The Principle of Least Action

Pick a thrown ball or a mass on a spring, set how long the motion lasts, and bend the true motion with the two sliders, keeping its start and end fixed. The page adds up the action, kinetic energy minus potential energy over time, for the true motion and the bent one, and shows which is smaller.

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Lesson

The theory — The Principle of Least Action

The action of a motion q(t) between fixed end positions at fixed times is S = ∫ (K − V) dt, kinetic minus potential energy, added up over the time. Hamilton's principle: the true motion is one for which S is stationary, so that any small change in the motion, with the ends held fixed, makes no first-order change in S. For a bend a sin(nπt/T) on a mass m with V = ½mω²q², the change is exactly (mT/4)((nπ/T)² − ω²) a². In uniform gravity, V = mgq, it is the same with ω = 0: a potential that is straight in q adds nothing to it.

What each symbol means

S
the action, ∫(K − V) dt, in joule-seconds
aₙ
the size of a bend with n humps, aₙ sin(nπt/T)
cₙ
what that bend costs per square metre, (mT/4)((nπ/T)² − ω²)
ω
the spring's angular frequency, 2π/P; zero for uniform gravity

The ball's arc has less action than the straight line 🖖

Press A ball thrown 20 m. The action is the kinetic energy minus the potential energy, added up over the flight: 67.9432 J·s for the true arc. The straight line between the same two ends, staying level all the way, scores 100.0000 J·s, more. Press The same, bent 1 m up: rising higher adds 12.4862 J·s to the potential-energy part, ∫V dt, but moving fast enough to get there and back in 2 s adds 13.7199 J·s to the kinetic part, ∫K dt, and the action, their difference, goes up by 1.2337 J·s. Set the bend to −1 and the action is 69.1769 J·s again, up by the same amount. Nothing about the change is first order: it is exactly 1.2337 times the square of the bend, so the true arc sits at the bottom of a bowl, as the lifeguard's crossing point did. Physicists call this Hamilton's principle, and in uniform gravity the bowl never turns over: the table shows every bend raising the action, for a throw of any duration.

On a spring, least action lasts only half a period 🖖

Switch to the spring and press Spring for 1.5 s, one hump. The spring's period is 2 s, so the motion runs past half a period. Now the one-humped bend lowers the action: bent 0.2 m, it falls from 0.0000 to −0.0822 J·s, and c₁ is −2.0562. Press Spring for 1.5 s, two humps and the action rises instead, to 0.1151 J·s. The true motion is lower than some neighbours and higher than others: a saddle, the same shape as the mirage ray in the hot road. Even the straight line between the same ends scores −2.1341 J·s, far below the true motion. The reason is the pull back to the centre: a bend with n humps costs (mT/4)((nπ/T)² − ω²), and once T passes half a period, π/T drops below ω and the first bend turns over. Gray and Taylor's paper on this is titled, fairly, "When action is not least"; they show it is a minimum for short enough motions and never a true maximum. The principle that is true at every duration is that the action is stationary.

At exactly half a period, a bend costs nothing 🖖

Press Spring for half a period. The mass goes from 1 m out to 1 m out on the other side in exactly 1 s. Bend it 0.2 m with one hump and the action does not move at all, 0.0000 J·s, and the same at a bend of any size: the parabola in the second panel lies flat. The bent motion is not an approximation to a true one. It is a true one: a mass on this spring let go at 1 m with a push reaches −1 m after half a period whatever the push was, because the period does not depend on the size of the swing. Every such motion joins the same two ends in the same time with the same action. Gray and Taylor call the second end the kinetic focus of the first; beyond it, as the 1.5 s presets showed, the action is a saddle.

Problem solved in full

  1. Falling through the Earth 4 steps

    Drill a straight hole through a uniform Earth and drop a ball in. Inside, gravity is proportional to the distance from the centre, so the ball moves like a mass on a spring with ω² = g/R, R = 6,371 km. For how long a trip is the ball's action still a minimum, and where is the ball then? Press Spring for 1.5 s, one hump to check the first step.

    Open this problem in the tool

    1. On the page's 2 s spring, past half a period the one-humped bend already lowers the action:

    2. c₁ changes sign where π/T = ω, which is half a period:

    3. For the tunnel, with g = 9.80665 m/s²:

    4. Half of that:

    Answer

    The ball's action is a minimum for any trip shorter than 42.20 minutes, and 42.20 minutes is exactly the time it takes to fall from one side of the Earth to the other: x = R cos ωt reaches −R at half a period. Trips that end on the far side are at the kinetic focus, and longer ones, on the way back, are saddles. The ball passes the centre at ωR = 7.90 km/s.

Learning path

Least time, least action

References (2)