16 Jul 2025

Relativistic Doppler Effect

Longitudinal, angular, and transverse relativistic Doppler shifts with explicit sign and angle conventions.

bsc semester-i special-relativity relativistic-doppler-effect

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.

The second light ray starts farther from the observer; the plotted intercepts give $T_{\rm obs}=\gamma T_0(1+\beta)$.

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 $\theta$.

Solved Problems

1. Recession speed from a spectral-line redshift

A source emits the hydrogen line at rest wavelength $\lambda_0=656.28\ \mathrm{nm}$. A stationary observer in frame $S$ measures $\lambda_{\rm obs}=820.35\ \mathrm{nm}$ while the source recedes along the line of sight. Find its speed. Take positive radial velocity to mean recession.

For longitudinal recession,

\[R\equiv\frac{\lambda_{\rm obs}}{\lambda_0} =\sqrt{\frac{1+\beta}{1-\beta}}.\]

Here $R=820.35/656.28=1.2500$. Squaring and solving,

\[R^2(1-\beta)=1+\beta,\] \[\beta=\frac{R^2-1}{R^2+1} =\frac{1.25^2-1}{1.25^2+1} =\boxed{0.21951}.\]

Therefore

\[v=+0.21951c =\boxed{6.58\times10^7\ \mathrm{m\,s^{-1}}}.\]

The sign is positive because the observed wavelength is longer than the rest wavelength. For $R\to1$, $v\to0$; at small redshift, $\beta\simeq R-1$, whereas the exact expression remains subluminal for every finite $R$.

2. Viewing angle for zero net Doppler shift

A source moves at $v=+0.600c$ along $+x$. At what observer-frame angle $\theta$ between $+x$ and the light propagation direction is the observed frequency equal to the source rest frequency?

The angular formula is

\[\frac{\nu_{\rm obs}}{\nu_0} =\frac{1}{\gamma(1-\beta\cos\theta)}.\]

For $\beta=0.600$, $\gamma=1.25$. Setting the ratio to one gives

\[\gamma(1-\beta\cos\theta)=1,\] \[\cos\theta =\frac{1-1/\gamma}{\beta} =\frac{1-0.800}{0.600} =\frac13.\]

Hence

\[\boxed{\theta=\cos^{-1}\!\left(\frac13\right) \simeq70.53^\circ}.\]

The photon has a positive $x$ component because $\cos\theta>0$. At smaller angles the longitudinal approach contribution dominates and produces a blueshift; at larger angles time dilation dominates and produces a redshift. This zero-shift angle tends to $90^\circ$ as $\beta\to0$, although the shift itself then vanishes for every angle.

Descriptive Questions

1. Derive the longitudinal relativistic Doppler factor for a receding source from successive emission and reception events.
2. Obtain the approaching-source result and show that the classical first-order shift is recovered for small radial speed.
3. Derive the observer-angle formula by Lorentz transforming the wave frequency and wave-vector component.
4. Explain the transverse Doppler redshift and why the reference frame used to define the angle must be stated.

Numerical Problems

1. A receding source moves at $0.800c$ and emits at $600\ \mathrm{THz}$. Find the observed frequency. Answer: $\nu_{\rm obs}=200\ \mathrm{THz}$.
2. A source approaches at $0.600c$ and emits wavelength $500\ \mathrm{nm}$. Find the observed wavelength. Answer: $\lambda_{\rm obs}=250\ \mathrm{nm}$.
3. A source moves transversely in the observer frame at $0.800c$ and emits at $500\ \mathrm{THz}$. Find the observed frequency. Answer: $\nu_\perp=300\ \mathrm{THz}$.
4. A source with $\beta=0.600$ is seen at $\theta=60^\circ$ and has $\nu_0=400\ \mathrm{THz}$. Find the observed frequency. Answer: $\nu_{\rm obs}=(8/7)\nu_0=457.14\ \mathrm{THz}$.
5. Two identical beacons move along the line of sight at equal unknown speeds, one approaching and one receding. Their observed frequencies are $800\ \mathrm{THz}$ and $200\ \mathrm{THz}$, respectively. Find their common speed and rest frequency. Answer: $v=0.600c$ and $\nu_0=\sqrt{\nu_{\rm app}\nu_{\rm rec}}=400\ \mathrm{THz}$.
6. A star recedes at $30.0\ \mathrm{km\,s^{-1}}$ and emits a $600\ \mathrm{nm}$ line. Compare the first-order and exact wavelength shifts using $c=3.00\times10^8\ \mathrm{m\,s^{-1}}$. Answer: $\Delta\lambda_{\rm first}=0.0600\ \mathrm{nm}$; $\Delta\lambda_{\rm exact}=0.060003\ \mathrm{nm}$.

The derivations and all problem values are checked in the Maxima worksheet; every printed residual is zero.

References

1. “Relativistic Doppler effect,” Wikipedia.
2. A. P. French, Special Relativity, 1st ed., MIT Introductory Physics Series, W. W. Norton, 1968, Chapter 5, “Relativistic Kinematics.”
3. David J. Griffiths, Introduction to Electrodynamics, 4th ed., Cambridge University Press, 2017, Chapter 12, §12.1, “The Special Theory of Relativity.”