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Using the approximate large-angle CMB relation $T/S\simeq -7n_T$ gives $T/S\simeq 7{\sigma }^4/(96M^4{N}^3)$ in quartic hilltop inflation. It is suppressed by the fourth power of the small-field scale. This large-angle temperature ratio is not the primordial tensor-to-scalar ratio $r=16{ϵ}_V$; their transfer functions and normalization differ.

// Target: cosmology.bigb

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.

For the monomial inflation potential on $\varphi >0$, $V_{,\varphi }/V=n/\varphi$ and $V_{,\varphi \varphi }/V=n(n-1)/{\varphi }^2$. Thus, to leading slow-roll approximation,For $n=2,3,4$, $2n\ge n^2/2$. Therefore the loose order-one requirements $ϵ\le 1$ and $|\eta |\le 1$ giveThis is the interpretation under which the printed threshold follows. It is not a sufficient criterion for genuine slow roll, which requires much smaller parameters and hence ${\varphi }^2/M_{\mathrm{Pl}}^2\gg 2n$ for these fixed powers. Nor is its equality the actual end of acceleration for $n=2,3$: there $ϵ\simeq n/4<1$. For $n=4$ the leading estimate has $ϵ\simeq 1$. The potential-based order-one criterion $|{\eta }_V|<1$ also differs from the Hubble-flow criterion, especially at $n=4$. We use the endpoint the paper explicitly stipulates, rather than treating this boundary as a controlled slow-roll approximation.

Let ${\varphi }_e^2=2n{M}_{\mathrm{Pl}}^2$ by that prescription. The slow-roll e-fold count, using $\dot{\varphi }/H\simeq -n{M}_{\mathrm{Pl}}^2/\varphi$, isFor $N=60$, the monomial inflation with a prescribed endpoint result isApplying the leading scalar spectral index in Hubble slow-roll parameters and tensor-to-scalar ratio formulas givesThese are predictions at the chosen 60-e-fold initial point, normally identified with the pivot scale's horizon exit. A physical endpoint determined by $ϵ=1$ would instead have the leading estimate ${\varphi }_e^2\simeq n^2{M}_{\mathrm{Pl}}^2/2$; its change to the e-fold count and predictions must not be silently mixed with the stipulated $122n$ calculation.

Using the leading predictions just obtained givesThe quadratic and cubic monomial inflation potentials satisfy the quoted interval for the scalar spectral index, whereas the quartic model lies below its lower endpoint. All three exceed the quoted bound on the tensor-to-scalar ratio. ThusThis conclusion uses the exam's numerical bounds and the specified 60-e-fold endpoint convention, not a substitution of later observational constraints. Even a modest endpoint adjustment does not reduce the predicted $r$ by the roughly factor of two or more needed here.

For $V=\lambda {\varphi }^4$,Thus $n_s-1=-6{ϵ}_V+2{\eta }_V=-3{ϵ}_V$. Combining this with the tensor-to-scalar ratio $r=16{ϵ}_V$ gives the quartic-inflation tilt-tensor relationThe measured tilt $n_s-1=-0.04$ would require $r\simeq 0.213$, whereas $r=0.001$ predicts $n_s-1\simeq -1.88\times 10^{-4}$. Therefore quartic slow-roll inflation is inconsistent with the stated measurements.

The expressions in part c implySince the tensor-to-scalar ratio is $r=16ϵ$,For positive $\alpha$, this class predicts a red scalar tilt andA blue tilt with positive $r$, or $r\ge 8(1-n_s)$, is incompatible. An allowed pair fixes $\alpha =2r/[8(1-n_s)-r]$, subject to slow roll and observational bounds.

Taking the ratio of the supplied primordial tensor power spectrum and scalar spectrum cancels the common factor $(H/2\pi {)}^2$:To first slow-roll order $ϵ={ϵ}_V$, so the tensor-to-scalar ratio is

Use the geodesic radial comoving coordinate, in which the Friedmann-Lemaître-Robertson-Walker metric has the formOn a slice of constant cosmic time, the radial proper-length element is $R(t)\mathrm{d}\chi$. ThereforeIf one starts from a curvature coordinate with radial element $R\mathrm{d}r/\sqrt{1-k{r}^2}$, then $\chi ={\int }_0^r\mathrm{d}{r}^{\prime }/\sqrt{1-k{r}^{\prime 2}}$; the curvature coordinate $r$ itself is not generally the $\chi$ used above.

For a comoving galaxy, $\dot{\chi }=0$, so $\dot{D}=\dot{R}\chi =HD$. The definitions areThis recession law concerns proper distances on common cosmic-time slices. It is not a local special-relativistic relative velocity measurement; a large recession rate is not a signal exceeding the local speed of light.

For a comoving volume $V=V_0{R}^3$, total fluid energy is $E=\rho c^{2}V$. Adiabatic energy conservation gives $\mathrm{d}E+P\mathrm{d}V=0$. Expanding,henceHere $\rho$ is mass-equivalent energy density. Radiation loses energy both by dilution and by pressure work; pressureless matter only dilutes in this description.

