1. The problem statement, all variables and given/known data A high-speed train is traveling from Capital City to Shelbyville. According to an observer at rest on the ground, the clocks at the railroad stations in Capital City and Shelbyville both strike noon at the same time. According to a passenger on the train, when the Capital City clock strikes noon, what time is it in Shelbyville? a. noon b. before noon c. after noon d. the answer depends on the speed of the train 2. Relevant equations This is pre-Lorentz transformations.. so just basic stuff. 3. The attempt at a solution I interpreted this problem in two ways. One is simply the answer is C (after noon) because the train is moving towards Shelbyville, i.e. it reaches the light pulse from Shelbyville first. So when the noon pulse from CC arrives, it's past noon for Shelbyville. However, I was told that even if you correct for those differences it should be "after noon." My prof said this: "Important to think of the train as stationary, and the 2 cities as moving - time lead increases in the direction opposite to the direction of travel of the cities." Does that mean time dilation is direction dependent (i.e. if it's approaching or receding)? Doing this, take the frame of the train. Then Shelbyville (S) is approaching at velocity v, and Capital City (CC) is retreating at velocity v. If it takes time "t1" for the noon light from S to reach the train, when the train is a distance d from the city, then: t1 * (c - v) = d1. Similarly, t2 * (c - v) = d2 describes how long it takes for the light from Capital City to arrive, since the light pulse and velocity are in opposite directions. So the difference in clock synchronization is t2 - t1 = d1/(c+v) - d2/(c-v). Not really sure if that shows anything though.... If you switch to the frame of the cities and account for time dilation and length contraction, then I think you get: t2' - t1' = (d1/(c+v) - d2/(c-v)) * (1-v^2/c^2), which I think is the clock asynchronization in the frame of the cities. Not really sure if that is what the problem means or not though. Thanks!