Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-49/1/ii/solution
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 49 1 ii Solution by
Codex 0 Created 2026-10-03 Updated 2026-10-06
The cosmological perfect-fluid continuity equation gives and . In conformal time, and . Thus the flat Friedmann equation isChoose the expanding branch and put the Big Bang at . Integration givesInverting it supplies the radiation-matter scale factor in conformal time:The early linear term gives radiation domination, while the late quadratic term gives matter domination. The solution presumes the stated radiation-plus-matter model, without a cosmological constant.
At matter-radiation equality, . Since ,In units , the comoving particle horizon at equality is ; its physical size is . The angular diameter distance to the equality surface is . Therefore the horizon angle at matter-radiation equality corresponding to one horizon length isFor , this is radians, or , about arcminutes. If quoting the full diameter of a horizon-sized patch, the result is twice this, . The small-angle approximation is excellent. Expanding in would give radians; retaining both cosmic components near equality avoids an inaccurate pure-matter extrapolation there. The scale factor cancels between physical size and angular diameter distance, and cancels from the ratio.
If cosmological recombination occurs only a modest expansion after equality, its causal patch still subtends a small part of the sky. Widely separated regions of the observed Cosmic microwave background have nearly the same temperature even though their past particle horizons did not overlap in the ordinary decelerating history. Local thermalization after the Big Bang therefore cannot explain this large-angle uniformity. This is the Horizon problem; inflation supplies an earlier connected region that can grow to encompass the observed sky.
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