Problem Statement
Taking into account that the wave phase $\phi$ is a Lorentz invariant (i.e., the same in all inertial frames), derive the relativistic formula for the Doppler effect. A source of light moves with velocity $v$ at an angle $\theta$ to the direction toward a stationary observer. Find the frequency of light received by the observer.
Given Information
- $\nu_0$ — proper frequency emitted by the source (in source rest frame)
- $v$ — speed of source relative to observer
- $\theta$ — angle between $\vec{v}$ and the direction from source toward observer (in lab frame)
- $\beta = v/c$, $\quad\gamma = 1/\sqrt{1-\beta^2}$
Physical Concepts & Formulas
The phase of an electromagnetic wave $\phi = \omega t – \vec{k}\cdot\vec{r}$ counts wave crests, so it is a Lorentz scalar. The four-vector $(\omega/c,\,\vec{k})$ transforms exactly like the energy-momentum four-vector. Applying a Lorentz boost along the direction of source motion and setting $|\vec{k}| = \omega/c$ (light dispersion) gives the relativistic Doppler formula directly, without any model-dependent assumptions.
- $(\omega/c,\,\vec{k})$ — wave four-vector; transforms like $(E/c,\,\vec{p})$ under Lorentz boosts
- $\omega’ = \gamma(\omega – v k_x)$ — Lorentz transformation of angular frequency
- $|\vec{k}| = \omega/c$ — dispersion relation for light in vacuum
Step-by-Step Solution
Step 1 — Set up the geometry and four-vector transformation: Take the source moving along the $x$-axis with speed $v$. The wave travels from source to observer; the component of $\vec{k}$ along $\vec{v}$ is $k_x = -(\omega/c)\cos\theta$ (negative because the wave travels toward the observer, opposite to $\vec{v}$). Applying the Lorentz boost to the frequency component:
$$\omega_0 = \gamma(\omega – v k_x) = \gamma\!\left(\omega + \frac{v\omega\cos\theta}{c}\right) = \gamma\omega(1+\beta\cos\theta)$$
Step 2 — Solve for observed frequency $\nu$: Since $\omega_0 = 2\pi\nu_0$ and $\omega = 2\pi\nu$:
$$\nu_0 = \gamma\nu(1+\beta\cos\theta) \implies \nu = \frac{\nu_0}{\gamma(1+\beta\cos\theta)}$$
Step 3 — Write the final formula and verify limiting cases: Substituting $\gamma = 1/\sqrt{1-\beta^2}$:
$$\nu = \nu_0\,\frac{\sqrt{1-\beta^2}}{1+\beta\cos\theta}$$
Worked Calculation
Three important special cases confirm the formula:
$$\theta=\pi \;\text{(approaching)}:\quad \nu = \nu_0\sqrt{\frac{1+\beta}{1-\beta}} \quad \text{(blueshift)}$$
$$\theta=0 \;\text{(receding)}:\quad \nu = \nu_0\sqrt{\frac{1-\beta}{1+\beta}} \quad \text{(redshift)}$$
$$\theta=\pi/2 \;\text{(transverse)}:\quad \nu = \nu_0\sqrt{1-\beta^2} = \nu_0/\gamma \quad \text{(pure time-dilation redshift)}$$
Answer
$$\boxed{\nu = \nu_0\,\frac{\sqrt{1-\beta^2}}{1+\beta\cos\theta}}$$
This is the general relativistic Doppler formula. For $v \ll c$ it reduces to the classical result $\nu \approx \nu_0(1-\beta\cos\theta)$. The transverse case gives a frequency shift $\nu = \nu_0/\gamma$ with no classical counterpart — purely a consequence of time dilation.
Physical Interpretation
The derivation demonstrates that the Doppler effect is a direct consequence of Lorentz invariance of the wave phase, not a mechanical wave-medium effect. The transverse Doppler shift was confirmed by Ives and Stilwell (1938) using fast hydrogen canal rays — the first direct experimental verification of time dilation. In modern applications: cosmological redshift of distant galaxies uses $\theta = 0$; GPS satellites require second-order Doppler corrections (together with gravitational redshift) of $\sim 38\,\mu\text{s/day}$ to maintain accuracy; Mossbauer spectroscopy exploits the extreme sharpness of nuclear gamma lines to measure the second-order term $\delta\nu/\nu_0 \approx -v^2/(2c^2)$ from thermal motion of atoms in a crystal.
Leave a Reply