All easy explanations of Brownian motion that I have found are all totally wrong in that they just essentially say something like "motion of the pollen is being moved by individual water molecules" which in today's culture says absolutely nothing. If the fluid particles bombarded the Brownian particle equally from all directions, it is clear the particle would not observably move at all due to there being typically something like $10^{20}$ collisions per second which each alone have zero observable effect to the particle.

What one thus surely needs to explain the phenomenon is some kind of fluctuations of density/pressure of the fluid. I have not, however, found any kind of intuitive explanation of why any kind of fluctuations would occur and to me it some way seems much more intuitive that any kind of macro phenomena would not arise, but instead the fluid would act some way uniformly over time.

Is there really no easy explanation for emergence of the necessary fluctuations or does one really need to study Stochastic processes to understand it at all?

  • $\begingroup$ Have you checked Einstein's papers on Brownian motion? $\endgroup$
    – Bob D
    Commented Jan 7, 2020 at 17:42

2 Answers 2


Here is the part you are missing:

The velocities of the water molecules that strike the pollen particle from all directions are not all the same. They exist in a distribution, with some velocities being slower than average and some faster than average. Since their energy depends on the square of their velocity, the faster-than-average molecules pack significantly more punch than the rest of the population, and when one of these "outliers" hits the pollen particle, the particle recoils in response.

  • $\begingroup$ That certainly sounds very simple explanation. Do you know if there is some definite derived say probability distribution on frequency of collisions with fluid particles of certain velocity from which one would then be able to see that the probability of collision with so fast particles which have observable effect in motion is large enough? $\endgroup$
    – user128534
    Commented Jan 9, 2020 at 13:58

You mention the example of pollen, so I'll work with that. There is a quantity known as the mean free path - which is the average distance between consecutive collisions between the pollen and some other molecule in the water. I'll denote this $a$. I'll also assume we're only interested in two dimensions for now.

Each collision scatters the pollen in some random direction. Each "step" between collisions can be represented by a vector of magnitude $a$ and components $x_{i}$ and $y_{i}$ in the x and y directions respectively. After say, $n$ collisions, the total displacement of the pollen grain will be the vector sum of all of the steps:

$\vec{r} = (x_{1} + x_{2} + \dots + x_{n})\vec{i} + (y_{1} + y_{2} + \dots + y_{n})\vec{j}$

What is the magnitude of this vector? If each step is random, the cross terms will average to zero and our magnitude will turn out to be $|\vec{r}| = \sqrt{(x_{1}^{2} + y_{1}^{2}) + (x_{2}^{2} + y_{2}^{2}) + \dots}$ (try proving it yourself!).

Notice, however that we already defined the magnitude of each step, $\sqrt{x_{i}^{2} + y_{i}^{2}}$, to be $a$! So if we count $N$ steps, the expectation of the magnitude of the pollen grain's total displacement will be $a\sqrt{N}$. Clearly, if we leave the system for a reasonable amount of time, the effect on the displacement of the pollen grain will not be negligible!

Of course, this model is greatly simplified. However, it helps to show that even though we might "expect" that the influences of all of the collisions will cancel out, we instead derive that they do not!


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