Lagrange & Hamiltonian mech => Newtonia mech.

  • #1
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My question is simple is every classical mechanics problem which is solvable by Lagrangian & Hamiltonian methods also solvable by Newtonian methods of forces and torques?

And why does it seem that LH make solutions to be a lot more easier than Newtonian methods, and is it always this way?
 

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  • #2
vanhees71
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The principle of least action is the common description of all physical theories at a fundamental level. All fundamental theories are most simply described as such a variational principle (standard model of elementary particle physics, classical field theory (e.g., classical electromagnetics, general relativity), and point-particle mechanics).

The reason for this is that it most easily allows to study symmetries of the fundamental laws of nature, such as space-time symmetries (Galileo, Poincare, general covariance) and external symmetries leading to conservation laws for engery, momentum, angular momentum, center-of-mass velocity, and various charge-like quantities (electrical charge, baryon number, lepton number, etc.), respectively.

Of course, on the classical (i.e., non-quantum) level of point-particle mechanics, the action principle is equivalent to the Newton (or relativistic if necessary) equations of motion for the particle, and you can of course write down this equations from the very beginning.

What makes the action principle more convenient from a practical point of view is that it is way easier to express the action (or the Lagrangian/Hamiltonian) of the system in general coordinates, adapted to the problem at hand, than using directly the equations of motion and performing the transformation from Cartesian to generalized coordinates.
 
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My question is simple is every classical mechanics problem which is solvable by Lagrangian & Hamiltonian methods also solvable by Newtonian methods of forces and torques?

And why does it seem that LH make solutions to be a lot more easier than Newtonian methods, and is it always this way?
If by "classical" you mean non-relativistic classical, then the three formulations are formally equivalent. You can, however, find computational or analytical difficulties. That is why one chooses the formulation more adequate to the problem.
 
  • #4
dextercioby
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It only <seems> that way, well said. Indeed, the lagrangian methods seem simpler, but in some cases, they are really not (here I mean dissipative systems, where the task to find a lagrangian is not simple). It normally all boils down to solving ODE's. To get to them can be simpler using one method or another, but it's not a general rule.
 

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