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Posts Tagged ‘phase space volume’

An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 3, 2013

That compilation now all of the following too (no further updates will be made to any of these) :

February 28, 2013 Rotation of diatomic molecules

February 28, 2013 Helmholtz free energy

February 26, 2013 Statistical and thermodynamic connection

February 24, 2013 Ideal gas

February 16, 2013 One dimensional well problem from Pathria chapter II

February 15, 2013 1D pendulum problem in phase space

February 14, 2013 Continuing review of thermodynamics

February 13, 2013 Lightning review of thermodynamics

February 11, 2013 Cartesian to spherical change of variables in 3d phase space

February 10, 2013 n SHO particle phase space volume

February 10, 2013 Change of variables in 2d phase space

February 10, 2013 Some problems from Kittel chapter 3

February 07, 2013 Midterm review, thermodynamics

February 06, 2013 Limit of unfair coin distribution, the hard way

February 05, 2013 Ideal gas and SHO phase space volume calculations

February 03, 2013 One dimensional random walk

February 02, 2013 1D SHO phase space

February 02, 2013 Application of the central limit theorem to a product of random vars

January 31, 2013 Liouville’s theorem questions on density and current

January 30, 2013 State counting

PHY452H1S Basic Statistical Mechanics. Lecture 7: Ideal gas and SHO phase space volume calculations. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on February 5, 2013

Disclaimer

Peeter’s lecture notes from class. May not be entirely coherent.

Review. Classical phase space calculation

\begin{aligned}E_{\mathrm{ideal}} = \sum_i \frac{\mathbf{p}_i^2}{2 m}\end{aligned} \hspace{\stretch{1}}(1.2.1)

From this we calculated $\gamma(E)$, and

\begin{aligned}\frac{d\gamma(E)}{dE} = \Omega_{\mathrm{classical}}(E)\end{aligned} \hspace{\stretch{1}}(1.2.2)

Fudging with a requirement that $\Delta x \Delta p \sim h$, we corrected this as

\begin{aligned}\Omega_{\mathrm{quantum}}(E) = \frac{\Omega_{\mathrm{classical}}(E)}{N! h^{3N}}\end{aligned} \hspace{\stretch{1}}(1.2.3)

Now let’s do the quantum calculation.

Quantum calculation

Recall that for the solutions of the Quantum free particle in a box, as in (Fig 1), our solutions are

Fig 1: 1D Quantum free particle in a box

\begin{aligned}\Psi_n(x) = \sqrt{\frac{2}{L}} \sin\left( \frac{ n \pi x}{L} \right),\end{aligned} \hspace{\stretch{1}}(1.2.4)

where $n = 1, 2, \cdots$, and

\begin{aligned}\epsilon_n = \frac{n^2 h^2}{8 m L^2}\end{aligned} \hspace{\stretch{1}}(1.2.5)

.

In three dimensions, with $n_i = 1, 2, \cdots$ we have

\begin{aligned}\Psi_{n_1, n_2, n_3}(x, y, z) = \left( \frac{2}{L} \right)^{3/2} \sin\left( \frac{ n_1 \pi x}{L} \right)\sin\left( \frac{ n_2 \pi x}{L} \right)\sin\left( \frac{ n_3 \pi x}{L} \right)\end{aligned} \hspace{\stretch{1}}(1.2.6)

and

\begin{aligned}\epsilon_{n_1, n_2, n_3} = \frac{h^2}{8 m L^2} \left( n_1^2 + n_2^2 + n_3^2 \right)\end{aligned} \hspace{\stretch{1}}(1.2.7)

\begin{aligned}\gamma^{3d}_{\mathrm{classical}}(E) = \underbrace{V}_{L^3}\int d^3 p \theta \left( E - \frac{\mathbf{p}^2}{2m} \right)= V \frac{4 \pi}{3} (2 m E)^{3/2}\end{aligned} \hspace{\stretch{1}}(1.2.8)

so that

\begin{aligned}\gamma^{3d}_{\mathrm{corrected}}(E) = V \frac{4 \pi}{3} \frac{(2 m E)^{3/2}}{h^3}\end{aligned} \hspace{\stretch{1}}(1.2.9)

\begin{aligned}\gamma^{3d}_{\mathrm{quantum}}(E) = \sum_{n_1, n_2, n_3} \Theta(E - \epsilon_{n_1, n_2, n_3} ).\end{aligned} \hspace{\stretch{1}}(1.2.10)

