= Solution
For one <Fourier mode>, the trace-free differential operator in the <Einstein field equations> is $-k^2Q_{ij}$, with $Q_{ij}=\hat k_i\hat k_j-\delta_{ij}/3$. Part (ii) gives $\Pi_{ij}=(4/5)\bar\rho_\nu\Theta_2Q_{ij}$. The two minus signs in the field equation therefore imply
$$
k^2(\phi-\psi)=\frac{32\pi G}{5}a^2\bar\rho_\nu\Theta_2.
$$
Insert $\Theta_2=\psi_0 k^2\eta^2/6$. During <radiation domination>, $a\propto\eta$, $\mathcal H=1/\eta$, and the <Friedmann equation> gives $8\pi Ga^2\bar\rho=3\mathcal H^2$. Thus
$$
\phi_0-\psi_0=\frac{16\pi G}{15}a^2\bar\rho_\nu\eta^2\psi_0
=\frac25 f_\nu\psi_0,
\qquad
\boxed{\phi_0=\left(1+\frac25f_\nu\right)\psi_0}.
$$
These are the <radiation-era adiabatic potentials with free-streaming neutrinos>; photons are taken to have negligible <scalar anisotropic stress>.
To normalize the potentials, use $\bar P=\bar\rho/3$. The denominator in the <comoving curvature perturbation> becomes
$$
4\pi Ga^2(\bar\rho+\bar P)=\frac{16\pi G}{3}a^2\bar\rho=2\mathcal H^2.
$$
With $\dot\phi=0$, the conserved curvature therefore obeys $\mathcal R=-\phi_0-\psi_0/2$. Combining it with the potential ratio gives
$$
\boxed{\psi_0(\mathbf k)=-\frac{10}{15+4f_\nu}\mathcal R(\mathbf k),\qquad
\phi_0(\mathbf k)=-\frac{10+4f_\nu}{15+4f_\nu}\mathcal R(\mathbf k)}.
$$
As a sign and normalization check, $f_\nu=0$ gives $\phi_0=\psi_0=-2\mathcal R/3$. Increasing the free-streaming radiation fraction increases $\phi_0/\psi_0$ while preserving the supplied curvature convention. The TeX's malformed definition of $\mathcal H$ is restored by the PDF: $\mathcal H=\dot a/a$.
Back to article page
