Relativistic Action and Proper-Time Principle

An electron crosses 2.00 m of laboratory in 10.0 ns. Between those two events there are infinitely many timelike worldlines it could have taken, each carrying its own proper time. The straight one reads $7.4494\ \mathrm{ns}$ . A rival that sprints out at $0.9c$ and coasts back reads $6.6834\ \mathrm{ns}$ — three-quarters of a nanosecond less, and it is always less. Nature picks the longest. To turn that into dynamics you must say what "picks" means, and the honest answer is not "minimises". It is stationary: the physical worldline is where the action stops changing to first order under any small deformation with fixed endpoints. Write that down covariantly — with an arbitrary parameter along the curve, because relativity forbids privileging $t$ — and something unasked-for happens. The theory develops a gauge symmetry. Relabel the worldline any way you like and the action does not notice. And a gauge symmetry always costs something: here, the four canonical momenta turn out not to be independent. They satisfy $p_\mu p^\mu = m^2c^2$ identically — before any equation of motion, for any curve, physical or not. The mass shell was never an input. It is the receipt for a symmetry you did not ask for.

Key takeaways

Signature $(+,-,-,-)$ : $\eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1)$ , $u\!\cdot\!u = +c^2$ , $p\!\cdot\!p = +m^2c^2$ . Timelike $\Rightarrow$ positive norm, so there is no minus under the root. Re-derived here, never transcribed.
The action is $S = -mc^2\!\int\!\mathrm{d}\tau = -mc\!\int\!\sqrt{\eta_{\mu\nu}\dot x^\mu\dot x^\nu}\,\mathrm{d}\lambda$ — prefactor $mc$ , not $mc^2$ : the root times $\mathrm{d}\lambda$ is a length, so $mc\times$ length is J·s.
Stationary, not least. $\delta S = 0$ to first order under any variation with fixed endpoints. Least has to be earned separately: for a free particle it can be (`aq-sr-relativistic-lagrangian-free-particle`), and in a gravitational field it generally cannot.
Vary with $\lambda$ arbitrary; gauge-fix to $\lambda = \tau$ only afterwards. $\mathrm{d}\tau$ depends on the path you are varying, so fixing it first holds the answer still while you look for it. And the Lagrangian is homogeneous of degree one in $\dot x^\mu$ — that is reparametrisation invariance, and it is a gauge symmetry.
Owned result. Euler's homogeneous-function theorem gives $\dot x^\mu\,\partial\mathcal{L}/\partial\dot x^\mu = \mathcal{L}$ , so the canonical Hamiltonian $H = \dot x^\mu p_\mu - \mathcal{L}$ vanishes identically; and $p_\mu = -mc\,\dot x_\mu/\sqrt{\dot x\!\cdot\!\dot x}$ contracts to $\boxed{p_\mu p^\mu = +m^2c^2}$ identically, for every parametrisation, on and off shell. The mass shell is a constraint, not an equation of motion.
Key idea: the covariant action does not tell you where the particle goes and then hand you a mass shell as a bonus. It hands you the mass shell for free — as the price of a symmetry — and only then, in the leftover degrees of freedom, tells you where the particle goes.

The core insight

Two sentences carry the article. (1) The action is $-mc$ times the length of the worldline, and a length does not care how you label points along a curve. (2) A Lagrangian that does not care about its own parameter is homogeneous of degree one in the velocities — and Euler's theorem then forces both $H \equiv 0$ and $p\!\cdot\!p = m^2c^2$ , identically. Everything in relativistic dynamics is downstream of that one homogeneity.

Where this lives today

The structure derived here — a gauge symmetry, a vanishing Hamiltonian, a first-class constraint — is not a quirk of the relativistic point particle. It is the template for every reparametrisation-invariant theory in physics, and each of them repeats these three lines with a different object on the worldline. Take the worldline to a worldsheet and the same square-root action becomes Nambu–Goto, whose two-dimensional reparametrisation invariance buys the Virasoro constraints instead of a single mass shell. Take the parametrised object to be spacetime itself and diffeomorphism invariance buys the Hamiltonian constraint $\mathcal{H} \approx 0$ of general relativity — which is why the Wheeler–DeWitt equation of canonical quantum gravity has no time in it. The "problem of time" in quantum gravity is this article's $H \equiv 0$ , one level up. And the shortest path from here to quantum mechanics is the constraint itself. Promote $p_\mu p^\mu = m^2c^2$ to an operator statement with $p_\mu \to i\hbar\partial_\mu$ and it reads $(\hbar^2\partial_\mu\partial^\mu + m^2c^2)\psi = 0$ : the Klein–Gordon equation, which is therefore not a new postulate but the quantised mass shell. That step is `aq-sr-relativistic-hamiltonian-formalism`'s.

