Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 70 (Early Universe Cosmology)

Write the reduced Planck mass as $M=(8\pi G{)}^{-1/2}=m_{\mathrm{pl}}/\sqrt{8\pi }$, and use natural units. Near the flat maximum, the quartic hilltop inflation trajectory has $V\simeq \alpha {\sigma }^4$ and $V^{\prime }\simeq -12\alpha {\varphi }^3$. The slow-roll approximation gives $H^2\simeq V/(3M^2)$ and $3H\dot{\varphi }\simeq -V^{\prime }$. Therefore the number of e-folds isThe lower field value dominates this integral. ThusThis is far below $\sigma$ for the assumed small symmetry-breaking scale. Strict slow roll actually fails before the minimum: $|M^2{V}^{\prime \prime }/V|\sim 1$ at ${\varphi }^2\sim {\sigma }^4/(36M^2)$. The last approach to the minimum is not described by slow roll. Using that earlier upper limit changes the dominant fifty-e-fold estimate by an order-one number of e-folds, not its leading scale dependence. No extrapolation of slow roll all the way to the minimum is needed.

A comoving wavenumber crosses the inflationary Hubble scale at $k=aH$. Consequently $d\ln k=d\ln a+d\ln H\simeq Hdt$ when $H$ changes slowly. Combining with the slow-roll approximation givesBoth $k$ and $\varphi$ increase along the rolling branch. If $N$ denotes the remaining number of e-folds rather than elapsed expansion, then $d\ln k\simeq -dN$. More precisely $d\ln k=(1-{ϵ}_H)d\ln a$; the correction is beyond the constant-Hubble approximation used here.

At leading order the scalar curvature power and total tensor power areApplying the preceding $d/d\ln k$ relation gives the scalar spectral index $n-1=-6{ϵ}_V+2{\eta }_V$ and the tensor spectral index $n_T=-2{ϵ}_V$. For this hilltop,Thus, fifty e-folds before the end,The tiny $-6{ϵ}_V$ contribution and finite-endpoint corrections have been omitted in the quoted leading scalar tilt. The tensor tilt is nearly zero but slightly negative; treating $H$ as exactly constant in its spectrum would incorrectly discard its leading nonzero value.

The given large-angle temperature transfer relation gives the low-multipole tensor fraction in quartic hilltop inflationThis is not the primordial tensor-to-scalar ratio $r=16{ϵ}_V$, because the scalar and tensor contributions to low CMB multipoles have different transfer functions. With $\sigma \ll m_{\mathrm{pl}}$, the tensor contribution is far below the scalar temperature signal and its cosmic variance, and the corresponding primordial gravitational-wave or polarization signal is extremely small.

For COBE normalization of quartic hilltop inflation, the historical scalar horizon amplitude is of order ${\delta }_H\simeq 2\times 10^{-5}$, or ${\mathcal{P}}_S\simeq 25{\delta }_H^2/4\sim 2.5\times 10^{-9}$ in the usual matter-era convention. The slow-roll spectrum simplifies toHenceLate-time transfer and tilt corrections to the COBE normalization change the order-one coefficient, not the conclusion that the dimensionless coupling must be extraordinarily small. The leading normalization is independent of $\sigma$: reducing the height of the potential also changes the slope and field value at fixed $N$, causing the scale factors to cancel.

Normalize $a_0=1$ and use the open dust model, with no cosmological constant and negligible present radiation. The Friedmann equation at the present epoch reads $H_0^2=H_0^2{\Omega }_m-k{c}^2$. Since $k=-r_{\mathrm{curv}}^{-2}$, the curvature radius of an open FLRW universe isIn the metric's $c=1$ convention the numerator is one. With $a_0=1$ this is also the present physical curvature radius; at scale factor $a$ the physical radius is $a{r}_{\mathrm{curv}}$. Inferring curvature from ${\Omega }_m$ alone would require this matter-only assumption; an additional dark-energy density would also enter the present Friedmann equation.

