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In "The Fluctuation Theorem" by Evans and Searles, they derive the transient fluctuation theorem from Liouville's theorem (pg 1541). Following their notation $ \Gamma = (\vec{q}, \vec{p}) $, they use Liouville's theorem and the continuity equation to conclude that

\begin{align} \frac{df}{dt} &= \frac{\partial f}{\partial t} + \frac{d f}{d \Gamma} \cdot \dot{\Gamma} \\ 0 &= \frac{\partial f}{\partial t} + \frac{d}{d \Gamma} \cdot (f \dot{\Gamma}) \\ \therefore \frac{df}{dt} &= -f \frac{d}{d \Gamma} \cdot \dot{\Gamma} \, . \end{align}

They then define $ \Lambda(\Gamma) = \frac{d}{d \Gamma} \cdot \dot{\Gamma} $ and derive the fluctuation theorem defining the dissipation function as

\begin{align} \Omega_t t &= \int_0^t \Omega(\Gamma(s))~ds \equiv \ln \left[{{\frac {{f(\Gamma(0) ,0)}}{{f(\Gamma (t),0)}}}}\right] - \int_0^t \Lambda(\Gamma(s))~ds \, . \end{align}

Wikipedia defines, \begin{align} \Omega _{t}(\Gamma )=\int _{0}^{t}{ds\;\Omega (\Gamma ;s)}\equiv \ln \left[{{\frac {{f(\Gamma(0) ,0)}}{{f(\Gamma (t),0)}}}}\right]+{\frac {{\Delta Q(\Gamma ;t)}}{kT}} \, . \end{align}

What magic permits this jump? Why is Wikipedia missing the factor of $t$ in front?

One thing I noticed is that for non-conservative systems

\begin{align} \dot{q}_i = \frac{\partial \mathcal{H}}{\partial p_i} + F_q ~&;~ \dot{p}_i = -\frac{\partial \mathcal{H}}{\partial q_i} + F_p \end{align}

\begin{align} \Lambda(\Gamma(s)) &= \frac{d}{d \Gamma} \cdot \dot{\Gamma} = \sum_i \left( \frac{\partial^2 \mathcal{H}}{\partial q_i \partial p_i} - \frac{\partial^2 \mathcal{H}}{\partial p_i \partial q_i} + \left(\frac{\partial F_q}{\partial q_i} + \frac{\partial F_p}{\partial p_i}\right)\right) = \sum_i \left(\frac{\partial F_q}{\partial q_i} + \frac{\partial F_p}{\partial p_i}\right) \\ {\frac {{\Delta Q(\Gamma ;t)}}{kT}} &= - \int_0^t \Lambda(\Gamma(s))~ds = -\sum_i \int_0^t \left(\frac{\partial F_q}{\partial q_i} + \frac{\partial F_p}{\partial p_i}\right) \, . \end{align}

Thus, for purely conservative systems $ F_q = F_p = 0 $, \begin{align} \frac{Pr(\Omega_t = A)}{Pr(\Omega_t = -A)} &= 1 \, . \end{align}

So entropy generation comes from forcing or dissipation terms.

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  • $\begingroup$ The compressibility of phase space (i.e., $df/dt \neq 0$) itself implies irreversibility, which is often interpreted as entropy production (though I should note that entropy and irreversibility may not be synonymous). $\endgroup$ – honeste_vivere Jul 29 '17 at 18:28
  • $\begingroup$ @honeste_vivere Thanks! How would you compute entropy in this case? The normal Boltzmann/Shannon entropy $ -\int f \log f $ doesn't seem like it could ever be related to $ \int \Lambda(\Gamma(s)) ds $ because $ \Lambda = \frac{d}{d\Gamma} \cdot \dot{\Gamma} $ doesn't have any factors of $ f $ $\endgroup$ – aidan.plenert.macdonald Jul 30 '17 at 2:56
  • $\begingroup$ I ask because if you need to compute it with some kind of summation $ \sum_i g(q_i, p_i) $ over the particles (ie. particle entropy, not over similarly prepared ensembles distributed like $ f(\Gamma) $), then I could see myself possibly working out a connection. $\endgroup$ – aidan.plenert.macdonald Jul 30 '17 at 3:07
  • $\begingroup$ You can use something similar to perturbation theory and/or mean field theory where you construct/assume $f \approx \langle f \rangle + \delta f$ and then compute the normal $f \ \log{f}$ to first order. Technically, by imposing/assuming the fluctuation dissipation theorem, you have "inserted" irreversibility into the equations. It's one of the consequences of the random phase approximation. $\endgroup$ – honeste_vivere Jul 30 '17 at 18:37
  • $\begingroup$ How does this help if there is no $f$ in $\Gamma$? Or does $f$ somehow cancel out to get the $\Gamma$ only expression? $\endgroup$ – aidan.plenert.macdonald Jul 30 '17 at 18:50
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Boltzmann, Vlasov, and Entropy

