Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 71 / 5

The numerical contour length is $L=N{b}_0=10^{10}\times 10^{-8}\mathrm{m}=100\mathrm{m}$, and $k_{B}T=4.14\times 10^{-21}\mathrm{J}$. Begin with a three-dimensional freely jointed chain in which each chemical monomer is also an independent statistical link of length $b_0$. This neglects excluded volume, hydrodynamic effects on equilibrium statistics, and stretching of the chemical backbone.

An applied force $F$ gives a one-link canonical partition functionThe mean extension is $z=k_{B}T{\partial }_F\log Z_1^N=N{b}_0(\coth \xi -1/\xi )$. Expanding the Langevin function at small force gives $z=N{b}_0^{2}F/(3k_{B}T)+O(F^3)$. The three-dimensional entropic chain stiffness is thusThe unforced root-mean-square end-to-end distance is $\sqrt{N}{b}_0=10^{-3}\mathrm{m}$, not the contour length. This response is entropic straightening of a fluctuating coil. Axial stretching of a straight material rod instead has stiffness $EA/L$, and is a different phenomenon. Aqueous solvent does not by itself prove ideal-chain statistics: a good-solvent self-avoiding walk has a different coil size and susceptibility. The estimate above states explicitly the model under which the supplied link data suffice.

For a three-dimensional worm-like chain, the tangent correlation length is $l_p=B/(k_{B}T)$, where $B$ is the filament bending modulus. The measured persistence length therefore givesRepresent the actin filament as an isotropic solid circular cylinder with diameter $d=10^{-9}\mathrm{m}$. Its second moment of area is $I=\pi d^4/64=4.909\times 10^{-38}{\mathrm{m}}^4$. Since $B=EI$,This is an effective Young's modulus within the given continuum cross-sectional model; the numerical diameter and persistence length must not be replaced by unrelated typical biological values.

The same measurement corrects the independent-link approximation: the Kuhn length is $b_K=2l_p=40\mathrm{nm}$, with $N_K=L/b_K=2.5\times 10^9$. Thus the long-chain entropic stiffness is $3k_{B}T/(2l_{p}L)=3.105\times 10^{-15}\mathrm{N}{\mathrm{m}}^{-1}$, and the root-mean-square end-to-end distance is $2\mathrm{mm}$. For comparison, the material axial stiffness is $EA/L=1.325\times 10^{-11}\mathrm{N}{\mathrm{m}}^{-1}$, using $A=\pi d^2/4$. The corrected coil response remains much softer than axial backbone stretching.

An optical tweezers force requires the change in optical momentum. Define the optical momentum efficiency $Q$ by $F=QP/c$, with all ray geometry and medium-index factors absorbed in $Q$ and $c$ the vacuum light speed. The specified relative refractive index does not determine $Q$ without bead size and the beam's convergence, waist, and ray directions. A unique exact force is therefore not supplied by these data.

Retaining the entropic stiffness from part (a), the parameterized answer isUsing the persistence-length-corrected worm-like chain instead gives $z=1.074Q\mathrm{m}$ in its small-force regime.

For a definite scale estimate, take the weak-contrast optical momentum change to be $Q\sim m_r-1=0.2$. This gives $F\sim 6.67\times 10^{-16}\mathrm{N}$, $z\sim 5.37\mathrm{cm}$ for part (a)'s model, or $z\sim 21.5\mathrm{cm}$ for the corrected chain. These are estimates with an explicitly chosen ray-momentum efficiency, not a universal formula for a transparent trapped sphere.

Another illustrative optical model makes the uncertainty visible: normal-incidence Fresnel reflectance is $\mathcal{R}=[(m_r-1)/(m_r+1){]}^2=1/121$. An effective reflector with $Q=2\mathcal{R}$ in the simple $P/c$ momentum convention gives $F=5.51\times 10^{-17}\mathrm{N}$ and extensions $4.44\mathrm{mm}$ or $17.7\mathrm{mm}$, respectively. A refracting bead in a focused beam is not just that reflector; entry and exit rays also transfer momentum. Neither estimate eliminates the missing optical geometry.

The extension must be compared with the actual contour length, $L=100\mathrm{m}$. For the illustrative $Q=0.2$ force and the persistence-length-corrected chain, $z/L\simeq 2.15\times 10^{-3}$ and $F{b}_K/(k_{B}T)\simeq 6.44\times 10^{-3}$. Both are small, so the small-force Langevin function expansion and entropic-extension approximation are consistent. Although the mean extension greatly exceeds the unforced millimetre-scale coil size, it remains far below the maximum contour length; that does not violate the model.

The material axial elongation at this force would be only $F/(EA/L)\simeq 5.0\times 10^{-5}\mathrm{m}$, illustrating that most of the predicted coil extension is orientation and straightening rather than stretching the backbone. Thus the order-of-magnitude entropic response is internally reasonable within the ideal-chain assumptions, but its numerical optical-force prediction is not unique. Solvent-dependent excluded volume, the unusually large stated contour length, actual trap geometry, and surface interactions all matter to an experimental interpretation.

The viscosity converts to $\eta =0.01\mathrm{Pa}\mathrm{s}$, and the translation speed is $U=10^{-8}\mathrm{m}{\mathrm{s}}^{-1}$. Mechanical power is $P_{\mathrm{motor}}={\zeta }_{\mathrm{chain}}{U}^2$. The hydrodynamic drag models for a polymer show why a conformation assumption is necessary.

For a largely aligned slender filament translating along its axis in unbounded fluid, useWith $L=100\mathrm{m}$, $\log (2L/a)=\log (4\times 10^{11})=26.715$, so ${\zeta }_{\parallel }\simeq 0.240\mathrm{kg}{\mathrm{s}}^{-1}$. The required force is about $2.4\times 10^{-9}\mathrm{N}$, andTransverse translation has roughly twice the leading logarithmic drag. This estimate treats the entire contour as an aligned slender body, assumes creeping flow, and omits bead drag, wall corrections, and additional motor losses. It is not a drag formula for an unstretched coil.

For an alternative nondraining coil, approximate the hydrodynamic radius by its radius of gyration, $R_h\sim R_g\simeq \sqrt{L{l}_p/3}=8.16\times 10^{-4}\mathrm{m}$. The Stokes drag law gives $\zeta \sim 6\pi \eta R_h=1.54\times 10^{-4}\mathrm{kg}{\mathrm{s}}^{-1}$ and $P\sim 1.54\times 10^{-20}\mathrm{W}$. A freely draining toy model that sums $N$ spherical-monomer drags of radius $a$ instead gives $P=N6\pi \eta a{U}^2\simeq 9.42\times 10^{-17}\mathrm{W}$. Hydrodynamic interactions are suppressed in this last model. These distinct answers explain the need to specify conformation and hydrodynamic coupling; the stated data do not select a unique one.