2
$\begingroup$

The thing that I can’t understand about Lagrangian mechanics is that to arrive at $T$ and $V$ you need to consider the work done, and that uses $\text{force} \times \text{distance}$ which uses $f=ma$ which uses Newtonian mechanics.

So is Newtonian mechanics therefore always necessary to understand the $T$ and $V$ of the foundational stuff (springs, projectiles, electrostatic force, gravity), and therefore is Lagrangian mechanics only a computational trick to deal with complex systems?

The least action principal seems like a fundamentally new feature of nature, but can we derive the equations of motion from a pure Lagrangian approach from first principles ie $T$ and $V$ need to be derived?

$\endgroup$
9
  • 3
    $\begingroup$ FWIW, the Lagrangian formulation extends far beyond Newtonian mechanics: Take e.g. the Lagrangian for the standard model of particle physics. $\endgroup$
    – Qmechanic
    Commented Nov 9 at 5:50
  • 2
    $\begingroup$ @Qmechanic There could be different meanings of "extends." From my point of view, a theory is an extension of another one if it includes it. Does Lagrangian formulation include the dynamical behavior in the presence of static friction? $\endgroup$ Commented Nov 9 at 6:30
  • $\begingroup$ Comment to the OP: the work concept is independent of the work-energy theorem. Its definition only requires the force $\times$ distance, not $f=ma$. $\endgroup$ Commented Nov 9 at 6:33
  • $\begingroup$ @GiorgioP-DoomsdayClockIsAt-90 - yes, work is fxd but you need to invoke f=ma to calculate the basic kinetic energy formulae to define T $\endgroup$ Commented Nov 9 at 7:02
  • $\begingroup$ I see things the other way: once the kinetic energy is defined in the usual way, one gets f=ma from the Lagrangian T-V and the Lagrange equations. $\endgroup$ Commented Nov 9 at 7:16

4 Answers 4

1
$\begingroup$

Energy, $T + V$, which is a conserved quantity resulting from the time-translation invariance of physical reality, is in my view the fundamental quantity. One does not need force to arrive at energy, but you can equate work with energy in the Newtonian sense.

It is also true that all physical theories that we know arise from the principle of least action. It just so happens that the thing one has to integrate to arrive at the correct equations of motion is the Lagrangian, $T - V$. There is a very good recent video on Veritasium about the principle of least action, check it out.

$\endgroup$
1
  • $\begingroup$ It sounds good in theory. The principle of least action requires us to know the state of the system in the future. But is this always possible? In Newton's theory we can model the motion of a mechanical system based on its current state. To get from city A to city B we do not optimize the Lagrangian. We are simply moving according to Newton's theory. We can try to optimize travel time for current traffic. But we never optimize the Lagrangian. :) $\endgroup$ Commented Nov 10 at 11:41
1
$\begingroup$

To set the stage for what I want to discuss let me first discuss some specific examples.

I want to discuss at one hand Kepler's laws of celestial mechanics, and on the other hand newtonian mechancis plus the inverse square law of gravity.

We have that the move from Kepler's laws to the inverse square law of gravity was fundamental progress.

In doing so the description of physics taking place was moved to a higher level of abstraction.

Later there was Faraday's introduction of the concept of a field. Electric field and magnetic field. In terms of such a field concept the Coulomb force and magnetic force are thought of as being mediated by a field.

The existence of that field is a supposition. The field itself is not a measurable entity. What is measurable is how the supposed field is affecting motion of objects.

Example: the trajectories of particles as they show up in a cloud chamber. Usually a magnetic field is applied, so that charged particles are pulled to curvilinear motion. From the amount of effect that the magnetic field has on the particle properties of the particle can be inferred. The more precise the strength of the magnetic field is calibrated the more precise the inferences about the particle.



We see a recurring pattern. In many cases achievement of progress entailed moving the description to a higher level of abstraction. Correlated with that: a recurring pattern is that at higher levels of abstraction unification is achieved.

But the things that are accessible to measurement are not at the level of those abstractions.


The units of the SI system are - obviously - defined in terms of measurable quantities.

As we know, in 2019 a new standard for defining the kilogram was ratified. The machine that is specific to establishing the value of Planck's constant is called Kibble balance (Named after Bryan Kibble)

A Kibble balance performs a force measurement. Force is a measurable quantity, energy isn't a measurable.

