Answer to Question #294226 in Mechanics | Relativity for Delightful

Question

Define the Hamiltonism and derive C anonical Equations of Motion

Answer 1

Hamiltonian approach is a reformulation of Lagrangian mechanics which consists in replacing the velocities $\dot q$ with momenta $p$. However, the interest of such reformulation is that Hamiltonian mechanics give a much more general approach to mechanics, as generalized coordinates $q$ and generalized momenta $p$ play much more symmetric (and thus less constrained) role than the pairs $(q, \dot q)$ of Lagrangian mechanics. In particular, Hamiltonian approach can be generalized for quantum mechanics, but neither Newtonian nor Lagrangian mechanics can not.

To derive the formalism and the canonical equations let us consider the lagrangian of a free particle :

$\mathcal L = \frac{m \dot x^2}{2} + \frac{m \dot y^2}{2}+ \frac{m \dot z^2}{2}$

We see that the momentum is given by

$p_i = \frac{\partial \mathcal L}{\partial \dot q_i}$

Which motivates the definition for the general case :

$p:= \frac{\partial \mathcal L}{\partial q}$

We also define the Hamiltonian function as

$\mathcal H := (\sum p\dot q ) - \mathcal L$

Strictly speaking, from a mathematical point of view we are performing a Legendre transformation to pass from the coordinates $(q,\dot q)$ to the coordinates $(p,q)$.

Now minimizing the action

$S = \int [(\sum p\dot q) - \mathcal H] dt$

$\delta S =\sum \left( \int (\dot q\delta p -\partial _p \mathcal H \delta p) dt+\int (p\delta \dot q - \partial_q \mathcal H \delta q) dt \right)$, and integrating by parts the $\delta \dot q = \dot{(\delta q)}$ term gives us a set of equations, called canonical equations of motion :

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