The physical picture

Hamilton's principle in Newtonian mechanics is a statement about a function of time: among all paths $\mathbf{x}(t)$ from $\mathbf{x}_1$ at $t_1$ to $\mathbf{x}_2$ at $t_2$ , the physical one makes $\int L\,\mathrm{d}t$ stationary. Relativity cannot accept that sentence as written, because it privileges $t$ — and $t$ is a coordinate, not a fact. The repair is to stop parametrising by time and start parametrising by anything. A worldline is a curve $x^\mu(\lambda)$ ; $\lambda$ is an arbitrary label running from $\lambda_1$ at event $A$ to $\lambda_2$ at $B$ . The action must be built from the curve and not from the label, and the only object with that property is the curve's length:

S \;=\; -mc\!\int_{\lambda_1}^{\lambda_2}\!\sqrt{\eta_{\mu\nu}\,\dot x^\mu\dot x^\nu}\;\mathrm{d}\lambda \;=\; -mc\!\int\!\mathrm{d}s \;=\; -mc^2\!\int\!\mathrm{d}\tau.

Two things follow, and the second is the whole article.

First, "stationary" is the honest word, not "least". The variational statement is $\delta S = 0$ to first order — that and nothing more. For a free particle one can go further and prove a genuine minimum (`aq-sr-relativistic-lagrangian-free-particle` does); the moment gravity enters, the extremal worldline can be a saddle, and "least" stops being true while $\delta S = 0$ goes on holding.

Second, the theory has a redundancy. Reparametrise, $\lambda \to \lambda'(\lambda)$ with $\mathrm{d}\lambda'/\mathrm{d}\lambda > 0$ : the integrand picks up a Jacobian from $\dot x^\mu$ and loses exactly the same Jacobian from $\mathrm{d}\lambda$ . $S$ is unchanged, exactly — not approximately, not up to a boundary term. Four functions $x^\mu(\lambda)$ describe a curve with only three physical degrees of freedom, because one of the four is pure labelling. That redundancy is a gauge symmetry, and gauge symmetries always leave a mark. The mark here is that the four canonical momenta cannot be independent — and the relation among them is the mass shell.

Figure

Reparametrisation invariance: one worldline, three labels, ONE action. The SAME electron worldline from event A to event B (2.00 m in 10.0 ns) is shown three times. What differs is only the LABEL λ: uniformly spaced ticks of coordinate time t; of proper time τ (bunched, because the moving clock runs slow); and of a deliberately perverse λ = (t/T)³ (crushed at one end). The curve is identical. The physics is identical. And the action is identical — S = −mc²τ = −6.0989×10⁻²² J·s in all three, to fifty digits, because √(η ẋẋ) dλ = ds is a LENGTH and a length does not know how you counted your way along it. THAT is the gauge symmetry, and its receipt is printed in the next figure: p·p = +m²c², identically, for any λ you choose

The core equations

\boxed{\;S \;=\; -mc\!\int_{\lambda_1}^{\lambda_2}\!\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 \eta_{\mu\nu} = \mathrm{diag}(+1,-1,-1,-1) \qquad (1)

where: the prefactor is $mc$ , not $mc^2$ . $\sqrt{\eta_{\mu\nu}\dot x^\mu\dot x^\nu}\,\mathrm{d}\lambda = \mathrm{d}s$ is a length, so $mc\times$ length has units $\mathrm{kg\,m\,s^{-1}}\cdot\mathrm{m} = \mathrm{J\,s}$ ✓, while $mc^2\times$ length is J·m — not an action. This article's own base prints $-mc^2\!\int\!\sqrt{\eta\dot x\dot x}\,\mathrm{d}\lambda$ , and is wrong by a factor of $c$ — while its own derivation card, two lines later, uses the correct $mc$ . The base disagrees with itself in print.

\delta S = 0 \quad\text{for all}\quad \delta x^\mu(\lambda) \text{ with } \delta x^\mu(\lambda_1) = \delta x^\mu(\lambda_2) = 0 \qquad (2)

where: this is the whole principle, and the word is stationary. Nothing here says minimum.

\boxed{\;p_\mu \;=\; \frac{\partial \mathcal{L}}{\partial \dot x^\mu} \;=\; -\,\frac{mc\,\dot x_\mu}{\sqrt{\dot x\!\cdot\!\dot x}}\;} \qquad\Longrightarrow\qquad \boxed{\;p_\mu p^\mu \;=\; \frac{m^2c^2\;\dot x_\mu\dot x^\mu}{\dot x\!\cdot\!\dot x} \;=\; +\,m^2c^2\;} \qquad (3)

where: the second box is an identity. The $\sqrt{\dot x\cdot\dot x}$ cancels completely, so it holds for every curve and every parametrisation, whether or not the equation of motion is satisfied. Positive, because a timelike four-momentum has positive norm in this signature.