Potential energy does not have an intrinsic zero point. There is freedom to put the zero point anywhere in the range. Whenever something involving a potential energy is measured the measurement is one of difference of potential. Difference of potential is a measurable, potential itself is not a measurable. Whenever a potential is used the aspect that is actually used is the derivative of the potential, not (the value of) the potential itself.

Kinetic energy is defined in terms of velocity, and we have the principle of relativity of inertial motion: velocity is relative. Kinetic energy of linear motion does not have an intrinsic zero point; the kinetic energy that we attribute to an object depends on the velocity that we attribute to the object. Kinetic energy of linear motion does not have an intrinsic zero point.


Of course, the reason that this whole arrangement works is that the inferred entities stand in a well defined relation to the measurable entities.

In the case of classical mechanics: key is that you can transform between on one hand representation in terms of force and acceleration and on the other hand representation in term of potential energy and kinetic energy.

The work-energy theorem provides that transformation.



Some months ago I went to the earliest PSE question about stationary action, and I submitted an answer in which I show how to go from F=ma to Hamilton's stationary action

The sequence from F=ma to Hamilton's stationary action has two stages:
-Derivation of the work-energy theorem from F=ma
-Demonstration that in cases where the work-energy theorem holds good Hamilton's stationary action will hold good also.

$\endgroup$
0
$\begingroup$

If you are interested in doing classical physics, then Newtonian mechanics and Lagrangian mechanics are equivalent formulations. (At least when their domains of applicability overlap -- there are equations of motion you can write down which don't follow from a Lagrangian so in that sense Newtonian mechanics allows for more general equations, and that can matter in practical applications, but as far as we know to date, at a fundamental level, physics can be described with a Lagrangian). Historically it may have been difficult for humans to discovery Lagrangian mechanics without knowing Newtonian mechanics, but logically you can start from either formulation and show the other is equivalent.

If you are interested in philosophical interpretations of classical physics, then indeed Lagrangian mechanics and Newtonian mechanics might suggest different philosophical viewpoints. However, this is not a physics question, in that this kind of interpretation of the formalism cannot be decided by experiment. In terms of observable quantities, both formulations give the same predictions.

If you are interested in generalizing beyond classical physics, then Lagrangian mechanics naturally leads you into the path integral formulation of quantum mechanics. There's not really a clean, direct way to go from Newtonian mechanics to quantum mechanics. This illustrates a point made by Feynman (and probably others), that having equivalent formulations of a theory is useful when you try to go beyond that theory, because one of the formulations might generalize to the next layer of abstraction more easily than another.

$\endgroup$
0
$\begingroup$

It depends a bit but in general, unless you work with simple problems, it’s quite difficult to track down the Newtonian formulation of a problem whereas the Lagrangian formulation is immediate. In this sense the Lagrangian formulation does not rely on the Newtonian formulation, which is too complicated to even attempt.

The Lagrangian formulation is certainly more powerful but also requires more sophisticated mathematical tools, such as partial derivatives, total derivatives, and integration to obtain the potential energy from forces using first principles. It is difficult to imagine that this machinery (especially the calculus part) could historically have come before the simpler Newtonian mechanics and the advent of calculus.

One of the advantages of the Lagrangian formulation is the ability to build in constraints so you can easily deal with problem such as the conical pendulum which - even though it’s a fairly simple system - would be quite complicated using Newtonian mechanics. The action-reaction pairs never appear in the Lagrangian formalism and in fact can become terribly awkward in the Newtonian formalism. Also the Lagrangian formulation can easily handle fields - never mind electric fields but think of fluid flow - and these are almost impossible to manage with Newtonian mechanics beyond simple cases.

This doesn’t mean Lagrangians are the do-all be-all. The Lagrangian formulation can get complicated with non-conservative systems and more generally forces that are not obtained by the gradient of a potential - the generalized forces must be added by hand and in these situations the Lagrangian approach is clearly a variation on the Newtonian approach.

So it is more a case that both formulations agree for simple problems, and are linked through the principle of virtual work, which itself relies on forces (assuming you leave out the variational approach, which again does not always apply to problem where forces are not conservative).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.