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A new derivation of the relationship between diffusion coefficient and entropy in classical Brownian motion by the ensemble method
by Yi Liao and XiaoBo Gong
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Submission summary
As Contributors:  Yi Liao 
Preprint link:  scipost_202010_00022v1 
Date submitted:  20201023 03:10 
Submitted by:  Liao, Yi 
Submitted to:  SciPost Physics 
Academic field:  Physics 
Specialties: 

Approach:  Theoretical 
Abstract
The diffusion coefficienta measure of dissipation, and the entropya measure of fluctuation are found to be intimately correlated in many physical systems. Unlike the fluctuation dissipation theorem in linear response theory, the correlation is often strongly nonlinear. To understand this complex dependence, we consider the classical Brownian diffusion in this work. Under certain rational assumption, i.e. in the bicomponent fluid mixture, the mass of the Brownian particle $M$ is far greater than that of the bath molecule $m$, we can adopt the weakly couple limit. Only considering the firstorder approximation of the mass ratio $m/M$, we obtain a linear motion equation in the reference frame of the observer as a Brownian particle. Based on this equivalent equation, we get the Hamiltonian at equilibrium. Finally, using canonical ensemble method, we define a new entropy that is similar to the KolmogorovSinai entropy. Further, we present an analytic expression of the relationship between the diffusion coefficient $D$ and the entropy $S$ in the thermal equilibrium, that is to say, $D =\frac{\hbar}{eM} \exp{[S/(k_Bd)]}$, where $d$ is the dimension of the space, $k_B$ the Boltzmann constant, $\hbar $ the reduced Planck constant and $e$ the Euler number. This kind of scaling relation has been wellknown and welltested since the similar one for single component is firstly derived by Rosenfeld with the expansion of volume ratio.
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Anonymous Report 2 on 2021216 (Invited Report)
 Cite as: Anonymous, Report on arXiv:scipost_202010_00022v1, delivered 20210216, doi: 10.21468/SciPost.Report.2565
Report
The authors report a derivation of a relation between
entropy and diffusion coefficient of Brownian pointlike
particles using Canonical ensemble.
This is a well studied problem, as the authors have said.
Similar scaling relations are mentioned in Eqs. 16 & 17.
But they have proposed a new way in deriving such scaling,
which could be interesting.
Before recommending, I expect the authors to address the
following points:
Requested changes
1. What is, if any, the advantage of this method over the
other methods? Is there any experimental comparison where
this method seem to work better?
2. The only comparison made so far is in the case of hard
sphere model (Eq. 18), where it is argued that for massive
Brownian particles (compared to that of the bath molecules),
the expressions of entropies are comparable. Is there a
quantitative measure of it possible (say, with typical
parameter values)?
3. Finally, the organization of the paper is somewhat confusing.
Earlier attempts (Eqs. 16 & 17) should not come in the
results section but should be moved to the introduction.
Anonymous Report 1 on 2020128 (Invited Report)
 Cite as: Anonymous, Report on arXiv:scipost_202010_00022v1, delivered 20201208, doi: 10.21468/SciPost.Report.2268
Report
The 1977 work of Rosenfeld exerted a strong influence on the literature of computer simulation of liquids, which is however accessible only the researchers with expertise in this specific field.
The issue of entropy and transport is of more general interest.
More recently, in the late 90’s there has been some attention devoted to explore the role of KomogorovSinai entropy in the dynamics of liquid. This corresponds to a transition from the mere technical level to a more important level addressing the fundamental principle of statistical mechanics.
