The relativistic Doppler shift describes how the observed frequency and wavelength of light change when the source and observer move at relativistic speeds relative to each other.
In contrast to the classical Doppler effect, the relativistic version incorporates:
Time dilation from Lorentz transformations.The invariance of the speed of light.
Relativistic transformation of angles between frames.
Consider a source moving at constant velocity v v x x S S
The source rest frame is S ′ S ′ v v S S
The source emits monochromatic light of frequency f ′ f ′ λ ′ λ ′
In the observer frame S S f f λ λ
The angle between the source velocity and the observed light direction is θ θ S S
The corresponding angle in the source rest frame is θ ′ θ ′
Our goals:
Derive expressions for f f λ λ β = v / c β = v / c θ θ
Understand how θ θ θ ′ θ ′
Apply these results to an AGN jet example .
Consider two successive wave crests emitted by the source.
In the source frame S ′ S ′
Δ t emit ′ = 1 f ′ Δ t emit ′ = f ′ 1
In the observer frame S S
Δ t emit = γ Δ t emit ′ = γ f ′ Δ t emit = γ Δ t emit ′ = f ′ γ
While the second crest is being emitted, the source moves forward by:
Δ x = v Δ t emit Δ x = v Δ t emit
The projection of this motion along the observer's line of sight is:
Δ x ∥ = v Δ t emit cos θ Δ x ∥ = v Δ t emit cos θ
Thus, the second crest travels an extra path length :
Δ ℓ = v Δ t emit cos θ Δ ℓ = v Δ t emit cos θ
This causes an extra time delay in arrival:
Δ t path = v Δ t emit cos θ c Δ t path = c v Δ t emit cos θ
Now, the total time between arrival of wave crests is:
Δ t obs = Δ t emit − Δ t path Δ t obs = Δ t emit − Δ t path
Why subtraction? Because if the source moves toward the observer, the extra path length is negative — the second crest arrives sooner.
Putting this together:
Δ t obs = Δ t emit ( 1 − v cos θ c ) Δ t obs = Δ t emit ( 1 − c v cos θ )
It is important to understand why we can combine the effects this way:
The time dilation factor affects how frequently wave crests are emitted in the observer frame.
The geometric delay affects how those emitted crests propagate to the observer.
Since these are independent stages of the process, their contributions to Δ t obs Δ t obs
The observed frequency is:
f = 1 Δ t obs f = Δ t obs 1
Substituting Δ t emit = γ / f ′ Δ t emit = γ / f ′
f = 1 ( γ f ′ ) ( 1 − v cos θ c ) f = f ′ 1 γ ( 1 − β cos θ ) f f = ( f ′ γ ) ( 1 − c v c o s θ ) 1 = f ′ γ ( 1 − β cos θ ) 1
f = f ′ 1 − β 2 1 − β cos θ f = f ′ 1 − β cos θ 1 − β 2
Using f λ = c f λ = c
λ = λ ′ 1 − β cos θ 1 − β 2 λ = λ ′ 1 − β 2 1 − β cos θ
Define the wave 4-vector :
k μ = ( ω c , k ⃗ ) k μ = ( c ω , k )
where:
ω = 2 π f ω = 2 π f k ⃗ k
∣ k ⃗ ∣ = ω c ∣ k ∣ = c ω
For a boost along the x x
