How could they arrange to do that, since the doors are far apart? Remember, though, that the train is itself going at a speed close to that of light, so they have to be quite precise about this timing! Assuming the light was positioned correctly in the middle of the tunnel, that should ensure that the two doors close simultaneously. Now consider this door-closing operation from the point of view of someone on the train. Of course , the train is a perfectly good inertial frame , so he sees these two flashes to be traveling in opposite directions, but both at c, relative to the train.
Meanwhile, he sees the tunnel itself to be moving rapidly relative to the train. So the two flashes of light going down the tunnel in opposite directions do not reach the two doors simultaneously as seen from the train. The concept of simultaneity, events happening at the same time, is not invariant as we move from one inertial frame to another. The man on the train sees the back door close first, and, if it is not quickly reopened, the front of the train will pile into it before the front door is closed behind the train.
This can be seen not to be the case through a symmetry argument, also due to Einstein. Suppose two trains traveling at equal and opposite relativistic speeds, one north, one south, pass on parallel tracks. Suppose two passengers of equal height, one on each train, are standing leaning slightly out of open windows so that their noses should very lightly touch as they pass each other.
Afterwards, then, N will have a bruised chin plus nose , S a bruised forehead plus nose. But this is a perfectly symmetric problem, so S would say N had the bruised forehead, etc. They can both get off their trains at the next stations and get together to check out bruises.
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They must certainly be symmetrical! The only consistent symmetrical solution is given by asserting that neither sees the other to shrink in height i. Therefore, the Lorentz contraction only operates in the direction of motion, objects get squashed but not shrunken.
Perhaps the most famous of the paradoxes of special relativity, which was still being hotly debated in national journals in the fifties, is the twin paradox. The scenario is as follows. One of two twins—the sister—is an astronaut. She sets off in a relativistic spaceship to alpha-centauri, four light-years away, at a speed of, say, 0. When she gets there, she immediately turns around and comes back. So as she steps down out of the spaceship, she is 32 months younger than her twin brother.
But wait a minute—how does this look from her point of view? She sees the earth to be moving at 0. The key to this paradox is that this situation is not as symmetrical as it looks.
The two twins have quite different experiences. The one on the spaceship is not in an inertial frame during the initial acceleration and the turnaround and braking periods. To get an idea of the speeds involved, to get to 0.
To try to see just how the difference in ages might develop, let us imagine that the twins stay in touch with each other throughout the trip. Each twin flashes a powerful light once a month, according to their calendars and clocks, so that by counting the flashes, each one can monitor how fast the other one is aging. If the brother, on earth, flashes a light once a month, how frequently, as measured by her clock, does the sister see his light to be flashing as she moves away from earth at speed 0. Once we have answered these questions, it will be a matter of simple bookkeeping to find how much each twin has aged.
In some ways, that was a very small scale version of the present problem. As the astronaut, conveniently moving at 0. As she passed the second ground clock, her own clock read 8 seconds and the first ground clock, which she photographed at that instant, she observed to read 4 seconds. That is to say, after 8 seconds had elapsed on her own clock, constant observation of the first ground clock would have revealed it to have registered only 4 seconds.
This effect is compounded of time dilation and the fact that as she moves away, the light from the clock is taking longer and longer to reach her. We know that as she passes that clock, it reads 10 seconds and her own clock reads 8 seconds. We must figure out what she would have seen that second ground clock to read had she glanced at it through a telescope as she passed the first ground clock, at which point both her own clock and the first ground clock read zero. But at that instant, the reading she would see on the second ground clock must be the same as would be seen by an observer on the ground, standing by the first ground clock and observing the second ground clock through a telescope.
Since the ground observer knows both ground clocks are synchronized, and the first ground clock reads zero, and the second is 6 light seconds distant, it must read -6 seconds if observed at that instant. In other words, she sees the clock she is approaching at 0. Finally, back to the twins. Evidently, the time dilation effect does not fully compensate for the fact that each succeeding flash has less far to go to reach her.see
How Einstein's general theory of relativity killed off common-sense physics
During the outward trip, then, she will see 32 flashes from home, she will see her brother to age by 32 months. As he watches for flashes through his telescope, the stay-at-home brother will see his sister to be aging at half his own rate of aging as long as he sees her to be moving away from him, then aging at twice his rate as he sees her coming back.
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