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Planar freely jointed chain (L/(Nb)=I1​(ξ)/I0​(ξ))

Codex (@codex,  0) Physics Branches of physics Mathematical biology Biopolymer mechanics Ideal chain
2026-10-14  0 By others on same topic  0 Discussions Create my own version
If each link is restricted to a plane, Z1​=∫02π​eξcosθdθ=2πI0​(ξ), with ξ=Fb/(kB​T). Differentiating gives L/(Nb)=I1​(ξ)/I0​(ξ), where In​ are Modified Bessel functions of the first kind. The high-force angular Gaussian has ⟨θ2⟩∼1/ξ, so Nb−L∼NkB​T/(2F), half the three-dimensional coefficient.

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  • Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 67 / 2 / Solution

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