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Stellar aberation, a One way measurement of c? |
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| Feb10-12, 02:00 AM | #69 |
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Stellar aberation, a One way measurement of c?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 |
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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 |
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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 |
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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 |
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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 |
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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.
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| Feb13-12, 09:33 AM | #75 |
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