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Derivation

Relativistic Doppler Effect and Aberration

D-101 Home PU-105 Threads light · waves · symmetry Depends on Lorentz Transformation from the Two Postulates, The Four-Wavevector and Phase Invariance
Statement

For a source emitting monochromatic light of proper (rest-frame) angular frequency ω0, moving with constant velocity v = βc along the x-axis relative to an observer, the frequency received by the observer is ω = ω0 / [γ(1 − β cos θ)], where θ is the angle in the observer's frame between the photon's propagation direction and the source velocity, and γ = (1 − β2)−1/2. The same boost of the null four-wavevector fixes the aberration of the ray direction, cos θ′ = (cos θ − β)/(1 − β cos θ).

Why it matters

The relativistic Doppler formula unifies three effects that are separate in Newtonian physics: the ordinary longitudinal shift of an approaching or receding source, a second-order transverse shift with no classical analogue, and the fact that "who is moving" is symmetric between source and observer. It is the working tool of astrophysical spectroscopy (radial velocities, relativistic jets, redshifts) and of precision tests of special relativity such as the Ives–Stilwell experiment.

Aberration is the same boost read on the spatial part of the wave-vector rather than the time part. It explains stellar aberration, relativistic beaming ("headlight effect") of moving emitters, and the distortion of the sky seen by a fast observer — all of which follow from a single Lorentz transformation of one null four-vector.

Assumptions
The source emits a single sharp proper frequency ω0 in its own inertial rest frame.Drop it and there is no one frequency to shift; you must Doppler-transform each Fourier component of the spectrum separately and the line acquires a shape as well as a shift. Both frames are inertial and the relative velocity is constant during emission and reception.Drop it and the boost is only instantaneous; an accelerating source needs the instantaneous velocity at emission, plus light-travel-time bookkeeping for how θ changes across the signal. Propagation is in vacuum, so the wave-vector is null: ω = c|k| in every frame.Drop it (a dispersive medium) and the phase four-vector is time-like or space-like; the phase speed is not c, the null condition fails, and the shift picks up the medium's refractive index (and Cherenkov geometry). Spacetime is flat; gravity is negligible over the light path.Drop it and a gravitational (frequency) shift adds to the kinematic one; near compact objects the two must be separated carefully before a velocity can be read off.
Derivation
1
k μ = (ω/c, k) = (ω/c)(1, cos θ, sin θ, 0)
The four-wavevector of light is null, |k| = ω/c; write it in the observer frame S with propagation in the x–y plane at angle θ to the boost axis (prior result: four-wavevector of light). A
2
k′ 0 = γ(k 0 − β k 1),   k′ 1 = γ(k 1 − β k 0),   k′ 2 = k 2
k μ is a four-vector, so it boosts with the same matrix as the coordinates; here S′ (the source rest frame) moves at +v relative to S (prior result: Lorentz transformation from the postulates). A
3
ω′/c = γ(ω/c)(1 − β cos θ)  ⟹   ω′ = γ ω(1 − β cos θ)
Insert the components of step 1 into the time-part of step 2. The primed frequency is the frequency measured in the source's own frame. A
4
ω′ = ω0  ⟹   ω = ω0 / [ γ(1 − β cos θ) ]
In its rest frame the source emits its defining proper frequency, ω′ = ω0; solve step 3 for the observed ω. This is the full Doppler law. B
5
(ω′/c) cos θ′ = γ(ω/c)(cos θ − β),   (ω′/c) sin θ′ = (ω/c) sin θ
Do the same with the space-parts of step 2, writing k′ = (ω′/c)(cos θ′, sin θ′, 0). These carry the direction information. B
6
cos θ′ = (cos θ − β)/(1 − β cos θ),   sin θ′ = sin θ / [ γ(1 − β cos θ) ]
Divide each equation of step 5 by ω′/c and use step 3, ω/ω′ = 1/[γ(1 − β cos θ)]. The two combine consistently since cos2+sin2=1. This is the aberration of light. B
7
tan(θ′/2) = √[(1 + β)/(1 − β)] · tan(θ/2)
Form tan(θ′/2) = sin θ′/(1 + cos θ′) from step 6; the factors of (1 − β cos θ) cancel and γ(1 − β) = √[(1 − β)/(1 + β)] leaves the clean half-angle map. C
Result
ω = ω0 / [ γ(1 − β cos θ) ]      cos θ′ = (cos θ − β)/(1 − β cos θ)

Reading. The bracket (1 − β cos θ) is the ordinary (classical) Doppler factor built from the line-of-sight velocity component; the extra 1/γ in front is pure time dilation of the moving source and is what makes the effect relativistic. At θ = 90° in the observer's frame the classical factor is 1 and only the 1/γ survives — the transverse Doppler redshift. The aberration relation says the same boost tips every ray toward the forward direction: an isotropic emitter beams its light into a forward cone.

