For either tensor <helicity>, <neutrino free streaming> produces the memory equation
$$
\ddot h+2\mathcal H\dot h+k^2h
=-24\mathcal H^2f_\nu\int_0^\eta
K(k(\eta-\eta'))\dot h(\eta')\,d\eta',
\qquad
f_\nu=\frac{\bar\rho_\nu}{\bar\rho_{\rm tot}}.
$$
The <neutrino tensor free-streaming kernel> encodes the angular propagation of the perturbation. Energy transferred to neutrino directional anisotropy damps the oscillation amplitude of a <gravitational wave> after horizon entry. This effect was calculated by https://arxiv.org/abs/astro-ph/0306304[Weinberg].

Gravitational-wave damping by free-streaming neutrinos

Write $m_{ij}=m_{ij}^{(\pm2)}$ and $h=h^{(\pm2)}$. Since $m_{ij}e^ie^j=\tfrac12(1-\mu^2)e^{\pm2i\varphi}$, the line-of-sight solution is
$$
\Theta=-\frac{e^{i\mathbf k\cdot\mathbf x}}{2\sqrt2}
m_{ij}e^ie^j
\int_0^\eta d\eta'\,\dot h(\eta')e^{-ik(\eta-\eta')\mu}.
$$
The angular pattern $m_{ij}e^ie^j$ is proportional to $Y_{2,\pm2}$. Projecting onto this <spherical harmonic> multiplies its coefficient by
$$
\frac{\int_{-1}^{1}(1-\mu^2)^2e^{-ix\mu}\,d\mu}
{\int_{-1}^{1}(1-\mu^2)^2\,d\mu}
=15\frac{j_2(x)}{x^2},
$$
where $x=k(\eta-\eta')$ and $j_2$ is a <Spherical Bessel function>. Thus the quadrupole is
$$
\boxed{\Theta\big|_{\ell=2}
=-\frac{15}{2\sqrt2}m_{ij}e^ie^j e^{i\mathbf k\cdot\mathbf x}
\int_0^\eta d\eta'\,\dot h(\eta')\frac{j_2(k(\eta-\eta'))}{[k(\eta-\eta')]^2}}.
$$
Use the <isotropic tensor integral>
$$
\int\frac{d\Omega_e}{4\pi}e^ie^je^ae^b
=\frac1{15}(\delta^{ij}\delta^{ab}+\delta^{ia}\delta^{jb}+\delta^{ib}\delta^{ja}).
$$
Because $m$ is symmetric and trace-free, contracting this with $m_{ab}$ gives $2m^{ij}/15$. Substitution into the paper's signed <neutrino tensor anisotropic stress> therefore gives
$$
\boxed{\Pi^{ij}=2\sqrt2\,\bar\rho_\nu\,e^{i\mathbf k\cdot\mathbf x}m^{ij}
\int_0^\eta d\eta'\,\dot h(\eta')K(k(\eta-\eta'))},
\qquad K(x)=\frac{j_2(x)}{x^2}.
$$
The <neutrino tensor free-streaming kernel> is finite at the origin, with $K(0)=1/15$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 312 / 4 / iii / Solution

{title2=$K(x)=j_2(x)/x^2$}

Projecting free-streaming <neutrinos> onto the tensor <spherical harmonic> gives
$$
K(x)=\frac1{16}\int_{-1}^{1}(1-\mu^2)^2e^{-ix\mu}\,d\mu
=\frac{j_2(x)}{x^2},\qquad K(0)=\frac1{15}.
$$
Here $j_2$ is a <Spherical Bessel function>. The continuous value at zero follows by integrating $(1-\mu^2)^2$, whose integral is $16/15$.

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