H \;=\; \dot x^\mu p_\mu - \mathcal{L} \;\equiv\; 0 \qquad\text{(identically)} \qquad (4)

where: $\mathcal{L}$ is homogeneous of degree one in $\dot x^\mu$ , so Euler's theorem gives $\dot x^\mu\,\partial\mathcal{L}/\partial\dot x^\mu = \mathcal{L}$ and $H$ collapses. Not "zero on shell" — zero as an algebraic identity. This looks fatal for the Hamiltonian formalism; the rescue is `aq-sr-relativistic-hamiltonian-formalism`. Fixing the gauge $\lambda = \tau$ (legal after varying, never before) gives $\sqrt{\dot x\!\cdot\!\dot x} = c$ and

u^\mu = \frac{\mathrm{d}x^\mu}{\mathrm{d}\tau} = \gamma(c,\mathbf{v}), \qquad p^\mu = m\,u^\mu = \Big(\frac{E}{c},\;\mathbf{p}\Big), \qquad \frac{\mathrm{d}p^\mu}{\mathrm{d}\tau} = 0 \qquad (5)

where: $u\!\cdot\!u = +c^2$ and $p\!\cdot\!p = +m^2c^2 \Leftrightarrow E^2 = (pc)^2 + (mc^2)^2$ — which is (3) with the gauge fixed, and therefore was never a separate assumption.

Working it out

Derivation AThe variation, done covariantly — with lambda arbitrary to the last possible moment

The goal is to derive the free-particle equation of motion from $\delta S = 0$ without ever choosing a frame or a parameter. This is more delicate than it looks: the tempting shortcut — set $\lambda = \tau$ at the start, because proper time is "natural" — is wrong, and wrong in an instructive way. Proper time is a functional of the path you are varying. Fixing it first is holding the answer still while you search for it.

Step 1. Vary. With $\mathcal{L} = -mc\sqrt{\dot x\!\cdot\!\dot x}$ and $\dot x\!\cdot\!\dot x \equiv \eta_{\alpha\beta}\dot x^\alpha\dot x^\beta$ ,

\delta S \;=\; -mc\!\int\! \frac{\eta_{\mu\nu}\,\dot x^\mu\,\delta \dot x^\nu}{\sqrt{\dot x\!\cdot\!\dot x}}\;\mathrm{d}\lambda \;=\; -mc\!\int\! \frac{\dot x_\nu}{\sqrt{\dot x\!\cdot\!\dot x}}\;\frac{\mathrm{d}}{\mathrm{d}\lambda}\big(\delta x^\nu\big)\;\mathrm{d}\lambda.

Why: $\delta\dot x^\nu = \mathrm{d}(\delta x^\nu)/\mathrm{d}\lambda$ because variation and $\lambda$ -differentiation commute — the variation is taken at fixed $\lambda$ , which is exactly the freedom preserved by not gauge-fixing.

Step 2. Integrate by parts. $\delta x^\nu$ vanishes at both endpoints, so the boundary term dies and

\delta S \;=\; +\,mc\!\int\! \frac{\mathrm{d}}{\mathrm{d}\lambda}\!\left(\frac{\dot x_\nu}{\sqrt{\dot x\!\cdot\!\dot x}}\right)\delta x^\nu\;\mathrm{d}\lambda.

Why: the endpoint condition is not decoration. It is what makes the boundary term vanish — and it is precisely what will fail in `aq-sr-charged-particle-lagrangian-em-field`, where a gauge transformation leaves a surviving total derivative at the endpoints, which is exactly why gauge invariance leaves the equations of motion alone.

Step 3. Set $\delta S = 0$ for arbitrary $\delta x^\nu$ . The fundamental lemma gives

\frac{\mathrm{d}}{\mathrm{d}\lambda}\!\left(\frac{\dot x_\nu}{\sqrt{\dot x\!\cdot\!\dot x}}\right) \;=\; 0.

Why: now, and only now, is it safe to choose a parameter. That equation is true for every $\lambda$ — which is what reparametrisation invariance guarantees.

Step 4. Gauge-fix. Choose $\lambda = \tau$ , so $\sqrt{\dot x\!\cdot\!\dot x} = c$ and $\dot x^\mu = u^\mu$ ; then $\mathrm{d}u_\nu/\mathrm{d}\tau = 0$ , i.e. $\mathrm{d}p^\mu/\mathrm{d}\tau = 0$ , i.e. $\mathrm{d}^2x^\mu/\mathrm{d}\tau^2 = 0$ . Why: a straight worldline, uniform four-velocity, conserved four-momentum — all three the same statement. And note what the calculation never used: no frame, no $t$ , no assumption that $u\!\cdot\!u = c^2$ . That normalisation is a consequence of the gauge choice, not an input. *Verified in verify_aq-sr-relativistic-action-proper-time.py: sympy varies $S$ for a symbolic worldline with generic $\lambda$ , reproduces the Euler–Lagrange expression above, and confirms that gauge-fixing afterwards gives $u\!\cdot\!u = +c^2$ and $\mathrm{d}^2x^\mu/\mathrm{d}\tau^2 = 0$ .*

Go deeper · Derivation B · 2 minThe owned result: reparametrisation invariance HANDS you the mass shell

Derivation BThe owned result: reparametrisation invariance HANDS you the mass shell

Here is the article's structural claim, and it is a strong one. You never impose $E^2 = (pc)^2 + (mc^2)^2$ . It is not an extra postulate, it is not a definition of mass, and it is not an equation of motion. It is an algebraic identity forced by the fact that the action is a length — and lengths are indifferent to how you label points along a curve. We prove it in three lines, then show that the same homogeneity kills the canonical Hamiltonian, and explain why that is a feature.

