# How do we know if a formulation of classical mechanics is correct?

For example, the Lagrangian formulation. I may be missing something, i.e. not having done it in enough detail, but here is my issue:

from the definition of the lagrangian ($\mathcal{L}$) and from the principle of least action ($S(t)$), we get that $$S(t) = \int \mathcal{L} \ dt$$ has to be minimised.

From this, one can derive the Euler-Lagrange equation, $$\frac{d}{dt} \left(\frac{\partial L}{\partial \dot{x}}\right) = \frac{\partial L}{\partial x}$$

and using $\mathcal{L} = T - V = m\dot{x}^2 - V(x)$ we get $$m\ddot{x} = -V'(x),$$ i.e. Newton's second law of motion.

Now: Does the fact that we get Newton's second law mean that the Lagrangian formulation (and therefore the principle of least action) is correct?

I mean, could it not be a coincidence that we get this equation? But then again, is there a way that this equation could not hold? The only difference between systems is the potential $V(x)$ (right?), so as long as it is included in a general form the equation should hold. (?).

In light of this, is there any physical meaning as to why the Lagrangian is defined as the difference between kinetic and potential energy, $\mathcal{L} = T - V$?

• Possible duplicates: physics.stackexchange.com/q/78138/2451 and links therein. Related question by OP: physics.stackexchange.com/q/106786/2451 – Qmechanic Apr 29 '14 at 22:43
• The Lagrangian is normally chosen, not derived, specifically so that the minimization condition is equivalent to the law you want to describe. So it's not really 'coincidental', but it's also not necessarily a direct physical quantity either. The idea is simply to turn physics into a minimization problem because you can then use new tools and certain manipulations become easier. It has a physical interpretation in path-integral QM but I haven't seen one classically. – Robert Mastragostino May 8 '14 at 18:35
• If you have access to Goldstein's book on Classical Mechanics, there is a derivation of the Euler-Lagrange equations in the first chapter motivated from D'Alembert's principle, which rests on results from Newtonian mechanics. The full derivation shows that the combination $\mathcal{L}=T-V$ appears naturally, and is not a simple coincidence. – physguy May 9 '14 at 6:06

As I understand, the main question that you ask is: "Does the fact that we get Newton's second law mean that the Lagrangian formulation (and therefore the principle of least action) is correct?"

To my taste this question can not be asked of the physics as we know it. What the physics must do is to describe experimental phenomena in such terms as to be able to predict new ones. We all know that this can be done by making use of some maths and by assigning some measurable quantities to physical processes, such as velocity of a particle, its mass, charge and so on.

If we say that a movement of a particle is described (and predicted given the initial data) by $F=m\ddot{x}$, this does not mean that the particle actually calculates its own acceleration and moves according to what sir Newton formulated. So the Newton's law is just a way to encode all the experimental data, past and future, in one nice formula. The same about other physical laws.

Now, we have a set of different laws: mechanics, electromagnetism, GR, etc. In a search of some principle that unifies all these laws we discover a prescription, that by doing very similar steps allows to reproduce all of them. Since, these laws themselves are just prescriptions to predict phenomena we are just one step higher. However, we have something more universal, less model specific and more flexible. This is what we call "more fundamental".

So, the answer is: no, it is not a coincidence, this is an intentionally designed prescription that allows to reproduce Newtons law from just one function $\mathcal{L}$. Why we like this prescription among others? Well, because there are basically no others. The principle of least action is the most fundamental (read universal and flexible) prescription to unify physical laws, that we in classical physics. In quantum physics we go further and come to the prescription of path integral.

Hence, what about the physical meaning. Many years ago I used to read books by Arnold about geometry of mechanical systems and it seems that there was some geometric sense of $\mathcal{L}$, but I do not remember exactly. You may try to find yourself. However in general there is nothing particular in the form $\mathcal{L}=T-V$ of classical mechanics as in field theory this is already not so straighforward.

My point is, that all physical laws including the least action principle are just prescriptions that allow us to predict phenomena. We can not ask if a prescription is correct or not: as long as it gives good predictions it is "correct" at that stage. In this sense the least action principle is correct as is any other law of physics.

is there any physical meaning as to why the Lagrangian is defined as the difference between kinetic and potential energy, $\mathcal{L}=T−V$?

The physical meaning to define the Lagrangian this way is that the resulting Euler-Lagrange equations are equivalent to the equations you would get from using Newton's laws.

Does the fact that we get Newton's second law mean that the Lagrangian formulation (and therefore the principle of least action) is correct?

The fact that you can derive Newton's second law from the Lagrangian formulation means that the Newtonian formulation is a subset of the Lagrangian formulation. I'll argue below that the equivalency goes both ways. To verify that these two formulations are correct, you would need to take a real physical system, write down its forces or Lagrangian, get the equations of motion, and compare those equations to the actual motion of the system.

is there a way that this equation could not hold?

This equation could fail to hold if it were the case that no functional existed that, when minimized, led to Newton's second law. Every step in the derivation from the principle of least action to the Euler-Lagrange equation is reversible, and every step from the Euler-Lagrange equation to $m\ddot x = -V'(x)$ is reversible. So as long as there exists some function $F(x)$ such that $-V'(x) = F(x)$, then there must be a function $\mathcal{L}$ such that the principle of least action is equivalent to Newton's laws. In one dimension $-V'(x) = F(x)$ isn't a particularly strong condition, but in three dimensions, it becomes $-\nabla \vec{V}(\vec{x}) = \vec{F}(\vec{x})$. That is a much more restrictive condition, specifically that the force be conservative.