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Clarification on symplectic forms and Hamiltonian – Rovelli

  1. May 4, 2015 #1
    Please refer to p. 99 and 100 of Rovelli’s Quantum Gravity book (here).

    I wonder what is the signification of the “naturalness” of the definition of ##\theta_0=p_idq^i##? If I take ##\theta_0'=q^idp_i## inverting the roles of the canonical variables and have the symplectic 2-forms of the Darboux’s theorem changing sign, ##\omega_0=d\theta_0=-d\theta_0'=-\omega_0'##, I would get ##(d\theta'_0)(X_0)=dH_0## the opposite of eq. 3.7. Or maybe I am not allowed to invert the roles of the variables because I need to work in the cotangent space and the only "natural" 1-form ##\theta_0## must be ##p_idq^i## where first you have the covector ##p_i##? What is the signification of this inversion of sign of the ##dH_0##? Could this be related to the assumption that the energy of a system is in reality the difference with an energy taken as zero? I would like to understand better the symplectic formulation of mechanics as it looks crucial when you go from the classical to the general-relativistic Hamiltonian formulation.

    Thank you in advance!
  2. jcsd
  3. May 9, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
  4. May 10, 2015 #3
    Sorry, but somehow the page that you've linked can't be viewed to me.
    But perhaps this is what you're looking for: There is a natural pairing between tangent and cotangent vectors. But in Hamiltonian mechanics, you're working on the cotangent bundle. Every tangent vector to the configuration space ([itex](q,v) = (q, v^i \partial_{i}) \in T_q (M)[/itex]) can be naturally treated as a tangent vector to the phase space [itex](q, p; v^i \partial_{q^i}) \in T(T^* (M))[/itex]. Conversely, every tangent vector to the phase space [itex]z = (q, p; v^{q^i} \partial_{q^i} + v^{p_i} \partial_{p_i}[/itex] has a component that can be treated as a tangent vector to the configuration space. If [itex]\Pi: T^*(M) \to M[/itex] be the canonical projection, then [itex]v = T(\Pi )\cdot z[/itex] (T(Pi) is the derivative, other common notations are Pi' or dPi). So you can assign to every point of the phase space a cotangent vector that, paired with a tangent vector to the phase space, gives the pairing of the momentum component of that point with the tangent vector, treated as a tangent vector to the configuration space. This is exactly the canonical 1-form. [itex]\Theta_{(q,p)} (z) := (p \vert T(\Pi)\cdot z)[/itex].
    Of course, [itex]q^i\,dp_i = d(q^i\,p_i) - p_i\,dq^i[/itex].
    Hope this helps.
  5. May 10, 2015 #4
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