Step 1. Establish the homogeneity. Under $\dot x^\mu \to k\,\dot x^\mu$ (any $k>0$ ),

\mathcal{L}(k\dot x) = -mc\sqrt{k^2\,\dot x\!\cdot\!\dot x} = k\,\mathcal{L}(\dot x).

So $\mathcal{L}$ is homogeneous of degree one in the four-velocity. Why: this is reparametrisation invariance in one line. Rescaling $\dot x$ is what a change of parameter does, and degree-one homogeneity is exactly the property that lets the $\mathrm{d}\lambda$ absorb it.

Step 2. Compute the momentum, and notice its degree.

p_\mu = \frac{\partial\mathcal{L}}{\partial\dot x^\mu} = -\frac{mc\,\dot x_\mu}{\sqrt{\dot x\!\cdot\!\dot x}}.

Numerator and denominator each scale as $k$ , so $p_\mu$ is homogeneous of degree zero — utterly blind to the parametrisation. Why: the momentum knows the direction of the worldline in spacetime, and nothing whatever about how fast you count your way along it. It is a property of the curve, not of the label — which is exactly what a physical quantity has to be.

Step 3. Contract it with itself. The $\sqrt{\dot x\!\cdot\!\dot x}$ cancels completely:

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

Why: read what this is. No equation of motion has been used. No parametrisation has been chosen. This holds for a straight worldline, for a wiggly one, for one that no particle would ever take. The four momenta are not four independent numbers. In the $(+,-,-,-)$ signature it is $+m^2c^2$ , positive — a base printing $-m^2c^2$ has silently switched metric.

Step 4. Watch the Hamiltonian die. Euler's theorem on homogeneous functions says that for $\mathcal{L}$ of degree one, $\dot x^\mu\,\partial\mathcal{L}/\partial\dot x^\mu = \mathcal{L}$ . Hence

H \;=\; \dot x^\mu p_\mu - \mathcal{L} \;=\; \mathcal{L} - \mathcal{L} \;\equiv\; 0.

Why: identically zero, not zero on shell. At first sight this destroys the Hamiltonian formalism — a vanishing $H$ generates no evolution at all. It does nothing of the kind: it is the signature of a constrained system, and the constraint is precisely Step 3. Dirac's 1950 machinery is built for exactly this, and `aq-sr-relativistic-hamiltonian-formalism` runs it: the constraint, promoted to a generator, is what pushes the particle along its worldline, and the same constraint, quantised, is the Klein–Gordon equation.

Step 5. State the moral, because it generalises far beyond this article. A gauge symmetry always buys a constraint. Worldline reparametrisation invariance buys $p\!\cdot\!p = m^2c^2$ ; the same structure one dimension up gives the Virasoro constraints of the string; and diffeomorphism invariance of spacetime itself gives the Hamiltonian constraint of general relativity, $\mathcal{H} \approx 0$ — the reason the Wheeler–DeWitt equation has no time in it. Why: relativists do not say "the particle happens to satisfy $E^2 = p^2c^2 + m^2c^4$ ". They say the mass shell is the constraint that generates time evolution. That is not a rephrasing; it is the correct causal order. *Verified in verify_aq-sr-relativistic-action-proper-time.py: sympy checks $\mathcal{L}(k\dot x) = k\mathcal{L}(\dot x)$ symbolically; confirms $p_\mu$ is degree-zero homogeneous; contracts $p_\mu p^\mu$ and gets $+m^2c^2$ identically in $\dot x$ , with no equation of motion imposed; verifies Euler's theorem and $H \equiv 0$ ; and re-checks all of it numerically for three different parametrisations of the same worldline ( $\lambda = t$ , $\tau$ , $t^3$ ), which give identical $p_\mu$ and identical $S$ to 50 digits.*

Go deeper · Derivation C · 1 minPromote eta to g and the same variation gives the geodesic equation

Derivation CPromote eta to g and the same variation gives the geodesic equation

This action does not merely survive the passage to curved spacetime — it is the theory of curved spacetime, essentially unchanged. Replace $\eta_{\mu\nu}$ by a general $g_{\mu\nu}(x)$ , repeat the variation verbatim, and out comes the geodesic equation with the Christoffel symbols appearing from nowhere but the $x$ -dependence of $g$ . Free fall is free motion. (A bridge, not a claim: the full theory is general relativity's.)

