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Independent link orientations make the canonical partition function $Z_1^N$, with the one-link integral defining the Langevin function. Differentiating its logarithm with respect to force gives $L=Nb\mathcal{L}(Fb/(k_{B}T))$. At high extension, $Nb-L\sim N{k}_{B}T/F$, so the force diverges inversely with remaining length deficit. This requires rigid links and excludes axial bond stretching.

The Langevin function is the mean projection of a freely orientable three-dimensional unit vector weighted by $e^{\xi \cos \theta }$. Integrating over solid angle gives $Z_1=4\pi \sinh \xi /\xi$, and differentiation gives $⟨\cos \theta ⟩={\partial }_{\xi }\log Z_1=\coth \xi -1/\xi$. It behaves as $\xi /3$ at small argument and $1-1/\xi$ at large positive argument.

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.

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.

Neglecting cell rotational inertia, the light and viscous torque sum to zero. Therefore the angular velocity is $\mathbf{\Omega }=\mathbf{\omega }/2+(\mathbf{p}\times \mathbf{L})/\alpha$. A material orientation evolves by $\dot{\mathbf{p}}=\mathbf{\Omega }\times \mathbf{p}$. The vector triple-product identity givesThe last term is the projection of $\mathbf{L}/\alpha$ onto the tangent plane to the unit sphere. In particular, $\mathbf{p}\cdot \dot{\mathbf{p}}=0$, as required to preserve the unit orientation. It aligns the swimmer towards the light; the vorticity rotates that orientation.

For rotational diffusion coefficient $D_r$, the phototactic orientation Fokker-Planck equation balances deterministic orientation drift with diffusion on the sphere. Its density is normalized with respect to solid angle, not $d\theta d\varphi$. When $\epsilon =0$, there is no azimuthal drift and the zero-current condition is $f_{0,\theta }=-\lambda \sin \theta f_0$. Thus $f_0=\eta e^{\lambda \cos \theta }$, and normalization givesThe limiting density at $\lambda =0$ is $1/(4\pi )$.

To justify the angular dependence of the first correction, define the operatorsThe steady Fokker-Planck equation is $\mathcal{A}f=-\epsilon \mathcal{R}f$. Its regular expansion $f=f_0+\epsilon f^{(1)}+O({\epsilon }^2)$ therefore givesThe coefficients of $\mathcal{A}$ are independent of $\varphi$, so each azimuthal Fourier harmonic is preserved. The forcing has only the sine harmonic with azimuthal number one. There can be no extra unforced harmonic in a smooth normalized first correction: writing a homogeneous solution as $f_{0}h$ gives $\mathcal{A}(f_{0}h)={\nabla }_p\cdot (f_0{\nabla }_{p}h)$. Multiplication by $h$ and integration by parts on the sphere gives $\int f_0|{\nabla }_{p}h{|}^{2}d\Omega =0$, hence $h$ is constant. Normalization $\int f^{(1)}d\Omega =0$ removes that multiple of $f_0$. It follows thatNo explicit calculation of $g$ is needed. Its dependence is on $\theta$ and the parameter $\lambda$, and its behavior at the poles must make the density smooth.

In the stated spherical coordinates, $\mathbf{p}=\cos \theta {\mathbf{e}}_x+\sin \theta \cos \varphi {\mathbf{e}}_y+\sin \theta \sin \varphi {\mathbf{e}}_z$. Azimuthal integration makes the leading $y,z$ components and the first-order $x,y$ corrections vanish. The surviving coefficients areThus the small-vorticity phototactic orientation response isBoth coefficients depend only on $\lambda$. In particular, $K_0$ is the Langevin function of $\lambda$ and is positive for positive light strength; its integral representation above suffices.

For $D_r>0$, the steady polar flux satisfies ${\partial }_{\theta }(\sin \theta j_{\theta })=0$. Regularity at both poles makes its integration constant zero. ThusNormalize with $x=\cos \theta$:The zero-flow steady gyrotactic orientation distribution is thereforeDifferentiating the logarithm of its normalization integral gives the Langevin function meanAs $\Lambda \to 0$, $f_0\to 1/(4\pi )$ and the mean tends to zero; as $\Lambda \to \infty$, the distribution concentrates at the upward pole and the mean tends to $V_s\mathbf{k}$. At zero diffusivity the pole-concentrated steady state is a probability measure, not a regular density of this formula.

Introduce a constant filament bending modulus $B$, which sets the mechanical force scale. The filament has intrinsic curvature of an elastic filament ${\kappa }_0(s)$, and its actual signed curvature in a small-slope Monge representation is $\kappa \simeq {\zeta }_{xx}$. To quadratic order one may identify material arc length $s$ with $x$ in ${\kappa }_0$, whileAt fixed positive extension force, the mechanical potential is bending energy minus $FL$. Replacing $L$ by $\mathcal{L}-(\mathcal{L}-L)$ leaves, up to a force-dependent constant,Its first variation isThe higher-order Euler-Lagrange equation is thereforeFor a Fourier transform relation, take a periodic bulk segment, or a long-filament bulk approximation with negligible endpoint contributions; these are necessary boundary conditions because the question specifies no filament end conditions. Using $f(x)=\int \hat{f}(q)e^{iqx}dq/(2\pi )$, every nonzero wavenumber obeysA zero displacement mode is an arbitrary translation. In a periodic representation a constant intrinsic-curvature mode contributes a constant bending cost but no bulk slope, so it does not contribute to the following deficit. The Parseval identity givesFor clarity about finite normalization, on a periodic interval of length $L$ define ${\kappa }_n=L^{-1}{\int }_0^L{\kappa }_0(x)e^{-i{q}_{n}x}dx$, $q_n=2\pi n/L$. Then the equivalent exact quadratic-order formula isThis is force-extension of an intrinsically curved filament, determined by its particular fixed shape rather than by a thermal orientation entropy.