The Vlasov equation is the collisionless form of the Boltzmann equation. The Vlasov equation can be written as: $$ \frac{ \partial f_{s} }{ \partial t } + \mathbf{v} \cdot \nabla f_{s} + \frac{ \mathbf{F} }{ m_{s} } \cdot \nabla_{\mathbf{v}} f_{s} = 0 \tag{1} $$ where $f_{s} = f_{s}\left( \mathbf{x}, \mathbf{v}, t \right)$ is the particle velocity distribution function of species $s$ (e.g., Maxwellian), $\mathbf{F}$ is an external force, and the kth component of $\nabla_{\mathbf{v}}$ is given by $\hat{\mathbf{k}} \ \partial / \partial v_{k}$. Let us modify Equation 1 by using the Lorentz force for $\mathbf{F} \rightarrow \mathbf{F}_{em}$ and linearizing it, i.e., assume $Q \rightarrow \langle Q \rangle + \delta Q$, where $\langle \ \rangle$ is an ensemble average and $\langle \delta Q \rangle = 0$. From this, we can construct an average and fluctuating form of Equation 1. I will drop the subscript $s$ out of laziness for the rest of the derivation.

The average form is given by: $$ \frac{ \partial \langle f \rangle }{ \partial t } + \mathbf{v} \cdot \nabla \langle f \rangle + \frac{ \langle \mathbf{F}_{em} \rangle }{m} \cdot \nabla_{\mathbf{v}} \langle f \rangle = \mathcal{C} \tag{2a} $$ where $\mathcal{C}$ is a proxy for a collision term given by: $$ \mathcal{C} = - \langle \frac{ \delta \ \mathbf{F}_{em} }{ m } \cdot \nabla_{\mathbf{v}} \delta f \rangle \tag{2b} $$ The fluctuating form is given by: $$ \frac{ \partial }{ \partial t } \delta f + \mathbf{v} \cdot \nabla \delta f + \langle \mathbf{F}_{em} \rangle \cdot \nabla_{\mathbf{v}} \delta f + \delta \mathbf{F}_{em} \cdot \nabla_{\mathbf{v}} \langle f \rangle = - \delta \mathbf{F}_{em} \cdot \nabla_{\mathbf{v}} \delta f - \mathcal{C} \tag{3} $$

There is an important distinction between the collision term, $\mathcal{C}$, and the classical binary collision term found in Boltzmann's equation. $\mathcal{C}$ does not conserve the local momentum and energy density of the particles. More importantly, $\mathcal{C}$ does not cause the plasma to relax to a Maxwellian [e.g., see Tidman and Krall, 1971].

Gibbs' entropy can be written as: $$ S = - k_{B} \ \int_{all \ d\mathbf{x} \ d\mathbf{v}} d\mathbf{x} \ d\mathbf{v} \ \langle f \rangle \ \ln{\lvert \langle f \rangle \rvert} \tag{4} $$ Gibbs realized that Equation 4 can diverge to negative infinity if one does not use the one-particle probability distribution, which is an average over ensemble states (thus the $\langle \ \rangle$'s). An interesting observation is that Liouville's equation also predicts that Equation 4 diverges to negative infinity without the use of ensemble averages [e.g., Evans and Morriss, 1990]. The reason for the divergence is that perturbations go to smaller and smaller scales in phase space, which require higher and higher dimensional forms of $f\left( \mathbf{x}, \mathbf{v}, t \right)$. One of the consequences (possibly indirect?) of this intuitive leap was the development of things like mean field theory.

Liouville's Equation

The first part of the following is adapted from my answer at https://physics.stackexchange.com/a/177972/59023.