Step 1. The only change. $S = -mc\!\int\!\sqrt{g_{\mu\nu}(x)\,\dot x^\mu\dot x^\nu}\,\mathrm{d}\lambda$ . Now $\mathcal{L}$ depends on $x^\mu$ too, so $\partial\mathcal{L}/\partial x^\sigma \ne 0$ and Euler–Lagrange acquires its second term. Why: in flat spacetime $x^\mu$ was cyclic and the momentum conserved. Curvature is exactly the statement that it is not.

Step 2. Gauge-fix (having varied) to $\lambda = \tau$ , so $g_{\mu\nu}u^\mu u^\nu = c^2$ , and Euler–Lagrange reads $\mathrm{d}(g_{\sigma\nu}u^\nu)/\mathrm{d}\tau = \tfrac12(\partial_\sigma g_{\mu\nu})u^\mu u^\nu$ . Expand the left side, symmetrise in $(\mu\nu)$ — legal, because $u^\mu u^\nu$ is symmetric — and raise the index:

\boxed{\;\frac{\mathrm{d}^2x^\sigma}{\mathrm{d}\tau^2} \;+\; \Gamma^{\sigma}_{\ \mu\nu}\,\frac{\mathrm{d}x^\mu}{\mathrm{d}\tau}\frac{\mathrm{d}x^\nu}{\mathrm{d}\tau} \;=\; 0\;}, \qquad \Gamma^{\sigma}_{\ \mu\nu} = \tfrac12 g^{\sigma\rho}\big(\partial_\mu g_{\rho\nu} + \partial_\nu g_{\rho\mu} - \partial_\rho g_{\mu\nu}\big).

Why: the Christoffel symbols were not put in. They fell out of a derivative of the metric — the only new object available. Set $g \to \eta$ : all derivatives vanish, $\Gamma = 0$ , and we recover $\mathrm{d}^2x^\sigma/\mathrm{d}\tau^2 = 0$ , the flat result of the first card. *Verified in verify_aq-sr-relativistic-action-proper-time.py: sympy varies the metric action for a symbolic $g_{\mu\nu}(x)$ in 2D and confirms the result equals $\ddot x^\sigma + \Gamma^\sigma_{\ \mu\nu}\dot x^\mu\dot x^\nu$ with $\Gamma$ as defined, and that $\Gamma \to 0$ when $g \to \eta$ .*

Figure

PUNCHLINE — the mass shell is a RECEIPT, not a postulate. The hyperbola E² − (pc)² = (mc²)² in the energy-momentum plane. Our electron (2.00 m in 10.0 ns) sits at pc = 457.62 keV, E = 685.96 keV — ON the curve, to fifty digits: √(E² − (pc)²) = 510.999 keV = mₑc². Nobody put it there. The three coloured markers are the SAME electron with the three parametrisations of the previous figure (λ = t, τ, t³): they land on the IDENTICAL point, because p_μ = −mc ẋ_μ/√(ẋ·ẋ) is homogeneous of DEGREE ZERO in ẋ — it cannot see the label. Contract that momentum with itself, the √(ẋ·ẋ) cancels, and out drops p·p = +m²c². No equation of motion was used. The shell is what a gauge symmetry costs

Limiting cases

Case	Result	Meaning
$\lambda = t$	$\mathcal{L}\,\mathrm{d}\lambda \to -mc^2\sqrt{1-v^2/c^2}\,\mathrm{d}t$	the coordinate-time Lagrangian of `aq-sr-relativistic-lagrangian-free-particle`
$\lambda = \tau$	$\sqrt{\dot x\!\cdot\!\dot x} = c$ , $\;p^\mu = mu^\mu$	the proper-time gauge — legal only after varying
$m \to 0$	$S \to 0$ — the action is empty	the square-root form dies on a null curve; use the einbein, whose $m=0$ limit enforces $\dot x\!\cdot\!\dot x = 0$
$\eta \to g_{\mu\nu}(x)$	$\ddot x^\sigma + \Gamma^\sigma_{\ \mu\nu}\dot x^\mu\dot x^\nu = 0$	the geodesic equation. The same action. Free fall is free motion

Common trapVary FIRST, gauge-fix SECOND — and this article's base gets the prefactor wrong

(a) The order matters. It is tempting to set $\lambda = \tau$ at the start, because proper time feels natural. It is illegal: $\tau$ is defined by the worldline you are varying, so imposing $\mathrm{d}\tau^2 = \eta_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu/c^2$ on every competing path constrains the variation, and you will get the right answer only by luck. Vary with $\lambda$ arbitrary; gauge-fix afterwards.

(b) The prefactor. This article's base prints $S = -mc^2\!\int\!\sqrt{\eta_{\mu\nu}\dot x^\mu\dot x^\nu}\,\mathrm{d}\lambda$ , which has units of J·m — it is not an action. The correct prefactor is $mc$ , because the root times $\mathrm{d}\lambda$ is a length. The base's own derivation card, two lines below, correctly writes $-mc\!\int\!\ldots$ : it contradicts itself in print, and the contradiction is invisible unless somebody checks the dimensions. Sister defect next door: base 1974 writes $-mc\!\int\!\mathrm{d}\tau$ , units kg·m. Same error, opposite direction.

