Hamilton's Equations Of Motion

Hamiltonian mechanics emerged in 1833 as a reformulation of Lagrangian mechanics. Introduced by Sir William Rowan Hamilton, Hamiltonian mechanics replaces (generalized) velocities $\dot q^i$ used in Lagrangian mechanics with (generalized) ''momenta''. Both theories provide interpretations of classical mechanics and describe the same physical phenomena.

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 $(M, \mathcal L)$ be a mechanical system with the configuration space $M$ and the smooth Lagrangian $\mathcal L.$ Select a standard coordinate system $(\boldsymbol,\boldsymbol)$ on $M.$ The quantities $\textstyle p_i(\boldsymbol,\boldsymbol,t) ~\stackrel~ /$ are called ''momenta''. (Also ''generalized momenta'', ''conjugate momenta'', and ''canonical momenta''). For a time instant $t,$ the Legendre transformation of $\mathcal$ is defined as the map $(\boldsymbol, \boldsymbol) \to \left(\boldsymbol,\boldsymbol\right)$ which is assumed to have a smooth inverse $(\boldsymbol,\boldsymbol) \to (\boldsymbol,\boldsymbol).$ For a system with $n$ degrees of freedom, the Lagrangian mechanics defines the ''energy function''
$E_(\boldsymbol,\boldsymbol,t)\, \stackrel\, \sum^n_ \dot q^i \frac - \mathcal L.$

The inverse of the Legendre transform of $\mathcal$ turns $E_$ into a function $\mathcal H(\boldsymbol,\boldsymbol,t)$ known as the . The Hamiltonian satisfies
$\mathcal H\left(\frac,\boldsymbol,t\right) = E_(\boldsymbol,\boldsymbol,t)$
which implies that
$\mathcal H(\boldsymbol,\boldsymbol,t) = \sum^n_ p_i\dot q^i - \mathcal L(\boldsymbol,\boldsymbol,t),$
where the velocities $\boldsymbol = (\dot q^1,\ldots, \dot q^n)$ are found from the ($n$-dimensional) equation $\textstyle \boldsymbol = /$ which, by assumption, is uniquely solvable for $\boldsymbol.$ The ($2n$-dimensional) pair $(\boldsymbol,\boldsymbol)$ is called ''phase space coordinates''. (Also ''canonical coordinates'').

From Euler-Lagrange equation to Hamilton's equations

In phase space coordinates $(\boldsymbol,\boldsymbol),$ the ($n$-dimensional) Euler-Lagrange equation
$\frac - \frac\frac = 0$
becomes ''Hamilton's equations'' in $2n$ dimensions

From stationary action principle to Hamilton's equations

Let $\mathcal P(a,b,\boldsymbol x_a,\boldsymbol x_b)$ be the set of smooth paths \boldsymbol q: ,b\to M for which $\boldsymbol q(a) = \boldsymbol x_a$ and $\boldsymbol q(b) = \boldsymbol x_.$ The action functional $\mathcal S : \mathcal P(a,b,\boldsymbol x_a,\boldsymbol x_b) \to \Reals$ is defined via
\mathcal S boldsymbol q= \int_a^b \mathcal L(t,\boldsymbol q(t),\dot(t))\, dt = \int_a^b \left(\sum^n_ p_i\dot q^i - \mathcal H(\boldsymbol,\boldsymbol,t) \right)\, dt,
where $\boldsymbol = \boldsymbol(t),$ and $\boldsymbol = \partial \mathcal L/\partial \boldsymbol$ (see above). A path $\boldsymbol q \in \mathcal P(a,b,\boldsymbol x_a,\boldsymbol x_b)$ is a stationary point of $\mathcal S$ (and hence is an equation of motion) if and only if the path $(\boldsymbol(t),\boldsymbol(t))$ in phase space coordinates obeys the Hamilton's equations.

Basic physical interpretation

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

In this example, the time derivative of 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 equals the ''Newtonian force'', and so the second Hamilton equation means that the force equals the negative gradient of potential energy.