The authors focus on the microscopic derivation of the Langevin equation with the method of Mazur and Oppenheim, which is a sort projection method. They derive the Langevin equation and address the important issue of establishing the correct Hamiltonian structure corresponding to this Langevin equation. To do that they define a reference system evaluated by neglecting the stochastic force and establishes the time distance between two Brownian particles that has the same initial position but different velocities. This is central point of their work that is based on the work of Ref. [22]. They define another dynamical system leading to the same solution and sharing a common phase curve. It is not clear to me what is the physical meaning of this statement. On the basis of this crucial technique they define the Hamiltonian of Eq. (11) and adopting the canonical partition function they establish a connection with thermodynamics, the resulting entropy being interpreted by them as a sort of KolmgorovSinai entropy. This is not unreasonable since Dzugutov and Vulpiani, not quoted in the paper, made the assumption that the KolmogorovSinai entropy can be connected to the conventional thermodynamic entropy.
I would be very happy to recommend this paper for publication, if the authors will be able to explain their derivation of Hamiltonian (11).
Thanks for your kind considerations and referee’s detailed comments and suggestions to improve this work.
We thank the referee's approvement. All the revised parts are written in bold.
In the revised paper, we have explained our derivation of Hamiltonian (11) in APPENDIX A.
We introduce the KolmogorovSinai entropy defined as
\begin{eqnarray}
S&=&\underset{Q}{\sup} h(Q)\equiv \underset{Q}{\sup}\{\underset{n\rightarrow \infty}{\lim}\frac{1}{n\tau}\underset{\omega}{\sum}\mu(\omega)\ln \mu(\omega) \},
\\\nonumber
\omega&=&\{\textbf{X}(t)=(t_i,\textbf{X}_i),t_i=i\tau,i=0,1,2\cdots,n1\}.
\end{eqnarray}
Here, $\omega$ denotes a path of the particle, and $\mu(\omega)$ is the probability.
A Brownian motion particle can be described by the Langevin equation which reads
\begin{equation}
\frac{d\textbf{p}}{d t}+\gamma \textbf{p}=\boldsymbol{\zeta}_0(t), \boldsymbol{\zeta}_{0}(t)=e^{L_{0}t}\boldsymbol{\zeta}(0).
\end{equation}
In the 1st and 2nd ensemble, the path of the Brownian particle is $\textbf{X}_{00}(t)$ and $\textbf{X}_{11}(t)$, respectively. In the two ensembles, the Brownian particles have different initial velocities being $\textbf{v}_{00}$ and $\textbf{v}_{11}$, but the bath molecules have the same initial velocity distributions. Due to $\boldsymbol{\zeta}_{0}(t)=e^{L_{0}t}\boldsymbol{\zeta}(0)$, the forces $\boldsymbol{\zeta}_{0}(t)$ are the same. One can get
\begin{equation}
\underset{n\rightarrow \infty}{\lim}[\textbf{X}_{11}(t)\textbf{X}_{00}(t)]=\underset{t\rightarrow \infty}{\lim}[\textbf{X}_{11}(t)\textbf{X}_{00}(t)]=\frac{\textbf{v}_{11}\textbf{v}_{00}}{\gamma}.
\end{equation}
Assuming that the systems are in thermal equilibrium, one can get the probability ratio of two paths which reads
\begin{equation}
\frac{\mu(\omega_2)}{\mu(\omega_1)}\propto \exp[\frac{M(\textbf{v}_{11}\textbf{v}_{00})^2}{2k_B T}]=\underset{t\rightarrow \infty}{\lim}\exp\{\frac{M\gamma^2[\textbf{X}_{11}(t)\textbf{X}_{00}(t)]^2}{2k_B T}\}.
\end{equation}
So, when $t\rightarrow \infty$, the probability of all possible paths satisfies
\begin{equation}
\mu\propto \exp\{\frac{M\textbf{v}^2_{00}}{2k_B T}\frac{M\gamma^2[\textbf{X}_{11}(t)\textbf{X}_{00}(t)]^2}{2k_B T}\}
\end{equation}
Therefore, based on the definition of KolmogorovSinai entropy, one can obtain the final Hamiltonian of the ensemble system which reads
\begin{equation}
H_{\rm total}(t=\infty)=\underset{i}{\overset{n}{\sum}}[\frac{1}{2M}\textbf{p}_i^{2}+\frac{M}{2}\gamma^{2}(\textbf{x}_i\textbf{x}^{o})^{2}+\phi(0)].
