can do better, and this is the structural heart of the article: for a free relativistic particle the straight worldline is a strict minimum of $S$ — because it is a strict maximum of proper time. Prove the second derivative's sign, then show the maximum is global.

Step 1. Set up a one-parameter family. Fix $A = (0,\mathbf{0})$ and $B = (T,\mathbf{0})$ , timelike separated. Let the particle depart at constant speed $w$ , turn round at $t = T/2$ , and return at speed $w$ , arriving at $B$ on time. Both endpoints are fixed for every $w \in [0,c)$ ; $w = 0$ is the straight worldline. Why: a legitimate variation holds the endpoints fixed, and this family does — and one number is all a second-derivative test needs.

Step 2. Compute the proper time. $\gamma$ is constant on each leg, so

\tau(w) \;=\; \int\!\frac{\mathrm{d}t}{\gamma} \;=\; T\sqrt{1 - \frac{w^2}{c^2}}.

Why: no integral is needed — $\gamma$ is piecewise constant, and smoothing the kink changes $\tau$ arbitrarily little.

Step 3. Two derivatives at $w = 0$ . From $\tau(w) = T\big(1 - \tfrac{w^2}{2c^2} - \tfrac{w^4}{8c^4} - \cdots\big)$ ,

\tau'(0) = 0, \qquad\qquad \boxed{\;\tau''(0) = -\frac{T}{c^2} \;<\; 0\;}.

Why: a vanishing first derivative says only stationary — all a variational principle ever gives you. The negative second derivative is the content: a strict maximum of $\tau$ . And it is $-T/c^2$ : the longer the trip, the sharper the maximum.

Step 4. Translate to the action. $S(w) = -mc^2\tau(w)$ , so $S'(0) = 0$ and

S''(0) \;=\; -mc^2\cdot\Big(\!-\frac{T}{c^2}\Big) \;=\; +\,mT \;>\; 0.

Why: the minus sign earns its keep here. Positive second derivative $\Rightarrow$ strict minimum. For the free relativistic particle, "least action" is not a historical courtesy but a theorem — and $-mc^2$ is what makes it one.

Step 5. Upgrade to a global statement. For any timelike worldline joining $A$ to $B$ , $\tau = \int_0^T\!\mathrm{d}t/\gamma \le \int_0^T\!\mathrm{d}t = T$ , with equality iff $v(t) \equiv 0$ . So $S = -mc^2\tau \ge -mc^2T$ always. Why: the inequality is trivial ( $\gamma \ge 1$ ); its consequence is not. The physical worldline is the global minimiser over all timelike competitors — stronger than anything Newtonian mechanics offers. *Verified in verify_aq-sr-relativistic-lagrangian-free-particle.py: sympy differentiates $\tau(w) = T\sqrt{1-w^2/c^2}$ twice and asserts $\tau'(0) = 0$ and $\tau''(0) = -T/c^2$ exactly, and $S''(0) = +mT$ exactly; the global bound $\tau \le T$ is checked on a random ensemble of timelike worldlines.*

Go deeper · Derivation C · 2 minThe covariant action: prefactor mc, no minus under the root, and a mass shell you never assumed

Derivation CThe covariant action: prefactor mc, no minus under the root, and a mass shell you never assumed

The lab-time Lagrangian is convenient and it is a lie of omission — it singles out $t$ , the one thing relativity says not to do. The covariant form treats all four coordinates alike, at the price of a gauge symmetry: reparametrisation invariance. The goal is to derive the covariant momentum from scratch in our signature and watch $p\!\cdot\!p = m^2c^2$ appear as an identity, not an input.

Step 1. Write the action with the right prefactor, and check the root. Since $\int\!\mathrm{d}\tau = \tfrac1c\!\int\!\sqrt{\eta_{\mu\nu}\dot x^\mu\dot x^\nu}\,\mathrm{d}\lambda$ ,

S = -mc\!\int\!\sqrt{\eta_{\mu\nu}\dot x^\mu\dot x^\nu}\,\mathrm{d}\lambda \;\equiv\; \int\!\mathcal{L}\,\mathrm{d}\lambda, \qquad \mathcal{L} = -mc\sqrt{\dot x\!\cdot\!\dot x}.

