# Track of the light in a moving light clock.

1. Feb 23, 2014

### rajeshmarndi

There is a moving light clock at a certain velocity. For an outside observer, the light just seemed to be traveling in a slant path. So far, it is right for the outside observer, the light is going in a slant path and hitting the top bar and then again returning and hitting the bottom bar.

Now, suppose the moving light clock changes its velocity periodically, then if I'm right, the track of the light now for the outside observer, will be, as if it is following and changing its track, as if the light knows when to change the track. How would the outside observer explain, the changing of the track of light in the space?

I hope I have made my question clear.

What is it, I'm not understanding here?

Thanks very much.

2. Feb 23, 2014

### Bill_K

It doesn't. As long as the clock is moving uniformly, everything is fine. A photon from mirror A aimed at mirror B will hit mirror B. But if you launch the photon, and then speed up, then obviously the photon will miss the mirror!

3. Feb 23, 2014

### terberculosis

The light clock thought experiment was used as an example of special relativity, which only covers inertial reference frames. Inertial reference frames are situations which involve constant velocity. No acceleration or gravitational fields allowed.

Your proposed situation falls under the domain of General Relativity, because of the changing velocity. The photon would miss the second mirror if the acting force did not also act on the photon.

If I recall correctly, general relativity indicates that if the clock is accelerated by a constant gravitational field perpendicular to the travel path of the photon, the clock would still function with the path you indicated.

4. Feb 23, 2014

### Bill_K

Special Relativity does not just cover inertial reference frames. It is perfectly adequate to deal with acceleration. General Relativity is a curved space theory of gravity.

5. Feb 23, 2014

### Staff: Mentor

As Bill_K said this is a common misconception. Here is a good FAQ on the topic.
http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html

6. Feb 23, 2014

### rajeshmarndi

Is the light track independent of the moving clock for the outside observer, is what I mean. That is, when the moving clock accelerates, the light will miss the mirror, as you said.

Does it mean, now the photon would continue straight in the slant path missing the mirror? Or it's path is also affected as observed by the outside observer, due to acceleration of the moving light clock, which I think shouldn't?

7. Feb 23, 2014

### Bill_K

Once it's launched, the photon continues in a straight path in the same direction it started, and the outside observer would see it miss the mirror. Even though the clock gets accelerated, along with the mirrors, the photon just continues on its way.

8. Feb 23, 2014

### phyti

Assume the photon misses mirror.
Replace mirror with a one that extends from front to back on ceiling of moving spaceship. Now we can measure our speed relative to light by amount of deviation from vertical.

Is this possible?

9. Feb 23, 2014

### Bill_K

No, this only measures our change in velocity.

You know, this depends in no way on Special Relativity. If you and I ride a streetcar, and sit across the aisle and throw a ball back and forth, if the streetcar suddenly speeds up the ball will miss you.

10. Feb 23, 2014

### jstephens716

this is all based on a vertical clock... what if it was horizontal (parallel with trajectory)? Does the whole experiment fall apart or will there still be time dilation? In other words, is time dilation just another way of saying, "we can't measure time with light while it's moving."

11. Feb 23, 2014

### Staff: Mentor

In a parallel light clock you get both time dilation and length contraction.

12. Feb 23, 2014

### jstephens716

I was under the impression that time dilation was a result of the "hypotenuse" created by the bouncing vertical light while in motion. If the clock is horizontal, we don't get that effect. Is it just a flawed analogy for the theory?

13. Feb 23, 2014

### Staff: Mentor

Time dilation is easiest to show for such a clock, but the principle of relativity ensures that all clocks show time dilation, regardless of their mechanism.

14. Feb 24, 2014

### ghwellsjr

As DaleSpam said, the vertical light clock is easiest to show on a pair of diagrams such as these from the wikipedia article on Time Dilation. First the one for the stationary light clock:

And then the one for the moving light clock:

However the horizontal light clock can also be shown easily on a pair of spacetime diagrams. First the one for the stationary light clock:

Now this diagram may look more like the previous diagram for the moving vertical light clock but it is for the stationary horizontal light clock. The two thick lines are the two mirrors. The photon starts at the bottom leaving the blue mirror and moving at c (1 foot per nanosecond) toward the red mirror which is five feet away so it takes 5 nsecs as shown going up the vertical axis. At 5 nsecs it reflects off the red mirror and heads back toward the blue mirror at c taking another 5 nsecs so the total time is 10 nsecs for the photon to make the round trip.

