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For a uniform elastic filament clamped below and free above, its own weight gives filament tension $T(z)=-\lambda g(L-z)$. With filament bending modulus $B=EI$, the static slope $u$ satisfies $B{u}^{\prime \prime }+\lambda g(L-z)u=0$, $u(0)=0$, and $u^{\prime }(L)=0$. Setting $\eta =(2/3)\sqrt{\lambda g/B}(L-z{)}^{3/2}$ gives $u={\eta }^{1/3}{J}_{-1/3}(\eta )$ up to amplitude; the $J_{1/3}$ branch violates the zero-moment condition. The first positive zero $j_{-1/3,1}$ therefore yields $\lambda g{L}_c^3/B\simeq 7.83735$. This is a self-buckling of a vertical rod threshold, not a constant-end-load Euler threshold.

The intrinsic signed curvature is the curvature of a stress-free filament, measured along material arc length. An elastic filament with actual curvature $\kappa$ stores bending energy $\frac{B}{2}\int (\kappa -{\kappa }_0{)}^{2}ds$, with filament bending modulus $B$. In a planar small-slope Monge representation, $\kappa \simeq {\zeta }^{\prime \prime }$ and the leading energy is quadratic in ${\zeta }^{\prime \prime }-{\kappa }_0$.

Write $B$ for the filament bending modulus and $C$ for the filament torsional stiffness, so that the energy per unit length is $(B{\kappa }^2+C{\Omega }^2)/2$. For one circular filament,where $I_1$ is the second moment of area, $J_1$ the polar moment, $G_s$ the shear modulus, and $\nu$ Poisson's ratio. Numerical twisting constants cannot be fixed by Young's modulus alone without specifying $\nu$.

For a filled sliding bundle, $N\sim (R/r{)}^2$. Filaments share the centerline curvature but slide instead of accumulating the additional axial strains of a bonded cross-section. Their bending energies add: $B\sim N{B}_1\sim E{r}^2{R}^2$. If end constraints impose the same material twist on every filament, their torsional energies also add: $C\sim G_s{r}^2{R}^2$.

For a bonded, one-filament-thick circular tube, the wall thickness $h$ is of order $r$. Its moments are $I\sim \pi R^{3}h$ and $J_p\sim 2\pi R^{3}h$, giving $B\sim Er{R}^3$ and $C\sim G_{s}r{R}^3$. For a bonded filled rod, $I\sim \pi R^4/4$ and $J_p\sim \pi R^4/2$, giving $B\sim E{R}^4$ and $C\sim G_s{R}^4$. Packing fractions and the convention for tube radius affect constants, not these powers.

There is an important qualification to bending and torsion of filament bundles: free axial sliding does not specify whether each filament can also spin freely. If both spin and slip are unconstrained, the common-material-twist calculation is not forced by the geometry. A helix of small twist rate has centerline curvature of order $\rho {\Omega }^2$, so its bending energy starts at ${\Omega }^4$, not ${\Omega }^2$. A nonzero linear collective torsional stiffness then requires additional end, contact, or spin constraints.

The compressive tip force is limited by Euler buckling of an elastic filament. With unsupported length $\ell$ and filament bending modulus $B=EI$,For a hollow microtubule, use $I=\pi (r_o^4-r_i^4)/4$ rather than the solid-cylinder value. For a clamped seed and a moment-free tip free to slide along the wall, linearized angle balance is $B{\theta }^{\prime \prime }+F\theta =0$, with $\theta (0)=0$ and ${\theta }^{\prime }(\ell )=0$. Its first nonzero solution has $\sqrt{F/B}\ell =\pi /2$, hence $F_c={\pi }^{2}B/(4{\ell }^2)$. A different tip constraint changes this coefficient.