Why: the $c^2$ became a $c$ — one factor was consumed converting $\mathrm{d}\tau$ into $\mathrm{d}s$ , and $[mc]\cdot\mathrm{m} = \mathrm{J\,s}$ ✓. This article's base writes $-mc\!\int\!\mathrm{d}\tau$ and 1975's writes $-mc^2\!\int\!\sqrt{\eta\dot x\dot x}\,\mathrm{d}\lambda$ ; both are off by a factor of $c$ , in opposite directions.

Step 2. And there is no minus under the root: on a timelike worldline $\eta_{\mu\nu}\dot x^\mu\dot x^\nu = c^2\dot t^2 - |\dot{\mathbf{x}}|^2 > 0$ here. Why: the most-transcribed error in the subject. The mostly-plus literature writes $\sqrt{-\eta_{\mu\nu}\dot x^\mu\dot x^\nu}$ because in that signature the timelike norm is negative and the minus rescues reality. Copy it here and the action is imaginary on every physical worldline.

Step 3. Vary. $\displaystyle p_\mu = \frac{\partial\mathcal{L}}{\partial\dot x^\mu} = -\,\frac{mc\,\dot x_\mu}{\sqrt{\dot x\!\cdot\!\dot x}}$ . Why: $p_\mu$ is homogeneous of degree zero in $\dot x$ — blind to how fast you run along the curve. Reparametrisation invariance made visible, and about to have a consequence.

Step 4. Contract, and watch the mass shell fall out.

p_\mu p^\mu \;=\; \frac{m^2c^2\,\dot x_\mu \dot x^\mu}{\dot x\!\cdot\!\dot x} \;=\; +\,m^2c^2.

Why: an identity, not an equation of motion — it holds for every parametrisation and every worldline, physical or not. The four momenta are not independent; they are constrained. And positive, because a massive particle's four-momentum is timelike and timelike means positive norm here: a base printing $p\!\cdot\!p = -m^2c^2$ has flipped its metric, whatever it says on the tin. The consequences — vanishing canonical Hamiltonian, first-class constraint, Klein–Gordon — are `aq-sr-relativistic-hamiltonian-formalism`'s.

Step 5. Fix the gauge. Choose $\lambda = \tau$ : then $\sqrt{\dot x\!\cdot\!\dot x} = c$ , the physical four-momentum is $p^\mu = mu^\mu = \gamma m(c,\mathbf{v}) = (E/c,\,\mathbf{p})$ with spatial part $\gamma m\mathbf{v}$ — equation (3) — and Euler–Lagrange gives $\mathrm{d}^2x^\mu/\mathrm{d}\tau^2 = 0$ : a straight worldline, covariantly. Why: the same physics, no frame chosen anywhere. In curved spacetime the right side picks up the Christoffel term and this becomes the geodesic equation — which is why this action, and not $F = ma$ , is what generalises. *Verified in verify_aq-sr-relativistic-lagrangian-free-particle.py: sympy differentiates $\mathcal{L} = -mc\sqrt{\eta_{\mu\nu}\dot x^\mu\dot x^\nu}$ with $\eta = \mathrm{diag}(+1,-1,-1,-1)$ and confirms $p_\mu p^\mu = +m^2c^2$ identically in $\dot x$ , degree-zero homogeneity of $p_\mu$ , both prefactors dimension-checked (J·s vs J·m), and $e_* = \sqrt{\dot x\cdot\dot x}/(mc) > 0$ back-substituting exactly — while the base's mostly-plus einbein gives $e^2 < 0$ .*

Figure

PUNCHLINE — "least action" is not a figure of speech. The family: a particle leaves event A, darts out at speed w, turns round and returns to event B. w = 0 is the straight worldline. LEFT: the worldlines. RIGHT: the two curves that matter. Proper time τ(w) = T√(1−w²/c²) has a strict MAXIMUM at w = 0 — τ′(0) = 0 and τ″(0) = −T/c² ＜ 0 — so the straight path is the one that AGES MOST. The action S(w) = −mc²τ(w) therefore has a strict MINIMUM there: S″(0) = +mT ＞ 0. The minus sign in S = −mc²∫dτ is exactly what turns maximal ageing into least action. And the maximum is GLOBAL: τ = ∫dt/γ ≤ T for every timelike competitor, with equality only for the straight line

