Extended Phase Space¶

Introduction¶

The usual formalism for classical mechanics treats time as a special variable, making it difficult to effect coordinate transforms that mix space and time such as the Lorentz transform. The idea of extended phase space is replace time with an arbitrary parameter, then to introduce time as an additional generalized coordinate. Here is the idea motivated by [Johns:1989] and [Struckmeier:2005].

[Struckmeier:2005]: https://doi.org/10.1088/0305-4470/38/6/006 'Jürgen Struckmeier, "Hamiltonian dynamics on the symplectic extended phase space for autonomous and non-autonomous systems", J. Phys. A 38(6), 1257 (2005)' [Johns:1989]: https://doi.org/10.1119/1.16086 'Oliver Davis Johns, "Canonical transformations with time as a coordinate", Am. J. Phys. 57(3), 204-215 (1989)'

Lagrangian Formulation¶

We start by expressing the variational principle for a single coordinate $q(t)$: the physical solution will extremize the action:

$$ S[q] = \int_{t_0}^{t_1}\d{t}\; L[q(t), \dot{q}(t), t]. $$

We maintain this principle, but introduce an additional parametrization of time $t(s)$ in terms of an (arbitrary) parameter $s$, with the new coordinates $q(s) = q\bigl(t(s)\bigr)$, $t(s)$, and their derivatives $q'(s) = \dot{q}t'(s)$ and $t'(s)$. Here we use the shorthand of a dot for derivatives with respect to time, $\dot{q} = \d{q}(t)/\d{t}$, and of primes for derivatives with respect to $s$, $q'(s) = \d{q}/\d{s} = \dot{q}t'(s)$.

$$ S[q] = \int_{s_0}^{s_1}\d{s}\; L_e[q(s), q'(s), t(s), t'(s), s] = \int_{s_0}^{s_1}\d{s}\; L[q(t), \dot{q}(t), t]\diff{t(s)}{s},\\ L_e[q(s), q'(s), t(s), t'(s), s] = L\left[q(t), \frac{q'(s)}{t'(s)}, t(s)\right]t'(s). $$

This extended Lagrangian is referred to as the "symmetric Lagrangian" in [Johns:1989]. The equations of motion from the new variational principle are:

$$ p_q = \pdiff{L_e}{q'} = \pdiff{L}{\dot{q}} = p, \qquad p_t = \pdiff{L_e}{t'} = L - \dot{q}\pdiff{L}{\dot{q}} = -H,\\ p'_{q} = \pdiff{L_e}{q} = \pdiff{L}{q}t', \qquad p'_{t} = \pdiff{L_e}{t} = \pdiff{L}{t}t'. $$

From these, we see that the original equations of motion are satisfied. The new momentum $p_q = p$ is the same as the old momentum, and the equation of motion is the same as before: $\dot{p} = L_{,q} = p'/t'$. The new equation of motion is simply an identify: $p'_t = -H' = -\dot{H}t'$ following from the original time-dependence of the Hamiltonian. Thus, the parametrization $t(s)$ remains arbitrary – the form is not fixed by the equations of motion.

[Johns:1989]: https://doi.org/10.1119/1.16086 'Oliver Davis Johns, "Canonical transformations with time as a coordinate", Am. J. Phys. 57(3), 204-215 (1989)'

Exercise: Remind yourself that that $\dot{H} = -\partial{L}/\partial{t}$ by explicit computation using the equations of motion.

Solution:

$$ \diff{H}{t} = \diff{p\dot{q} - L}{t} = \diff{p\dot{q}}{t} - \dot{L} = \overbrace{\diff{p\dot{q}}{t} \underbrace{- \overbrace{\pdiff{L}{q}}^{\diff{}{t}\pdiff{L}{\dot{q}}}\dot{q} - \pdiff{L}{\dot{q}}\ddot{q}}_{- \diff{}{t}\Biggl(\overbrace{\pdiff{L}{\dot{q}}}^{p}\dot{q}\Biggr)}}^{0} - \pdiff{L}{t} = - \pdiff{L}{t}. $$

