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Main Question or Discussion Point
see the attchments
Referring to fig 1
We have a clock that is used to set the time on a second clock. At point A is a laser. At B there is a photon detector. The laser at A fires a photon, shown as the red broken line. The photon makes its way from point A to point B.
At A there is also an LED (the green circle). When the laser fires the LED emits a pulse of light. At B there is a second LED (the green circle). When the detector at B receives a photon from the laser at A the LED at B emits a pulse of light.
There is an observer at the point Z and he has a stop watch. When he sees the first pulse of light he starts the stop watch when he sees a second pulse of light he stops the stop watch. The time difference on the stop watch is used to set another clock.
The entire apparatus and the observer are in a space ship that can be stationary ie V = 0 or moving V > 0
The length from A to B is L
While the space ship is stationary a test is done. The laser at A is fired and the observer records the time for the first and second pulse of lights from the LEDs.
The time would be t = L/c
If the clock was rotated thru 180 degrees the same result would be obtained.
Fig 2
The space ship is now moving at a constant velocity V in the direction of the arrow shown. I note before you all go ballistic I have not shown the space ship at time zero and at some time in the future as it makes the picture to messy. This is the same for fig 3.
At time zero the clock will be in position A and B. At some time in the future the clock will have moved with the space ship and be in the new position A hat and B hat.
At time zero a pulse of light is emitted from the laser at point A. The LED at A also emits a pulse of light which the observer observes and starts his stop watch.
At some time in the future the pulse of light emitted from the laser will strike the detector at B hat. And a pulse of light will be emitted from the LED at point B hat. The observer sees this pulse and stops his stop watch.
Analysis
The length from A to B is L
Vt is the distance the clock travelled between the pulse leaving the laser at A and the pulse from the laser striking the detector at point B hat.
The length from A to B hat is L + vt
Time difference = (L + Vt)/ c c is the speed of light.
I note two things
1. As expected as the clock is moving the time difference is greater than the stationary clock.
2. If the observer had been standing at position B hat he would have observed something very different to this result.
Fig 3
As in Fig 2 the space ship is moving at a constant velocity V in the direction of the arrow shown.
The clock is now rotated thru 180 degrees compared with fig 2.
At time zero the clock will be in position A and B. At some time in the future the clock will have moved with the space ship and be in the new position A hat and B hat.
At time zero a pulse of light is emitted from the laser at point A. The LED at A also emits a pulse of light which the observer observes and starts his stop watch.
At some time in the future the pulse of light emitted from the laser will strike the detector at B hat. And a pulse of light will be emitted from the LED at point B hat. The observer sees this pulse and stops his stop watch.
Analysis
The length from A to B is L
Vt is the distance the clock travelled between the pulse leaving the laser at A and the pulse from the laser striking the detector at point B hat.
The length from A to B hat is L  vt
Time difference = (L  Vt)/ c c is the speed of light.
I note two things
1. As the clock is moving by theory the time difference must be greater than the stationary clock. The stationary clocks time difference is L/c (fig 1). In fig 3 the time difference is (L – Vt)/c which will always be less than L/c. As the observer uses this difference to set another clock the set clock will run faster than the stationary clock so a moving clock is running faster than a stationary clock.!
2. If the observer had been standing at position B he would have obtained a different result.
Is this analysis right?
How is it that an observer at rest with a clock can get a different result from the clock simply by changing his position of observation? Time is a vector
Surely this apparatus can be used as a test to identify if you are in a moving frame of reference? If so you can demonstrate you are in a moving frame of ref by an experiment internal to the frame of ref!
Referring to fig 1
We have a clock that is used to set the time on a second clock. At point A is a laser. At B there is a photon detector. The laser at A fires a photon, shown as the red broken line. The photon makes its way from point A to point B.
At A there is also an LED (the green circle). When the laser fires the LED emits a pulse of light. At B there is a second LED (the green circle). When the detector at B receives a photon from the laser at A the LED at B emits a pulse of light.
There is an observer at the point Z and he has a stop watch. When he sees the first pulse of light he starts the stop watch when he sees a second pulse of light he stops the stop watch. The time difference on the stop watch is used to set another clock.
The entire apparatus and the observer are in a space ship that can be stationary ie V = 0 or moving V > 0
The length from A to B is L
While the space ship is stationary a test is done. The laser at A is fired and the observer records the time for the first and second pulse of lights from the LEDs.
The time would be t = L/c
If the clock was rotated thru 180 degrees the same result would be obtained.
Fig 2
The space ship is now moving at a constant velocity V in the direction of the arrow shown. I note before you all go ballistic I have not shown the space ship at time zero and at some time in the future as it makes the picture to messy. This is the same for fig 3.
At time zero the clock will be in position A and B. At some time in the future the clock will have moved with the space ship and be in the new position A hat and B hat.
At time zero a pulse of light is emitted from the laser at point A. The LED at A also emits a pulse of light which the observer observes and starts his stop watch.
At some time in the future the pulse of light emitted from the laser will strike the detector at B hat. And a pulse of light will be emitted from the LED at point B hat. The observer sees this pulse and stops his stop watch.
Analysis
The length from A to B is L
Vt is the distance the clock travelled between the pulse leaving the laser at A and the pulse from the laser striking the detector at point B hat.
The length from A to B hat is L + vt
Time difference = (L + Vt)/ c c is the speed of light.
I note two things
1. As expected as the clock is moving the time difference is greater than the stationary clock.
2. If the observer had been standing at position B hat he would have observed something very different to this result.
Fig 3
As in Fig 2 the space ship is moving at a constant velocity V in the direction of the arrow shown.
The clock is now rotated thru 180 degrees compared with fig 2.
At time zero the clock will be in position A and B. At some time in the future the clock will have moved with the space ship and be in the new position A hat and B hat.
At time zero a pulse of light is emitted from the laser at point A. The LED at A also emits a pulse of light which the observer observes and starts his stop watch.
At some time in the future the pulse of light emitted from the laser will strike the detector at B hat. And a pulse of light will be emitted from the LED at point B hat. The observer sees this pulse and stops his stop watch.
Analysis
The length from A to B is L
Vt is the distance the clock travelled between the pulse leaving the laser at A and the pulse from the laser striking the detector at point B hat.
The length from A to B hat is L  vt
Time difference = (L  Vt)/ c c is the speed of light.
I note two things
1. As the clock is moving by theory the time difference must be greater than the stationary clock. The stationary clocks time difference is L/c (fig 1). In fig 3 the time difference is (L – Vt)/c which will always be less than L/c. As the observer uses this difference to set another clock the set clock will run faster than the stationary clock so a moving clock is running faster than a stationary clock.!
2. If the observer had been standing at position B he would have obtained a different result.
Is this analysis right?
How is it that an observer at rest with a clock can get a different result from the clock simply by changing his position of observation? Time is a vector
Surely this apparatus can be used as a test to identify if you are in a moving frame of reference? If so you can demonstrate you are in a moving frame of ref by an experiment internal to the frame of ref!
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