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I noticed that in deriving Hamilton equations from the total deriveative of the Hamiltonian with respect to time, for the first equation $$\frac{dx_k}{dt}=\partial_{p_k}H$$ we do not need Lagrange's equation, which are needed to derive $$\frac{dp_k}{dt}=-\partial_{x_k}H.$$

It is clear to me that only the vector $(x_k(t),p_k(t))$ can describe the evolution of the system in its phase space, and hence the first equation alone does not make much sense, but is there any phyisical/mathematical reason behind this asymmetry?

Francesco Bilotta
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1 Answers1

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We can restore symmetry between the Lagrangian and Hamiltonian formalisms as follows:

  1. The first half $v=\frac{\partial H}{\partial p}$ of Hamilton's equations is the Legendre transformed relation of $p=\frac{\partial L}{\partial v}$.

  2. The second half $\frac{dp}{dt}=-\frac{\partial H}{\partial x}$ of Hamilton's equations corresponds to Lagrange equations $\frac{dp}{dt}=\frac{\partial L}{\partial x}$.

  3. In both Hamiltonian & Lagrangian approaches we identify $v=\frac{dx}{dt}$.

For more details, see e.g. this Phys.SE post.

Qmechanic
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