Hamiltonian Formulation¶

Unfortunately, as pointed out in [Johns:1989], one cannot present a similar "symmetric Hamiltonian" formulation. Naïvely trying to form the extended Hamiltonian from a Legendre transform fails as it yields $H_e = 0$:

$$ H_e = p_qq' + p_tt' - L_e = (p\dot{q} - L)t' - Ht' = 0. $$

In principle, this is not a problem (consider i.e. the Hamilton-Jacobi equations), but in this case, $p_q' = q' = 0$ is not generally a solution. What fails is that the coordinate transformation is singular:

$$ \begin{vmatrix} \pdiff{p_q}{q'} & \pdiff{p_q}{t'}\\ \pdiff{p_t}{q'} & \pdiff{p_t}{t'} \end{vmatrix} = \begin{vmatrix} \pdiff[2]{L_e}{q'} & \pdiff{L_e}{t'\partial q'}\\ \pdiff{L_e}{t'\partial q'} & \pdiff[2]{L_e}{t'} \end{vmatrix} = 0. $$

Hence the equations $p_x(x', t')$ and $p_t(x', t')$ cannot be inverted to obtain $x'(p_x, p_t)$ and $t'(p_x, p_t)$ as required to form the symmetric Hamiltonian.

[Johns:1989]: https://doi.org/10.1119/1.16086 'Oliver Davis Johns, "Canonical transformations with time as a coordinate", Am. J. Phys. 57(3), 204-215 (1989)'

Exercise: Prove that the determinant above is zero.

Solution:

To deal with this, we must revert back to the non-symmetric Hamiltonian formulation with the equations of motion:

$$ p_q' = -\pdiff{H}{q}t', \qquad q' = \pdiff{H}{p_q}t',\\ p_t' = -\pdiff{H}{t}t' = \diff{-H}{t}t' = \diff{-H}{s} = \diff{p_t}{s}, \qquad t' = -\pdiff{H}{p_t}t' = \pdiff{(-H)}{(-H)}t' = t'. $$

The first line are just the Hamilton's original equations with $s$ as the dependent variable, while the second are simply a tautologies indicating that the function $t(s)$ is arbitrary (not constrained by the equations of motion). As discussed in [Johns:1989], there are many different non-symmetric Hamiltonian formulations. Obtaining the equations of motion from these can be challenging due to issues of invertibility (that do not appear for this simplest formulation), but must be considered when more complex transformations are consided.

[Johns:1989]: https://doi.org/10.1119/1.16086 'Oliver Davis Johns, "Canonical transformations with time as a coordinate", Am. J. Phys. 57(3), 204-215 (1989)'

Canonical Transformations¶

Despite the complications presented above, one can still employ extended canonical transformations. For example, transformations of type 2 can be implemented by specifying a function $G_2(q, P_q, t, P_t, s)$ with:

$$ p_q = \pdiff{G_2}{q}, \qquad p_t = \pdiff{G_2}{t}, \\ Q = \pdiff{G_2}{P_q}, \qquad T = \pdiff{G_2}{P_t}. $$

Example: Lorentz Transformations¶

For example, the Lorentz transformation can be implemented via:

$$ G_2(q, P_q, t, P_t) = P_q(\overbrace{q\cosh\eta - ct\sinh\eta}^{Q}) + \frac{P_t}{c}(\overbrace{ct\cosh\eta - x\sinh\eta}^{cT}) $$

Lagrangian Formulation¶

Consider the following Lagrangian representing the 1D motion of a relativistic particle with (rest) mass $m$ acted on by a constant force $F$:

$$ L[x, \dot{x}, t] = -mc^2\sqrt{1-\frac{\dot{x}^2}{c^2}} + Fx. $$

This gives rise to the extended Lagrangian (using $\dot{x} = x'/t'$ etc.):

$$ L_e[x, x', t, t', s] = \left(-mc^2\sqrt{1-\frac{(x')^2}{(ct')^2}} + Fx\right)t' = -mc\sqrt{(ct')^2-(x')^2} + Fxt'. $$

Now we can implement a Lorentz boost to a frame moving with velocity $v$:

$$ \begin{pmatrix} ct_v\\ x_v \end{pmatrix} = \begin{pmatrix} \cosh\eta & -\sinh\eta\\ -\sinh\eta & \cosh\eta \end{pmatrix} \cdot \begin{pmatrix} ct\\ x \end{pmatrix},\qquad \begin{pmatrix} ct'_v\\ x'_v \end{pmatrix} = \begin{pmatrix} \cosh\eta & -\sinh\eta\\ -\sinh\eta & \cosh\eta \end{pmatrix} \cdot \begin{pmatrix} ct'\\ x' \end{pmatrix}. $$

