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mananvpanchal
mananvpanchal is offline
#3
Feb15-12, 03:14 AM
P: 215
Thanks PAllen.

But, I need some detail to understand it. So I am going to describe what is my understanding about the paradox.

A is stationary and B is moving with constant speed.
There is no acceleration involved (Just assume because we trying to solve it only with asymmetry).
(The x and t is in A's reference frame)
Suppose A at (x=0, t=0) and B (x=0, t=0).
A flashes a beam each per second. A knows the speed of B. So A can calculate at which (x=?, t=?) beam will meet to B.
At t=1 A at (x=0, t=1) and B at (x=1, t=1).
At t=1 A calculate that beam and B will meet at (x=a, t=b). where a > 1, b > 1.
At t=2 A at (x=0, t=2) and B at (x=2, t=2).
At t=2 A calculate that beam and B will meet at (x=c, t=d). where c > 2, d > 2 ,(c-2) > (a-1), (d-2) > (b-1).
As time elapse, B sees A's clock slowing down more. Because B gets beam more late.
(The frequency of receiving beam increases until B is in motion relative to A. Whenever B stops it receives quick signals for some time because of signals coming behind it. But after some time B starts receiving signals in time which is flashed after stopping B.)

In inward journey
At t=10 A at (x=0, t=10) and B at (x=10, t=10).
At t=10 A calculate that beam and B will meet at (x=p, t=q). where p < 10, q > 10.
At t=11 A at (x=0, t=11) and B at (x=9, t=11).
At t=11 A calculate that beam and B will meet at (x=r, t=s). where r < 9, s > 11, (r-9) < (p-10), (s-11) < (q-10).
So, as time elapsed, B sees A's clock running more fast. Because B gets beam more quickly.
(If B doesn't stay motionless for some time described in above round braces, then B get much more signals in starting phase of returning journey which is flashed before returning of B. when B starts receiving signals flashed after returning, we can describe above scenario)

Is my understanding of the paradox is right? If no then please give me some detail.