Classical mechanics

In classical mechanics a conservative dynamical system is represented by one sole function, the Hamiltonian, which represents the total energy of the system. The system indeed $''is''$ its Hamiltonian, in the sense that all the information we have about the system is reflected in the form of the Hamiltonian, from which we can also determine how the system changes.

>From the point of view of the Hamiltonian, "process" means nothing but a continuous redistribution of the relative amounts of kinetic and potential energy in the total energy. At some stage of the process a greater part of the energy is kinetic, a moment later some kinetic energy is transformed to potential energy: energy is so to speak poured to and fro between the pots labelled "potential energy" and "kinetic energy".

Whereas Newton's law $F$ $=$ $ma$ comprehends second derivatives, namely ${\ddot {x}}$ $=$ $a$, where the dot represents the time derivative, the Hamiltonian formalism only contains first derivatives, and instead of the one coordinate $x$, there are two variables, the generalized coordinate $q$ and momentum $p$, treated as independent parameters. The Hamiltonian is thus

$$H=H(q,p)=T(q,p)+V(q)$$ (1)

where $T$ and $V$ are the kinetic and the potential energies, respectively.

The development of the system is given by the equation of motion, which is determined by means of the principle of least action. It says that to each system we can ascribe an action

$$S=\int_{t_{1}}^{t_{2}}L(q,\dot{q})dt$$ (2)

such that the action is stationary as the system moves between the positions $q(t_{1})$ and $q(t_{2})$ at the instants $t_{1}$ and $t_{2}$. The function $L$ is the Lagrangian of the system, which is related to the Hamiltonian through a Legendre transformation

$$S=\int L(q,\dot{q})dt=\int[\dot{q}p-H(q,p)]dt$$ (3)

The action being stationary we have

$$\delta S= \delta \int_{t_{1}}^{t_{2}}L(q,\dot{q})dt = 0\nonumber\\$$

which leads to Euler-Lagrange equations

$$\frac{\partial L}{\partial q}-\frac{d}{dt}\frac{\partial L} {\partial \dot{q}}=0$$ (4)