- #1
Alfred Cann
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This analysis uses an AND gate, the very embodiment of simultaneity, for clarity.
A 1km long train travels through a 1km tunnel at a speed of 0.6c=180m/microsecond. At each end of the tunnel is a photo-electric beam sensor, which can be blocked by the train. They are connected to an AND gate at the center of the tunnel, which drives a light. The light will be on whenever both beams are unblocked. (We can even dispense with electrical transmission by using 45° mirrors and placing the photocells at the AND gate.)
In the tunnel reference frame, the train is shrunk to a length of 800mhttps://www.physicsforums.com/file:///C:/Users/fred/Desktop/Fredmisc/Train%20in%20tunnel.docx#_ftn1 . Let us say the rear beam is unblocked at t=0. Then the front beam is blocked at t=200m/180m/us=1.11..us. Since the transmission times from the two beam sensors are equal, the light is turned on for a duration of 1.11..us.
Assume the train has an observation car so that an observer can see the output light. In the train reference frame, the tunnel is shrunk to a length of 800m. Say the front beam is blocked at t=0. As seen from the train, that signal travels toward the AND gate at a speed of 300m/us. But the gate is traveling away from it at a speed of 180m/us, so the net speed is 300-180=120m/us (even though an observer at the gate would measure a speed of 300m/us). The distance is half the shrunken tunnel length, 400m. Thus, the travel time is 400/120=3.33..us.
The rear beam is unblocked at t=200m/180m/us=1.11..us. That signal travels toward the gate at a speed of 300m/us. But the gate is traveling toward it at a speed of 180m/us, yielding a total speed of 480m/us (even though an observer at the gate would measure a speed of 300m/us). The distance is400m; thus, the travel time is 400m/480m/us=0.833..us, and the arrival time is 1.11..+0.833..=1.944..us. The light is on (both beams unblocked) for a duration of 3.33..-1.944..=1.388..us.
As viewed from the train, all tunnel clocks run slow by a factor of 0.8. The light is such a ‘clock’, and its ‘on’ time is stretched to 1.11../0.8=1.388..us, in agreement with the previous paragraph.https://www.physicsforums.com/file:///C:/Users/fred/Desktop/Fredmisc/Train%20in%20tunnel.docx#_ftnref1 Using L'=L√(1-(v/c)Λ2)
A 1km long train travels through a 1km tunnel at a speed of 0.6c=180m/microsecond. At each end of the tunnel is a photo-electric beam sensor, which can be blocked by the train. They are connected to an AND gate at the center of the tunnel, which drives a light. The light will be on whenever both beams are unblocked. (We can even dispense with electrical transmission by using 45° mirrors and placing the photocells at the AND gate.)
In the tunnel reference frame, the train is shrunk to a length of 800mhttps://www.physicsforums.com/file:///C:/Users/fred/Desktop/Fredmisc/Train%20in%20tunnel.docx#_ftn1 . Let us say the rear beam is unblocked at t=0. Then the front beam is blocked at t=200m/180m/us=1.11..us. Since the transmission times from the two beam sensors are equal, the light is turned on for a duration of 1.11..us.
Assume the train has an observation car so that an observer can see the output light. In the train reference frame, the tunnel is shrunk to a length of 800m. Say the front beam is blocked at t=0. As seen from the train, that signal travels toward the AND gate at a speed of 300m/us. But the gate is traveling away from it at a speed of 180m/us, so the net speed is 300-180=120m/us (even though an observer at the gate would measure a speed of 300m/us). The distance is half the shrunken tunnel length, 400m. Thus, the travel time is 400/120=3.33..us.
The rear beam is unblocked at t=200m/180m/us=1.11..us. That signal travels toward the gate at a speed of 300m/us. But the gate is traveling toward it at a speed of 180m/us, yielding a total speed of 480m/us (even though an observer at the gate would measure a speed of 300m/us). The distance is400m; thus, the travel time is 400m/480m/us=0.833..us, and the arrival time is 1.11..+0.833..=1.944..us. The light is on (both beams unblocked) for a duration of 3.33..-1.944..=1.388..us.
As viewed from the train, all tunnel clocks run slow by a factor of 0.8. The light is such a ‘clock’, and its ‘on’ time is stretched to 1.11../0.8=1.388..us, in agreement with the previous paragraph.https://www.physicsforums.com/file:///C:/Users/fred/Desktop/Fredmisc/Train%20in%20tunnel.docx#_ftnref1 Using L'=L√(1-(v/c)Λ2)
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