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SHO Phase Space Volume


Consider an $$N$$ particle system in three dimensions, and focus on only the evolution of $$\mathrm{d}\mathcal{N}$$ particles. Within phase space, these $$\mathrm{d}\mathcal{N}$$ particles occupy an infinitesimal volume given by


 * $$\mathrm{d}\Gamma = \displaystyle\prod_{i=1}^Nd^3p_id^3q_i.$$

We want $$\frac{\mathrm{d}\mathcal{N}}{\mathrm{d}\Gamma}$$ to remain the same throughout time, so that $$\rho(\Gamma, t)$$ is constant along the trajectories of the system. If we allow our particles to evolve by an infinitesimal time step $$\delta t$$, we see that each particle phase space location changes as


 * $$\begin{cases}

q_i' = q_i + \dot{q_i}\delta t\\ p_i' = p_i + \dot{p_i}\delta t, \end{cases}$$

where $$\dot{q_i}$$ and $$\dot{p_i}$$ denote $$\frac{dq_i}{dt}$$ and $$\frac{dp_i}{dt}$$ respectively, and we have only kept terms linear in $$\delta t$$. Extending this to our infinitesimal hypercube $$\mathrm{d}\Gamma$$, the side lengths change as


 * $$\begin{cases}

dq_i' = dq_i + \frac{\partial\dot{q_i}}{\partial q_i}dq_i\delta t\\ dp_i' = dp_i + \frac{\partial\dot{p_i}}{\partial p_i}dp_i\delta t. \end{cases}$$

To find the new infinitesimal phase space volume $$\mathrm{d}\Gamma'$$, need the product of the above quantities. To first order in $$\delta t$$, we get the following.


 * $$dq_i'dp_i' = dq_idp_i\left[1 + \left( \frac{\partial\dot{q_i}}{\partial q_i} + \frac{\partial\dot{p_i}}{\partial p_i}\right) \delta t\right]$$

So far, we have yet to make any specifications about our system. Let us now specialize to the case of $$N$$ $$3$$-dimensional isotropic harmonic oscillators. That is, each particle in our ensemble can be treated as a simple harmonic oscillator. The Hamiltonian for this system is given by


 * $$H = \displaystyle\sum_{i = 1}^{3N}\left(\frac{1}{2m}p_i^2 + \frac{m\omega^2}{2}q_i^2\right)$$

By using Hamilton's equations with the above Hamiltonian we find that the term in parentheses above is identically zero, thus yielding


 * $$dq_i'dp_i' = dq_idp_i.$$

From this we can find the infinitesimal volume of phase space.


 * $$\mathrm{d}\Gamma' = \displaystyle\prod_{i=1}^N d^3q_i'd^3p_i' = \prod_{i=1}^N d^3q_id^3p_i = \mathrm{d}\Gamma$$

Thus we have ultimately found that the infinitesimal phase space volume is unchanged, yielding


 * $$\rho(\Gamma', t + \delta t) = \frac{\mathrm{d}\mathcal{N}}{\mathrm{d}\Gamma'} =  \frac{\mathrm{d}\mathcal{N}}{\mathrm{d}\Gamma} = \rho(\Gamma, t),$$

demonstrating Liouville's Theorem holds for this system.

The question remains of how the phase space volume actually evolves in time. Above we have shown that the total volume is conserved, but said nothing about what it looks like. For a single particle we can see that its trajectory in phase space is given by the ellipse of constant $$H$$. Explicitly, one can solve Hamilton's equations for the system and find


 * $$\begin{align}

q_i(t) &= Q_i\cos{\omega t} + \frac{P_i}{m\omega}\sin{\omega t}\\ p_i(t) &= P_i\cos{\omega t} - m\omega Q_i\sin{\omega t}, \end{align}$$

where $$Q_i$$ and $$P_i$$ denote the initial position and momentum of the $$i^\mathrm{th}$$ particle. For a system of multiple particles, each one will have a phase space trajectory that traces out an ellipse corresponding to the particle's energy. The frequency at which the ellipse is traced is given by the $$\omega$$ in the Hamiltonian, independent of any differences in energy. As a result a region of phase space will simply rotate about the point $$(\mathbf{q}, \mathbf{p}) = (0,0)$$ with frequency dependent on $$\omega$$. This can be seen in the animation above.

Damped Harmonic Oscillator


One of the foundational assumptions of Liouville's Theorem is that the system obeys the conservation of energy. In the context of phase space, this is to say that $$\rho$$ is constant on phase space surfaces of constant energy $$E$$. If we break this requirement by considering a system in which energy is not conserved, we find that $$\rho$$ also fails to be constant.

