Can one non-circularly derive Newton's 2nd law from Lagrangian?

Question

My question emerged as I read this answer and comment by a user under it which says

How do you define lagrangian at first place in form of kinetic and potential energy. Newton's second law is axiomatic and which came from definition of inertia or state of motion

I think this user makes a good point, because as far as I understand, force is defined through the equation $F=ma$. In English one would say "Force is something which causes a change in motion". So, how would it be possible to define force for defining something like potential energy (*) without actually having $F=ma$ estabilished already?

*: I take the conventional definition of PE as $\nabla U = -F$.

Answer 1

I guess this is all about definitions. The Euler-Lagrange equations are $$\dfrac{d}{dt}\dfrac{\partial L}{\partial \dot{q}^i}=\dfrac{\partial L}{\partial q^i}\tag{1}$$

so basically you can define force and momentum to be $$F_i=\dfrac{\partial L}{\partial q^i},\quad p_i=\dfrac{\partial L}{\partial \dot{q}^i}\tag{2}$$

so that the Euler-Lagrange equations take the form of Newton's second law $$F=\dfrac{dp}{dt}\tag{3}.$$

In that case, considering a particle in $d$-dimensional space with generalized coordinates being the position components $q^i = x^i$ and considering a Lagrangian of the form $$L=\dfrac{1}{2}mv^2-V(x^i),\quad v^2\equiv \sum_{i=1}^d (\dot{x}^i)^2\tag{4}$$

it immediately follows that $p_i = m\dot{x}^i$ and you get Newton's second law for the conservative force $F_i = -\frac{\partial V}{\partial x^i}$.

You might say: "ah but how one would know (4) without starting from Newtonian mechanics?", but that is then more a point about pedagogical presentation. There is absolutely nothing circular here: we can postulate (4) together with the variational principle inasmuch as we can postulate $F = ma$. Indeed, in Lagrangian mechanics you have the variational principle as basic axiom and you postulate a Lagrangian to define a theory, with the ultimate justification for that specific Lagrangian being that "it works".

Answer 2

I think you have a point.

I start this answer with a quote from an answer by stackexchange contributor Kevin Zhou:

[...] derivations in physics work very differently than proofs in mathematics.

For example, in physics, you can often run derivations in both directions: you can use X to derive Y, and also Y to derive X. That isn't circular reasoning, because the real support for X (or Y) isn't that it can be derived from Y (or X), but that it is supported by some experimental data D. This two-way derivation then tells you that if you have data D supporting X (or Y), then it also supports Y (or X).

The recurring pattern is: when there is some derivability we find it is bi-directional derivability.

Then again: the two representations connected by the bi-directional derivation are not necessarily on equal par.

On the status of 'Force' and 'Potential energy'.

Force is a measurable.

The concept of force is applicable both in statics and in dynamics.

Derek Muller (Veritasium) has a video about the calibration of force transducers.

One application of force transducer measurement is to measure the thrust of a rocket engine in a static setup. With the thrust calibrated: when the rocket is in space the onboard accelerometers double check whether the engine is delivering the expected thrust.

As we know, potential energy is defined as the negative of work done.

Work done is the integral of force acting over distance.

(Conversely: to recover the force from a potential we differentiate the potential with respect to the spatial coordinate.)

As we know: Potential energy does not have an intrinsic zero point. When using potential energy the choice of zero point is arbitrary. What is used in calculations is difference of potential between point A and point B.

In that sense potential energy is not a measurable. Force is a measurable, potential energy isn't.

This has implications for defining the concepts, since in physics any definition must connect back to physical measurement.

The thing about potential energy is: to connect back to physical measurement you go through the concept of force. That's the way it is.

Potential energy is a meaningful concept by virtue of potential energy being defined as the integral of force over distance.

I concur: I submit that in order to define the concept of the Lagrangian in the Euler-Lagrange equation one must have a definition of potential energy in place, and in order to have a definition of potential energy in place $F=ma$ must be granted.

Calculus of Variations in Classical Mechanics

Nowadays is it customary to refer to 'the Lagrangian' in terms of the Lagrangian that enters the Euler-Lagrange equation.

However: this is not how things went down historically, and it makes a difference. Joseph Louis Lagrange developed his mechanics decades prior to the introduction of Hamilton's stationary action.

Lagrange did use calculus of variations in his work 'Mecanique Analytique', but only in the course of treating statics problems. The dynamics in the Mecanique Analytique does not involve calculus of variations.

The powerful mathematical approach that Lagrange introduced was:

• Using generalized coordinates (when applicable)
• Representing the physics taking place in terms of kinetic energy and potential energy

The above two are sufficient; if Hamilton's stationary action would never have been introduced then the mechanics of Joseph Louis Lagrange would be as powerful and versatile as it is today.

Anyway, nowadays we have that in the perception of most physicists 'Lagrangian mechanics' is strongly associated with Hamilton's stationary action, to the point of the two being regarded as the one and the same.

To derive Hamilton's stationary action from $F=ma$ does not require additional assumption.

The path from F=ma to Hamilton's stationary action is a two-stage process:

1. The Work-Energy theorem is derived from $F=ma$
2. It is demonstrated: in circumstances where the Work-Energy theorem holds good: Hamilton's stationary action will hold good also.

The scope of generalized coordinates

To avoid possible misunderstanding I add the following:

The concept of generalized coordinates extends to $F=ma$. That is, there is a meaningful concept of 'generalized force'

I have seen authors assert that $F=ma$ supports only cartesian coordinates, making generalized coordinates exclusive to Lagrangian mechanics. Obviously that assertion is wrong. There is no mathematical obstacle, or self-consistency obstacle, to using a 'generalized force' concept.