The principle of stationary action

The same start and finish, different histories

A point particle of mass 2 kg travels on a fixed straight coordinate line, with right positive. Fix x(0) = 0 m and x(2 s) = 2 m. Use an inertial Newtonian model: a force-free particle has constant velocity, so the model’s motion between these endpoints is x₀(t) = (1 m/s)t for 0 ≤ t ≤ 2 s.

Now draw other smooth position-versus-time histories with those same endpoints. They may speed up or slow down along the way. These are mathematical trials for comparison; they are not additional force-free motions. A history specifies a position at every time in the interval, rather than only a geometric route through space.

We will assign one number to each history and ask how that number changes when the history changes. The first comparison needs only algebra and units. The later general-variation calculation uses derivatives and integrals, introduced briefly where they occur; it can be revisited after the linked calculus preparation.

Accumulate an energy difference over time

Velocity v = dx/dt is the time rate of position change. For our constant-mass classical model, kinetic energy is K = mv²/2. A potential energy V(x) describes the conservative interaction; for the free particle choose V = 0. The Lagrangian used here is L = K − V, with energy units J. It differs from mechanical energy K + V.

The action of a history is S[x] = ∫₀ᵀ L(x(t), v(t)) dt. The brackets remind us that the input is an entire function x(t). The integral accumulates L times short durations over the fixed interval; when L is constant, it is simply L × T. Thus the free constant-velocity path has L = 1 J and S[x₀] = (1 J)(2 s) = 2 J s.

Action has units J s, not J. It is not work done along the path and need not be positive when V is present. Adding the same constant C to the potential changes every compared action by −CT; because the duration is fixed, this does not change which histories are stationary.

Let one parameter select a whole history

Choose the original trial family xβ(t) = (1 m/s)t + β(1 m/s²)t(2 s − t). Here β is a dimensionless constant; −0.5 ≤ β ≤ 0.5 is the range drawn. The added term vanishes at both fixed endpoint times. Positive β puts the path ahead of x₀ in the middle; negative β puts it behind.

Hold β constant while differentiating in time: vβ(t) = 1 m/s + β(1 m/s²)(2 s − 2t). For the drawn range, vβ is nonnegative throughout 0 ≤ t ≤ 2 s. Only β = 0 has constant velocity and solves the force-free model. An allowed trial need not solve that model.

To calculate action, write τ = t/(1 s), a dimensionless numerical time. Then S(β) = (1 J s)∫₀²[1 + β(2 − 2τ)]² dτ. Expand the square. The term linear in β integrates to zero, because ∫₀²(2 − 2τ)dτ = 0. The squared term has integral ∫₀²(2 − 2τ)²dτ = 8/3. Consequently S(β) = 2 J s + (8/3 J s)β².

Read time and path parameter on different axes

In the upper plot, follow one line from t = 0 to t = 2 s while its β stays fixed. Circles and a solid line identify β = 0; squares and a dashed line identify β = −0.5; triangles and a dotted line identify β = +0.5. Both endpoint diamonds are shared. The table supplies the exact sampled positions.

The lower plot has β on its horizontal axis and S on its vertical axis. Each point here stands for an entire upper-panel history. The action curve is lowest at β = 0 for this family: S(−0.5) = S(+0.5) = 8/3 J s, while S(0) = 2 J s. Moving horizontally in this lower plot is a comparison between histories, not the passage of time.

The smooth curves are exact drawings of the supplied quadratics inside the labeled windows. Markers indicate selected exact values; they are not measurements. Their geometric construction points are not extra samples. Comparing three markers alone would not establish what happens for every allowed variation.

What “stationary” requires

Consider a chosen history x*(t) and a nearby family xε(t) = x*(t) + εη(t). The constant ε is dimensionless. The function η(t) has position units and describes the shape of the change. Require η(0) = η(T) = 0, so positions and times at both ends stay fixed. Here we allow η with a continuous first time derivative throughout the finite interval; this keeps the trial velocity and action well defined.

For each such shape, vary ε around zero and calculate the rate dS[xε]/dε at ε = 0. Stationary action means that rate is zero for every allowed η. This derivative compares whole histories. It is not dS/dt, and it does not assert that the particle stops or that every finite trial has the same action.

For our β family, dS/dβ = (16/3 J s)β, which vanishes at β = 0. This is one required check, not the full every-variation condition. The next calculation establishes the stronger statement for this free-particle model.

Why the free-particle example really is a minimum

Let xε(t) = (1 m/s)t + εη(t), using any continuously differentiable η with zero endpoint values over 0 ≤ t ≤ 2 s. Write η′(t) = dη/dt, which has velocity units. Expanding K gives S[xε] − S[x₀] = m(1 m/s)ε∫₀^(2 s)η′(t)dt + (mε²/2)∫₀^(2 s)[η′(t)]²dt.

