Here’s what I don’t understand about time dilation. Alice is at rest and Bob is moving with velocity V with respect to Alice. Let’s say Alice measures two events separated in time by $\Delta t_A$ according to her own watch, and she measures those events in the same place, so she has measured her own proper time.
By applying LTs, we get that $\Delta t_B’ = \gamma \Delta t_A$. What does this time mean? Is this the proper time elapsed between the two events as measured by another observer Bob who is measuring time with his own clock at the same place? I mean, what we compute by using LTs, is Bob’s proper time or is it something different? And if it IS something different, how can we compute Bob’s proper time?
Also, since $\Delta t_B’ > \Delta t_A$, I don’t understand why people usually state that “moving clocks tick slower’. Alice being at rest thinks her proper time $\Delta t_A$ is less than $\Delta t_B’$. For examples, Alice thinks she has measured one hour and Bob measured 2 hours, right? If I were Alice, I would conclude that Bob’s moving clock is running faster, rather than slower.
Best Answer
Technically, the proper time is only defined along the worldline of an observer, so for the time between two events to be Alice's proper time it is necessary not only that they be at the same place, but also that place must be Alice's position. (This is a nit-picking requirement that is not important in SR but becomes important in GR). So I will assume that the two events are indeed on Alice's worldline so that it is indeed her proper time.
It cannot be Bob's proper time because Bob is moving relative to Alice, so at most one of those events could be on both Bob and Alice's worldline. So it is instead Bob's coordinate time, meaning the time as shown on a lattice of clocks at rest relative to Bob and synchronized using the Einstein synchronization convention.
It is true that this phrasing is a bit confusing. Since whether a clock is moving or not depends on the reference frame and since no frame is specified, it is ambiguous.
What is unambiguous is the following: $\Delta t_A$ is measured by a single clock which was present at both events, but $\Delta t'_B$ is measured by a pair of synchronized clocks each of which was present at only one event. The time difference for the single clock is always less than the time difference for the pair of synchronized clocks. "Moving clocks tick slower" means that whenever you are comparing a single clock to a lattice of synchronized clocks (in the lattice clock's frame) then the single clock will be slower whenever it is moving (in the lattice frame).
But in that comparison Alice is not looking at a single Bob clock. Any one Bob clock will run slow compared to Alice's lattice of synchronized clocks. But Bob's clocks are not synchronized in Alice's frame due to the relativity of simultaneity. So subtractions of times on different Bob clocks is essentially meaningless for Alice.