Why are position and momentum independent with respect to the Hamiltonian Action $S_H$ given by $$ S_H = \int_{t_1}^{t_2} (p \dot q - H) dt \ \ \ ? \tag{1} $$

While deriving Hamilton's equations from this action by varying the path, we assume that the variation in positions $\delta q$ is independent of the variation in momentum $\delta p$ and hence, we get 2$n$ equations. However, in the Lagrangian Action $$ S_L = \int_{t1}^{t_2} L dt \tag{2} $$ we show that the variations in position and velocity are related by $$ \delta \dot q = \frac{d}{dt} \delta q \tag{3} $$ How can position and momentum be independent but not position and velocity in the same set-up? Aren't velocity and momentum intrinsically the same thing?

(The Hamiltonian action was referred to in Qmechanics' answer to a similar question, but I couldn't show that the position and momentum are independent for the Hamiltonian action. Any help proving this would be much appreciated.)

  • $\begingroup$ $q$ and $\dot q$ as well as $p$ and $q$ are both just manifold coordinates, so they are a priori independent variables. This question has been asked many times on the site. $\endgroup$
    – Charlie
    Jun 3, 2021 at 20:23
  • $\begingroup$ Hi Drishti Gupta. Welcome to Phys.SE. Hypothetically, how do you plan to use formula (1) if position and momentum are not independent? [On the other hand, if we assume that position and momentum are independent in formula (1), then we can readily derive Hamilton's equations by standard reasoning.] $\endgroup$
    – Qmechanic
    Jun 3, 2021 at 20:49
  • $\begingroup$ Hi @Qmechanic. Ideally, I think I would write $p$ as a function of $q$ and $\dot q$ (using $p = \frac{\partial L}{\partial \dot q}$) and then effect a variation $\delta p$ by varying $q$ in the function. I would use this $\delta p$ in (1). But all the standard texts indicate that this isn't the right thing to do, though I don't understand why. $\endgroup$
    – Dris
    Jun 3, 2021 at 21:40
  • $\begingroup$ In (1) we are in the Hamiltonian formulation, so one cannot readily use a Lagrangian formula $p = \frac{\partial L}{\partial \dot q}$. $\endgroup$
    – Qmechanic
    Jun 5, 2021 at 4:17

1 Answer 1

  • Lagrangian mechanics takes place in configuration space with coordinates $q$. The Lagrangian is a function of $q$ and $\dot q$, the latter being the time derivative of the former.

  • Hamiltonian mechanics takes place in phase space, with coordinates $(q,p)$. The relationship between the two is one of Hamilton's equation of motion. When deriving these equations of motion from an action principle, you have no choice but to consider them independent variables. For example, with the usual $H(p,q) = \frac{p^2}{2m} + V(q)$, one of Hamilton's equation is : $$\dot q = \frac{\partial H}{\partial p} = \frac{p}{m}$$

  • There is a connection between the two formalism (when the kinetic term is nice enough) :

    • from a Lagrangian $L(q,\dot q)$, taking the Legendre transform with respect to $\dot q$ gives a Hamiltonian $H(q,p)$ such that, taken together, both of Hamilton's equations are equivalent to the Euler-Lagrange equation. This means that from the configuration space with coordinate $q$ and Lagrangian function $L(q,\dot q)$, we can build a phase space with coordinates $(q,p)$ and a Hamiltonian function on it $H(q,p)$ such that the dynamics of $q$ under Hamilton's equations of motion is identical to the one under Euler-Lagrange.
    • The converse is a bit subtler : from a phase space $(q,p)$ and a Hamiltonian $H(q,p)$, we take the Legendre transform and define a function $L(q,v) = vp - H(q,p)$. The new variable introduced by the Legendre transform is fixed by $ v = \partial H/ \partial p$ so by Hamilton's equation we see that $v= \dot q$. Therefore, we can choose to consider only the configuration space $(q)$ and let the dynamics be governed by the Euler-Lagrange equations.
  • $\begingroup$ Oh, so the Hamiltonian Formulation doesn't assume any relation between the velocities and momentum and in fact finds it. So since we aren't actually using the definition of momentum we define in the Lagrangian, does this mean that $ H = p\dot q - L $ is just a method to figure out the form of the Hamiltonian and the Hamiltonian definitions actually have nothing to do with Lagrangian definitions? $\endgroup$
    – Dris
    Jun 4, 2021 at 7:05
  • $\begingroup$ There is a relationship between the two (I edited some explanations into my answer). $\endgroup$ Jun 4, 2021 at 9:17
  • $\begingroup$ It feels a little misleading to say that Lagrangian mechanics takes place on configuration space and then going on to say Hamiltonian mechanics takes place on phase space (sort of implying it is unrelated to configuration space), since they are both just real functions on the tangent and cotangent bundles of configuration space respectively. Neither of them really takes place on configuration space alone. $\endgroup$
    – Charlie
    Jun 5, 2021 at 14:57
  • $\begingroup$ Well, Hamiltonian dynamics can be expressed on a general symplectic manifold, of which the cotangent bundle of a configuration space is just one example. One important consequence is that canonical transformation can mix position and momentum coordinates, which is not possible in Lagrangian mechanics. $\endgroup$ Jun 5, 2021 at 16:03

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