The integral of the derivative is η(2 s) − η(0) = 0: accumulation of a rate recovers the endpoint change. The remaining term is nonnegative because it is an integral of a square and m is positive. Therefore S[xε] ≥ S[x₀], and there is no first-order term. In this particular free-particle problem, the constant-velocity history is stationary and a minimum among these endpoint-fixed histories.

The reasoning uses the supplied kinetic-energy form, a fixed duration and no varying potential. It does not say that motion generally minimizes kinetic energy, time, distance or action. To see the difference, change the physical model.

A stationary spring history that is not a minimum

Supply a different ideal conservative model: a 2 kg point mass on a line has spring force F = −kx with k = 2 N/m and V = kx²/2, measured from x = 0. Newton’s equation ma = F makes x*(t) = 0 a valid rest history. Fix x(0) = x(4 s) = 0, with duration 4 s. Initial trial velocities are not fixed in this endpoint comparison.

At rest, K and V both vanish. For any smooth endpoint-fixed perturbation xε = εη, the action is ε²∫₀^(4 s)[(m/2)(η′)² − (k/2)η²]dt. There is no term linear in ε, so the rest history is stationary for every such variation.

Now choose xλ(t) = λ(1 m/s²)t(4 s − t), with λ dimensionless. Its endpoints fit, though it is not asserted to solve the spring equation. Exact integration gives S(λ) = −(64/5 J s)λ². At λ = 0.1, S = −0.128 J s, less than the rest history’s zero. Arbitrarily small nonzero λ has the same strict sign, so rest is not a minimum. The negative curvature along this selected family does not prove a maximum over all histories.

Keep the principle and its domain clear

For the smooth conservative classical models considered here, the principle of stationary action states that a physical history has zero first-order action change under every admissible variation with the prescribed fixed endpoint positions and times. “Least action” is a common name, but stationarity is the condition that survives our spring counterexample.

Our examples use constant positive mass, an inertial frame and a smooth position-dependent potential. Extending the method to constraints, friction, relativistic motion or fields requires their appropriate formulation; simply inserting every force into L = K − V is not a general rule. Boundary conditions must also be stated before comparing actions.

The next lesson, The Euler–Lagrange equations, turns this every-variation condition into a local equation of motion under explicit regularity assumptions. Here we checked the supplied models directly. The histories and numbers are exact teaching assumptions, not an experiment showing how a real particle chooses a path.

Worked examples

Try it yourself

1. In the free-particle family, use β = 0.25. Find x(1 s) and S.

   Show the answer and reasoning

   x(1 s) = 1.25 m and S = 13/6 J s.

   The midpoint perturbation is β × 1 m. The action is 2 J s + (8/3 J s)(1/16) = 13/6 J s.

2. May x(t) = (1 m/s)t + 0.1 m be used as an allowed trial in our original fixed-endpoint problem?

   Show the answer and reasoning

   No. It changes both endpoint positions.

   It gives x(0) = 0.1 m and x(2 s) = 2.1 m. A constant offset is not a zero-endpoint variation, even though its velocity is unchanged.

3. A learner calls dS/dβ an energy because S was built from energy. What variable is being changed, and what is the derivative’s unit here?

   Show the answer and reasoning

   β changes the whole selected history; dS/dβ has units J s.

   β is dimensionless, so differentiating with respect to it preserves action units. This is not differentiation with respect to elapsed time.

4. Three trial histories have action values 8/3, 2 and 8/3 J s. Does this list alone prove stationary action under every admissible variation?

   Show the answer and reasoning

   No. It checks only those three histories.

   Other smooth zero-endpoint changes remain possible. Our general free-particle expansion establishes the every-variation condition and minimum; a finite list alone does neither.

5. In the separate 4 s spring comparison, change λ from 0.1 to 0.2. What is the trial action, and what follows about the rest history?

   Show the answer and reasoning

   S = −0.512 J s; the stationary rest history is not a minimum.

   The action scales as λ², so doubling λ multiplies −0.128 J s by four. More generally every arbitrarily small nonzero λ in this family gives negative action. This does not establish a maximum over all histories.

6. For the same spring and trial form xλ(t) = λ(1 m/s²)t(1 s − t), now fix zero endpoints at 0 and 1 s. The kinetic and potential action coefficients are 1/3 J s and 1/30 J s. Find S(λ). Does this one-family check prove a minimum against every variation?

   Show the answer and reasoning

   S(λ) = (3/10 J s)λ². This one-family check does not prove a minimum against every variation.

   Subtract 1/30 from 1/3. The positive coefficient contrasts with the 4 s comparison: action comparisons depend on the fixed duration and admissible histories. Establishing a minimum for all variations would require a stronger argument.

What to carry forward

• Action assigns a number with units J s to an entire history over a stated interval.
• An admissible endpoint-fixed trial need not satisfy the physical equation of motion.
• A time derivative follows one history; a parameter derivative compares histories.
• Stationary action requires zero first-order change for every allowed variation.
• A minimum occurs in the supplied free-particle model, but the spring example disproves a universal minimum.