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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.

Community
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Steve
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2 Answers2

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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 $$ x = \gamma (x' + \beta ct') \\ ct = \gamma (ct' + \beta x') $$

Your measurements are as follows:

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 $\gamma$: $$ c\Delta t'_C = \gamma c\Delta t_D = L/\beta $$

As you can see, the station sees the ship clocks run slowly, while the ship sees the station clock run slowly. Both are right in their own frames and observe consistent sequences of events.

udrv
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You're driving the ship at speed $3/5$. I'm on the station.

What we agree on:

  • When the front of your ship comes through my station, your front clock and my station clock both say 0:00.
  • When the back of your ship comes through my station, your rear clock says 20:00 and my station clock says 16:00.

Your explanation:

  • The ship is 12 light-hours long (we're not pretending these numbers are realistic). So the trip through the space station, at speed $3/5$, took 20 hours.
  • The ship's clocks, which are both accurate, show that the journey started at 0:00 and ended at 20:00.
  • But during that time, the station clock advanced only 16 hours, from 0:00 to 16:00. That station clock runs slow!

My explanation:

  • The ship is 9.36 light-hours long. So the trip through the space station, at speed $3/5$, took 16 hours.
  • When the front of your ship passed through, its clock showed the correct time of 0:00. But your rear clock was showing 7:12
  • When the rear of your ship arrived, its clock showed 20:00. Over the course of a 16 hour journey, it had advanced by only 12:48. That rear clock of yours runs slow! (So does your front clock, by the way).

Edited to add: I saw urdv's answer after I had written this. His answer and mine are essentially the same, though if you want to solve problems like this on your own, you'll learn more from his.

WillO
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