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This paper is about deriving hamilton's equations from Euler-Lagrange equation, what i don't understand is equation 19. In equation 18 the process involved is if we substitute lagrangian $L$ for function $f$ we get one of hamilton's equation but in equation 19 i don't know what process is involved to obtain another hamilton's equation?

Reference: P. Gutierrez, Physics 5153 Classical mechanics Canonical transformation

Qmechanic
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Keshav Shrestha
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1 Answers1

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  1. The function $$f(q,\dot{q},p,t)~\equiv~ L_H(q,\dot{q},p,t)~\equiv~ p\dot{q}-H(q,p,t)$$ is the so-called Hamiltonian Lagrangian. Note in particular that its arguments are independent variables.

  2. Its EL equations lead to Hamilton's equations (HE): The 1st half of HE is eq. (18). The 2nd half of HE is eq. (19). To derive eq. (19), use the fact that $f$ does not depend on $\dot{p}$, so that the EL eq. (19) simplifies to $\frac{\partial f}{\partial p}=0$.

  3. For more details, see e.g. this and this related Phys.SE posts.

Qmechanic
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  • #Qmechanic I still don't get a proof when i use fact that $f$ is not a function of $\dot p$. Will you please provide me a proof of it? – Keshav Shrestha Aug 31 '22 at 07:37
  • I updated the answer. – Qmechanic Aug 31 '22 at 07:50
  • #Qmechanic yes but after substituting $L$ for $f$ finally i get $p$ $\frac {\partial \dot q} {\partial p}$ + $\dot q$ - $\frac {\partial H} {\partial p}$ =0 Where the first term $p$ $\frac {\partial \dot q} {\partial p}$ is not equal to zero what to do after that? – Keshav Shrestha Aug 31 '22 at 08:17
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    $\dot{q}$ and $p$ are here independent variables. – Qmechanic Aug 31 '22 at 08:31
  • Since we have assume that $f$ being a function of q(t) , p(t) , $\dot q(t)$, $\dot p(t)$ all these variables are function of some parameter $t$ but they do not depend on each other as $\dot q$ is not a function of p, is that correct understanding ? – Keshav Shrestha Aug 31 '22 at 08:46
  • Also from lagrangian we assume that $L$ (q(t), p(t), $\dot q(q(t), p(t)), $\dot p(t) that's why $\dot q$ is not a function of $p$ but it is a function of $t$ is that right? – Keshav Shrestha Aug 31 '22 at 09:11
  • @Qmechanic as I've only run into the name Hamiltonian Lagrangian in your posts, I wonder if there is some Classical Mechanics book you are referencing or you invented this name for such function. – Mr. Feynman Sep 17 '22 at 18:15
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    It's my invention :) – Qmechanic Sep 17 '22 at 18:27