What is the derivation for the exponential energy relation and where does it apply?

Very often when people state a relaxation time $\tau_\text{kin-kin}, \tau_\text{rot-kin}$,, etc. they think of a context where the energy relaxation goes as $\propto\text e^{-t/\tau}$. Related is an approach to compute it via $$\tau=E(0)/\left(\tfrac{\text d E}{\text d t}\right)_{E=0}.$$

Both are justified from considering dynamics for which

$$\frac{\text d E}{\text d t}=-\frac{1}{\tau}(E-E(0)).$$

My question is: What fundamentally leads to this relation?

I conjecture it relates to a Master equation, which mirrors the form "$\dot x=Ax+b$". But I'm not sure how the degrees of freedom in the Master equation translate to the time dependence of the macroscopic energy value. There will also be a derivation from the Boltzmann equation somehow, for some conditions, but what is the general argument and where does it work?

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Re Boltzmann equation: the relaxation time approximation to the Boltzmann collision term is called the BGK approximation which sees wide use. But I'm not sure what exactly are its conditions of validity. Starting from the BGK equation it is a simple matter to get your energy equation with the assumption that external forces vanish (otherwise there is a power input term on the rhs). –  Michael Brown May 15 '13 at 20:36

This kind of exponential decay toward "equilibrium" can be derived when one looks at a Markov process. In this case, if we call $S_t$ the state of the system at time $t$ and $S_{t+1}$ the state at time $t+1$, one has for the evolution:

$S_{t+1} = T S_t$

where $T$ is called the transition matrix. This implies that $S_t = T^t S_0$. The idea is then to introduce the set of eigenstates $E_i$ such that $T E_i = \lambda_i E_i$. The set of $\{E_i\}_{i=1...}$ is a mathematical set of vectors and they need not always corresponds to a probability state. In fact, since the solution is unique for any given $S_0$ it implies that there can only be one probability state such that $T E_p = E_p$ i.e. such that $\lambda_p = 1$. Now, starting from any state $S_0 = \sum_i S_i^0 E_i$, one has then $S_t = \sum_i S_i^0 \lambda_i^t E_i = S_{eq} + \sum_{i \neq p} S_i^0 \lambda_i^t E_i$. $T$ is a positive definite matrix and in spectral theory, one can show that $\lambda_p = 1$ is the heighest eigenvalue, it therefore means all other eigenvalues are smaller than $1$. Let us call $\lambda_2$ the second highest eigenvalue of $T$, we then have:

$S_t-S_{eq} \approx S_2^0 \lambda_2^t \approx S_2^0 e^{t \ln \lambda_2} \sim e^{-t/\tau_2}$

where $\tau_2 = -1/\ln \lambda_2$. $\hspace{0cm}$

At the end of the day the idea is that the initial state can be projected on eigenstates among which only one is physical and happens to have the highest eignevalue of value 1, this one corresponds to the equilibrium state.

The major assumption here is that the dynamics is markovian.

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I like your derivation. Still, where is the justification for $S_{t+1}=TS$ for our physical system? $S'\propto S$ of course sneaks in the exponential behaviour. –  NikolajK May 16 '13 at 11:56
Modeling the system's dynamics by a Markov chain is in most cases an assumption for collective variables. However, you can also most of the time find microvariables that obey a markovian dynamics. Liouville's theorem corresponds to a Markovian dynamics. When you look at slower collective variables, this microscopic Markovian dynamics generates a memory kernel for the slow variable that, over sufficiently long time gives back a markovian behaviour if there is adiabatic separation of time scales. –  gatsu May 16 '13 at 16:06
What is a memory kernel? Is is a dynamics generating function which, compared to the evolution of the Markovian variables, isn't just constant? –  NikolajK May 16 '13 at 19:34
A general definition in discrete time of what memory kernel means is that $S_t = f(S_{t-1}, S_{t-2},..,S_{t-m})$ where $m$ is the size of the memory. –  gatsu May 16 '13 at 20:18

This form of $dE/d\tau$ is valid only when the system is not too far from equilibrium and linear response assumption is valid. The fact that $dE/d\tau$ depends on the difference $E - E(0)$ alone is a consequence of assuming a linear response.

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