Limiting cases

Case	Result	Meaning
$v \ll c$	$L \to -mc^2 + \tfrac12 mv^2$	Newton, plus an inert constant
next order	$L = -mc^2 + \tfrac12 mv^2 + \tfrac18\dfrac{mv^4}{c^2} + \cdots$	the first correction Newton misses. Coefficient $\tfrac18$ , not $\tfrac38$
$v \to c^-$	$L \to 0^-$ , $\;\gamma m\mathbf{v} \to \infty$	the Lagrangian degenerates while the momentum diverges
$m \to 0$	$-mc^2\!\int\!\mathrm{d}\tau \to 0$ — useless	the square-root action dies for a photon. The einbein survives: at $m=0$ , varying $e$ enforces $\dot x\!\cdot\!\dot x = 0$

Common trapSignature: this base is split against itself. Re-derive every metric; transcribe none

The AllQFT base prints, cards apart: eq[0] $c^2\mathrm{d}\tau^2 = c^2\mathrm{d}t^2 - \mathrm{d}\mathbf{x}^2$ ✓ ours and eq[5] $p_\mu p^\mu = +m^2c^2$ ✓ ours — but also eq[6] $S = -mc\!\int\!\sqrt{-\eta_{\mu\nu}\dot x^\mu\dot x^\nu}\,\mathrm{d}\lambda$ , whose symbol table declares $(-,+,+,+)$ ; eq[7] the mostly-plus einbein $+\tfrac12\!\int\!(e^{-1}\dot x^2 - e\,m^2c^2)$ ; and a derivation card with $p_\mu = mc\,\dot x_\mu/\sqrt{-\dot x^\nu\dot x_\nu}$ .

The base contradicts itself in print. In $(+,-,-,-)$ a timelike $\eta_{\mu\nu}\dot x^\mu\dot x^\nu$ is positive, so $\sqrt{-\eta_{\mu\nu}\dot x^\mu\dot x^\nu}$ is imaginary on every physical worldline, and the mostly-plus einbein gives $e_*^2 < 0$ — an imaginary einbein. Never transcribe a metric expression. Re-derive it, test it on a timelike vector, then write it down.

Worked examples

1 · What does it cost, in action, to not go straight?

A 1 kg mass floats free for $T = 1\ \mathrm{s}$ between two events at the same place in the lab. A rival worldline darts out and back at $w = 1000\ \mathrm{m/s}$ — a rifle bullet — arriving at the same event at the same time. Which has the greater action, and by how much?

Step 1 — identify. Both are timelike worldlines with identical endpoints, so both are legal competitors. The rival is the $w = 1000\ \mathrm{m/s}$ member of the family above; we need $\tau(w) = T\sqrt{1-w^2/c^2}$ and $S = -mc^2\tau$ .

Step 2 — the small parameter, and a warning. $\beta = w/c = 3.33564\times10^{-6}$ , so $\beta^2 = 1.11265\times10^{-11}$ . The deficit $\Delta\tau = T\big(1-\sqrt{1-\beta^2}\big)$ is a difference of two numbers agreeing to eleven decimal places. Evaluate it in the algebraically identical, numerically stable form:

\Delta\tau \;=\; \frac{T\beta^2}{1+\sqrt{1-\beta^2}} \;=\; \mathbf{5.56325\times10^{-12}\ s}.

Naive float64 subtraction returns $5.56333\times10^{-12}$ — wrong in the fifth digit, and it looks perfectly respectable. Runbook trap 26, met on a number small enough to be harmless.

Step 3 — the action difference. $S_{\text{straight}} = -mc^2T = \mathbf{-8.98755\times10^{16}\ J\,s}$ , and

\Delta S \;=\; S(w) - S_{\text{straight}} \;=\; +\,mc^2\,\Delta\tau \;=\; \mathbf{5.00000\times10^{5}\ J\,s} \;>\; 0.

Step 4 — the boundary number. The sign: $\Delta S > 0$ — the straight worldline really does have the smaller action. Least action, literally. The magnitude: $\Delta S = 5.00000\times10^{5}\ \mathrm{J\,s}$ , while $\tfrac12 mw^2T = 5\times10^{5}\ \mathrm{J\,s}$ exactly — the relativistic action penalty for wandering is the Newtonian kinetic action, to a relative accuracy of $2.8\times10^{-12}$ . That is the $c\to\infty$ limit doing its job, and the cleanest demonstration that this bizarre negative Lagrangian contains Newton whole. The ratio: $\Delta S/|S| = 5.56\times10^{-12}$ — nature discriminates between worldlines on the twelfth decimal place of the action, and gets it right every time.

