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##\frac{\partial{H}}{\partial{t}} = - \frac{\partial{L}}{\partial{t}}##, where ##L## is the Lagrangian.

But how can this be? This assumes the generalized coordinates are independent of t, but that's not the case for dynamic systems.

- Thread starter Nikitin
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- #1

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##\frac{\partial{H}}{\partial{t}} = - \frac{\partial{L}}{\partial{t}}##, where ##L## is the Lagrangian.

But how can this be? This assumes the generalized coordinates are independent of t, but that's not the case for dynamic systems.

- #2

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- #3

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- #4

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- #5

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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)## ?

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)## ?

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- #6

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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 $$ \left(\partial H \over \partial t \right) \mathrm d t = - \left(\partial L \over \partial t \right) \mathrm d t. $$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?

Hmm. You are reading about Lagrangian and Hamiltonian formulations and yet do not understand what ##q## and ##p## are?and could you please tell me what exactly the q's and p's represent? Are they some kind of starting coordinates describing the system at time=0?

- #7

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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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