# Synchronization of clocks and observations of time in special relativity

I have thought up a situation that I cannot understand with my understanding of special relativity. I don't know general relativity, but as the situation doesn't involve gravity or acceleration, I'm not sure if it is needed.

Imagine there are 2 digital count up timers, A and B, separated by 100 light days in the same frame of reference. There is a person at each timer. Both timers are turned on at the same time (this could be done via a signal at a point C which is the same distance away from both A and B).

At this point, a person at A will see the timer at B being 100 days less than the timer at A (due to the time light takes to travel from A to B). Vice versa for a person at B.

Now imagine a ship is traveling at 0.99c through A to B. The ship also has a timer on it, which is initially off.

When it passes past A the pilot notes timer A is at 1000 days. The pilot then starts the ship's timer.

Eventually, the ship will go past B. My questions are:

1) What time will people A and B see on the their timer and the ship's timer when the ship passes past B?

2) What time will the ship see on A and B when it passes past B?

Here are the issues I have with this:

From B's perspective, the ship passes past A when B's timer is 1100 days. If you ignore relativity, you would expect a ship traveling at 0.99c to take 100 / 0.99 = 101.01 days to get to B. Wouldn't that mean from B's perspective, it only takes 1.01 days for the ship to get from A to B? I thought length contraction would explain that, but A and B aren't moving relative to each other. S would appear to be moving faster than the speed of light, which doesn't make sense.

From A's perspective, S will experience time 7.088 slower than A. So when it passes B, the ships timer will only be 101.01 / 7.088 = 14.25 days. Taking 101.01 days to get to B, it will see B's timer being 1101.01 days. It will see this time when it's timer is 1201.01 days. That means it will take 201.01 days for the ship's timer to advance by 14.24 days, a time dilation of 201.01 / 14.24 = 14.12, which is different to 7.088? I must be double counting somewhere, I don't know.

-
I like your question and it's obvious you put a lot of effort into it. Unfortunately I think you're asking way too may questions about your scenario. I think you should cut it down to just one core question that if you get the answer to, would answer others. If it still doesn't answer your question you should ask follow-up questions on the site. – Brandon Enright May 29 '13 at 6:10
I agree with what Brandon said. I'm closing this for now, but once you condense it down to ask about the one main conceptual issue you're having, flag it for moderator attention and I'll be happy to reopen it. – David Z May 29 '13 at 6:16
@BrandonEnright the problem is I'm not sure where my understanding of the situation is breaking down. Is it due to the events not being simultaneous at point A? Or is the difference between times at A and B not what I think they mean? I am finding the answer's I get appear correct by themselves, but I get weird conflicts when consider the answers together. I suppose thought the answers regarding point C are the least important. – David Miani May 29 '13 at 6:17
I'd suggest dropping point C from the setup, then. Only ask about A and B for now. You could bring up your last issue, about point C, in a followup question. – David Z May 29 '13 at 6:18
@DavidMiani I think you should cut down your scenario a bit, skip the first questions and get straight to your 7.088 discrepancy. You may want to ask it as a homework-like question too (essentially saying please help me understand the steps). – Brandon Enright May 29 '13 at 6:20

Yep, from point B it'll look like the ship travels from A in 1.01 days, if you don't account for the travel time of light. It is possible for something to appear to move faster than light this way - in fact we've actually seen gas jets from distant galaxies doing it. But when you correct for the travel time of light, you always find a speed less than $c$. – David Z May 29 '13 at 7:56