2 · Where Newton's Lagrangian actually breaks: a benchtop electron microscope

A scanning electron microscope accelerates electrons through 30 kV. If you designed its optics with the Newtonian Lagrangian $\tfrac12 mv^2$ , how wrong would the momentum be?

Step 1 — identify. $K = 30.0\ \mathrm{keV}$ , $m_ec^2 = 511.00\ \mathrm{keV}$ . The true Lagrangian gives $\mathbf{p} = \gamma m\mathbf{v}$ ; the Newtonian one gives $m\mathbf{v}$ . Their ratio is exactly $\gamma$ , so the fractional error is $\gamma - 1 = K/(m_ec^2)$ .

Step 2 — the Lorentz factor. $\gamma = 1 + 30.0/511.00 = \mathbf{1.058709}$ , and $\beta = \sqrt{(\gamma-1)(\gamma+1)}/\gamma = \mathbf{0.328376}$ (the stable form — never $\sqrt{1-1/\gamma^2}$ by naive subtraction), so $v = \mathbf{9.84447\times10^{7}\ m/s}$ : a third of the speed of light, in a machine that sits on a bench.

Step 3 — apply. $L = -m_ec^2/\gamma = \mathbf{-482.66\ keV}$ , while $\tfrac12 m_ev^2 = \tfrac12 m_ec^2\beta^2 = \mathbf{27.551\ keV}$ . Opposite signs, a factor of eighteen apart. What is comparable: $L + m_ec^2 = \mathbf{28.336\ keV}$ against $27.551$ keV — the Newtonian kinetic term is 2.85 % low.

Step 4 — the boundary number. The momentum error is worse, and it is the momentum that focuses the beam:

\frac{p_{\text{rel}}}{p_{\text{Newt}}} \;=\; \frac{\gamma m_e v}{m_e v} \;=\; \gamma \;=\; 1.058709 \qquad\Longrightarrow\qquad \textbf{5.87 \% too small.}

A 30 kV electron microscope — an ordinary instrument — is already 5.9 % relativistic in its momentum, which is why electron-optics codes carry $\gamma$ . Run it backwards: the Newtonian momentum is 1 % wrong as soon as $\gamma = 1.01$ , i.e. $\beta = 0.140$ , $v = 4.21\times10^{7}\ \mathrm{m/s}$ — an electron of kinetic energy just $\mathbf{5.11\ keV}$ . Relativistic dynamics is not a high-energy speciality. It starts in the kilovolt range.

Physics of the portrait

Least action is the oldest idea in theoretical physics that is still load-bearing — and relativity is where it stopped being a convenience and became the only way to write the theory down.

Origins · 2 minPrinciple of least action and how this was discovered

Principle of least action

1744–1950 · Maupertuis and Euler independently formulate it (1744) · Hamilton casts dynamics as a variational problem (1834-35) · Planck writes the relativistic kinetic potential −mc²√(1−v²/c²) (1906) · Minkowski's proper time makes the action a LENGTH (1908) · Dirac's constrained dynamics explains the vanishing Hamiltonian (1950)

The principle arrives twice in 1744, from opposite directions: Maupertuis's Accord de différentes lois de la nature, arguing on grounds as much theological as physical that nature economises a quantity he called action, and Euler's Methodus inveniendi, which gave the mathematically serious version. Hamilton, in 1834–35, reduced the motion of any system of attracting or repelling points to one characteristic function, and produced the canonical equations as a by-product. Relativity then changed the principle's status. In Newtonian mechanics $L = T - V$ is a recipe — elegant, derivable from $F = ma$ , therefore optional. In relativity the recipe fails outright, there is no $T - V$ , and the variational principle becomes the only frame-independent way to specify dynamics: an action can be a scalar when an equation of motion cannot be. Planck saw it at once. His 1906 Das Prinzip der Relativität und die Grundgleichungen der Mechanik, presented in March, was the first public defence of Einstein's theory by a senior physicist — and it did not merely defend. Planck wrote relativistic dynamics as Hamilton's principle with kinetic potential $H = -mc^2\sqrt{1-q^2/c^2}$ and read off $\xi = \partial H/\partial\dot x = m\dot x/\sqrt{1-q^2/c^2}$ : the relativistic momentum, from a Lagrangian, two years before anyone had spacetime to interpret it in. Minkowski supplied the interpretation in 1908: with Eigenzeit, the strange Lagrangian becomes obvious — the action is $-mc$ times the length of the worldline. What Planck found by algebra, Minkowski turned into geometry. The last piece is Dirac's: the covariant action's canonical Hamiltonian vanishes identically, which looks like a catastrophe until you see it is a constraint — his 1950 Generalized Hamiltonian Dynamics.

