Twin Paradox Can anyone clarify and or correct the following for me? A space ship is flying at speed v equal to 0.8 times the speed of light. Within the ship are three stations, a transmitter at station A, station B directly across the ship from station A and station C directly forward of station A. The distance AB is noted as d and the distance AC equals AB. A photon is transmitted from A towards B at the speed of light c. At the same instant another photon is transmitted from A towards C. The transit time t for the photons is measured inside the space craft as d divided by c in both cases. A stationary observer notes the transit of the photon from A to B and measuring the transit calculates the distance to be Tab multiplied by the speed of light c and also calculates the distance travelled by the space craft as v * Tab and concludes that : d = Tab * (c^2-v^2)^0.5 And as d = t*c t = Tab * (1- (v/c)^2)^0.5 For v/c = 0.8, (1- (v/c)^2)^0.5 = 0.6 t = 0.6 * Tab
This is the Lorentz equation for time dilation on which the twin paradox is based. Now consider what the observer sees of the photon travelling from A to C. Firstly Lorentz would contend that there is a shortening of lengths in the direction of travel and as a consequence the observer sees the distance A to C as d multiplied by (1- (v/c)^2)^0.5 hence the observed distance of travel is v*Tab + d*(1- (v/c)^2)^0.5 and: Tac = (v*Tab + d*(1- (v/c)^2)^0.5)/c Tac = 1.93333*t If this sum is done for a photon travelling from C to A the numbers are even harder to understand Tca = (-v*Tab + d*(1- (v/c)^2)^0.5)/c Tac = -0.73333*t As the time of arrival within the space ship is the same in all three cases then should not the observed time of arrival also be the same for all three cases not three different times and particularly the later one cannot be negative. Please will someone resolve my paradox.