I ask this question on math stackexchange but got no answer. Not sure how to move the post so I'm reposting it here.

I was arguing with my friend that Brownian motion, in the sense of a pollen moving in the fluid, could be explained by physics laws (such as $F=ma$) and statistics laws.

To check it out I found Albert Einstein's paper, "Investigations on the theory of Brownian motion", however I found it a bit hard to understand. It starts with osmotic pressure which i have no idea, and the paper is 114 pages long, though in big font.

I'm wondering if there's an easier iteration? hopefully shorter?

  • $\begingroup$ Your question is not very clear to me. I guess what you are probably looking for is something like the Langevin equation: Basically you have a new force term due to (thermal) noise (which you can relate to temperature through the fluctuation-dissipation theorem). To solve equations of this stochastic type analytically, you'll need to learn stochastic calculus (or e.g. transform to Fokker-Planck form to solve PDEs). $\endgroup$ – alarge Aug 5 '14 at 5:30

So, whenever I want to find a nice introduction to a concept in physics, I check the American Journal of Physics, as it is full of articles with clever descriptions of phenomenon appropriate for presentation in university courses. In this case, this yields many results. In particular, I found the following three articles very helpful:

  • The mathematics of Brownian motion and Johnson Noise - Daniel Gillepsie [doi][pdf]
  • Two Models of Brownian Motion - David Mermin [doi][pdf]
  • Fluctuation and Dissipation in Brownian Motion - Daniel Gillepsie [doi][pdf]

For completeness, I'll give my version of a modern approach to describing brownian motion, where I will borrow heavily from the above.

If you want to think in terms of Newton's laws, we will take an approach that in spirit is the same that Langevin gave three years after Einstein's paper (a translation of which also appeared in the American Journal of Physics [doi] [pdf]), that gives the same result.

If we imagine a pollen particle suspended in a liquid, we can assume that the forces on the pollen particle are given by a dissipative friction force, and some random jostling by impacts from the water molecules, we'll write

$$ m \dot v = -\gamma v + f \Gamma(t) $$ where $\gamma$ is the drag coefficient, $f$ is some constant we will have to determine, and $\Gamma(t)$ represents a random gaussian process. That is, we will assume the effect of all of the jostling by the water molecules amounts to drawing a random variable at every instant of time. Then, the next step of the argument is to make this whole deal consistent with statistical mechanics, namely the equipartition theorem, so in particular, if we look at long times, we should have $$ \frac 12 m \left\langle v^2(\infty) \right \rangle = \frac 12 k T $$ or in words, the average kinetic energy at long times should be a half $kT$ if we are going to be consistent with statistical mechanics.

So, we need only compute the average fluctuations in our velocity for long times. You can follow the papers to see a detailed mathematical account, but for just a taste, we can get the answer from dimensional analysis.

We are interested in determining $ \langle v^2(\infty) \rangle $, and this answer should only depend on the parameters in our equation for the forces the particle feels, namely $m$, $\gamma$ and $f$. The dimensions of $m$ and $\gamma$ are easy to read off of the equation

$$ [m] = [M] \qquad [\gamma] = [M T^{-1}] $$

but what about that $f$? Well, it depends on the dimensions of our random gaussian noise term, which is a bit tricky. But, the way we tried to describe it, the noise was supposed to be completely uncorrelated in time, so though I didn't detail it, this means that

$$ \langle \Gamma(t) \Gamma(t') \rangle = \delta(t-t') $$ in detail. And since we know that $\int dt\, \delta(t) = 1$, we have the dimensions $$ [ \delta(t) ] = [ T^{-1} ] \qquad [\Gamma(t) ] = [T^{-1/2}] $$ which tells us that $$ [f^2] = [ M L T^{-3} ] $$ which seems funny, but it enables us to determine that $$ \langle v^2 \rangle \propto \frac{ f^2 }{ \gamma m } $$ and in particular, we will assume that the proportionality constant is 1, which using equipartition, gives us $$ m \langle v^2 \rangle = f^2/\gamma = k T $$ or $$ f = \sqrt{ \gamma k T } $$ if we had done all of the math properly, the real answer turns out to be $$ \boxed{ f = \sqrt{ 2 \gamma k T } } $$ which is pretty darn close.

