The Lagrangian and the Condition for Stationary Action
The initial and final positions and times are fixed, and a single number called the action is calculated for each path.
Adjust conditions:
The sliders adjust the values
時間幅 3π/2 s。どちらの経路の作用が小さいか？
時間幅を π/2 s にすると、作用が小さい経路は？
The equations explain the phenomenon
1Constructing the LagrangianL=T − U=½ m x² − U(x)……
2Integrating L along the path gives the action S.S=∫t₁t₂ L(x, x) dt……
3A ball thrown vertically upward illustrates how varying the path changes the action.x(t)→x(t)+ε η(t)，η(t₁)=η(t₂)=0……
4Expanding L gives the 1st-order change in action.δS=ε ∫t₁t₂(∂L/∂x · η+∂L∂x · η)dt……
The aim is to factor out η inside the integral to obtain a condition that holds for every η. The term containing η prevents this.
5Integrating the term containing η by parts allows η to be factored out.δS=ε ∫t₁t₂[∂L/∂x− ddt(∂L/∂x)]η dt……
6The condition for δS = 0 for every variation follows.ddt (∂L∂x)=∂L∂x……
7The stationarity condition yields Newton's equation of motion.m x=−dUdx……
8Calculating the 2nd-order change in action for a springS(A) − S₀=(m A² τ4)[(πτ)² − km]……
9Describing pendulum motion using an angleL=½ m ℓ² θ²+m g ℓ cosθ→θ=−(gℓ) sinθ……
Further applications to other types of motion
10The method also applies to a double pendulum.L(θ₁, θ₂, θ₁, θ₂)→2 equations……
11Rewriting the description using the HamiltonianH=p x − L，x=∂H∂p，p=−∂H∂x……
12Plotting motion in the plane of angle and generalized momentumH(θ, p)=Constant……
What this chapter has shown
The stationarity condition requires the first-order change in the action to be 0 for any small variation of the path with its endpoints held fixed. In this chapter, this condition is equivalent to Newton's equation of motion. For a pendulum, using the angle as the coordinate incorporates the constraint of a fixed string length, allowing the equation of motion to be formulated without determining the tension.
−6.54 mJ·sThis is the difference in action for τ=3π/2 s
Stationary action is not necessarily a minimum. For a time interval of 3π/2 s, the trial path's action is 6.54 mJ·s smaller than that of the path at rest.
In everyday lifeThe path of refracted light can also be found by requiring the 1st-order change in travel time to be 0 when the point of passage is varied slightly (Fermat's principle).
Detailed summary
Motion can be determined either from the force at each instant or by requiring the action S for the entire path to be stationary (its 1st-order variation is 0). These 2 formulations are equivalent, but the latter allows a free choice of coordinates and eliminates the need to consider constraint forces. Stationary does not mean minimal: over a sufficiently long time interval, some paths have a smaller action than the actual path. Moving from the Lagrangian to the Hamiltonian leads to a formulation that serves as a starting point for quantum mechanics and statistical mechanics.
Recall
出発・到着 x = 0、候補 A sin(πt/τ)。k = 4.0 N/m、m = 1.0 kg、τ = π s で作用が小さいのは？