Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 71 / 4

For a homogeneous canonical scalar, ${\rho }_{\varphi }={\dot{\varphi }}^2/2+V$ and $P_{\varphi }={\dot{\varphi }}^2/2-V$. The acceleration equation requires $V>{\dot{\varphi }}^2$ for cosmic inflation, equivalently ${ϵ}_H=-\dot{H}/H^2<1$ in the flat scalar-dominated limit. Inflation need not be exactly de Sitter.

A standard sufficient regime for a long period is positive potential energy dominating the kinetic energy, physical gradient energy and curvature. Homogeneity over an expanding approximately Hubble-sized patch allows the gradient term to redshift away. In comoving coordinates the gradient energy is $|{\nabla }_{\mathrm{com}}\varphi {|}^2/(2a^2)$; the gradient in the displayed energy formula must be interpreted physically if no $a^{-2}$ is written there. RequireThe slow-roll equations are then $3H\dot{\varphi }\simeq -V^{\prime }$ and $H^2\simeq 8\pi V/(3m_{\mathrm{pl}}^2)$. Sufficient potential conditions areThe potential must remain sufficiently flat over a large enough field interval to produce the required expansion. These slow-roll conditions are stronger than the bare necessary condition for acceleration.

Write $V=V_0{e}^{-A\varphi /m_{\mathrm{pl}}}$, where the stated model has $V_0=m_{\mathrm{pl}}^4$, and take $A>0$. Slow roll givesThus $d[e^{A\varphi /(2m_{\mathrm{pl}})}]/dt=A^2\sqrt{V_0}/(\sqrt{96\pi }{m}_{\mathrm{pl}})$. Absorb the integration constant into $b$ to obtainIt follows that $H=16\pi /[A^2(t+b)]$, soThe exponential potential has constant ${ϵ}_V=A^2/(16\pi )$ and ${\eta }_V=A^2/(8\pi )$. Inflation requires $p>1$, whereas reliable slow-roll approximation requires $A^2/(8\pi )\ll 1$. This is the slow-roll normalization for exponential inflation.

The stated field normalization is a slow-roll result, not an exact solution of the original full equations. Keeping the acceleration and kinetic terms yields the same $p$ but $C^2=A^4{V}_0/[2m_{\mathrm{pl}}^2(48\pi -A^2)]$, as follows by substituting $\dot{\varphi }=2m_{\mathrm{pl}}/[A(t+b)]$ into both full equations. It agrees to leading order for small slope and is the exponential-potential power-law inflation solution.

There is no minimum about which this scalar can undergo the damped oscillations normally used for reheating. Its slow-roll parameters are constant, so an inflating exponential solution has no automatic exit. An independent stopping and energy-transfer mechanism is needed.

At the assumed exit, $V_R=m_{\mathrm{pl}}^4{e}^{-30}$. Equating this to the thermal radiation density givesfor $g_{\ast }=10^3$. Corrections from the small kinetic fraction do not change this rough estimate.

Curvature obeys $|\Omega -1|\propto (aH{)}^{-2}$. For constant ${ϵ}_H=A^2/(16\pi )$ and $N=\log (a_R/a_i)$, its suppression during cosmic inflation is $e^{-2(1-{ϵ}_H)N}$. Starting from order-one curvature and requiring present curvature no larger than order one givesAfter instantaneous reheating, entropy conservation implies $a_R/a_0=(T_0/T_R)(g_{\ast s,0}/g_{\ast s,R}{)}^{1/3}$, with $H_R\simeq 1.66\sqrt{g_{\ast }}{T}_R^2/m_{\mathrm{pl}}$. Present scales make the numerator roughly $60$; using the literal approximate units from question2 gives about $63$. A tighter present curvature tolerance adds its corresponding half-logarithm. In slow-roll approximation this is the conventional requirement of roughly sixty e-folds.

Because the field increases toward ${\varphi }_R$, enough expansion requiresThis is the flatness bound on the initial field in exponential inflation. It is an upper bound on the initial field, not a nonzero minimum: starting earlier at a smaller field produces more expansion. If one additionally restricts the initial potential below $m_{\mathrm{pl}}^4$, then ${\varphi }_i\ge 0$ supplies a lower bound, but that is a separate restriction. Without initial-curvature and final-tolerance choices there is no unique numerical initial-field value.