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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.

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.