# Noncanonical Hamiltonian mechanics¶

## Lagrangian dynamics¶

Let $$\mathbf{x}(t)$$ describe the evolution of a mechanical system. Then is well known that the equations of motions determining the evolution can be derived by finding the minimum of the action integral

$S[\mathbf{x}] = \int_{t_0}^{t_1} L(\mathbf{x},\dot{\mathbf{x}},t) dt$

where $$L(\mathbf{x},\dot{\mathbf{x}},t)$$ is the Lagrangian of the system. Note that the action can be computed for any given path $$\mathbf{x}(t)$$. However, if $$\mathbf{x}(t)$$ is the path that minimizes the action integral then perturbing it to $$\mathbf{x}(t)+\delta\mathbf{x}(t)$$ will not change the action to first order. This condition, combined with the boundary conditions $$\delta\mathbf{x}(t_0)=\delta\mathbf{x}(t_1)=0$$, leads to the Euler-Lagrange equations

(1)$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}^i}\right) = \frac{\partial L}{\partial x^i}$

which, given the initial conditions $$\mathbf{x}(0)$$ and $$\mathbf{\dot{x}}(0)$$, determine the time-evolution of the system.

For example, the Lagrangian for the motion of a particle in a time-dependent electromagnetic field is given by

(2)$L(\mathbf{x},\dot{\mathbf{x}},t) = \frac{m}{2}|\dot{\mathbf{x}}|^2 + \frac{e}{c}\dot{\mathbf{x}}\cdot\mathbf{A}(\mathbf{x},t) - e \Phi(\mathbf{x},t)$

where $$m$$ and $$e$$ are the particle mass and charge respectively and the scalar and vector potentials, $$\Phi$$ and $$\mathbf{A}$$, determine the electromagnetic fields via

$\begin{split}\mathbf{E} &= -\nabla\Phi - \frac{1}{c}\frac{\partial \mathbf{A}}{\partial t} \\ \mathbf{B} &= \nabla\times\mathbf{A}.\end{split}$

As the Lagrangian is a scalar it can be transformed to generalized coordinates $$\mathbf{q}(t)$$ via the transformation $$\mathbf{x}(\mathbf{q},t)$$ and

$\dot{\mathbf{x}} = \frac{\partial \mathbf{x}}{\partial t} + \dot{q}^i \frac{\partial \mathbf{x}}{\partial q^i}.$

Here and below summation over repeated indices is assumed. This transformation changes the Lagrangian to $$L(\mathbf{q},\dot{\mathbf{q}},t)$$, however, does not change the form of the Euler-Lagrange equations, Eq. (1), themselves.

## Hamiltonian dynamics¶

Instead of using a Lagrangian formulation a Hamiltonian formulation can be used. For this the canonical momentum is introduced via

$p_i \equiv \frac{\partial L}{\partial \dot{q}^i}.$

This can be used to eliminate the $$\dot{q}^i$$ in favor of the canonical momentum to introduce the Hamiltonian function via a Legendre transformation as follows

(3)$H(\mathbf{q},\mathbf{p},t) = \mathbf{p}\cdot\dot{\mathbf{q}} - L(\mathbf{q},\dot{\mathbf{q}}(\mathbf{q},\mathbf{p},t),t)$

For example, for the single-particle motion (2) the canonical momentum is

$p_i = m \dot{x}_i + \frac{e}{c}A_i$

and the Hamiltonian becomes

(4)$H(\mathbf{x},\mathbf{p},t) = \frac{1}{2m} \left| \mathbf{p} - \frac{e}{c}\mathbf{A} \right|^2 + e \Phi(\mathbf{x},t).$

To derive the equations of motion in terms of the Hamiltonian consider a variation of (3)

$\begin{split}\delta H &= \frac{\partial H}{\partial q^i}\delta q^i + \frac{\partial H}{\partial p_i}\delta p_i + \frac{\partial H}{\partial t}\delta t \\ &= -\frac{\partial L}{\partial q^i}\delta q^i + \dot{q}^i \delta p_i - \frac{\partial L}{\partial t}\delta t\end{split}$

From which we get Hamilton’s equations for the canonical coordinates $$q^i$$ and $$p_i$$

(5)$\dot{q}^i = \frac{\partial H}{\partial p_i} \quad\mathrm{and}\quad \dot{p}_i = -\frac{\partial H}{\partial q^i}.$

## Phase-space Lagrangian¶

An advantage of the Lagrangian formulation is that one can make arbitrary coordinate transformations without changing the form of the Euler-Lagrange equations. Hence, it would be useful to look for a Lagrangian that yields, as the Euler-Lagrange equations, Hamilton’s equations. This would allow determining Hamilton’s equations under an transform of both $$\mathbf{q}$$ and $$\mathbf{p}$$. The phase-space Lagrangian provides just that. It reads

(6)$\mathcal{L}(\mathbf{q},\mathbf{p},\dot{\mathbf{q}},\dot{\mathbf{p}},t) = \mathbf{p}\cdot\dot{\mathbf{q}} - H(\mathbf{q},\mathbf{p},t).$

It is easy to verify that the Euler-Lagrange equations for this Lagrangian yield Hamilton’s equations, Eq. (5).

