 Solution

Assume homogeneous space and time, isotropic space, inertial frames related linearly, the relativity principle and a common isotropic speed $c$ for light. Take coincident origins at zero and parallel axes, with $S^{\prime }$ moving at $v$ along $x$. The primed origin requires $x^{\prime }=A(x-vt)$. Write $t^{\prime }=Bx+Dt$; mapping each light ray $x=\pm ct$ into $x^{\prime }=\pm c{t}^{\prime }$ gives $D=A$ and $B=-Av/c^2$. Rotational symmetry about the boost axis gives a common transverse scale $B(v)$; reversing the boost direction by spatial isotropy gives $B(v)=B(-v)$. Reciprocity then gives $B(v{)}^2=1$, and continuity and the common orientation select $B(v)=1$. Thus transverse directions are unchanged. Reciprocity requires the inverse to have the same scale $A$ with $v$ replaced by $-v$. Composing gives $A^2(1-v^2/c^2)=1$; continuity from $A=1$ at $v=0$ selects the positive root. Hence the Lorentz transformation isThe printed wavelength formula uses the angle of the line of sight from observer towards star, opposite to photon propagation. Choose the transverse axis so the ray lies in the $xy$ plane. With that directed-angle convention the photon four-momentum isThe inverse coordinate boost takes a rest-frame four-vector to $S$, giving $p^0=\gamma (p^{\prime 0}+\beta p^{\prime 1})$ and $p^1=\gamma (p^{\prime 1}+\beta p^{\prime 0})$, where $\beta =v/c$. Thus $\nu =\gamma {\nu }^{\prime }(1-\beta \cos {\theta }^{\prime })$, and the aberration of light relation is $\cos \theta =(\cos {\theta }^{\prime }-\beta )/(1-\beta \cos {\theta }^{\prime })$. More directly, the boost from $S$ to $S^{\prime }$ gives $p^{\prime 0}=\gamma (p^0-\beta p^1)$, hence ${\nu }^{\prime }=\gamma \nu (1+\beta \cos \theta )$. Since wavelength is $c/\nu$, the relativistic Doppler effect givesIf the angle were instead measured along the actual photon momentum, both cosine signs would reverse and the same physical relation would have a minus sign. Specifying this orientation avoids confusing the two versions.

For $\cos \theta =1$, $\lambda /{\lambda }^{\prime }=\sqrt{(1+\beta )/(1-\beta )}$: a receding star is redshifted. For $\cos \theta =-1$, the ratio is $\sqrt{(1-\beta )/(1+\beta )}$: an approaching star is blueshifted. For $\cos \theta =0$, $\boxed{\lambda /{\lambda }^{\prime }=\gamma }$: the transverse redshift is the time-dilation effect. This transverse condition refers to the observer's angle, not ${\theta }^{\prime }=\pi /2$ in the source frame.