We know that $\langle f \rangle$ satisfies Liouville's equation, or more appropriately, $\partial \langle f \rangle$/$\partial t = 0$. In general, the equation of motion states: $$ \frac{ \partial f }{ \partial t } = f \left[ \left( \frac{ \partial }{ \partial \mathbf{q} } \frac{ d\mathbf{q} }{ dt } \right) + \left( \frac{ \partial }{ \partial \mathbf{p} } \frac{ d\mathbf{p} }{ dt } \right) \right] + \left[ \frac{ d\mathbf{q} }{ dt } \cdot \frac{ \partial f }{ \partial \mathbf{q} } + \frac{ d\mathbf{p} }{ dt } \cdot \frac{ \partial f }{ \partial \mathbf{p} } \right] \tag{5} $$ where I have defined the canonical phase space of $(\mathbf{q}, \mathbf{p})$. If I simplify the terms $dQ/dt$ to $\dot{Q}$ and let $\boldsymbol{\Gamma} = (\mathbf{q}, \mathbf{p})$, then I find: $$ \begin{align} \frac{ \partial f }{ \partial t } & = - f \frac{ \partial }{ \partial \boldsymbol{\Gamma} } \cdot \dot{\boldsymbol{\Gamma}} - \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial f }{ \partial \boldsymbol{\Gamma} } \tag{6a} \\ & = - \frac{ \partial }{ \partial \boldsymbol{\Gamma} } \cdot \left( \dot{\boldsymbol{\Gamma}} \ f \right) \tag{6b} \end{align} $$ where one can see that the last form looks like the continuity equation. If I define the total time derivative as: $$ \frac{ d }{ dt } = \frac{ \partial }{ \partial t } + \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial }{ \partial \boldsymbol{\Gamma} } \tag{7} $$ then I can show that the time rate of change of the distribution function is given by: $$ \begin{align} \frac{ d f }{ dt } & = \frac{ \partial f }{ \partial t } + \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial f }{ \partial \boldsymbol{\Gamma} } \tag{8a} \\ & = - \left[ f \ \frac{ \partial }{ \partial \boldsymbol{\Gamma} } \cdot \dot{\boldsymbol{\Gamma}} + \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial f }{ \partial \boldsymbol{\Gamma} } \right] + \dot{\boldsymbol{\Gamma}} \cdot \frac{ \partial f }{ \partial \boldsymbol{\Gamma} } \tag{8b} \\ & = - f \ \frac{ \partial }{ \partial \boldsymbol{\Gamma} } \cdot \dot{\boldsymbol{\Gamma}} \tag{8c} \\ & \equiv - f \ \Lambda\left( \boldsymbol{\Gamma} \right) \tag{8d} \end{align} $$ where $\Lambda \left( \boldsymbol{\Gamma} \right)$ is called the phase space compression factor [e.g., Evans and Morriss, 1990]. Note that Equations 8a through 8d are different forms of Liouville's equation, which have been obtained without reference to the equations of motion and they do not require the existence of a Hamiltonian. I can rewrite Equation 8d in the following form: $$ \frac{ d }{ dt } \ln \lvert f \rvert = - \Lambda\left( \boldsymbol{\Gamma} \right) \tag{9} $$

If the equations of motion can be generated from a Hamiltonian, then $\Lambda \left( \boldsymbol{\Gamma} \right) = 0$, even in the presence of external fields that act to drive the system away from equilibrium. Note that the existence of a Hamiltonian is a sufficient, but not necessary condition for $\Lambda \left( \boldsymbol{\Gamma} \right) = 0$.

Entropy Production

Recall that we defined $\partial \langle Q \rangle / \partial t = 0$ and we know from Equation 9 that $d/dt \ln{ \langle f \rangle } = 0$ for conservative systems (i.e., those with incompressible phase space). If we define $\mathcal{F} \equiv \langle f \rangle \ \ln{\lvert \langle f \rangle \rvert}$, then we can show that: $$ \begin{align} \frac{ \partial \mathcal{F} }{ \partial t } \equiv \frac{ \partial }{ \partial t } \left( \langle f \rangle \ln \lvert \langle f \rangle \rvert \right) & = \left( 1 + \ln \lvert \langle f \rangle \rvert \right) \frac{ \partial \langle f \rangle }{ \partial t } = 0 \tag{10a} \\ \frac{ \partial \mathcal{F} }{ \partial t } + \nabla \cdot \left( \mathbf{v} \mathcal{F} \right) & = 0 \tag{10b} \\ & = \frac{ d \mathcal{F} }{ d t } + \mathcal{F} \left( \nabla \cdot \mathbf{v} \right) \tag{10c} \\ 0 + \nabla \cdot \left( \mathbf{v} \mathcal{F} \right) & = \langle f \rangle \ln \lvert \langle f \rangle \rvert \left( \nabla \cdot \mathbf{v} \right) + \left( 1 + \ln \lvert \langle f \rangle \rvert \right) \left[ \mathbf{v} \cdot \nabla \langle f \rangle \right] \tag{10d} \end{align} $$ Using these relations, one can show that: $$ \begin{align} \left( \nabla \cdot \mathbf{v} \right) \mathcal{F} & = - \mathbf{v} \cdot \nabla \mathcal{F} \tag{11a} \\ \left( \nabla \cdot \mathbf{v} \right) \left( \langle f \rangle \ln \lvert \langle f \rangle \rvert \right) & = - \mathbf{v} \cdot \nabla \left( \langle f \rangle \ln \lvert \langle f \rangle \rvert \right) \tag{11b} \\ & = - \langle f \rangle \mathbf{v} \cdot \nabla \ln \lvert \langle f \rangle \rvert - \ln \lvert \langle f \rangle \rvert \mathbf{v} \cdot \nabla \langle f \rangle \tag{11c} \\ & = - \mathbf{v} \cdot \nabla \langle f \rangle - \ln \lvert \langle f \rangle \rvert \mathbf{v} \cdot \nabla \langle f \rangle \tag{11d} \\ & = - \left( 1 + \ln \lvert \langle f \rangle \rvert \right) \mathbf{v} \cdot \nabla \langle f \rangle \tag{11e} \end{align} $$

