Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 74 / 2 / c

Choose real normalized eigenfunctions, including the two zero-energy filament modes. The expansion isIntegration by parts with the natural endpoints gives $\int W_m^{\prime \prime }{W}_n^{\prime \prime }dx=k_n^4{\delta }_{mn}$. HenceFor each positive mode, the equipartition theorem givesDistinct modal amplitudes are independent centered Gaussian random variables in the canonical bending ensemble.

The rigid translation and tilt amplitudes have no energy cost. Their integrals in the canonical ensemble are unbounded, so a completely free filament has no normalizable equilibrium height distribution and no finite height variance determined by this energy alone. This is a physical qualification of the requested variance, not a reason to omit the bending fluctuations.

Fixing translation and tilt, for example by imposing $\int hdx=\int (x-L/2)hdx=0$, leaves the rigid-motion-projected thermal covariance of a free filament:For unnormalized modes, divide each summand by $\int W_n^{2}dx$. Any prescribed independent variance of the two rigid amplitudes must be added separately; equipartition theorem does not assign it.

The free-filament variance with fixed translation and tilt also has a closed form. Integrate a white noise curvature field twice, and subtract its affine least-squares projection. With $u=x/L$, $t=s/L$, the dimensionless kernel isIt has zero mean and first spatial moment, while its second $u$ derivative is the delta source. Since the covariance is $(k_{B}T/A)\delta (s-s^{\prime })$, integrate its squared kernel to obtainThis conditional profile is symmetric about the midpoint: $P(1-u)=P(u)$, with endpoint value $1/105$ and midpoint value $1/320$. It refers to removing the free rigid motion, not to clamping either end.