How do the multiplicities scale by energy? We’ll have expect something like (Fig 2).

Provided the energies $E \gg 3h^2/(8 m L)$ are large enough, we can approximate the sum with

\begin{aligned}\sum_{n_1, n_2, n_3} \sim \int_0^\infty dn_1 dn_2 dn_3\end{aligned} \hspace{\stretch{1}}(1.2.11)

So

\begin{aligned}\gamma^{3d}_{\mathrm{quantum}} \left( E \gg \frac{h^2}{8 m L^2} \right) \approx\int_0^\infty dn_1 dn_2 dn_3 \Theta \left( E - \frac{h^2}{8 m L^2} \left( n_1^2 + n_2^2 + n_3^2 \right) \right)=\frac{1}{{8}}\frac{4 \pi}{3} \left( \frac{8 m L^2 E}{h^2} \right)^{3/2}=L^3\frac{4 \pi}{3} \frac{\left( 2 m E \right)^{3/2}}{h^3}\end{aligned} \hspace{\stretch{1}}(1.2.12)

Harmonic oscillator in 1D.

Our phase space is of the form (Fig 3).

Fig 3: 1D classical SHO phase space

Where the number of states in this classical picture are found with:

\begin{aligned}\gamma^{\mathrm{classical}}(E) = \int dx dp \theta\left( E - \left( \frac{1}{{2}} k x^2 + \frac{1}{{2m }} p^2 \right) \right).\end{aligned} \hspace{\stretch{1}}(1.2.13)

Rescale

\begin{aligned}\tilde{x} = x \sqrt{ \frac{k}{2}}\end{aligned} \hspace{\stretch{1}}(1.0.14a)

\begin{aligned}\tilde{p} = \frac{p}{\sqrt{2m}}\end{aligned} \hspace{\stretch{1}}(1.0.14b)

so that we have

\begin{aligned}\gamma^{\mathrm{classical}}(E) = \int d\tilde{x} d \tilde{p} \sqrt{\frac{2 \times 2 m}{k}} \theta\left( E - \tilde{x}^2 - \tilde{p}^2 \right)=2 \sqrt{\frac{m}{k}} \pi E= 2 \pi \sqrt{\frac{m}{k}} E.\end{aligned} \hspace{\stretch{1}}(1.0.15)

\begin{aligned}\gamma^{\mathrm{SHO}}_{\mathrm{corrected}}(E) = 2 \pi \sqrt{\frac{m}{k}} \frac{E}{h}.\end{aligned} \hspace{\stretch{1}}(1.0.16)

We have for the energy

\begin{aligned}E_n^{\mathrm{SHO}} = \left( n + \frac{1}{{2}} \right) \hbar \omega\end{aligned} \hspace{\stretch{1}}(1.0.17a)

\begin{aligned}\omega = \sqrt{\frac{k}{m}}\end{aligned} \hspace{\stretch{1}}(1.0.17b)

\begin{aligned}\hbar = \frac{h}{2 \pi}\end{aligned} \hspace{\stretch{1}}(1.0.17c)

graphing the counts (Fig 4), we again have stepping as a function of energy, but no multiplicities this time

Fig 4: 1D quantum SHO states per energy level

\begin{aligned}\gamma_{\mathrm{quantum}}(E) = \sum_{n = 0}^\infty \Theta\left( E - \left( n + \frac{1}{{2}} \hbar \omega \right) \right)\end{aligned} \hspace{\stretch{1}}(1.0.18)

we make the continuous approximation for the summation again, and throwing away the zero point energy, we have

\begin{aligned}\gamma_{\mathrm{quantum}}(E \gg \hbar \omega) \approx\int_{0}^\infty dn \Theta\left( E - n \hbar \omega \right)= 2 \pi \frac{E}{h} \sqrt{\frac{m}{k}}\end{aligned} \hspace{\stretch{1}}(1.0.19)

Why $N!$?

