The dreaded question from last year...

Hi all. I promised myself that I would look at this question again sometime this semester and prove to myself that I could do it... Of course, I am talking about the ratchet questions from last year's BIPH2000 exam.

The first part of the question goes like this: if I have a G-ratchet, draw the free energy landscape. This will be one-dimensional. Since the rod is so long and has only one degree of freedom (the axial direction) the entropy is constant. If we choose to measure energies such that G=0 when none of the springs is depressed we will find that the graph looks just like Nelson's plot of the potential.

The next question asks for the steady-state probability distribution. Consider a single step of length L in either direction: neither gives a net free energy change, no matter what the starting point was. Thus there is no tendency to move in either direction more often than the other, which is the kinetic interpretation of equilibrium. We know that this means the distribution is just the Boltzmann distribution.

Since the rod is semi-infinite the probability of being found in any position approaches zero in the equilibrium state. However, for each length [0,L] we can write

.

Now if the width of the springed ratchet head is l, and the maximum (compressed) energy is epsilon, the potential takes the form

 

Since this is equilibrium we have the expectation that j = 0.

The final part of the question asks what would happen if the ratchet were tightly coupled to an ion transporter. Since this is tight coupling the rod is in equilibrium when the ions are, and since we have already seen that the rod alone has no net driving force we need only consider the ions. (By the way, the free energy landscape is again one-dimensional, because of the tight coupling, and drops by some amount with every rightward step).

So, to halt the rod we just need to apply the Nernst voltage. If this is the voltage across the membrane (there are no counterions mentioned, so no need to worry about depletion layers or screening) then we find the field to be

,

(t is the thickness of the membrane) and this must be directed from left to right in order to stop the flux of ions.

Any takers? I get E~10^7 V/m.

Reflections...

For those who are interested: my reflections on BIPH3001 (the assignment from last week).
Seth: I noticed a mistake in equation four and just had to fix it (there was a missing factor of mu)!

Policy for late assignments

No more assignments will be accepted past 9am Monday morning. I will be sending out emails to let people know if they still have outstanding work.

The Mystical Transfer Matrix

Hi all. The blog has been very quiet (I hope exams are going well for you all).

Anyhow, I decided I'd sit down and figure out what on earth the transfer matrix method was all about (this is given on pg 360 of Nelson). I had previously glossed over this portion of the text because I didn't understand at all what the transfer matrix did (or why it's even called the transfer matrix) and hadn't the time to ponder it deeply.

So... here's my take on things, with some assistance.

Suppose I have a single link in my polymer chain (not a very exciting polymer!). By Nelson's notation, my partition function is

 Now, let's write this as

just for the fun of it (this will be useful in a second!).

Suppose I add a second monomer, so my `polymer' is now a dimer. There are four ways this can happen, depending on the states of the existing monomer and the new one (call them 'up' and 'down', or u and d). Thus I can have uu, ud, du and dd. Each of these options contributes to the partition function of the dimer. Since two options go with having the first monomer 'up' (uu and ud), and two with 'down' (du and dd) I can write a matrix equation, which for now I'll represent as
.
This needs filling in. Let's do the uu term. If the first monomer is 'up' and I add another 'up' then the term needs to be multiplied by . Repeating the process for the other combinations yields
.
We call the matrix in the middle the `transfer matrix', T, but it would probably be more sensible to call it the `extension matrix', since it accounts for all possible ways in which the system may be extended by one monomer.

Now, suppose I take my dimer and add another monomer, to make a trimer. Since the interactions are local (they only depend on the nearest-neighbours) there are again only `four' ways in which I can do this (the rest of the possible configurations are already counted in the partition function of the dimer). Thus I simply get
.
It's now quite simple to see that for a polymer of N residues the partition function is
,
which is a really neat result.

To summarise: the transfer (extension) matrix finds the terms which must be added to the partition function every time a new monomer is added, so the polymer partition function may be built up by repeated application of T. Enjoy!

By the way, the equations here were rendered with an online LaTeX interpreter which outputs images: very useful and fast!

Microscopic Model of Osmosis

Hi all. BIPH study is now underway...
I was re-reading the section in Nelson about osmotic pressure as the rectification of Bronwian motion (remember all of us walked away from that chapter feeling a little nonplussed?) and decided that I'd like to understand what on earth he was going on about.

My biggest issue was the way that osmotic flow was ascribed to entrainment of fluid allowing solvent to be 'sucked' across the membrane: surely the situation is symmetrical, because the particles also entrain fluid as they head toward the membrane (therefore they presumably pull fluid toward the membrane just as fast as they suck it away).

I found a very nice, readable and detailed paper which goes through the same sort of reasoning as Nelson, but I think a bit more maths helps to understand it a lot better (this is also a good reference for basic low-Re fluid mechanics).

The basic ideas are all the same as Nelson, but there was one line that really grabbed me;
"... the membrane acts on the fluid using the particles as its agent[s]."
The argument goes like so: the fluid is responsible for excitation of Brownian motion (creation of fluctuations) and the dissipation of energy, so every force which acts on the particle also induces fluid motion. At the membrane there is (on average) only a net force on the fluid when the particles bounce off the membrane, so they really do 'suck' solute through the membrane by rectifying the Brownian motion of the fluid using the particles as agents!

Neglecting Inertia

I was thinking about the differences between classical and quantum stat mech...

When we talk about a bunch of particles diffusing under the influence of an external force we use the Smoluchowski equation, with the force corresponding to the negative gradient of some potential energy. We then proceed to say that the concentration at equilibrium is proportional to exp(-U/kT). I had always thought that by doing this we were essentially assuming that the particles have negligible inertia i.e. we are operating in the low-Re regime (since I would have thought that the concentration would be proportional to exp(-(K+U)/kT) if this weren't the case, K being the kinetic energy).

It turns out that I was wrong:
concentration = integral of the distribution function over all momenta,
so the momentum is `integrated out', contributing only to the constant sitting out the front of exp(-U/kT).

For those doing statmech: why don't I need a degeneracy function i.e. the density of states?
The answer lies in the interpretation;
N = integral of g*f over energies = integral of concentration over space = integral of f over phase space;
so the degeneracy is automatically accounted for by the fact that one integrates over a volume in phase space, rather than over a line (in semi-classical mechanics this is where the density of states, g, comes from in the first place: it is the volume of a thin slice of phase space with energy e to e+de, divided by the number of uncertainty-limited cells that fit in that slice).

It's worth noting that we will still be in the low-Re regime anyway: if the force-velocity response is linear it implies the drift velocity is less than or comparable to the thermal velocity, which means that for most things (proteins in water, etc.) Re will be small.

Good luck with exams!

Announcement courtesy of Seth

To the people who haven't finished marking the assignments (including yours truly), we should submit the marks before the exam period begins which is before Monday. The same applies to our end of course summary, blog comments and so on, which was meant to be due today.

Here's his reply to me:

"The sooner the better, but I absolutely want it before the exam period starts.....anything submitted after the beginning of the exam period will not be counted."

Good luck everyone! :D