In the small-slope approximation the <worm-like chain> bending energy is
$$
E[h]=\frac A2\int_0^L h_{xx}^2\,dx.
$$
The clamped end obeys $h(0)=h_x(0)=0$, while a force-free and torque-free tip has the natural conditions $h_{xx}(L)=h_{xxx}(L)=0$. A static transverse tip force $f$ produces the cantilever compliance $h(L)/f=L^3/(3A)$. The <equipartition theorem>, or equivalently the static fluctuation--response relation, therefore gives the exact tip variance
$$
\boxed{\langle h(L)^2\rangle=\frac{k_BT L^3}{3A}
=\frac{L^3}{3L_p}.}
$$

For the dynamics, <resistive-force theory> gives the transverse drag per length
$$
\zeta_\perp\simeq\frac{4\pi\mu}{\log(L/a)+1/2}
$$
at logarithmic accuracy and the stochastic beam equation
$$
\zeta_\perp\partial_t h=-A\partial_x^4h+f_T.
$$
Let $\phi_n$ be clamped--free bending modes with $q_nL=\alpha_n$, where
$$
\cos\alpha_n\cosh\alpha_n=-1,
\qquad \alpha_1\simeq1.875.
$$
Writing $N_n=\int_0^L\phi_n^2dx$, independent thermally driven modes give
$$
\boxed{\langle h(L,t)h(L,0)\rangle
=\sum_{n=1}^\infty
\frac{k_BT\phi_n(L)^2}{Aq_n^4N_n}
e^{-|t|/\tau_n},
\qquad
\tau_n=\frac{\zeta_\perp}{Aq_n^4}.}
$$
The first mode contains almost all of the tip variance, so
$$
\langle h(L,t)h(L,0)\rangle
\simeq\frac{k_BT L^3}{3A}e^{-|t|/\tau_1},
\qquad
\tau_1=\frac{\zeta_\perp L^4}{A\alpha_1^4}.
$$

For a microtubule, $A=k_BT L_p$. With $L=10\,\mu\mathrm m$ and $L_p=3\,\mathrm{mm}$,
$$
\boxed{\sqrt{\langle h(L)^2\rangle}
=\sqrt{\frac{L^3}{3L_p}}
\simeq0.33\,\mu\mathrm m.}
$$
Taking water at room temperature and $a=0.5\,\mu\mathrm m$ gives $\zeta_\perp\simeq3.6\times10^{-3}\,\mathrm{Pa\,s}$ and
$$
\boxed{\tau_1\simeq0.24\,\mathrm s,}
$$
with an order-one uncertainty from the logarithmic slender-body drag approximation.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 355 / 2 / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-355/2/solution