We have a problem with out counting here. Consider some particles in a box as in (Fig 5).

Fig 5: Three particles in a box

1. particle $1$ at $\mathbf{x}_1$
2. particle $2$ at $\mathbf{x}_2$
3. particle $3$ at $\mathbf{x}_3$

or

1. particle $1$ at $\mathbf{x}_2$
2. particle $2$ at $\mathbf{x}_3$
3. particle $3$ at $\mathbf{x}_1$

This is fine in the classical picture, but in the quantum picture with an assumption of indistinguishability, no two particles (say electrons) cannot be labelled in this fashion.

\begin{aligned}\underbrace{S_{\mathrm{ideal}}^{(\mathrm{E}, \mathrm{N}, \mathrm{V})}}_{\text{Statistical entropy}}= k_{\mathrm{B}} \ln \left( \frac{\Omega_{\mathrm{classical}}}{h^{3N}} \right)\underbrace{\approx}_{N \gg 1} k_{\mathrm{B}} \left( N \ln V + \frac{3 N}{2} \ln \left( \frac{4 \pi m E }{3 N h^2} \right) + \frac{3 N}{2} \right)\end{aligned} \hspace{\stretch{1}}(1.0.20)

Suppose we double the volume as in (Fig 6), then our total entropy for the bigger system would be

Fig 6: Gibbs volume doubling argument. Two identical systems allowed to mix

\begin{aligned}S_{\mathrm{total}}^{(\mathrm{E}, \mathrm{N}, \mathrm{V})}= k_{\mathrm{B}} \ln \left( \frac{\Omega_{\mathrm{classical}}}{h^{3N}} \right)\approx k_{\mathrm{B}} \left( (2 N) \ln (2 V) + \frac{3 (2 N)}{2} \ln \left( \frac{4 \pi m (2 E) }{2 ( 2 N) h^2} \right) + \frac{3 (2 N)}{2} \right).\end{aligned} \hspace{\stretch{1}}(1.0.21)

We have

\begin{aligned}S_{\mathrm{total}} = S_1 + S_2 + k_{\mathrm{B}} (2 N) \ln 2= S_1 + S_2 + k_{\mathrm{B}} \ln 2^{2 N}.\end{aligned} \hspace{\stretch{1}}(1.0.22)

This is telling us that each particle could be in either the left or the right side, but we know that this uncertainty shouldn’t be in the final answer. We must drop this $k_{\mathrm{B}}$ term.

So, if we assume that these particles are identical, and divide $\Omega$ by $N!$, then we find

\begin{aligned}S_{\mathrm{ideal}} = k_{\mathrm{B}} \left( N \ln \frac{V}{N} + \frac{3 N}{2} \ln \left( \frac{4 \pi m E }{3 N h^2} \right) + \frac{5 N}{2} \right)\end{aligned} \hspace{\stretch{1}}(1.0.23)

PHY452H1S Basic Statistical Mechanics. Lecture 6: Volumes in phase space. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on January 29, 2013

Disclaimer

Peeter’s lecture notes from class. May not be entirely coherent.

