B.5.TU:5.2 - Use variational mechanics to answer a motion question

An engineer wants to understand what a variational account predicts for an ideal carriage coasting along a straight horizontal track. Adopt a one-dimensional nonrelativistic free-particle model: positive mass m, negligible resistance and no applied driving force during the interval. Its Lagrangian is L(q, v) = m*v*v/2. For this case, the action is the time integral of that quantity along a proposed path. The physical model supplies this choice of Lagrangian.

The mechanical rule selects paths of stationary action: for every small path variation that leaves the endpoint positions fixed, the first-order change of action must vanish. The calculation below finds that path and shows that, for this free particle, it minimizes the action. Reproducing the calculation uses differentiation and definite integration; the displayed action identity can also be obtained as a supplied mathematical result.

Let the carriage pass q=0 at time 0 and q=d at positive elapsed time T. Recover the operations hidden by “calculate the action”: choose a position function q(t), differentiate it to obtain velocity, evaluate L on those values, then integrate over time. This is the constructive use developed in SICM, §§1.3–1.4. The following one-dimensional case uses it.

Start with q_0(t)=d*t/T, and try paths q_a(t)=d*t/T + a*(t/T)*(1-t/T), where a is a length. Every path has the specified endpoint positions. Its velocity is v_a(t)=(d+a*(1-2*t/T))/T. Substitution and integration give S_a = m*(d*d+a*a/3)/(2*T). For m=1 kg, d=2 m and T=1 s, the straight path has action 2 J s; a=1 m gives 13/6 J s.

This comparison favors the straight path within the chosen family. The general argument is also short. For any continuously differentiable added displacement eta(t) that is zero at both endpoints, the cross term integrates to (m*d/T)*(eta(T)-eta(0))=0. Thus the change in action is (m/2)*integral(eta'(t)^2 dt), which is nonnegative. The straight path minimizes this action among those paths. For a nonzero eta, write this positive change as K. Along the paths q_0+c*eta, the action is S[q_0]+c*c*K, whose derivative at c=1 is 2*K>0. Thus a nonzero displacement from q_0 cannot be stationary. The selected motion has constant velocity d/T.

The application now has an interpreted answer: under the adopted model, passing those endpoint positions in that time entails constant velocity. A computational implementation must evaluate the path derivative in the velocity argument before integrating. It can reproduce the numerical action comparison; the displayed argument supplies the wider conclusion.

Suppose the same two-metre trip must instead begin and end at rest. The constant-velocity answer fails that condition. The useful return is that the undriven free-particle account cannot supply this trip: the design needs acceleration and deceleration, and an account of the forces producing them. The next contribution is that driven-motion model. Increasing the resolution of the same free-particle calculation would retain the missing physical contribution.