Relativistic beaming refers to the dramatic enhancement in the observed brightness and directional concentration of radiation emitted by objects moving at speeds close to the speed of light. It is a direct consequence of special relativity and plays a critical role in high-energy astrophysics, particularly in the study of jets from active galactic nuclei (AGN), blazars, pulsars, and gamma-ray bursts.
At its core, relativistic beaming arises from two effects:
Aberration of light : The apparent direction of emitted radiation is compressed toward the direction of motion.
Doppler boosting : The observed intensity and frequency of radiation are increased when the emitter moves toward the observer.
Together, these lead to a strong anisotropy in how radiation is received by different observers depending on their orientation relative to the source's motion.
Before getting into the derivation, first recall that the specific intensity I ν I ν ν ν
I ν = d E d t d A d Ω d ν . I ν = d t d A d Ω d ν d E .
That is:
d E d E d t d t d A d A d Ω d Ω d ν d ν
It measures energy per time, per area, per solid angle, per frequency .
Since the specific intensity plays a foundational role in calculating all forms of observed radiation including intensity, flux, luminosity, etc, our goal in the derivation below is to understand how I ν I ν under a Lorentz boost.
Since our goal is to determine how I ν I ν S S S ′ S ′
d E d E Doppler shift d t d t d A d A d Ω d Ω d ν d ν
Tracking all of these one by one is possible, but cumbersome.
Instead, we can take a more elegant approach by use of the invariance of the phase-space volume.
The key principle we will be using for this derivation is that d 3 x d 3 p d 3 x d 3 p
d 3 x d 3 x d 3 p d 3 p
This means that the volume of phase space occupied by photons is the same in all inertial frames .
Let us now define the photon distribution function f ( x ⃗ , p ⃗ ) f ( x , p ) d N d N
d N = f ( x ⃗ , p ⃗ ) d 3 x d 3 p = Number of photons in d 3 x with momenta in d 3 p . d N = f ( x , p ) d 3 x d 3 p = Number of photons in d 3 x with momenta in d 3 p .
Since the number of photons is conserved during a Lorentz transformation:
d N = d N ′ , d N = d N ′ ,
we have the relation,
f ( x ⃗ , p ⃗ ) d 3 x d 3 p = f ′ ( x ⃗ ′ , p ⃗ ′ ) d 3 x ′ d 3 p ′ , f ( x , p ) d 3 x d 3 p = f ′ ( x ′ , p ′ ) d 3 x ′ d 3 p ′ ,
but since d 3 x d 3 p = d 3 x ′ d 3 p ′ d 3 x d 3 p = d 3 x ′ d 3 p ′
f ′ ( x ⃗ ′ , p ⃗ ′ ) = f ( x ⃗ , p ⃗ ) . f ′ ( x ′ , p ′ ) = f ( x , p ) .
Which simply states that the photon distribution function is also Lorentz invariant
The total energy in a phase-space element of photons is:
d E = h ν d N = h ν f ( x ⃗ , p ⃗ ) d 3 x d 3 p . d E = h ν d N = h ν f ( x , p ) d 3 x d 3 p .
Now divide by d t d A d Ω d ν d t d A d Ω d ν
I ν = d E d t d A d Ω d ν = h ν f ( x ⃗ , p ⃗ ) d 3 x d 3 p d t d A d Ω d ν . I ν = d t d A d Ω d ν d E = d t d A d Ω d ν h ν f ( x , p ) d 3 x d 3 p .
Next, observe that the spatial volume element d 3 x d 3 x
d 3 x = c d t d A , d 3 x = c d t d A ,
where:
d t d t d A d A photons travel a distance c d t c d t d t d t
Therefore:
I ν = h ν f ( x ⃗ , p ⃗ ) c d t d A d 3 p d t d A d Ω d ν . I ν = d t d A d Ω d ν h ν f ( x , p ) c d t d A d 3 p .
Canceling d t d t d A d A
I ν = h ν c f ( x ⃗ , p ⃗ ) d 3 p d Ω d ν . I ν = h ν c f ( x , p ) d Ω d ν d 3 p .
In spherical coordinates:
d 3 p = p 2 d p d Ω p . d 3 p = p 2 d p d Ω p .
For photons:
p = h ν c , d p = h c d ν . p = c h ν , d p = c h d ν .
Thus:
d 3 p = ( h ν c ) 2 h c d ν d Ω p = h 3 ν 2 c 3 d ν d Ω p . d 3 p = ( c h ν ) 2 c h d ν d Ω p = c 3 h 3 ν 2 d ν d Ω p .
Putting this all together:
I ν = h ν c f ( x ⃗ , p ⃗ ) h 3 ν 2 / c 3 d Ω d ν d ν d Ω p . I ν = h ν c f ( x , p ) d Ω d ν h 3 ν 2 / c 3 d ν d Ω p .
