# Trying to understand relativity

1. Jul 26, 2012

### kr75

I am a novice in relativity as of now and I'm still learning about it. Currently reading Feynman's lectures and an example has confused me. So here it is,

Consider a spaceship moving at a velocity u. If a light source is placed equidistant from two detectors at opposite ends, one in front of it and the other behind it (with respect to the direction of motion). Then for a person inside the spaceship, a ray of light reaches both detectors at the same time but for a person observing from outside the spaceship the ray of light takes more time to reach the detector in front of it than the one behind it (as the detector in front is seemingly moving away from the source while the one behind is moving towards it).

I do not fully understand this concept. I'd be glad if someone could explain it to me

2. Jul 26, 2012

### PAllen

It is called relativity of simultaneity. The concept of 'at the same time' is not a universal truth, and cannot even be perceived over any significant distance without some measurement convention. For a reasonable convention, we find, that in our universe, if two observers are in relative motion, they will disagree on which 'events' are 'at the same time'. Thus, using normal measurement conventions, both rocket observer and home observer see light as traveling the same speed, but they disagree on which events are 'at the same time'. For the home observer, the 'arrival at back' event occurs before the 'arrival at front event. Inside the rocket, these same two events are considered simultaneous.

3. Jul 26, 2012

### kr75

Ok but i'm still confused as to why for a person outside the spaceship, the ray of light reaches the detector behind it before it reaches the detector in front of it. Since the spaceship is moving at a velocity 'u' wouldn't the light source and the detectors be moving with the same velocity too? Or is my approach to understanding this wrong fundamentally?

4. Jul 26, 2012

### Staff: Mentor

- The spaceship is moving at a velocity u relative to the observer outside the ship. That's a critical detail, so important that it cannot be left unsaid.
- The speed of the source is irrelevant to the speed of the ray of light and to the speed of the detector relative to the observer outside the ship. The different-time effect is caused by the fact that both observers see the ray of light traveling at the same speed (speed of light in vacuum is the same for all observers) but covering different distances. Both agree that the ray of light was emitted at the same place and time, right under the noses of both observers - but because one of the observers sees the detectors changing position while the ray is flight, and the other doesn't, they measure different distances traversed. Same speed, different distance... Different travel time.

5. Jul 26, 2012

### Jimmy

Relativity of Simultaneity:

Last edited by a moderator: Sep 25, 2014
6. Jul 26, 2012

### kr75

ahh now I understand better. I was making a rudimentary mistake while thinking about it. Thanks

7. Jul 26, 2012

### kr75

8. Jul 26, 2012

### ghwellsjr

For the person inside the spaceship, the reason the ray of light reaches both detectors at the same time is purely and only because the person has defined the same time to be when the two rays hit the two detectors. Let's back up a bit. In order for the spaceman to determine when the light rays hit the two detectors, he must have timing devices installed in the two detectors that record the times when the rays hit them. He performs his experiment and then he goes to one detector and sees what time is recorded on it. He goes to the other detector and sees what time is recorded on it. Chances are, the first time he does this, they will not record exactly the same time. So what he does is tweak the time on one of them by whatever amount the difference in the two readings was and then he repeats the experiment, and whatayahknow, the times are now the same.

Now what about the guy outside the spaceship. He can't do the same experiment, because he's not on the spaceship so he sets up his own detectors with timers that he adjusts by the same technique. The only problem is that he needs a bunch more detectors because he doesn't know where the spaceship will be when it flies by. And he also needs to set up all the timers so that no matter where the light rays start from, each pair of detectors at equal distances from the light source will have to read the same time.

Now there is one more added complication. He needs to have camcorders placed at every detector because he doesn't just want to know when the light rays get to each detector, he also wants to know which two of his detectors were adjacent to the two detectors on the spaceship when the rays hit them.

It will take him awhile to do this but when he gets done he is ready for the spaceship to come flying by. When it does, and the light source emits the two rays in opposite directions, this will be recorded by whichever camcorder is closest to the light source. We'll call this camcorder 'S' (for source). Then as the rays travel outwards in both directions, each pair of his detectors that are equal distant from the first one will record equal times, all the way out to the end. However, since the spaceship is moving while this is happening, the detector at the rear of the spaceship will see the light ray first and it will be recorded on a stationary camcorder nearby. We'll call this camcorder 'R' (for rear). Later on, the other ray will eventually catch up to the detector at the front of the spaceship and this will be recorded by another camcorder which we will call 'F' (for front).

Now it should be obvious that the time on 'R' will be earlier than the time on 'F' even though the times on the spaceships two detectors will read the same time as each other, but not the same times as what the stationary detectors record. Each camcorder records both the time on the stationary detector and the time on the adjacent spaceship detector but only two camcorders capture the events of the light rays hitting the spaceships detectors. And although they will record the spaceship's two detectors having the same time on them, the two stationary detectors will have different times on them.

9. Jul 27, 2012

### kr75

yes that does help. Of course i will have to go over it again till I understand perfectly by myself, but this helps me put it into perspective in a better way. Thanks George

10. Jul 27, 2012

### phyti

No one needs clocks.
Using a basic space-time diagram, the outsider concludes a long path, short path to the front and the reverse to the back, in equal times.
The anaut is only coincident with the emission and detection of the return signals which are simultaneous for him. Not knowing his speed in space, he cannot determine when or where the reflections occurred. Using the SR clock convention, he assumes the paths to be equal.

11. Jul 27, 2012

### ghwellsjr

The scenario under consideration in this thread does not have any return signals and there are no reflections.