We also know that in the limit of $x \rightarrow \pm \infty$, the term $\langle \mathbf{F}_{em} \rangle \rightarrow 0$ because we assume all gradients asymptotically approach zero in this limit and the $\mathcal{C}$-term goes to zero as well. Therefore, we find that Equation 2a operating on the quantity: $$ \int \ d\mathbf{v} \ \left( 1 + \ln \lvert \langle f \rangle \rvert \right) \tag{12} $$ results in the following: $$ \begin{align} \left[ \frac{ \partial }{ \partial t } + \mathbf{v} \cdot \nabla + \langle \mathbf{F}_{em} \rangle \cdot \nabla_{\mathbf{v}} \right] \int d\mathbf{v} \ \left( 1 + \ln \lvert \langle f \rangle \rvert \right) & = \int d\mathbf{v} \ \mathcal{C} \left( 1 + \ln \lvert \langle f \rangle \rvert \right) \tag{13a} \\ \left[ \mathbf{v} \cdot \nabla \right] \int d\mathbf{v} \ \left( 1 + \ln \lvert \langle f \rangle \rvert \right) & = \int d\mathbf{v} \ \mathcal{C} \left( 1 + \ln \lvert \langle f \rangle \rvert \right) \tag{13b} \\ \nabla \cdot \int d\mathbf{v} \ \mathbf{v} \langle f \rangle \ln \lvert \langle f \rangle \rvert & = \int d\mathbf{v} \ \mathcal{C} \left( 1 + \ln \lvert \langle f \rangle \rvert \right) \tag{13c} \end{align} $$ Now we insert the form for $\mathcal{C}$ from Equation 2b to find: $$ \nabla \cdot \int \ d\mathbf{v} \ \mathbf{v} \ \langle f \rangle \ln \lvert \langle f \rangle \rvert = \frac{e}{m} \int d\mathbf{v} \ \langle \delta \mathbf{F}_{em} \ \delta f \rangle \cdot \nabla_{\mathbf{v}} \ln \lvert \langle f \rangle \rvert \tag{14} $$ where the term on the left-hand side is the divergence of the entropy flux.

Equation 14 is an example of how entropy can be produced in a collisionless medium with incompressible phase space, which arises from the time-dependence in $\langle f \rangle$ introduced by the non-zero fluctuations given by the $\langle \delta \mathbf{F}_{em} \ \delta f \rangle$-term. So although $\langle f \rangle$ is conserved along phase trajectories from the incompressible Liouville equation, it develops finer and finer features corresponding to phase mixing. The introduction of irreversibility (i.e., also entropy here) into this system is largely a result of the approach, which is analogous to the coarse graining procedure used to derive the the Boltzmann equation from the reversible Liouville equation [e.g., Evans and Morriss, 1990; Tidman and Krall, 1971].

A similar approach can be used if we do not assume $d/dt \ln{ \langle f \rangle } = 0$ to derive a form for entropy.

References

  • Evans, D.J., and G. Morriss Statistical Mechanics of Nonequilibrium Liquids, 1st Edition, Academic Press, London, 1990.
  • Tidman, D.A., and N.A. Krall Shock waves in collisionless plasmas, Wiley Series in Plasma Physics, New York: Wiley-Interscience, 1971.
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  • $\begingroup$ Awesome answer! It'll take me a while to get through it in detail. Thanks for the references too $\endgroup$ – aidan.plenert.macdonald Jul 31 '17 at 1:16

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