# A question about a relativity light clock

1. May 3, 2007

### Taremos

I'm learning about relativity and how going close to the speed of light can distort time, but I have questions/need for confirmation on a few things. After seeing a little video about I thought of some things and came to a few logical conclusions.

Okay. A light clock as seen at http://www.geocities.com/physics_world/sr/light_clock.htm tells time by having two mirrors .5 light second apart. Thus every time a photo originates from one plate, bounces off the far plate, and comes back and hits the close plate, one second has passed. This is used to show how time would become warped when approaching c. So, for example, if there is a clock like this attacked to a ship traveling at .99c... If the plates are perpendicular to the ship, it will register .14 time (every second in stationary space is .14 seconds here). But what if the plates are positioned parallel to the ship, out in front of it for example? It would take 50 seconds for the photon to move away from the ship and hit the far plate (which is traveling at .99c), and then about .251 seconds to come back if the plates are .5 second c apart at stationary. So according to that clock one second would take 50.251 seconds.

Does this mean that when you are traveling at a velocity near c that time in one direction is different than a time looking in another direction? Many places seem to say that you would interpret a 10 year voyager at .99c as 1.4 years, but why wouldn't you interpret it as 0.02 years?

2. May 3, 2007

### neutrino

I assume you mean when the mirrors are placed such that they are perpendicular to the direction of motion. In the case, you have to take into account the "length contraction." The distance between the mirrors will not be 0.5 light-seconds any more. It would 0.5/gamma light-seconds. With that in mind, now calculate the factor by which time would be "dilated."

BTW, 'stationary space' is really not a good term to use in relativity. Try something like 'rest frame'.

3. May 3, 2007

### JesseM

You need to take into account Lorentz contraction, which contracts lengths parallel to the direction of motion. So if the mirrors are oriented parallel to the direction of motion, and they're 0.5 light seconds apart in their own rest frame, in the frame where the ship is moving at 0.99c they'll only be $$0.5*\sqrt{1 - v^2/c^2}$$ = $$0.5*\sqrt{1 - 0.99^2}$$ = 0.0705 light seconds apart. So in the frame where the ship is moving, the photon will take 7.05 seconds to travel from the back plate to the front plate, and 0.0354 seconds to return from the front plate to the back plate, for a total of 7.09 seconds. This means that in 1 second of our time, this light clock only registers 1/7.09 = 0.14 seconds, just as with the light clock that's oriented perpendicular to the axis of motion.

Last edited: May 3, 2007
4. May 3, 2007

### robphy

To see the Lorentz contraction in the longitudinal direction, as neutrino and JesseM highlighted, and how it arises from the principle of relativity in the "light clock", take a look at my animated visualizations: http://www.phy.syr.edu/courses/modules/LIGHTCONE/LightClock/ and my paper http://arxiv.org/abs/physics/0505134 . You can also read about in, say, Mermin's Space and Time in Special Relativity or Griffiths' Electrodynamics text. What's different about what I did above is that I draw spacetime diagrams (as opposed to spatial diagrams) and uncover what I think is a nice geometrical interpretation of proper-time.

Last edited: May 3, 2007
5. May 25, 2007

### petm1

What would I see if I were traveling in this ship?

6. May 25, 2007

### JesseM

The light would go straight up and straight down and the distance between the mirrors would be 0.5 light-seconds, so you'd measure the light to take 0.5 seconds to go from the bottom mirror to the top and another 0.5 seconds to go from the top mirror to the bottom.

7. May 25, 2007

### petm1

I would think that I would see the first flash, if I were standing next to the emitter in the ship, and the second flash when the light returned from the front of the ship one second later and another flash every second as the light returned and changed directions. For the observer in the other frame I would see the flash when the light was emitted and 7.09 seconds later I would see a second flash as it returned from the front then .0354 of a second later I would see a third flash as it again changed toward the front. With this timing of the light clock I would think that the stationary observer could tell which of the two of us was moving, and his speed, or am I missing something?

8. May 26, 2007

### JesseM

Yes, that's what I said, you'd measure the time to go from back to front as 0.5 seconds and the time from front to back as 0.5 seconds, so it would return to the back every 0.5 + 0.5 = 1 second.
That's what the observer in the other frame would measure for the time-coordinates of the light hitting each mirror, although it's not what they'd see with their eyes since what they see is affected by the fact that the distance from the mirrors to their eyes is constantly changing.
The timing only tells the stationary observer that the light clock is moving relative to himself, not that it's moving in any absolute sense. The situation is completely symmetrical--if the "stationary" observer had his own light clock at rest in his frame, then the "moving" observer would measure the time for light to go from one mirror to another of that clock as 7.09 seconds in one direction and 0.0354 seconds in the other, just like what the "stationary" observer measures for the "moving" light clock.

9. May 26, 2007

### petm1

Thanks Jesse.

10. May 27, 2007

### phyti

The moving observer cannot see the event 'reflection of signal'.
He is only aware of the emission and reception of the signal.
By symmetry, the reflected signal perpendicular to the motion occurs half way.
The (x,t) for reflection of the signal sent parallel to the motion is unknown,
because the observer cannot detect his own motion.
He arbitrarily assigns it half way.

11. May 27, 2007

### JesseM

Sure he can, he can just look through a telescope at the distant mirror, and when he sees a flash in that mirror, he's just seen the event of the reflection of the signal. Of course if he is standing next to the other mirror then he'll see this event at exactly the same moment he sees the light returning to that mirror, but he can calculate the actual time of reflection in his coordinate system by subtracting (his distance from that mirror)/c from the time he saw the reflection in his telescope. He could also just have a friend standing next to the distant mirror with a clock synchronized with his own using the Einstein synchronization convention, and the friend could note the local time on his clock that he saw the flash.
In relativity there is no truth about "his own motion", in his inertial rest frame he defines himself to be at rest, and he synchronizes different clocks using the Einstein synchronization convention which is based on the assumption that light travels at c in every direction in his frame. The physical reason for making this assumption is that if every inertial observer defines their own rest frame in an analogous way, then they will find that the laws of physics obey the same equations in each rest frame, because the laws of physics have the property of Lorentz-invariance.

12. May 28, 2007

### phyti

The moving observer cannot see the event 'reflection of signal'.
A poor choice of words, just ignore that.
As for the parallel signals,if one of two or more r.frames are moving at different speeds, and one is at rest, the others can't place the reflection event halfway.
If you put a device at the mirror to record the time of reception of the signal, you will record the one-way speed of light, and thus determine the absolute velocity of the ref.frame.

Last edited: May 28, 2007