At fixed conformal time and fixed radial coordinate, the metric gives transverse physical separation $D=a({\tau }_1)r_1\theta$ to first order in the angle. Thus the angular diameter distance isThe coordinate $r_1$ is the transverse comoving radius appearing in the angular metric coefficient. It is not the radial geodesic distance $r_{\mathrm{curv}}\mathrm{arsinh}(r_1/r_{\mathrm{curv}})$; replacing one by the other would omit the curvature dependence of angular projection.

Adiabatic photon cooling gives $a_{\ast }\simeq 1/1200$. During matter domination, $H(a)\simeq H_0\sqrt{{\Omega }_m}{a}^{-3/2}$, so the comoving sound distance traveled before last scattering isMultiplication by $a_{\ast }$ gives the physical sound horizon radius:This is a radius, or one-way sound-propagation length, rather than its diameter. The integral treats the early radiation interval as negligible under the specified late-equality approximation; it also neglects curvature at last scattering, requiring ${\Omega }_m{a}_{\ast }^{-3}\gg (1-{\Omega }_m)a_{\ast }^{-2}$. A very small matter fraction would violate those approximations.

A radial null ray obeys $cd\tau =dr/\sqrt{1+r^2/r_{\mathrm{curv}}^2}$. HenceFor the open matter solution, let $\beta =\sqrt{1-{\Omega }_m}$. The Friedmann equation gives $a^{\prime }=H_0\sqrt{{\Omega }_{m}a+{\beta }^2{a}^2}$, and the age in conformal time isThusHere the stated approximation ${\tau }_{\ast }\ll {\tau }_0$ was used only in the last distance estimate. The flat limit is smooth: ${\tau }_0\to 2/H_0$ and $r_{\ast }\to 2c/H_0$. Retaining ${\tau }_{\ast }$ in the preceding null-ray formula gives the finite-emission-time correction.

Divide the physical sound radius by the angular diameter distance $a_{\ast }{r}_{\ast }$. The sound-horizon angle in an open matter universe isThe characteristic acoustic multipole is approximately half a wavelength across this angle:Thus negative curvature moves the acoustic feature to smaller angular scales and larger multipoles. The first temperature peak is of order $200/\sqrt{{\Omega }_m}$ in this approximation; precise peak positions include the sound-speed history, radiation effects and acoustic phase shifts, so the acoustic-scale multipole is not an exact peak-location formula.

For instantaneous complete hydrogen reionization, take $n_{e0}={\Omega }_b{\rho }_{\mathrm{crit},0}/(m_p{c}^2)=3{\Omega }_b{H}_0^2/(8\pi G{m}_p)$ and $n_e(z)=n_{e0}(1+z{)}^3$. Helium or incomplete ionization would multiply this by the appropriate free-electron count per baryon. In the matter-plus-curvature model,The reionization optical depth in an open matter universe is thereforeSet $u=\sqrt{1+{\Omega }_{m}z}$ and ${\beta }^2=1-{\Omega }_m$. The integral is $2{\Omega }_m^{-2}[u^3/3-{\beta }^{2}u{]}_1^{u_{\mathrm{re}}}$, givingAt high redshift where matter dominates over curvature, this approaches $\kappa \simeq {\Omega }_b{C}_H(1+z_{\mathrm{re}}{)}^{3/2}/\sqrt{{\Omega }_m}$. The exact expression should be retained when ${\Omega }_m{z}_{\mathrm{re}}$ is not large.

For ${\Omega }_m=1$ the preceding result reduces to $\kappa ={\Omega }_b{C}_H[(1+z_{\mathrm{re}}{)}^{3/2}-1]$. Using $C_H=0.032$ and ${\Omega }_b=0.05$ givesAt fixed $H_0$ and ${\Omega }_b$, a lower matter density lowers $H(z)$ relative to the flat dust value at positive redshift. The same electron density then persists for a longer scattering time, increasing the optical depth. Hence the reionization redshift giving unit depth is lower in an open universe.

In the high-redshift matter regime the scaling is $1+z_{\mathrm{re}}\propto {\Omega }_m^{1/3}$ at fixed optical depth; for ${\Omega }_m=0.3$ this suggests a redshift near fifty rather than seventy. This comparison holds the baryon density fixed, not the baryon fraction ${\Omega }_b/{\Omega }_m$, and is not an estimate of the actual observed reionization epoch.

