Bending energy and persistence length

Persistence length $L_p$ is a basic mechanical property quantifying the bending stiffness of a polymer and is defined as the relaxation of $\langle\cos(\theta)\rangle$ of bond angles of the polymer chain:

$$ \langle\cos(\theta(s))\rangle = \exp(-sl/L_p) $$

The problem of calculating $L_p$ becomes calculating $\langle\cos(\theta(s))\rangle$, if we choose the $\theta$ as the angle of adjacent bonds, i.e., $s=1$, we have:

$$ \langle\cos(\theta)\rangle = \exp(-l/L_p) $$

in simulations, the stiffness constant is given by a bending energy $U_b$, when $U_b$ is large, where excluded volume effect is negligible, we can calculate $\langle\cos(\theta)\rangle$ as

$$\langle\cos(\theta)\rangle=\frac{\int_0^\pi \cos(\theta)\sin(\theta)\exp(-\beta U_b)\mathrm{d}\theta}{\int_0^\pi\sin(\theta)\exp(-\beta U_b)\mathrm{d}\theta}$$

the $\sin(\theta)$ is a geometric weight that when 2 bonds are at an angle of $\theta$, then in 3-D space, the number of bonds is proportion to $\sin(\theta)$.

Here is an example of harmonic bending energy:

$$U_b(\theta)=-\frac{1}{2}k \theta^2$$

where $k$ is the stiffness constant in unit of $k_BT$, the $\langle\cos(\theta)\rangle$ is simply

$$\frac{e^{\frac{3}{4 k}} \left(-2 \text{erf}\left(\frac{1}{\sqrt{k}}\right)+\text{erf}\left(\frac{1+i \pi  k}{\sqrt{k}}\right)+\text{erf}\left(\frac{1-i \pi  k}{\sqrt{k}}\right)\right)}{2 \left(-2 \text{erf}\left(\frac{1}{2 \sqrt{k}}\right)+\text{erf}\left(\frac{1+2 i \pi  k}{2 \sqrt{k}}\right)+\text{erf}\left(\frac{1-2 i \pi  k}{2 \sqrt{k}}\right)\right)}$$

For large $k$ ($k\ge 6$), $\langle\cos(\theta)\rangle\simeq L(k)$, with $L(k)$ being the Langevin function.

Steady compliance (Linear viscoelasty)

Boltzmann superposition: assume that there is no stress when $t\le 0$, so the integral starts from $0$ and $\sigma(0^{-})=0$:

$$\begin{equation} \gamma(t)=\int_0^t J(t-t^\prime)\dot{\sigma}(t^\prime)\mathrm{d}t^\prime \end{equation}$$

 Performing Laplace Transform yields: 

$$\begin{align} \hat{\gamma}(s)&=\hat{J}(s)\hat{\dot{\sigma}}(s)\\ &=\hat{J}(s)\left(s\hat{\sigma}(s)-\sigma(0^{-})\right) \\ &=\hat{J}(s)\left(s\hat{G}(s)\hat{\dot{\gamma}}(s)-\sigma(0^{-})\right)\\ &=\hat{J}(s)\left(s\hat{G}(s)(s\hat{\gamma}(s)-\gamma(0^-))-\sigma(0^{-})\right)\\ &=s^2\hat{J}(s)\hat{\gamma}(s)\hat{G}(s)  \end{align}$$

Here we let $\hat{f}(s):=\mathcal{L}\lbrace f(t)\rbrace(s)$ be the Laplace transformation and since the stress starts at time $0$, $\gamma(0^-)$ and $\sigma(0^-)$ are simply $0$. The 2nd to 3rd step is derived from the convolution relation $\sigma(t)=\int_0^t G(t-t^\prime)\dot{\gamma}(t^\prime)\mathrm{d}t^\prime$. Cancelling out the $\hat{\gamma}(s)$ and rearranging the last equation give

$$\begin{equation} \frac{1}{s^2}=\hat{J}(s)\hat{G}(s) \end{equation}$$

The inverse Laplace of above equation gives the convolution relation

$$\begin{equation} t=\int_0^{t}J(t-t^\prime)G(t^\prime)\mathrm{d}t^\prime \end{equation}$$

