Past exam of the mathematics course of the University of Cambridge / 2011 / ii / Paper 3 / 15E (Cosmology)

A fixed comoving volume $d^{3}q$ labels a fixed collection of cold dark matter particles. Their background mass is $a^3\bar{\rho }{d}^{3}q$, which is constant by mass conservation. At the perturbed positions it is $\rho d^{3}r$. The Jacobian determinant of the Zeldovich approximation map isTaking its reciprocal and using mass conservation givesFor the perturbed field the Poisson equation must be ${\nabla }_r^2\Phi =4\pi G\rho =4\pi G\bar{\rho }(1+\delta )$. The displayed background-only right-hand side in the PDF would give no clumping force; it is the background equation, not the full perturbed equation.

Write $\Phi ={\Phi }_b+\phi$, where ${\nabla }_r{\Phi }_b=(4\pi G/3)\bar{\rho }\mathbf{r}$. To linear order,For the longitudinal, curl-free displacement used in the growing Zel'dovich mode, this integrates to ${\nabla }_q\phi =-4\pi G\bar{\rho }{a}^2\mathbf{\psi }$, after removing external homogeneous harmonic fields. More generally the force involves only the longitudinal part ${\mathbf{\psi }}_{\parallel }$, which has the same divergence as $\mathbf{\psi }$.

Differentiate $\mathbf{r}=a(\mathbf{q}+\mathbf{\psi })$ twice at fixed $\mathbf{q}$ and substitute $\ddot{\mathbf{r}}=-{\nabla }_r\Phi$. The terms proportional to $\mathbf{q}+\mathbf{\psi }$ cancel using $\ddot{a}=-(4\pi G/3)\bar{\rho }a$, leavingFor a longitudinal displacement the right side is $4\pi G\bar{\rho }\mathbf{\psi }$. Taking minus the comoving divergence gives the general scalar density contrast equation

Both background components are pressure-free, so their densities scale as $a^{-3}$ and ${\Omega }_H$ is constant. Since $\dot{a}/a=2/(3t)$ and ${\bar{\rho }}_C=(1-{\Omega }_H)/(6\pi G{t}^2)$, a power law $\delta =t^{\alpha }$ satisfiesThusFor $0\le {\Omega }_H<1$ the plus sign gives a growing mode and the minus sign a decaying mode. At ${\Omega }_H=1$ the growing exponent becomes zero, since no gravitating clumping component remains.

When ${\Omega }_H=2/3$, ${\alpha }_{+}=1/3$, so $\delta \propto t^{1/3}\propto a^{1/2}$. The redshift relation gives $a_{\mathrm{today}}/a_{\mathrm{eq}}=1+z_{\mathrm{eq}}\simeq 10^5$. Even an entirely growing initial mode therefore givesThe perturbation remains far below order unity, so it cannot produce the observed nonlinear gravitational clumping within the stated model. A decaying admixture does not provide the missing sustained growth.