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A strongly stretched inextensible worm-like chain has two small transverse tangent components. Its quadratic energy is $\frac{1}{2}\int [A|{\partial }_s{\mathbf{t}}_{\perp }{|}^2+f|{\mathbf{t}}_{\perp }{|}^2]ds-fL$. For a long bulk chain, the equipartition theorem assigns each Fourier mode covariance $k_{B}T/(A{q}^2+f)$. Integrating over $q$ and summing both components gives $⟨|{\mathbf{t}}_{\perp }{|}^2⟩=k_{B}T/\sqrt{Af}$, hence the extension deficit is half that value. With persistence length $L_p=A/(k_{B}T)$, the deficit is $\sqrt{k_{B}T/(4L_{p}f)}$. It requires small slopes and negligible stretching of the backbone.

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.

Use the arc length $s$ and the unit tangent $\mathbf{t}={\partial }_s\mathbf{r}$ of an inextensible filament. In the high-force state choose its positive longitudinal branch. The constraint givesThe extension is $z={\int }_0^L{t}_{z}ds=L-\frac{1}{2}{\int }_0^L|{\mathbf{t}}_{\perp }{|}^{2}ds+\cdots$. The curvature of a space curve obeys ${\kappa }^2=|{\partial }_s\mathbf{t}{|}^2$. Since ${\partial }_s{t}_z=-{\mathbf{t}}_{\perp }\cdot {\partial }_s{\mathbf{t}}_{\perp }+\cdots$ is second order in small fluctuations, its squared contribution is fourth order. Thus the worm-like chain energy through quadratic order isPositive force suppresses transverse orientation fluctuations, while the filament bending modulus penalizes their spatial variation.

To apply the equipartition theorem without missing either component or a mode-counting factor, expand each component in a real orthonormal basis of eigenfunctions of $-{\partial }_s^2$, $t_{\alpha }(s)={\sum }_q{a}_{\alpha q}{w}_q(s)$, with $\int w_q{w}_{q^{\prime }}ds={\delta }_{q{q}^{\prime }}$ and $\alpha =x,y$. The quadratic energy is $\frac{1}{2}{\sum }_{\alpha ,q}(A{q}^2+f)a_{\alpha q}^2$. Each real quadratic degree of freedom contributes $k_{B}T/2$, soFor a chain long compared with the force-dependent correlation length, the interior is translation invariant and the mode sum per length becomes $\int dq/(2\pi )$. Equivalently, for the Fourier transform ${\tilde{t}}_{\alpha }(q)=\int e^{-iqs}{t}_{\alpha }(s)ds$,The negative and positive wavenumbers here represent the real-field mode count consistently, rather than two independent complex modes. Summing both transverse components gives the high-force worm-like chain elasticitywhere the integral follows by setting $q=\sqrt{f/A}u$ and using ${\int }_{-\infty }^{\infty }(1+u^2{)}^{-1}du=\pi$. ThereforeHere $L_p$ is the three-dimensional zero-force persistence length. The result requires $f{L}_p\gg k_{B}T$, negligible backbone stretching, and a long-chain bulk approximation. It does not describe the eventual bond-stretching regime. For a continuum polymer description, the force-dependent length should also exceed the microscopic cutoff.

For comparison, a three-dimensional freely jointed chain has contour length $L=Nb$ and independent link orientations. At force $f$, a link of polar angle $\vartheta$ has Boltzmann factor $e^{x\cos \vartheta }$ with $x=fb/(k_{B}T)$. Its angular canonical partition function isThe total canonical partition function is $Z_1^N$. Differentiating with respect to force gives the force-extension of a three-dimensional freely jointed chain:ConsequentlyThe extension deficits scale as $f^{-1}$ and $f^{-1/2}$ respectively. Inverting them gives force divergences proportional to the first and second inverse powers of the deficit. Freely jointed link rotations cost no bending energy; the worm-like chain instead has a continuum of coupled bending modes whose contributing wavelengths change with force. Setting a long-wavelength coarse-grained link length $b=2L_p$ can match weak-force statistics, but does not make these high-force laws identical.

For the transverse tangent correlation of a stretched worm-like chain, use a separation $r$ consistently: the source writes $C(y)$ but integrates a separation denoted $r$. In the translation-invariant bulk the spatial average is redundant andTo evaluate the integral without an assumed decay law, let $G(r)$ be the inverse Fourier transform of $(A{q}^2+f{)}^{-1}$. Then $(-A{\partial }_r^2+f)G=\delta (r)$. Away from zero, decay at both infinities requires $G=B{e}^{-|r|/\xi }$ with $\xi =\sqrt{A/f}$. Integrating across zero yields $-A[G^{\prime }(0^{+})-G^{\prime }(0^{-})]=1$, so $B=1/(2\sqrt{Af})$. ThusAt $r=0$ this agrees with the two-component equipartition theorem calculation. The correlation length decreases with force and is distinct from the zero-force persistence length. The full tangent correlation has a nonzero aligned background; the exponentially decaying quantity here is specifically the transverse correlation.

