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Hamiltonian mechanics

Formulation of classical mechanics using momenta

  • 15 min read
  • 7 chapters
Sir William Rowan Hamilton
Unknown author Unknown author · Public domain

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Through Physics, Action (physics) and Potential energy

Hamiltonian mechanics has a close relationship with geometry (notably, symplectic geometry and Poisson structures) and serves as a link between classical and quantum mechanics.

Overview

Phase space coordinates (p, q) and Hamiltonian H

Let (M,L) be a mechanical system with configuration space M and smooth Lagrangian L. Select a standard coordinate system (q,q˙) on the tangent bundle TM. The quantities pi(q,q˙,t)=def∂L/∂q˙i are called momenta. (Also generalized momenta, conjugate momenta, and canonical momenta). For a time instant t, the Legendre transformation of L is defined as the map (q,q˙)→(p,q) which is assumed to have a smooth inverse (p,q)→(q,q˙). For a system with n degrees of freedom, the Lagrangian mechanics defines the energy function EL(q,q˙,t)=def∑i=1nq˙i∂L∂q˙i−L.

The Legendre transform of L turns EL into a function H(p,q,t) known as the Hamiltonian. The Hamiltonian satisfies H(∂L∂q˙,q,t)=EL(q,q˙,t) which implies that H(p,q,t)=∑i=1npiq˙i−L(q,q˙,t), where the velocities q˙=(q˙1,…,q˙n) are found from the (n-dimensional) equation p=∂L/∂q˙ which, by assumption, is uniquely solvable for ⁠q˙⁠. The (2n-dimensional) pair (p,q) is called phase space coordinates. (Also canonical coordinates).

From Euler–Lagrange equation to Hamilton's equations

In phase space coordinates ⁠(p,q)⁠, the (n-dimensional) Euler–Lagrange equation ∂L∂q−ddt∂L∂q˙=0 becomes Hamilton's equations in 2n dimensions

dqdt=∂H∂p,dpdt=−∂H∂q.

Proof

The Hamiltonian H(p,q) is the Legendre transform of the Lagrangian L(q,q˙), thus one has

where p=∂L/∂q˙.

By rearranging the equation p=∂L/∂q˙, we can write q˙ in terms of q and p as q˙(q,p). So 1 becomes an equation in two variables: q and p.

Taking the partial derivative of both sides of 1 with respect to p (i.e. keeping q fixed) gives ∂L∂q˙∂q˙∂p+∂H∂p=q˙+p∂q˙∂p⟹∂H∂p=q˙ Taking the partial derivative of both sides of 1 with respect to q instead (i.e. keeping p fixed) gives ∂L∂q+∂L∂q˙∂q˙∂q+∂H∂q=p∂q˙∂q⟹∂L∂q=−∂H∂q,

Now the Euler–Lagrange equations yield p˙=ddt∂L∂q˙=∂L∂q=−∂H∂q.

From stationary action principle to Hamilton's equations

Let P(a,b,xa,xb) be the set of smooth paths q:[a,b]→M for which q(a)=xa and q(b)=xb. The action functional S:P(a,b,xa,xb)→R is defined via S[q]=∫abL(t,q(t),q˙(t))dt=∫ab(∑i=1npiq˙i−H(p,q,t))dt, where ⁠q=q(t)⁠, and p=∂L/∂q˙ (see above). A path q∈P(a,b,xa,xb) is a stationary point of S (and hence is an equation of motion) if and only if the path (p(t),q(t)) in phase space coordinates obeys the Hamilton equations.

Basic physical interpretation

A simple interpretation of Hamiltonian mechanics comes from its application on a one-dimensional system consisting of one nonrelativistic particle of mass m. The value H(p,q) of the Hamiltonian is the total energy of the system, in this case the sum of kinetic and potential energy, traditionally denoted T and V, respectively. Here p is the momentum mv and q is the space coordinate. Then H=T+V,T=p22m,V=V(q) T is a function of p alone, while V is a function of q alone (i.e., T and V are scleronomic).

In this example, the time derivative of q is the velocity, and so the first Hamilton equation means that the particle's velocity equals the derivative of its kinetic energy with respect to its momentum. The time derivative of the momentum p equals the Newtonian force, and so the second Hamilton equation means that the force equals the negative gradient of potential energy.

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