Simplifying:
I ν = h 4 ν 3 c 2 f ( x ⃗ , p ⃗ ) . I ν = c 2 h 4 ν 3 f ( x , p ) .
Why Momentum Space Solid Angle is the same as Real Space Solid Angle
In the step where we rewrite d 3 p = p 2 d p d Ω p d 3 p = p 2 d p d Ω p d Ω p d Ω p I ν I ν d Ω d Ω
For photons, however, these two are equivalent. This is because a photon’s velocity is always exactly parallel to its momentum vector: v ⃗ = c p ^ v = c p ^ p ⃗ p d Ω p = d Ω d Ω p = d Ω
This equivalence is special to massless particles. For massive particles, momentum and velocity are not strictly parallel. The momentum is given by p ⃗ = γ m v ⃗ p = γ m v γ = 1 / 1 − v 2 / c 2 γ = 1 / 1 − v 2 / c 2 γ γ p ⃗ p v ⃗ v
While the directions of p ⃗ p v ⃗ v
In the Lorentz boosted frame:
I ν ′ = h 4 ν ′ 3 c 2 f ′ ( x ′ ⃗ , p ′ ⃗ ) . I ν ′ = c 2 h 4 ν ′ 3 f ′ ( x ′ , p ′ ) .
Since f ( x ⃗ , p ⃗ ) f ( x , p )
I ν ν 3 = Lorentz invariant . ν 3 I ν = Lorentz invariant .
Explicitly:
I ν ν 3 = I ν ′ ν ′ 3 . ν 3 I ν = ν ′ 3 I ν ′ .
This is the central result that will allow us to derive the full relativistic beaming formula.
Under a Lorentz boost, we already know how the frequency changes due to a relativistic doppler shift:
ν = 1 − β 2 1 − β cos θ ν ′ = D ν ′ ν = 1 − β cos θ 1 − β 2 ν ′ = D ν ′
where D is the Doppler factor :
D = 1 γ ( 1 − β cos θ ) . D = γ ( 1 − β cos θ ) 1 .
Solid angle is defined as:
d Ω = sin θ d θ d ϕ . d Ω = sin θ d θ d ϕ .
But we can equivalently write this as:
d Ω = − d ( cos θ ) d ϕ , d Ω = − d ( cos θ ) d ϕ ,
since
d ( cos θ ) d θ = − sin θ → − d ( cos θ ) = sin θ d θ . d θ d ( cos θ ) = − sin θ → − d ( cos θ ) = sin θ d θ .
Now we can compute:
d Ω d Ω ′ = d ( cos θ ) d ϕ d ( cos θ ′ ) d ϕ ′ . d Ω ′ d Ω = d ( cos θ ′ ) d ϕ ′ d ( cos θ ) d ϕ .
Since the azimuthal angle ϕ ϕ y y z z around the x x , it is unaltered by the lorentz boost along the x-axis. Thus ϕ = ϕ ′ ϕ = ϕ ′
d Ω d Ω ′ = d ( cos θ ) d ( cos θ ′ ) . d Ω ′ d Ω = d ( cos θ ′ ) d ( cos θ ) .
Remember the relativistic abberation formaula:
cos θ = cos θ ′ + β 1 + β cos θ ′ . cos θ = 1 + β cos θ ′ cos θ ′ + β .
Taking the derivative:
d cos θ d cos θ ′ = ( 1 + β cos θ ′ ) − β ( cos θ ′ + β ) ( 1 + β cos θ ′ ) 2 = 1 − β 2 ( 1 + β cos θ ′ ) 2 . d cos θ ′ d cos θ = ( 1 + β cos θ ′ ) 2 ( 1 + β cos θ ′ ) − β ( cos θ ′ + β ) = ( 1 + β cos θ ′ ) 2 1 − β 2 .
Therefore:
d Ω d Ω ′ = 1 − β 2 ( 1 + β cos θ ′ ) 2 . d Ω ′ d Ω = ( 1 + β cos θ ′ ) 2 1 − β 2 .
Recall the definition of Doppler factor:
D = 1 γ ( 1 − β cos θ ) . D = γ ( 1 − β cos θ ) 1 .
Now by simply manipulating the inverse aberation formula (not shown here), you can get it into the form shown here:
1 − β cos θ = 1 − β 2 1 + β cos θ ′ . 1 − β cos θ = 1 + β cos θ ′ 1 − β 2 .
Therefore:
D = 1 γ ⋅ 1 + β cos θ ′ 1 − β 2 . D = γ 1 ⋅ 1 − β 2 1 + β cos θ ′ .
Now invert this:
1 − β 2 ( 1 + β cos θ ′ ) = 1 γ D . ( 1 + β cos θ ′ ) 1 − β 2 = γ D 1 .
