How is it possible for binary stars to not show changes in aberration angles?

Click For Summary
Binary stars can exhibit significant changes in their velocity components, yet no corresponding changes in their aberration angles are observed, raising questions about the consistency with Special Relativity Theory (SRT). The discussion highlights that aberration is influenced by the observer's velocity relative to a chosen baseline frame, rather than the relative velocity between the observer and the light source. The absence of "active" aberration in binary stars aligns with predictions of SRT, as established in various literature, including works by Liebscher and Brosche. Observers measure the direction of light from stars based on their own motion, which complicates the understanding of how changes in the source's velocity affect perceived angles. Overall, the conversation emphasizes the nuanced relationship between motion, observation, and the theoretical frameworks governing aberration.
  • #61
mfb said:
It boils down that you use a definition of aberration no one else does, I think. And one that fails as soon as the motion is not uniform any more.

So, according to you, tan \theta'=\frac{v}{c} is no longer a valid expression for aberration if v=v(t)?

How do you compare angles and directions in frames with a relative motion?

I am starting to see your problem, this is not about "comparing angles". This is not about "frameS" , it is about ONE frame, the Earth (lab) frame and ONE angle (variable in time), the angle \theta'(t)=arctan \frac{v(t)}{c}.

Certainly not, see the binary stars.

Can you provide the mathematical analysis? This is the third time I am asking for it.

Just view everything in the frame of the sun and there is absolutely no reason to expect any aberration effect (and nothing to calculate). Light travels in a straight line and does not care about the velocity of the emitter.

But the binary stars (and other stars as well) MOVE wrt. the Sun. Even worse, they MOVE wrt. the Earth observer.
 
Physics news on Phys.org
  • #62
PAllen said:
Where do you get that? From the very first post, this as been about stellar aberration in the context of SR.

The references in Histspec's #5 cover all the issues in this thread and more. In some sense, all after that is redundant.

Yes, you're probably right. And casting the classical formula into a distance traveled problem rather than a velocity problem is an unecessary complication. I deleted post 56 but can no longer delete 19, 21, 28, 51 and 52. If a moderator can do that, I agree they should be deleted.
 
  • #63
xox said:
All formulas are based on the relative speed between source and observer.
This is not true. Einstein's derivation related change observed light angle between any two inertial frames. The only velocity term is that between the two frames. The observed light source need not be at rest in either frame, and its velocity in either frame does not enter the formula at all. All confusion on this, in the SR case, is related to the historic convention from the era of Bradley of using the rest frame of the source as one of the frames, and the rest frame of the observer as the other. But Einstein's derivation and formula have no such requirements.
xox said:
Are you talking about the formula cos \theta_{obs}=\frac{cos \theta_{src}-v/c}{1 -v/ccos \theta_{src}}? Because I am talking about something totally different: \tan \theta_{obs}=\frac{v}{c}, where v=v(t) is the relative speed between the source and the observer. See here. Even in cos \theta_{obs}=\frac{cos \theta_{src}-v/c}{1 -v/ccos \theta_{src}} v=v(t) is the relative speed between the source and the observer.
That formula is Bradley' and it is derived from the ballistic theory of light. It is approximately correct to within observational limits, when both are valid.
xox said:
Yes, obviously. But this is not what I am talking about.
You can't avoid it.
xox said:
The "speed of the source" (the star) in the baricentric reference frame of the Sun is a definite contributor to the aberration, just as much as the speed of the Earth , in the Sun reference frame. See here
The formula you give there is not valid unless you factor in that the the θ to which θ' is related is ever changing for an accelerating source, in just such a way as to cancel the effect of the source motion on observed angular position.
xox said:
Yes, ..which is exactly what I posted earlier.




Well, the source-angular position is nothing but a function of the source velocity in the Sun - anchored frame. So, to say that "aberration is not a function of the speed of the star wrt. the Sun-based frame" is just a misnomer. As an aside, could you put what you said in words into math, the way I did it? This would make things a lot clearer.

What I said is true, and very well known.

Rather than repeat what has been derived elsewhere, very clearly, I will simply point you to references already provided in this thread:

For the relativistic proof that a binary star's motion has no affect on aberrations see last sections of:

http://www.mathpages.com/rr/s2-05/2-05.htm

Note also, the link you provided: http://en.wikipedia.org/wiki/Stellar_aberration_%28derivation_from_Lorentz_transformation%29#Application:_Aberration_in_astronomy

makes no mention of source velocity.