For single-particle motion in an electromagnetic field the phase-space Lagrangian is

$\mathcal{L} = \mathbf{p}\cdot\dot{\mathbf{x}} - \frac{1}{2m} \left| \mathbf{p} - \frac{e}{c}\mathbf{A} \right|^2 - e \Phi(\mathbf{x},t).$

As an example of a transformation, consider using the particle velocity instead of the canonical momentum

$\mathbf{v} \equiv \frac{1}{m}\left( \mathbf{p} - \frac{e}{c}\mathbf{A} \right).$

This gives the transformed phase-space Lagrangian

$\mathcal{L} = \left( m\mathbf{v} + \frac{e}{c}\mathbf{A}(\mathbf{x},t) \right)\cdot\dot{\mathbf{x}} - \frac{m}{2}|\mathbf{v}|^2 - e\Phi(\mathbf{x},t).$

As $$\partial \mathcal{L}/\partial \dot{\mathbf{v}} = 0$$ we get the kinematic relation $$\dot{\mathbf{x}}=\mathbf{v}$$. The other Euler-Lagrange equation leads to

$\frac{d}{dt}\left( m\mathbf{v} + \frac{e}{c}\mathbf{A} \right) = \nabla\cdot \left( \frac{e}{c}\mathbf{A} - e\Phi \right).$

Using $$d \mathbf{A}/dt = \partial \mathbf{A}/\partial t + \mathbf{v}\cdot\nabla\mathbf{A}$$ and some vector identities leads to the well known equation of motion

$m\dot{\mathbf{v}} = e\mathbf{E} + \frac{e}{c}\mathbf{v}\times\mathbf{B}.$

## Transformation of the phase-space Lagrangian¶

The phase-space Lagrangian Eq. (6) leads to Hamilton’s equations, Eq. (5), which have a very specific form. Consider a general set of coordinates $$z^\alpha$$, $$\alpha=1,\ldots,2N$$, in terms of which we can write $$q^i(\mathbf{z},t)$$ and $$p_i(\mathbf{z},t)$$. Note that for a N degree of freedom system we need 2N general coordinates $$z^\alpha$$. The question we now ask is: what are the equations of motion in these new coordinates? Note that the transformation to the coordinates $$\mathbf{z}$$ is completely arbitrary: we can not, in general, pick out half of the coordinates as “positions” and other half as “generalized momentum”, and the phase-space may be completely mixed in the new coordinate system.

In these new coordinates we have

$\dot{q}^i = \frac{\partial q^i}{\partial t} + \dot{z}^\alpha \frac{\partial q^i}{\partial z^\alpha}.$

Using this in the phase-space Lagrangian we get

(7)$\mathcal{L}(\mathbf{z},t) = \Lambda_\alpha \dot{z}^\alpha - \mathcal{H}$

where

$\Lambda_\alpha \equiv p_i\frac{\partial q^i}{\partial z^\alpha}$

and

$\mathcal{H} \equiv H - p_i \frac{\partial q^i}{\partial t}.$

The first term Eq. (7) is called the symplectic part and the second term is called the Hamiltonian part.

The Euler-Lagrange equations corresponding to this transformed Lagrangian is

$\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{z}^\alpha}\right) = \frac{\partial \mathcal{L}}{\partial z^\alpha}.$

We have

$\begin{split}\frac{\partial \mathcal{L}}{\partial \dot{z}^\alpha} &= \Lambda_\alpha \\ \frac{\partial \mathcal{L}}{\partial z^\alpha} &= \frac{\partial \Lambda_\beta}{\partial z^\alpha}\dot{z}^\beta - \frac{\partial \mathcal{H}}{\partial z^\alpha}.\end{split}$

From this the Euler-Lagrange equations give

$\frac{d\Lambda_\alpha}{dt} \equiv \frac{\partial \Lambda_\alpha}{\partial t} + \dot{z}^\beta \frac{\partial \Lambda_\alpha}{\partial z^\beta} = \frac{\partial \Lambda_\beta}{\partial z^\alpha}\dot{z}^\beta - \frac{\partial \mathcal{H}}{\partial z^\alpha}.$

This gives the equations of motion

$\omega_{\alpha\beta}\thinspace\dot{z}^\beta = \frac{\partial \mathcal{H}}{\partial z^\alpha} + \frac{\partial \Lambda_\alpha}{\partial t}.$

Here the Lagrange matrix is defined as

$\omega_{\alpha\beta} \equiv \frac{\partial \Lambda_\beta}{\partial z^\alpha} - \frac{\partial \Lambda_\alpha}{\partial z^\beta}.$

Assuming that $$\det(\boldsymbol{\omega})$$ is non-singular the explicit form of the equations of motion can be written as

(8)$\dot{z}^\beta = \Pi^{\beta\alpha}\left(\frac{\partial \mathcal{H}}{\partial z^\alpha} + \frac{\partial \Lambda_\alpha}{\partial t} \right)$

where the Poisson structure $$\mathbf{\Pi} = \boldsymbol{\omega}^{-1}$$. This are the equations of motions we have been seeking.

