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Hamiltonian mechanics: ∂H/∂t = ?

  1. Sep 6, 2014 #1
  2. jcsd
  3. Sep 6, 2014 #2
    The derivation that you referenced does not in fact require that the gen. coord's be independent of time.
     
  4. Sep 6, 2014 #3
    But ##H = \sum_i (p_i \dot{q_i}) - L##, so when you take the time-partial derivative ,##\frac{\partial{H}}{\partial{t}} = - \frac{\partial{L}}{\partial{t}}## can only be true if the ##(p_i \dot{q_i})##s are not functions of time. So at least the product of them must be time-independent.
     
  5. Sep 6, 2014 #4
    Yes, the equation $$ {\partial H \over \partial t} = - {\partial L \over \partial t} $$ implies that $${\partial \sum p_i \dot q_i \over \partial t } = 0. $$ But the latter does not necessarily require that ##q_i## be independent of ##t##, if that is what you find problematic (it is unclear to me what your real difficulty is here).
     
  6. Sep 6, 2014 #5
    I'm sorry, I am really confused about all of this at the moment.

    What I mean to ask, is why $${\partial \sum p_i \dot q_i \over \partial t } = 0. $$ is true? My statistical physics book (which sucks) just took it for granted.

    and could you please tell me what exactly the q's and p's represent? Are they some kind of states representing particles? So particle i is represented by the vector ##(q_i,p_i)## ?
     
    Last edited: Sep 6, 2014
  7. Sep 6, 2014 #6
    Let's take the starting equation you linked earlier: $$ \mathrm d H = \sum \left[ \left(\partial H \over \partial q_i \right) \mathrm d q_i + \left(\partial H \over \partial p_i \right) \mathrm d p_i \right]+ \left(\partial H \over \partial t \right) \mathrm d t .$$ This is the total differential of function ##H(q, p, t)## and this equation holds true regardless of what ##q## and ##p## are; in particular, they may also have dependence on ##t##. What happens then is re-derivation of this total differential by taking the definition of ##H## and using Leibniz's rule and the chain rule: $$ \mathrm d H = \sum \left[ \dot q_i \mathrm d p_i + p_i \mathrm d \dot q_i - \left(\partial L \over \partial q_i \right) \mathrm d q_i - \left(\partial L \over \partial \dot q_i \right) \mathrm d \dot q_i \right] - \left(\partial L \over \partial t \right) \mathrm d t .$$ This is again a total differential and is valid no matter what ##q ## and ##\dot q## we have. Then we note that by definition of ##p##, the second and fourth term in the square brackets cancel each other, leaving $$ \mathrm d H = \sum \left[ \dot q_i \mathrm d p_i - \left(\partial L \over \partial q_i \right) \mathrm d q_i \right] - \left(\partial L \over \partial t \right) \mathrm d t .$$ Now we compare eq. 1 and eq. 3 and obtain immediately $$ \left(\partial H \over \partial t \right) \mathrm d t = - \left(\partial L \over \partial t \right) \mathrm d t. $$

    Hmm. You are reading about Lagrangian and Hamiltonian formulations and yet do not understand what ##q## and ##p## are?
     
  8. Sep 7, 2014 #7
    Thank you. Yes it's embarrassing, though I am just trying to get a superficial overview over for now.

    But,, are the ##q_i##'s the coordinates for the particles in the system, and ##p_i##'s their "generalized momentum" (can't say I full understand that term..)? So the ##(q,p)## vector, which contains all the ##q_i##'s and ##p_i##'s defines the state of the system, right?
     
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