Worked examples

1 · An electron crosses the lab: which worldline, and by how much?

An electron is seen at event $A$ and again at $B$ , $\Delta x = 2.00\ \mathrm{m}$ away and $\Delta t = 10.0\ \mathrm{ns}$ later. What is the proper time along the physical worldline, and along a rival that sprints out at $0.9c$ and coasts back?

Step 1 — identify. Light covers $c\Delta t = 3.00\ \mathrm{m}$ in 10.0 ns and the events are 2.00 m apart, so the separation is timelike and a massive particle can connect them. The physical worldline is the stationary point of $S = -mc^2\!\int\!\mathrm{d}\tau$ — here, the straight line.

Step 2 — the straight worldline. $\beta = \Delta x/(c\Delta t) = \mathbf{0.667128}$ and $\gamma = 1/\sqrt{(1-\beta)(1+\beta)} = \mathbf{1.342385}$ (the factorised form — never $1/\sqrt{1-\beta^2}$ by naive subtraction). So

\tau_{\text{straight}} = \frac{\Delta t}{\gamma} = \frac{10.0\ \mathrm{ns}}{1.342385} = \mathbf{7.4494\ ns},

which is exactly $\Delta s/c$ with $\Delta s = \sqrt{(c\Delta t - \Delta x)(c\Delta t + \Delta x)} = \mathbf{2.2333\ m}$ — the invariant length of the worldline, as it must be.

Step 3 — the rival. Out at $\beta_1 = 0.9$ for 5.0 ns (covering $1.3491\ \mathrm{m}$ ), then back at whatever speed lands it on $B$ : $\beta_2 = \mathbf{0.434256}$ , comfortably subluminal, so it is a legal competitor. Its proper time is the sum of two straight legs:

\tau_{\text{rival}} = 5.0\sqrt{1-0.9^2} + 5.0\sqrt{1-0.434256^2} \;=\; 2.1794 + 4.5040 \;=\; \mathbf{6.6834\ ns}.

Step 4 — the boundary number. The straight worldline ages by $\Delta\tau = \mathbf{0.7660\ ns}$ more. In action,

S_{\text{straight}} = -m_ec^2\,\tau_{\text{straight}} = \mathbf{-6.0989\times10^{-22}\ J\,s}, \qquad \Delta S = m_ec^2\,\Delta\tau = \mathbf{+6.2716\times10^{-23}\ J\,s} \;>\; 0.

The rival's action is larger, so the straight worldline is the stationary point and, here, the minimum. Two things to take away. First, $|S_{\text{straight}}| = 5.78\times10^{12}\,\hbar$ — twelve orders of magnitude above the quantum of action, which is why the electron behaves classically over a metre of lab and not over an atom. Second, the whole calculation is a statement about lengths of curves: no force, no field, no equation of motion. A stopwatch and a ruler.

2 · The mass shell you never assumed

Take that same electron. Compute its four-momentum — then compute it again in two other parametrisations, and check whether anything moves.

Step 1 — identify. $p_\mu = -mc\,\dot x_\mu/\sqrt{\dot x\!\cdot\!\dot x}$ , equation (3). This is a degree-zero object: rescale $\dot x^\mu$ by any $k>0$ and $p_\mu$ does not budge. So the answer must be the same for $\lambda = t$ , $\lambda = \tau$ and $\lambda = (t/T)^3$ .

Step 2 — apply, in the lab gauge. With $\gamma = 1.342385$ and $\beta = 0.667128$ ,

E = \gamma m_ec^2 = \mathbf{685.96\ keV}, \qquad pc = \gamma\beta\,m_ec^2 = \mathbf{457.62\ keV}.

Step 3 — contract. $\sqrt{(E-pc)(E+pc)} = \mathbf{510.999\ keV}$ — which is $m_ec^2$ , to the last digit carried. Nobody imposed that. It fell out of the ratio $\dot x_\mu\dot x^\mu/(\dot x\!\cdot\!\dot x) = 1$ , true by inspection for any $\dot x$ .

Step 4 — the boundary number, and it is a null result. Repeat with $\lambda = \tau$ and with the perverse $\lambda = (t/T)^3$ : the components of $\dot x^\mu$ change by factors of $\gamma$ and of $3(t/T)^2$ , and $p_\mu$ does not change at all — the same $(685.96/c,\ 457.62/c)$ keV, to 50 digits. The output does not move when the label moves. Anywhere else that would be the tell-tale of an artefact (runbook trap 26); here it is the whole point, and it is why the shell is an identity. The verify script checks both readings: $p_\mu$ is invariant under relabelling, and perturbing a genuine physical input — $\Delta t$ — moves $E$ , $pc$ and $S$ while leaving the shell exactly where it was.

Physics of the portrait

The phrase in every textbook is "the principle of least action". The phrase that is actually true is stationary action — and relativity is where the difference stops being pedantry.

