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A Lagrange v. Hamilton

  1. Jun 30, 2016 #1

    When doing a little internet search today on generalized coordinates I stumbled on this document:


    If you are willing, would you be so kind as to open it up and look at the top of (numbered) page 6?

    OK, so the very existence of this table tells me that these men formulated different ways to structure classical mechanics

    I can accept Newton and d'Alembert as two different approaches (and in this post, when I use the name of the person, I assume the equation itself)

    But I have difficulty seeing why this author lists Lagrange and Hamilton SEPARATELY.

    (forget Gauss as that is not really relevant to my question.)

    It seems to me in order to progress in mechanics, one MUST use Lagrange and Euler TOGETHER.

    In other words, Hamilton provided a, well, blanket or superset to cover Lagrange. It really was not different (setting aside the Hamiltonian here and just looking at the two formulations of Lagranges equation and Least Action). Are those two not really to be taken TOGETHER? Am I missing something? Are Lagrange and Hamilton as distinct from each other as Newton is from d'Alembert?
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  3. Jun 30, 2016 #2


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    Don't confuse Equivalence with Equality. The distinctness is in how one is expressing the dynamics of a system. I can give you the Lagrangian and Euler-Lagrange Equations follow, or I can give you the Hamiltonian and Hamilton's equations follow. Via the Legendre transformations the dynamical equations should be equivalent but the formulation is by no means identical.
  4. Jun 30, 2016 #3
    Something wrong in attached above pdf, bottom of page 3.

    Actually, the generalized forces ##Q_i## are obtained from the active forces, not from external ones as it has been written there.
    Some of active forces may also be internal.
  5. Jul 1, 2016 #4


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    The Hamilton principle of least action is different for the Lagrangian and the Hamilton version. In the Lagrange formulation the action functional is a functional of trajectories in configuration space,
    $$S[q]=\int_{t_1}^{t_2} \mathrm{d} t L(q,\dot{q},t).$$
    The equations of motion are given by the stationary point of this functional for trajectories in configuration space with fixed boundaries, i.e., for ##\delta q(t_1)=\delta q(t_2)=0##, i.e., the Euler-Lagrange equations,
    $$\frac{\delta S}{\delta q}=0 \; \Rightarrow \; \frac{\partial L}{\partial q}-\frac{\mathrm{d}}{\mathrm{d} t} \frac{\partial L}{\partial \dot{q}}.$$
    The formalism is form-invariant under arbitrary diffeomorphisms in configuration space, i.e., the EL equations are of the same form in any generalized configuration-space variables.

    In the Hamiltonian formalism of the least-action principle, you consider trajectories in phase space and the variational principle is for variations of phase-space trajectories. The action functional reads
    $$A[q,p]=\int_{t_1}^{t_2} \mathrm{d} t [\dot{q} \cdot p-H(q,p,t)].$$
    The equations of motion are the stationary points of this functional under variations of the phase-space trajectories ##(q,p)##, with the ##\delta p## arbitrary and ##\delta q(t_1)=\delta q(t_2)=0##. The equations of motion are the Hamilton canonical equations,
    $$\dot{q}=\frac{\partial H}{\partial p}, \quad \dot{p}=-\frac{\partial H}{\partial q}.$$
    The transformations that leave these equations form-invariant are the much larger set of canonical transformations (symplectomorphisms on phase space), i.e., those transformations, which leave the canonical Poisson-bracket relations invariant,
    $$\{q^j,q^k \}=\{p_j,p_k \}=0, \quad \{q^j,p_k\}={\delta^j}_k,$$
    where the Poisson bracket of any pair of phase-space functions ##A,B## is defined as (Einstein summation convention implied)
    $$\{A,B \}=\frac{\partial A}{\partial q^k} \frac{\partial B}{\partial p_k} - \frac{\partial B}{\partial q^k} \frac{\partial A}{\partial p_k}.$$
    The great thing with the Hamilton formalism in phase space is that together with the Poisson brackets the function space of phase-space functions becomes a Lie algebra, and from the point of view of modern physics it's the most fundamental way to describe classical mechanics. With a little "deformation" (in the mathematical sense) you get quantum theory for (almost) free!
  6. Jul 1, 2016 #5
    By the way, the change ##(p,q)\mapsto (P,2Q)## keeps the shape of Hamilton equations but this transformation is not a symplectomorphism
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