Consider a star moving with velocity v at an angle θ with respect to its line of sight to Earth. Show that the relativistic Doppler shift is
λobs=1−vccos(θ)√1−v2c2λem
in which c is the speed of light, λobs is the observed wavelength, and λem is the emitted wavelength.
Can someone show me to derive this equation? So far, I have been using a reference frame S′ for a certain angle θ′ in which the y′=ct′sin(θ′) and x′=ct′cos(θ′). I used the Lorentz transformation to find that
x=x′+vt′√1−v2c2=ct′cos(θ′)√1−v2c2=ct′(cos(θ′)+vc)√1−v2c2
I am not sure what do from here. Also, what happens for velocities that are much smaller than c? How can I use this equation to write how at v≪c the equation reduces to the usual expression for a Doppler shift such that
λobs=(1+vrc)λem
in which vr is the radial velocity?
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