How can I express an arbitrary Hamiltonian w.r.t. to his Lagrangian.


$ H = \sum{\dot{q} p} - L $


$ p = \frac{\partial L}{\partial \dot{q}} $


$ \dot{p} = -\frac{\partial H}{\partial q} $


$ \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} =-\frac{\partial H}{\partial q} $

We can derivate:

$ \frac{\partial}{\partial q}\left( \sum_k{\dot{q_k}\frac{\partial L}{\partial \dot{q_k}}} - L\right) =-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} $

$ \frac{\partial}{\partial q}\left( \dot{q}\frac{\partial L}{\partial \dot{q}} - L\right) $

$ \frac{\partial \dot{q}}{\partial q}\frac{\partial L}{\partial \dot{q}} + \dot{q}\frac{\partial L}{\partial q \partial \dot{q}}-\frac{\partial L}{\partial q}=-\frac{d}{dt}\frac{\partial L}{\partial \dot{q}} $

So simply:

$ \dot{q}\frac{\partial L}{\partial q \partial \dot{q}}-\frac{\partial L}{\partial q}+\frac{d}{dt}\frac{\partial L}{\partial \dot{q}}= 0 $

It can be written as Euler Lagrange équation offseted.

$ \frac{d}{dt}\frac{\partial L}{\partial \dot{q}} - \frac{\partial L}{\partial q} + \dot{q}\frac{\partial L}{\partial q \partial \dot{q}} = 0 $

Is that mean $\dot{q}\frac{\partial L}{\partial q \partial \dot{q}}$ is necessary 0?

Is that formulation correct? Am I mistaken Somewhere?

  • $\begingroup$ Question is worth answering. $\endgroup$ Commented Jan 1 at 19:01
  • $\begingroup$ @naturallyInconsistent thanks for your relevant feedback. $\endgroup$
    – chkone
    Commented Jan 1 at 19:14

1 Answer 1


I believe the problem arises because you are quite carelessly switching between two different sets of variables. The derivative $\partial_q \mathcal L(q, \dot q)$ is not the same as $\partial_q \mathcal L(q, p_q)$. In particular, in terms of variables $(q,p_q)$, $\partial_q \dot q(q,p_q)$ is not necessarily $0$.

You should repeat this calculation with attention to the coordinates in which you are performing each differentiation. A convenient notation trick borrowed from thermodynamics is to write all the independent variables in a subscript for each differentiation $\left( \frac{\partial \mathcal L}{\partial q}\right)_{\dot q}$ is the derivative performed in coordinates $(q,\dot q)$ i.e. the derivative of the function $\mathcal L(q, \dot q)$, while $\left( \frac{\partial \mathcal L}{\partial q}\right)_{p_q}$ is performed in the coordinates $(q, p_q)$, i.e. the derivative of the function $\mathcal{L}(q,p_q)$. To be clear, in the coordinates $(q, p_q)$, $\dot q$ is not independent - it becomes a function $\dot q(q,p_q)$ and the Lagrangian can be expressed via the composition $\mathcal{L}(q,p_q) = \mathcal L(q,\dot q(q, p_q))$.

  • 1
    $\begingroup$ Yes, this is it. When one writes the Hamiltonian thus, $H=\sum p\dot q-L$, one is in fact doing a Legendre transformation like the ones in thermodynamics. $\endgroup$ Commented Jan 1 at 19:47

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