 Solution

ID: past-exam-of-the-short-options-paper-of-the-physics-prelims-of-the-university-of-oxford/2022/s02/3/c/solution

Normalize the scale factor to

a (t_{0}) = 1

and set

a_{d} = 1 0^{- 3}

. In the same matter-only model

a = (t / t_{0})^{2/3}

, so

t_{d} = t_{0} a_{d}^{3/2} ≃ 2.94 \times 1 0^{5} yr, r_{p} (t_{d}) = 3 c t_{d} ≃ 0.271 Mpc .

Its present proper size after expansion with the scale factor is

r_{p, expanded} = \frac{r _{p} ( t _{d} )}{a _{d}} = \frac{2 c}{H _{0}} a_{d} ≃ 271 Mpc .

χ_{d} = \int_{t_{d}}^{t_{0}} \frac{c d t}{a ( t )} = \frac{2 c}{H _{0}} (1 - a_{d}) ≃ 8295 Mpc .

D_{A} = a_{d} χ_{d}

, so the angular size of a transverse length equal to one particle horizon radius is

θ ≃ \frac{r _{p} ( t _{d} )}{D _{A}} = \frac{a _{d}}{1 - a _{d}} = \frac{1}{1000 - 1} ≃ 0.0327 rad ≃ 1.8 7^{\circ} .

If “the horizon” denotes the diameter of the whole causally connected ball, its angular diameter is twice this, approximately

3.7 4^{\circ}

. The degree-scale conclusion is the same. Neglecting early radiation is an explicit approximation inherited from the question.

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