Subsections

• Relaxation Time
• Time Correlation Function
• Multi-Variate Case
  • Kinetic Equations
  • Symmetry properties

Relaxation & Time Correlation Function

Relaxation Time

Suppose we have a thermodynamic variable x with $\left\langle x\right\rangle=0$. What happens if at some moment $x=x_0\ne0$?

Thermodynamics Language:

We have a restricted ensemble: a set of systems with the given x.

Example:

I prepared a set of beakers at the temperature T₀ and put it in the thermostat with the temperature T.

Kinetics and Equilibrium:

A beaker at the temperature T is not in the equilibrium with the environment--technically speaking, we cannot use thermodynamics!

A Trick:

we say that it is partially equilibrated--everything is equilibrated for this temperature.

General situation:

we have a slowly changing variable x, and say, that every other degree of freedom is fast.

Then we have a function $\left\langle x(t)\right\rangle$, such as  

$$\left\langle x(0)\right\rangle=x_0$$ (1)

At $\left\langle x\right\rangle=0$ we have $d\left\langle x\right\rangle/dt=0$ (why?). For small deviations $d\left\langle x\right\rangle/dt$ is small. We can expand it in series in $\left\langle x\right\rangle$, and  

$$\frac{d\left\langle x\right\rangle}{dt} = -\lambda\left\langle x\right\rangle,\quad \lambda\gt$$ (2)

Solution:  

$$\left\langle x(t)\right\rangle=x_0e^{-\lambda t}, \quad t\gt$$ (3)

Small fluctuations decay exponentially with the relaxation time $\tau=1/\lambda$.

This formula is valid only for t>0: we can prepare an ensemble with given x₀ and observe it, but cannot go backward! This is another manifestation of time irreversibility.

Time Correlation Function

For space-dependent problems we defined the correlation function $\left\langle x(\mathbf{0})x(\mathbf{r})\right\rangle$. For time-dependent case we define time correlation function $\varphi(t)=\left\langle x(0)x(t)\right\rangle$. We want to calculate $\varphi(t)$. This is the averaging over the ensemble of systems with arbitrary x(0). Note this difference with the previous case: there we had x(0) fixed, and used restricted ensembles. Now we return to the general case and consider equilibrium ensemble.

Translation along the time axis:

$$\left\langle x(t_1)x(t_2)\right\rangle=\left\langle x(0)x(t_2-t_1)\right\rangle=\varphi(t_2-t_1)$$

and therefore  

$$\varphi(-t)=\varphi(t)$$ (4)

Double averaging method:

1.

Divide the ensemble into subensembles with given x(0)=x_i. Each subensemble has many microscopic states with given macroscopic value x(0).

2.

Average x(0)x(t) in a subensemble:

$$\Phi(x_i,t) = \left\langle x_ix(t)\right\rangle_{\text{sub}}$$

3.

Average $\Phi(x_i,t)$ over all subensembles:

$$\varphi(t) = \left\langle \Phi(x_i,t)\right\rangle$$

Since in a subensemble x(0) is given,

$$\Phi(x_0,t) = \left\langle x_0x(t)\right\rangle_{\text{sub}... ...left\langle x(t)\right\rangle_{\text{sub}}=x_0^2e^{-\lambda t}$$

Second averaging:

$$\left\langle x_0^2e^{-\lambda t}\right\rangle = \left\langle... ...right\rangle e^{-\lambda t} = \frac{kT}{\gamma}e^{-\lambda t}$$

with generalized susceptibility $\gamma$.

We obtained for the correlation function:

$$\varphi(t)=\frac{kT}{\gamma}e^{-\lambda t},\quad t\gt$$

From (4) we obtain for negative t that $\varphi(t)\sim e^{\lambda t}$, or  

$$\varphi(t)=\frac{kT}{\gamma}e^{-\lambda \left\lvert t\right\rvert}$$ (5)

This function has different derivatives at $t\pm0$: time irreversibility!

Multi-Variate Case

Kinetic Equations

We have several variables x_i. Kinetic equations:

$$\frac{d\left\langle x_i\right\rangle}{dt}=-\sum_k \lambda_i \left\langle x_k\right\rangle$$

Now we have multi-variate correlation functions:

$$\varphi_{ij}(t) = \left\langle x_i(0)x_j(t)\right\rangle$$

Symmetry properties

1.

Translation of the time axis:

$$\left\langle x_i(t_1)x_j(t_2)\right\rangle=\left\langle x_i(0)x_j(t_2-t_1)\right\rangle=\varphi_{ij}(t_2-t_1)$$

We see:  

$$\varphi_{ij}(t)=\varphi_{ji}(-t)$$ (6)

2.

Equations of motion are symmetric with respect to time inversion $t\to-t$:  

$$\varphi_{ij}(t)=\varphi_{ij}(-t)$$ (7)

3.

From (6) and (7)  

$$\varphi_{ij}(t)=\varphi_{ji}(t)$$ (8)