= Solution
A small direction-dependent temperature change $T=\bar T(1+\Theta)$ shifts a thermal spectrum according to $\bar f(\epsilon/(1+\Theta))=\bar f(\epsilon)-\Theta\epsilon\bar f'(\epsilon)$ at first order. Hence $\boxed{\Theta=\delta T/\bar T}$ is the angular <photon temperature perturbation>. For a general spectrum the same ansatz describes an energy-independent brightness dilation; a literal thermodynamic temperature interpretation additionally assumes a thermal shape without spectral distortions.
Use $E=\epsilon/a$ in the tetrad <stress-energy tensor>. The background <photon energy density> is
$$
\boxed{\bar\rho_\gamma=\frac{4\pi}{a^4}\int_0^\infty d\epsilon\,\epsilon^3\bar f(\epsilon).}
$$
Integration by parts gives $-\int d\epsilon\,\epsilon^4\bar f'=4\int d\epsilon\,\epsilon^3\bar f$, assuming the boundary term vanishes. Thus each directional energy perturbation is four times its temperature perturbation. Writing $\langle\cdot\rangle_\Omega=(4\pi)^{-1}\int d\Omega$, we have
$$
\delta T^{\hat0\hat0}=4\bar\rho_\gamma\langle\Theta\rangle_\Omega,\qquad
\delta T^{\hat0\hat i}=4\bar\rho_\gamma\langle e^i\Theta\rangle_\Omega.
$$
In the paper's convention for <photon angular temperature moments>, there is no $(2\ell+1)$ factor multiplying $\Theta_\ell$. <Legendre polynomial> orthogonality gives $\langle\Theta\rangle_\Omega=\Theta_0$ and $\langle e^i\Theta\rangle_\Omega=-i\hat k^i\Theta_1/3$. Comparing with the stated density and velocity conventions yields
$$
\boxed{\delta_\gamma=4\Theta_0,\qquad v_\gamma=-\Theta_1.}
$$
For completeness, $\langle e^ie^jP_2(\hat{\boldsymbol k}\cdot\boldsymbol e)\rangle_\Omega=(\hat k^i\hat k^j-\delta^{ij}/3)/5$. The $(-i)^2$ quadrupole phase then gives $\boxed{\Pi_\gamma=-3\Theta_2/5}$. These moment relations are <photon angular temperature moments> and keep the <anisotropic stress> convention consistent with the subsequent hierarchy.
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