# 物理代写|统计物理代写Statistical Physics of Matter代考|Transverse Dynamics

## 物理代写|统计物理代写Statistical Physics of Matter代考|Transverse Dynamics

The effective Hamiltonian can also be expressed as $\mathcal{F}=-\int d s \hat{\lambda}(s) \cdot \boldsymbol{h}(s)$, where the transverse force induced on the segment $s$ is obtained by (11.50):
$$\lambda(s)=-\frac{\delta \mathcal{F}}{\delta \boldsymbol{h}(s)}=f\left(\frac{\partial^2 \boldsymbol{h}}{\partial s^2}\right)-\kappa\left(\frac{\partial^4 \boldsymbol{h}}{\partial s^4}\right) .$$
Via the hydrodynamic interaction, this force at a segment $s^{\prime}$ affects the fluid velocity at another segment at $s$ (Fig. 20.4), which is same as the average undulation velocity therein, $\partial \boldsymbol{h}(s, t) / \partial t$, by the boundary condition. Thus we have
$$\frac{\partial \boldsymbol{h}(s, t)}{\partial t}=\int d s^{\prime} \boldsymbol{\Lambda}\left(s-s^{\prime}\right) \cdot \lambda\left(s^{\prime}, t\right)+\boldsymbol{\xi}{\boldsymbol{h}}(t) .$$ where $\xi_h(t)$ is the appropriate Gaussian white noise. For a nearly straight chain, we restrict the above equation to be linear in $\boldsymbol{h}$, and the Oseen tensor $\boldsymbol{\Lambda}\left(s-s^{\prime}\right)$ can be simplified to: $$\boldsymbol{\Lambda}\left(s-s^{\prime}\right)=\frac{1}{8 \pi \eta\left|s-s^{\prime}\right|}(I+\widehat{\boldsymbol{x}} \widehat{\boldsymbol{x}})$$ Because $\lambda$ is vertical to $\hat{\boldsymbol{x}}$, (20.67) is expressed as \begin{aligned} \frac{\partial \boldsymbol{h}(s, t)}{\partial t} & =-\frac{1}{8 \pi \eta} \int d s^{\prime} \frac{1}{\left|s-s^{\prime}\right|} \frac{\delta \mathcal{F}}{\delta \boldsymbol{h}\left(s^{\prime}, t\right)}+\boldsymbol{\xi}{\boldsymbol{h}}(t) \ & =\frac{1}{8 \pi \eta} \int d s^{\prime} \frac{1}{\left|s-s^{\prime}\right|}\left{f\left(\frac{\partial^2 \boldsymbol{h}}{\partial s^{\prime 2}}\right)-\kappa\left(\frac{\partial^4 \boldsymbol{h}}{\partial s^{\prime 4}}\right)\right}+\boldsymbol{\xi}_{\boldsymbol{h}}(t) . \end{aligned}
To facilitate solving for $\boldsymbol{h}(s, t)$ we apply the one-dimensional Fourier transform $$\boldsymbol{h}(q, t)=\int_0^L d s e^{-i q \cdot s} \boldsymbol{h}(s, t)$$
and obtain the equation of motion:
\begin{aligned} \frac{\partial \boldsymbol{h}(q, t)}{\partial t} & =-\Lambda_h(q)\left{f q^2+\kappa q^4\right} \boldsymbol{h}(q, t)+\xi_{\boldsymbol{h}}(q, t) \ & =-\tau_T(q)^{-1} \boldsymbol{h}(q, t)+\xi_{\boldsymbol{h}}(q, t) \end{aligned}
where
$$\tau_T(q)=\frac{1}{\left(\kappa q^4+f q^2\right) \Lambda_h(q)}$$

## 物理代写|统计物理代写Statistical Physics of Matter代考|Chain Longitudinal Dynamics

Now we turn our attention to dynamics of the longitudinal length $\mathcal{X}(t)$ by starting with its equilibrium time correlation function $\langle\Delta \mathcal{X}(t) \Delta \mathcal{X}(0)\rangle_0$. To this end, it is more straightforward to use the linear response theory in which the time correlation is directly related by the response $\langle\Delta \mathcal{X}(t)\rangle$ to a small time-dependent tension as was done for the static case in Chap. 11. We consider that the time-dependent tension, $\delta f(t)$, is applied at an chain end additionally beginning from $t=0$.

