# Stellar aberration – again

1. May 13, 2014

### exmarine

I thought I understood how stellar aberration conformed to Special Relativity. The CHANGE in that angle comes from the CHANGE in our orbital velocity direction about the sun over six months. And it is the same for all stars. That is fine if there are no significant changes in a star’s state of motion relative to our sun over short periods of time. “Significant” is relative to our orbital velocity; and “short” is relative to six months.

Now I read that there are binary stars out there that DO have significant velocity component changes parallel to the plane of our orbit during short periods of time. Yet we observe no corresponding changes in the aberration angles from those stars.

How can that be consistent with SRT? Only SOME changes in relative velocities between source and observer cause changes in aberration angles?

2. May 13, 2014

### jbriggs444

The source-relative direction of the light that is emitted from a source and detected at a receiver will vary with the motion of the source at the time of transmission. It will not vary with the motion of the receiver at the the time of reception.

[Change the frame of reference that the source measures angle against and you change the measured angle. Change the speed of the target all you like and it doesn't matter as long as it's in the right place at the right time when the light hits it].

The receiver-relative direction of the light that is emitted from a source and detected at a receiver will vary with the motion of the receiver at the time of reception. It will not vary with the motion of the source at the time of transmission.

[Change the frame of reference that the receiver measures angle against and you change the measured angle. Change the speed of the source all you like and it doesn't matter as long as it was in the right place at the right time when it emitted the light]

3. May 13, 2014

### clem

The difference between the aberration angle for SR is so close to that for a classical non-relativistic calculation,
that the difference would not be noticed for a moving star.

4. May 16, 2014

### exmarine

quote…

And, after a month or two, when the binary’s contribution to the change in relative velocity with us is significant? But still the only effect we observe in the aberration angle is due to our contribution to that change in relative velocity. How do you explain that? And we are talking about first order effects, not some exotic higher orders.

The only treatment of aberration with binaries I’ve been able to find in the literature is in Kevin Brown’s book “Reflections on Relativity”. In an otherwise fine chapter, he finally gets to this issue, and then ignores the binary’s change in orbital velocity. I am quite surprised. It would be nice if someone else could verify my understanding of his analysis. You can find that chapter of his book on MathPages.

5. May 16, 2014

### Histspec

I think you are referring to
http://www.mathpages.com/rr/s2-05/2-05.htm
There is another chapter in this book where this question of aberration of binary stars is discussed in more detail
http://www.mathpages.com/home/kmath160/kmath160.htm

Here is another treatment:
Liebscher, D.-E.; Brosche, P. (1998): Aberration and relativity. In: Astronomische Nachrichten. 319,
See fig. 10 on p. 313 where the aberration of double stars is discussed, and where they showed that there is no "active" aberration due to the motion of the source alone. This refutes the claims of some "anti-relativists" of the 1920ies who thought that the lack of "active" aberration of double stars contradicts relativity.

So even though there seem to exist different descriptions of stellar aberration (depending on the interpretation of velocity "v" in the aberration formula) - all of them agree that the absence of "active" aberration in the appearance of double stars is in agreement with the predictions of special relativity.

PS: Liebscher et al. also remarked that the erroneous concept of "active" aberration of double stars was already discussed in the 19th century, until it was refuted in the following paper by Herschel in 1844:

Last edited: May 16, 2014
6. May 16, 2014

### jartsa

Here are two true sentences:

1: Aberration depends on the velocity of the observer.
2: Observer does not know his velocity.

And here is a false statement:

3: Aberration depends on the relative velocity of the observer and the light source.

I guess there may exist some aberration formula, where the relative velocity of observer and light source is one parameter. My comment regarding that formula is: Ok, but sentence number 3 is still false.

Last edited: May 16, 2014
7. May 16, 2014

### Bill_K

From the point of view of an observer riding on the binary star, the position of Earth in the sky varies back and forth. If he had to aim his photons, he would need to take this motion into account. But he does not - he sprays photons in all directions. The photon that winds up hitting Earth goes straight to Earth. His motion changes only which photon this is.

8. May 16, 2014

### jbriggs444

Aberration relative to a baseline measurement, real or imagined, depends on velocity relative to the baseline frame.

An earth-bound observer can know his velocity relative to many things that could be used as baseline frames.

9. May 16, 2014

### exmarine

I thought aberration depended on the Lorentz transform. And does that transform not depend on the relative velocity?

I need to see if I can obtain those papers mentioned by the other fellow. And the other chapter by Brown doesn't seem to be in my printed book??? Will look at it online.

10. May 16, 2014

### jartsa

It depends on observer's velocity relative to a baseline frame. (Thank you jbriggs444)

(baseline frame = whatever frame the observer decides to pick as a baseline frame)

11. May 18, 2014

### exmarine

The most rigorous treatment I can find - by far - is in a book by Thomas Phipps, and he credits Aharoni. It appears to be a perfectly orthodox treatment. I suppose I could type the derivation here, but I assume many of you have access to that original source - I don’t. So I’ll just type the result from my rework of Phipps’ rendition.

$cos\left(\alpha\right)=1-\frac{1-\ell^{2}}{1-\beta\ell}\left(1-\sqrt{1-\beta^{2}} \right)$

The expansion of the inverse cosine of the aberration angle to second order: (tricky!)

$\alpha=\sqrt{1-\ell^{2}}\beta+\frac{\ell\sqrt{1-\ell^{2}}}{2}\beta^{2}+...$

el is the direction cosine of the light ray from the source to that inertial axis of the source parallel to the relative velocity with the sink. And beta is of course the relative velocity divided by c. I can’t find any questionable or buried assumptions in the derivation.

