Stellar aberration (derivation from Lorentz transformation)

Derivation of the formula for motion along the x-axis

For the derivation it is assumed, that the light signal only travels through space regions where the gravitation field is negligible. Hence is suffices to use special relativity and the path of the light signal is a straight line in any inertial frame of reference.

Observation in the rest frame S of the center of mass of our Solar System

The rest frame of the center of mass (barycenter) is a very good quasi-inertial frame of reference for periods of time in the order of thousands of years, since our solar system needs about 230 million years (galactic year) to move completely around the center of the Milky Way. The space coordinates of this frame of reference form a Cartesian coordinate system.

In the reference frame S ( x 1 | y 1 | 0 | c t 1 ) with c t 1 < 0 and ( x 1 | y 1 ) ≠ ( 0 | 0 ) are the (space-time) coordinates at which the star emits a light signal and ( 0 | 0 | 0 | 0 ) are the coordinates at which the astronomer receives the light signal.

In reference frame S the light signal starts at t 1 < 0 and stops at time t 2 = 0 and therefore the light signal did cover the distance d = c ⋅ ( 0 − t 1 ) = − c t 1 .

In S the path of the light signal is a straight line and it forms an angle δ with the x-axis with: sin ⁡ δ = y 1 d = y 1 − c t 1   and cos ⁡ δ = x 1 − c t 1

Observation in the inertial frame of reference S' which is in uniform motion (relative to S) along the x-axis

The origin of the reference frame S' is in uniform motion relative to S with ( v | 0 | 0 ) , i.e. moves along the x-axis, and the x-,y- und z-axes of S' and S are parallel to each other and at time t = t ′ = 0 the origins of S and S' coincide. Let β = v c , γ = 1 1 − β 2

S' now is an equally good quasi-inertial frame of reference as S: the space coordinates form a Cartesian coordinate system and the path of the light signal is a straight line.

According to the Lorentz transformation one gets: x 1 ′ = γ ⋅ ( x 1 − β ⋅ c t 1 ) , y 1 ′ = y 1 , z 1 ′ = 0 , c t 1 ′ = γ ⋅ ( c t 1 − β ⋅ x 1 )

In reference frame S' the light signal starts at t 1 ′ < 0 and stops at time t 2 ′ = 0 and therefore the light signal did cover the distance d ′ = c ⋅ ( 0 − t 1 ′ ) = − c t 1 ′ .

In S' the path of the light signal is a straight line, too. It forms an angle δ ′ with the x'-axis and one gets:

sin ⁡ δ ′ = y 1 ′ d ′ = y 1 − c t 1 ′ = y 1 − γ ⋅ ( c t 1 − β ⋅ x 1 ) = y 1 − c t 1 ⋅ γ ⋅ ( 1 − β ⋅ x 1 c t 1 ) = y 1 − c t 1 γ ⋅ ( 1 + β ⋅ x 1 − c t 1 ) = sin ⁡ δ γ ⋅ ( 1 + β ⋅ cos ⁡ δ )

cos ⁡ δ ′ = x 1 ′ − c t 1 ′ = γ ⋅ ( x 1 − β ⋅ c t 1 ) − γ ⋅ ( c t 1 − β ⋅ x 1 ) = − c t 1 ⋅ γ ⋅ ( x 1 − c t 1 + β ) − c t 1 ⋅ γ ⋅ ( 1 + β ⋅ x 1 − c t 1 ) = x 1 − c t 1 + β 1 + β ⋅ x 1 − c t 1 = cos ⁡ δ + β 1 + β ⋅ cos ⁡ δ

Hence: tan ⁡ δ ′ = sin ⁡ δ ′ cos ⁡ δ ′ = sin ⁡ δ γ ⋅ ( cos ⁡ δ + β )

These are the same formulas as in aberration of light#Explanation.

Approximate formula for motion along the x-axis in case of v/c <<1

△ δ = δ ′ − δ is the change of the angle δ. As β<<1 this change is also very small.