Square both sides:
( 1 − β 2 ) 2 ( 1 + β cos θ ′ ) 2 = 1 γ 2 D 2 . ( 1 + β cos θ ′ ) 2 ( 1 − β 2 ) 2 = γ 2 D 2 1 .
But look at your original solid angle factor:
d Ω d Ω ′ = 1 − β 2 ( 1 + β cos θ ′ ) 2 . d Ω ′ d Ω = ( 1 + β cos θ ′ ) 2 1 − β 2 .
Therefore:
d Ω d Ω ′ = ( 1 − β 2 ( 1 + β cos θ ′ ) 2 ) = 1 ( 1 − β 2 ) γ 2 D 2 = 1 D 2 . d Ω ′ d Ω = ( ( 1 + β cos θ ′ ) 2 1 − β 2 ) = ( 1 − β 2 ) γ 2 D 2 1 = D 2 1 .
d Ω = 1 D 2 d Ω ′ d Ω = D 2 1 d Ω ′
We know:
I ν ν 3 = invariant . ν 3 I ν = invariant .
Thus:
I ν = I ν ′ ′ ( ν ν ′ ) 3 . I ν = I ν ′ ′ ( ν ′ ν ) 3 .
But since we know how the frequency transforms:
I ν = I ν ′ ′ D 3 I ν = I ν ′ ′ D 3
Another important quantity transformation we should understand is I ν d Ω I ν d Ω
I ν d Ω = I ν ′ ′ D 3 d Ω ′ D 2 . I ν d Ω = I ν ′ ′ D 3 D 2 d Ω ′ .
I ν d Ω = I ν ′ ′ D d Ω ′ . I ν d Ω = I ν ′ ′ D d Ω ′ .
Let us now compute how the bolometric flux transforms under a Lorentz boost when the source is moving at velocity v v
The bolometric flux is the total energy per unit time per unit area received by the observer, integrated over all frequencies and all directions :
F = ∫ 0 ∞ ∫ 4 π I ν ( θ , ϕ ) cos θ d Ω d ν F = ∫ 0 ∞ ∫ 4 π I ν ( θ , ϕ ) cos θ d Ω d ν
We want to express this in terms of quantities measured in the rest frame S ′ S ′
From relativistic beaming:
The specific intensity transforms as:I ν ( ν , θ ) = D 3 I ν ′ ′ ( ν ′ , θ ′ ) I ν ( ν , θ ) = D 3 I ν ′ ′ ( ν ′ , θ ′ )
The frequency transforms as:ν = D ν ′ ν = D ν ′
The solid angle transforms as:d Ω = d Ω ′ D 2 d Ω = D 2 d Ω ′
The direction cosine transforms as (from aberration):cos θ = cos θ ′ + β 1 + β cos θ ′ cos θ = 1 + β cos θ ′ cos θ ′ + β
We now change variables in the flux integral to primed quantities using these transformations.
Substitute the transformation rules into the original expression:
F = ∫ 0 ∞ ∫ 4 π I ν ( ν , θ ) cos θ d Ω d ν F = ∫ 0 ∞ ∫ 4 π I ν ( ν , θ ) cos θ d Ω d ν
Change variables ν = D ν ′ ν = D ν ′ d ν = D d ν ′ d ν = D d ν ′
F = ∫ 0 ∞ ∫ 4 π ( D 3 I ν ′ ′ ( ν ′ , θ ′ ) ) D ( cos θ ′ + β 1 + β cos θ ′ ) ( d Ω ′ D 2 ) D d ν ′ F = ∫ 0 ∞ ∫ 4 π ( D 3 I ν ′ ′ ( ν ′ , θ ′ ) ) D ( 1 + β cos θ ′ cos θ ′ + β ) ( D 2 d Ω ′ ) D d ν ′
Simplify the powers of D D
F = ∫ 0 ∞ ∫ 4 π I ν ′ ′ ( ν ′ , θ ′ ) ⋅ D 2 ⋅ ( cos θ ′ + β 1 + β cos θ ′ ) d Ω ′ d ν ′ F = ∫ 0 ∞ ∫ 4 π I ν ′ ′ ( ν ′ , θ ′ ) ⋅ D 2 ⋅ ( 1 + β cos θ ′ cos θ ′ + β ) d Ω ′ d ν ′
This is the general relativistic transformation law for bolometric flux , valid for arbitrary angular and spectral emission patterns in the comoving frame.
D = 1 γ ( 1 − β cos θ ) D = γ ( 1 − β c o s θ ) 1 θ ′ θ ′ This result is essential when computing the apparent fluxes of relativistic jets, pulsar beams, or any moving astrophysical source.