ω = γ ( ω ′ + β c k x ′ ) k x = γ ( k x ′ + β ω ′ c ) k y = k y ′ k z = k z ′ ω k x k y k z = γ ( ω ′ + β c k x ′ ) = γ ( k x ′ + c β ω ′ ) = k y ′ = k z ′
In the source frame S ′ S ′ θ ′ θ ′
k x ′ = ω ′ c cos θ ′ k y ′ = ω ′ c sin θ ′ k z ′ = 0 k x ′ k y ′ k z ′ = c ω ′ cos θ ′ = c ω ′ sin θ ′ = 0
Substituting k x ′ k x ′ ω ω
ω = γ ( ω ′ + β c ω ′ c cos θ ′ ) = ω ′ γ ( 1 + β cos θ ′ ) ω = γ ( ω ′ + β c c ω ′ cos θ ′ ) = ω ′ γ ( 1 + β cos θ ′ )
Thus:
ω ω ′ = f f ′ = γ ( 1 + β cos θ ′ ) ω ′ ω = f ′ f = γ ( 1 + β cos θ ′ )
Substitute the relativistic abberation formula into the frequency expression:
f f ′ = γ ( 1 + β cos θ − β 1 − β cos θ ) = γ 1 − β 2 1 − β cos θ = 1 − β 2 1 − β cos θ f ′ f = γ ( 1 + β 1 − β cos θ cos θ − β ) = γ 1 − β cos θ 1 − β 2 = 1 − β cos θ 1 − β 2
Thus we recover the same result:
f = f ′ 1 − β 2 1 − β cos θ f = f ′ 1 − β cos θ 1 − β 2
f = f ′ 1 + β 1 − β , λ = λ ′ 1 − β 1 + β f = f ′ 1 − β 1 + β , λ = λ ′ 1 + β 1 − β
Blueshift — wave crests arrive more frequently.
f = f ′ 1 − β 1 + β , λ = λ ′ 1 + β 1 − β f = f ′ 1 + β 1 − β , λ = λ ′ 1 − β 1 + β
Redshift — wave crests arrive less frequently.
f = f ′ γ , λ = γ λ ′ f = γ f ′ , λ = γ λ ′
Pure time dilation — no classical Doppler shift.
Example: Doppler Shift from Relativistic Jet
A relativistic jet in an AGN is moving at speed v = 0.95 c v = 0 . 9 5 c θ = 2 0 ∘ θ = 2 0 ∘
The jet emits an emission line with rest wavelength λ ′ = 500 λ ′ = 5 0 0
What is the observed wavelength of this line?
What is the corresponding observed frequency shift?
Compute β β γ γ
β = 0.95 , γ = 1 1 − β 2 ≈ 3.202 β = 0 . 9 5 , γ = 1 − β 2 1 ≈ 3 . 2 0 2
Compute 1 − β cos θ 1 − β cos θ
cos 2 0 ∘ ≈ 0.9397 cos 2 0 ∘ ≈ 0 . 9 3 9 7
1 − β cos θ = 1 − 0.95 × 0.9397 = 1 − 0.8927 = 0.1073 1 − β cos θ = 1 − 0 . 9 5 × 0 . 9 3 9 7 = 1 − 0 . 8 9 2 7 = 0 . 1 0 7 3
Observed wavelength:
λ = λ ′ 1 − β cos θ 1 − β 2 = 500 n m × 0.1073 0.3122 ≈ 171.9 n m λ = λ ′ 1 − β 2 1 − β cos θ = 5 0 0 n m × 0 . 3 1 2 2 0 . 1 0 7 3 ≈ 1 7 1 . 9 n m
Observed frequency ratio:
f f ′ = 1 − β 2 1 − β cos θ = 0.3122 0.1073 ≈ 2.909 f ′ f = 1 − β cos θ 1 − β 2 = 0 . 1 0 7 3 0 . 3 1 2 2 ≈ 2 . 9 0 9
Observed wavelength: 171.9 nm (strong blueshift into the UV).
Observed frequency: 2.91× higher than rest-frame.
We have rigorously derived the relativistic Doppler shift :
Time interval method — intuitive.Lorentz transformation of wave 4-vector — covariant and general.
We also derived the angle transformation between frames and applied the result to an AGN jet example .
Einstein, A. (1905). Zur Elektrodynamik bewegter Körper .
Rindler, W. (2006). Introduction to Special Relativity .
Urry, C. M., & Padovani, P. (1995). Unified Schemes for Radio-Loud AGN .
Blandford, R. D., & Königl, A. (1979). Relativistic Jets as Compact Radio Sources .