Finally, Histspec also provided a link to Herschel's proof in 1844 using pre-relativistic aberration, than no effect from binary stars would be expected.

http://articles.adsabs.harvard.edu/full/1844AN...22..249H
 
  • #64
PAllen said:
This is not true. Einstein's derivation related change observed light angle between any two inertial frames.

I made it quite clear, several times, that this is not the formula I am talking about, I am talking about tan \theta'=\frac{v}{c}

The only velocity term is that between the two frames.

I know that very well. This is precisely why the formula is totally useless in orienting the telescopes, I pointed this out several times as well.

But Einstein's derivation and formula have no such requirements.

I am not talking about the Einstein aberration formula.

For the relativistic proof that a binary star's motion has no affect on aberrations see last sections of:

http://www.mathpages.com/rr/s2-05/2-05.htm

Let's concentrate on the above because we can all see the complete derivation. First off, the derivation is an ugly mess. The mess aside, what Brown is saying is that under certain circumstances, the aberration of the "circling star" is close enough to the aberration of the "central star". The aberration formulas are clearly not the same, they become the same only after assuming R<<L (see his notation) AND neglecting some of the higher powers of \frac{v}{c}.
 
Last edited:
  • #65
xox said:
I made it quite clear, several times, that this is not the formula I am talking about, I am talking about tan \theta&#039;=\frac{v}{c}
I know that very well. This is precisely why the formula is totally useless in orienting the telescopes, I pointed this out several times as well.
This is false. Derivations based on Einstein's formula are the modern foundation for all aberration theory, and underlie orienting telescopes.
xox said:
I am not talking about the Einstein aberration formula.

But the only accepted derivations today are based on Einstein's formula. We accept SR don't we?

[edit: Why don't you actually read the links, then come back with questions. They all concern aberration for the purpose of orienting telescopes.]
 
  • #66
PAllen said:
This is false. Derivations based on Einstein's formula are the modern foundation for all aberration theory, and underlie orienting telescopes.

The \theta in cos \theta&#039;=\frac{cos \theta +\beta}{1+ \beta cos \theta} is unknown , so the formula used is the one I posted in post 18. The discussion is not about the validity (it is valid), it is about its practical use (it isn't).


But the only accepted derivations today are based on Einstein's formula. We accept SR don't we?

Sure, we all accept SR. You are missing the point, the discussion is not about the validity of SR, it is about whether or not there are expected effects of the varying relative speed between the source and the receiver.


[edit: Why don't you actually read the links,

I actually read them, this is how I could detect all the typos and hacks in the mathpages link. I listed the conditions under which the aberrations of the two stars "become" similar (they are not identical), have you missed that?

then come back with questions.

I do not have questions, I am quite clear on the subject. Please stop talking down to me, I have a level of understanding that is equal to yours.

They all concern aberration for the purpose of orienting telescopes.

Correct. cos \theta&#039;=\frac{cos \theta +\beta}{1+ \beta cos \theta} while perfectly valid, is not the used formula , nor is it useful, for reasons explained above.
 
Last edited:
  • #67
[Edit: comments below fixed for confusing β and θ]

xox said:
The \theta in cos \theta&#039;=\frac{cos \theta +\beta}{1+ \beta cos \theta} is unknown , so the formula used is the one I posted in post 18. The discussion is not about the validity (it is valid), it is about its practical use (it isn't).
You don't need to know θ. If you apply this [Einstein] formula for the history of Earth frames relative to any given θ in the solar frame, using Earth velocity relative to the sun, you get the aberration pattern. You can allow for θ(t) for a star with significant movement. You don't need to incorporate its velocity. Given the aberration pattern, you know that if you found an object at one location in January, where to look for it in June.
xox said:
Sure, we all accept SR. You are missing the point, the discussion is not about the validity of SR, it is about whether or not there are expected effects of the varying relative speed between the source and the receiver.
There aren't. Every reputable author says no. Do you have any reference that says yes?
xox said:
I actually read them, this is how I could detect all the typos and hacks in the mathpages link. I listed the conditions under which the aberrations of the two stars "become" similar (they are not identical), have you missed that?

I did miss discussion of typos in the mathpages links. Can you indicate the posts in this now long thread?