The last step is substituting $J(t)=J_e + t/\eta$ into above convolution relation and making $t\to\infty$: 

$$\begin{equation} \begin{aligned} \color{blue}{t} &=\int_0^{t}\left(J_e+\frac{t-t^\prime}{\eta}\right)G(t^\prime)\mathrm{d}t^\prime\\ &= J_e\int_0^\infty G(t^\prime)\mathrm{d}t^\prime + {\color{blue}{\frac{t}{\eta}\int_0^\infty G(t^\prime)\mathrm{d}t^\prime}} - \int_0^\infty G(t^\prime)\frac{t^\prime}{\eta}\mathrm{d}t^\prime \end{aligned} \end{equation}$$

Since $\eta:=\int_0^\infty G(t^\prime)\mathrm{d}t^\prime$, the blue terms cancelled out, which gives \begin{equation} J_e\int_0^\infty G(t^\prime)\mathrm{d}t^\prime = J_e\eta= \int_0^\infty G(t^\prime)\frac{t^\prime}{\eta}\mathrm{d}t^\prime = \frac{1}{\eta} \int_0^\infty G(t^\prime)t^\prime\mathrm{d}t^\prime \end{equation} Therefore we have 

$$J_e = \frac{1}{\eta^2}\int_0^\infty tG(t)\mathrm{d}t$$


Saddle point method and end-to-end distribution of polymer models

Saddle point methods are widely used in estimation of integrals with form
$$ I = \int \exp\left(-f(x)\right)\mathrm{d}x $$
where function $f$ can be approximated by first 2 terms of its Taylor series around some $x_0$, i.e.
$$ f(x)\approx f(x_0) + f^\prime (x_0)(x-x_0) + \frac{1}{2}f^{\prime\prime}(x_0)(x-x_0)^2 $$
The integral is thus approximated by its saddle point, where $f^\prime (x_0)=0$ and $f^{\prime\prime}(x_0)>0$:
$$ \begin{align} I&\approx \int \exp\left(-f(x_0) - \frac{1}{2}f^{\prime\prime}(x_0)(x-x_0)^2\right) \mathrm{d}x\\ &=\exp(-f(x_0))\sqrt{\frac{2\pi}{f^{\prime\prime}(x_0)}} \end{align}$$

  • Stirling's formula:
With the knowledge of $\Gamma$ function we know that
$$N!=\int_0^\infty \exp(-x)x^N\mathrm{d}x$$
let $f(x):=x-N\ln(x)$, with large $N$, the negative part is negligible, solving $f^\prime (x) = 0$, we have:
  • Partition function:

$$Z = \int \exp(-\beta U(\mathbf{x})) \mathrm{d}\mathbf{x}$$
$$U(\mathbf{x})\approx U(\mathbf{x}_0) + \frac{1}{2}(\mathbf{x}-\mathbf{x}_0)^T H[U](\mathbf{x}-\mathbf{x}_0)$$
where $H$ represents hessian matrix.

End-to-end distribution function of random walk model of polymer chains
For an $N$-step random walk model, the exact end-to-end vector distribution is
$$ \begin{align} P(\mathbf{Y})&=\frac{1}{(2\pi)^3}\int \mathrm{d}\mathbf{k} \exp(-i\mathbf{k}\cdot\mathbf{Y})\tilde{\phi}^N\\ &=\int_{0}^{\infty} k\sin(kY) \left(\frac{\sin(kb)}{kb}\right)^N \mathrm{d}k \end{align} $$
with $\phi(\mathbf{x})=\frac{1}{4\pi b^2}\delta(|\mathbf{x}|-b)$ is the distribution of one step vector (length=$b$) and $\tilde{\phi}$ is the characteristic function of $\phi$; $\mathbf{Y}:=\sum_{i=1}^N \mathbf{x}_i$ is the end-to-end vector. Let $s=kb$ and $f(s):=i\frac{Y}{Nb}s -\ln\frac{\sin(s)}{s}$ then we have:
$$ \begin{align} P&=\frac{i}{4\pi^2 b^2 Y}\int_{-\infty}^{+\infty} s \exp\left(-is\frac{Y}{b}\right)\left(\frac{\sin(s)}{s}\right)^N\mathrm{d}s\\ &=\frac{i}{4\pi^2 b^2 Y} \int s \exp(-Nf(s)) \mathrm{d} s\end{align}$$
in this step, the integral is extened to $(-\infty, +\infty)$ due to the symmetry of sin/cos function: the first sin function is replaced with form of $\exp(-ix)$ by Eular's equation: $\exp(ix)=\cos(x)+i\sin(x)$.

