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yogi
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According to the dictates of Special Relativity, the rate of a clock is unaffected by motion of the frame at which it is at rest. But a puzzle arises when one considers the reciprocity of the frames in a case where a precedent determination of the temporal interval is given for each frame.
Here there are two trains, each 5 light years in length from beginning to end
- they are approaching each other on parallel tracks at velocity approximately c.
E* ---------------train 2---->B* ........ S* frame
S frame......B <--------train 1---------E
B is the beginning of the train at rest in the S frame and E is its end,
B* is the beginning of the train at rest in the S* frame and E* is its end.
E and B represent clocks that are synchronized in S,
E* and B* represent clocks that are synchronized in S*
As beginning points BB* pass each other, each takes note of the others time -
we will assume this to be “0" for convenience.
This is the position shown in the Fig 1. At this time, the operator of train 1 conducts an experiment that produces a muon M, and the operator of train 2 does a similar experiment and produces a muon M* Both muons have a lifetime of 2 μsec in
their own frame and M remains at rest with respect to clock B and M* remains at rest with respect to B*
If frame S is considered at rest, the time required for B* to reach E will be 5 years as measured by clocks B and E.
In the S* frame, the muon M* will be at rest relative to B* and if the relative
velocity is sufficiently close to c, will not yet have decayed when B* reaches E.
So, B* clock will read only 2 μsec when the S clock has logged 5 years.
Now consider the S* frame at rest. The time that has passed in S* as measured by the synchronized clocks B* and E* when B reaches E* should be 5 years (not 2 usec) since the proper distance in S* is 5 light years. But muon M will
only have aged 2 usec in its rest frame S - so clock B should read 2 usec, not 5 years.
Question: When B* is opposite E and B is opposite E*
Does clock B read 2 μsec or 5 years? Does clock B* read 2 μsec or 5 years?
Here there are two trains, each 5 light years in length from beginning to end
- they are approaching each other on parallel tracks at velocity approximately c.
E* ---------------train 2---->B* ........ S* frame
S frame......B <--------train 1---------E
B is the beginning of the train at rest in the S frame and E is its end,
B* is the beginning of the train at rest in the S* frame and E* is its end.
E and B represent clocks that are synchronized in S,
E* and B* represent clocks that are synchronized in S*
As beginning points BB* pass each other, each takes note of the others time -
we will assume this to be “0" for convenience.
This is the position shown in the Fig 1. At this time, the operator of train 1 conducts an experiment that produces a muon M, and the operator of train 2 does a similar experiment and produces a muon M* Both muons have a lifetime of 2 μsec in
their own frame and M remains at rest with respect to clock B and M* remains at rest with respect to B*
If frame S is considered at rest, the time required for B* to reach E will be 5 years as measured by clocks B and E.
In the S* frame, the muon M* will be at rest relative to B* and if the relative
velocity is sufficiently close to c, will not yet have decayed when B* reaches E.
So, B* clock will read only 2 μsec when the S clock has logged 5 years.
Now consider the S* frame at rest. The time that has passed in S* as measured by the synchronized clocks B* and E* when B reaches E* should be 5 years (not 2 usec) since the proper distance in S* is 5 light years. But muon M will
only have aged 2 usec in its rest frame S - so clock B should read 2 usec, not 5 years.
Question: When B* is opposite E and B is opposite E*
Does clock B read 2 μsec or 5 years? Does clock B* read 2 μsec or 5 years?
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