Deriving Newton's 1st Law for One-Dimensional Motion

Here is my attempt at using the classical least-action principle to derive Newton's 1st law of motion for straight-line motion.

Suppose a particle goes from x=0 x = 0 to x=D x = D between time t1t_1 and time t2t_2 . Suppose also that space is potential-free (and thus force-free). The action is defined as:

S=∫t1t2(E−U) dt=∫t1t2E dtS = \int_{t_1}^{t_2} ( E - U) \, dt = \int_{t_1}^{t_2} E \, dt

Suppose we discretize the path into NN constant-velocity periods. N N can be made arbitrarily large. For each period, the velocity is vkv_k . The action becomes:

S=m2 Σk=1k=Nvk2 ΔtΔt=t2−t1N S = \frac{m}{2} \, \Sigma_{k=1}^{k=N} v_k ^2 \, \Delta t \\ \Delta t = \frac{t_2 - t_1}{N}

The actual path must minimize the action while satisfying the constraint that the changes in xx position must sum to +D +D . We can use the method of Lagrange multipliers with the following Lagrangian:

L=m2 (Σk=1k=Nvk2 Δt)+λ((Σk=1k=Nvk Δt)−D) L = \frac{m}{2} \, (\Sigma_{k=1}^{k=N} v_k ^2 \, \Delta t) + \lambda \Big ((\Sigma_{k=1}^{k=N} v_k \, \Delta t) - D \Big)

Taking partial derivatives with respect to the period-specific velocities and setting to zero (per standard practice) results in:

∂L∂v1=m v1 Δt+λ Δt=0  ⟹  v1=−λm∂L∂v2=m v2 Δt+λ Δt=0  ⟹  v2=−λmetc. \frac{\partial{L}}{\partial{v_1}} = m \, v_1 \, \Delta t + \lambda \, \Delta t = 0 \implies v_1 = -\frac{\lambda}{m} \\ \frac{\partial{L}}{\partial{v_2}} = m \, v_2 \, \Delta t + \lambda \, \Delta t = 0 \implies v_2 = -\frac{\lambda}{m} \\ \text{etc.}

Thus, for motion along a straight line in potential-free (force-free) space, the least-action principle dictates that the velocity must be constant.

#Mechanics

Note by Steven Chase
1 year, 11 months ago

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Comments

I like how you have approached a problem involving the calculus of variations by discretizing the integral. In addition to proving that velocity remains constant, another important principle can be drawn attention to, using this analysis. That being the principle of conservation of momentum. On inspecting the Lagrange multiplier λ\lambda, one can see that it can be interpreted as linear momentum which is a constant. This validates the notion that momentum is conserved in the absence of external forces, which is an inference usually made using Newton's second law of motion.

An alternate approach to this proof would be to avoid discretization of the integral altogether. The action SS becomes minimum when Euler - Lagrange's equation is satisfied. This is based on the Calculus of variations.This would lead to:

ddt∂E∂x˙−∂E∂x=0\frac{d}{dt}\frac{\partial{E}}{\partial{\dot{x}}} - \frac{\partial{E}}{\partial{x}} = 0

Now, using the fact that in 1-D motion, E=12mx˙2E = \frac{1}{2}m\dot{x}^2, this can be plugged into Lagrange's equation giving us: mx¨=0m\ddot{x} = 0. Integrating this leads to:

mx˙=constantm\dot{x} = constant

Which is the desired result.

An interested reader can delve into the concepts of calculus of variations. Wikipedia is a good starting point.

Karan Chatrath - 1 year, 11 months ago

@Karan Chatrath Do you believe this line of argument?

Steven Chase - 1 year, 11 months ago

Good point about the lambdas. I had neglected to consider their physical meaning since I was only using them as a bridge to equate velocities from different infinitesimal periods. And the first law does indeed come effortlessly from Lagrangian Mechanics, once LM has been established as a valid principle. I have become proficient in the application of LM to dynamics problems, but I am not yet very familiar with its derivation from variational calculus (that will be some homework). The discretized approach appeals to me on an intuitive level. On that subject, I have posted a three-dimensional version at the link below, which should easily extend to any number of dimensions.

Thanks

https://brilliant.org/discussions/thread/deriving-newtons-1st-law-for-three-dimensional/

Steven Chase - 1 year, 11 months ago
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