Solving for $f^\prime(s)=0$, one could find that $is$ satisfies
i.e. the Langevin function, $is_0=L^{-1}(\frac{Y}{Nb})$. We therefore have:
$$\begin{align} P &\approx\frac{s_0}{4\pi^2b^2Y}\sqrt{\frac{2\pi}{Nf^{\prime\prime}(s_0)}}e^{-Nf(s_0)}\\ &=\frac{1}{(2\pi{}Nb^2)^{3/2}}\frac{L^{-1}(x)^2}{x(1-(L^{-1}(x)\csc\left(L^{-1}(x)\right))^2)^{1/2}}\\&\times\left(\frac{\sinh\left(L^{-1}(x)\right)}{L^{-1}(x)\exp\left(xL^{-1}(x)\right)}\right)^N\end{align}$$
with $x:=\frac{Y}{Nb}$.

Fancy way to calculate polymer chain model characteristics via NumPy

NumPy is the fundamental package for scientific computing with Python. It contains among other things:
  • a powerful N-dimensional array object
  • sophisticated (broadcasting) functions
  • tools for integrating C/C++ and Fortran code
  • useful linear algebra, Fourier transform, and random number capabilities
Besides its obvious scientific uses, NumPy can also be used as an efficient multi-dimensional container of generic data. Arbitrary data-types can be defined. This allows NumPy to seamlessly and speedily integrate with a wide variety of databases.

The ndarray is also the basic data structure of JIT tools like Numba, which boosts python dramatically. Recently, I wrote a program about calculation of the gyration tensor, center of mass and end to end vectors of polymer chain models under periodic boundary condition, and I found that using NumPy and Numba would finish the task in very few lines of codes, and also applicable to various situations. Here is the code:
def pbc(r, d):
    return r - d * np.round(r / d)