Typos or not, the content of the mathpages are correct and accepted (except for stupid history debates about whether or not Einstein had a misunderstanding at some point).
 
Last edited:
  • #68
exmarine said:
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?

The above statement has a key qualifier, the effect effect is negligible if the distance between the binary stars and Earth , L, is much larger than the radius of their orbits, R. IN ADDITION, the terms in \frac{v^2}{c^2} and larger NEED to be neglected as well. For a fairly muddled, typo-ladden proof, see here, at the bottom of the page.
 
  • #69
PAllen said:
You don't need to know what you have renamed β.

I did not mention anything about \beta, my point was about not knowing the angle \theta.





Typos or not, the content of the mathpages are correct and accepted (except for stupid history debates about whether or not Einstein had a misunderstanding at some point).

I did not dispute the correctness of Brown's derivation, I simply pointed out that the aberrations for the two stars are NOT identical, the difference in their speeds DOES make a difference. That's all.
 
  • #70
xox said:
The above statement has a key qualifier, the effect effect is negligible if the distance between the binary stars and Earth , L, is much larger than the radius of their orbits, R. IN ADDITION, the terms in \frac{v^2}{c^2} and larger NEED to be neglected as well. For a fairly muddled, typo-ladden proof, see here, at the bottom of the page.

Those qualifiers simply rule out the case of a star that not only has some substantial velocity, it also changes position substantially (on scale of observation, e.g. a day or a year). The fact that you see it change position is not called aberration. Aberration would be the difference between how a platform stationary with respect to the sun would see it move versus how an Earth observer sees it move. It is just the seasonal ripple on such a substantial motion.
 
  • #71
xox said:
I did not mention anything about \beta, my point was about not knowing the angle \theta.
Sorry, I misread your formula which used beta where I use v. Substitute θ for β in what I said and it stands.
xox said:
I did not dispute the correctness of Brown's derivation, I simply pointed out that the aberrations for the two stars are NOT identical, the difference in their speeds DOES make a difference. That's all.

Only if you include in aberration something no one else does: seeing an object move. A speed effect of aberration is the claim that two colocated objects with different speeds or changing speed (while remaining colocated) - are seen as not colocated from earth. That Earth sees a star that moves a lot relative to another star move, is obvious and is not aberration.
 
  • #72
PAllen said:
Sorry, I misread your formula which used beta where I use v. Substitute θ for β in what I said and it stands.

This is the closest you'll come to admitting that you were wrong. Let me try to explain to you one more time: θ is not known and cannot be known. By contrast , β is known with very high accuracy (see , again, post 18). So, contrary to your claim, the two are not interchangeable.
Only if you include in aberration something no one else does: seeing an object move.

Has nothing to do with "seeing an object move". Has everything to do with angling the telescopes in such a way as to be able to "see" an object. Since I am getting tired of exchanging prose with you, let me put this in mathematical terms:

Say that in the frame of the emitter a ray of light propagates with velocity \vec{c}.
In the frame of the Earth, moving with velocity V wrt the source, the velocity of the light ray is not \vec{c} but:

\vec{c&#039;}=\frac{1}{\gamma(V)(1-\beta cos \theta)}(\vec{c}+\frac{(\gamma(V)-1)cos \theta -\beta \gamma(V)}{\beta}\vec{V})

The departure of \vec{c&#039;} from \vec{c} is the rigorous, mathematical definition of light aberration. The presence of \gamma(V) is what makes it the definition of relativistic light aberration. In the above,

\beta=\frac{V}{c}
\gamma=\frac{1}{\sqrt{1-\beta^2}}
cos \theta =\frac{\vec{V} \cdot \vec{c}}{Vc}

A speed effect of aberration is the claim that two colocated objects with different speeds or changing speed (while remaining colocated) - are seen as not colocated from earth.
Definition by enumeration is a very poor way of doing physics, I gave you the exhaustive, rigorous definition.
That Earth sees a star that moves a lot relative to another star move, is obvious and is not aberration.

The Earth doesn't see" anything, the observers note that the position of the stars may be aberrated differently, as expressed by the formula given above.
 
  • #73
xox said:
This is the closest you'll come to admitting that you were wrong. Let me try to explain to you one more time: θ is not known and cannot be known. By contrast , β is known with very high accuracy (see , again, post 18). So, contrary to your claim, the two are not interchangeable.

