Relativistic Doppler effect

Remarks

Proof idea: the wave vector and angular frequency form a four-vector $(\omega / c, \mathbf{k})$, because the phase $\mathbf{k} \cdot \mathbf{x} - \omega t$ is the same in all frames. Apply the Lorentz transformation to the source frame: $\omega_s = \gamma (\omega_r - \mathbf{v} \cdot \mathbf{k}_r) = \gamma \omega_r (1 - \beta \cos\theta)$, using $\abs{\mathbf{k}_r} = \omega_r / c$. Example: at $\theta = \pi / 2$ in the receiver frame one gets $f_r = f_s / \gamma < f_s$, the transverse Doppler effect, which has no classical counterpart and is a form of time dilation. For $\beta = 0.6$ and $\theta = 0$, $f_r = f_s \sqrt{1.6 / 0.4} = 2 f_s$. To first order in $\beta$ the formula agrees with the classical Doppler formula $f_r \approx f_s (1 + \beta \cos\theta)$.

Going deeper

For a student

The cleanest derivation treats $(\omega / c, \mathbf{k})$ as a null four-vector and transforms it like any other four-vector. The same transformation gives the aberration of light, the change of the direction $\theta$ between frames. The longitudinal formula is used to measure the speeds of distant galaxies and of fast particles, and its square-root form makes it symmetric between source and receiver.