Now the diagram for the case where the light clock is moving to the right at 60% of the speed of light:

You can tell that the mirrors are moving at 60%c because, for example, in 10 nsec the blue mirror has moved to the right 6 feet. But because of that, the photon that leaves at the bottom takes longer to reach the red mirror but not as long as it would have taken if it weren't for length contraction which shortens the distance between the two mirrors from 5 feet to 4 feet. You can see this by looking at the Coordinate Time of 10 nsec where the blue mirror is at 6 feet and the red mirror is at 10 feet.

But after the reflection, the photon takes only 2.5 nsecs to get back to the blue mirror because it is moving toward the photon making the total time 12.5 nsecs for the round trip.

So in the stationary case, the light clock took 10 nsecs to make one cycle of the photon but in the moving case it took 12.5 nsecs, illustrating the time dilation for the horizontal light clock.

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15. Feb 24, 2014

### phyti

If a light clock emits a single photon that reflects in a mirror located vertically above the emitter/detector, while in a rest frame, will it still work if it is moving at a constant speed past this same frame?
If yes, how does the photon adjust to a different angle?
If no, does the angle have to be set for a change in speed?

16. Feb 24, 2014

### WannabeNewton

If the photon emerges from one mirror and bounces off the other mirror in the rest frame of the light clock then it must do so in all frames. If this constraint is not imposed then all bets are off. In a given frame we could instead arrange for the photon to miss the opposite mirror by emitting it straight up at the left edge of the bottom mirror and having the light clock move sufficiently fast to the right but if it misses it in this frame then it must miss it in all frames-in the rest frame of the light clock the path of the photon will now be angled to the left. This constitutes a different constraint from the original one.

The photon doesn't need to adjust anything-the only thing the photon cares about is its path through space-time which is the absolute geometric characterization of its trajectory. The coordinate dependent spatial path of the photon is not at all frame-invariant, and changes from frame to frame in accordance with the frame-invariant constraint we've set that the photon must emerge from one mirror and reflect off the opposite mirror composing the light clock. This constraint must be met in all frames and so the spatial path of the photon changes from frame to frame in order to hold true to the constraint.

Last edited: Feb 24, 2014
17. Feb 25, 2014

### jstephens716

Well, this is not exactly a parallel clock, right? They've changed the way the clock operates. This kind of illustrates that it wouldn't be possible with a parallel clock where the light reflects between only two points, a & B. They're version of a parallel clock operates by bouncing the light off three different points... I just don't think it translates, or maybe I missed something.

18. Feb 25, 2014

### Staff: Mentor

It is a parallel light clock. Note that the vertical axis is time, not space.

19. Feb 25, 2014

### ghwellsjr

...it might look like the light is bouncing off of three different points, but A and C are the same mirror, just shown at two different positions in time. Actually, it would have been better to show the top mirror in three places side by side and the bottom mirror in three places side by side where each pair of mirrors above and below represent both mirrors in three different places at three different times. So at all times, wherever the photon is, there is a mirror above and a mirror below.

Does that make sense to you? Or did I misunderstand your concern?

20. Feb 25, 2014

### jstephens716

Just curious, how can this be demonstrated on a clock that doesn't tick with light?

And... say you have two synchronized clocks, send one whizzing past the other in space. According to relativity, they will each perceive the other's clock to run slow, right? So what happens when you compare the times of these two clocks? They'll be the same, won't they? Clock A can't register slower than Clock B if Clock B registers slower than clock A.

I think the problem with light in motion is that, by representing light speed as x=t, there will be tiny instants in time during which things aren't actually where they appear to be in physical space... the light refracting off of those objects will pass through your reference frame according to the given equations, but it's not where the objects actually are at that point in time.

I can't digest the fact that everything is relative in this model, except light. If you think of light, one photon at a time moving through a giant 3D grid, it has to be relative... it can't be in two places at once, and we know it doesn't form an instantaneous bridge between it's source and the observer.