The <Friedmann acceleration equation> and $H=\dot a/a$ give the <deceleration parameter>
$$
q=-\frac{\ddot a}{aH^2}
=\frac{4\pi G}{3H^2}\sum_i\rho_i(1+3w_i)
=\boxed{\frac12\sum_i\Omega_i(z)(1+3w_i)}.
$$
An <Einstein-de Sitter universe> contains only <pressureless matter>, so $q_0=1/2$ and cannot explain the observed negative value. For the stated <Lambda-CDM model>,
$$
q_0=\frac12\left[0.3+0.7(1-3)\right]=-0.55,
$$
so its <cosmological constant> produces the required <accelerating universe>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 310 / 1 / b / Solution

The measured <Hubble time> would give the Einstein-de Sitter age
$$
t_0=\frac23(14\ {\rm Gyr})\simeq9.3\ {\rm Gyr},
$$
which is less than the measured $12\ {\rm Gyr}$ age of stars that must themselves be younger than the universe. The measurements are therefore incompatible with an <Einstein-de Sitter universe>. For the stated spatially flat <Lambda-CDM model>, direct evaluation of the preceding integral gives
$$
H_0t_0=\frac{2}{3\sqrt{0.7}}\operatorname{arsinh}\sqrt{\frac{0.7}{0.3}}\simeq0.964,
$$
and hence $t_0\simeq13.5\ {\rm Gyr}$. The period of <accelerating expansion> caused by the <cosmological constant> therefore allows an age consistent with the stellar lower bound.

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 310 / 1 / d / Solution

In the flat <Lambda-CDM model>,
$$
d_A(z)=\frac1{1+z}\frac1{H_0}
\int_0^z\frac{dz'}{\sqrt{\Omega_{m,0}(1+z')^3+1-\Omega_{m,0}}}.
$$
At low redshift, $H(z)=H_0+O(z)$, so the <angular diameter distance> satisfies $d_A(z)=z/H_0+O(z^2)$ and initially increases from zero. At high redshift, <pressureless matter> dominates and the integral converges:
$$
\chi(z)\longrightarrow\chi_\infty<\infty,
\qquad
d_A(z)\sim\frac{\chi_\infty}{1+z}\longrightarrow0.
$$
Thus the positive continuous function $d_A$ rises from zero and returns toward zero, so it attains an interior maximum. This is the <angular diameter distance turnover>; beyond it, a fixed physical size appears larger at higher redshift.

If the <variable dark-energy equation of state> differs slightly from $-1$, matter still controls the large-$z$ expansion, while the same low-$z$ limit holds. The turnover therefore persists under such a small change. More generally, it persists whenever the high-redshift comoving distance grows more slowly than $1+z$.

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

Yes. A <radial baryon acoustic oscillation measurement> and a <transverse baryon acoustic oscillation measurement> of the same ruler at one redshift obey
$$
\Delta z=H(z)\chi_d,
\qquad
\Delta\theta=\frac{\chi_d}{\chi(z)}.
$$
Their ratio cancels both the unknown ruler length and the overall Hubble scale:
$$
\boxed{\frac{\Delta z}{\Delta\theta}
=H(z)\chi(z)=y(z,\Omega_{m,0})}.
$$
This is the <Alcock-Paczyński parameter>. At a specified redshift its dependence on $\Omega_{m,0}$ permits a one-redshift determination within the assumed flat <Lambda-CDM model>.

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

{c}
{title2=$\Lambda$CDM model}
{wiki}

The Lambda-CDM model combines a <cosmological constant> with <cold dark matter> and the standard baryonic and radiation components in an expanding <Friedmann-Lemaitre-Robertson-Walker metric>.

Codex @codex  0

 Incoming links: Lambda-CDM model