Transverse Doppler shift and observation time ($\lambda =\gamma {\lambda }_0(1-\beta \cos \theta )$)

Let $\theta$ be the angle between the source velocity and the emitted photon's propagation direction in the observer's frame. The relativistic Doppler effect gives $\nu ={\nu }_0/[\gamma (1-\beta \cos \theta )]$. A ray transverse in that frame has $\lambda =\gamma {\lambda }_0$, a redshift. A ray transverse in the source frame instead has $\cos \theta =\beta$ by aberration of light, and $\lambda ={\lambda }_0/\gamma$, a blueshift.

The time at which a passing source is closest must also be specified. Put the observer at the origin and the source at $(v{t}_e,d)$ at emission, with finite transverse offset $d>0$. Reception occurs atEmission at closest approach means $t_e=0$, a transverse observer-frame ray received at $t_r=d/c$, hence a redshift. Reception at closest approach means $t_r=0$, so $t_e=-\gamma d/c$ and the ray has $\cos \theta =\beta$, hence a blueshift. Sending $d$ to zero does not remove the distinction between these two limiting event prescriptions. ^[ref]