6
$\begingroup$

Supposing that I have linear chain with polymer of $N$ identical particles (interacting harmonically with adjacent particle) with position of first and last particle fixed, how do I find the partition function of the polymer?

Here is what I thought, $\vec p_i$ being the momentum of the polymer and $\vec r_i$ being the position of $i$th element of polymer, the Hamiltonian of each particle is given by $$H = \sum_{i=1}^N\frac {p_i^2} {2m} + k\sum_{i=1}^{N-1}(r_{i+1}-r_i)^2$$ The partition function (for discrete canonical system) is given by $$Q_N = \sum_{\{x\}}^n e^{-\beta H_{x}} = \prod_{i=2}^{N-1} e^{-\beta \frac{p_i^2}{2m}}\prod_{i=1}^{N-1} e^{-\beta k (r_{i+1}-r_i)^2}$$ I can't go beyond this. Any comment is appreaciated.

$\endgroup$
  • 1
    $\begingroup$ Your expression for potential energy should be ½kx². You can make a reasonable guess that each of the terms in the multiplications in the partition function are going to be the same, so factorise them. Then it becomes much easier to integrate over phase space. You'll just have two Gaussian functions. $\endgroup$ – Max Tyler Aug 18 '16 at 8:10
  • $\begingroup$ @MaxTyler My goal is to find the thermodynamic properties such as sp. heat capacity of the system via Helmholtz free energy. Could you provide answer for it? $\endgroup$ – Santosh Linkha Aug 18 '16 at 9:15
  • $\begingroup$ I'll just write one $\endgroup$ – Max Tyler Aug 18 '16 at 9:20
  • $\begingroup$ You need to find the normal modes of this system first. In that basis your energy will become sum of N independent harmonic oscillators. The rest is evaluating Gaussian integrals. $\endgroup$ – Hosein Aug 18 '16 at 11:07
3
$\begingroup$

The hamiltonian for the whole system can be given by: $$H_{total}=\sum_{i=1}^{N-2}\frac{p_i^2}{2m}+\sum_{j=1}^{N-1}\frac{k(\gamma-x_j)^2}{2}$$ Where the terms for momentum come from the masses in the chain and the potential comes from the springs. The $\gamma-x$ term comes from the deviation of each spring from their equilibrium position, with $x=\gamma$ giving the point with $0$ potential.

The probability of the system (in thermal contact with surroundings at temperature $T$) being at energy $E$ is given by: $$q(E)=\frac{1}{Z}e^{-\beta H}$$ Where $\beta=\frac{1}{k_B T}$. The partition function $Z$ is given by integration over phase space of the total hamiltonian of the system. Luckily this hamiltonian can be factorised quite easily. $$Z=\int_{p,x}e^{-\beta\sum_{1}^{N-2}\frac{p^2}{2m}}e^{-\beta\sum_{1}^{N-1}\frac{k(\gamma-x)^2}{2}}dp\ dx=\int_{-\infty}^{\infty}e^{\frac{\beta(2-N)}{2m}p^2}dp\int_{0}^{\infty}e^{\frac{\beta(1-N)k}{2}(\gamma-x)^2}dx$$

The first is a gaussian ($\int_{-\infty}^{\infty} e^{-ax^2}=\sqrt{\frac{\pi}{a}}$), and the second needs a little massaging.

After integrating out the momentum we get: $$Z=\sqrt{\frac{2\pi m}{\beta(N-2)}}\int_0^{\infty}e^{-\frac{\beta(N-1)k}{2}(x-\gamma)^2}dx$$

Change the variable $x-\gamma$ to $q$, we get $dx=dq$ and the limits are $\int_{-\gamma}^{\infty}$. This second integral needs the error function to calculate, because of the non-zero lower limit. Change the constants in the exponential into an easier to handle form - $\frac{\beta(N-1)k}{2}=\alpha$:

$$Z=\sqrt{\frac{2\pi m}{\beta(N-2)}}\int_{-\gamma}^{\infty}e^{-\alpha q^2}dq=\sqrt{\frac{2\pi m}{\beta(N-2)}}\sqrt{\frac{\pi}{4\alpha}}(1-erf(-\gamma\sqrt\alpha))$$ Finally: $$Z=\frac{\pi}{\beta}\sqrt{\frac{m}{k(N-1)(N-2)}}(1-erf(-\gamma\sqrt{\frac{\beta(N-1)k}{2}}))$$

There are some approximations that can be made here. If $N$ is large, then we have $(N-1)(N-2)\approx N^2$ and $N-1\approx N$: $$Z_{large\ N}\approx \frac{\pi}{\beta N}\sqrt{\frac{m}{k}}(1-erf(-\gamma\sqrt{\frac{\beta N k}{2}}))$$ Using an approximation for the error function from wikipedia, we can get $Z$ into analytic functions. Using $x=-\gamma\sqrt{\frac{\beta N k}{2}}$, and assuming $\gamma$ is always positive, we get: $$Z=\frac{\pi}{\beta N}\sqrt{\frac{m}{k}}(1-\sqrt{1-exp(-x^2\frac{\frac{4}{\pi}+ax^2}{1+ax^2})})$$ Where $a=\frac{8(\pi-3)}{3\pi(4-\pi)}$.

$\endgroup$
  • 1
    $\begingroup$ I have a small query. Why can we write $\sum_{i} p_i = (N-2)p$?? If I separate each $p_i$ from exponential, then I get $ \left ( \dots \right )^{N-2}$ instead of $\sqrt{ \frac {\dots } {(N-1)(N-2)}}$ $\endgroup$ – Santosh Linkha Aug 18 '16 at 17:22
  • 1
    $\begingroup$ @SantoshLinkha it's because $e^{\sum_N p}=\prod_N e^p=(e^p)^N=e^{Np}$ It's factorisation of the partition function. $\endgroup$ – Max Tyler Aug 18 '16 at 17:45

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.