Time dilation: period of a clock

[ephedyn]

Relevant equations

t' = t * sqrt(1 - v^2/c^2),

where t' is the proper time interval, and t is the measured time interval in another reference frame.

Homework Statement /

Q: A spaceship is moving past us at a velocity of v = 3c/5. If, in our frame of reference, we measure a time interval of 1 second, what will be the time measured by clocks within the moving ship?

Actually, I'm questioning the solution that I'm given. I can get the same answer in both ways, but I want to confirm if I am correct in theory. So my question would be: should I assign t = 1 or t' = 1? The solution I was given suggests t = 1 such that t' = 0.8s.

The attempt at a solution

In my opinion, it's t' = 1 because each event (successive positions of the second hand; let's suppose it's analog) of the clock with us is observed to occur at the same position, while that of the clock(s) within the moving ship are observed to occur at different positions. Hence 1 second is the proper time interval. This gives me t = 1.25s, which is the measured period of the second hand's motion on the ship. Then taking 1/1.25 = 0.8, we have 0.8 of a cycle of the second hand's motion on the moving ship - the time showed on the clock on the ship is 0.8s.

 

[CompuChip]

Of course, they are equivalent.
Usually, one takes the unprimed system to be the "rest" system and the primed system the one moving with respect to it.
In this question, we are given that the elapsed time in our rest system is 1 second, and we are asked for the corresponding time in the moving system.

What you are doing is considering an observer on the spaceship and asking the question: if one second passes in the space ship, how much time will pass on ephedyns clock. The answer is indeed 1.25 second, so if this observer would time 1/1.25 seconds instead of 1/1 second, then 1 second would pass on your watch.

 

[ephedyn]

OK, I understand now. Just to make sure, so am I correct to say that the proper time interval is the time on the clock in the ship as measured by the observer in the unprimed system (me)?

Thanks a lot!

 

[ephedyn]

On second thoughts, forget that confirmation. I've got it. The proper time interval is the time on the clock in the ship as measured by the observer in the ship, since it would occur to the observer on the ship that the positions remain unchanged.

But thanks very much nonetheless for sorting out the idea of primed and unprimed frames of reference; I was referring to Halliday and it avoids this phrasing, which I find a little harder to grasp.