Origins · 2 minStationary action and how this was discovered

Stationary action

1834–1950 · Hamilton makes stationarity, not minimality, the principle (1834-35) · Planck writes the relativistic kinetic potential (1906) · Minkowski turns the action into the LENGTH of a worldline (1908) · Klein, Fock and Gordon quantise the constraint (1926) · Dirac explains the vanishing Hamiltonian (1950)

Hamilton's two Philosophical Transactions papers of 1834 and 1835, both titled On a General Method in Dynamics, are where the modern statement crystallises — and Hamilton was careful about what he claimed. The condition is that the action be stationary; the eighteenth-century rhetoric of nature "economising" was quietly retired. Two centuries of textbooks have quietly un-retired it. Relativity makes the distinction load-bearing rather than decorative. Planck, in March 1906, wrote relativistic point mechanics as Hamilton's principle with the kinetic potential $H = -mc^2\sqrt{1-q^2/c^2}$ and extracted the relativistic momentum from it — the first time the variational principle did work that $F = ma$ could not. Minkowski, in 1908, explained why that Lagrangian: with Eigenzeit in hand, the action is simply $-mc$ times the invariant length of the worldline, and a length is manifestly indifferent to how you parametrise it. That indifference is reparametrisation invariance, and it is the source of everything here — including the two things that look like failures: the canonical Hamiltonian vanishes, and the four momenta are constrained. Both were understood as structure rather than pathology only when Dirac published Generalized Hamiltonian Dynamics in 1950 and gave the general theory of constrained systems, in which a first-class constraint generates a gauge transformation and a vanishing Hamiltonian is exactly what a reparametrisation-invariant theory must have. In between, in 1926, Klein (April), Fock (July) and Gordon (September) had each written down what the constraint becomes when you quantise it — the relativistic wave equation that carries two of their names, which is nothing but $p_\mu p^\mu = m^2c^2$ with $p_\mu \to i\hbar\partial_\mu$ . The mass shell of a free classical particle, promoted to an operator equation. The chain from Hamilton's characteristic function to Klein–Gordon is unbroken.

History

1834

— Hamilton, "On a General Method in Dynamics" (Phil. Trans. R. Soc., part II for 1834, pp. 247-308), reduces dynamics to a single characteristic function; the Second Essay (part I for 1835, pp. 95-144) gives the canonical equations. The condition is stationarity, not minimality.

1906

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

1908

— Minkowski's "Raum und Zeit" introduces proper time, making the action minus mc times the invariant length of the worldline, and reparametrisation invariance manifest.

1926

— Klein (received 28 April), Fock (30 July) and Gordon (29 September) publish the relativistic wave equation obtained by promoting the mass-shell constraint to an operator equation.

1950

— Dirac, "Generalized Hamiltonian Dynamics" (Can. J. Math. 2, 129-148), gives the theory of constrained Hamiltonian systems, explaining the identically vanishing 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. · Misner, Thorne & Wheeler, Gravitation, ch. 13 · Dirac, Can. J. Math. 2, 129 (1950) · Hamilton, Phil. Trans. R. Soc. (1834, 1835).

Provenance. Signature $(+,-,-,-)$ ; timelike norm positive, $p\!\cdot\!p = +m^2c^2$ , no minus under the root. Every equation is re-derived from that metric and every number recomputed at 50-digit precision in verify_aq-sr-relativistic-action-proper-time.py, which greps this file for each printed value and carries an artefact guard.

Base defects found and repaired. Three. (i) eq[3] prints the covariant action as $-mc^2\!\int\!\sqrt{\eta_{\mu\nu}\dot x^\mu\dot x^\nu}\,\mathrm{d}\lambda$ — units J·m, wrong by a factor of $c$ — while the base's own derivation card, immediately below, uses the correct $-mc$ . It contradicts itself. (ii) eq[7] prints the mostly-plus einbein, whose $e$ -equation gives an imaginary einbein in our signature. (iii) eq[6] writes the EM coupling with a Gaussian $q/c$ ; in SI it is $-q\!\int\!A_\mu\,\mathrm{d}x^\mu$ with no $1/c$ . All three corrected above and asserted in the verify script. The base's derivation card "Lorentz force from the charged-particle action" is `aq-sr-charged-particle-lagrangian-em-field`'s owned payload: recalled by name, not re-derived.

Scope. The coordinate-time Lagrangian $L = -mc^2/\gamma$ , the Euler–Lagrange crank, and the proof that the free-particle stationary point is a genuine minimum are `aq-sr-relativistic-lagrangian-free-particle`. The Legendre transform, the covariant Hamiltonian that repairs $H\equiv0$ , and the Klein–Gordon step are `aq-sr-relativistic-hamiltonian-formalism`. Minimal coupling is `aq-sr-charged-particle-lagrangian-em-field`. The geodesic card is a bridge; the full curved-spacetime theory is general relativity's.

