Stellar aberation, a One way measurement of c?

[PAllen]

I respectfully disagree with you. I would like to provide additional details of the experiment which was proposed. However, If doing so will cause me to be locked out or banned I will just shut up and go away. What I have proposed is not speculation, a new theory, or an attempt to disprove anything. I am only proposing a test of which I think you misunderstood my description.

If you clarify your proposed experiment, you may get useful responses. It is fine to propose experiments for analysis.

This is the key phrase I was responding to is:

"then if the angle of aberration is the same for reversed positions of the viewing device"

What is angle of aberration? It is a difference from what is expected. But what is expected is simply either the result of a measurement in a different frame, or (in the actually used convention) the derived position imputed to the solar system frame based on collection angles observed over a year (or by applying a formula based on known speeds - but then you have not a measurement of aberration but a computation of aberration which is computed from the assumption of c). Thus, in one frame, all you can measure is 'where it is'. The minimum needed to measure aberration is two frames at different relative speed (you then have two angular positions to compare).

So, if you have something else in mind, you should clearly specify it.

 

[Tracer]

If you clarify your proposed experiment, you may get useful responses. It is fine to propose experiments for analysis.

This is the key phrase I was responding to is:

"then if the angle of aberration is the same for reversed positions of the viewing device"

What is angle of aberration? It is a difference from what is expected. But what is expected is simply either the result of a measurement in a different frame, or (in the actually used convention) the derived position imputed to the solar system frame based on collection angles observed over a year (or by applying a formula based on known speeds - but then you have not a measurement of aberration but a computation of aberration which is computed from the assumption of c). Thus, in one frame, all you can measure is 'where it is'. The minimum needed to measure aberration is two frames at different relative speed (you then have two angular positions to compare).

o, if you have something else in mind, you should clearly specify it.

Mount a device like a super sniper scope onto a large refracting telescope which is on an equatorial mount and is compensating for the Earth’s rotation. Adjust the sniper scope and its mount so that it is aimed at the same object as the main refracting telescope. Build the sniper scope’s mount such that it can be turned to accurately reverse its viewing direction by exactly 180 degrees. Mount a system of mirrors on the main telescope to reflect the image being observed by the sniper scope by exactly 180 degrees so that when the sniper scope is turned from a looking forward to a looking backward position the direction of the light from the viewed object will also be reversed by 180 degrees.
Select a star for which the amount of aberration is well known. The angle of aberration is not important but the larger it is the better. Aim the main telescope at the targeted star and finely adjust the telescope so that the star image is in the center of its viewing field. Similarly, finely adjust the sniper scope so that the targeted star’s image is centered in the scope’s crosshairs. Note that this only amounts to a calibration of the measurement device and the actual angle of aberration being experience by that star is totally unimportant.
Now reverse the viewing direction of the sniper scope by 180 degrees so that it is now viewing the reflected image of the targeted star. This is in effect is similar to a measurement taken six months later than the first. If the speed of light is the same for the direct and reflected views then the image of the targeted star will be remain centered in the sniper scope’s cross hairs in both viewing directions. Any change in the speed of light between the two measured directions will cause the targeted star’s position to be in different positions in the viewing field of the sniper scope.
If this test proves that the speed of light is the same for both the direct (forward) and reflected (backward) direction through the sniper scope, then all of the many two way measurements of the average speed of light, can be known to be the average of two one way passes in which each pass is exactly c.
Admittedly, this method does not produce a direct measurement of c. However it should remove all doubt that the forward and backward speed of light in a two way measurement is at exactly c.

 

[PAllen]

...

Now reverse the viewing direction of the sniper scope by 180 degrees so that it is now viewing the reflected image of the targeted star. This is in effect is similar to a measurement taken six months later than the first.

This part is wrong. It is not at all the same as 6 months later. What characterizes 6 months later is that the Earth's direction of motion has reversed (relative to 6 months earlier). Your measurement will produce a null result, always, giving no information at all on the speed of light in any direction. In fact, your measurement now has nothing to do with the aberration or the speed of light. It asks: given an image, if I reflect it 180°, will it be reflected 180°?

 

[Tracer]

Oops you're right. OK then, let's make a start. I think that it has already been clarified that aberration is caused by the velocity difference due to the Earth's orbit. The Earth's orbital speed is simply its speed relative to the Sun; and the Sun does not designate an absolute reference frame. Instead we could for example choose the combined speed of Earth + Sun through the Milky way. That would yield a very different one-way speed of light if we use the same t.

