Random walk
Take 10 minutes to prepare this exercise.
Then, if you lack ideas to begin, look at the given clue and start searching for the solution.
A detailed solution is then proposed to you.
If you have more questions, feel free to ask them on the forum.
Because of the thermal motion, the particles in a gas or a liquid have broken lines trajectories.
Starting from the initial point O at \(t = 0\), a particle makes \(N\) moves \({\vec \delta _i}\) (\(i=1\) to \(N\)).
After each collision, it goes towards another direction, independent from the previous one.
Question
What is the average distance \(r\) between O and the particle after \(t=N\tau\) (\(N\) is huge), where \(\tau\) is the average time between to following collisions ?
Express \(r\) with \(t\), \(L\) (free mean path) and \(u\) (mean square velocity).
Solution
After \(N\) moves \({\vec \delta _i}\), the particle is in :
\(\vec r = \sum\limits_{i = 1}^N {{{\vec \delta }_i}}\)
So, in modulus :
\({\vec r\;^2} = {r^2} = \sum\limits_{i = 1}^N {\vec \delta _i\;^2} + 2\sum\limits_i {\sum\limits_{j < i} {{{\vec \delta }_i}.{{\vec \delta }_j}} }\)
And, as the definition of the free mean path states :
\({L^2} = \frac{1}{N}\sum\limits_{i = 1}^N {\vec \delta _i\;^2}\)
Furthermore, for \(N\) really huge, the random characteristic of scattering makes cosinus equally distributed between \(-1\) and \(+1\), so :
\(\sum\limits_i {\sum\limits_{j < i} {{{\vec \delta }_i}.{{\vec \delta }_j}} } = 0\)
It remains, with \(L = u\tau\) :
\({r^2} = N{L^2} = \frac{t}{\tau }u\tau L = Lut\)
So :
\(r = \sqrt {Lut}\)
Question
Numerical application : estimate the time required by a fragrance to diffuse after the opening to a distance \(d = 10\; cm\) ; comment on it.
Solution
Numericale application :\( t=3 \;min \;20 \;s\).
Scattering phenomenons are really slow.
In reality, the molecules are heavier than air, so \(u\) is smaller and \(t\) is longer, and here is the interest of convection.