Liouville’s theorem

We’ve looked at the continuity equation of phase space density

\begin{aligned}0 = \frac{\partial {\rho}}{\partial {t}} + \sum_{i_\alpha} \left(\frac{\partial {}}{\partial {p_{i_\alpha}}} \left( \rho \dot{p}_{i_\alpha} \right) + \frac{\partial {\left( \rho \dot{x}_{i_\alpha} \right) }}{\partial {x_{i_\alpha}}}\right)\end{aligned} \hspace{\stretch{1}}(1.2.1)

which with

\begin{aligned}\frac{\partial {\dot{p}_{i_\alpha}}}{\partial {p_{i_\alpha}}} + \frac{\partial {\dot{x}_{i_\alpha}}}{\partial {x_{i_\alpha}}} = 0\end{aligned} \hspace{\stretch{1}}(1.2.2)

led us to Liouville’s theorem

\begin{aligned}\\ boxed{\frac{d{{\rho}}}{dt}(x, p, t) = 0}.\end{aligned} \hspace{\stretch{1}}(1.2.3)

We define Ergodic, meaning that with time, as you wait for $t \rightarrow \infty$, all available phase space will be covered. Not all systems are necessarily ergodic, but the hope is that all sufficiently complicated systems will be so.

We hope that

\begin{aligned}\rho(x, p, t \rightarrow \infty) \implies \frac{\partial {\rho}}{\partial {t}} = 0 \qquad \mbox{in steady state}\end{aligned} \hspace{\stretch{1}}(1.2.4)

In particular for $\rho = \text{constant}$, we see that our continuity equation 1.2.1 results in 1.2.2.

For example in a SHO system with a cyclic phase space, as in (Fig 1).

Fig 1: Phase space volume trajectory

\begin{aligned}\left\langle{{A}}\right\rangle = \frac{1}{{\tau}} \int_0^\tau dt A( x_0(t), p_0(t) ),\end{aligned} \hspace{\stretch{1}}(1.2.5)

or equivalently with an ensemble average, imagining that we are averaging over a number of different systems

\begin{aligned}\left\langle{{A}}\right\rangle = \frac{1}{{\tau}} \int dx dp A( x, p ) \underbrace{\rho(x, p)}_{\text{constant}}\end{aligned} \hspace{\stretch{1}}(1.2.6)

If we say that

\begin{aligned}\rho(x, p) = \text{constant} = \frac{1}{{\Omega}},\end{aligned} \hspace{\stretch{1}}(1.2.7)

so that

\begin{aligned}\left\langle{{A}}\right\rangle = \frac{1}{{\Omega}} \int dx dp A( x, p ) \end{aligned} \hspace{\stretch{1}}(1.2.8)

then what is this constant. We fix this by the constraint

\begin{aligned}\int dx dp \rho(x, p) = 1\end{aligned} \hspace{\stretch{1}}(1.2.9)

So, $\Omega$ is the allowed “volume” of phase space, the number of states that the system can take that is consistent with conservation of energy.

What’s the probability for a given configuration. We’ll have to enumerate all the possible configurations. For a coin toss example, we can also ask how many configurations exist where the sum of “coin tosses” are fixed.

A worked example: Ideal gas calculation of $\Omega$

• $N$ gas atoms at phase space points $\mathbf{x}_i, \mathbf{p}_i$
• constrained to volume $V$
• Energy fixed at $E$.

\begin{aligned}\Omega(N, V, E) = \int_V d\mathbf{x}_1 d\mathbf{x}_2 \cdots d\mathbf{x}_N \int d\mathbf{p}_1 d\mathbf{p}_2 \cdots d\mathbf{p}_N \delta \left(E - \frac{\mathbf{p}_1^2}{2m}- \frac{\mathbf{p}_2^2}{2m}\cdots- \frac{\mathbf{p}_N^2}{2m}\right)=\underbrace{V^N}_{\text{Real space volume, not N dimensional volume''}} \int d\mathbf{p}_1 d\mathbf{p}_2 \cdots d\mathbf{p}_N \delta \left(E - \frac{\mathbf{p}_1^2}{2m}- \frac{\mathbf{p}_2^2}{2m}\cdots- \frac{\mathbf{p}_N^2}{2m}\right)\end{aligned} \hspace{\stretch{1}}(1.10)

