Understanding Doppler Shift for Light in Special Relativity

[SteveDC]

How do Doppler shifts work for light if, according to special relativity, light is constant velocity for all observers? So if c is unchanged, then surely wavelength and frequency don't change.

I appreciate that I must be misunderstanding something, because redshift on stars occurs, but I am struggling to explain why.

Thanks in advance for any help :)

 

[WannabeNewton]

The frequency of a light wave corresponds to the time-like component of the wave 4-vector of a light wave, and components of 4-vectors are not invariant under non-trivial Lorentz transformations. More generally, energy is a frame dependent quantity. Why should wavelength and frequency of light waves be unchanged between inertial frames just because $c$ is unchanged between inertial frames? They can scale inversely under Lorentz transformations so as to cancel out any change in their ratio.

 

[Mentz114]

$c=\lambda\nu$ is the equation relating velocity and wavelength and frequency. So both can change while keeping c constant.

[edit. got it wrong first time]

 

[Bill_K]

How do Doppler shifts work for light if, according to special relativity, light is constant velocity for all observers? So if c is unchanged, then surely wavelength and frequency don't change.

One thing which is the same in all rest frames is the phase of the wave. That is, for example, the wave crest always remains a wave crest. A typical wave is exp(i(kx - ωt), where the phase is kx - ωt, or equivalently since ω = ck, the phase is k(x - ct). This quantity must be an invariant.

Under a Lorentz transformation,

x' = γ(x - vt)
t' = γ(t - v/c² x)

implying that x - ct just picks up an overall factor:

x' - ct' = γ(1 - v/c)(x - ct) = √((1 - v/c)(1 + v/c)) (x - ct)

Invariance requires that the wave vector k picks up the inverse factor:

k' = √((1 + v/c)(1 - v/c)) k

which represents the Doppler shift.