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Reading any classical mechanics book which introduces the Lagrangian formalism of mechanics, a one particle system is introduced to show that we obtain the euler-lagrange equations from Newton's second law when we introduce the lagrangian function. Then we proceed to generalize to Hamilton's principle but I have a few questions about this generalization.

  • It was said in this answer that, in classical mechanics a particle takes a definite path and it happens to be the path which makes the action stationary. But, how do we guarantee, mathematically, that the functional will have only one function that makes it stationary? How do we guarantee that there wouldn't be multiple extrema? (I know physically there has to be only one solution but I do not see how that is guaranteed in the derivation.

  • This answer suggests that there in QFT some functionals have multiple extrema, but since I have not taken QFT yet it would be nice if someone can give a gentle introduction as to why this is the case.

This answer, which was suggested to answer my question, does not answer my q as it does not motivate why for most classical systems we get only one extrema, my question is why was this guranteed and if not what was the factor which led to only one extrema?

realanswers
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  • The second link provides a classical counterexample that contradicts the title (v3). Possible duplicates: https://physics.stackexchange.com/q/203493/2451 and links therein. – Qmechanic Aug 06 '22 at 21:42

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Even in the simplest classical mechanics problems, there are often multiple paths which make the action stationary. As a simple example, consider a simple harmonic oscillator, $V(x) = \frac{1}{2} m \omega^2 x^2$, and consider the set of paths which start at the origin at $t = 0$ and return to the origin at $t=\pi / \omega$. Clearly there are multiple paths which satisfy Hamilton's principle: one solution is for the particle to sit at the origin the whole time, but more generally any oscillation $x(t) = A \sin \omega t$ for any value of $A$ will extremize the action for the given boundary value problem. Of course, we physically know that you need to specify an initial velocity to obtain a unique solution.

The lesson here is that Hamilton's principle is sometimes not enough to determine the full motion of the classical system. In practice, it is better to regard Hamilton's principle as a way of generating the equations of motion: by demanding that the action is stationary, you can derive a second-order differential equation which the classical motion must satisfy. Then, given proper initial conditions (ie, the initial position and velocity), you arrive at a unique classical trajectory.

Zack
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  • my answer is even given initial conditions, how do we know we have one unique "family" of solutions? – realanswers Aug 06 '22 at 22:03
  • If you have given initial conditions, then your unique solution follows immediately from the uniqueness theorem of ODEs. But this does not tell you there is a unique trajectory extremizing an action, it tells you there's a unique solution to the equations of motion given (say) a particle's initial position and velocity. – Zack Aug 06 '22 at 23:36