Codex @codex  0

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.