Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 3 / 15D (Cosmology)

Mass conservation in the Zeldovich approximation gives $\rho d^{3}r=\bar{\rho }{a}^3{d}^{3}q$. Expanding the trajectory Jacobian determinant to first order therefore yields $\boxed{\delta =-{\nabla }_q\cdot \Psi }$. At the same order, ${\nabla }_r=a^{-1}{\nabla }_q$ andThe last equality uses ${\nabla }_q\times \Psi =0$ and the vector calculus $\nabla \mathrm{div}={\nabla }^2+\nabla \times \nabla \times$.

Split the gravitational potential into its homogeneous-background and perturbed parts. The background has ${\nabla }_r{\Phi }_0=(4\pi G\bar{\rho }/3)r$. For the perturbation, Poisson equation and irrotationality givewith a harmonic or uniform-acceleration contribution set to zero by the perturbation boundary conditions. Its divergence is $-4\pi G\bar{\rho }{\nabla }_q\cdot \Psi =4\pi G\bar{\rho }\delta$, as required. Insert this and the pressure gradient into $\ddot{r}=\ddot{a}(q+\Psi )+2\dot{a}\dot{\Psi }+a\ddot{\Psi }$. The background acceleration cancels using $\ddot{a}/a=-4\pi G\bar{\rho }/3$, leavingTake minus the comoving divergence, and expand $\delta (q,t)={\sum }_k{\delta }_k(t)e^{ik\cdot q}$. Since ${\nabla }_q^2$ acts on a Fourier mode by $-k^2$, each mode satisfiesHere the expansion coordinate is comoving; confusing it with the physical coordinate would lose the factor $a^{-2}$.

For $P=\beta {\rho }^{4/3}$, the sound speed obeys $c_s^2=(4\beta /3){\bar{\rho }}^{1/3}$. Using $a=(t/t_0{)}^{2/3}$ and $\bar{\rho }=(6\pi G{t}^2{)}^{-1}$ givesThe mode equation is an Euler differential equation. Substituting ${\delta }_k=t^n$ produces $n(n-1)+(4/3)n-2/3+{\bar{v}}_s^2{k}^2=0$. Thus, for distinct roots,A genuinely growing mode occurs only for ${\bar{v}}_s^2{k}^2<2/3$; at $k=0$ the exponents are $2/3$ and $-1$. At ${\bar{v}}_s^2{k}^2=2/3$ the larger root is zero. Between $2/3$ and $25/36$, both roots are negative. At the repeated-root value ${\bar{v}}_s^2{k}^2=25/36$, the independent solutions are $t^{-1/6}$ and $t^{-1/6}\log t$. Beyond it, write $\nu =\sqrt{{\bar{v}}_s^2{k}^2-25/36}$ to obtain $t^{-1/6}[A\cos (\nu \log t)+B\sin (\nu \log t)]$. Thus the phrase “growing and decaying” applies to the long-wavelength regime; pressure-stabilized modes decay instead.