Units check. γ and (1 − β cos θ) are dimensionless, so ω carries the units of ω0 (rad s−1); the formula is identical for ordinary frequency f = ω/2π or, via λ = c/f, for wavelength. The aberration equation sets a dimensionless cosine equal to a ratio of dimensionless quantities. Both reduce to identities at β → 0.

Limiting cases
  • Longitudinal approach (θ = 0): ω = ω0 √[(1 + β)/(1 − β)] — blueshift, diverging as β → 1.
  • Longitudinal recession (θ = 180°): ω = ω0 √[(1 − β)/(1 + β)] — redshift, → 0 as β → 1.
  • Transverse (observer frame) (θ = 90°): ω = ω0/γ — a pure second-order redshift, ≈ ω0(1 − β2/2).
  • Transverse (source frame) (θ′ = 90°): ω = γ ω0 — a blueshift; the "transverse" answer depends on which frame's right angle you mean.
  • Non-relativistic (β ≪ 1): ω ≈ ω0(1 + β cos θ + …), recovering the classical first-order shift, with the β2/2 transverse term as the leading relativistic correction.
  • Aberration at β → 1: nearly all rays crowd into θ ≈ 0; the forward half-cone has half-angle ≈ 1/γ (relativistic beaming).
Breaks when
  • Dispersive medium. If the light travels through matter with refractive index n ≠ 1, the phase four-vector is no longer null (ω = c|k|/n), so step 1 fails; the shift then depends on n and can even change sign, and for v > c/n Cherenkov geometry takes over.
  • Non-inertial or gravitational settings. An accelerating source or curved spacetime invalidates the single global boost of step 2; only the instantaneous velocity enters locally, and a gravitational frequency shift (governed by the metric, not β) must be added separately.
  • Broad or incoherent source. If ω0 is not sharp (finite linewidth, thermal broadening) the single-frequency premise of step 4 breaks; each spectral component shifts by its own factor and the observed line is a convolution.
Failure modes
  • Forgetting the 1/γ. Using the classical factor (1 − β cos θ) alone drops the transverse effect entirely and gives a first-order-correct but second-order-wrong result.
  • Angle-frame confusion. Plugging the source-frame emission angle θ′ into the observer-frame formula (or vice versa). The formula ω = ω0/[γ(1 − β cos θ)] requires the observer-frame θ; with the emission angle use ω = γ ω0(1 + β cos θ′).
  • Wrong sign of the transverse shift. Asserting a transverse blueshift for the observer-frame right angle. When θ = 90° in the observer frame the result is a redshift, ω0/γ; the blueshift belongs to θ′ = 90°.
  • Propagation vs. line-of-sight direction. Confusing the photon's travel direction (source → observer) with the direction the observer looks (observer → source); they are antiparallel, so mixing them flips cos θ and swaps blue for red.
  • Confusing kinematic and cosmological redshift. Applying this flat-space formula to compute distances from galactic redshifts; cosmological redshift comes from metric expansion, not a global boost, and 1 + z ≠ √[(1+β)/(1−β)] in general.
Discussion

The entire result is one statement: the phase of a plane wave, φ = kμxμ, is a Lorentz scalar, so its gradient kμ is a genuine four-vector. Everything — longitudinal shift, transverse shift, aberration — is then just the four components of that vector seen in two frames. Doppler is the time component, aberration the space components; they are not independent phenomena but two projections of the same boost. This is why the number of photons is frame-independent yet their frequency and direction both change.

The transverse effect is the cleanest signature that time dilation is real and symmetric. A source at rest but a moving observer, or a moving source and a resting observer, are related by the same γ, and the observer-frame right angle always yields ω0/γ. The apparent paradox ("each sees the other's clock slow") is resolved by the angle: the right angle is defined in a particular frame, and specifying it breaks the symmetry cleanly.

Aberration and the intensity transformation together produce relativistic beaming. Because solid angle contracts forward and each photon is blueshifted, the specific intensity of an approaching relativistic jet is boosted by a large power of the Doppler factor 𝒟 = 1/[γ(1 − β cos θ)] (typically 𝒟3 or 𝒟4), which is why one lobe of an astrophysical jet can appear far brighter than its receding twin.

Read covariantly, the observed frequency is ω = kμuμ, the contraction of the photon wave-vector with the observer's four-velocity uμ = γ(c, v). Writing this out in components reproduces ω0/[γ(1 − β cos θ)] immediately, and it generalizes without change to curved spacetime — there kμuμ evaluated along a null geodesic, with a locally transported uμ, gives the combined kinematic-plus-gravitational shift. The flat-space Doppler law is thus the tangent-space shadow of a single geometric object.

Common misconceptions. "Transverse Doppler is negligible so it can't be tested" — it is exactly what Ives–Stilwell measured in 1938 by averaging forward and backward shifts, and modern ion-storage-ring versions confirm it to parts in 109. And "redshift means moving away" is only true longitudinally: a source moving purely sideways still redshifts.