Increasing the unsupported length decreases that threshold. For growth against a wall at fixed projected gap $d$, polymerization-driven cantilever postbuckling gives a more direct quantitative statement. The inextensible elastica obeys $B{\theta }^{\prime \prime }+F\sin \theta =0$. Multiplying by ${\theta }^{\prime }$ and using the zero tip moment givesPut $k=\sin ({\theta }_t/2)$ and substitute $\sin (\theta /2)=k\sin \varphi$. Integrating arc length and its horizontal projection giveswhere the complete elliptic integrals are $K(k)={\int }_0^{\pi /2}(1-k^2{\sin }^2\varphi {)}^{-1/2}d\varphi$ and $E(k)={\int }_0^{\pi /2}(1-k^2{\sin }^2\varphi {)}^{1/2}d\varphi$. Using $K=(\pi /2)(1+k^2/4+\cdots )$ and $E=(\pi /2)(1-k^2/4+\cdots )$ gives $\ell /d=1+k^2+O(k^4)$, and thereforeThus the initial postbuckling force decreases as more contour is inserted. This result describes the weakly buckled branch before overhang, further wall contact, or altered tip constraints invalidate the model. It is a mechanical force law; a chemical force limit from polymerization can impose a smaller force.

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.

The filament bending modulus has units force times length squared. The perpendicular drag coefficient in this equation is drag per unit length divided by velocity, so its units are force times time per length squared. Hence $A/(\zeta \omega )$ has units length to the fourth power. Balancing oscillatory viscous drag $\zeta \omega h$ against elastic force density $Ah/{\ell }^4$ gives the elastohydrodynamic penetration lengthIncreasing frequency confines the response nearer the actuator. The chosen coefficient one in this characteristic length is a convention; the exact decay lengths acquire the root-dependent factors found below.

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.

Put $B=E{I}_2$, the filament bending modulus in the easier bending direction; $I_2$ is the relevant second moment of area. Take filament tension as positive in extension. A cut at height $z$ supports the weight of the material above it, so its axial force is compressive:The small transverse displacement induces a transverse tension force density $(T{X}_z{)}_z$, while bending produces $-B{X}_{zzzz}$. Their balance givesThe same result follows from elastic energy and gravitational energy. To quadratic order, local inextensibility shortens the height of the material point labelled $z$ by $\frac{1}{2}{\int }_0^z{X}_z(s{)}^{2}ds$. Integrating its weight and interchanging the integrals gives the second-order potential energyTwo integration by parts steps in its first variation giveThe clamped boundary conditions fix displacement and slope at the base, and the unconstrained tip must have zero bending moment and transverse end force. Thus the complete four conditions areSince the filament tension vanishes at the tip, the last condition is $X_{zzz}(L)=0$. Integrate the ordinary differential equation once: $B{X}_{zzz}-T{X}_z$ is constant, and its tip value makes that constant zero. For the slope $u=X_z$, it follows thatThe other tip condition $u_{zz}(L)=0$ already follows from this equation for a regular solution; it is not a third independent condition on the second-order slope problem. Finally reconstruct $X(z)={\int }_0^{z}u(s)ds$ to impose the base position.

Set $x=L-z$, $\alpha =\lambda g/B$, and $\eta =(2/3)\sqrt{\alpha }{x}^{3/2}$. Then $d\eta /dx=3\eta /(2x)$ and $d^2\eta /d{x}^2=3\eta /(4x^2)$, so $u_{xx}+\alpha xu=0$ becomesSubstituting $u={\eta }^{1/3}F(\eta )$ givesThis is the Bessel differential equation of order $1/3$, whose two independent Bessel functions of the first kind giveThe tip is $x=\eta =0$. To decide which branch is admissible, retain both leading behaviours, not just the fact that $u$ stays finite:Since ${\eta }^2\propto x^3$ and ${\eta }^{2/3}\propto x$, the first branch has zero tip derivative, while the second has a nonzero tip derivative. The moment-free condition therefore sets $b=0$, and it does not require the tip slope itself to vanish. The base condition now requiresLet $j_{-1/3,1}$ be the first positive zero. The Bessel threshold for self-buckling of a vertical rod isTo identify this as the first loss of stability, write the quadratic energy in terms of the slope as $\frac{1}{2}[B\int u_z^{2}dz-\lambda g\int (L-z)u^{2}dz]$. On functions with $u(0)=0$, its critical coefficient is the minimum Rayleigh quotient $\int u_z^{2}dz/\int (L-z)u^{2}dz$; variation supplies the moment-free condition at the tip and the same Sturm-Liouville problem. The first zero gives the lowest eigenvalue, so the straight state is stable below this load, neutral at it, and unstable above it, with the easier $I_2$ direction buckling first.

