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Background

In physics, there is the concept of "mean free path," which is the expected value for the distance a molecule (for example) can travel before it hits another one.

If they're all the same size, and the neighbor molecules are motionless, this can be calculated as $\frac{1}{\pi d^2 n_v}$ (where $d$ is diameter of a molecule, $n_v$ is number of molecules per-volume).

That all makes sense to me.

What I want to determine is: what is the probability of collision, for a given molecule which travels some distance $D$ ?

Initial Stumblings

I would expect the probability graph to look something like this:

enter image description here

That is: as the particle travels further and further, the odds of it hitting a neighbor approach 1 (assuming infinitely large surroundings of of uniform density).

I'm trying to find a formula for this, but am a little stuck.

Also, I'm trying to find the relationship between this and the mean free distance formula, which also seems elusive.

Could you help me?


Update:

I made some more progress: I think I was looking for the Poisson Distribution, and in particular the graph I drew is the CDF for that distribution.

The general formula for Poisson is:

$$ \frac{\lambda^k e^{-\lambda}}{k!} $$

And in my case, $k = 1$ (one collision).

I did find this source, which explicitly states the formula for the answer to my exact question (but does not explain it):

$$ \frac{1}{ l_{free} } exp(\frac{-r}{l_{free}}) $$

i.e.

$$ \frac{1}{ l_{free} } e^{\frac{-r}{l_{free}}} $$

where $r$ is distance traveled and $l_{free}$ is the length of the mean free path.

That looks very close to a Poisson distribution, except for that $r$ term in the exponent.

There is also a formula buried in the Wikipedia page for mean free path:

... the probability that a particle is absorbed between x and x + dx is given by $$ \frac{1}{L}e^{-x/L}{dx} $$

which again is very close to the stated "correct answer," but if I try to plug in "$x=0, dx=D$" to see the probability of absorption after traveling distance D, it is wrong. Presumably it is actually an infintesimal formula (maybe the wikipedia description is not entirely accurate?).

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Let's assume that other particles are motionless and "located independently of each other".1) Under this assumption, the probability of a molecule freely traveling an infinitesimally small distance $\mathrm{d}l$ is $1 - \pi d^2 n_v \mathrm{d}l$, and this event is independent of what will happen in the remaining travelling path of that molecule.

So, if $L$ denotes the distance the molecule travels before hitting another one, then $\mathbb{P}(L > l)$, the probability of traveling a distance of $l > 0$ freely, is given by the following product integral:

$$ \mathbb{P}(L > l) = \prod_{0}^{l}(1 - \pi d^2 n_v \, \mathrm{d}l) = \exp\left(-\int_{0}^{l} \pi d^2 n_v \, \mathrm{d}l\right) = e^{-\pi d^2 n_v l} = e^{-l / l_{\text{MFP}}} $$

where $l_{\text{MFP}} = \frac{1}{\pi d^2 n_v}$ stands for the mean free path. Hence $L$ is distributed according to the exponential distribution with mean $l_{\text{MFP}}$:

$$ \mathbb{E}[L] = \int_{0}^{\infty} \mathbb{P}(L > l) \, \mathrm{d}l = l_{\text{MFP}} $$

Then the probability density function $f_L(\cdot)$ of $L$ is given by

$$ f_L(l) = \frac{\mathbb{P}(\text{$L$ is between $l$ and $l+\mathrm{d}l$})}{\mathrm{d}l} = \frac{\mathrm{d}}{\mathrm{d}l} \mathbb{P}(L \leq l) = \frac{1}{l_{\text{MFP}}} e^{-l/l_{\text{MFP}}} $$

for $l > 0$. This coincides with what is explained in OP's link. Also, the the probability of collision,

$$ \mathbb{P}(L \leq l) = 1 - e^{-l/l_{\text{MFP}}}, $$

as a function of $l / l_{\text{MFP}}$ then looks like:

Graph of collision probability


1) Mathematically, this is the same as assuming that other particles are located according to the Poisson point process in $\mathbb{R}^3$ with intensity $n_v$.

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  • $\begingroup$ Thanks for this. The various subtly different formulae for the Beer-Lambert derivation on Wikipedia ($\frac{1}{l}e^{-x/l}{dx}$), Poisson distribution ($\lambda e^{-\lambda}$) and others were confusing to navigate. $\endgroup$
    – jwd
    Commented Mar 28 at 18:43
  • $\begingroup$ @jwd, Glad it helped. Exponential distribution and Poisson distribution are clearly related, perhaps too closely that it looks confusing for those who see them at the first time. Explaining them in the above context of (highly simplified) molecule collision model, exponential distributions are for "inter-collision times" and Poisson distributions are for "number of collisions in a given distance". $\endgroup$ Commented Mar 28 at 18:47
  • $\begingroup$ One thing I'm not quite grasping: what is the difference between the first formula ("the probability of traveling a distance of $l>0$ freely") and the probability density function, that comes later. How does the leading $\frac{1}{l_{MFP}}$ term appear, in that final formula? Also, doesn't the first formula answer the original question already? Why the subsequent work? $\endgroup$
    – jwd
    Commented Mar 28 at 20:43
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    $\begingroup$ @jwd, The probability density function (PDF) $f_L(l)$ is literally the "density of probability" (per unit length, in this context). Saying differently, $f_L(l)$ is the quantity satisfying the following infinitesimal relation $$ \mathbb{P}(\text{collision occurs between $l$ and $l+\mathrm{d}l$})\approx f_L(l)\,\mathrm{d}l.$$ Then, the prefactor $\frac{1}{l_{\text{MFP}}}$ in $f_L(l)$ is picked up by differentiating the exponential. $\endgroup$ Commented Mar 28 at 21:23
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    $\begingroup$ @jwd, Also, you are correct that the first formula contains all the necessarily information. I just wanted to show how other, related quantities are derived from the first one, because I was worried that other sources only mentions PDF. (By the way, the first formula corresponds to what is called "complementary cumulative distribution function", or abbreviated as CCDF.) $\endgroup$ Commented Mar 28 at 21:25

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