As an example of this, consider again the system of $$N$$ particles each in a $$3$$-dimensional isotropic harmonic potential, the Hamiltonian for which is given in the previous example. This time, we add the condition that each particle experiences a frictional force. As this is a non-conservative force, we need to extend Hamilton's equations as


 * $$\begin{align}

\dot{q_i} &= \frac{\partial H}{\partial p_i}\\ \dot{p_i} &= -\frac{\partial H}{\partial q_i} - \gamma p_i, \end{align}$$

where $$\gamma$$ is a positive constant dictating the amount of friction. Following a very similar procedure to the undamped harmonic oscillator case, we arrive again at


 * $$dq_i'dp_i' = dq_idp_i\left[1 + \left( \frac{\partial\dot{q_i}}{\partial q_i} + \frac{\partial\dot{p_i}}{\partial p_i}\right) \delta t\right].$$

Plugging in our modified Hamilton's equations, we find


 * $$\begin{align}

dq_i'dp_i' &= dq_idp_i\left[1 + \left( \frac{\partial^2 H}{\partial q_i\partial p_i} - \frac{\partial^2 H}{\partial p_i\partial q_i} - \gamma\right) \delta t\right]\\ &= dq_idp_i\left[1 -\gamma \delta t\right]. \end{align}$$

Calculating our new infinitesimal phase space volume, and keeping only first order in $$\delta t$$ we find the following result.


 * $$\mathrm{d}\Gamma' = \displaystyle\prod_{i=1}^N d^3q_i'd^3p_i' = \left[1-\gamma\delta t\right]^{3N}\prod_{i=1}^N d^3q_id^3p_i = \mathrm{d}\Gamma\left[1-3N\gamma\delta t\right]$$

We have found that the infinitesimal phase space volume is no longer constant, and thus the phase space density is not conserved. As can be seen from the equation as time increases, we expect our phase space volume to decrease to zero as friction affects the system.

As for how the phase space volume evolves in time, we will still have the constant rotation as in the undamped case. However, the damping will introduce a steady decrease in the radii of each ellipse. Again we can solve for the trajectories explicitly using Hamilton's equations, taking care to use the modified ones above. Letting $$\alpha \equiv \frac{\gamma}{2}$$ for convenience, we find


 * $$\begin{align}

q_i(t) &= e^{-\alpha t}\left[Q_i\cos{\omega_1 t} + B_i\sin{\omega_1 t}\right] & &\omega_1 \equiv \sqrt{\omega^2 - \alpha^2}\\ p_i(t) &= e^{-\alpha t}\left[P_i\cos{\omega_1 t} - m(\omega_1 Q_i + 2\alpha B_i)\sin{\omega_1 t}\right] & &B_i \equiv \frac{1}{\omega_1}\left(\frac{P_i}{m} + 2\alpha Q_i\right), \end{align}$$

where the values $$Q_i$$ and $$P_i$$ denote the initial position and momentum of the $$i^\mathrm{th}$$ particle. As the system evolves the total phase space volume will spiral in to the origin. This can be seen in the figure above.

Symplectic Geometry
We can also formulate Liouville's Theorem in terms of symplectic geometry. For a given system, we can consider the phase space $$(q^\mu, p_\mu)$$ of a particular Hamiltonian $$H$$ as a manifold $$(M,\omega)$$ endowed with a symplectic 2-form


 * $$\omega = dp_\mu\wedge dq^\mu.$$

The volume form of our manifold is the top exterior power of the symplectic 2-form, and is just another representation of the measure on the phase space described above.

On our phase space symplectic manifold we can define a Hamiltonian vector field generated by a function $$f(q,p)$$ as


 * $$X_f = \frac{\partial f}{\partial p_\mu}\frac{\partial}{\partial q^\mu} - \frac{\partial f}{\partial q^\mu}\frac{\partial}{\partial p_\mu}.$$

Specifically, when the generating function is the Hamiltonian itself, $$f(q,p) = H$$, we get


 * $$X_H = \frac{\partial H}{\partial p_\mu}\frac{\partial}{\partial q^\mu} - \frac{\partial H}{\partial q^\mu}\frac{\partial}{\partial p_\mu} = \frac{d q^\mu}{d t}\frac{\partial}{\partial q^\mu} + \frac{d p^\mu}{dt}\frac{\partial}{\partial p_\mu} = \frac{d}{dt}$$

where we utilized Hamilton's equations of motion and the definition of the chain rule.

In this formalism, Liouville's Theorem states that the Lie derivative of the volume form is zero along the flow generated by $$X_H$$. That is, for $$(M,\omega)$$ a 2n-dimensional symplectic manifold,


 * $$\mathcal{L}_{X_H}(\omega^n) = 0.$$

In fact, the symplectic structure $$\omega$$ itself is preserved, not only its top exterior power. That is, Liouville's Theorem also gives


 * $$\mathcal{L}_{X_H}(\omega) = 0.$$