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.

A positive-eigenvalue normal mode satisfies $K{W}_n=A{k}_n^4{W}_n$. Solving the constant-coefficient equation gives the sine, cosine, hyperbolic sine and hyperbolic cosine family. To avoid confusing modal coefficients with the filament bending modulus, write those coefficients as $c_1,c_2,c_3,c_4$.

The free conditions at zero require $c_3=c_1$ and $c_4=c_2$. With $q=kL$, the remaining conditions areIts determinant is $2(1-\cos q\cosh q)$. Therefore the free-free bending spectrum isA convenient corresponding eigenfunction iswhere $N_n$ gives unit $L^2$ norm. The stable characteristic equation for free-free bending modes is $\cos q=\mathrm{sech}q$. Intersections or numerical bracketing giveFor the entire positive sequence, let $q_n^0=(n+\frac{1}{2})\pi$. Expanding $\cos q_n$ near its zero and using $\mathrm{sech}{q}_n\sim 2e^{-q_n}$ givesThere are also two zero-energy filament modes, not captured by substituting $k=0$ into the finite-coefficient trigonometric expression. Solving $W^{(4)}=0$ with free endpoints gives the affine kernel. An orthonormal basis of that kernel isThe positive normal modes are orthogonal to both. Including these two rigid-motion modes is essential for completeness; the regular self-adjoint operator on a finite interval has a complete discrete eigenbasis.

For each of the four standard pairs, integration by parts givesConsequently the eigenvalues are nonnegative. For a positive eigenvalue ${\mu }_n=A{k}_n^4$, the eigenvalue equation is $W_n^{\prime \prime \prime \prime }=k_n^4{W}_n$. Its four characteristic roots are $\pm k_n,\pm i{k}_n$, soHere $C_j$ are coefficients, avoiding a collision with the filament bending modulus $A$. The regular finite-interval self-adjoint operator has compact resolvent; applying the spectral theorem for compact self-adjoint operators to a shifted inverse supplies a complete orthonormal basis of eigenfunctions.

For clamped boundary conditions at zero, a clamped--clamped bending mode can be writtenAt $L$, writing $\beta =kL$, the two remaining boundary conditions areThe determinant is $2(1-\cos \beta \cosh \beta )$. Hence the positive wavenumbers obeyEquivalently, intersect $\cos \beta$ with $\mathrm{sech}\beta$. Numerical root bracketing givesThe entire sequence has the useful large-$n$ descriptionIndeed, put $\beta =(n+\frac{1}{2})\pi +\delta$ in $\cos \beta =\mathrm{sech}\beta$ and use $\cos \beta =(-1{)}^{n+1}\delta +O({\delta }^3)$ and $\mathrm{sech}\beta =2e^{-\beta }+O(e^{-3\beta })$.

The apparent root $\beta =0$ in the determinant equation is spurious for the clamped problem: at zero eigenvalue, $W$ is a cubic polynomial, and its four clamped conditions force $W=0$. For other endpoint choices, zero-energy filament modes must be treated separately from the trigonometric formula. The free-free kernel consists of affine functions, the torqued-torqued kernel consists of constants, and the hinged-hinged kernel is trivial. Including those kernels is necessary for a complete eigenfunction expansion.

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 $A/(k_{B}T)$.