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RossBlenkinsop
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- TL;DR Summary
- train carriages in a MFR
In the attached fig 1 you have a moving frame of reference (MFR) moving in the direction V at velocity V
Looking from above in the MFR there are 2 identical train carriages (car 1 and car 2), Car 1 is on a long set of tracks. Car 2 cannot move but is on train tracks.
There are also 2 Identical clocks, clock 1 and clock 2 which are attached to the train tracks. Clock 2 can receive electrical signals from car 2 thru the tracks . Clock 1 can receive electrical signals from car 1 thru the tracks
Car 1 can move along the train tracks in the direction v1 at a constant velocity v1
Fig 2 shows inside one of the train carriages, in it is a light tube of length L. The light tube has a light and detector at one and and a mirror at the other. The light tube has a wire that is connected to the train tracks . When photons are emitted by the light source in the light tube, an electric pulse is sent down the wire to the train tracks. The electric pulse flows along the train track back to the clock which starts the clock running .
When an electric pulse is sent it makes a mark on the train track, so the position of a train at the time the electric pulse was sent is known..
Back at the light tube, the photons are reflected by the mirror, in the light tube, back to the detector, in the light tube. When the photons arrive at the detector, in the light tube, a second electric pulse is sent down the wire, along the train tracks and back to the clock, which stops the clock, thus measuring the time it takes for light to traverse the length L. the length of the light tube
Both train carriages are identical in every way and both train carriages are set up in an identical fashion. They are identical in every way except one can move along the tracks, the other cant.
if the MFR is traveling in direction V at velocity V. When the velocity of Car 1 ( velocity V1) is equal to the velocity of the MFR (V) and when the direction of car 1 (direction of V1) is exactly opposite direction of the MFR (direction of V), the time to traverse the length of the light tube, in Car 1, will be the shortest.
Is that correct ?
From the above the velocity and direction of the MFR can be calculated
Looking from above in the MFR there are 2 identical train carriages (car 1 and car 2), Car 1 is on a long set of tracks. Car 2 cannot move but is on train tracks.
There are also 2 Identical clocks, clock 1 and clock 2 which are attached to the train tracks. Clock 2 can receive electrical signals from car 2 thru the tracks . Clock 1 can receive electrical signals from car 1 thru the tracks
Car 1 can move along the train tracks in the direction v1 at a constant velocity v1
Fig 2 shows inside one of the train carriages, in it is a light tube of length L. The light tube has a light and detector at one and and a mirror at the other. The light tube has a wire that is connected to the train tracks . When photons are emitted by the light source in the light tube, an electric pulse is sent down the wire to the train tracks. The electric pulse flows along the train track back to the clock which starts the clock running .
When an electric pulse is sent it makes a mark on the train track, so the position of a train at the time the electric pulse was sent is known..
Back at the light tube, the photons are reflected by the mirror, in the light tube, back to the detector, in the light tube. When the photons arrive at the detector, in the light tube, a second electric pulse is sent down the wire, along the train tracks and back to the clock, which stops the clock, thus measuring the time it takes for light to traverse the length L. the length of the light tube
Both train carriages are identical in every way and both train carriages are set up in an identical fashion. They are identical in every way except one can move along the tracks, the other cant.
if the MFR is traveling in direction V at velocity V. When the velocity of Car 1 ( velocity V1) is equal to the velocity of the MFR (V) and when the direction of car 1 (direction of V1) is exactly opposite direction of the MFR (direction of V), the time to traverse the length of the light tube, in Car 1, will be the shortest.
Is that correct ?
From the above the velocity and direction of the MFR can be calculated
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