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During the introductory class on Hamiltonian mechanics after we had learnt the Lagrangian formalism, our professor said

"We cannot use Lagrangian mechanics to find the state of a system at any particular instant (he did not give any explanation), so that is why we use Hamiltonian mechanics."

But I think he could be incorrect, because when I did problems using Lagrangian as well as Hamiltonian I could find the generalized coordinates as a function of time. Can anybody explain where is the conceptual loop hole?

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  • $\begingroup$ Id be interested as well. Afaik, they should be equivalent. Even if there are some superfluous DOF it should still be equivalent $\endgroup$
    – Confuse-ray30
    Commented Dec 7 at 13:16
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    $\begingroup$ Perhaps he meant directly finding points in phase space? $\endgroup$
    – BioPhysicist
    Commented Dec 7 at 13:18
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    $\begingroup$ It's OK to ask here, but you should be comfortable asking the professor for clarification - that's the whole point of having a professor (as opposed to, say, studying on your own from a book). $\endgroup$
    – Filip Milovanović
    Commented Dec 7 at 21:29

2 Answers 2

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  1. OP's professor is possibly referring to the fact that the stationary action principle is a boundary value problem (BVP) [as opposed to an initial value problem (IVP)].

    In contrast, EOMs together with initial conditions (ICs) can in principle be integrated from an initial time to find the trajectory as a function of time.

  2. The issues of ICs vs. BCs for the principle of stationary action are already covered in this & this related Phys.SE posts and this related Math.SE post.

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  • $\begingroup$ In classical mechanics, how is the principle of least action any sort of problem? (IC, BC) What I mean is, the principle of least action does not constrain the paths considered at all. It only constrains the variations at the end points. But this does not imply we can only have boundary value problems as a result. We can still ask the for two IC and solve the EOM $\endgroup$
    – Confuse-ray30
    Commented Dec 7 at 13:57
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    $\begingroup$ It seems OP's professor should clarify what they meant. $\endgroup$
    – Qmechanic ♦
    Commented Dec 7 at 14:32
  • $\begingroup$ How do you find the EOM without (edit ...) the action principle and a Lagrangian? The answer could be 'experimentally' but that would make the theory semiempirical. The statement of the OP's professor does appear sensible to me. $\endgroup$
    – my2cts
    Commented Dec 7 at 21:23
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The action principle and the Lagrangian give you the equation of motion. The EOM tells you which states of the system are possible. By applying conditions, for example initial ones, a subset of these states is selected. The boundary conditions of the action principle are unrelated to the conditions imposed on the solution of the EOM.

As far as I understand the question, the Hamiltonian method assumes the EOM as a starting point. This approach is part of the Lagrangian approach.

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  • $\begingroup$ The hamiltonian approach does not start from EOM, in so far you accept (just as in the lagrangian formalism the least action) that H generates time translation, i.e. it is the hamiltonian function of the flow of phase space. From this, the EOM follow IN PARTICULAR for p and q. $\endgroup$
    – Confuse-ray30
    Commented Dec 7 at 16:43

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