Principle of Stationary Action

Equivalent forms

\delta \int_{t_1}^{t_2} (T - V)\, dt = 0

\frac{\delta S}{\delta q(t)} = 0

All of classical mechanics compressed into one sentence: δS = 0. Newton's laws are a corollary.

Symbol	Name	SI unit	Dimension	Range
$S$	Actionoutput Time integral of the Lagrangian along a path	J·s	ML^2T^-1	0 – 100
$m$	Mass Mass of the free particle	kg	M	0.1 – 10
$v$	Speed Constant speed of the free particle	m/s	LT^-1	0.1 – 20
$t$	Travel time Duration of the motion	s	T	0.1 – 10

Unit systems

Symbol	SI	CGS	Imperial
$S$	J·s	erg·s	ft·lbf·s
$m$	kg	g	lb
$v$	m/s	cm/s	ft/s
$t$	s	s	s

Where it holds

All of classical mechanics with holonomic constraints and monogenic forces; generalizes to fields and to quantum mechanics via the path integral.

Dimensional analysis

S \to [ML^{2}T^{-1}]

\int L dt \to [ML^{2}T^{-2}]\cdot [T] = [ML^{2}T^{-1}]

Action has units of energy

\times

time — the same units as ħ.

\checkmark

Discovery

Pierre-Louis Maupertuis / William Rowan Hamilton · 1834

Maupertuis proposed a least-action idea in 1744 on near-theological grounds (nature is 'economical'). Hamilton made it precise in 1834: the true path makes the time-integral of L = T − V stationary. Feynman later built quantum mechanics on top of it — every path contributes a phase e^{iS/ħ}.

Try this

Why does a thrown ball trace a parabola and not a zigzag? Nature is auditioning every path — and picking the laziest one.

Compare the action S of the true projectile path against wiggled trial paths and watch S bottom out exactly on the real trajectory.

Research status: stable

Real-world applications

Spacecraft trajectory optimization (variational methods)
Feynman path integrals in quantum field theory
Optimal control theory and robotics (Pontryagin's principle is its descendant)
Computer graphics physics engines using variational integrators

Common misconceptions

Action is always a minimum — it is only stationary; for long paths it can be a saddle point
The particle 'tries' paths in time order — the principle is a global statement about the whole path
Least action is mysterious teleology — locally it is exactly equivalent to $F = ma$

Experimental verification

Every verified prediction of Newtonian mechanics doubles as a test, since

\delta S = 0

reproduces

F = ma

. Fermat's least-time principle for light is the optical sibling, verified to extreme precision in interferometry.

Derivation

Demand that

S[q + \varepsilon \eta ]

is stationary

at \varepsilon = 0

for every variation

\eta (t)

vanishing at the endpoints.

Expanding to first order and integrating

the \eta

̇ term by parts gives

\int \eta [\partial L/\partial q - d/dt(\partial L/\partial q

̇

)] dt = 0

.

Since

\eta is

arbitrary, the bracket must vanish — the Euler–Lagrange equation.

Limiting cases

V = 0

(free particle)⟶ Straight line at constant speedWith

L = \tfrac{1}{2}mv^{2}

, any wiggle in space or speed only adds kinetic energy, raising S.

S ≫ ħ⟶ Classical limitQuantum phases e^{iS/ħ} from neighboring paths cancel except near the stationary one — classical mechanics emerges.

S \approx

ħ⟶ Quantum regimeMany paths contribute comparably; the particle no longer has a single trajectory.

What if…

What if I wiggle the true path slightly?

S changes only at second order in the wiggle — that's what 'stationary' means.

What if ħ were huge?

Neighboring paths would stop cancelling and objects would visibly 'spread' over many trajectories.

What if L depended on extra time derivatives (q̈)?

Ostrogradsky instability: the energy becomes unbounded below — nature avoids such Lagrangians.

1

Action of a free particle

Given ·

m:: 1
v:: 5
t:: 2

Find · S

Steps

Free particle: $L = T = \tfrac{1}{2}mv^{2} (V = 0)$
Along the true path v is constant, so $S = \tfrac{1}{2}mv^{2} \times t$
$S = 0.5 \times 1 \times 5^{2} \times 2 = 25 J\cdot s$

Result ·

S = \tfrac{1}{2}mv^{2}t = 0.5 \times 1 \times 25 \times 2 = 25 J\cdot s

2

Why constant speed wins

Given ·

m:: 1
v:: 5
t:: 2

Find · S for an uneven-speed path covering the same distance

Steps

Same endpoints, same total time, same 10 m distance
$S = \tfrac{1}{2}\times 1\times 4^{2}\times 1 + \tfrac{1}{2}\times 1\times 6^{2}\times 1 = 8 + 18 = 26 J\cdot s$
Any deviation from uniform speed raises S — the straight, steady path is stationary

Result · Going 4 m/s for 1 s then 6 m/s for 1 s gives

S = \tfrac{1}{2}(16) + \tfrac{1}{2}(36) = 26 J\cdot s

>

25 J\cdot s

Euler–Lagrange Equation→The Hamiltonian (Legendre Transform)→Kinetic Energy→Projectile Range→