3
$\begingroup$

A model for protein in 2D can be formed by adding bonds of fixed length $l\sqrt{2}$ on a square lattice along the diagonal, ie $\hat{\mathbf{b}}_i=\frac{1}{\sqrt{2}}(\pm \hat{\mathbf{x}}\pm \mathbf{y})$. Consider a protein made up of $N$ bonds $\hat{\mathbf{b}}_i=\frac{1}{\sqrt{2}}(x_i,y_i)$ where $x_i=\pm 1, y_i=\pm 1$ and $i \in \{1,\dots,N\}$.

The energy of the underformed protein is given by

$$\mathcal{H}_0=-\sum_{i \neq j}K \hat{\mathbf{b}}_i \cdot \hat{\mathbf{b}}_j=-\sum_{i \neq j}\frac{K}{2}(x_ix_j+y_iy_j)$$

where the sum above is over all pairs $i, j \in \{1,\dots,N \}$. The displacement between the two ends of the protein is $\mathbf{R}=l\sqrt{2}\sum_{i=1}^N \hat{\mathbf{b}}_i$.

Under a tensile force $\hat{\mathbf{f}}=f\hat{\mathbf{n}}$, $f>0$, is the force aligned along the vector $\hat{\mathbf{n}}=(\cos \theta, \sin \theta)$, the system is described by the Hamiltonian $$\mathcal{H}=\mathcal{H}_0-\mathbf{f} \cdot \mathbf{R}=\mathcal{H}_0 -fl \sum_{i=1}^N (x_i \cos \theta + y_i \sin \theta)$$

Finite stiffness. Consider the case $K \neq 0$ for the long protein made of $N \gg 1$ bonds under a tensile force aligned along the x-axis:

It can be shown that the Hamiltonian can be written in the form $$\mathcal{H}(\{x_i,y_i \})=\mathcal{H}_0-\mathbf{f} \cdot \mathbf{R}=-l\sqrt{2}\sum_{i=1}^N \mathbf{f}_i \cdot \hat{\mathbf{b}}_i+E_0$$

where $\displaystyle \mathbf{f}_i=\sum_{j=1}^N \frac{K}{l\sqrt{2}}\hat{\mathbf{b}}_j-\mathbf{f}$

and it is easy to show that $\langle \mathbf{f}_i\rangle=\mathbf{f}+\Delta \mathbf{f}$ and $\Delta \mathbf{f}=KA \langle \mathbf{R} \rangle$ where $A=\frac{1}{2l^2}$.

A mean field Weiss approximation involves replacing the instantaneous force felt by each bond by its average. I am then asked to find the mean field Hamiltonian.

I get the mean field Hamiltonian will be $$\begin{align}\mathcal{H}_{mf} &= -l\sqrt{2}\sum_{i=1}^N (\mathbf{f}+\Delta \mathbf{f})\cdot \hat{\mathbf{b}}_i+E_0 \\ &= -l\sqrt{2} \left(\frac{K}{2l^2}\langle \mathbf{R} \rangle + \mathbf{f}\right)\cdot \sum_{i=1}^N \hat{\mathbf{b}}_i+E_0 \\ \end{align}$$

I then try to calculate the Helmholtz free energy $F=-k_BT \ln Z$

We may drop the $E_0$ so

$$\begin{align}Z &= \sum_{\Gamma}e^{-\beta \mathcal{H}_{MF}(\Gamma)} \\ &= \sum_{\Gamma}e^{-\beta \left(-l\sqrt{2} \left(\frac{K}{2l^2}\langle \mathbf{R} \rangle + \mathbf{f}\right)\cdot \sum_{i=1}^N \hat{\mathbf{b}}_i\right)} \end{align}$$

but this isnt very tractable. Where are am I going wrong?

$\endgroup$
1
+100
$\begingroup$

You're actually dealing with the Potts model, which is a slight generalization of Ising. Not that it really matters, as you won't need any results from Potts.

The point of mean field theory is typically to make each site independent of their neighbors, which allows you to evaluate the partition function by only iterating through the possible states of one site, instead of through all the combinations.

This is to say that given that the protein is long so you need not care about boundary effects, you can write the partition function corresponding to your mean field as $$Z_\text{force} = \left(\sum_\gamma e^{-\beta \left(-l\sqrt{2} \left(\frac{K}{2l^2}\langle \mathbf{R} \rangle + \mathbf{f}\right)\cdot \mathbf{b}\right)}\right)^N$$ where $\gamma$ iterates through the four states of $\mathbf{b}$ (and "force" denotes the fact that I've dropped the cooperative effects, $E_0$).

Now if the force is into the x-direction, this reduces to $$Z_\text{force} = 2^N\left(e^{-\beta \left(-l\sqrt{2} \left(\frac{K}{2l^2}\langle R \rangle + f\right)\right)} + e^{\beta \left(-l\sqrt{2} \left(\frac{K}{2l^2}\langle R \rangle + f\right)\right)}\right)^N = 4^N\cosh^N\left(\beta l\sqrt{2} \left(\frac{K}{2l^2}\langle R \rangle + f\right)\right)$$

Another alternative is, if you want a continuum model, to replace the sum over $\mathbf{b}_i$ by $\mathbf{R}$ and the state sum then by an integral. Doing this, though, you should really make the pair interactions behave in a continuum way as well (which introduces the stiffness).

$\endgroup$
  • $\begingroup$ Whats the rationale for dropping $E_0$? $\endgroup$ – Permian May 25 '15 at 17:06
  • $\begingroup$ @sandstone Well you dropped it so I figured I would too. That said, I'm expecting that you're going to differentiate the free energy wrt f, so the $E_0$ term doesn't matter. If this is not the case, though, then you can't drop it. $\endgroup$ – alarge May 25 '15 at 21:45

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.