Choose a vacuum order parameter ${\Phi }_0$ and let $H$ be its stabilizer in $G$. Acting with $G$ produces all symmetry-related minima; two elements produce the same minimum precisely when they differ by an element of $H$. The vacuum manifold is thereforeThis assumes a single transitive orbit of minima; additional independent vacuum sectors would require more than one orbit. For a local gauge symmetry, $G/H$ describes the vacuum directions used in finite-energy boundary conditions and defect classification, rather than distinct observable vacua after quotienting by every gauge redundancy.

Finite-energy configurations approach the vacuum manifold outside defect cores. The surrounding boundary in the transverse directions is a sphere, and noncontractible boundary data prevent a smooth deformation to the uniform vacuum.

Nontrivial ${\pi }_0(M)$ means disconnected vacuum components. Choosing different components on the two sides of an interface produces domain walls. Here ${\pi }_0$ labels components and need not itself carry a group structure. Nontrivial ${\pi }_1(M)$ means noncontractible loops: a circle surrounding a line can wind around the vacuum manifold, producing cosmic strings. Nontrivial ${\pi }_2(M)$ means noncontractible maps from an enclosing two-sphere, producing pointlike magnetic monopoles in three spatial dimensions.During the transition, initially uncorrelated vacuum choices in causally separated regions can generate these boundary classes through the Kibble mechanism. Nontrivial topology allows stable defects; it does not specify their exact formation efficiency or abundance.

The Abelian Higgs model has local $G=U(1)$, and a nonzero minimally charged scalar expectation value has trivial stabilizer $H=\{1\}$. Thus $M\simeq U(1)\simeq S^1$ and ${\pi }_1(M)=\mathbb{Z}$, while ${\pi }_0$ and ${\pi }_2$ are trivial. The defects are local strings, or Nielsen-Olesen vortices, whose winding fixes flux $2\pi n/e$.

Put $R=\eta +\varphi /\sqrt{2}$ and $\vartheta =\psi /(\eta \sqrt{2})$. The specified gauge-field shift is $A=A^{\prime }+\partial \vartheta /e$, soExpanding the potential gives $V=\lambda {\eta }^2{\varphi }^2/2+O({\varphi }^3)$. Hence the broken-phase spectrum of the Abelian Higgs model follows fromThe phase is not an additional physical massless scalar: the Higgs mechanism supplies the vector's longitudinal polarization. In the symmetric phase the complex scalar has two real degrees of freedom and the massless vector has two polarizations, a total of four. In the broken phase there are one scalar and three massive-vector polarizations, again four. The symmetric phase is the thermally restored phase; the origin of the given zero-temperature potential is unstable, not a separate stable particle vacuum. The local polar change of variables cannot remove global vortex winding everywhere, so the massive spectrum is consistent with string defects.

The SU(5) grand unified theory has simply connected space $G=SU(5)$ and a connected Standard Model subgroup. More precisely that subgroup is $(SU(3)\times SU(2)\times U(1))/{\mathbb{Z}}_6$; the product notation suppresses a finite central quotient. The long exact sequence of homotopy groups of a fibration $H\to G\to G/H$ givesThus stable magnetic monopoles form, whereas this transition does not require walls or strings. The monopole core and mass scales in an SU(5) transition are set by the massive gauge and scalar fields:For order-one couplings, the core size is of order ${\eta }^{-1}$ and the mass is of order $\eta$ up to the often substantial factor $4\pi /g$. These heavy objects are nonrelativistic soon after formation and can survive as stable relics.

Take the transition temperature $T_i\sim \eta$ and radiation-dominated universe. With $g_{\ast }\simeq 100$ relativistic degrees of freedom, ${\rho }_{r,i}=({\pi }^2/30)g_{\ast }{T}_i^4$ and the Friedmann equation givesA saturated causal correlation length is of order the radiation-era horizon $H_i^{-1}=2t_i$. Write $C_n\sim 1$ for the geometric/formation efficiency and $C_M=M_M/\eta$. One monopole per horizon volume gives the horizon-density monopole abundance bound estimateThe first quantity is number density and the second is defect energy density. Relative to the initial radiation critical density,This can initially be small even when the surviving monopole abundance is unacceptable today. Causality supplies an order-of-magnitude lower abundance estimate under the stipulated saturation; it is not a precise statistical count of defects.