Tracer asks: "Yes but why would anyone bother to do that. Many years would be required before differences in aberration due to galactic motion could be observed. In the meantime many cycles of stellar aberration of galactic objects could be observed using just the Earth's relative velocity with the sun. Even if the correct value is used for t for the Earth's relative velocity with the milky way wouldn't c still be the same?"

It always boils down to the same thing: if we choose a certain reference system as "rest" frame, then we will measure the speed of light relative to it as c, but else we won't. And then it depends on what one means with "one way measurement of c".

In a two mirror measurement of c, I would define a two way measurement of c as the distance (d) between the two mirrors divided by one half of the time (t) taken for light to travel from one mirror to the other and be reflected and arrive back at the first mirror.
c= d/0.5t.

A one way measurement would be for either half of light's travel in just one direction between the two mirrors. At least that is my understanding of the term.

 

[harrylin]

Yes but why would anyone bother to do that. Many years would be required before differences in aberration due to galactic motion could be observed. In the meantime many cycles of stellar aberration of galactic objects could be observed using just the Earth's relative velocity with the sun. Even if the correct value is used for t for the Earth's relative velocity with the milky way wouldn't c still be the same?

Evidently you agree that it is not the Earth's velocity that matters for the phenomenon, but the difference of its velocities between two measurements - else no time would be required to observe the effect, just the instantaneous velocity would do.

However, that is not the right information if you try to determine the velocity of light propagation; your method would only be correct if you could assume that the average velocity of the Earth is truly or absolutely zero: only then the differences correspond to twice the (absolute!) velocities of Earth and light. And as you know, there is no reason to think such a thing. Thus, again: please redo your analysis without assuming that the Sun is in rest, so that you take into account that the corresponding velocities of the Earth are for not +v and -v but for example V+v and V-v.

In a two mirror measurement of c, I would define a two way measurement of c as the distance (d) between the two mirrors divided by one half of the time (t) taken for light to travel from one mirror to the other and be reflected and arrive back at the first mirror.
c= d/0.5t.

A two way measurement is simply defined as the return distance divided by the return time.

A one way measurement would be for either half of light's travel in just one direction between the two mirrors. At least that is my understanding of the term.

Yes. I'm afraid that you misunderstood a remark of mine. c is a constant of nature, and you may be able to determine it with great accuracy, for example with the help of one-way light rays. However, you now indicate that you are not interested in trying to determine that constant of nature called c, but in trying to measure the one-way velocities of light rays that arrive from outer space on Earth, relative to the Earth.

 

[Tracer]

Evidently you agree that it is not the Earth's velocity that matters for the phenomenon, but the difference of its velocities between two measurements - else no time would be required to observe the effect, just the instantaneous velocity would do.

However, that is not the right information if you try to determine the velocity of light propagation; your method would only be correct if you could assume that the average velocity of the Earth is truly or absolutely zero: only then the differences correspond to twice the (absolute!) velocities of Earth and light. And as you know, there is no reason to think such a thing. Thus, again: please redo your analysis without assuming that the Sun is in rest, so that you take into account that the corresponding velocities of the Earth are for not +v and -v but for example V+v and V-v.QUOTE]

Why wouldn't the change in velocities over a six month period regardless of the value of V simply be:

delta v = (V + v) - (V-v) = 2v

 

[harrylin]

Why wouldn't the change in velocities over a six month period regardless of the value of V simply be:

delta v = (V + v) - (V-v) = 2v

Your equation relates to velocities, not differences of velocities - or at least, that is how you apparently apply it. In order to be able to truly measure ("confirm") the one-way speed of light relative tot the earth, you need to know the total speed of the Earth (V+v). However, as you remarked yourself, the effect that you measure corresponds to 2v.

According to relativity you can pick any V you like (but <c) and the experiment will not show you wrong (here relativistic corrections come at play that I did not mention as they are of less importance).

 

[PAllen]

Thinking more about aberration measurement, I notice something important:

1) If you use the relativistic derivation, you are Lorentz transforming light propagation angle from e.g. E1 (earth January) to E2 (earth June). The relation ship between angle, relative velocity of E1 and E2, and c is a consequence of the Lorentz transform (the only involvement of the light from the source is that it is light, thus follows a lightlike path). Thus, the only thing you are validating is the Lorentz transform. It seems to me, you are not even really measuring the velocity of light at all! You are just using light from one source to validate the form of the transform, and the constant c within it. Also, note, that in a correct relativistic treatment, motion of the source (star) is irrelevant.