Builds on. `aq-sr-relativistic-lagrangian-free-particle` · `aq-sr-proper-time` · `aq-sr-four-velocity` · `aq-sr-four-momentum` · `aq-sr-minkowski-metric`. Leads to. `aq-sr-relativistic-hamiltonian-formalism` · `aq-sr-charged-particle-lagrangian-em-field` · `aq-sr-bmt-equation`.

Is it least action or stationary action? The two are used interchangeably everywhere.

They are not interchangeable, and relativity is where the difference bites. The principle is $\delta S = 0$ — the action is unchanged to first order under any small variation with fixed endpoints. That is all a variational principle ever asserts. Whether the stationary point is a minimum, a maximum or a saddle is a separate question, answered by the second variation. For a free relativistic particle in flat spacetime it is a minimum, and a global one (`aq-sr-relativistic-lagrangian-free-particle` proves it). Add gravity and it can stop being one: between two events there may be several extremal worldlines — that is gravitational lensing — and past a conjugate point the extremum is no longer a maximum of proper time. Through all of it, $\delta S = 0$ holds without amendment.

Why can't I just set lambda equal to tau at the start? It always gives the right answer.

It often does, and that is what makes it dangerous. $\mathrm{d}\tau = \tfrac1c\sqrt{\eta_{\mu\nu}\mathrm{d}x^\mu\mathrm{d}x^\nu}$ is built out of the path you are varying, so imposing " $\lambda = \tau$ " on every competitor is a constraint on the space of variations, not a change of notation — and a constrained variation is a different problem. You get away with it here because the constraint happens to be preserved by the true equation of motion, but the habit fails the moment it is not, and it makes the geodesic card (where $g_{\mu\nu}$ depends on $x$ , and $\tau$ depends on $g$ ) simply incorrect. Vary with $\lambda$ free; gauge-fix afterwards. One extra line, always right.

A Hamiltonian that is identically zero generates no time evolution. Isn't the theory dead?

It would be, if $H$ were an ordinary Hamiltonian. It is the canonical Hamiltonian of a gauge theory, and $H \equiv 0$ is the standard signature of one. Dirac's resolution: the dynamics is generated not by $H$ but by the constraint $\phi \equiv p_\mu p^\mu - m^2c^2 \approx 0$ , multiplied by a Lagrange multiplier whose arbitrariness is the freedom to reparametrise. Take the multiplier to be $1/2m$ and you get $\mathcal{H} = (p\!\cdot\!p - m^2c^2)/2m$ , which generates translation in $\tau$ — the covariant Hamiltonian of `aq-sr-relativistic-hamiltonian-formalism`. What looks like the theory dying is the theory telling you that one of your four coordinates was never a degree of freedom.

If the mass shell is an identity, what exactly does the equation of motion do?

Less than you think. The identity $p\!\cdot\!p = m^2c^2$ fixes the magnitude of the four-momentum — it confines it to a hyperboloid — and says nothing about its direction or how it changes along the worldline. The equation of motion $\mathrm{d}p^\mu/\mathrm{d}\tau = 0$ supplies exactly that: the four-momentum is not merely on the shell, it is constant on it. Turn on an electromagnetic field and the momentum starts to move — $\mathrm{d}p^\mu/\mathrm{d}\tau = qF^{\mu\nu}u_\nu$ — but it moves along the shell and never off it, because $F^{\mu\nu}$ is antisymmetric and $p_\mu F^{\mu\nu}p_\nu = 0$ identically. The constraint says where the particle may live; the equation of motion says where it goes.

Does this action work for a photon?

No, and the failure is instructive. $S = -mc\!\int\!\mathrm{d}s$ has $m = 0$ out front and $\mathrm{d}s = 0$ along the curve — it is $0\times0$ , and there is no proper time to parametrise by either. The einbein action $S_e = -\tfrac12\!\int\!(e^{-1}\dot x\!\cdot\!\dot x + e\,m^2c^2)\,\mathrm{d}\lambda$ is polynomial in $\dot x$ , finite at $m=0$ , and there the $e$ -equation reads $\dot x\!\cdot\!\dot x = 0$ — the null constraint, produced by the formalism rather than imposed on it. The light ray then satisfies $\ddot x^\mu = 0$ with respect to an affine parameter. Reparametrisation invariance survives all of this: the massless shell is $p\!\cdot\!p = 0$ , still an identity, still the receipt for the same gauge symmetry.

Why does the article insist that p dot p equals PLUS m squared c squared?

Because a large fraction of the literature — and 31 of the 151 AllQFT base articles behind this corpus — works in $(-,+,+,+)$ , where the same statement reads $p\!\cdot\!p = -m^2c^2$ and the action carries a minus under the root. Both conventions are correct; mixing them is not, and mixing them is what happens when a formula is transcribed rather than re-derived. The tell is immediate: in $(+,-,-,-)$ a timelike vector has positive norm, so $\sqrt{-\eta_{\mu\nu}\dot x^\mu\dot x^\nu}$ is imaginary along every physical worldline. If a source hands you an imaginary action it is not wrong — it is in the other convention, and you imported it without translating. Re-derive; never transcribe.