= Doppler dipole
{c}
{title2=$\Delta T/T_0=(v/c)\cos\theta+O(v^2/c^2)$}
An observer moving through isotropic <blackbody radiation> sees a direction-dependent <temperature>. With the sky angle measured toward the velocity apex in the observer frame, $T(\theta)=T_0/[\gamma(1-\beta\cos\theta)]$, where $\beta=v/c$ and $\gamma=(1-\beta^2)^{-1/2}$. First-order expansion gives the displayed dipole. The forward sky is hotter. The <cosmic microwave background dipole> is an application.
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