def rgTensor(x, mass, box):
    bond_vecs = pbc(np.diff(x, axis=-2, prepend=x[..., :1, :]),
    ree = bond_vecs[...,-1,:]
    cm = np.sum(bond_vecs * np.expand_dims(mass, axis=-1), axis=-2) / mass.sum(
        axis=-1, keepdims=True)
    bond_vecs -= np.expand_dims(cm, -2)
    cm = pbc(x[..., 0, :] + cm, box)
    return np.einsum('...pi,...pj->...ij', bond_vecs,
                     bond_vecs) / bond_vecs.shape[-2], cm, ree
This simple code could handle lots of situations, for example, suppose that we have N trajectories, each trajectory contains M time frames, and C chains in each time frame with L be the chain length, in D dimensional space, then we have an array of shape (N, M, C, L, D). Due to PBC is applied, we also need information of box, if NVT simulation is performed, then box would be, generally, a 1-D NumPy array of shape (D,). For NPT simulations, the box should be given in (N, M, D) means that for N trajectories and M time frames for each. The broadcasting property of NumPy array tells us if we want function pbc work, we need box be, for example, either (D,) or (N, M, 1, 1, D), otherwise, we cannot broadcast through the chain length and chain count dimension. Therefore, we need to expand dimensions to box if we perform NPT simulations, by calling box = np.expand_dims(box, axis=(-2, -3)), this is because that if we have different box for each frame, then except for the last axis (D) of box, the other axes must exactly match number of trajectories, number of frames..., etc. The box information is only irrelevant to chain information, which contains 2 dimensions (C, L), therefore, expand box twice from the last axis transforms the shape of box from (..., D) to (..., 1, 1, D) to align with the sample, whose shape is (..., C, L, D). The algorithm of calculation of gyration tensor is straightforward, 1. we unwrap the chain from pbc box; 2. set center of mass of the chain to 0; 3. the gyration tensor is a (D, D) matrix which is the product of Cartesian coordinate matrix, gyration tensor $T=R^TR/L$ for $R$ is $(L,D)$ the Cartesian coordinate matrix that represents $D$ dimensional space and $L$ coordinates. In the 1st step, we simply assume that no bond vector contains components larger than corresponding box components. Therefore, using function pbc(bond_vector, box) will unwrap bond vectors. After we unwrap all bond vectors, a cumulation sum is performed on bond vectors, we then have an unwrapped chain with its 1st monomer at $\mathbf{0}$. Now the center of mass of unwrapped chain is simply the mean of coordinates. After removing the center of mass, we calculate the gyration tensor by using the matrix production mentioned above. In the above code, let x be the coordinates, which has shape of (..., C, L, D), the 2nd (axis=-2) axis from last is always the chain length axis. The bond vectors are the difference of the positions of chain monomers, generally, the difference of an n-array is (n-1)-array due to the boundary condition, here we use prepend option to set the left boundary of the array with shape (..., C, 1, D) which equals to the position of 1st monomer of each chain, :1 used here to keep the dimension, if we set prepned=x[..., 0, :], an error will be raised, for x[..., 0, :] has shape of (..., C, D), which is 1-dimensional lesser than x. After the difference operation, the bond vectors are obtained, for each chain, the "1st" bond is always $\mathbf{0}$, we now could call pbc function, as mentioned above, pbc function takes bond vectors (..., C, L, D) and boxes (..., 1, 1, D) or just (D,), after pbc function being called, a cumulation sum is called along axis -2, which is the chain length axis, bond vectors on each chain were cumulatively summed, with 1st coordinate be $\mathbf{0}$ and last, i.e. bond_vecs[..., -1, :] be the end-to-end vector. The center of mass is just the weighted mean of mass of each monomer; mass, which is generally a 1-D array with shape of chain length (L,), we need to extend its shape to (L, 1) to broadcast to the -2 nd L-axis. After removing the center of mass, we calculate the matrix product of (..., D, L),(..., L, D)->(...,D, D), this part is a little bit tricky, we could use np.matmul which is essentially np.einsum: the matmul function will automatically perform on last 2 axis, for example, for an array A consists of N matrices of (a, b), and B consists of N matrices of (b, c), np.matmul(A, B) will be a new (N, a, c) array with each (a, c) matrix be the corresponding (a, b) and (b, c) matrices, i.e., $A_i\cdot B_i=C_i$ for $i=0,...,N-1$. Therefore, using np.matmul, it will be np.matmul(np.swapaxes(bond_vecs, -2, -1), bond_vecs), which is $T=R^TR$. Using np.einsum will bypass the transpose, we just summing up along the L-axis ---- in the code, it's the duplicated indices p. This program will cover almost all cases as long as the chain number, chain length and spatial dimension are the last 3 axis of data. This code could be further optimized by Numba, Numba allows one to define generalized ufuncs by @guvectorize decorator of Numba, e.g.,
from numba import float64
from numba import guvectorize
@guvectorize([(float64[:, :], float64[:, :], float64[:, :])],
             '(n,p),(p,m)->(n,m)', target='parallel')  # target='cpu','gpu'
def batch_dot(a, b, ret):
    for i in range(ret.shape[0]):
        for j in range(ret.shape[1]):
            tmp = 0.
            for k in range(a.shape[1]):
                tmp += a[i, k] * b[k, j]
            ret[i, j] = tmp
and simply replace the np.einsum... with batch_dot(np.swapaxes(bond_vecs, -2, -1), bond_vecs). There would be ~20x faster using batch_dot than np.einsum.

Not SQ again...

Structure factor (SQ) is a characterization quantity which is frequently calculated in molecular simulations. It is easily to code the calculation program according to the definition. In most molecular simulations, periodic boundary condition is adopted therefore
is actually a circular convolution, where FFT can give a dramatic boost in contrast to calculate directly. The steps using FFT are:
  1. Calculate $\rho(\mathbf{r})$, which is a summation of Dirac-Delta function that can be estimated as a hisotgram;
  2. Calculate $\hat{\rho}(\mathbf{q})$ by FFT;
  3. $S(\mathbf{q})=|\hat{\rho}(\mathbf{q})|^2$;
  4. Calculate mean over modulus: $S(q)=\int S(\mathbf{q})\delta(|\mathbf{q}|-q)\mathrm{d}\mathbf{q}/\int \delta(|\mathbf{q}|-q)\mathrm{d}\mathbf{q}$
If the simulation box is divided into $N$ (in 3D systems, $N=N_xN_yN_z$ for example) bins, the FFT gives $O(N\log(N))$ complexity and step 4 is $O(N)$. Generally, in comparison with the direct method, one needs at least loop over number of particles and $N$ bins for $\mathbf{q}$, the number of particles is obviously, way larger than $\log(N)$. For data of most simulations are real numbers, rFFT designed for real inputs could further boost the program. Here is code for a 3D example, i, j, k represent $N_x$, $N_y$, $N_z$ respectively.
fftn(a) == np.concatenate([rfftn(a), conj(rfftn(a))[-np.arange(i),-np.arange(j),np.arange(k-k//2-1,0,-1)]], axis=-1)

