# Question on special relativity

1. Jul 7, 2006

### kknull

hello,
imagine 2 source of light. In the middle there is an observer who is moving towards one source at a determined speed s. The 2 sources send a ray of light when the observer is perfectly in the middle. In the observer's point of view, the 2 rays of light travel towards him at the relative speed of c, so they meet exactly at the same point on the observer; this is impossible because in the meantime the observer has travelled for some space, so the 2 rays have to meet together not in the middle, but near a source.
I cannot solve this problem, can you help me?
thank you.
regards, kknull

P.S. Excuse me for my bad english!

2. Jul 7, 2006

### MeJennifer

Well it all depends on from which frame of reference you measure this.
I am not sure what the problem is, there is no problem actually!

3. Jul 7, 2006

### kknull

well, if you see the events from sources' point of view, the ray of lights meet together at the middle. If you see the situation from observer's point of view, they meet next to a source, it's impossible!

4. Jul 7, 2006

### Janus

Staff Emeritus
No, from both the sources' view and the observer's point of view the rays meet at the middle. What they will[ disagree about is whether or not the rays both started from their sources at the same time. According to the sources, they do. according to the observer, they do not. This is known as the Relativity of Simultaniety.

5. Jul 7, 2006

### MeJennifer

It seems kknull that you have not as of yet accepted the view that distance and duration are not absolute properties but properties that depend on the frame of reference. Until you do the effects of relativity will remain to you impossible.

6. Jul 7, 2006

### kknull

!!!

You're trying to tell me this "the rays will not meet at the observer beacause the rays will not meet at the observer!( according to the observer, they do not)" But that is the point, I do not understand the relativity of simultaneity: if the rays start simultaneosly (according to the sources), the rays will have the same speed c for the observer, so they must meet at the observer!

P.S. I do not understand also the famous Einstein's example of the train, this is the point.

7. Jul 7, 2006

### clj4

Stop. Right here. This is wrong. If the observer travels towards a light soure he will:

1. FIRSTLY encounter the wavefront from that particular light source
2. LATER encounter the wavefront from the other light source.

This is called "closing speed" and it is well known in physics in general, relativistic or classical.

The rays will still meet SIMULTANEOUSLY in the middle of the distance between the light sources, they do NOT meet simultaneously on the moving observer.

8. Jul 7, 2006

### kknull

so in this case it is no longer valid the rule that relative speed of light is always c?

9. Jul 7, 2006

### Parlyne

It is always valid that the relative speed of light is c. However, if the observer was in the middle at the time the light was emitted (measuring time in the source frame), and he is moving, then he will no longer be in the middle when the light gets there. Hence, the light rays cannot meet where he is.

If we consider the observer's frame, we have the added complication that the relativity of simultaneity means that the light rays were not emitted at the same time - at least, not as the observer measures time.

10. Jul 7, 2006

### MeJennifer

Not at all. The speed of light is not inifinite, hence it takes time to reach the observer.
If the observer moves, well then one will reach him before the other. T
his is not even relativity but a simple observation we can make with things like balls.

Have two people one left and one right of you throw a ball with the same speed at you.
At the same time you move in the direction of one of them. Then surely you woud understand that one of the balls will reach you sooner.
If the speed of light were infinite then you would be correct. But it is not.

11. Jul 7, 2006

### kknull

how can you demonstrate it?
you're trying to tell "the two rays don't meet at the observer because when "the observer measures time" the two rays don't meet at the observer (they are not simultaneous) we are in a closed circle.

12. Jul 7, 2006

### kknull

mmm.... classical mechanics told me that if I am 10m far from an object and our relative speed is 1m/s, then we will reach in 10s.
So, if the observer is 10m away from both sources, and have the same relative speed with the 2 rays of light (c), then the 2 rays will meet at the observer.

Last edited: Jul 7, 2006
13. Jul 7, 2006

### MeJennifer

That is not what you wrote at all:

So, do you understand it now?

14. Jul 7, 2006

### kknull

the source and the light aren't the same things. you have to delete the sources from your imagination. There are only two rays of light which start at the same distance from the observer. So it has no importance that I am moving towards a ray of light, simply because I CAN'T, the relative speed is always c, the sources have no importance in this part of the problem.

15. Jul 7, 2006

### MeJennifer

So do you want help in understanding a problem you have with relativity or are you here to explain that it is all wrong?

16. Jul 7, 2006

### Staff: Mentor

If at the start of this problem you are at the center and you start to move when the light beams are fired (according to preset, synchronized clocks), obviously you will no longer be at the center when the light beams get there.

So then, is your question really about how you can still measure both beams of light approach at C? Ie, if they are both baseballs and not beams of light, fired at 100m/s toward the center and you are moving at 10m/s, you will measure one at 110m/s and the other at 90m/s.

17. Jul 7, 2006

### kknull

exactly, if I have balls they change relative speed, (ie 110,90), but if I have light, the relative speed is always c!

18. Jul 7, 2006

### nrqed

The key point is this: when you say that the the two sources send a light beam when the observer is in the middle, you are implicitly saying that they send a light beam at the same time . But who decides that they are emitted at the same time? Someone in the frame of the light sources or the oberver moving at some speed relative to the light sources?

That is the key issue. If it's as seen from the light sources, then yes, the light beams will cross at the middle of the two sources and they will reach the observer at different times.
SO how does thing look like from the point of view of the observer? What happens is that he/she will see the two light beams emitted at different times . He will see one beam emitted before the other and that's what will explain from his/her point of view that the beams don't reach him/her at the same time.

Hope this clarifies things.

Patrick

19. Jul 7, 2006

### kknull

Yes, this is the key point: is there a logical explanation that for the observer the 2 rays are not simultaneous, or it is like "it is because it is"?. As I say before, we are in closed circle, we are trying to demonstrate that the 2 rays of lights are not simultaneous because they're not.

Last edited: Jul 7, 2006
20. Jul 7, 2006

### clj4

Are you taking physics? Or are you just reading things at random? Of course light speed is still c. "Closing speed" is the combined speed of TWO systems, the observer and the lightwave front. The closing speed is c+v for the incoming front and c-v for the front that "chases" the observer.

The incoming fron meets the observer after the time T1;

c*T1+v*T1=L , therefore T1=L/(c+v)

The chasing wavefront catches the observer after T2:

L+v*T2=c*T2 , therefore T2=L/(c-v)

You can see that T2>T1.
This problem has very little do do with relativity except the fact that light always travels at c.

Last edited: Jul 7, 2006