With $\gamma$ defined implicitly by

\begin{aligned}\frac{d\gamma}{dE} = \Omega\end{aligned} \hspace{\stretch{1}}(1.3.11)

so that with Heavyside theta as in (Fig 2).

\begin{aligned}\Theta(x) = \left\{\begin{array}{l l}1 & \quad x \ge 0 \\ 0 & \quad x < 0\end{array}\right.\end{aligned} \hspace{\stretch{1}}(1.0.12a)

\begin{aligned}\frac{d\Theta}{dx} = \delta(x),\end{aligned} \hspace{\stretch{1}}(1.0.12b)

Fig 2: Heavyside theta

we have

\begin{aligned}\gamma(N, V, E) = V^N \int d\mathbf{p}_1 d\mathbf{p}_2 \cdots d\mathbf{p}_N \Theta \left(E - \sum_i \frac{\mathbf{p}_i^2}{2m}\right)\end{aligned} \hspace{\stretch{1}}(1.0.13)

In three dimensions $(p_x, p_y, p_z)$, the dimension of momentum part of the phase space is 3. In general the dimension of the space is $3N$. Here

\begin{aligned}\int d\mathbf{p}_1 d\mathbf{p}_2 \cdots d\mathbf{p}_N \Theta \left(E - \sum_i \frac{\mathbf{p}_i^2}{2m}\right),\end{aligned} \hspace{\stretch{1}}(1.0.14)

is the volume of a “sphere” in $3N$– dimensions, which we found in the problem set to be

\begin{aligned}V_{m} = \frac{ \pi^{m/2} R^{m} }{ \Gamma\left( m/2 + 1 \right)}.\end{aligned} \hspace{\stretch{1}}(1.0.15a)

\begin{aligned}\Gamma(x) = \int_0^\infty dy e^{-y} y^{x-1}\end{aligned} \hspace{\stretch{1}}(1.0.15b)

\begin{aligned}\Gamma(x + 1) = x \Gamma(x) = x!\end{aligned} \hspace{\stretch{1}}(1.0.15c)

Since we have

\begin{aligned}\mathbf{p}_1^2 + \cdots \mathbf{p}_N^2 \le 2 m E\end{aligned} \hspace{\stretch{1}}(1.0.16)

\begin{aligned}\text{radius} = \sqrt{ 2 m E}.\end{aligned} \hspace{\stretch{1}}(1.0.17)

This gives

\begin{aligned}\gamma(N, V, E) = V^N \frac{ \pi^{3 N/2} ( 2 m E)^{3 N/2}}{\Gamma( 3N/2 + 1) }= V^N \frac{2}{3N} \frac{ \pi^{3 N/2} ( 2 m E)^{3 N/2}}{\Gamma( 3N/2 ) },\end{aligned} \hspace{\stretch{1}}(1.0.17)

and

\begin{aligned}\Omega(N, V, E) = V^N \pi^{3 N/2} ( 2 m E)^{3 N/2 - 1} \frac{2 m}{\Gamma( 3N/2 ) }\end{aligned} \hspace{\stretch{1}}(1.0.19)

This result is almost correct, and we have to correct in 2 ways. We have to fix the counting since we need an assumption that all the particles are indistinguishable.

• Indistinguishability. We must divide by $N!$.
• $\Omega$ is not dimensionless. We need to divide by $h^{3N}$, where $h$ is Plank’s constant.

In the real world we have to consider this as a quantum mechanical system. Imagine a two dimensional phase space. The allowed points are illustrated in (Fig 3).

Fig 3: Phase space volume adjustment for the uncertainty principle

Since $\Delta x \Delta p \sim \hbar$, the question of how many boxes there are, we calculate the total volume, and then divide by the volume of each box. This sort of handwaving wouldn’t be required if we did a proper quantum mechanical treatment.

