Thread: Twin Paradox again View Single Post
 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.