\end{equation}
Author: Yi Liao on 20210313
(in reply to Report 2 on 20210216)Thanks for your kind considerations and referee’s detailed comments and suggestions to improve this work.
We thank the referee's approvement. All the revised parts are written in bold.
Point by point: Reply to the referee comment
Answer: We thank the referee's approvement. In the revised paper, we have performed more details to address the three points based on the referee's suggestions.
Answer: The derivation of Eq.(13) based on the KolmogorovSinai entropy would be showed in APPENDIX A. in APPENDIX B, the formula of the thermodynamic entropy of Brownian particle is derived, but it is hard to analytically solve. Fortunately, Dzugutov et al. have point that KolmogorovSinai entropy, when expressed in terms of the atomic collision frequency, is uniquely related to the thermodynamic excess entropy by a universal linear scaling law (Dzugutov M., Aurell E., and Vulpiani A. 1998, Phys. Rev. Lett. 81, 1762). The linear law is not influence the exponential relationship between the diffusion coefficient and the entropy. Our method can give the analytic formula of KolmogorovSinai entropy and make it possible to calculate some more complex model. The KolmogorovSinai entropy is regarded as a measure, for the loss of information about the state of the system, per unit of time. This quantity is more mathematical than physical, so there are not any experimental comparison.
Answer: One can assume that a system labelled as System $1$ with the volume $V$, only includes $N$ bath particles,which Hamiltonian reads,
\[ H= \frac{\textbf{p}^{N}\cdot\textbf{p}^{N}}{2m}+U(\textbf{r}^{N}). \]The partition function of this system under canonical ensemble is
\[ Z_{1}=\frac{1}{N!h^{dN}}\int e^{\beta H}d\textbf{p}_{1}... d\textbf{p}_{N}d\textbf{r}_{1}...d\textbf{r}_{N}.\\ \]When one introduces a heavier Brownian particle to join in the system, it is labelled as System $2$, which partition function is
\[ Z_{2}=\frac{1}{N!h^{dN}h^{d}}\int e^{\beta H_{s}}d\textbf{p}_{1}... d\textbf{p}_{N}d\textbf{r}_{1}...d\textbf{r}_{N}d\textbf{p}d\textbf{x} \]one can define the entropy of Brownian particle which equals the difference of entropy of System 2 and System 1. One can obtain $\Delta \ln Z\equiv \ln Z_{2}\ln Z_{1}$, based on the formula of the thermodynamic entropy, the thermodynamic entropy of Brownian particle $S_{T}$ reads
\[ S_{T}=\frac{kd}{2}[\ln(\frac{2\pi M}{h^{2}\beta})+1] k\ln[\frac{<e^{\beta\Phi}>}{V}]+k\beta \frac{\partial}{\partial\beta }\ln(<e^{\beta\Phi}>). \]Because
\[ <e^{A}>=<1+A+\frac{1}{2}A^{2}+\frac{1}{6}A^{3}+\ldots> = e^{\langle A\rangle + \frac{1}{2}(\langle A^2 \rangle  \langle A \rangle^2) + O(A^3)} \]\[ \frac{\partial}{\partial\beta }<\Phi>=<\Phi><H_{0}><\Phi H_{0}> =<\Phi><U>+<\Phi><\Phi><\Phi>^{2}<\Phi U>\\ =<\Phi><\Phi><\Phi>^{2}\\ \]\[ <\phi(\textbf{x}\textbf{r}_{i}) U(\textbf{r}_{i}\textbf{r}_{j})> =<\phi><U>(first ~integral ~with~\textbf{r}_{j}). \]So, there is a quantitative measure of it possible but still hard if one can know the values of $<\Phi>$ and $<U>$.
Answer: We thank the referee for the very comprehensive suggestions. These suggestions contribute to improving the paper. We have turn the Eqs. 16 & 17 into Eqs. 1 & 2 in the Introduction in the revised manuscript.
Thanks and best regards
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