Without substantial annihilation or decay, monopole number in a comoving volume is conserved. Hence $n_M\propto a^{-3}$ and, for constant nonrelativistic mass, ${\rho }_M\propto a^{-3}$. Radiation redshifts as $a^{-4}$, so the monopole-to-radiation ratio grows with $a$.

Use comoving entropy conservation rather than assuming the same relativistic content at every epoch:With $g_{\ast s,i}\simeq 100$, $g_{\ast s,0}\simeq 3.91$ and $T_0\simeq 2.35\times 10^{-13}\mathrm{GeV}$, the surviving energy density isTaking $m_{\mathrm{pl}}\simeq 1.22\times 10^{19}\mathrm{GeV}$ and $H_0\simeq 70\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ gives ${\rho }_{\mathrm{crit},0}\simeq 4.0\times 10^{-47}{\mathrm{GeV}}^4$. Requiring ${\rho }_{M0}<0.01{\rho }_{\mathrm{crit},0}$ yieldsUnknown geometric and mass prefactors change this estimate only through a fourth root. Conventional grand-unification scales near $10^{15}$–$10^{16}\mathrm{GeV}$ are far above the bound: this is the cosmological monopole problem, not a universal prohibition on unification. Inflation after monopole formation can dilute the relics; reheating must then stay below their regeneration temperature. Later entropy production, efficient annihilation, or a symmetry-breaking history avoiding stable monopoles can also weaken the bound. These mechanisms change the assumed conserved initial relic abundance.

Each separately conserved component obeys the constant-equation-of-state density scaling $\rho \propto a^{-3(1+w)}$. For the flat Friedmann-Lemaître-Robertson-Walker metric, the Friedmann equations therefore giveWrite $H(z)/H_0=1+h_{1}z+O(z^2)$, where $h_1=[3{\Omega }_m+4{\Omega }_r+3(1+w){\Omega }_X]/2$. Expanding the reciprocal inside the luminosity distance integral givesMatching its quadratic coefficient gives the deceleration parameter $q_0=h_1-1$. Using flatness in this part,A negative deceleration parameter signifies accelerated expansion; negative pressure must overcome the positive matter and radiation terms. Restoring units replaces $H_0{d}_L$ by $H_0{d}_L/c$ in the expansion, leaving $q_0$ unchanged.

In the observational setting of this paper, the cosmic microwave background favours nearly flat geometry, while cluster dynamics and large-scale structure favour a matter density near ${\Omega }_m\sim 0.3$. The missing contribution to the critical density is then roughly ${\Omega }_X\sim 0.7$. The supernova luminosity distance measurements require recent accelerated expansion. With present radiation negligible, $q_0<0$ requiresFor an X-dominated universe the threshold would instead be $w<-1/3$. Thus X cannot be ordinary pressureless matter or radiation. It must be comparatively smooth on cluster scales, so that it contributes to the background expansion without being counted as clustered dynamical matter. Its negative pressure makes its density dilute more slowly than matter, allowing late dominance. A cosmological constant, with $w=-1$, or sufficiently potential-dominated quintessence can have these properties.

The comparison with a perfect fluid in general relativity needs a distinction between background stress and perturbations. A homogeneous canonical scalar has canonical scalar stress as a perfect fluid:It has isotropic pressure and no background shear stress, just as a perfect fluid in general relativity does. Negative pressure therefore does not, by itself, prevent a perfect-fluid description. However, an adiabatic barotropic fluid with the constant closure $p=w\rho$ would have $c_s^2=dp/d\rho =w$. For negative $w$, a short-wavelength density mode then has ${\omega }^2\simeq w{k}_{\mathrm{phys}}^2<0$, the negative-w barotropic-fluid gradient instability. Such a closure cannot give a stable, smooth propagating X component.