The corrected result is

\begin{aligned}\boxed{\Omega_{\mathrm{correct}} = \frac{V^N}{N!} \frac{1}{{h^{3N}}} \frac{( 2 \pi m E)^{3 N/2 }}{E} \frac{1}{\Gamma( 3N/2 ) }}\end{aligned} \hspace{\stretch{1}}(1.0.20)

To come

We’ll look at entropy

\begin{aligned}\underbrace{S}_{\text{Entropy}} = \underbrace{k_{\mathrm{B}}}_{\text{Boltzmann's constant}} \ln \underbrace{\Omega_{\mathrm{correct}}}_{\text{phase space volume (number of configurations)}}\end{aligned} \hspace{\stretch{1}}(1.0.21)

PHY452H1S Basic Statistical Mechanics. Lecture 5: Motion in phase space. Liouville and Poincar’e theorems. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on January 22, 2013

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

Disclaimer

Peeter’s lecture notes from class. May not be entirely coherent.

Motion in phase space

Classical system: $\mathbf{x}_i, \mathbf{p}_i$ with dimensionality

\begin{aligned}\underbrace{2}_{x, p}\underbrace{d}_{\text{space dimension}}\underbrace{N}_{\text{Number of particles}}\end{aligned} \hspace{\stretch{1}}(1.2.1)

Hamiltonian $H$ is the “energy function”

\begin{aligned}H = \underbrace{\sum_{i = 1}^N \frac{\mathbf{p}_i^2}{2m} }_{\text{Kinetic energy}}+ \underbrace{\sum_{i = 1}^N V(\mathbf{x}_i) }_{\text{Potential energy}}+ \underbrace{\sum_{i < j}^N \Phi(\mathbf{x}_i - \mathbf{x}_j)}_{\text{Internal energy}}\end{aligned} \hspace{\stretch{1}}(1.2.2)

\begin{aligned}\mathbf{\dot{p}}_i = \mathbf{F} = \text{force}\end{aligned} \hspace{\stretch{1}}(1.0.3a)

\begin{aligned}\mathbf{\dot{x}}_i = \frac{\mathbf{p}_i}{m}\end{aligned} \hspace{\stretch{1}}(1.0.3b)

Expressed in terms of the Hamiltonian this is

\begin{aligned}\dot{p}_{i_\alpha} = - \frac{\partial {H}}{\partial {x_{i_\alpha}}}\end{aligned} \hspace{\stretch{1}}(1.0.4a)

\begin{aligned}\dot{x}_{i_\alpha} = \frac{\partial {H}}{\partial {p_{i_\alpha}}}\end{aligned} \hspace{\stretch{1}}(1.0.4b)

In phase space we can have any number of possible trajectories as illustrated in (Fig1).

Fig1: Disallowed and allowed phase space trajectories

Liouville’s theorem

We are interested in asking the question of how the density of a region in phase space evolves as illustrated in (Fig2)

Fig2: Evolution of a phase space volume

We define a phase space density

\begin{aligned}\rho(p_{i_\alpha}, x_{i_\alpha}, t),\end{aligned} \hspace{\stretch{1}}(1.0.5)

and seek to demonstrate Liouville’s theorem, that the phase space density does not change. To do so, consider the total time derivative of the phase space density

