# Implications of orthogonal clocks in rockets

1. Apr 16, 2012

### BOYLANATOR

Hi.
If two light clocks are put on a rocket at rest and then accelerated to relativistic velocities with one of the light clocks parallel to the direction of motion and one perpendicular, will one clock continue to measure the rate of change of time in the rest plane while the other one measures a different time?
If instead grandfather clocks were used, would the result be the same?
What about mechanical clocks (i.e. with cogs of a certain orientation)
Any responses to my wondering are appreciated.

2. Apr 16, 2012

### ghwellsjr

After the rocket achieves terminal velocity, both clocks will tick at the same rate.
A pendulum clock requires gravity and will tick at a different rate depending on the gravity.
Balance clocks will tick the same regardless of orientation under any constant velocity.

3. Apr 16, 2012

### BOYLANATOR

I thought the reason a light clock orientated perpendicular to the direction of movement measured time as slower than a rest frame was due to the idea that the light beam follows the hypotenuse of a triangle as viewed by the observer at rest. What causes the parallel light clock to relatively slow?

4. Apr 16, 2012

### Staff: Mentor

Length contraction.

5. Apr 16, 2012

### ghwellsjr

You're right about the perpendicularly oriented light clock but think about one that was moving very close to the speed of light. The triangle would be almost a flat line, wouldn't it? And if you approximated it to the parallel version, the two mirrors would have to be very close together in order to tick at the same rate, wouldn't they? But at slower speeds, you would have to move them farther apart to maintain the same rate until finally, with no motion at all, they would be the same distance apart as the perpendicular version, wouldn't they?

6. Apr 17, 2012

### jartsa

Well surely not. In a parallel light clock, moving at speed 0.9 c, light can travel across the clock in about half the normal time, because the clock is shortened to about half the length.

But the rate that the light - mirror distance changes is 30000 km/s (0.1 c).

So it takes about five times longer to travel across a light clock that is travelling at speed 0.9 c.

(Note: no incorrect velocity addition here)

7. Apr 17, 2012

### BOYLANATOR

I don't really follow this part. Is it possible to describe it in another way?

8. Apr 17, 2012

### BOYLANATOR

I didn't even think about length contraction. This seems like the obvious solution.

9. Apr 17, 2012

### Staff: Mentor

Don't forget that in a parallel light clock the "forward" half of the tick takes significantly longer than the "backward" half. If you work it out, it gives the wrong time unless there is length contraction.

10. Apr 17, 2012

### jartsa

Let us calculate one example case.

Rest length of a parallel light clock = 1 light second
Velocity of the clock = 0.866 c (relativistic factor is 2)

Light travels from rear to front a distance of half light seconds at speed 0.134 c.
That takes 3.73 seconds.
Then light travels from front to rear a distance of half light seconds at speed 1.866 c
That takes 0.268 seconds.

Back and forth travel takes 3.998 seconds.

We expected it to take twice the time of the time that it takes when the clock is standing still, which is 2 seconds.

11. Apr 17, 2012

### ghwellsjr

Don't you agree that if there was no motion, there would be no difference between the perpendicular and the parallel versions? And so the mirrors would be the same distance apart. Then as the speed of the motion increases, the mirrors in the parallel case will have to move closer together in order to maintain the same tick rate as the perpendicular case? In the limit, where the speed is almost as fast as light, the mirrors will have to be very close together because the shape of the triangle will be almost a straight line and if the mirrors were as far apart in the parallel case as they were in the perpendicular case (and by that I mean the distance apart along the direction of motion, which is zero) then it would take a very long time for the light to make the round trip.

12. Apr 17, 2012

### BOYLANATOR

Thanks jartsa. It's now clear to me that I have not found a flaw in the theory haha

13. Apr 17, 2012

### Staff: Mentor

If there were no length contraction then we would expect the forward trip to take 7.46 s and the backwards trip to take .54 s, for a round trip duration of 8 s.

But with the length contraction factor of 2 the forward trip actually takes 3.73 s and the backward trip actually takes 0.27 s, for a round trip duration of 4 s, as expected.

Including length contraction is essential. That is the geometric piece which is different for the parallel and perpendicular clocks that BOYLANATOR was asking about.