Stellar aberation, a One way measurement of c?

In summary: Earth around the sun you need to have clocks at both the start and finish!"harrylin, I see there's another long thread elsewhere on this same topic, so further discussion should probably be done there." -I agree. There's a lot of discussion about this topic on other threads, and I don't think it's worth it to continue discussing it here.
  • #71
Tracer said:
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
 
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  • #72
harrylin said:
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?
 
  • #73
Tracer said:
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.
 
  • #74
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.
 
  • #75
PAllen said:
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.
 

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