Thanks for all the responses, but many of you - or at least some of you - seem to be pointing at a straw man. Of course we don’t know alpha. We only observe changes in alpha. Note the number of times I tried to emphasize the word "CHANGES" in my original post. And I see nothing in the derivation that limits those changes in alpha to being caused only by changes in the observer’s state of motion.

I am going away now. I remain puzzled by this.

12. May 18, 2014

### jartsa

Change of velocity of a light source does not cause a change of position of the light source according to an inertial observer.

There's a good intuitive explanation what happens when the velocity of light source changes in post #7.

13. May 18, 2014

### Meir Achuz

$$\tan\theta'=\frac{\sin\theta}{\gamma(\cos\theta+v/c)}$$,
where v is the velocity of the star wrt the observer, $\theta$ is the angle between v and the direction from the star to the observer, $\theta'$ is the angle that the telescope should be set at.
The aberration angle is $\theta'-\theta$.
This relativistic is close to the non-relativistic result
$$\tan(\theta'-\theta)=\frac{v}{c}\sin\theta$$,
which was originally derived, and measured, in 1729 by James Bradley.

14. May 19, 2014

### jartsa

How does the observer measure the direction from the star to the observer? (By aiming his telescope so that the star is in the view, I guess)

$\theta'-\theta$ that might be the angle between the apparent direction of a star and the "real" direction of the star.

The apparent direction is the direction the telescope should be aimed at.

The "real" direction is the direction to use if the observer becomes static reletive to the star, but not if the star accelerates to the observer frame, observer must accelerate to the star frame.

Last edited: May 19, 2014
15. May 20, 2014

### PhilDSP

Liebscher and Brosche seem to have done an excellent job of compiling the historical publications and debates about the subject. Of all of their references, I'd have to give most credence to Pauli, who as they say disagrees with their assessment and solution.

But aren't Liebscher and Brosche missing the crucial observation that's it's the relative transverse velocity or motion between the source and observer, in principle, that determines the aberration?

Last edited: May 20, 2014
16. May 20, 2014

### Meir Achuz

Theta is the angle for the direction of the star would be seen at with no aberration. Theta' is the actual angle of the telescope to best see the star. Theta' varies as the relative transverse velocity varies. Bradley measured the wobble in a stars apparent direction in half year intervals, with the transverse velocity being the speed of the earth in its orbit. For a binary star, the transverse velocity causing the wobble is the velocity of one star in its orbit. Theta is unimportant, as it is the variation of theta' that is measure.

17. May 20, 2014

### PAllen

I don't think so. If the observer is moving inertially, there is no aberration detectable, period. To detect aberration you need to compare observations from two states of motion (e.g. times of the year - different velocity). Thus, the key is observations in different states of motion, and the motion of the source is irrelevant except to the extent that it changes position of the distant source in relation to other distant sources.

18. May 20, 2014

### xox

I would like to add to the excellent post above that $\theta$ cannot be known, whereas $\theta'$ is what we measure, therefore, we do not need to spend too much time dwelling on the relationship between $\theta'$ and $\theta$. More exactly, astronomers need to incline the telescope by the angle:

$$\theta'(t)=arctan\frac{v(t)}{c}$$

as seen in this picture.

The angle $\theta'(t)$ is not constant, it varies with time because the speed of the Earth (where the telescope is located) $v=v(t)$ varies during the astronomical year in a rather complicated way:

$$v=v_e sin(\Omega_e t) cos \Phi_e +v_d sin(\Omega_d t) cos \Phi_d$$

where:

$v_e=30km/s$ is the Earth orbital speed
$v_d=0.355km/s$ is the Earth rotational speed
$2\pi/\Omega_e=$1 year
$2 \pi/\Omega_d=$1 day
$\Phi_e=$Earth axle inclination with respect to the orbital plane
$\Phi_d=$angle between the location on the Earth and the equatorial plane

The above is valid for the case of a star fixed with respect to the Sun. The case of a star moving with respect to the Sun gets more complicated, we need to adjust $v(t)$ in order to incorporate the relative motion between the star and the Sun:

$$v(t)=v_s(t)+v_e sin(\Omega_e t) cos \Phi_e +v_d sin(\Omega_d t) cos \Phi_d$$

where $v_s(t)$ is the relative speed between the star and the Sun. I hope this answers the OP question to his satisfaction.

Last edited: May 20, 2014
19. May 20, 2014

### PhilDSP

It will be necessary, won't it?, to take the integral of the relative change of position between the source and observer over the time that the photon is in flight. That is, once the photon has been emitted, any further motion of the source is immaterial.

That means that any one instantaneous relative velocity contributes only infinitesimally. But the calculation can't be performed properly without determining the relative motion based on where the source was at the time of photon emission.

If r is the position of the observer relative to the point of photon emission then
$$\Delta r = \int_{t_e}^{t_a} v dt$$
where $t_e$ is the time that the photon was emitted, $t_a$ is the time that the photon was absorbed by the observer and v, i.e. v(t) is the relative velocity between the point of emission and the observer at time t

Last edited: May 21, 2014
20. May 21, 2014

### PAllen

Suppose all distant objects and observer are mutually stationary. Then, the observer moves and takes observations in its new state of motion. There is observable aberration. Contrast with observer remains inertial, distant sources move in various way. There is no observable aberration. Thus, the determining feature is clearly comparing observations at relative motion to each other.