By construction, this leaves the quadratic form $(ct_v')^2 - (x_v')^2 = (ct')^2 - (x')^2$ invariant, so the extended Lagrangian becomes:

$$ L_e[x_v, x_v', t_v, t_v', s] = -mc\sqrt{(ct_v')^2-(x_v')^2} + \frac{F}{c}(ct_v\sinh\eta + x_v\cosh\eta)(ct_v'\cosh\eta + x_v'\sinh\eta). $$

Manipulation of the last term yields:

$$ L_e[x_v, x_v', t_v, t_v', s] = -mc\sqrt{(ct_v')^2-(x_v')^2} + Fx_vt_v' + \frac{F}{c}\diff{}{s}\left(\frac{c^2t_v^2 + x_v^2}{2}\cosh\eta\sinh\eta + cx_vt_v\sinh^2\eta\right) \equiv -mc\sqrt{(ct_v')^2-(x_v')^2} + Fx_vt_v'. $$

The last term is a total derivative, hence it does not affect the equations of motion. Thus, the original equations are Lorentz covariant.

Hamiltonian Formulation¶

In the Hamiltonian formulation we start with:

$$ p = \diff{L}{\dot{x}} = \overbrace{\frac{1}{\sqrt{1-\frac{\dot{x}^2}{c^2}}}}^{\gamma}m\dot{x} = \gamma m\dot{x} $$

where the time dilation factor is

$$ \gamma = \frac{1}{\sqrt{1-\frac{\dot{x}^2}{c^2}}} = \sqrt{1+\frac{p^2}{m^2c^2}}. $$

Effecting the Legendre transform, we have:

$$ H[q, p, t] = p\dot{x} - L = mc^2 \sqrt{1+\frac{p^2}{m^2c^2}} - Fx. $$

With the extended Lagrangian, we have:

$$ p_x = p, \qquad p_t = -H. $$

We now implement the Lorentz boost to transform to coordinates $x_v$ and $t_v$ with momenta $p_v$ and $p_{t_v} = -H_v$:

$$ G_2(x, p_v, t, -H_v) = p_v(\overbrace{x\cosh\eta - ct\sinh\eta}^{x_v}) - \frac{H_v}{c}(\overbrace{ct\cosh\eta - x\sinh\eta}^{ct_v}),\\ \overbrace{p = \pdiff{G_2}{x}}^{p_q = \pdiff{G_2}{q}} = p_v\cosh\eta + \frac{H_v}{c}\sinh\eta, \qquad \overbrace{x_v = \pdiff{G_2}{p_v}}^{Q = \pdiff{G_2}{P_q}} = x\cosh\eta - ct\sinh\eta,\\ \overbrace{ct_v = c\pdiff{G_2}{(\underbrace{-H_v}_{p_{t_v}})}}^{T = \pdiff{G_2}{P_t}} = ct\cosh\eta - x\sinh\eta, \qquad \overbrace{\underbrace{-H}_{p_t} = \pdiff{G_2}{t}}^{p_t = \pdiff{G_2}{t}} = - cp_v\sinh\eta - H_v\cosh\eta, \qquad $$

In summary, inverting to find $x$ and $p$ we have:

$$ x = x_v\cosh\eta + ct_v\sinh\eta,\qquad ct = ct_v\cosh\eta + x_v\sinh\eta, \qquad p = p_v\cosh\eta + \frac{H_v}{c}\sinh\eta,\\ H_v = \frac{H - cp_v\sinh\eta}{\cosh\eta} = \gamma H - \beta \gamma cp_v. $$

For this problem, we have:

$$ (c H_v + F(cx_v+sc_0t_v) + sc_0p_v)^2 = c_0^2[m^2c_0^2 + (cp_v + sH_v/c_0)^2] $$$$ H_v = \frac{\overbrace{\gamma m c^2 - Fx}^{H} - \beta \gamma c p_v}{\gamma} = m c^2 - Fx_v - c\beta(Ft_v - p_v) $$

References¶

• [Struckmeier:2005]: "Hamiltonian dynamics on the symplectic extended phase space for autonomous and non-autonomous systems".
• [Johns:1989]: "Canonical transformations with time as a coordinate".

[Struckmeier:2005]: https://doi.org/10.1088/0305-4470/38/6/006 'Jürgen Struckmeier, "Hamiltonian dynamics on the symplectic extended phase space for autonomous and non-autonomous systems", J. Phys. A 38(6), 1257 (2005)' [Johns:1989]: https://doi.org/10.1119/1.16086 'Oliver Davis Johns, "Canonical transformations with time as a coordinate", Am. J. Phys. 57(3), 204-215 (1989)'