Least Action / Newton's 1st Law

Classical Mechanics Level 4

Returning to the Principle of Least Action (see the first part here ), we recall that the "action" S S is defined as follows ( t 2 − t 1 = T ) (t_2 - t_1 = T) :

S = ∫ t 1 t 2 ( E − U ) d t \large{S = \int_{t_1}^{t_2} (E - U) \, dt}

Here, E E is the kinetic energy and U U is the potential energy. Suppose we have a free particle, which is not subjected to any forces / potentials. In this case, we can, by convention, declare the potential zero ( U = 0 ) (U = 0) and simplify the action definition.

S = ∫ t 1 t 2 E d t \large{S = \int_{t_1}^{t_2} E \, dt}

We know from Newton's 1st Law that if a moving particle is not subjected to any forces, it's velocity will remain constant. In the potential-free case (with no forces), if a particle moves between two specific points in a definite amount of time, we therefore expect constant-velocity motion to minimize the action.

Suppose there are two points in space, separated by a distance D D , and a free particle moves between them in a straight line in a period of time T T . Evaluate the following actions ( t = 0 t = 0 is the beginning of the integration window):

1) S 1 = S_1 = Action for constant-velocity straight-line motion: v = D T \large{v = \frac{D}{T}}
2) S 2 = S_2 = Action for variable-velocity straight-line motion: v = D T ⋅ [ 1 + s i n ( 2 π T t ) ] \large{v = \frac{D}{T} \cdot \Big[1 + sin(\frac{2 \pi}{T}t) \Big] }

Both velocity profiles get the particle from one point to the other in a time period T T , but we expect S 2 S_2 to be larger than S 1 S_1 . Determine the ratio S 2 S 1 \large{\frac{S_2}{S_1}} .

Note: Of course, the behavior associated with S 2 S_2 is non-physical, but we can still calculate S 2 S_2 and demonstrate that the behavior is non-physical, in light of the least action principle.


The answer is 1.5.

Steven Chase by Steven Chase , 37, USA

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1 solution

Arjen Vreugdenhil
Dec 28, 2017

The first integral is S 1 = ∫ 0 T ( D T ) 2 d t = D 2 T . S_1 = \int_0^T \left(\frac D T\right)^2\:dt = \frac{D^2}T. The second integral may be written as S 2 = ( D T ) 2 ∫ 0 T ( 1 + sin ⁡ 2 π t T ) 2 d t = D 2 T ∫ 0 1 ( 1 + sin ⁡ 2 π τ ) 2 d τ , S_2 = \left(\frac D T\right)^2 \int_0^T \left(1 + \sin \frac{2\pi t}T\right)^2\:dt = \frac{D^2}{T} \int_0^1 (1 + \sin 2\pi\tau)^2\:d\tau, where we set τ = t / T \tau = t/T ; expanding further and using sin ⁡ 2 θ = 1 2 − 1 2 cos ⁡ 2 θ \sin^2 \theta = \tfrac12 - \tfrac12\cos 2\theta , S 2 S 1 = ∫ 0 1 ( 1 + sin ⁡ 2 π τ ) 2 d τ = ∫ 0 1 ( 1 + 2 sin ⁡ 2 π τ + 1 2 − 1 2 cos ⁡ 4 π τ ) d τ = ∫ 0 1 ( 3 2 + 2 sin ⁡ 2 π τ − 1 2 cos ⁡ 4 π τ ) d τ = 3 2 . \frac{S_2}{S_1} = \int_0^1 (1 + \sin 2\pi\tau)^2\:d\tau = \int_0^1 \left(1 + 2\sin2\pi\tau + \tfrac12 - \tfrac12\cos 4\pi\tau\right)\:d\tau \\ = \int_0^1 \left(\tfrac32 + 2\sin2\pi\tau - \tfrac12\cos 4\pi\tau\right)\:d\tau = \tfrac32. (The second and third term do not contribute to the integral because integration of sine or cosine over a multiple of 2 π 2\pi results in zero.)

Since the action of the second motion is greater than that of the first motion, the second motion is non-physical.

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