The key transformation results for relativistic beaming are:
I ν = I ν ′ ′ D 3 d Ω = 1 D 2 d Ω ′ I ν d Ω = I ν ′ ′ D d Ω ′ ν = D ν ′ I ν d Ω I ν d Ω ν = I ν ′ ′ D 3 = D 2 1 d Ω ′ = I ν ′ ′ D d Ω ′ = D ν ′
where:
D = 1 γ ( 1 − β cos θ ) . D = γ ( 1 − β cos θ ) 1 .
These relations allow you to transform the radiation pattern and observed flux between frames.
Radiation is beamed forward in the direction of motion.
Observers looking into the forward direction see a brightened and blueshifted beam.
Observers looking at large angles see diminished and redshifted radiation.
This is why relativistic jets appear bright and collimated .
Example: Relativistic Beaming of a Jet
Suppose a relativistic jet is moving at v=0.99c relative to the observer.
Assume that in the jet comoving frame S′:
Specific intensity at rest-frame frequency ν′ is:
I ν ′ ′ = 1 unit . I ν ′ ′ = 1 unit .
The emission is isotropic in S′:
I ν ′ ′ independent of angle . I ν ′ ′ independent of angle .
D = 1 γ ( 1 − β cos θ ) = 1 7.09 ( 1 − 0.99 ⋅ 1 ) = 1 7.09 × 0.01 ≈ 14.1. D = γ ( 1 − β cos θ ) 1 = 7 . 0 9 ( 1 − 0 . 9 9 ⋅ 1 ) 1 = 7 . 0 9 × 0 . 0 1 1 ≈ 1 4 . 1 .
ν = D ν ′ ≈ 14.1 ν ′ . ν = D ν ′ ≈ 1 4 . 1 ν ′ .
I ν = I ν ′ ′ D 3 = 1 × ( 14.1 ) 3 ≈ 2800. I ν = I ν ′ ′ D 3 = 1 × ( 1 4 . 1 ) 3 ≈ 2 8 0 0 .
d Ω = 1 D 2 d Ω ′ = 1 ( 14.1 ) 2 d Ω ′ ≈ 1 198.8 d Ω ′ . d Ω = D 2 1 d Ω ′ = ( 1 4 . 1 ) 2 1 d Ω ′ ≈ 1 9 8 . 8 1 d Ω ′ .
I ν d Ω = I ν ′ ′ D d Ω ′ = 1 × 14.1 d Ω ′ = 14.1 d Ω ′ . I ν d Ω = I ν ′ ′ D d Ω ′ = 1 × 1 4 . 1 d Ω ′ = 1 4 . 1 d Ω ′ .
I ν ≈ 2800 I ν ≈ 2 8 0 0 I ν d Ω ≈ 14.1 d Ω ′ I ν d Ω ≈ 1 4 . 1 d Ω ′ ν ≈ 14.1 ν ′ ν ≈ 1 4 . 1 ν ′
D = 1 γ ( 1 − β cos 9 0 ∘ ) = 1 γ = 1 7.09 ≈ 0.141. D = γ ( 1 − β cos 9 0 ∘ ) 1 = γ 1 = 7 . 0 9 1 ≈ 0 . 1 4 1 .
ν = D ν ′ ≈ 0.141 ν ′ . ν = D ν ′ ≈ 0 . 1 4 1 ν ′ .
I ν = I ν ′ ′ D 3 = 1 × ( 0.141 ) 3 ≈ 0.0028. I ν = I ν ′ ′ D 3 = 1 × ( 0 . 1 4 1 ) 3 ≈ 0 . 0 0 2 8 .
d Ω = 1 D 2 d Ω ′ = 1 ( 0.141 ) 2 d Ω ′ ≈ 50.3 d Ω ′ . d Ω = D 2 1 d Ω ′ = ( 0 . 1 4 1 ) 2 1 d Ω ′ ≈ 5 0 . 3 d Ω ′ .
I ν d Ω = I ν ′ ′ D d Ω ′ = 1 × 0.141 d Ω ′ = 0.141 d Ω ′ . I ν d Ω = I ν ′ ′ D d Ω ′ = 1 × 0 . 1 4 1 d Ω ′ = 0 . 1 4 1 d Ω ′ .
I ν ≈ 0.0028 I ν ≈ 0 . 0 0 2 8 I ν d Ω ≈ 0.141 d Ω ′ I ν d Ω ≈ 0 . 1 4 1 d Ω ′ ν ≈ 0.141 ν ′ ν ≈ 0 . 1 4 1 ν ′
This explains why relativistic jets in AGN and GRBs appear so bright and collimated even if the intrinsic emission is isotropic in the jet frame.
We have rigorously derived relativistic beaming using both phase space invariance and explicit transformations of frequency and solid angle. This provides a robust physical understanding of why moving sources of radiation can appear so strongly beamed to observers.