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

In a local orthonormal frame, a massless neutrino has $p^{\hat{a}}=E(1,\mathbf{e})$ and $E=ϵ/a$. With degeneracy $g_{\nu }$, kinetic theory gives the stress-energy tensorCoordinate components follow by inserting the orthonormal-frame vectors on the indices.

For the perturbation $\delta f=-\Theta ϵ{\bar{f}}^{\prime }(ϵ)$, integration by parts givesassuming the boundary term vanishes. Consequently the physical trace-free spatial stress isThe paper uses the opposite signed source, ${\Pi }^{\hat{i}\hat{j}}=-{\pi }_{\nu }^{\hat{i}\hat{j}}$, consistently with its tensor equation having $-16\pi G{a}^2{\Pi }_{ij}$ on the right. In that convention the neutrino tensor anisotropic stress is

The Collisionless Boltzmann equation is $df/d\eta =0$. At first order, directional deflections act only on the isotropic background and hence give no angular contribution. Differentiating $f=\bar{f}-\Theta d\bar{f}/d\ln ϵ$ along the unperturbed trajectory givesUsing the tensor redshift of the comoving energy,The method of characteristics follows the backward ray $\mathbf{x}({\eta }^{\prime })=\mathbf{x}-(\eta -{\eta }^{\prime })\mathbf{e}$. With zero initial perturbation, the line-of-sight solution isTo this order the frame and coordinate direction components may be identified inside a quantity already proportional to $h$.

Write $m_{ij}=m_{ij}^{(\pm 2)}$ and $h=h^{(\pm 2)}$. Since $m_{ij}{e}^i{e}^j=\frac{1}{2}(1-{\mu }^2)e^{\pm 2i\phi }$, the line-of-sight solution isThe angular pattern $m_{ij}{e}^i{e}^j$ is proportional to $Y_{2,\pm 2}$. Projecting onto this spherical harmonic multiplies its coefficient bywhere $x=k(\eta -{\eta }^{\prime })$ and $j_2$ is a Spherical Bessel function. Thus the quadrupole isUse the isotropic tensor integralBecause $m$ 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 givesThe neutrino tensor free-streaming kernel is finite at the origin, with $K(0)=1/15$.

Insert the result for the neutrino tensor anisotropic stress into the tensor wave equation and divide out $m_{ij}{e}^{i\mathbf{k}\cdot \mathbf{x}}/\sqrt{2}$. Since the Fourier transform replaces $-{\nabla }^2$ by $k^2$, the source coefficient is $-64\pi G{a}^2{\bar{\rho }}_{\nu }$. The flat Friedmann equation givesso $64\pi G{a}^2{\bar{\rho }}_{\nu }=24{\mathcal{H}}^2{\bar{\rho }}_{\nu }/{\bar{\rho }}_{\mathrm{tot}}$. HenceThis memory term transfers coherent tensor motion into directional neutrino perturbations. Free-streaming neutrinos damp the tensor oscillation amplitude after horizon entry, suppressing the gravitational-wave contribution to cosmic microwave background temperature and polarization spectra. The kernel acts on $\dot{h}$, so a constant superhorizon tensor mode is unaffected at leading order.