# 物理代写|流体力学代写Fluid Mechanics代考|FY828

## 物理代写|流体力学代写Fluid Mechanics代考|The Free Energy of Polymer Translocation

Consider a flexible polymer of $N$-segments during translocation through a narrow pore in a membrane with $n$ segments on the tran-side and $N-n$ on the cis-side as shown in then Fig. 10.5a. What is the free energy $\mathcal{F}(n ; N)$ required to maintain the chain in this configuration, namely, the free energy of the translocation?

Focusing the chain on the trans side, we consider that $n$ is its primary degree of freedom; integrating over all accessible configurations given $n$ yields the chain free energy $\mathcal{F}(n)$ :
$$e^{-\beta \mathcal{F}(n)}=\sum_{\boldsymbol{r}} e^{-\beta \mathcal{F}(\boldsymbol{r} / n)}=\int_{x>0} d \boldsymbol{r} P(\boldsymbol{r} ; n),$$
where $P(\boldsymbol{r} ; n)$ is given by (10.28) and the integration is performed over the position in the half space where $x>0$. The integration yields
$$\int_o^{\infty} d x\left(2 \pi n l^2 / 3\right)^{-1 / 2} \frac{6 x \epsilon}{n l^2} \exp \left(-\frac{3 x^2}{2 n l^2}\right) \sim n^{-1 / 2} \int_0^{\infty} d \tilde{x} \tilde{x} e^{-3 \tilde{x}^2 / 2} \sim n^{-1 / 2}$$
where $\tilde{x}=\left(n l^2\right)^{-1 / 2} x$ is a dimensionless coordinate. Consequently, the free energy given by $(10.31)$ is
$$\mathcal{F}(n)=\frac{1}{2} k_B T \ln n$$
apart from a irrelevant constant. Considering the chain in the cis side is anchored also at the origin, its free energy is obtained similarly as $\mathcal{F}(N-n)=\frac{1}{2} k_B T \ln (N-n) .$

## 物理代写|流体力学代写Fluid Mechanics代考|A Flexible Chain Under External Fields

In many situations a polymer is subject to external forces, confinements, and intra-chain interactions. An important problem is to find the chain conformations and thermodynamic behaviors under such conditions. Due to the chain connectivity a polymer under such constraints manifests many interesting entropic behaviors that are not seen in ordinary particle systems.

We consider an approximation in which each segment can be treated as if it is under an effective external potential, called the self-consistent field, similar to the mean field in the Debye-Hückel theory. A central object to find first is the polymer Green’s function, $G\left(\boldsymbol{r}, \boldsymbol{r}^{\prime} ; N\right)$, which is the probability density of finding the chain end at the distance $\boldsymbol{r}$ given the initial segment position at $r^{\prime}$. We consider the $N$-step random walk with each step influenced by an effective external potential energy $u$. Given the probability $G\left(\boldsymbol{r}{N-1}, \boldsymbol{r}^{\prime} ; N-1\right)$ for the $N-1$ th step to be at $\boldsymbol{r}{N-1}$, it can jump to $\boldsymbol{r}$ at the $N$ th step with the probability density $p\left(\boldsymbol{r}-\boldsymbol{r}{N-1}\right)$, satisfying the recurrence relation, $$G\left(\boldsymbol{r}, \boldsymbol{r}^{\prime} ; N\right)=\int d \boldsymbol{r}{N-1} e^{-\beta u(\boldsymbol{r})} p\left(\boldsymbol{r}-\boldsymbol{r}{N-1}\right) G\left(\boldsymbol{r}{N-1}, \boldsymbol{r}^{\prime} ; N-1\right),$$
where the $u(\boldsymbol{r})$ is the effective potential energy at $\boldsymbol{r}$.

## 物理代写|流体力学代写Fluid Mechanics代考|The Free Energy of Polymer Translocation

$$e^{-\beta \mathcal{F}(n)}=\sum_r e^{-\beta \mathcal{F}(r / n)}=\int_{x>0} d \boldsymbol{r} P(r ; n),$$

$$\int_o^{\infty} d x\left(2 \pi n l^2 / 3\right)^{-1 / 2} \frac{6 x \epsilon}{n l^2} \exp \left(-\frac{3 x^2}{2 n l^2}\right) \sim n^{-1 / 2} \int_0^{\infty} d \tilde{x} \tilde{x} e^{-3 \tilde{x}^2 / 2} \sim n^{-1 / 2}$$

$$\mathcal{F}(n)=\frac{1}{2} k_B T \ln n$$

## 物理代写|流体力学代写Fluid Mechanics代考|A Flexible Chain Under External Fields

$$G\left(\boldsymbol{r}, \boldsymbol{r}^{\prime} ; N\right)=\int d \boldsymbol{r} N-1 e^{-\beta u(\boldsymbol{r})} p(\boldsymbol{r}-\boldsymbol{r} N-1) G\left(\boldsymbol{r} N-1, \boldsymbol{r}^{\prime} ; N-1\right),$$

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