New Reply

Stellar aberation, a One way measurement of c?

 
Share Thread Thread Tools
Feb10-12, 02:00 AM   #69
 

Stellar aberation, a One way measurement of c?


Quote by harrylin View Post
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).
Quote by PAllen
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).
Feb10-12, 02:13 PM   #70
 
Quote by harrylin View Post
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?
Feb10-12, 03:06 PM   #71
 
Quote by Tracer View Post
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
Feb10-12, 03:43 PM   #72
 
Quote by harrylin View Post
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?
Feb11-12, 01:48 AM   #73
 
Quote by Tracer View Post
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.
Feb11-12, 07:42 AM   #74
 
Mentor
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.
Feb13-12, 09:33 AM   #75
 
Quote by PAllen View Post
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.
New Reply
Thread Tools


Similar Threads for: Stellar aberation, a One way measurement of c?
Thread Forum Replies
Actual earth measurement contradicts measurement predicted by special relativity Special & General Relativity 93
Stellar Remnants Astrophysics 9
about time measurement vs measurement of movement Special & General Relativity 21
How to get a stellar GPA General Discussion 2
Stellar Evolution Advanced Physics Learning Materials 0