User:AbiLtoC/Stellar aberration (Derivation with Lorentz transformation)

According to the disambiguation page, stellar aberration (Aberration of light) is an astronomical phenomenon "which produces an apparent motion of celestial objects". This article demonstrates that the stellar aberration is due to the change of the astronomer's Inertial frame of reference. The formula is derived with the use of Lorentz transformation of the star's coordinates.

As the astronomer John Herschel has already explained in 1844, the stellar aberration does not depend on the relative velocity of the star towards the earth. Otherwise eclipsing binary stars would appear to be separated, in stark contrast to observation.

Stellar aberration is only due to the change of the astronomer's inertial frame of reference
In the year 1926 the astrophysicist Robert Emden has published the article Aberration und Relativitätstheorie in the journal Naturwissenschaften. In this article he states, that the direction of a light ray isn't influenced by the motion of the star and neither by the motion of the earth. At that time, the opponents of the special theory of relativity reasoned that the theory must be flawed, because it would state that the stellar aberration would depend on the relative velocity of the star — which would be in contradiction to observation — and R. Emden's article explains that the special theory of relativity does not predict this. Today, the special theory of relativity isn't contested anymore but there are still articles that suggest that the aberration would depend on the relative velocity of the star. It is nevertheless straightforward to show that the aberration doesn't depend on the star's motion. And the observed position of the star wouldn't depend on the earth's motion either, if the astronomer would use the same inertial frame of reference all the time. But of course the astronomer uses his current rest frame and these current rest frames are different at different times as the earth orbits around the sun. It is mathematically convenient to declare the position of star in a rest frame of the sun (more exactly: the barycenter of the solar system) as the "real" position and to declare the difference to this "real" position the "aberration".

Sample calculation
The rest frame S of the center of mass of our solar system is a very good quasi-Inertial frame of reference for time periods in the order of thousands of years as the circulation period of our solar system around the center of the Milky Way is about 230 million years (Galactic year). The reference frame S' now is in uniform motion with vx = 0,5c relative to S, such that in the future the star comes nearer to the origin of the coordiante system (and consequently further afar in the past). The x-,y- and z-axes of both systems ought to be parallel and at time t=t'=0 the origins of both systems ought to coincide. Therefore one gets according to Lorentz transformation: $$\scriptstyle \beta=v/c $$, $$\scriptstyle \gamma=1/\sqrt{1-\beta^{2}}$$: y'=y ; z=z';  $$\scriptstyle x'=\gamma\cdot(x-\beta\cdot ct)$$ Suppose now that the star emitted a light lignal at time $$\scriptstyle c\,t_1=-5\,Ly$$ at the location $$\scriptstyle x_1=4\,Ly\;,\;y_1= 3\,Ly\;,\;z_1=0 $$ and that this light signal is received by an astronomer at time $$\scriptstyle c\,t_2=0$$ at the location $$\scriptstyle x_2=0\;,\;y_2=0\;,\;z_2=0$$.

In S the star's position and the x-axis form an angle $$\delta$$ with $$\scriptstyle \tan \delta =3/4 \; \rightarrow \;\delta =36.87^\circ $$ But in S' $$\scriptstyle x_1'=\gamma \cdot (x_1-\beta \cdot c\,t_1)=1,1547\cdot (4 Lj - 0,5\cdot(-5 Lj))=7,50555 Lj; \quad y_1'=y_1=3 Lj$$ and therefore the angle $$\delta'$$ between the star's position and x'-axis is $$\scriptstyle \tan \delta'=y_1'/x_1'=3/7,50555\;\rightarrow\;\delta' = 21,79^\circ$$ Now the calculation with help of the formula of in Aberration of light gives the same result $$\scriptstyle \tan (\delta'/2) = \tan (\delta /2) \cdot \sqrt {(1-0,5)/(1+0,5)} = \tan (36,87^\circ/2) \cdot \sqrt{1/3} = 0,19245\;\rightarrow\;\delta'/2= 10,89^\circ\;\rightarrow\;\delta'=21,79^\circ$$.

Derivation of the formula
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 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 $$(x_1|y_1) \ne (0|0)$$ 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 path of the light signal forms an angle &delta; with the x-axis with $$\tan \delta = \frac{y_1}{x_1}$$.