The point of all of that, and of Einstein's original paper is that we've shown that the fluctuations ($f$) causes by the jostling of the unseen water molecules is directly related to the dissipation ($\gamma$) you can observe in ordinary fluid experiments. This is the major result of Einsteins and Langevin's papers. With a bit more work, we can relate this to the diffusion constant, which tells us how the root mean square position increases linearly with time:

$$ \langle x^2(\infty) \rangle = D t $$

doing our dimensional analysis again, we discover we need a relation of the form

$$ D \propto \frac{f^2}{\gamma^2} $$ or, getting rid of this silly $f$ thing, using our other result above $$ \boxed{ D = \frac{ kT }{\gamma } } $$ which turns out to be right even if you do the math right (the proportionality constant is 1).

This was the actual formula Einstein got famous for in his paper, relating the diffusion constant, something you can measure in experiment, to the drag coefficient, something you can also measure from a different set of experiments. Giving in the end a quantitative theory of brownian motion that worked to help solidify the atomic hypothesis and some of the early results of statistical mechanics.

  • $\begingroup$ thanks for the reply. In $\langle v^2(\infty) \rangle$ , what does $\langle \cdot \rangle$ mean? Seems it's not Quadratic variation, as Quadratic variation is defined as $[X]_t = \lim_{\| P\| \to 0} \sum_{k=1}^n (X_{t_k} - X_{t_{k-1}})^2$ so the $\langle v^2(\infty) \rangle$ doesn't fit due to the $\infty$... $\endgroup$ – athos Aug 6 '14 at 4:44
  • $\begingroup$ The brackets just were to stand for average. Technically I guess it would be the time average, and the infinity meant to mark it as the long time limit. You could take it to be $$ \langle v^2(\infty) \rangle \equiv \lim_{t_0\to\infty} \lim_{T\to \infty} \frac{1}{T} \int_{t_0}^{t_0 + T} dt\, (v(t))^2 $$ But I think the words are clearer: it is the average square velocity in the limit of long times. $\endgroup$ – alemi Aug 6 '14 at 4:54
  • $\begingroup$ Thanks a lot for the explanation. May I summarize it as: 1. assuming random jostling from the water molecules $\Gamma(t)$ is a random gaussian process $\Rightarrow$ 2. $ \langle x^2(\infty) \rangle = Dt$ $\Rightarrow$ 3. $x(t)$ is Brownian motion. May I ask, Q1: from 2 to 3 is due to Levy characterization (en.wikipedia.org/wiki/…)? Q2: what's the base of 1, is it due to Levy Lindeberg Central Limit Theorem (en.wikipedia.org/wiki/Central_limit_theorem)? $\endgroup$ – athos Aug 13 '14 at 2:21
  • $\begingroup$ @athos Given that the papers in question were published in 1908 and 1911, and Levy didn't become a professor until 1917, its hard to say that it is because of Levy's mathematical justification. These started as physical theories, and much is the practice in physics, the initial papers were not mathematically rigorous. Detailed questions about the mathematical justifications is probably a better fit for Mathematics, I'm afraid I can't help you. $\endgroup$ – alemi Aug 13 '14 at 3:05
  • $\begingroup$ @alemi Dear Alemi, your input on this post: physics.stackexchange.com/questions/339121/… would be very valuable, look forward. $\endgroup$ – user929304 Jun 13 '17 at 15:12

A Brownian particle, 2.79 microns in diameter is 10,000 times a water molecule’s diameter and thus, a trillion times its volume and mass; a water molecule cannot budge a Brownian particle by random collisions. Thus Brownian motion is independent of the kinetics of the type of motion of individual water molecules.

If Einstein’s version of the cause of Brownian motion is incorrect, then there is no reason to consider that the motion of individual water molecules must be in the scale of Brownian particle motion and may be markedly less. There might be another easily understood mechanism of molecular kinesis that would still account for diffusion.

Consider oxygen molecules in a fixed container at normal temperature and pressure. Imagine that each molecule exhibits simple harmonic elastic oscillation about its center of mass (i.e., the frequency of oscillation would be like a pendulum, independent of amplitude). Drifting molecules, bumping against neighboring molecules, would be bumped away from their original positions in random directions. An added coherently oscillating molecule of nitrogen would similarly be bumped around randomly by collisions with oscillating oxygen molecules, it’s changing position (i.e., diffusiion) being similar to Einstein’s random walk analogy.

If molecular kinesis in fluids originates from simple harmonic elastic oscillation around molecular centers of mass (nuclei), then the laminae of laminar flow may approach molecular dimensions. As liquid helium approaches absolute zero, laminae would approach molecular dimensions, would become increasingly smooth (non-oscillating molecules might act like ball bearings) and slippery, accounting for ultra-cold helium’s viscosity approaching zero.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.