For an ensemble with quenched intrinsic curvature, first minimize each realization and then average the last expression, replacing $|{\kappa }_n{|}^2$ by its ensemble mean. For stationary bulk disorder, let $S_{\kappa }(q)$ be the power spectrum, normalized so that its correlation is $⟨{\kappa }_0(x){\kappa }_0(x+r)⟩=\int S_{\kappa }(q)e^{iqr}dq/(2\pi )$. The measurable mean deficit per unit projected length isIf the curvature is smooth enough that the second spectral moment is finite, one can take the large-force limit under this integral, obtaining the smooth-curvature high-force extension lawFor a stationary ensemble the integral is $L⟨({\kappa }_{0,x}{)}^2⟩$. Equivalently, the bulk slope is ${\zeta }_x\simeq -(B/F){\kappa }_{0,x}$; a plot of the mean deficit against $F^{-2}$ measures the mean squared intrinsic-curvature gradient, if $B$ is known. The $F^{-2}$ law is a smooth-curvature bulk result, not a universal law for every random filament. For example, ideal white intrinsic-curvature disorder with $S_{\kappa }(q)=S_0$ has an infinite second spectral moment; retaining the full integral instead gives $⟨\mathcal{L}-L⟩/L=S_0\sqrt{B}/(8\sqrt{F})$ in that ideal continuum model.

Endpoint conditions also matter. If each end is moment-free and has no transverse applied force, the natural conditions from the variation areat each end. These can produce a free-end curvature boundary layer of width $\ell =\sqrt{B/F}$. For a constant intrinsic filament curvature $\kappa$, integration of the higher-order Euler-Lagrange equation with zero transverse force gives $B({\zeta }_x{)}_{xx}-F{\zeta }_x=0$ and henceThe last formula follows by integrating ${\sinh }^2[(x-L/2)/\ell ]$, and tends to ${\kappa }^2{\ell }^3/2\propto F^{-3/2}$, despite the vanishing bulk curvature gradient. Thus a finite free-ended measurement must include or suppress endpoint bending before interpreting a bulk $F^{-2}$ fit.

For the freely jointed chain, rigid links have independent orientations, so the restoring force is entropic. Let $T$ now denote absolute temperature, and $\beta =1/(k_{B}T)$ with Boltzmann constant $k_B$. In the usual three-dimensional chain, a segment at polar angle $\theta$ has force energy $-Fb\cos \theta$. Its angular canonical partition function and the full chain partition function areDifferentiating $\log Z_N$ with respect to $\beta F$ yields the exact mean projected displacementHere $\mathcal{L}$ is the Langevin function; this is force-extension of a three-dimensional freely jointed chain, in the fixed-force canonical ensemble. At high extension, $\coth \xi =1+2e^{-2\xi }+\cdots$, soThus the smooth quenched-curvature bulk deficit falls as $F^{-2}$, whereas the three-dimensional freely jointed chain deficit falls as $F^{-1}$ and depends explicitly on temperature. The distinction is mechanical straightening of a fixed intrinsic shape versus orientational entropy; thermal bending fluctuations of a stiff filament have not been included in the first calculation.

If the comparison chain is also restricted to two dimensions, it is a planar freely jointed chain rather than the three-dimensional model. In this case,The last identity follows directly by differentiating the integral defining $I_0$; $I_0$ and $I_1$ are Modified Bessel functions of the first kind. At large $\xi$ the angular weight is proportional to $e^{-\xi {\theta }^2/2}$, giving $⟨\cos \theta ⟩\simeq 1-⟨{\theta }^2⟩/2=1-1/(2\xi )$. Consequently $Nb-⟨L⟩\sim N{k}_{B}T/(2F)$: the same inverse-force exponent, with half the three-dimensional coefficient. Both versions assume inextensible links and forces below the regime where the links themselves stretch.

For isotropic rotational diffusion, orientation drift and probability density obey the displayed Fokker-Planck equation, normalized with respect to solid angle. Without flow, constant light $L{\mathbf{e}}_x$ gives the zero-current density $f_0=\lambda e^{\lambda p_x}/(4\pi \sinh \lambda )$, $\lambda =L/(\alpha D_r)$. Integration over the sphere gives unit mass; its mean projection is the Langevin function. The zero-light limit is the uniform density $1/(4\pi )$.

For aligning rate $1/B$, positive rotational diffusivity $D_r$ and $\Lambda =(B{D}_r{)}^{-1}$, the zero-current steady Fokker-Planck equation gives $f_{0,\theta }=-\Lambda \sin \theta f_0$. Normalization over the unit sphere gives the displayed density per unit solid angle. Its mean direction is $⟨\mathbf{p}⟩=(\coth \Lambda -1/\Lambda )\mathbf{k}$, the Langevin function. As $B{D}_r\to 0$, it concentrates weakly into a point mass at the upward pole; the limit is not an ordinary bounded probability density.