Question

It is well known that the Lagrangian of a classical free particle equal to kinetic energy. This statement can be derived from some basic assumptions about the symmetries of the space-time. Is there any similar reasoning (eg. symmetry based or geometrical) why the Lagrangian of a classical system is equal kinetic energy minus the potential energy? Or it is just because we can compare the Newton's equations with the Euler-Lagrange equation and realize how they can match?

Answer 1

Well, Lagrangian is what one usually starts with, even in the modern physics, and everything else is derived. So I don't believe you can explain it any better than it is the way it is because it matches observation.

But there are two points about the above worth taking.

First symmetries can give you a nudge into the right direction of constructing Lagrangian. E.g. in classical mechanics in free space you want to have laws that are translation and rotation invariant, etc., so that excludes lots of strange Lagrangians first-hand. Nevertheless, you could also ask what is the symmetry derived from and for that there is no good answer. At some point you have to stop with the chain of deductions and realize that physics can't really be derived mathematically but has to be observed. It's actually fascinating that the world can be reduced into few simple laws that appear to be purely mathematical.

Second, there can sometimes be a microscopic model of the given system which kind of "explains" why the macroscopic description takes the given form in a limit. One can e.g. derive the principle of least action from quantum mechanics. So you can explain some things. But I don't believe you can really explain Lagrangian.