The mean equilibrium end-to-end distance, $X=\langle\mathcal{X}\rangle$ was given by (11.32) in terms of the transverse undulation. The corresponding quantity at a time $t,\langle\mathcal{X}(t)\rangle$, is given by
$$\frac{\langle\mathcal{X}(t)\rangle}{L}=1-\frac{1}{2 L^2} \sum_q q^2\langle\boldsymbol{h}(q, t) \cdot \boldsymbol{h}(-q, t)\rangle$$
Solving the Langevin equation (20.71) for $\boldsymbol{h}(q, t)$ with $f=f_0+\delta f(t)$ to the linear order in $\delta f(t)$, we can obtain
$$\langle\boldsymbol{h}(q, t) \cdot \boldsymbol{h}(-q, t)\rangle=\langle\boldsymbol{h}(q, t) \cdot \boldsymbol{h}(-q, t)\rangle_0-\int_0^t d t^{\prime} m\left(q, t-t^{\prime}\right) \delta f\left(t^{\prime}\right),$$
where $$m\left(q, t-t^{\prime}\right)=\frac{4 k_B T L A_h(q)}{\left(\kappa q^2+f_0\right)} e^{-\frac{t-1}{\tau_L(q)}} .$$
Here we define the longitudinal relaxation time by
$$\tau_L(q)=\frac{\tau_T(q)}{2}=\frac{1}{2}\left(\kappa q^4+f_0 q^2\right) \Lambda_h(q) .$$

# 统计物理代考

## 物理代写|统计物理代写Statistical Physics of Matter代考|Transverse Dynamics

$$\lambda(s)=-\frac{\delta \mathcal{F}}{\delta \boldsymbol{h}(s)}=f\left(\frac{\partial^2 h}{\partial s^2}\right)-\kappa\left(\frac{\partial^4 h}{\partial s^4}\right)$$

$$\frac{\partial \boldsymbol{h}(s, t)}{\partial t}=\int d s^{\prime} \boldsymbol{\Lambda}\left(s-s^{\prime}\right) \cdot \lambda\left(s^{\prime}, t\right)+\boldsymbol{\xi} \boldsymbol{h}(t) .$$

$$\boldsymbol{\Lambda}\left(s-s^{\prime}\right)=\frac{1}{8 \pi \eta\left|s-s^{\prime}\right|}(I+\widehat{\boldsymbol{x}} \widehat{\boldsymbol{x}})$$

$\langle$ begin{aligned} $}$ frac{へartial $\backslash$ boldsymbol{h}$}(s, t)} \backslash$ partial t $} \&=-\mid$ frac ${1}{8 \backslash p i \backslash$ leta} $}$ int $d s^{\wedge}{$ prim

$$\boldsymbol{h}(q, t)=\int_0^L d s e^{-i q \cdot s} \boldsymbol{h}(s, t)$$

$\backslash$ begin{aligned $} \backslash$ frac $\backslash$ partial $\backslash$ boldsymbol ${h}(q, t)} \backslash$ partial $t} \&=-\backslash$

$$\tau_T(q)=\frac{1}{\left(\kappa q^4+f q^2\right) \Lambda_h(q)}$$

## 物理代写|统计物理代写Statistical Physics of Matter代考|Chain Longitudinal Dynamics

$\langle\Delta \mathcal{X}(t) \Delta \mathcal{X}(0)\rangle_0$. 为此，使用时间相关性与响应直接相关的线性响应理论更为直接
$\langle\Delta \mathcal{X}(t)\rangle$ 到一个小的时间相关的张力，就像在第 1 章的静态案例中所做的那样。11. 我 们认为时间依赖性紧张， $\delta f(t)$ ，应用于链端另外从 $t=0$.

$$\frac{\langle\mathcal{X}(t)\rangle}{L}=1-\frac{1}{2 L^2} \sum_q q^2\langle\boldsymbol{h}(q, t) \cdot \boldsymbol{h}(-q, t)\rangle$$

$$\langle\boldsymbol{h}(q, t) \cdot \boldsymbol{h}(-q, t)\rangle=\langle\boldsymbol{h}(q, t) \cdot \boldsymbol{h}(-q, t)\rangle_0-\int_0^t d t^{\prime} m\left(q, t-t^{\prime}\right) \delta f\left(t^{\prime}\right)$$

$$m\left(q, t-t^{\prime}\right)=\frac{4 k_B T L A_h(q)}{\left(\kappa q^2+f_0\right)} e^{-\frac{t-1}{\tau_L(q)}}$$

$$\tau_L(q)=\frac{\tau_T(q)}{2}=\frac{1}{2}\left(\kappa q^4+f_0 q^2\right) \Lambda_h(q)$$

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