Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 312 / 3

Let $q^i=e^i-\frac{1}{2}{h}^i{}_j{e}^j$, so $p^0=ϵ/a^2$ and $p^i=(ϵ/a^2)q^i$. The orthonormal tetrad gives ${\delta }_{ij}{e}^i{e}^j=1$. To first order,Hence the time-index Christoffel symbols contract toThe $\mathcal{H}{h}_{ij}$ term cancels the perturbation in the coordinate direction norm. This cancellation is why the local tetrad direction must not be mistaken for the coordinate direction.

Along the photon null geodesic, $d/d\lambda =p^{0}d/d\eta$, andThe time component of the geodesic equation now gives the photon energy redshift from a tensor metric perturbation:The derivative on the left follows the ray; the dot on $h$ is a partial conformal time derivative. The background redshift has already been removed by using comoving energy. The calculation only needs ${\Gamma }^0{}_{00}$ and ${\Gamma }^0{}_{ij}$; the unrelated printed ${\Gamma }^i{}_{j0}$ should contain $\frac{1}{2}{\dot{h}}^i{}_j$, rather than the full derivative.

To first order, propagate along the unperturbed ray $\mathbf{x}(\eta )={\mathbf{x}}_0-({\eta }_0-\eta )\mathbf{e}$ and keep its orthonormal tetrad direction fixed. Corrections to direction or trajectory multiply an already first-order temperature source and are second order. Define the angular scattering sourceUsing the photon energy redshift from a tensor metric perturbation, the Photon Boltzmann equation with Thomson scattering along this ray becomesMultiply by the survival probability $e^{-\tau }$. The cosmological visibility function is $g(\eta )=-\dot{\tau }{e}^{-\tau }$, so integration givesEvery source here is evaluated on the same ray, and $\tau ({\eta }_0)=0$.

The instantaneous-visibility approximation makes $e^{-\tau }$ a step from zero to one at ${\eta }_{\ast }$. It suppresses the early boundary term and localizes the scattering source to $S_{\ast }$. To obtain only the requested tensor CMB line-of-sight source, additionally neglect this last-scattering angular source: in leading tight coupling, the tensor-induced incident quadrupole is small, and a tensor mode has no scalar monopole or dipole. Ignore later reionization and its scattering as well. ThenInstantaneous visibility alone would leave the term $S_{\ast }$; neglect of that term is an additional approximation. The time derivative on $h$ is essential and is present in the PDF, although it was lost in the extracted TeX.

Choose the arbitrary normalization of the complex plane wave polarization so that $h_{ij}=u_i{u}_{j}h(\eta )e^{ikz}$ with $\mathbf{u}=(1,i,0)$. For $\mathbf{e}=(\sin \theta \cos \phi ,\sin \theta \sin \phi ,\cos \theta )$, contraction givesAt the observer origin, write $\mu =\cos \theta$. The tensor CMB line-of-sight source isThe rest of the expression is independent of $\phi$. Projection onto $Y_{\ell m}^{\ast }\propto e^{-im\phi }$ therefore gives zero unless $m=2$. This is the helicity selection for tensor CMB multipoles. It applies to the specified complex helicity mode; taking a physical real field also introduces the conjugate mode and its conjugate azimuthal dependence.

For the very long wavelength solution during matter domination, differentiation gives $\dot{h}=-h_{\mathrm{prim}}{k}^2\eta /5+O(k^4{\eta }^3)$. Expand the plane wave factor once to obtainwhereLet ${\mathcal{N}}_{22}=\frac{1}{4}\sqrt{15/(2\pi )}$. Then $Y_{22}={\mathcal{N}}_{22}{\sin }^2\theta e^{2i\phi }$ and $Y_{32}=\sqrt{7}{\mathcal{N}}_{22}\mu {\sin }^2\theta e^{2i\phi }$. Reading off the long-wavelength tensor CMB quadrupole and octupole yieldsIn the limit ${\eta }_{\ast }/{\eta }_0\to 0$,A perfectly frozen tensor produces no redshift; the first contribution comes from its order-$k^2{\eta }^2$ evolution. The extra spatial phase makes the octupole one power of $k{\eta }_0$ smaller than the quadrupole.