Relaxation time confusion

Question

I was reading kinetic theory from these notes.

In the very first chapter there is a derivation for relaxation time, which is nothing but average time between collisions, denoted by $\tau$. If $P(t)$ is the probability that a molecule does $\textit{not }$ undergo collision in time interval $[0,t]$, and $w$ is the collision rate (i.e. probability of a collision in a small time $\delta t$ is $w\delta t$), then the derivation goes on to show that \begin{align} P(t)=w~\mathrm{e}^{-wt} \end{align} which is a exponential distribution. Then $\tau$ is found as \begin{align} \tau=\int_0^\infty dt ~t~P(t)=\frac{1}{w} \end{align} Now mathematically this is just another result, but physically it doesn't make sense. Here's my confusion: $P(t)$ is the probability that a collision does not occur within time $t$, so the integral in above expression gives the average time in which a collision does $\textit{not}$ occur. But $\tau$ by definition is the average time between collisions, and the two statements are not logically equivalent. What I mean to say is that, "a collision does not occur within time $t$" is not the same as saying "first collision occurs at time $t$ (or in some small neighborhood of $t$)", and the probability of the latter is what is required to calculate $\tau$.

Any thoughts?

Answer 1

[Related to v1 of the question] The distribution $P(t)$ is not a Poisson- but an exponential distribution, but they are closely related, because the number of collisions during time $t$ is Poisson. The Poisson distribution for $k=1$ events occurring would be $P(k=1) = \frac{(wt)^1}{1!}\, e^{-wt}$.

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Crucially, the distribution at hand describes the waiting time between two events. The parameter $w$ is the expected number of events during some unit time interval, e.g a (nano)second, day, month,...

It follows that its mean is the average time between two events, which by definition is $\tau$, and not surprisingly equal to $w^{-1}$.

The statement

$P(t)$ is the probability that a collision does not occur within time $t$

is not correct. In general PDFs do not represent probabilities directly! You can appreciate this by noting that a PDF may very well take values larger than one. The correct statement is

$P(t)dt$ is the probability of $t$ being in the interval $[t,t+dt]$; in your case: The probability that the time between two collision is in the interval $[t,t+dt]$ is $P(t)dt$.

What you need is the cumulative distribution function:

$$ P(t\leq x) = \int_0^x P(t)dt = 1 - e^{-wx}$$ which tells you the probability of an event occurring in $[0,x]$. Then $$1 - P(t\leq x) = e^{-wx} $$ is the probability of no event occurring during $[0,x]$.

Answer 2

I think it's a matter of choosing the normalization constant, in going from: $$ \frac{d P}{d t}=-w P,$$ to $$P = w e^{-wt},$$ because as you could see, with this choice $P(0)=w$, but we expect $P(0)$ to be $1$. So if you just set $P$ to be $e^{-wt}$, then the mean collision time can be obtained as: $$ \tau = \int_{0}^{\infty}t\, P(t)\, wdt,$$ where the $w \,dt$ term coming after $P(t)$ is resposible for a collision between $t$ and $t+\,dt.$