Example

A spherical pendulum consists of a mass ''m'' moving without friction on the surface of a sphere. The only forces acting on the mass are the reaction from the sphere and gravity. Spherical coordinates are used to describe the position of the mass in terms of (''r'', ''θ'', ''φ''), where is fixed, .
The Lagrangian for this system is

$L = \frac ml^2\left( \dot^2+\sin^2\theta\ \dot^2 \right) + mgl\cos\theta.$

Thus the Hamiltonian is
$H = P_\theta\dot \theta + P_\phi\dot \phi - L$
where
$P_\theta = \frac = ml^2\dot \theta$
and
$P_\phi=\frac = ml^2\sin^2 \!\theta \, \dot \phi .$
In terms of coordinates and momenta, the Hamiltonian reads
$H = \underbrace_+\underbrace_ = \frac + \frac - mgl\cos\theta$
Hamilton's equations give the time evolution of coordinates and conjugate momenta in four first-order differential equations,
$\begin \dot &=\\ \dot &=\\ \dot &=\cos\theta-mgl\sin\theta \\ \dot &=0. \end$
Momentum $P_\phi$, which corresponds to the vertical component of angular momentum $L_z = l\sin\theta \times ml\sin\theta\,\dot\phi$, is a constant of motion. That is a consequence of the rotational symmetry of the system around the vertical axis. Being absent from the Hamiltonian, azimuth $\phi$ is a cyclic coordinate, which implies conservation of its conjugate momentum.

Deriving Hamilton's equations

Hamilton's equations can be derived by a calculation with the Lagrangian $\mathcal L$, generalized positions , and generalized velocities , where $i = 1,\ldots,n$. Here we work off-shell, meaning $q^i, \dot^i, t$ are independent coordinates in phase space, not constrained to follow any equations of motion (in particular, $\dot^i$ is not a derivative of $q^i$). The total differential of the Lagrangian is: $\mathrm \mathcal = \sum_i \left ( \frac \mathrm q^i + \frac \mathrm \dot^i \right ) + \frac \mathrmt \ .$ The generalized momentum coordinates were defined as $p_i = \partial \mathcal/\partial \dot^i$, so we may rewrite the equation as:

$\begin \mathrm \mathcal &=& \displaystyle\sum_i \left( \frac \mathrm q^i + p_i \mathrm \dot^i \right) + \frac\mathrmt \\ &=& \displaystyle \sum_i \left( \frac \mathrmq^i + \mathrm( p_i \dot^i) - \dot^i \mathrm p_i \right) + \frac\mathrmt\,. \end$

After rearranging, one obtains:

$\mathrm\! \left ( \sum_i p_i \dot^i - \mathcal \right ) = \sum_i \left( - \frac \mathrm q^i + \dot^i \mathrmp_i \right) - \frac\mathrmt\ .$

The term in parentheses on the left-hand side is just the Hamiltonian $\mathcal H = \sum p_i \dot^i - \mathcal L$ defined previously, therefore:

$\mathrm \mathcal = \sum_i \left( - \frac \mathrm q^i + \dot^i \mathrm p_i \right) - \frac\mathrmt\ .$

One may also calculate the total differential of the Hamiltonian $\mathcal H$ with respect to coordinates $q^i, p_i, t$ instead of $q^i, \dot^i, t$, yielding:

$\mathrm \mathcal =\sum_i \left( \frac \mathrm q^i + \frac \mathrm p_i \right) + \frac\mathrmt\ .$

One may now equate these two expressions for $d\mathcal H$, one in terms of $\mathcal L$, the other in terms of $\mathcal H$:

$\sum_i \left( - \frac \mathrm q^i + \dot^i \mathrm p_i \right) - \frac\mathrmt \ =\ \sum_i \left( \frac \mathrm q^i + \frac \mathrm p_i \right) + \frac\mathrmt\ .$

Since these calculations are off-shell, one can equate the respective coefficients of $\mathrmq^i, \mathrmp_i, \mathrmt$ on the two sides:

$\frac = - \frac \quad, \quad \frac = \dot^i \quad, \quad \frac = - \ .$

On-shell, one substitutes parametric functions $q^i=q^i(t)$ which define a trajectory in phase space with velocities $\dot q^i = \tfracq^i(t)$, obeying Lagrange's equations:

$\frac \frac - \frac = 0\ .$

Rearranging and writing in terms of the on-shell $p_i = p_i(t)$ gives:

$\frac = \dot_i\ .$

Thus Lagrange's equations are equivalent to Hamilton's equations:

$\frac =- \dot_i \quad , \quad \frac = \dot^i \quad , \quad \frac = - \frac\, .$

In the case of time-independent $\mathcal H$ and $\mathcal L$, i.e. $\partial\mathcal H/\partial t = -\partial\mathcal L/\partial t = 0$, Hamilton's equations consist of first-order differential equations, while Lagrange's equations consist of second-order equations. Hamilton's equations usually do not reduce the difficulty of finding explicit solutions, but important theoretical results can be derived from them, because coordinates and momenta are independent variables with nearly symmetric roles.

Hamilton's equations have another advantage over Lagrange's equations: if a system has a symmetry, so that some coordinate $q_i$ does not occur in the Hamiltonian (i.e. a ''cyclic coordinate''), the corresponding momentum coordinate $p_i$ is conserved along each trajectory, and that coordinate can be reduced to a constant in the other equations of the set. This effectively reduces the problem from coordinates to coordinates: this is the basis of symplectic reduction in geometry. In the Lagrangian framework, the conservation of momentum also follows immediately, however all the generalized velocities $\dot q_i$ still occur in the Lagrangian, and a system of equations in coordinates still has to be solved.

The Lagrangian and Hamiltonian approaches provide the groundwork for deeper results in classical mechanics, and suggest analogous formulations in quantum mechanics: the path integral formulation and the Schrödinger equation.

Properties of the Hamiltonian

*The value of the Hamiltonian $\mathcal H$ is the total energy of the system if and only if the energy function $E_ \mathcal L$ has the same property. (See definition of $\mathcal H).$
*$\frac = \frac$ when $\mathbf p(t), \mathbf q(t)$ form a solution of Hamilton's equations. Indeed, $\frac = \frac\cdot \dot\boldsymbol + \frac\cdot \dot\boldsymbol + \frac,$ and everything but the final term cancels out.
*$\mathcal H$ does not change under ''point transformations'', i.e. smooth changes $\boldsymbol \leftrightarrow \boldsymbol$ of space coordinates. (Follows from the invariance of the energy function $E_$ under point transformations. The invariance of $E_$ can be established directly).
*$\frac = -\frac.$ (See Deriving Hamilton's equations).
*$-\frac = \dot p_i = \frac.$ (Compare Hamilton's and Euler-Lagrange equations or see Deriving Hamilton's equations).
*$\frac = 0$ if and only if $\frac=0.$A coordinate for which the last equation holds is called ''cyclic'' (or ''ignorable''). Every cyclic coordinate $q^i$ reduces the number of degrees of freedom by $1,$ causes the corresponding momentum $p_i$ to be conserved, and makes Hamilton's equations easier to solve.

Hamiltonian of a charged particle in an electromagnetic field

A sufficient illustration of Hamiltonian mechanics is given by the Hamiltonian of a charged particle in an electromagnetic field. In Cartesian coordinates the Lagrangian of a non-relativistic classical particle in an electromagnetic field is (in SI Units):
$\mathcal = \sum_i \tfrac m \dot_i^2 + \sum_i q \dot_i A_i - q \varphi$
where is the electric charge of the particle, is the electric scalar potential, and the are the components of the magnetic vector potential that may all explicitly depend on $x_i$ and $t$.

This Lagrangian, combined with Euler–Lagrange equation, produces the Lorentz force law
$m \ddot = q \mathbf + q \dot \times \mathbf \, ,$
and is called minimal coupling.

Note that the values of scalar potential and vector potential would change during a gauge transformation, and the Lagrangian itself will pick up extra terms as well; But the extra terms in Lagrangian add up to a total time derivative of a scalar function, and therefore won't change the Euler–Lagrange equation.

The canonical momenta

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