History

1744

— Maupertuis publishes "Accord de differentes lois de la nature qui avaient jusqu'ici paru incompatibles" in the Paris Memoires, proposing that nature extremises a quantity he calls action.

1744

— Euler, in "Methodus inveniendi lineas curvas maximi minimive proprietate gaudentes", gives the calculus-of-variations formulation and applies the action integral to a moving particle.

1834

— Hamilton's "On a General Method in Dynamics" (Phil. Trans. R. Soc., part II for 1834, pp. 247-308) reduces dynamics to one characteristic function; the Second Essay (part I for 1835) gives the canonical equations.

1906

— Planck, "Das Prinzip der Relativitat und die Grundgleichungen der Mechanik" (Verh. Dtsch. Phys. Ges. 8), writes relativistic mechanics as Hamilton's principle with kinetic potential H = -mc^2 sqrt(1 - q^2/c^2), obtaining p = m v / sqrt(1 - v^2/c^2).

1908

— Minkowski's "Raum und Zeit" introduces proper time, making the action -mc times the invariant length of the worldline.

1950

— Dirac, "Generalized Hamiltonian Dynamics" (Can. J. Math. 2, 129-148), develops constrained Hamiltonian dynamics, explaining the vanishing canonical Hamiltonian of a reparametrisation-invariant action.

Going further

Texts. Landau & Lifshitz, Classical Theory of Fields, §8 (note their signature) · Rindler, Introduction to Special Relativity, 2nd ed. · Goldstein, ch. 7 · Taylor & Wheeler, Spacetime Physics · Planck, Verh. Dtsch. Phys. Ges. 8, 136 (1906) · Dirac, Can. J. Math. 2, 129 (1950).

Provenance. Signature $(+,-,-,-)$ throughout; timelike norm positive, $p\!\cdot\!p = +m^2c^2$ , no minus under the root. Every equation is re-derived from that metric — none transcribed — and every number above is recomputed at 50-digit precision in verify_aq-sr-relativistic-lagrangian-free-particle.py, which also greps this file for each printed value.

Base defects found and repaired. The base is signature-split against itself: eq[0] and eq[5] are ours, but eq[6] prints $S = -mc\!\int\!\sqrt{-\eta_{\mu\nu}\dot x^\mu\dot x^\nu}\,\mathrm{d}\lambda$ and declares $(-,+,+,+)$ in its own symbol table; eq[7] prints the mostly-plus einbein, whose $e$ -equation gives $e^2 < 0$ ; and a derivation card carries $p_\mu = mc\,\dot x_\mu/\sqrt{-\dot x^\nu\dot x_\nu}$ . All three are repaired above and asserted symbolically. The base's card "Energy and the relativistic Hamiltonian" is 1977's payload and is handed over, not re-derived. And one defect of our own: this draft first printed the $v^4$ coefficient of $L$ as $\tfrac38$ — the kinetic-energy coefficient — and sympy refused it. It is $\tfrac18$ . Recorded, not quietly fixed.

Scope. Proper time and the interval are `aq-sr-proper-time` and `aq-sr-invariant-spacetime-interval`. The Legendre transform to $H = \gamma mc^2$ , the vanishing covariant Hamiltonian and the Klein–Gordon step are `aq-sr-relativistic-hamiltonian-formalism`. Reparametrisation invariance as a principle is `aq-sr-relativistic-action-proper-time`. Minimal coupling is `aq-sr-charged-particle-lagrangian-em-field`.

Builds on. `aq-sr-proper-time` · `aq-sr-invariant-spacetime-interval` · `aq-sr-four-velocity` · `aq-sr-four-momentum` · `aq-sr-minkowski-metric`. Leads to. `aq-sr-relativistic-action-proper-time` · `aq-sr-charged-particle-lagrangian-em-field` · `aq-sr-relativistic-hamiltonian-formalism` · `aq-sr-tachyons-superluminal-particles`.

If the physical path maximises proper time, why is it called the principle of LEAST action?

