Rigid-motion-projected thermal covariance of a free filament ($C(x,y)=\frac{k_{B}T}{A}{\sum }_{n\ge 1}\frac{W_n(x)W_n(y)}{k_n^4}$)

Free bending has two zero-energy filament modes, so it does not define an unrestricted normalizable thermal height distribution. Fix the mean displacement and linear tilt, or project out that affine kernel. On the positive complement, the equipartition theorem gives independent Gaussian mode variances $k_{B}T/(A{k}_n^4)$ for normalized modes. The resulting covariance is the bending operator pseudoinverse multiplied by $k_{B}T$. It depends on how the rigid motion is fixed.

The quadratic bending energy makes curvature a white noise field in the generalized thermal description. Integrate it twice, then subtract the affine least-squares projection to fix translation and tilt. With $u=x/L$ and $t=s/L$, the dimensionless kernel is $(u-t{)}_{+}-(1-t{)}^2/2-(1-3t^2+2t^3)(u-1/2)$. Its second spatial derivative is the source delta and its mean and first moment are zero. Its squared integral yields the free-filament variance with fixed translation and tilt.

For the frame fixed by zero mean height and zero first height moment, the variance profile is $P(u)=1/105-11u/105+13u^2/35-u^3/3-u^4/3+3u^5/5-u^6/5$. It obeys $P(1-u)=P(u)$, $P(0)=1/105$ and $P(1/2)=1/320$. Integrating a white noise curvature field twice, subtracting its affine projection and integrating the squared kernel proves the profile. These free-end conditions must not be confused with physically clamping the ends.