2) If you use Galilean relativity and a corpuscular light theory (the Bradley derivation), you are measuring one way c in one or the other frame. Theoretically, you will get deviations compared to (1), and your derivation is based on light speed being different in the two frames (rather than just angle being different). You would also expect to get a (slightly) different c if you used the moon over two weeks rather than the Earth over 6 months. You would also expect to see an effect of source motion. While many of these differences are too small to detect, source motion dependence has been rigorously ruled out by measurement of aberration from rapidly orbiting binary stars.

With (2) completely ruled out, we have the conclusion of (1) - within a relativistic framework, this doesn't measure actual light speed at all. It measures that light propagation direction transforms according to the Lorentz transform (which includes the constant c defined from the two way speed of light).

[EDIT: and thus we close the loop on how this is functionally the same measuring one way c with slow clock transport. If if measures something different from c, disproving SR equivalent theories, it can measure one way light speed between some source and target frame (light speed no longer being a universal constant). As long SR is confirmed, it doesn't provide any additional information about one way lightspeed.]

 

[harrylin]

Your equation relates to velocities, not differences of velocities - or at least, that is how you apparently apply it. In order to be able to truly measure ("confirm") the one-way speed of light relative tot the earth, you need to know the total speed of the Earth (V+v). However, as you remarked yourself, the effect that you measure corresponds to 2v.

According to relativity you can pick any V you like (but <c) and the experiment will not show you wrong (here relativistic corrections come at play that I did not mention as they are of less importance).

Theoretically [..] your derivation is based on light speed being different in the two frames (rather than just angle being different).

It may be useful to give a silly illustration with a better understood phenomenon.

You are on a huge cruise ship on which they even installed installed a giant wheel with a 50m diameter. As you have nothing else to do while the ship is cruising on the ocean, you decide to try it. Bad luck, it starts to rain. While stuck in that thing in the poring rain, you notice (thanks to your extremely developed senses) that when you are up high, the rain falls under a slightly different angle than when you are down below.

After pondering over this phenomenon, you think that you can determine the speed of the rain drops relative to you, simply by measuring the angles and the rotation frequency of the wheel. From that you first calculate your speed v and next you extract (or so you think!) the speed V of the raindrops from V= v/tan(Theta).

 

[Tracer]

It may be useful to give a silly illustration with a better understood phenomenon.

You are on a huge cruise ship on which they even installed installed a giant wheel with a 50m diameter. As you have nothing else to do while the ship is cruising on the ocean, you decide to try it. Bad luck, it starts to rain. While stuck in that thing in the poring rain, you notice (thanks to your extremely developed senses) that when you are up high, the rain falls under a slightly different angle than when you are down below.

After pondering over this phenomenon, you think that you can determine the speed of the rain drops relative to you, simply by measuring the angles and the rotation frequency of the wheel. From that you first calculate your speed v and next you extract (or so you think!) the speed V of the raindrops from V= v/tan(Theta).

I am not sure of what your point is here. But let me put some values on a sample problem so you can see if I am doing something illogical or my math is wrong.

A. Let the ships speed through the wind equal to 15 meters/sec directly fore to aft.


B. Let the rim velocity of the wheel be 10 meters/sec and its direction of rotation is such that the rim velocity adds to the ship’s wind velocity at the top of the wheel and subtracts from the ship’s wind velocity at the bottom of the wheel.


C. Let the rain drops fall vertically at 30 meters/sec when the wind velocity is zero. This will be treated as an unknown until it has been calculated based on its viewed angle of approach to the observer.

At the top of the wheel the wind velocity will be 25 meters/sec. If the angle of incidence (theta)to the observer at the top of the wheel is 50.194429 degrees, then the true vertical velocity of the rain drops is:

V = 25tan(theta)=25tan(50.194429) =25(1.2) = 30 meters/sec

The rain drops will strike the observer at the top of the wheel at:

V = 30/sin(theta) = 30/0.7682213 = 39.051248 meters/sec

At the bottom of the wheel the wind velocity will be -5 meters/sec. if theta at the bottom of the wheel is measured to be -80.537678 degrees, then the true vertical velocity of the rain drops is:

-5tan(theta) =-5tan(-80.537678) = -5(-6) = 30 meters/sec

The rain drops will strike the observer at the bottom of the wheel at:

V = 30/sin(theta) =30/0.9863939 = 30.413813 meters/sec

Is this correct? What can be determined from composite measurements from the top and the bottom of the wheel?