For a finite open chain, the source's integral samples $s+r$ beyond $L$ and supplies no endpoint tangent conditions. An exact finite-length correlation is therefore not uniquely specified. The result above is the intended bulk form when $L\gg \xi$. As one explicit finite convention, periodically extending the tangent fluctuations gives the periodic finite-length tangent correlation of a stretched worm-like chainwhere $d_L(r)$ is the distance from zero around the periodic interval. Then $C_L(0)=(k_{B}T/\sqrt{Af})\coth [L/(2\xi )]$, and the extension formula acquires this same finite-length factor. The bulk law is recovered as $L/\xi \to \infty$. This makes the endpoint assumption explicit rather than claiming an unspecified finite-chain correlation is exactly translationally invariant.

For one small transverse displacement, the elastic energy of the elastic filament iswhere $A$ is the filament bending modulus. Its first variation isThe clamped boundary conditions remove the left boundary terms. A force-free and torque-free tip permits independent $\delta h(L)$ and $\delta h^{\prime }(L)$, so the natural boundary conditions for a free endpoint are $h^{\prime \prime }(L)=h^{\prime \prime \prime }(L)=0$: zero bending moment and transverse shear.

There is an important distinction between an energy minimum and a normal mode. The unloaded higher-order Euler-Lagrange equation is $h^{\prime \prime \prime \prime }=0$, whose only solution with these four boundary conditions is $h=0$. A general fluctuating shape is a sum of modes, not one of the stated sinusoidal/hyperbolic functions. To obtain the clamped--free bending modes, extremize the bending Rayleigh quotient, or equivalently $U[W]-\frac{1}{2}A{k}^4\int W^{2}dx$ with a fixed norm. Its Euler-Lagrange equation isThis is the constrained variational characterization of bending modes. Its four characteristic roots are $\pm k,\pm ik$. Clamping gives $F=-D$ and $E=-B$, henceThe free-tip conditions reduce toIts determinant is $-2(1+\cos q\cosh q)$, so nontrivial modes requireNo zero mode exists, since a cubic satisfying the homogeneous clamp/free conditions is zero. The clamped-free bending spectrum starts with $q_1\in (\pi /2,2)$; bisection or Newton iteration gives $q_1\simeq 1.875104$, so $k_1\simeq 1.875104/L$. Choosing $D=N(\sin q_n+\sinh q_n)$ and $B=-N(\cos q_n+\cosh q_n)$ givesThe second boundary condition follows from the root equation, and $N$ is arbitrary until a normalization is chosen.

The bending operator with these boundary conditions is positive and self-adjoint. Twice integrating by parts givesSymmetry implies orthogonality when $n\ne m$. Expand $h(x)={\sum }_n{a}_n{W}_n(x)$ and use $I_n=L^{-1}{\int }_0^L{W}_n^{2}dx$. Thenby the equipartition theorem. The supplied endpoint identity therefore yields the thermal bending fluctuations of a clamped filament:To evaluate the sum, apply a tip force $f$ and minimize $U-fh(L)$. The modified free-end conditions are $h^{\prime \prime }(L)=0$ and $-A{h}^{\prime \prime \prime }(L)=f$, and $h^{\prime \prime \prime \prime }=0$ in the interior. Integration gives $h(x)=f{x}^2(3L-x)/(6A)$, so the tip-force compliance of a cantilever is $L^3/(3A)$. On the other hand, minimizing the modal energy minus $f{\sum }_n{a}_n{W}_n(L)$ gives $a_n=f{W}_n(L)/(AL{k}_n^4{I}_n)$ and henceComparison evaluates the fourth inverse-power sum of the cantilever spectrum without truncating the modes:The boxed variance is for the specified single transverse direction. An independent equilibrium check follows by differentiating the Gaussian partition function with respect to $f$: $\partial ⟨h(L)⟩/\partial f=⟨h(L{)}^2⟩/(k_{B}T)$ at zero load, reproducing the same result from the static compliance. Keeping only the first normal mode gives about $0.32356k_{B}T{L}^3/A$, roughly $97.1\%$ of the exact variance. With two independent equivalent transverse directions, their summed variance is twice the boxed result. The small-slope model requires $k_{B}TL/A\ll 1$, or $L$ small compared with the usual three-dimensional persistence length $A/(k_{B}T)$.

The quadratic energy isEach mode is a centered Gaussian, soThe $n=1$ variance diverges as $F\uparrow F_c$ because the first bending mode becomes soft; Gaussian theory then fails and the quartic term controls the fluctuations.

At zero force,With the common persistence length convention $L_p=A/(k_{B}T)$,Thermal bending therefore shortens the mean projected length even without compression. Under the strictly planar tangent-correlation convention $L_p^{(2D)}=2A/(k_{B}T)$, the same result is $L/(6L_p^{(2D)})$.

The bulk tangent covariance is twice the inverse Fourier transform of $k_{B}T/(A{q}^2+f)$, one term for each transverse direction. Equivalently the inverse of $-A{\partial }_r^2+f$ is $e^{-|r|/\xi }/(2\sqrt{Af})$. This gives the displayed correlation and identifies a force-dependent correlation length, different from the zero-force persistence length. Finite chains require endpoint conditions or a periodic convention.