Because $S = -mc^2\!\int\!\mathrm{d}\tau$ , and the minus sign converts the maximum of $\tau$ into a minimum of $S$ . They are the same statement. The sign is not free: it is fixed by demanding that $L$ 's low-speed expansion contain $+\tfrac12 mv^2$ — so the sign that makes "least action" true is the sign that makes kinetic energy positive. And for the free relativistic particle "least" is not a euphemism for "stationary": $S''(0) = +mT > 0$ , and $\tau \le T$ makes the minimum global. Newtonian mechanics cannot say that — a harmonic oscillator's action stops being a minimum past half a period.

Could I use L = one-half gamma m v squared instead? It has the right non-relativistic limit.

It does, and it is wrong. That is the whole point: the low-speed limit is necessary, never sufficient. Feed $\tfrac12\gamma mv^2$ to Euler–Lagrange and $\partial L/\partial\mathbf{v}$ is not $\gamma m\mathbf{v}$ — it carries extra powers of $\gamma$ — so what it conserves is not the momentum actually conserved in a collision. Substitute your candidate into the equation it is supposed to produce, before you believe it. That one habit would have caught most of the defects in this corpus, including the $\tfrac38$ this article's own first draft carried.

Why is the covariant momentum homogeneous of degree zero, and why does anyone care?

$p_\mu = -mc\,\dot x_\mu/\sqrt{\dot x\!\cdot\!\dot x}$ is unchanged under $\dot x^\mu \to k\dot x^\mu$ — numerator and denominator each pick up a factor $k$ . That is reparametrisation invariance made visible: the momentum knows the direction of the worldline and nothing about how fast you run along it, because $\lambda$ is a label you invented. The consequence is large: contract $p_\mu$ with itself and the $\sqrt{\dot x\cdot\dot x}$ cancels completely, leaving $p\!\cdot\!p = m^2c^2$ identically, on and off shell. The momenta are not independent; the theory is constrained — and a constrained theory has a vanishing canonical Hamiltonian, which is exactly why the covariant formalism looks so strange and why Klein–Gordon appears when you quantise it.

Why does the square-root action fail for a photon, and what does the einbein do about it?

$S = -mc^2\!\int\!\mathrm{d}\tau$ has an $m$ out front and a $\mathrm{d}\tau$ that vanishes on a null worldline: it collapses to $0\times\infty$ , and there is no proper time to parametrise by either. The einbein form is polynomial in $\dot x$ and perfectly finite at $m=0$ ; vary $e$ with the mass term gone and the equation of motion is simply $\dot x\!\cdot\!\dot x = 0$ — the null constraint, delivered by the formalism rather than imposed by hand. For $m \ne 0$ the two actions are exactly equivalent, so nothing is lost: the trick buys a smooth massless limit at the price of one gauge redundancy. Applied to a worldsheet, the same manoeuvre turns Nambu–Goto into Polyakov.

Is the additive constant minus mc squared in L physically meaningless?

Its value is inert and its sign is not — the two halves are usually conflated. Adding a constant to $L$ shifts $S$ by (constant $\times$ elapsed time) and changes no equation of motion, so $-mc^2/\gamma$ and $-mc^2/\gamma + mc^2$ are dynamically identical. But you cannot flip the sign of the $mc^2$ in front of the integral: that constant multiplies $\int\!\mathrm{d}\tau$ , and flipping it turns the minimum into a maximum, predicting a free particle that spontaneously wanders. The constant that is inert is the one you add; the load-bearing one is the one you multiply by.

Nothing here ever assumed E = gamma m c squared. Where did the energy go?

It is still coming. This variational principle produces the momentum directly, because momentum is $\partial L/\partial\mathbf{v}$ , and it produces the equation of motion. The energy is derived: the conserved quantity associated with $L$ 's lack of explicit time dependence — Noether — computed by the Legendre transform $H = \mathbf{p}\cdot\mathbf{v} - L$ . With $L = -mc^2/\gamma$ and $\mathbf{p} = \gamma m\mathbf{v}$ it collapses to $\gamma mc^2$ ; eliminating $\mathbf{v}$ gives $H = \sqrt{p^2c^2 + m^2c^4}$ . That is `aq-sr-relativistic-hamiltonian-formalism` — a separate article because the Legendre transform is where the canonical and mechanical momenta begin to part company, which they do the instant an electromagnetic field is switched on.