## Canonical transforms, Symplectic matrices¶

Let the vector $$\mathbf{M} = (q^1,\ldots,q^N, p_1,\ldots,p_N)$$ represent canonical coordinates. Then the equations of motion Eq. (5) can be written in the compact form

$\dot{M}^\alpha = \sigma^{\alpha \beta}\frac{\partial H}{\partial M^\beta}$

where the fundamental symplectic matrix $$\boldsymbol{\sigma}$$ is defined as

$\begin{split}\boldsymbol{\sigma} = \left( \begin{matrix} \mathbf{0} & \mathbf{I} \\ -\mathbf{I} & \mathbf{0} \end{matrix} \right)\end{split}$

where $$\mathbf{I}$$ is a $$2N\times 2N$$ unit matrix. For a time-independent transformation $$M^\alpha(\mathbf{z})$$ Eq. (8) shows that

$\dot{z}^\alpha = \Pi^{\alpha\beta}\frac{\partial H}{\partial z^\beta}.$

Substituting $$M^\alpha(\mathbf{z})$$ in the canonical equations of motion and comparing with the above equation it is clear that if

(9)$\mathbf{D}\thinspace\mathbf{\sigma}\thinspace\mathbf{D}^T = \mathbf{\sigma},$

where $$D^\alpha_\beta = \partial z^\alpha/\partial M^\beta$$ is the Jacobian matrix of the transformation, then the form of Hamilton’s equations remains unchanged. The class of all such transformation that preserve the form of the Hamilton’s equations is called canonical transforms. All matrices $$\mathbf{D}$$ that satisfy (9) are called symplectic matrices. For arbitrary time-independent transforms, however, the relation Eq. (9) will not hold, i.e. the Jacobian matrix of the transformation will not be symplectic.

## Poisson brackets¶

With the matrices $$\Pi^{\alpha\beta}$$ we can define the Poisson bracket of two functions $$f(\mathbf{z},t)$$ and $$g(\mathbf{z},t)$$ as

$\{f,g\} \equiv \frac{\partial f}{\partial z^\alpha} \Pi^{\alpha\beta} \frac{\partial g}{\partial z^\beta}.$

For canonical transforms this reduces to

$\begin{split}\{f,g\} &= \frac{\partial f}{\partial z^\alpha} \sigma^{\alpha\beta} \frac{\partial g}{\partial z^\beta} \\ &= \frac{\partial f}{\partial q^i}\frac{\partial g}{\partial p_i} - \frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q^i}.\end{split}$

Starting from the equations of motion in canonical coordinates $$\mathbf{M}$$ (defined in the previous section) for a time-independent transformation $$M^\alpha(\mathbf{z})$$ we can show that

$\dot{z}^\alpha = \Pi^{\alpha\beta}\frac{\partial H}{\partial z^\beta} = \frac{\partial z^\alpha}{\partial M^\delta} \sigma^{\delta\gamma} \frac{\partial z^\beta}{\partial M^\gamma} \frac{\partial H}{\partial z^\beta} = \{z^\alpha,z^\beta\} \frac{\partial H}{\partial z^\beta}.$

which indicates that

$\Pi^{\alpha\beta} = \{z^\alpha,z^\beta\}$

where the Poisson bracket is defined with respect to the canonical coordinates. Note that once the expression for the Poisson brackets are known the equation of motion can be written in the compact form

$\dot{z}^\alpha = \{z^\alpha,H\}.$

## Liouville theorem¶

Let $$\mathcal{J}=\det(\mathbf{D}^{-1})$$ be the Jacobian of the transformation, where $$D^\alpha_\beta = \partial z^\alpha/\partial M^\beta$$. For a time-dependent transformation the Jacobian satisfies

$\frac{\partial \mathcal{J}}{\partial t} + \frac{\partial} {\partial z^\alpha} \left(\dot{z}^\alpha\mathcal{J}\right) = 0.$

This indicates that the equations of motion satisfy the Liouville theorem, that is, the Hamiltonian flow conserves phase-space volume $$d\mathbf{M} = \mathcal{J}d\mathbf{z}$$.

For time-independent transforms this gives

$0 = \frac{\partial} {\partial z^\alpha}\left( \mathcal{J}\Pi^{\alpha\beta} \frac{\partial H}{\partial z^\beta} \right) = \frac{\partial} {\partial z^\alpha}\left( \mathcal{J}\Pi^{\alpha\beta} \right) \frac{\partial H}{\partial z^\beta}$

where the antisymmetry of the $$\Pi^{\alpha\beta}$$ was used, and which yields the Liouville identities

$\frac{\partial} {\partial z^\alpha}\left( \mathcal{J}\Pi^{\alpha\beta} \right) = 0.$

As can be verified, this allows writing the noncanonical Poisson bracket as a phase-space divergence

$\{f,g\} = \frac{1}{\mathcal{J}} \frac{\partial} {\partial z^\alpha}\left( f \mathcal{J}\Pi^{\alpha\beta} \frac{\partial g}{\partial z^\beta} \right).$