# Can observers in different reference frames agree on the time an event took after the fact?

Ok, I don't think I did a good job of communicating this, so here's a better way to phrase the experiment.

There is a spaceship traveling horizontally in the x-direction at near the speed of light (let's say .866c). On its bow is a clock composed of two mirrors, stacked vertically in the y direction, with a light pulse bouncing back and forth between them, each bounce counting as one "tick". On its stern is an identical clock. Engineers on the ship use a laser to calibrate the light clocks so that they are in sync (as described by Einstein).

The ship is observed from the reference frame of a donut shaped space station on which another light clock exists. The ship passes through the plane encircled by the space station. The following measurements are taken based on when the ship breaches and exits the plane:

A: The station records the number of ticks in its own clock between the breach and exit of the ship.

B: The station records the number ticks it perceives in both of the clocks on the ship between the breach and exit of the ship.

C: The ship records the number of ticks in its own clocks between the breach and exit of the plane.

D: The ship records the number of ticks it perceives in the clock on the station between the breach and exit of the plane.

How will these measurements compare? Is there something flawed in my logic or setup? It seems that both frames of reference will think that the other persons clock is slow. Which will be right based on the values of A and C?

This post seems like it answers a very similar question: The Four-Clock Special Relativity Conundrum

I can't quite make the interpolation from that scenario to mine though. My feeling is that it has something to do with the fact that the station will not perceive the clocks on the ship as in sync.

• – user12262 Sep 15 '15 at 5:35
• After the edits, I think is okay - it's been reduced to a conceptual question about connecting the results of the previous question to this situation. (The title is still not good though!) – David Z Sep 15 '15 at 6:32

Your setup and intuition are correct:

The station observes different times for the two clocks on the ship. In fact, if we extrapolate and imagine a whole series of synchronized clocks spanning the entire length of the ship, then the station will observe different times on each one of them. It will see the rear clock always ahead of the front clock by a constant time differential. Likewise, it will see the intermediate ones progressively ahead of the front clock, but lagging behind the rear clock. All this is due to relativity of simultaneity. If the station would have a finite dimension along x, then the ship would see a similar situation for clocks at different positions along x on the station. Both the ship and the station observe the other's clocks running slow, yet each of them sees a consistent time sequence in its own frame.

To see how this works let $L$ be the rest length of the ship and $v$ its velocity relative to the station. Denote $\beta = \frac{v}{c}$ and $\gamma = \frac{1}{\sqrt{1 - \beta^2}}$, as usual. Let the station be positioned at $x=0$ in its own frame, and let the ship breach the station's plane at station time $t=0$ and ship time $t'=0$. For future reference let's write the Lorentz transformations between the ship's frame $(x', ct')$ and the station frame $(x, ct)$: $$x' = \gamma (x - \beta ct) \\ ct' = \gamma (ct - \beta x)$$ and $$x = \gamma (x' + \beta ct') \\ ct = \gamma (ct' + \beta x')$$

A. The station observes the ship breaching at time $ct=0$, and exiting after its entire length crosses the station plane. Since the station observes the ship length contracted along $x$ by $1/\gamma$, the exit time must be $ct = (L/\gamma)/\beta = L/\beta\gamma$ and measurement A must read $$c\Delta t_A = L/\beta\gamma$$
B. For the clock at the front of the ship, the station sees a time $ct'=0$ when the ship breaches at $ct=0$. When the ship exits, the station will see the front clock at position $x = L/\gamma$ and time $ct = L/\beta\gamma$ in station frame. By the Lorentz transformations, this means the ship coordinates of the front clock are $x'= \gamma ( L/\gamma - \beta (L/\beta\gamma) ) = 0$ and $ct' = \gamma (L/\beta\gamma - \beta (L/\gamma) ) = L (1/\beta - \beta)) = L /\beta\gamma^2$. The front clock time difference as seen station-side is therefore $c\Delta t'_F = L/\beta\gamma^2 = c\Delta t_A/\gamma$. Equivalently, the ship's front clock runs slower than the station clock by a factor of $\gamma$, $c\Delta t_A = \gamma c\Delta t'_F$.
Since the time rate is the same for the rear ship clock, the station observes the same time differential on it as on the front clock. Both are time dilated by $\gamma$ relative to station time: $$c\Delta t'_R = c\Delta t'_F = c\Delta t_A/\gamma = L/\beta\gamma^2$$ or $$c\Delta t_A = \gamma c\Delta t'_R = \gamma c\Delta t'_F = L/\beta\gamma$$
C. The ship sees its bow breach at $ct'=0$ and its rear exit after the station plane passed its entire length L, so the time between breach and exit on the ship is $$c\Delta t'_C = L/\beta$$
D. A reasoning similar to that for measurement B shows that the station time lapse as seen on the ship is $$c\Delta t_D = L/\beta\gamma$$ which is time dilated relative to the ship lapse by a factor of $\gamma$: $$c\Delta t'_C = \gamma c\Delta t_D = L/\beta$$