Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 53 / 4

Use $\hslash =c=1$. For a canonical light scalar field with effective mass $m^2\ll H^2$, define the canonical Fourier mode $u_k=a\delta {\Phi }_k$. In the nearly de Sitter approximation it obeysDeep inside the Hubble radius, $k\gg aH$, the mode behaves like a Minkowski oscillator. Choosing the Bunch-Davies vacuum sets $u_k\to e^{-ik\eta }/\sqrt{2k}$. The normalized de Sitter solution isInflation stretches the physical wavelength until it exits the Hubble radius. For $|k\eta |\ll 1$, the growing scalar solution freezes with $|\delta {\Phi }_k{|}^2=|u_k/a{|}^2\to H^2/(2k^3)$. The dimensionless power spectrum is thereforeHere $H_k$ is evaluated when $k=aH$, since slow evolution near horizon exit fixes the frozen amplitude. Nearly constant $H$ and a light field make this spectrum almost scale-invariant; slow background evolution and small mass corrections produce a small tilt. The stretched vacuum modes supply the primordial stochastic fluctuations.

Vary the quadratic Einstein-Hilbert action within the transverse trace-free tensor sector. Integration by parts giveswith endpoint terms set to zero. Hence the tensor equation isA Fourier mode on super-Hubble scales has $k^2\ll {\mathcal{H}}^2$, so its leading equation is $(a^2{E}_{ij}^{T\prime }{)}^{\prime }=0$. Its solutions areThere is thus a constant tensor mode. During inflation the second mode decays; in exact de Sitter expansion it is proportional to $a^{-3}$ after shifting the constant of integration. Constancy is the leading super-Hubble behaviour, exactly realized for the zero-gradient solution.

Reality of $E_{ij}^T$ and the given polarization relation imply ${\psi }_{(p)}(-\mathbf{k})={\psi }_{(p)}(\mathbf{k}{)}^{\ast }$. Insert the Fourier expansion twice into each quadratic term. The spatial integral givesThis cancels the two factors of $(2\pi {)}^{-3/2}$. Set ${\mathbf{k}}^{\prime }=-\mathbf{k}$ and use $M_{ij}^{(q)}(-\mathbf{k})=M_{ij}^{(q)}(\mathbf{k}{)}^{\ast }$. The unit-normalization condition eliminates cross-polarization terms, givingFor the gradient term, $(i{k}_l)(i{k}_l^{\prime })$ becomes $k^2$ after the delta-function substitution. ThusThe full momentum-space integral and its reality relation are those of real fields; there is no additional independent copy of the negative-wavevector modes.

The tensor normalization with unit-norm polarization tensors identifies the canonically normalized real scalar for each helicity as $\delta {\Phi }_{(p)}=M_{\mathrm{Pl}}{\psi }_{(p)}$. The two polarization actions are independent and have precisely the massless scalar normalization, so the vacuum fluctuations derived in the introductory solution apply to each of them:In the isotropic Bunch-Davies vacuum, the independent helicities have no cross correlation. Translation invariance gives the wavevector delta function, and the dimensionless-spectrum convention givesThis is power per polarization in the printed metric convention $h_{ij}=2E_{ij}^T$ and unit-norm polarization basis. Summing polarizations or quoting a spectrum for $h_{ij}$ changes normalization factors; neither operation belongs in the displayed per-polarization ${\mathcal{P}}_{\psi }$.

Now use proper-time dots and $H=\dot{a}/a$. Differentiate the slow-roll relation $H^2\simeq V/(3M_{\mathrm{Pl}}^2)$, then substitute $\dot{\bar{\Phi }}\simeq -V_{,\Phi }/(3H)$:Thus the Hubble slow-roll parameter ${ϵ}_H$ agrees with ${ϵ}_V$ to this order. Since ${\mathcal{P}}_{\psi }\propto H_k^2$ and $k=aH$ at horizon exit,Consequently the tensor spectral index isThe final equality is the requested first-order slow-roll approximation; corrections beyond that order also affect the horizon-exit spectrum itself.