Problem set 2¶
Problem 2.1¶
There were very good suggestions and ideas during the work session. Some systems fall into multiple categories, depending on circumstances, and there may be other ways to classify them beyond the options that were mentioned. Here are some suggestions. Your arguments are more important than the actual answer.
Yogurt: can be liquid, can be gel, can be called solid
Diamond: (famous symmetry of a) crystal
The salt ions in vegetable soup: gas (a pinch of salt), liquid (a spoon), solid (a cup), a plasma (fluid of charged particles - are also abundant in outer space, or in fusion reactors, but also in vegetable soup)
a school of sardines: nematic liquid crystal, liquid, a ‘swirlonic state’ (yes). One person mentioned gel, and that seemed unexpectedly appropriate. These schools seem to behave visco-elastically when perturbed by predators.
actin networks: (active) gel, solid / glass when dry
a cell membrane: liquid with more solid-like rafts
a sample of proteins after Cryo-EM: glass, random aggregate
skin tissue: gel / solid / …
a soap bubble: generally liquid. May form foams.
muscle tissue: solid, active gel, fiber composite, …
Problem 2.2¶
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Problem 2.3¶
See notes: \(\mu = 0\), \(\sigma = 1\).
The mean \(\mu_N = N \mu = 0\) (it should be zero, because this random walk is symmetric - walking left or right is equally probable). The standard deviation \(\sigma_N = \sqrt{N}\sigma = 10\)
After \(N=10^4\) steps, \(\sigma_N = \sqrt{N}\sigma = 100\). The random walk describes diffusion, and this process is not efficient to explore space. The distance that one typically explores (\(\sigma_N\)) is not linear with the number of steps, but is proportional with \(\sqrt{N}\). To give a feeling, if 1 step is 1 m and takes 1 s, then it would take 100 s to explore 10 m, \(10^4\) s to explore 100 m (about 3 hours), and \(10^6\) s to explore 1 km (about half a month). Check how long it takes to explore 10 km or 100 km… This can be a serious issue if you want particles to diffuse to a template, or to diffuse over a large distance.
If one ends up \(y\) steps to the right, we are interested in all combinations of steps that lead to \(y\) after \(N\) steps, so \(X_1+X_2+...+X_N=Y\), where all the stochastic variables \(X\) can take values -1 (step to the left) or 1 (step to the right). This is very similar to the problem with the dice, except that we only have 2 possible outcomes here instead of 6. We can use the central limit theorem here to argue that
\[P(y) = C e^{-\frac{y^2}{2N}}\]with \(C\) a normalisation constant.
In this case one can also calculate the exact answer. Given that one ends up at position \(y\) after \(N_+\) steps to the right and \(N_-\) steps to the left, with a total number of \(N\) steps, we can figure out how many steps one has to take to the right:
\[\begin{split}\begin{aligned} N_+ - N_- &= y\\ N_+ + N_- &= N \end{aligned}\end{split}\]so
\[\begin{split}\begin{aligned} N_+ &= \frac{N+y}{2}\\ N_- &= \frac{N-y}{2} \end{aligned}\end{split}\]so the total number of ways in which one can walk to \(y\) is \(\binom{N}{N_+} = \binom{N}{N_-}\). The total number of random walks is \(2^N\) (for each step there are 2 possibilities), so the probability that one would end up at \(y\) after \(N\) steps is
\[ P(y) = 2^{-N}\binom{N}{\frac{N+y}{2}} = 2^{-N}\frac{N!}{\left(\frac{N+y}{2}\right)!\left(\frac{N-y}{2}\right)!} \]
In the next sessions we will see what these results imply for the shape of flexible polymers, and for association-dissociation equilibria.
Problem 2.4¶
A long polymer has \(N\) sites where certain types of particles can bind. These particles can bind to each site with equal probability. If a particle is bound to a site, no other particle can bind to that site.
There are \(N\) different sites, so one particle can bind in \(N\) ways.
The first particle in \(N\) ways, the second in \(N-1\) ways, and the \(n^\mathrm{th}\) in \(N-n+1\) ways, so the total number is
\[\frac{N!}{(N-n)!}\]If the particles are all identical, we can swap particles without changing the configuration, so the total number of ways is
\[\binom{N}{n}\]It is a binomial distribution, so the largest value is obtained for \(n=\frac{N}{2}\) (by knowledge of what a binomial distribution typically looks like, or by observing that the binomial distribution has a symmetry around \(n=\frac{N}{2}\) and is increasing with \(n<\frac{N}{2}\)), or via the Stirling approximation and seeking the extremum - see next session.)
One would get a binomial distribution, which looks very much like a Gaussian distribution for \(N=100\) (see previous exercise), with a mean of \(\mu_N=50\).
In fact, one can map this situation on a random walk. Introducing \(X_1+X_2+...X_N=Y\), where each \(X_i\) can take the value 0 (no particle at site \(i\)) or 1 (a particle bound at site \(i\)) with equal probability, the mean \(\mu_N=50\) and standard deviation \(\sigma_N=5\). The number of ways would be proportional to this distribution (\(2^N P(n)\)).
Bonus challenge (optional): The solutions are now: a) \(N\), b) \(N^n\), c) \(\binom{N+n-1}{n}\). If you were able to solve f) please contact me.
These results will be used again in the future. They are very meaningful if one is interested in the elastic properties of polymers, or the number of occupied binding sites on a polymer or membrane.