Has nothing to do with "seeing an object move". Has everything to do with angling the telescopes in such a way as to be able to "see" an object. Since I am getting tired of exchanging prose with you, let me put this in mathematical terms:

Say that in the frame of the emitter a ray of light propagates with velocity \vec{c}.
In the frame of the Earth, moving with velocity V wrt the source, the velocity of the light ray is not \vec{c} but:

\vec{c&#039;}=\frac{1}{\gamma(V)(1-\beta cos \theta)}(\vec{c}+\frac{(\gamma(V)-1)cos \theta -\beta \gamma(V)}{\beta}\vec{V})

The departure of \vec{c&#039;} from \vec{c} is the rigorous, mathematical definition of light aberration. The presence of \gamma(V) is what makes it the definition of relativistic light aberration. In the above,

\beta=\frac{V}{c}
\gamma=\frac{1}{\sqrt{1-\beta^2}}
cos \theta =\frac{\vec{V} \cdot \vec{c}}{Vc}

Definition by enumeration is a very poor way of doing physics, I gave you the exhaustive, rigorous definition.

The Earth doesn't see" anything, the observers note that the position of the stars may be aberrated differently, as expressed by the formula given above.

I claim this formula is the one that cannot (easily) be applied because it is relative to the emitter momentary rest frame. In the case of changing velocity of the emitter, you must deal with changing angle as expressed in emitter frames that change from one moment to the next. In your notation, the unprimed null vector is unknown, and the one that reaches Earth is changing moment to moment between changing emitter rest frames. Thus, for a tightly circling emitter, it is quite complicated to apply this formula in a way to predict the observed pattern of aberration for such a source.

Meanwhile, computing aberration relative to a solar frame, with only the velocity of Earth involved, does not have this problem, and does not (and has no place) for emitter speed (only emitter position in the solar frame). An observation on any night can be transformed to a solar frame with the very well known velocity of Earth relative to sun. From there, you can compute the effect of the Earth's motion at any future time. It is this mode of analysis which shows, in a very simple way, why nothing about observable aberration includes any effect due to the speed of the emitter.
 
Last edited:
  • #74
PAllen said:
Meanwhile, computing aberration relative to a solar frame, with only the velocity of Earth involved, does not have this problem, and does not (and has no place) for emitter speed (only emitter position in the solar frame).

What you claim above works if and only if the star is stationary wrt. the Sun. Does not work for stars moving wrt. the Sun.
 
  • #75
xox said:
What you claim above works if and only if the star is stationary wrt. the Sun. Does not work for stars moving wrt. the Sun.

Nonsense. Transforming an observed light angle from Earth frame to sun frame is not affected in any way by the speed of whatever emitted the light. It is a straight Lorentz transform, whose result is Einstein's formula (which applies between any two frames, and is true independent of the motion of the source in either frame).
 
  • #76
PAllen said:
Nonsense. Transforming an observed light angle from Earth frame to sun frame is not affected in any way by the speed of whatever emitted the light.

This is not what I said. What I told you is that, contrary to your claim, a light ray coming from a star moving wrt the Sun IS aberrated in the frame of the Sun. Nothing to do with any "Transforming an observed light angle from Earth frame to sun frame", nothing to do with the "Earth frame". Everything to do with the relative motion star-Sun.
 
  • #77
xox said:
This is not what I said. What I told you is that, contrary to your claim, a light ray coming from a star moving wrt the Sun IS aberrated in the frame of the Sun. Nothing to do with any "Transforming an observed light angle from Earth frame to sun frame", nothing to do with the Earth. Everything to do with the relative motion star-Sun.

That aberration, again, is between the sun's observation and an observation in a momentary rest frame of the star. Nobody can see this or cares about this. Given, over some viewing period, a description of the motion of moving, accelerating star (imagine a very large binary system) in the solar frame (computed by adjusting the raw observation for day by day changes in Earth's motion), you can then extrapolate the stars position in the solar inertial frame, and than compute where to observe it at any future date.

A priori, nobody knows the velocity of a star relative to Earth or to the sun. However, earth-sun velocity is very precisely known. You can then derive the pattern of its motion seen from a convenient inertial frame by adjusting raw observations for Earth motion.
 