Observation in the Inertial frame of reference S' which is in uniform motion (relative to S)
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 $$\scriptstyle t=t'=0$$ the origins of S and S' coincide. Let $$\beta = \frac{v}{c}, \gamma = \frac{1}{\sqrt{1-\beta^2}}$$

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. In S', the angle between the path of the light signal and the x'-axis is $$\delta'$$ with $$\tan \delta' = \frac{y_1'}{x_1'}$$

According to the Lorentz transformation one gets: $$ x_1' = \gamma \cdot(x_1-\beta \cdot c\,t_1), y_1'=y_1 , z_1'=0 , c\,t_1' = \gamma \cdot (c\,t_1-\beta \cdot x_1)$$ Therefore  $$\tan \delta' = \frac{y_1'}{x_1'} = \frac{y_1}{x_1'}= \frac {y_1}{\gamma \cdot (x_1-\beta \cdot c\,t_1)}$$

According to the Pythagorean theorem the distance d from $$(x_1|y_1|0)$$ to $$(0|0|0)$$ is $$ d = \sqrt{y_1^2+x_1^2}$$. In order to reach the astronomer at time $$t_2=0$$ at the location $$(0|0|0)$$, the light signal must therefore have been emitted from the star at time $$t_1 = -d/c$$.

Case 1: $$x_1>0$$: Now y 1 can be divided by x 1 and therefore $$\tan \delta'= \frac {y_1/x_1}{\gamma \cdot \left( 1 -\beta \cdot c\,t_1/x_1 \right) }$$.

Furthermore gilt $$c\,t_1 = -d = -\sqrt{y_1^2+x_1^2} = -x_1 \cdot \sqrt{(y_1/x_1)^2+1}$$ and hence: $$\tan \delta' =\frac{y_1/x_1}{\gamma \cdot \left( 1+\beta \cdot \sqrt{(y_1/x_1)^2+1} \right) } =\frac{\tan \delta}{\gamma \cdot \left(1+ \beta \cdot \sqrt{(\tan \delta)^2+1} \right) } =\frac{\tan \delta}{\gamma \cdot \left(1+ \frac{\beta}{\cos \delta} \right)} =\frac {\sin \delta}{\gamma \cdot \left( \cos \delta +\beta \right) }$$

Case 2: $$x_1<0$$: Now $$c\,t_1 = -\sqrt{y_1^2+x_1^2} = +x_1 \cdot \sqrt{(y_1/x_1)^2+1}$$ and consequently: $$\tan \delta' =\frac{y_1/x_1}{\gamma \cdot \left( 1-\beta \cdot \sqrt{(y_1/x_1)^2+1} \right)} =\frac{\tan \delta}{\gamma \cdot \left(1- \beta \cdot \sqrt{(\tan \delta)^2+1} \right)} =\frac{\tan \delta}{\gamma \cdot \left(1- \frac{\beta}{-\cos \delta} \right)} =\frac {\sin \delta}{\gamma \cdot \left( \cos \delta +\beta \right) }$$

Case 3a: $$x_1=0, y_1>0$$: Now $$c\,t_1 = -\sqrt{y_1^2+0^2} = -\sqrt{(y_1)^2}=-y_1$$ and therefore $$\tan \delta' = \frac{y_1}{\gamma \cdot \beta \cdot y_1} = \frac{1}{\gamma \cdot \beta}$$. In this case $$\delta=90^\circ$$ and hence $$\sin \delta =1$$ and also $$\cos \delta =0$$.

Case 3b: $$x_1=0, y_1<0$$: Es gilt $$c\,t_1 = -\sqrt{y_1^2+0^2} = -\sqrt{(y_1)^2}=+y_1$$ and therefore $$\tan \delta' = \frac{y_1}{-\gamma \cdot \beta \cdot y_1} = \frac{-1}{\gamma \cdot \beta}$$. In that case is $$\delta=-90^\circ$$ and hence $$\sin \delta =-1$$ and also $$\cos \delta =0$$.

Conclusion: All cases are therefore covered by the formula: $$\tan\delta'=\frac {\sin \delta}{\gamma(\cos \delta +\beta)}$$ and that is one of the two formulas as in Aberration of light.

Discussion and application
The stellar aberration is only an effect of the change of the reference frame. The astronomer orbits (with the earth) around the sun in one year and furthermore rotates around the axis of the earth in one day. 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 the earth is negligible in the period of observation That is correct for almost all stars: The amplitude of the Parallax of a star in a distance of ≥10 pc is ≤ 0,1".

