I am studing Brownian motion, in particular I am reading the book "Brownian Motion, Fluctuation, Dynamics and Application" by Mazo. Now I am dealing with Smoluchowski theory, but I am having some difficulties.

Smoluchowski's work is based on the fact that we can consider the system made up of hard spheres colliding (light ones with mass $m$ and a heavy one with mass $M$). Let's call then $C$ and $c$ the root-mean-square velocity of the heavy and light particles respectively; using the equipartition theorem we obtain $c/C = (M/m)^{1/2}$. Now let's indicate $\mathbf{v}$ and $\mathbf{V}$ the velocities of the light and hard particles; the velocity will be unprimed if they indicate the velocities before the collision and primed if they indicate those after the collision. Of course we have that $c^2=\langle \mathbf{v} \cdot \mathbf{v} \rangle$, and a similar relation for the velocity of the heavy particle. Now let $\mathbf{g}=\mathbf{v}-\mathbf{V}$; the kinematic of the collision of hard spheres tell that: \begin{equation} \mathbf{V'}=\mathbf{V}-2\frac{m}{M+m}(\mathbf{g}\cdot\mathbf{k})\mathbf{k} \end{equation} where $\mathbf{k}$ is the unit vector normal to the common tangent. And now I have some doubt: I have tried finding out where this relations came from and in the book "Kinetic theory" by Liboff it is said that this relation is true when the spheres have the same mass, that in this case is not true. Is this just and approximation or how can I justify the use of this relation?

Then in the book it is written that the equation above shows that $C=C' + O((m/M)^2)$ on average, so it is possible to neglect the effects of order $(m/M)^2$. My question now is why the mass ratio is squared: from the equation above I can see only $m/M$.

Anyway neglecting the effects the way is written, we obtain that $C=C'$ so the heavy particle's velocity only changes in direction. The book says that the angle $\epsilon$, between $\mathbf{V}$ and $\mathbf{V'}$, is given by $\sin \epsilon = (3/4)(m/M)(c/C)$ (in the footprint the author said that reproducing the calcuclation he obtain similar numerical factors like $0.708$ or $\pi/4$ ). Do you have any suggestion on how I can obtain this approximation or on the path I shoul do to obtain the value of $\sin \epsilon$. I tried calculating $\cos \epsilon = \frac{\langle \mathbf{V} \cdot \mathbf{V'}\rangle}{|\mathbf{V}||\mathbf{V'}|}$ but this became very complex and I give up.

I hope I have explained clearly my doubts. If someone has some suggestions I will be very happy to read them. Thanks in advance!!

  • $\begingroup$ I am not an expert on this topic. It makes sense to me that considering hydrogen bonding, brownian motion in water might not properly fit the model of hard spheres colliding. It might better be described as some sort of sticky spheres. $\endgroup$ – J Thomas Sep 21 '18 at 13:26
  • $\begingroup$ A long time ago I read that on a bacterial scale, water behaves like a liquid with a gigantic reynolds number. To them it is more viscous than molasses to us. The claim was that shear forces acted like giant fracture planes. Bacteria and their flagella must withstand those forces. The flagella work by rotating; I've never seen a description of how much the bacterial cell rotates instead, or a description of how the rotation moves the bacterium forward, but it obviously does. $\endgroup$ – J Thomas Sep 21 '18 at 13:32
  • $\begingroup$ Thank you so much for your answer, very interesting, I will also study this aspect of the Brownian motion and of the water behaviour. My question's aim anyway is just to understand the theory of Smolichowski, then I will of course dig deep in the topic you wrote about $\endgroup$ – Alessandro Pecile Sep 21 '18 at 13:37
  • $\begingroup$ @AlessandroPecile I saw a comment from you today on my answer asking for the details leading to your first equation, but midway through preparing these, the comment seems to have disappeared again. Would you like me to insert these few lines of algebra, or are you happy now? $\endgroup$ – user197851 Oct 5 '18 at 20:53
  • $\begingroup$ I actually was able to find them myself. I am very sorry for this misunderstanding. Actually now I am fine with this chapter. Thank you so much, your help was very very important. $\endgroup$ – Alessandro Pecile Oct 5 '18 at 20:57

You have asked several questions; I can only give a clear answer to the first one, and some comments on the others.

As far as I can tell, the expression for the change in velocity of the heavy particle is correct. In this kind of collision dynamics, one expresses the velocity changes of both particles in terms of the collisional impulse $\Delta \mathbf{P}$: $$ \mathbf{V}' = \mathbf{V} + \Delta\mathbf{P}/M, \qquad \mathbf{v}' = \mathbf{v} - \Delta\mathbf{P}/m $$ which guarantees conservation of momentum. Then one imposes the fact that $\Delta \mathbf{P}$ must lie along the line of centres $\mathbf{k}$ at contact (smooth hard spheres) so we only require to calculate its magnitude. This can be found by requiring conservation of total kinetic energy which can be written. $$ M({V'}^2-V^2) + m({v'}^2-v^2) = 0 $$ The problem can be made simpler by recognizing that the velocity components perpendicular to $\mathbf{k}$ are unchanged, and so their contributions to the change in kinetic energy can be dropped. We can resolve everything in the $\mathbf{k}$ direction and we just need to solve a scalar equation involving the components along $\mathbf{k}$, not a vector one. Anyway, I seem to be able to satisfy the energy conservation equation with the given expression for $\Delta \mathbf{P}/M$ appearing in the equation for $\mathbf{V}'$. You might like to check this yourself, it is just a few lines of algebra.

On the question of the order of magnitude of the change in $C$, I am as puzzled as you. Clearly, on average, the sign of the difference $C'-C$ must be zero, because the collisions must leave the velocity distributions of both the particles unchanged, on average. $C'$, like $C$, is determined by the mass $M$ and the temperature. The argument clearly relates to the magnitude of the typical change in speed of the heavy particle. It is small, and I can believe that it is $\mathcal{O}(m/M)$, but I haven't found a way of showing that it is $\mathcal{O}(m^2/M^2)$.

The general form for the formula for $\sin\epsilon$ looks like it comes from estimating the magnitude of $\Delta \mathbf{P}/M$: the mass prefactor gives you $m/M$ and the dependence on $\mathbf{g}$ gives you a proportionality to $c$. ($\mathbf{g}$ is a difference of velocities $\mathbf{v}-\mathbf{V}$, but the contribution of $\mathbf{V}$ is negligible compared with $\mathbf{v}$, which is of typical magnitude $c$.) Adding a vector of magnitude $(m/M)c$ to a vector $\mathbf{V}$, which is of approximate magnitude $C$, will rotate it by a typical angle $\sim (m/M)(c/C)$ which we identify with $\epsilon$, or $\sin\epsilon$. However, even Mazo, in his footnote, said that he had difficulty reproducing the numerical prefactor. I don't think that it is reasonable to expect a typical StackExchange reader to take on this derivation! I certainly won't be trying.

Good luck with the rest of that chapter, which looks fairly heavy going, to me.


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.