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First Let me write both the laws So Newton's law says that

$$\mathbf{F}=m\mathbf{a}$$

and least action principle says that a particle occupy, at the instants $t_1$ and $t_2$, positions defined by two sets of values of the co-ordinates, $q^{(1)}$ and $q^{(2)}$. Then the condition is that the system moves between these postions in such a way that the integral $$S=\int^{t_2}_{t_1}\mathcal{L}(q,\dot{q},t)dt$$

takes the least possible value (in general extremum).

Now we know How to derive Newton's law from the least action principle that first derives the Euler-Lagrange equation and then plug lagrangian into it.

But you see, I'm confused with the initial and boundary of these two problems. In Newton's law particle Just follow the equation at every instant of time. And in the least action principle the particle knows where to where he needs to go and then he follows this minimization condition so first, he decides and then goes.

I'm not asking the following. If you say that I can reduce this least action to Euler-Lagrange which particle follow at any instant and that's same as newton's law. for instance, just forget about this Euler-Lagrange equation so that I compute minimum action and then take a path.

Second I know the least action principle is coming from this Feynman integral thing which is totally quantum so that too I don't need an explanation.

Question: Can anybody give me an intuition (explanation with least math) that these two principles are actually equivalent and so the initial and boundary condition? So that both are doing the same thing. And please keep track of that boundary thing and global and local thing. That is newton's law is something local and the Least principle is global (you need to specify a global boundary).

I have seen these answers already so don't pull them over. 1. 2. I need something that is obvious or at least after some math.

Qmechanic
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Young Kindaichi
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1 Answers1

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To define a unique path $x_i(t)$ in configuration space the boundary conditions needed in Newtonian mechanics are the initial co-ordinates $x_i(0)$ and the initial velocities $\dot x_i(0)$. In the Lagrangian formulation we need to know the initial co-ordinates $x_i(0)$ and the co-ordinates $x_i(t’)$ at some later time $t=t’$ - but we don’t need to know any velocities. The amount of information carried by the boundary conditions is exactly the same in the two formulations.

A simple one-dimensional example is driving along a fixed road (the constraint) at a fixed speed. Newtonian mechanics answers the question “if I start from A and drive at speed $v$ where will I be in one hour’s time ?”. Lagrangian mechanics answers the question “If I start from A and want to be at B in one hour’s time, how fast do I need to drive ?”.

gandalf61
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  • If I understand you correctly you are pointing out the following: in the case of uniform velocity distinction between initial conditions and boundary conditions is moot. However, the question is about the case of accelerated motion. With accelerated motion the distinction is acute. – Cleonis Oct 31 '20 at 13:40
  • @Cleonis Initial conditions are boundary conditions. My illustration was the simplest example that I could think of. If you want an example involving accelerated motion, think of a bead on a smooth wire where the height of the wire varies. Once again, Newtonian mechanics says "if I start here with this speed where will I be at some future time" and Lagrangian mechanics says "if I start here and want to be there at some future time what does my initial speed have to be ?". – gandalf61 Oct 31 '20 at 14:12
  • Well, yeah, mathematically you can always interconvert between differential form and variational form (with the variational form subject to extremum constraint). In terms of physics taking place you need differential form. That is why in dynamics using the Euler-Lagrange equation is a necessity (instead of optional). You obtain an equation of motion if and only if the variational form is converted to EL-equation. – Cleonis Oct 31 '20 at 16:38