In the time-unit convention for the cosmological constant, it represents a uniform vacuum contribution with pressure equal to minus its energy density. There is a units convention in the printed field equations: their term is $\Lambda /3$, so the symbol $\Lambda$ there has units of inverse time squared. If the geometrical constant is instead given in inverse length squared, this symbol equals $c^2{\Lambda }_{\mathrm{geom}}$. The curvature parameter $k$ labels constant-curvature spatial slices; with a conventional scale-factor normalization it is $+1,0,-1$ for positively curved, flat and negatively curved slices. It fixes local curvature sign, not every possible global spatial topology.

For a flat fluid-only model without a cosmological constant and with $P=w\rho c^2$, the two field equations giveThe two printed cases and their scale-factor laws are worked under the literal inline labels below.

Inflation is accelerated expansion, $\ddot{R}>0$, equivalently a decreasing comoving Hubble radius $(RH{)}^{-1}$ on an expanding branch. An approximately constant positive vacuum-like energy can give nearly exponential expansion. Enough inflation can put regions now widely separated inside a common earlier causal patch, suppress $|k|/(RH{)}^2$ and dilute unwanted relics such as monopoles. These are the horizon, flatness and relic-abundance motivations. Reheating is needed to populate the subsequent hot universe. Quantum fluctuations can seed primordial perturbations, but inflation alone is not a universal solution to the initial singularity or a proof that all initial conditions inflate.

Now adopt the printed natural units $c=\hslash =1$. For the Homogeneous massive-scalar equation in an expanding universe,The continuity equation givesWhere $\dot{\varphi }\ne 0$, this implies the expanding-universe Klein-Gordon equationA regular canonical-field solution extends it through isolated turning points by continuity. Purely dividing the conservation equation does not determine a trajectory on an interval where $\dot{\varphi }$ is identically zero; canonical field dynamics, or variation of its action, supplies the equation there. In particular a constant nonzero massive field is not an exact freely evolving solution.

With $\Lambda =k=0$,ThereforeThis inflation condition for a homogeneous massive scalar means a potential-dominated field has sufficiently negative pressure to accelerate expansion. In quadratic chaotic inflation, a large initial field with $V=m^2{\varphi }^2/2$ can enter slow roll, with $3H\dot{\varphi }\simeq -m^2\varphi$ and $H^2\simeq 4\pi G{m}^2{\varphi }^2/3$. The potential slow-roll parameter is ${ϵ}_V=1/(4\pi G{\varphi }^2)$, so a large field relative to the gravitational scale can support many e-folds. The field eventually rolls far enough that this regime ends and oscillation and reheating become relevant. Large potential energy with arbitrary initial kinetic energy does not by itself guarantee an inflationary interval.

Differentiating the flat Friedmann equation and then using the scalar equation,For the expanding branch,This Nonincreasing Hubble rate for a canonical inflaton decreases strictly where the field rolls and can be momentarily stationary at a turning point; the statement “always decreasing” means nonincreasing. This does not conflict with acceleration, because $\ddot{R}/R=H^2+\dot{H}$ remains positive when ${ϵ}_H=-\dot{H}/H^2<1$.

At horizon exit, larger comoving wavelengths leave earlier, when $H$ is larger. For a canonical field in slow roll, with reduced Planck mass ${\overline{M}}_{\mathrm{Pl}}=(8\pi G{)}^{-1/2}$,The decreasing expansion rate gives a red tensor tilt at this order. The scalar curvature spectrum also contains the conversion from field fluctuations to curvature:Here $N=\ln R$ counts elapsed e-folds. Thus the Decreasing Hubble rate does not alone fix scalar tilt. Evolution of ${ϵ}_H$ matters. For the quadratic slow-roll model, ${\eta }_V={ϵ}_V$ and $n_s-1\simeq -6{ϵ}_V+2{\eta }_V=-4{ϵ}_V<0$. If $N_{\mathrm{rem}}$ e-folds remain until the end, leading estimates give $n_s\simeq 1-2/N_{\mathrm{rem}}$ and tensor-to-scalar ratio $r_{\mathrm{TS}}\simeq 8/N_{\mathrm{rem}}$, with small end-of-inflation corrections.

The relevant contemporary comparison is a nearly scale-invariant but red scalar spectrum. The February2015 Planck analysis reported $n_s=0.968\pm 0.006$ for its temperature, low-polarization and lensing combination, and an upper limit $r_{0.002}<0.11$ at95percent confidence. The scalar tilt is qualitatively consistent with a slowly evolving inflaton, but that result does not establish the quadratic potential; its relatively large tensor prediction was already disfavored against lower-tensor alternatives in that analysis.^[ref]

For canonical slow-roll inflation with $V(\varphi )=\lambda {\varphi }^4$, the leading scalar spectral index and tensor-to-scalar ratio obeyIt is a parameter-free consistency check of the quartic potential at leading slow-roll order.