  1. Binsize effect.
  2. The binsize in histogram should be smaller than half of the minimun of the pair distances in the system according to Nyquist-Shannon sampling theorem. For example, in Lennard-Jones systems, $0.4\sigma$ is a good choice. However, if values of only some small $q$ are concerned, i.e., in some self-assembly systems, there is no need to calculate in the "pair accuracy", the sampling distance smaller than half of the interested domain size is fine. 
  3. Randomness in the bin
  4. Particles in bins are randomly distributed, especially for large binsize mentioned above. The summation $\sum_i \exp(-\mathbb{I}\mathbf{q}\cdot\mathbf{r}_i)$ can be decomposed into $$\sum_\mathrm{bin} \exp(-\mathbb{I}\mathbf{q}\cdot\mathbf{r}_\mathrm{bin})\left(\sum_i \exp( -\mathbb{I}\mathbf{q}\cdot\delta\mathbf{r}_i)\right)$$ The idea is straightforward: for particles inside a bin, i.e., for particle $i$,  position $\mathbf{r}_i$ can be decomposed as $\mathbf{r}_i=\mathbf{r}_\mathrm{bin}+\delta\mathbf{r}_i$, and for particles in the bin, $\delta\mathbf{r}$ is assumed randomly distributed in the bin. The latter summation in the decomposition is thus represented as $n_\mathrm{bin}\overline{\exp(-\mathbb{I}\mathbf{q}\cdot\delta\mathbf{r})}$, the average is approximated according to the distribution of $\delta\mathbf{r}$. If we consider $\delta r$ (1D case, for example) uniformly distributed in $(-L/2, L/2)$ with $L$ be the binsize, the average is $\mathrm{sinc}(qL/2)$. Multiply the sinc function during step 4, the $S(q)$ will be corrected.

Free energy calculation: umbrella integration

Formulae of umbrella sampling method can be found on wikipedia. In umbrella sampling, a reaction coordinate $\xi$ is pre-defined from the atomic coordinates, a bias potential is added to the atoms of interest to keep $\xi$ of the system at a specific window $\xi_w$. The bias form is usually a harmonic potential:
Therefore, the energy of biased system $A^b_w = A^{ub}_w + u^b_w(\xi)$. The superscript $ub$ is short for "un-biased". In the simulations, we can sample the reaction coordinate in each window and evaluate their distribution $P^b(\xi)$, since the free energy $A=-k_BT\ln(P)$, we have:
$$A_w^{ub}(\xi) = -k_BT\ln(P^b_w(\xi))-u^b_w(\xi)-F_w$$
with $F_w$ is a reference free energy of each window and remains an unknown constant. One method to derive $F_w$ is WHAM, in year 2005, Kästner et al. (The Journal of Chemical Physics 123, no. 14 (October 8, 2005): 144104.) have proposed a new method whose idea is to take derivative of $A^u_w$ with respect to $\xi$ to eliminate $F_w$, In this method, $P(\xi)$ is assumed to be analytic, e.g., a Gaussian distribution. The general idea is to expand $A(\xi)$ into Taylor series (at $\langle \xi \rangle_w$):
$$ A(\xi)=a_1 \xi + a_2 \xi^2 + a_3 \xi^3 +\ldots$$
if $a_i=0$ with $i\ge 3$, the Gaussian form is restored. For systems with higher order terms, the distributions, for example, generally have a non-zero skewness. The crucial step is to determine $\lbrace a_i\rbrace$. In practice, the probability with form of $\exp(-\sum_i a_i \xi^i)$ is difficult to determine, in year 2012, Kästner et al. (The Journal of Chemical Physics 136, no. 23 (June 21, 2012): 234102) studied cases with order $4$ and small $a_3, a_4$. I have tried another more "numerical" method to calculate $\lbrace a_i\rbrace$ when datasets are poorer: fit $\exp(-\sum_i a_i \xi ^i)$ with a Gaussian KDE to find $\lbrace a_i \rbrace$. For some "poor" datasets, higher-order terms of expansion of free energy are required. The normalization factor of the distribution is estimated as $n=\int_{\xi_\mathrm{min}}^{\xi_\mathrm{max}} P(\xi)\mathrm{d}\xi$, the fitting range should be carefully chosen to ensure the convergence of $n$.
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