 

[harrylin]

I am not sure of what your point is here. But let me put some values on a sample problem so you can see if I am doing something illogical or my math is wrong.

A. Let the ships speed through the wind equal to 15 meters/sec directly fore to aft.

B. Let the rim velocity of the wheel be 10 meters/sec and its direction of rotation is such that the rim velocity adds to the ship’s wind velocity at the top of the wheel and subtracts from the ship’s wind velocity at the bottom of the wheel.

C. Let the rain drops fall vertically at 30 meters/sec when the wind velocity is zero. This will be treated as an unknown until it has been calculated based on its viewed angle of approach to the observer.

At the top of the wheel the wind velocity will be 25 meters/sec. If the angle of incidence (theta)to the observer at the top of the wheel is 50.194429 degrees, then the true vertical velocity of the rain drops is:

V = 25tan(theta)=25tan(50.194429) =25(1.2) = 30 meters/sec

The rain drops will strike the observer at the top of the wheel at:

V = 30/sin(theta) = 30/0.7682213 = 39.051248 meters/sec

At the bottom of the wheel the wind velocity will be -5 meters/sec. if theta at the bottom of the wheel is measured to be -80.537678 degrees, then the true vertical velocity of the rain drops is:

-5tan(theta) =-5tan(-80.537678) = -5(-6) = 30 meters/sec

The rain drops will strike the observer at the bottom of the wheel at:

V = 30/sin(theta) =30/0.9863939 = 30.413813 meters/sec

Is this correct? What can be determined from composite measurements from the top and the bottom of the wheel?

As you indicate, here the rain drop velocity is around 30 m/s relative to the guy in the giant wheel (note that this is also called closing speed), but it depends on the ship's velocity and varies over time (sorry: I did not check your calculations but it looks fine).

My point was, and still is: he doesn't know these rain drop velocities relative to him. Apparently you think/thought that it is but one velocity which he should be able to determine from the wheel's speed and the difference of observed angles (theta is in fact the angle between one inclination and the other one). Did you try if he can indeed achieve that feat?

Harald

 

[Tracer]

As you indicate, here the rain drop velocity is around 30 m/s relative to the guy in the giant wheel (note that this is also called closing speed), but it depends on the ship's velocity and varies over time (sorry: I did not check your calculations but it looks fine).

My point was, and still is: he doesn't know these rain drop velocities relative to him. Apparently you think/thought that it is but one velocity which he should be able to determine from the wheel's speed and the difference of observed angles (theta is in fact the angle between one inclination and the other one). Did you try if he can indeed achieve that feat?

Harald

The rain drop's closing velocity with the observer was calculated to be 39.051248meters/sec when he was at the top of the wheel and 30.413813 meters/sec when he was at the bottom of the wheel.

Since the closing velocity of the rain drops with the observer is different for measurements from the top and bottom of the wheel, what information would using the differences between velocities and angles between top and bottom of the wheel provide?

 

[harrylin]

The rain drop's closing velocity with the observer was calculated to be 39.051248meters/sec when he was at the top of the wheel and 30.413813 meters/sec when he was at the bottom of the wheel.

Since the closing velocity of the rain drops with the observer is different for measurements from the top and bottom of the wheel, what information would using the differences between velocities and angles between top and bottom of the wheel provide?

Again, that's the equivalence with your first post! Stellar aberration is the observation of the difference of two angles due to the different velocities of the Earth at two points of its orbit.

Do you think that based on the provided information, the guy in the giant wheel can determine the speed of the rain drops relative to himself, or relative to the wheel? I don't think so.

 

[Dale]

There is simply no way to measure the one way velocity of light using stellar aberration or any other means. In order to measure a one-way velocity of light you need to use a theory which allows it to vary (i.e. you cannot use special relativity), such as Edward's theory or the Mansouri-Sexl test theory. However, in both of those theories, the one way speed of light depends on the synchronization convention. So, any experimental result, including stellar abberation, is consistent with a range of one-way speed of light.

 

[PhilDSP]

While many of these differences are too small to detect, source motion dependence has been rigorously ruled out by measurement of aberration from rapidly orbiting binary stars.

Thanks for the nicely organized presentation of the different theoretically possible observed results. I'd like to see a good analysis of DeSitter's (very old) binary star observations and interpretations on their meaning. I spent a little time looking at his material and the impression I got was that his thinking was extremely crude - very, very, very far from either rigor or precision. I haven't seen any kind of real analytical reference to it - only the vague hand waving kind.