  • #78
PAllen said:
Einstein's derivation related change observed light angle between any two inertial frames. The only velocity term is that between the two frames. The observed light source need not be at rest in either frame, and its velocity in either frame does not enter the formula at all. All confusion on this, in the SR case, is related to the historic convention from the era of Bradley of using the rest frame of the source as one of the frames, and the rest frame of the observer as the other. But Einstein's derivation and formula have no such requirements.

Brilliant comment! I need to re-think with this in mind. Thanks!
 
  • #79
PAllen said:
That aberration, again, is between the sun's observation and an observation in a momentary rest frame of the star. Nobody can see this or cares about this. Given, over some viewing period, a description of the motion of moving, accelerating star (imagine a very large binary system) in the solar frame (computed by adjusting the raw observation for day by day changes in Earth's motion), you can then extrapolate the stars position in the solar inertial frame, and than compute where to observe it at any future date.
The rigorous thing to do is to compose the instantaneous velocity of the star wrt. the Sun with the instantaneous velocity of the Sun wrt. the Earth. Use the resultant speed to calculate the aberration. This is what I am showing in post 18.
What you are doing is composing the instantaneous position of the star with respect to the Sun with the instantaneous velocity of the Earth wrt. the Sun. This isn't rigorous, no wonder that applying this method enables you to claim that "see, the speed of the source (the star) does not affect the aberration".

A priori, nobody knows the velocity of a star relative to Earth or to the sun.

All you need to know is the angular velocity of one star about the other star, as Brown does in his derivation:

" The coordinates of the smaller star revolving at a radius R and angular speed w around the larger star in a plane perpendicular to the Earth are x2(t) = -vt + Rcos(q), y2(t) = Rsin(q), z2(t) = L, and t2(t) = t, where q = wt + f is the angular position of the smaller star in its orbit. Again, since light travels along null paths, a pulse of light arriving on Earth at time t = 0 was emitted at time t = T satisfying the relation
Dividing through by L2 and re-arranging terms, we have
Consequently, for L sufficiently great compared to R, the second two terms on the right side are negligible, so we have again T = , and hence the tangents of the angles of incidence in the x and y directions are
The leading terms in these tangents represent just the inherent "static" angular separation between the two stars viewed from the Earth, and these angles are negligibly small for sufficiently large L. Thus the tangent of the aberration angle is (again) essentially just , and so, as before, we have sin(a) = v, which of course is the same as for the central star. Incidentally, recall that Bradley's original formula for aberration was tan(a) = v, whereas the corresponding relativistic equation is sin(a) = v. The actual aberration angles for stars seen from Earth are small enough that the sine and tangent are virtually indistinguishable."

A collection of approximations, indeed, but he does use the angular velocity of one star as it orbits the "stationary" one. He does use a combination of the angular velocity of the "orbiting star" with the position of the "central star" wrt. the Sun.
 
Last edited:
  • #80
xox said:
The actual aberration angles for stars seen from Earth are small enough that the sine and tangent are virtually indistinguishable."

My appreciation for Brown's explanations is growing, but something inside does shutter a bit on encountering a crude approximation without a rigorous analysis of what the implied error entails. Is there any chance that the approximation error can be further analyzed? What does it mean with respect to other parameters? (A simple answer may suffice)

Then again, this may be a digression from the crucial issue...
 
  • #81
exmarine said:
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?

I am too left wondering about the usual explanation given for annual stellar aberration and can imagine some questions many could come up with, so maybe somebody could try and address these points.
It is normally accepted that the relativistic explanation follows the same reasoning as the classical Newtonian one, just giving a better approximation since it uses the relativistic velocity addition.

The first question would be, how can it be the same reasoning, classical Newtonian physics is based in an absolute space and absolute velocities, so it makes sense to attribute the aberration effect just to the observer's velocity, but it doesn't sound so right in the relativistic paradigm, more so if we consider the case of binaries mentioned by the OP.
So if we are using relativity and consider just relative velocities between the source and the observer one may wonder why shoul the sun's frame be important in the relative motion between the Earth and the stellar source. Surely the Earth is in relative motion wrt many objects in a periodic way that we do not care for in order to determine aberration wtt distant stars. So what is special sbout the sun's frame?
 
  • #82
FWIW, the light went on for me with the excellent comment by PAllen, #?.

