Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 320 1 Solution Created 2026-10-03 Updated 2026-10-06
For one target galaxy, an encounter with relative speed sweeps a cylinder of volume in time . Multiplication by the number density gives the encounter rate . Thus the geometric galaxy-encounter rate givesThis is the expected number of mergers in the geometric model. For independent encounters, the Poisson process probability of at least one is , so the displayed linear probability is valid for rare encounters, . There is no factor of one-half for a single target; that factor would enter a count of distinct pairs across the whole population.
For an illustrative group environment, take , , and a Hubble time of order years, or . The path length is about , using the stated distance conversion. HenceThe estimate scales linearly with environment density and quadratically with the adopted merger radius. It is an order-of-magnitude model estimate, not a universal observed merger fraction; reducing the illustrative density by a factor of one hundred reduces the estimate by the same factor.
For the tidal calculation, use the relative potential convention in which acceleration is . Expand the point-mass potential about the target centre:The constant does not exert a force. The linear term accelerates the target centre and is subtracted in its freely falling frame. The remaining leading quadrupolar point-mass tidal potential isThe condition permits the convergent expansion; using only this term is the leading tidal approximation, accurate when the target radius is small compared with the closest separation.
The given impact and velocity directions imply . This trajectory is in the plane, correcting the incompatible printed plane. In the impulse approximation, hold the stellar position fixed while integrating the tidal tensor:For , the needed integrals arewith the mixed integral zero by oddness. Thus the integrated tidal tensor of a straight-line flyby is andThere is stretching along the impact direction, compression in the other transverse direction and no net kick along the path.
Write for the star's initial velocity, distinguishing it from the perturber speed . Its specific energy change is . The no-correlation assumption removes the first term on averaging; it is not an identity for every individual star. The mean specific heating at a given position is thereforeFor a spherical target, the mass-weighted averages satisfy . The tidal impulse heating of a spherical galaxy is consequentlyHere is assumed finite. Notice the change from specific energy to total energy after multiplying by .
For two equal targets, each gains the same internal energy, so . Their relative-motion reduced mass is . If denotes their relative speed at infinity, their initial orbital energy is . Internal tidal heating comes from this orbital energy. In the tidal capture of galaxies model, capture occurs if , which gives the equal-mass tidal-capture thresholdCapture allows further passages and eventual merger in this simplified picture.
The encounter lasts roughly . The impulse approximation needs , so that a star barely moves during the tide. When , stellar orbits respond during the perturbation and the kicks can cancel. Adiabatic invariance of an orbital action produces adiabatic shielding of tidal encounters, rather than the impulsive heating used above. The capture inequality cannot be extrapolated into that slow-encounter regime.
