What does “an average over noise” mean in Zwanzig's book

Question

This is a very specific question about Robert Zwanzig's book Nonequilibrium Statistical Mechanics.

Specifically, what is he talking about in equation 1.25 on page 10 that he calls "an average over noise", and how is it calculated? $$\left<v\left(t\right)\right>_\text{noise}=e^{-\frac{\xi t}{m}}v\left(0\right)$$

Is he just saying that the noise averages out if you average long enough? If so, how exactly do you show it.

I realize this question is hard to answer without having the book, but I'm not sure how else to phrase it. I don't think he previously defined that term, and I'm not sure how to provide background without copying all the previous pages in the book.

FWIW, here are the two preceding paragraphs:

The first example is to obtain the velocity correlation function of a Brownian particle. In this example, it is instructive to calculate both the equilibrium ensemble average and the long-term average.

Calculating the equilibrium ensemble average involves both an average over noise and an average over the initial velocity. The noise average leads to

$$\left<v\left(t\right)\right>_\text{noise}=e^{-\frac{\xi t}{m}}v\left(0\right).$$

Answer 1

I do have the book, but not in front of me, so I am guessing from the form of equations. A Brownian particle can be represented by the stochastic differential equation $$m\dot{v} = -\xi v +\varepsilon$$ where the last term is the stochastic term, which is assumed to behave like $\langle\varepsilon\rangle = 0$, $\langle\varepsilon(t)\varepsilon(t')\rangle = \Gamma\delta(t-t')$. Then quite simply by taking the expectation value of both sides of the equation one ends up with the ordinary DE: $m\frac{d}{dt}\langle v\rangle = -\xi \langle v \rangle$, the solution of which is precisely what you wrote in your question.

Physically speaking this would mean that given the initial velocity, and a random (probably thermal) process pushing around a Brownian particle, you would expect the velocity to die out as described. Now if you repeat the experiment a thousand times, sometimes it dies out quicker, sometimes slower, the average corresponding to the solution. Now in the actuality experiment you'd probably also have a probability distribution over the initial velocity of the particle, so to truly fit your data with a theory, you'd next need to take that into account as well (here, a simple integral over the solution times the distribution of initial velocities).

Answer 2

Extending the other answers to see how "noise averages" are used elsewhere in the book as well, we can think of "noise average" of a dynamical variable ${\bf A}({\bf x},t)$ as

$ \langle {\bf A}({\bf x},t) \rangle_{noise} = \int_\Omega {\bf A}({\bf x},t) \rho({\pmb \epsilon}) d{\pmb \epsilon}$

where ${\pmb \epsilon}$ is a random noise distributed as $\rho({\pmb \epsilon})$ over $\Omega \in \mathbb{C}^n$. From the SDE, $m \dot{{\pmb v}} = -\xi {\pmb v}(t) + {\pmb \epsilon}(t)$, we get,

$\langle m \dot{{\pmb v}} \rangle_{noise} = -\langle \xi {\pmb v}(t)\rangle_{noise} + \langle {\pmb \epsilon}(t) \rangle_{noise}$.

If $\langle {\pmb \epsilon}(t) \rangle_{noise} = 0$, we obtain the expression $ \langle {\pmb v(t) }\rangle_{noise} = {\pmb v}(0) \exp{(-\xi t/m )}$.

Now in the actuality experiment you'd probably also have a probability distribution over the initial velocity of the particle, so to truly fit your data with a theory, you'd next need to take that into account as well

To get this complete picture, the "noise-averaged" distribution of velocities (or the state ${\bf x}$, which satisfies the Langevin equation, ${\bf \dot{x}} = {\pmb a}({\pmb x}) + {\pmb \epsilon}(t)$) is derived in Chapter 2 of the book. The distribution of states, $f({\pmb x},t)$ satisfies,

$\begin{align*} \dot{f}({\pmb x},t) &= -\frac{\partial}{\partial {\pmb x}} \cdot {\pmb a}({\pmb x}) f({\pmb x},t) - \frac{\partial}{\partial {\pmb x}} \cdot {\pmb \epsilon}({t}) f({\pmb x},0) + \\ &\frac{\partial}{\partial {\pmb x}} \cdot {\pmb \epsilon}({t}) \int_0^t ds \exp{\Bigg(-(t-s)\frac{\partial}{\partial {\pmb x}} \cdot {\pmb a}({\pmb x})\Bigg)} \frac{\partial}{\partial {\pmb x}} \cdot {\pmb \epsilon}({s}) f({\pmb x},s) \end{align*} $

We can use the definiton above to take the "noise average" of this equation. We get,

$\begin{align*} \frac{\partial}{\partial t}\langle {f}({\pmb x},t) \rangle_{noise} &= -\frac{\partial}{\partial {\pmb x}} \cdot {\pmb a}({\pmb x}) \langle f({\pmb x},t) \rangle_{noise} - \frac{\partial}{\partial {\pmb x}} \cdot \underbrace{\langle {\pmb \epsilon}({t}) \rangle_{noise} }_{= 0} f({\pmb x},0) + \\ &\frac{\partial}{\partial {\pmb x}} \cdot \int_0^t ds \exp{\Bigg(-(t-s)\frac{\partial}{\partial {\pmb x}} \cdot {\pmb a}({\pmb x})\Bigg)} \frac{\partial}{\partial {\pmb x}} \cdot \langle {{\pmb \epsilon}({t})\pmb \epsilon}({s}) f({\pmb x},s) \rangle_{noise} \end{align*} $

To get the noise average of the last term of the above equation, the property of delta-correlation of the noise is used, $\langle {\pmb \epsilon}(t) {\pmb \epsilon}(t') \rangle_{noise} = {\pmb \Gamma }{\delta}(t-t')$. In particular,

$\begin{align} \langle {{\pmb \epsilon}({t}) \pmb \epsilon}({s}) f({\pmb x},s) \rangle_{noise} = {\pmb \Gamma} \delta(t-s) \langle f({\pmb x},s) \rangle_{noise} \end{align}$

The above equation comes from an identity of Gaussian random variables : $\begin{align} \langle {\pmb \epsilon}(t_1) G({\pmb \epsilon}(t_2)) \rangle_{noise} = {\pmb \Gamma} \delta(t_1-t_2) \langle \frac{\partial}{\partial {\pmb \epsilon}(t_2)} G \rangle_{noise} \end{align}$ and when we recognize the fact that $f({\pmb x},s)$ depends on $ {\pmb \epsilon}({s'})$ only for $s' < s$. From here, we directly arrive at the Fokker Planck Equation as given in 2.42 of the book.

Answer 3

equally one could use the solution of the Langevin equation:

$$m\dot{v} = -\zeta v +\delta F(t)$$ witch is: $$v(t)=v(0)e^{-\zeta t \over m }+ \int \limits_0^t dt'e^{-\zeta (t-t') \over m } \delta F(t) $$ doing the ensamble average gives:

$$v(t)=\overbrace{\langle 1 \rangle}^{=1} v(0)e^{-\zeta t \over m }+ \int \limits_0^t dt'e^{-\zeta (t-t') \over m } \overbrace{\langle \delta F(t) \rangle}^{=0} $$ by the definition of the noise term as $\langle \delta F(t) \rangle=0$ and $\langle \delta F(t) \delta F(t')\rangle= \delta(t-t')$ giving us:

$$\left<v\left(t\right)\right>_\text{noise}=e^{-\frac{\xi t}{m}}v\left(0\right).$$

Why this is called noise avarage would be interesting to know. I think it is just referring to the fact that in this ensamble average the random variable we are looking at is the noise, making the average we take the ensable average over the noise thus noise average...