Case I: δ ≠ ± 90 ∘ → cos ⁡ δ ≠ 0

As Δδ<<1 one gets: tan ⁡ δ ′ − tan ⁡ δ △ δ ≈ d d δ tan ⁡ δ = 1 ( cos ⁡ δ ) 2 → △ δ = ( cos ⁡ δ ) 2 ⋅ ( tan ⁡ δ ′ − tan ⁡ δ )

As β<<1 one gets: tan ⁡ δ ′ = sin ⁡ δ γ ( cos ⁡ δ + β ) ≈ sin ⁡ δ cos ⁡ δ + β = sin ⁡ δ cos ⁡ δ ⋅ ( 1 + β cos ⁡ δ ) ≈ tan ⁡ δ ⋅ ( 1 − β cos ⁡ δ )

Therefore △ δ = ( cos ⁡ δ ) 2 ⋅ ( tan ⁡ δ ′ − tan ⁡ δ ) = ( cos ⁡ δ ) 2 ⋅ tan ⁡ δ ( 1 − β cos ⁡ δ − 1 ) = − cos ⁡ δ ⋅ tan ⁡ δ ⋅ β = − β ⋅ sin ⁡ δ

Case IIa: δ = 90 ∘ , hence: tan ⁡ δ ′ = sin ⁡ 90 ∘ γ ⋅ ( cos ⁡ 90 ∘ + β ) = 1 γ ⋅ β ≈ 1 β → cot ⁡ δ ′ = β → δ ′ = arccot ⁡ β ≈ π 2 − β

and therefore: △ δ = δ ′ − δ = − β ( = − β ⋅ sin ⁡ ( 90 ∘ ) )

Case IIb: δ = − 90 ∘ , and hence: tan ⁡ δ ′ = − 1 γ ⋅ β ≈ − 1 β → cot ⁡ δ ′ = − β → δ ′ = arccot ⁡ β ≈ − π 2 + β    

Hence: △ δ = δ ′ − δ = + β ( = − β ⋅ sin ⁡ ( − 90 ∘ ) )

Conclusion:The change of the angle Δδ = δ'-δ in the case of β = v/c << 1 can be described by the approximate formula △ δ = − v c ⋅ sin ⁡ δ resp. in the degree measure △ δ = − v c ⋅ sin ⁡ δ ⋅ 180 ∘ π

Symmetric form of the (exact) formula for motion along the x-axis

With help of tangent half-angle formula tan ⁡ ( α / 2 ) = sin ⁡ α / ( 1 + cos ⁡ α )   one can prove the symmetric form: tan ⁡ δ ′ 2 = 1 − β 1 + β ⋅ tan ⁡ δ 2   (derivation found in a SR-textbook)

tan ⁡ δ ′ 2 = sin ⁡ δ ′ 1 + cos ⁡ δ ′ = sin ⁡ δ γ ⋅ ( 1 + β ⋅ cos ⁡ δ ) 1 + β ⋅ cos ⁡ δ 1 + β ⋅ cos ⁡ δ + cos ⁡ δ + β 1 + β ⋅ cos ⁡ δ = sin ⁡ δ γ ⋅ ( 1 + β ⋅ cos ⁡ δ + cos ⁡ δ + β ) = sin ⁡ δ γ ⋅ ( 1 + β ) ⋅ ( 1 + cos ⁡ δ ) = tan ⁡ δ 2 γ ⋅ ( 1 + β )

And as 1 γ ⋅ ( 1 + β ) = 1 − β 2 1 + β = ( 1 + β ) ( 1 − β ) 1 + β = 1 − β 1 + β the symmetric form follows.

Formula for motion along the y-axis

Let θ be the angle between the light ray (=path of the light signal which is a straight line) and the y-axis whereby θ is positive if the y-axis would have to rotate counter-clockwise to coincide with the light ray. Then the derivation of the formula of angle θ' for the motion along the y-axis is the same as the derivation of the formula of angle δ' for the motion along the x-axis.

Hence: tan ⁡ θ ′ = sin ⁡ θ γ ⋅ ( cos ⁡ θ + β )

The symmetric form is: tan ⁡ θ ′ 2 = 1 − β 1 + β ⋅ tan ⁡ θ 2 and the approximate formula is: △ θ = − β ⋅ sin ⁡ θ

