For the transverse <baryon acoustic oscillation> ruler, $d_A=a\chi$ and the physical ruler length is $a\chi_d$. Its observed angle is therefore
$$
\theta(z)=\frac{a\chi_d}{d_A(z)}
=\frac{\chi_d}{\chi(z)}
=\frac{H_0\chi_d}{I(z;\Omega_{m,0})},
$$
where
$$
I(z;\Omega_{m,0})=
\int_0^z\frac{dz'}{\sqrt{\Omega_{m,0}(1+z')^3+1-\Omega_{m,0}}}.
$$
The unknown <cosmological standard ruler> length and $H_0$ occur only through the common amplitude $H_0\chi_d$. Measurements at two distinct redshifts give
$$
\frac{\theta(z_1)}{\theta(z_2)}
=\frac{I(z_2;\Omega_{m,0})}{I(z_1;\Omega_{m,0})},
$$
which generically determines the one shape parameter $\Omega_{m,0}$. Thus two redshifts suffice; additional redshifts overdetermine the model and improve precision.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 310 / 1 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-310/1/c/solution