\begin{aligned}\frac{d{{\rho}}}{dt} &= \frac{\partial {\rho}}{\partial {t}} + \sum_{i_\alpha} \frac{\partial {p_{i_\alpha}}}{\partial {t}} \frac{\partial {\rho}}{\partial {p_{i_\alpha}}} + \frac{\partial {x_{i_\alpha}}}{\partial {t}} \frac{\partial {\rho}}{\partial {x_{i_\alpha}}} \\ &= \frac{\partial {\rho}}{\partial {t}} + \sum_{i_\alpha} \frac{\partial {p_{i_\alpha}}}{\partial {t}} \frac{\partial {\rho \dot{p}_{i_\alpha}}}{\partial {p_{i_\alpha}}} + \frac{\partial {x_{i_\alpha}}}{\partial {t}} \frac{\partial {\rho \dot{x}_{i_\alpha}}}{\partial {x_{i_\alpha}}} - \rho \left(\frac{\partial {\dot{p}_{i_\alpha}}}{\partial {p_{i_\alpha}}} +\frac{\partial {\dot{x}_{i_\alpha}}}{\partial {x_{i_\alpha}}} \right) \\ &= \frac{\partial {\rho}}{\partial {t}} + \underbrace{\sum_{i_\alpha} \left(\frac{\partial {p_{i_\alpha}}}{\partial {t}} \frac{\partial {(\rho \dot{p}_{i_\alpha})}}{\partial {p_{i_\alpha}}} + \frac{\partial {x_{i_\alpha}}}{\partial {t}} \frac{\partial {(\rho \dot{x}_{i_\alpha})}}{\partial {x_{i_\alpha}}} \right)}_{\equiv \boldsymbol{\nabla} \cdot \mathbf{j}}- \rho \sum_{i_\alpha} \underbrace{\left(-\frac{\partial^2 H}{\partial p_{i_\alpha} \partial x_{i_\alpha}}+\frac{\partial^2 H}{\partial x_{i_\alpha} \partial p_{i_\alpha}}\right)}_{= 0}\end{aligned} \hspace{\stretch{1}}(1.0.6)

We’ve implicitly defined a current density $\mathbf{j}$ above by comparing to the continuity equation

\begin{aligned}\frac{\partial {\rho}}{\partial {t}} + \boldsymbol{\nabla} \cdot \mathbf{j} = 0\end{aligned} \hspace{\stretch{1}}(1.0.7)

Here we have

\begin{aligned}\rho(\dot{x}_{i_\alpha}, \dot{p}_{i_\alpha}) \rightarrow \mbox{current in phase space}\end{aligned} \hspace{\stretch{1}}(1.0.8)

Usually we have

\begin{aligned}\mathbf{j}_{\mathrm{usual}} \sim \rho \mathbf{v}\end{aligned} \hspace{\stretch{1}}(1.0.9a)

\begin{aligned}\mathbf{j} = -D \boldsymbol{\nabla} \rho\end{aligned} \hspace{\stretch{1}}(1.0.9b)

\begin{aligned}v \frac{\partial {\rho}}{\partial {t}} + \boldsymbol{\nabla} \cdot \mathbf{j} = 0\end{aligned} \hspace{\stretch{1}}(1.0.10)

The implication is that

\begin{aligned}\frac{d{{\rho}}}{dt} = 0.\end{aligned} \hspace{\stretch{1}}(1.0.11)

Flow in phase space is very similar to an “incompressible fluid”.

Time averages, and Poincar\’e recurrence theorem

We want to look at how various observables behave over time

\begin{aligned}\overline{A} = \frac{1}{{T}} \int_0^T dt \rho(x, p, t) A(p, x)\end{aligned} \hspace{\stretch{1}}(1.0.12)

We’d like to understand how such averages behave over long time intervals, looking at $\rho(x, p, t \rightarrow \infty)$.

This long term behaviour is described by the Poincar\’e recurrence theorem. If we wait long enough a point in phase space will come arbitrarily close to its starting point, recurring or “closing the trajectory loops”.

A simple example of a recurrence is an undamped SHO, such as a pendulum. That pendulum bob when it hits the bottom of the cycle will have the same velocity (and position) each time it goes through a complete cycle. If we imagine a much more complicated system, such as $N$ harmonic oscillators, each with different periods, we can imagine that it will take infinitely long for this cycle or recurrence to occur, and the trajectory will end up sweeping through all of phase space.