PAllen said:
Einstein's derivation related change observed light angle between any two inertial frames. The only velocity term is that between the two frames. The observed light source need not be at rest in either frame, and its velocity in either frame does not enter the formula at all. All confusion on this, in the SR case, is related to the historic convention from the era of Bradley of using the rest frame of the source as one of the frames, and the rest frame of the observer as the other. But Einstein's derivation and formula have no such requirements.

He is eggzackly right. Any number of passing inertial observers must see a perfect sphere of outgoing light from the very same flash. The Lorentz transform applies between any two, though obviously neither share the inertial reference frame of the light source. I, and perhaps some others, had a sort of semi-ballistic idea about the source without realizing it. All the examples and proofs, from Brown on down were quite unconvincing for me. In my opinion, this is the key to understanding stellar aberration: the state of motion of the light source has no bearing on anything.
 
  • #83
exmarine said:
FWIW, the light went on for me with the excellent comment by PAllen, #?.
.


That's a nice comment, but IMO it doesn't by itself clarify the specific questions I posed above: if indeed any two inertial frames(the Earth's and another) are valid, why the correct aberration is only obtained using the relative velocity of the Earth and the Sun frames, and not using the Earth and any other object's frame with motion periodically related to the Earth , what is special about the specific sun-earth velocity for the observed aberration of light coming from distants stars?
Note again that I'm sticking to relativistic reasoning to formulate these questions, they don't come up if one uses prerelativistic classical explanations(Bradley, Fresnel) as there the Earth's motion is absolute.
 
  • #84
TrickyDicky said:
...why the correct aberration is only obtained using the relative velocity of the Earth and the Sun frames, and not using the Earth and any other object's frame with motion periodically related to the Earth ...

I am probably in over my head here, but I don't think the correct aberration is ONLY obtained with the relative velocity between us and the sun. It is just the most convenient. I think any other "non-accelerating" frame would work. Or is that not precisely your concern?
 
  • #85
exmarine said:
I am probably in over my head here, but I don't think the correct aberration is ONLY obtained with the relative velocity between us and the sun. It is just the most convenient. I think any other "non-accelerating" frame would work. Or is that not precisely your concern?


For Bradley classical annual aberration only the earth-Sun relative velocity in the aberration formula gives the correct angle correction, have you ever found a different velocity used ?
 
Last edited:
  • #86
Bill_K said:
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.


I considered this approach at first, and even found a paper that tries to explain it using relative motion between photons sent from the source and the observer, instead of source-observer relative motion, but this clearly can't be correct, mostly because as we all know the photon's frame can't be employed like this in physics.
 
  • #87
Another point in this thread that I think deserves to be better explained is that of the relativistic composition of velocities, a formula was given by jartsa that involved addition of the velocity of photons coming from the stellar source to observer's velocity wrt sun, I was not aware that relativistic addition of velocities could be performed with photons velocities, always saw it done with objects moving below c. Again this seems to go counter the restriction to use the unphysical photon's frame.
 
  • #88
TrickyDicky said:
For Bradley classical annual aberration only the earth-Sun relative velocity in the aberration formula gives the correct angle correction, have you ever found a different velocity used ?

More carefully, isn't it the annual CHANGES in our relative velocity with the sun that gives the correct observed changes in the aberration angle? So wouldn't our annual CHANGES be the same with any other inertial frame?
 
  • #89
exmarine said:
More carefully, isn't it the annual CHANGES in our relative velocity with the sun that gives the correct observed changes in the aberration angle?
Yes.
So wouldn't our annual CHANGES be the same with any other inertial frame?
No, unless this change would casually amount to 30 km/s too.

What could seem puzzling if one considers aberration of sufficiently distant stars is why this particular change in velocity wrt the sun prevails over many other faster ones that depend on the Earth's motion, like its motion in the galaxy, or the local group..., and for closer ones any other relative motion wrt random objects orbiting near the earth.
 
Last edited:
  • #90
??

V_{earth}=V_{sun}\pm{V_{orbit}}
\Delta{V_{earth}}=2V_{orbit}

V_{earth}=V_{any other inertial frame}\pm{V_{orbit}}
\Delta{V_{earth}}=2V_{orbit}

What am I missing? Relativistic addition wouldn't change this would it?
 

Similar threads

Replies
6
Views
2K
Replies
11
Views
2K
Replies
8
Views
2K
  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 28 ·
Replies
28
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
27
Views
3K
  • · Replies 14 ·
Replies
14
Views
4K
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
4K