16 Jul 2025
Relativistic Doppler Effect
Longitudinal, angular, and transverse relativistic Doppler shifts with explicit sign and angle conventions.
Let a source emit successive wave crests with proper period $T_0=1/\nu_0$. The two emissions occur at the same source position in its rest frame, so an observer who sees the source move at speed $v$ measures the emission interval
\[\Delta t_{\rm e}=\gamma T_0.\]Receding source
During this interval the source moves away by
\[\Delta x_{\rm e}=v\Delta t_{\rm e}=v\gamma T_0.\]The second crest must traverse this additional distance, taking $\Delta x_{\rm e}/c$. The reception period is therefore
\[\begin{aligned} T_{\rm obs} &=\Delta t_{\rm e}+\frac{\Delta x_{\rm e}}{c}\\ &=\gamma T_0(1+\beta). \end{aligned}\]Consequently,
\[\begin{aligned} \nu_{\rm obs} &=\frac{\nu_0}{\gamma(1+\beta)}\\ &=\nu_0\frac{\sqrt{1-\beta^2}}{1+\beta}\\ &=\boxed{\nu_0\sqrt{\frac{1-\beta}{1+\beta}}}. \end{aligned}\]The frequency decreases: this is a redshift.
Approaching source
If the source approaches, its second emission is closer by $v\gamma T_0$. The reception interval becomes
\[T_{\rm obs}=\gamma T_0(1-\beta),\]and
\[\boxed{\nu_{\rm obs} =\nu_0\sqrt{\frac{1+\beta}{1-\beta}}}.\]The frequency increases: this is a blueshift. For purely longitudinal motion, both cases can be written with signed radial velocity $\beta_r=v_r/c$, positive for recession:
\[\boxed{\frac{\nu_{\rm obs}}{\nu_0} =\sqrt{\frac{1-\beta_r}{1+\beta_r}}}.\]For $\lvert\beta_r\rvert\ll1$, use $(1+x)^a\simeq1+ax$:
\[\frac{\nu_{\rm obs}}{\nu_0} =(1-\beta_r)^{1/2}(1+\beta_r)^{-1/2} \simeq\left(1-\frac{\beta_r}{2}\right) \left(1-\frac{\beta_r}{2}\right) \simeq1-\beta_r.\]Thus the classical first-order Doppler shift is recovered, while the exact square-root factor contains the relativistic correction.
Arbitrary viewing angle
Let $\theta$ be the angle, measured in the observer’s frame, between the source velocity $+\hat{\mathbf x}$ and the light propagation direction. A light wave has $k_x=(\omega/c)\cos\theta$. Transforming its frequency to the source rest frame gives
\[\omega_0 =\gamma(\omega-vk_x) =\gamma\omega(1-\beta\cos\theta).\]Therefore the observed frequency is
\[\boxed{ \nu_{\rm obs} =\frac{\nu_0}{\gamma(1-\beta\cos\theta)} }.\]Here $0\le\beta<1$ and $-1\le\cos\theta\le1$, so $1-\beta\cos\theta\ge1-\beta>0$; the frequency is finite and positive throughout its physical domain.
This convention reproduces the longitudinal results: $\theta=\pi$ for a receding source and $\theta=0$ for an approaching source. When the light is transverse in the observer’s frame, $\theta=\pi/2$, so
\[\boxed{\nu_\perp=\frac{\nu_0}{\gamma}}.\]This purely relativistic transverse Doppler shift is a redshift. The angle must always be stated because the emission angle in the source frame is not the same as $\theta$.
The longitudinal and angular shift forms, their limiting angles, and the low-speed coefficient are verified in the Maxima worksheet; every printed residual is zero.
Discussion