As θ = δ − 90 ∘ and as sin ⁡ ( δ − 90 ∘ ) = − sin ⁡ ( 90 ∘ − δ ) = − cos ⁡ δ , cos ⁡ ( δ − 90 ∘ ) = cos ⁡ ( 90 ∘ − δ ) = sin ⁡ δ and tan ⁡ ( δ − 90 ∘ ) = − tan ⁡ ( 90 ∘ − δ ) = − cot ⁡ δ   one gets:

tan ⁡ ( δ ′ − 90 ∘ ) = sin ⁡ ( δ − 90 ∘ ) γ ⋅ ( cos ⁡ ( δ − 90 ∘ ) + β ) and therefore:  − cot ⁡ δ ′ = − cos ⁡ δ γ ⋅ ( sin ⁡ δ + β ) and hence cot ⁡ δ ′ = cos ⁡ δ γ ⋅ ( sin ⁡ δ + β )

Since cot ⁡ δ ′ = 1 tan ⁡ δ ′   one also gets: tan ⁡ δ ′ = γ ⋅ ( sin ⁡ δ + β ) cos ⁡ δ

And the symmetric form is: tan ⁡ δ ′ − 90 ∘ 2 = 1 − β 1 + β ⋅ tan ⁡ δ − 90 ∘ 2

And as △ θ = ( δ ′ − 90 ∘ ) − ( δ − 90 ∘ ) = △ δ   the approximate formula is: △ δ = △ θ = − β ⋅ sin ⁡ ( δ − 90 ∘ ) = + β ⋅ cos ⁡ δ    

Formula for motion along a ray lying in the x-y-plane with direction vector (cos α | sin α)

With the same reasoning as above one gets the formula: tan ⁡ ( δ ′ − α ) = sin ⁡ ( δ − α ) γ ⋅ ( cos ⁡ ( δ − α ) + β )

The symmetric form is: tan ⁡ δ ′ − α 2 = 1 − β 1 + β ⋅ tan ⁡ δ − α 2

The approximate formula is: △ δ = − β ⋅ sin ⁡ ( δ − α )   and in the degree measure: △ δ = − v c ⋅ sin ⁡ ( δ − α ) ⋅ 180 ∘ π

Application: Aberration in astronomy

The stellar aberration is purely an effect of the change of the reference frame. The astronomer orbits (with Earth) around the Sun and furthermore rotates around the axis of Earth. His current rest frame S' therefore has different velocities relative to the rest frame S of the barycenter of the Solar System at different times. Hence the astronomer observes that the position of the star changes. The formula is derived under the condition that the change of the position of the star and of Earth is negligible in the period of observation. That is correct for almost all stars: the amplitude of the parallax of a star, for a distance of ≥ n parsec, is ≤ 1/n ".

Stellar aberration due to the orbit of Earth (around the Sun)

The mean orbital speed of Earth is v e = 2 ⋅ π ⋅ 1 A E 365 , 25 d = 29 , 78 k m / s , and therefore v e c = 0 , 00009935 .

-> v c ⋅ 180 ∘ π = 20 , 5 ′ ′ .

k A = 20 , 5 ′ ′ is dubbed constant of aberration for the annual aberration.

Stellar aberration due to Earth's rotation

An astronomer at the latitude φ rotates in 24 hours around the axis of Earth. His speed of rotation is therefore v r = cos ⁡ φ ⋅ 40000 k m 1 d = cos ⁡ φ ⋅ 463 m / s . Hence v r c = cos ⁡ φ ⋅ 1 , 54 ⋅ 10 − 6 . Form this so called diurnal aberration one gets an additional contribution of (at max.) cos ⁡ φ ⋅ 0 , 32 ′ ′ .

Stellar aberration due the orbit of our solar system around the center of the Milky Way

The rest frame of the center of mass of our solar system isn't a perfect inertial frame of reference since our solar system orbits around the center of the Milky Way. An estimation for the time of period of circulation is 230 million years (estimations vary between 225 and 250 million years). As the estimation for the distance between our solar system and the center of the Milky Way is about 28000 Ly, the assumed orbital speed of our solar system is v = 2 π ⋅ 280000 ⋅ 9 , 461 ⋅ 10 15 m 230 ⋅ 10 6 ⋅ 365 , 25 ⋅ 24 ⋅ 3600 ≈ 230 k m / s . This would cause an aberration ellipse with a major semiaxis of 2,6' (arcminutes). Therefore, in one year the aberration angle could change (at max.) 2 , 6 ′ ⋅ 2 π ⋅ 1 a 230000000 a = 4,3 µas (microarcseconds). This very small value isn't detectable now, perhaps it's possible with the planned mission of the Gaia spacecraft.