A canonical scalar instead has rest-frame sound speed squared equal to one: its spatial and temporal kinetic terms have the same coefficient. Its pressure perturbation need not obey the background relation $\delta p=w\delta \rho$. This nonadiabatic response suppresses small-scale clustering without the negative-sound-speed instability. A cosmological constant is a separate limiting case with no independent propagating fluid density mode. Thus the issue is the microphysics and perturbation closure, not whether the background stress-energy tensor has perfect-fluid form.

Varying the canonical scalar action gives $\square \varphi +V_{,\varphi }=0$. In a homogeneous flat Friedmann-Lemaître-Robertson-Walker metric written in conformal time, $\sqrt{-g}=a^4$ and $g^{00}=a^{-2}$, soThe quadratic potential therefore givesEquivalently $\ddot{\varphi }+3H\dot{\varphi }+m^2\varphi =0$ in cosmic time. The intrinsic roll/oscillation time is $m^{-1}$ and the damping time is $H^{-1}$. If $m\gg H_0$, the scalar has long since begun rapid oscillations: averaging over a quadratic oscillation gives equal kinetic and potential energies, $⟨p⟩\simeq 0$, so it behaves as matter rather than accelerating dark energy. To remain nearly frozen or only begin rolling today requires the mass scale of quadratic quintessenceIf its energy is predominantly potential, equating $m^2{\varphi }_0^2/2$ to ${\rho }_{X0}=3M^2{H}_0^2{\Omega }_X$, with $M$ the reduced Planck mass, givesTaking the natural onset estimate $m\sim H_0$ and ${\Omega }_X\sim 0.7$ gives $|{\varphi }_0|\sim 2M\sim 5\times 10^{18}\mathrm{GeV}$, of order the Planck mass. More strongly frozen, negative-pressure evolution requires a smaller mass and correspondingly larger field. The density fixes the product $m|{\varphi }_0|$ at this accuracy, rather than a unique mass and field separately.

There are two normalization problems in the printed energy identity. In conformal time, the physical density of a homogeneous scalar is $T_{00}/a^2$, and its kinetic contribution contains ${\varphi }^{\prime 2}$, not one power of ${\varphi }^{\prime }$. The conformal density of a canonical scalar field follows directly from the stress-energy tensor:Writing $K={\varphi }^{\prime 2}/(2a^2)$, the condition $K-V=-(K+V)/3$ gives $V=2K$, henceThe reconstruction is consistent with this corrected identity; the unsquared printed identity cannot hold as an energy relation.

The constant-equation-of-state density scaling for this coasting fluid gives ${\rho }_{\varphi }={\rho }_{X0}{a}^{-2}$, where ${\rho }_{X0}=3M^2{H}_0^2{\Omega }_X$. Choose the branch on which $\varphi$ increases. ThenKeeping all three density parameters explicitly, the Friedmann equations in conformal time giveFor ${\Omega }_X>0$, defineThe square root is $\sqrt{(a+b{)}^2-A^2}$. Integrating on the expanding branch giveswhere $a({\varphi }_0)=1$. Thus the scalar-field reconstruction of a coasting component isSubstitution into the potential gives the requested present-field parametrization:The domain is the branch where $a>0$ and the hyperbolic argument is positive; reversing the direction of the field changes the sign of $B$. The actual discriminant condition is ${\Omega }_m^2/4>{\Omega }_r{\Omega }_X$. The supplied stronger bound suffices for physical density fractions $0<{\Omega }_X\le 1$. No density parameter has been eliminated using the flatness sum.

As a check, ${\varphi }^{\prime }$ is constant and $V_{,\varphi }=-2V(a^{\prime }/a)/{\varphi }^{\prime }$. Since $a^{2}V={\varphi }^{\prime 2}$ along this solution, ${\varphi }^{\prime \prime }+2(a^{\prime }/a){\varphi }^{\prime }+a^2{V}_{,\varphi }=0$ identically. The reconstructed field therefore satisfies its equation of motion as well as the Friedmann equations and required equation of state.