Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 77 3 i Solution Created 2026-10-03 Updated 2026-10-14
Neglecting cell rotational inertia, the light and viscous torque sum to zero. Therefore the angular velocity is . A material orientation evolves by . The vector triple-product identity givesThe last term is the projection of onto the tangent plane to the unit sphere. In particular, , as required to preserve the unit orientation. It aligns the swimmer towards the light; the vorticity rotates that orientation.
For rotational diffusion coefficient , the phototactic orientation Fokker-Planck equation balances deterministic orientation drift with diffusion on the sphere. Its density is normalized with respect to solid angle, not . When , there is no azimuthal drift and the zero-current condition is . Thus , and normalization givesThe limiting density at is .
To justify the angular dependence of the first correction, define the operatorsThe steady Fokker-Planck equation is . Its regular expansion therefore givesThe coefficients of are independent of , so each azimuthal Fourier harmonic is preserved. The forcing has only the sine harmonic with azimuthal number one. There can be no extra unforced harmonic in a smooth normalized first correction: writing a homogeneous solution as gives . Multiplication by and integration by parts on the sphere gives , hence is constant. Normalization removes that multiple of . It follows thatNo explicit calculation of is needed. Its dependence is on and the parameter , and its behavior at the poles must make the density smooth.
In the stated spherical coordinates, . Azimuthal integration makes the leading components and the first-order corrections vanish. The surviving coefficients areThus the small-vorticity phototactic orientation response isBoth coefficients depend only on . In particular, is the Langevin function of and is positive for positive light strength; its integral representation above suffices.