# Special Relativity: Four possible time spans between events?

I've got another question regarding time in special relativity. I'll start with the standard configuration: there are two observers O and O', their x-axis are aligned, in the rest frame of O observer O' moves in direction of the x-axis with speed v. Event 1 shall be the observers meeting when their origin clocks (the ones they carry in their hands) read t = t' = 0. Event 2 a firecracker that goes off at some x0 and t0 in the rest frame of O. According to the Lorentz transformation, the event has the coordinates x0' and t0' in the rest frame of O'.

Now I've been told to think of the observers carrying a network of clocks with them. The clock network of O is synchronized in the rest frame of O and the clock network of O' is synchronized in the rest frame of O'. Due to whatnot, the clocks of O' are not in sync in the rest frame of O and the other way around. I don't have a problem with that. But does that imply that there are four time spans we can find between the occurrence of two events?

1) The time span t0 which is the time between the events as measured by the clocks of O in the rest frame of O.

2) The time span t0' which is the time between the events as measured by the clocks of O' in the rest frame of O'.

3) Another time span t0'' which tells us by how much the origin clock of O' ticks forward between the events when viewed from the rest frame of O. If I am not mistaken, that's the one I need when considering time dilation.

4) Another time span t0''' which tells us by how much the origin clock of O ticks forward between the events when viewed from the rest frame of O'.

Is that correct or is this a misconception? Are there even more time spans one could find between events 1 and 2?

Also: I'm not sure if I properly understood how to calculate case 3). The way I do it: since event 2 happens at x0', observer O' will use the clock located at x0' to determine the time of the event and this clock shows t0'. When it does so, the origin clock of O' should show t0' + x0' * v/c². So the origin clock of O', when seen from the rest frame of O, should go from t' = 0 to t' = t0' + x0' * v/c² between the events, hence t0'' = t0' + x0' * v/c².

This is not for homework or anything like that, I'm just trying to understand special relativity properly and I'm not a student (what I know is mostly self-taught). Hopefully you guys can help me out.

## 1 Answer

3) and 4) don't represent "the time between the events" 1 and 2 in either frame, but you can calculate them if you wish, and as you said these are times that you could apply the time dilation equation to. Basically, to apply the time dilation equation, you must pick the time between two events on the worldline of a single inertial clock, and then you compare the proper time $\tau$ between the two events according to the clock itself (or equivalently, the time in the frame where the clock is at rest) with the time t between these same two events in a frame where the clock is moving at velocity v. In terms of this notation, the times will then be related by $t = \tau / \sqrt{1 - v^2/c^2}$.

So in your 3), you're picking as your first event the original Event 1 of the two origin clocks passing one another, and you're picking as your second event the event on the worldline of the origin clock of O' that is simultaneous in the O frame with the firecracker exploding, not the event of the firecracker explosion itself. Here the time between these events according to the origin clock of O' is your t0'' (this fills the role of $\tau$ in the time dilation equation, since it's the time between two events on the worldline of a single clock according to the clock itself), and the time between these events in the O frame is t0 (which fills the role of t in the time dilation equation), so the time dilation equation says they should be related by $t_0 = t_0'' / \sqrt{1 - v^2/c^2}$. Likewise, in your 4), you're picking as your first event the Event 1 of the two origin clocks passing one another, and you're picking as your second event the event on the worldline of the origin clock of O that is simultaneous in the O' frame with the firecracker exploding. The time between these events according to the origin clock of O is your t0''' (which fills the role of $\tau$ since both events are on the worldline of the origin clock of O), and the time between these events in the O' frame is t0' (which fills the role of t), and the two times are related by $t_0' = t_0''' / \sqrt{1 - v^2/c^2}$.

Long story short, for any specific pair of events there are only two different times between them in two inertial frames, but in your 3) and 4) you're implicitly picking different pairs of events, the first being the original Event 1 of the two origin clocks passing, but with the second event in each case being an event that is simultaneous with the firecracker explosion in one of the two frames, and lies on the worldline of the origin clock belonging to the other frame. And since you've picked a pair of events that both lie on the worldline of a single clock, the time dilation equation can be applied.

• Thank you for making this so clear. The only thing I have to wrap my head around is that "the event on the worldline of the origin clock of O that is simultaneous in the O' frame with the firecracker exploding" is not the same as the firecracker event. I guess the word "worldline" throws me of as I have taken a purely algebraic approach to SR and consequently ignored all graphs. But with your info, I did manage to show that the formula for t0'' in my question is the same as the one you gave in your answer once you apply the Lorentz transformation. Thanks! – MeBe Nov 9 '14 at 11:43
• If it helps, in relativity an "event" is understood to be something that can be localized to a single point in spacetime--if two things happen at the same time-coordinate but different space-coordinates, than they are separate "events". And "worldline" is just the series of points/events in spacetime that an object passes through, for example my worldline in some frame might be defined by x(t)=t*(0.1c) + 35, y(t)=t*(0.05c), z(t)=2, which would give me a unique trio of x,y,z coordinates at each t-coordinate. – Hypnosifl Nov 9 '14 at 17:00