Worked examples
1
Longitudinal approach. A probe approaches Earth head-on and broadcasts at proper frequency f0 = 100 MHz; its speed is β = 0.50 (θ = 0). Find the received f.
Approaching source along the line of sight, so use the longitudinal limit. Symbols first. A
2
f = f0 √[(1 + β)/(1 − β)] = f0 √(1.50/0.50) = f0 √3
Set θ = 0 in the master formula and simplify γ(1 − β) = √[(1 − β)/(1 + β)]. A
3
f = 100 MHz × 1.7321 = 173.2 MHz
Insert the number √3 = 1.7321. A
f ≈ 173 MHz  (blueshift of +73%)

Reading. A half-light-speed head-on approach nearly doubles the frequency; the classical prediction f0(1 + β) = 150 MHz falls short because it omits the time-dilation factor.

Units check. MHz × (dimensionless) = MHz.

1
Aberration + transverse-source Doppler. A source moving at β = 0.80 emits a photon at θ′ = 90° in its own frame (proper frequency f0 = 5.00×1014 Hz). Find the observer-frame angle θ and frequency f.
Emission angle is given in the source frame, so use the inverse relations. Symbols first. B
2
cos θ = (cos θ′ + β)/(1 + β cos θ′) = (0 + 0.80)/(1 + 0) = 0.80  ⟹   θ = 36.9°
Invert the aberration relation of step 6 (swap primes and flip the sign of β) with cos θ′ = 0. B
3
f = γ f0(1 + β cos θ′) = γ f0,   γ = 1/√(1 − 0.64) = 1/0.60 = 1.667
Use the emission-angle form of the Doppler law; with cos θ′ = 0 only the γ (transverse-source blueshift) survives. B
4
f = 1.667 × 5.00×1014 Hz = 8.33×1014 Hz
Insert γ = 5/3. A
θ ≈ 36.9°,   f ≈ 8.3×1014 Hz (blueshift)

Reading. A photon sent sideways by the source is seen well forward of 90° (beaming) and blueshifted by γ. The sideways-in-source case is a blueshift — the opposite sign to the sideways-in-observer transverse redshift.

Units check. Angle in degrees; Hz × (dimensionless) = Hz.

Problems
  1. A star recedes directly from Earth at β = 0.10. Find the fractional wavelength shift Δλ/λ0 of one of its spectral lines and compare with the naive classical value β.
    Solution Recession, θ = 180°: f/f0 = √[(1 − β)/(1 + β)] = √(0.90/1.10) = 0.9045, so λ/λ0 = 1/0.9045 = 1.1055. Thus Δλ/λ0 = 0.106 (10.6% redshift), slightly above the classical 10% because of the extra time-dilation factor.
  2. A source crosses the field of view so that its light reaches the observer at exactly θ = 90° in the observer's frame, with β = 0.30. By what fraction is a line redshifted, and how does this compare to the leading small-β estimate?
    Solution Transverse (observer frame): f = f0/γ. γ = 1/√(1 − 0.09) = 1/√0.91 = 1.0483, so f/f0 = 0.9540 — a 4.6% redshift. The small-β estimate 1 − β2/2 = 1 − 0.045 = 0.955 (4.5%) matches to first non-vanishing order; classically the transverse shift is exactly zero.
  3. An observer moves at β = 0.50 straight toward a star that lies at θ′ = 90° (directly abeam) in the star's rest frame. At what angle from the direction of motion does the observer see the star?
    Solution Use cos θ = (cos θ′ + β)/(1 + β cos θ′) with cos θ′ = 0: cos θ = 0.50/1 = 0.50, so θ = 60°. The star appears shifted 30° forward of its rest-frame abeam position — the aberration of starlight.
  4. At what speed β must a source move so that its observer-frame transverse (θ = 90°) frequency is exactly half the proper frequency?
    Solution Require f/f0 = 1/γ = 1/2, so γ = 2. Then β = √(1 − 1/γ2) = √(1 − 1/4) = √0.75 = 0.866. The source moves at 0.866c.
  5. (Ives–Stilwell.) Hydrogen atoms in a beam move at β = 0.0050 and emit the H-α line, λ0 = 656.28 nm. Wavelengths are measured looking forward (θ = 0) and backward (θ = 180°) along the beam. Compute both, and show their mean equals γλ0, a net redshift that measures time dilation.
    Solution Forward (approaching, blueshift): λf = λ0√[(1 − β)/(1 + β)] = 656.28 × 0.995012 = 653.01 nm. Backward (receding, redshift): λb = λ0√[(1 + β)/(1 − β)] = 656.28 × 1.005013 = 659.57 nm. Mean = (λf + λb)/2 = (λ0/2)[√((1−β)/(1+β)) + √((1+β)/(1−β))] = (λ0/2)·(2/√(1 − β2)) = γλ0. Numerically γ = 1.0000125, so the mean is 656.289 nm versus λ0 = 656.280 nm — a +0.008 nm shift independent of the first-order Doppler terms, isolating the transverse (time-dilation) effect.