Approximation of the formula in case of v/c <<1
Let $$\triangle\delta\;=\;\delta' - \delta\; $$ be the change of the angle &delta; As &beta;<<1 this change is also very small.

Case I: $$\;\delta\ne\pm 90^\circ\; \rightarrow \;\cos \delta \ne 0$$

As &Delta;&delta;<<1 one gets: $$\frac{\tan \delta' - \tan \delta}{\triangle\delta} \approx\frac{d}{d\delta}\tan \delta = \frac{1}{(\cos \delta)^2}\; \rightarrow \; \triangle\delta\;=\;(\cos \delta)^2 \cdot (\tan \delta' - \tan \delta)$$

As &beta;<<1 one gets: $$\tan\delta'=\frac {\sin \delta}{\gamma(\cos \delta +\beta)} \approx \frac {\sin \delta}{\cos \delta +\beta} = \frac{\sin \delta}{\cos \delta \cdot \left(1+\frac{\beta}{\cos \delta}\right)} \approx \tan \delta \cdot \left(1-\frac{\beta}{\cos \delta}\right)$$

Therefore $$\triangle\delta\;=\;(\cos \delta)^2 \cdot (\tan \delta' - \tan \delta) = (\cos \delta)^2 \cdot \tan \delta \left(1-\frac{\beta}{\cos \delta}\;-1\right) = - \cos \delta \cdot \tan \delta \cdot \beta = -\beta \cdot \sin \delta $$

Case IIa: $$\;\delta\;=90^\circ\;$$, hence: $$\tan \delta' = \frac{1}{\gamma \cdot \beta}\;\approx \frac {1}{\beta} \quad \rightarrow \quad \cot \delta' = \beta \quad \rightarrow \quad \delta' = \arccot \beta \approx \frac{\pi}{2} - \beta $$

And therefore: $$\triangle\delta = \delta' - \delta = -\beta \quad \left(= -\beta \cdot \sin(90^\circ)\right)$$

Case IIb: $$\;\delta\;=-90^\circ\;$$, and hence: $$\tan \delta' = -\frac{1}{\gamma \cdot \beta}\;\approx -\frac {1}{\beta} \quad \rightarrow \quad \cot \delta' = -\beta \quad \rightarrow \quad \delta' = \arccot \beta \approx -\frac{\pi}{2} + \beta \quad$$

Hence: $$\triangle\delta = \delta' - \delta = +\beta\quad \left(= -\beta \cdot \sin(-90^\circ)\right)$$

Conclusion: The change of the angle &Delta;&delta; = &delta;'-&delta; in the case of &beta; = v/c << 1 can be described by the approximation formula $$\triangle\delta = -\frac{v}{c} \cdot \sin \delta\;$$ resp. in the degree measure $$\triangle\delta = -\frac{v}{c} \cdot \sin \delta \cdot \frac{180^\circ}{\pi}$$

Stellar aberration due to the orbit of the earth (around the sun)
The mean orbital speed of the earth is $$v_e = \frac{2 \cdot \pi \cdot 1\,AE}{365,25\,d} = 29,78\,km/s $$, and therefore $$\frac{v_e}{c} = 0,00009935$$.

-> $$\frac{v}{c} \cdot \frac{180^\circ}{\pi} = 20,5^{\prime\prime}$$.

$$ k_A= 20,5^{\prime\prime}$$ is dubbed constant of aberration for the annual aberration.

Stellar aberration due to the earth's rotation
An astronomer at the Latitude $$\varphi$$ rotates in 24 hours around the axis of the earth. His speed of rotation is therefore $$v_r= \frac{\cos \varphi \cdot 40000\,km}{1\,d} = \cos \varphi \cdot 463\,m/s$$. Hence $$\frac{v_r}{c} = \cos \varphi \cdot 1,54\cdot10^{-6}$$. Form this so called diurnal aberration one gets an additional contribution of (at max.) $$ \cos \varphi \cdot 0,32^{\prime\prime}$$.

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 $$\scriptstyle v = \frac{2\pi \cdot 280000 \cdot 9,461\cdot10^{15}\,m}{230\cdot 10^{6} \cdot 365,25 \cdot 24 \cdot 3600} \approx\, 230\,km/s$$. 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.) $$\scriptstyle 2,6^\prime \cdot \frac{2 \pi \cdot 1\,a}{230000000\,a}$$ = 4,3 µas (Microarcseconds). This very small value isn't detectable now, perhaps it's possible with the planned mission of the Gaia spacecraft.