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Topic 3 of 11

Hamiltonian Mechanics

The Hamiltonian formulation replaces the $n$ second-order Lagrange equations with $2n$ first-order Hamilton's equations. Phase space — the space of positions and momenta — is the natural arena. This geometric view reveals deep connections to quantum mechanics and statistical physics.

Key Concepts

Hamiltonian

The Legendre transform of

L

H(q,p,t)=\sum_i p_i\dot q_i - L

. For conservative systems with natural kinetic energy,

H

equals the total energy

T+V

Hamilton's Equations

\dot q_i=\partial H/\partial p_i

and

\dot p_i=-\partial H/\partial q_i

. These

2n

first-order ODEs replace the

n

second-order Euler-Lagrange equations.

Poisson Bracket

For functions

f(q,p)

and

g(q,p)

\{f,g\}=\sum_i\left(\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i}-\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}\right)

. The fundamental brackets are

\{q_i,p_j\}=\delta_{ij}

Liouville's Theorem

Phase space volume is conserved under Hamiltonian flow:

d\rho/dt=0

. This means the density of representative points in phase space is incompressible — a fundamental result for statistical mechanics.

Key Equations

Hamilton's equations

\dot{q}_i = \frac{\partial H}{\partial p_i}, \quad \dot{p}_i = -\frac{\partial H}{\partial q_i}

2n first-order equations replacing n second-order Euler-Lagrange equations.

Hamiltonian (natural)

H = \frac{p^2}{2m} + V(q)

For a particle with standard kinetic energy; equals total mechanical energy.

Poisson bracket

\{f,g\} = \frac{\partial f}{\partial q}\frac{\partial g}{\partial p} - \frac{\partial f}{\partial p}\frac{\partial g}{\partial q}

One-dimensional form; the fundamental bracket is {q,p}=1.

Worked Example

Harmonic Oscillator in Phase Space

Problem

For a harmonic oscillator $H=p^2/(2m)+\frac{1}{2}k x^2$ with $m=1.0$ kg and $k=4.0$ N/m, find: (a) the angular frequency $\omega$ , and (b) the amplitude if the total energy is $E=8.0$ J.

Solution

Hamilton's equations: $\dot x = p/m$ , $\dot p = -kx$ . Differentiating: $\ddot x = -kx/m = -\omega^2 x$ .

\omega = \sqrt{k/m} = \sqrt{4.0/1.0} = 2.0\text{ rad/s}

At maximum displacement ( $p=0$ ): $H = \frac{1}{2}kA^2 = E$ .

A = \sqrt{2E/k} = \sqrt{16/4} = 2.0\text{ m}

Answer

\omega=2.0

rad/s,

A=2.0

Practice

Exercises

7 problems

1 of 7

For a harmonic oscillator with $m=1.5$ kg, $k=100$ N/m, $p=3.0$ kg·m/s, $x=0.10$ m, compute the Hamiltonian $H$ (total energy, in J).

Your answer

2 of 7

For a harmonic oscillator with $k=4.0$ N/m and total energy $E=8.0$ J, find the amplitude $A$ (in m).

Your answer

3 of 7

From Hamilton's equations for $H=p^2/(2m)$ (free particle), find $\dot p$ (in N). Express your answer as a number.

Unlock Exercise 3