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Write ${\psi }_0=\psi (0,\mathbf{k})$. For regular adiabatic initial conditions during radiation domination, use ${\Theta }_0=-{\psi }_0/2+O(k\eta )$, ${\Theta }_2=O((k\eta {)}^2)$, and constant potentials. The dipole equation in the neutrino Boltzmann hierarchy givesRegularity removes a constant dipole, and integration yieldsThen the quadrupole equation givesIts regular initial value is zero, soFor the usual regular growing solution, the monopole has no linear correction and parity improves the dipole and quadrupole errors to $O((k\eta {)}^3)$ and $O((k\eta {)}^4)$, respectively. The weaker bounds above already follow from the assumptions explicitly given. Although the quadrupole is small on superhorizon scales, the trace-free Einstein field equations divide its stress by $k^2$, allowing a finite leading potential difference.

Let $Q^{ij}={\hat{k}}^i{\hat{k}}^j-{\delta }^{ij}/3$. Rotations around $\hat{\mathbf{k}}$ and the vanishing trace imply that the angular integral in the scalar neutrino anisotropic stress has the form $A_{\ell }{Q}^{ij}$. Contract with ${\hat{k}}_i{\hat{k}}_j$ and use ${\mu }^2-1/3=(2/3)P_2(\mu )$. By the Orthogonality of Legendre polynomials,Only the quadrupole in the neutrino Boltzmann hierarchy contributes. Its phase factor is $(-i{)}^2=-1$, so the supplied stress definition becomesComparing with the chosen normalization $-(4/3){\bar{\rho }}_{\nu }\Pi Q^{ij}$ givesThe sign and factor here follow jointly from the stress convention and the unweighted Legendre polynomial coefficients; they should not be transferred unchanged to a differently normalized hierarchy.

For one Fourier mode, put $\mu =\hat{\mathbf{k}}\cdot \mathbf{e}$. The collisionless equation becomesUse the unweighted Legendre polynomial expansion specified in the paper: $\Theta ={\sum }_{j\ge 0}(-i{)}^j{\Theta }_j{P}_j(\mu )$. The Legendre polynomial recurrence relation saysThe coefficient of $P_{\ell }$ in $ik\mu \Theta$ receives contributions from $j=\ell -1$ and $j=\ell +1$. Dividing by $(-i{)}^{\ell }$ gives respectively $-k\ell {\Theta }_{\ell -1}/(2\ell -1)$ and $k(\ell +1){\Theta }_{\ell +1}/(2\ell +3)$. The metric sources are $\dot{\varphi }{P}_0$ and $-ik\psi P_1$, the latter becoming $k\psi$ after division by $(-i)$. Hence the neutrino Boltzmann hierarchy isThe lower-neighbour term is absent for $\ell =0$. The low moments areThese coefficients depend on the expansion convention. In the convention with a $(2\ell +1)$ factor, the temperature multipoles are ${\Theta }_{\ell }/(2\ell +1)$; using those multipoles without converting them would give incorrect factors in the scalar neutrino anisotropic stress.

For regular adiabatic initial conditions on superhorizon scales during radiation domination, the unweighted neutrino Boltzmann hierarchy gives ${\Theta }_1=\psi k\eta /2$ and ${\Theta }_2=\psi (k\eta {)}^2/6$. With $f_{\nu }={\bar{\rho }}_{\nu }/({\bar{\rho }}_{\nu }+{\bar{\rho }}_{\gamma })$ and neutrinos the only source of scalar anisotropic stress, the trace-free Einstein field equations gives $\varphi =(1+2f_{\nu }/5)\psi$. For the sign convention $\mathcal{R}=-\varphi -\psi /2$,

Using the unweighted neutrino Boltzmann hierarchy convention and ${\Pi }_{ij}=-(4{\bar{\rho }}_{\nu }/3){\Pi }_{\nu }({\hat{k}}_i{\hat{k}}_j-{\delta }_{ij}/3)$, the scalar scalar anisotropic stress is ${\Pi }_{\nu }=-3{\Theta }_2/5$. Only the quadrupole contributes, since the trace-free tensor $e_i{e}_j-{\delta }_{ij}/3$ has angular degree two. Different stress and multipole sign conventions change this displayed coefficient.