Question on special relativity

In summary, the conversation is discussing the concept of relativity and the relativity of simultaneity. The main point of contention is the idea that two light rays will meet at the same point on an observer when the observer is moving towards one of the light sources. However, this is not possible because the observer will have moved from the middle point by the time the light rays reach him. This is due to the fact that the speed of light is not infinite. The conversation also brings up the example of throwing balls at an observer to illustrate this concept.
  • #1
kknull
39
0
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 traveled 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!
 
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  • #2
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
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
kknull said:
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!

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
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
!

Janus said:
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.

You're trying to tell me this "the rays will not meet at the observer because 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!

Thank you for your answers, Janus and Jennifer.
P.S. I do not understand also the famous Einstein's example of the train, this is the point.
 
  • #7
kknull said:
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;

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.


this is impossible because in the meantime the observer has traveled 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!

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
clj4 said:
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.

so in this case it is no longer valid the rule that relative speed of light is always c?
 
  • #9
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
kknull said:
so in this case it is no longer valid the rule that relative speed of light is always c?
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.
:smile:
 
  • #11
Parlyne said:
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.

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
MeJennifer said:
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.
:smile:

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.
 
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  • #13
kknull said:
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.
That is not what you wrote at all:

imagine 2 source of light. In the middle there is an observer who is moving towards one source at a determined speed s.

So, do you understand it now?
 
  • #14
MeJennifer said:
That is not what you wrote at all:



So, do you understand it now?

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
kknull said:
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.
So do you want help in understanding a problem you have with relativity or are you here to explain that it is all wrong? :confused:
 
  • #16
kknull said:
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.
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
russ_watters said:
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.

exactly, if I have balls they change relative speed, (ie 110,90), but if I have light, the relative speed is always c!
 
  • #18
kknull said:
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 traveled 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!

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
nrqed said:
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
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.
 
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  • #20
kknull said:
so in this case it is no longer valid the rule that relative speed of light is always c?

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.
 
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  • #21
Simple argument for the relativity of simultaneity

kknull said:
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"?.
You may find it helpful to read the discussion in this thread: https://www.physicsforums.com/showthread.php?t=121068
 
  • #22
kknull, are you Swedish? In that case, you've got a spectacular name...
 
  • #23
EL said:
kknull, are you Swedish? In that case, you've got a spectacular name...
Is it pronounced caca-null?
 
  • #24
EL said:
kknull, are you Swedish? In that case, you've got a spectacular name...

i'm not swedish. KKnull is referred to the japanese noise artist.
 
  • #25
clj4 said:
Is it pronounced caca-null?

no, i don't think so. Maybe you're reffering to the emperor of Austria in the '800, but "kk" are initials of a name, so they can be pronounced as will
 
  • #26
kknull said:
hello,
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 false. Yes, they both travel at c. No they don't meet at the same time.
 
  • #27
actionintegral said:
This is false. Yes, they both travel at c. No they don't meet at the same time.

That depends! If they were emitted at the same time according to the observer then they will reach him/her at the same time.
One has to be very careful with statements like this.

Patrick
 
  • #28
kknull said:
hello,
The 2 sources send a ray of light when the observer is perfectly in the middle.

This looks to me like the frame of reference of the sources
 
  • #29
mmm, the point is that I'm very confused on the concept of relativistic simultaneity, I'm going to read something about it...
 
  • #30
kknull said:
i'm not swedish. KKnull is referred to the japanese noise artist.
Ok. In Swedish it's an abreviation for having sex with a friend...:wink:
 
  • #31
understood. thank you! So my mistake was to consider my question as a paradox, while it was simply an empirical demonstration that simultaneity is relative.
 
Last edited:
  • #32
Exactly!..
 
  • #33
Again, from the Einstein train example, the staionary observer sees te light flashed at the same time. The one in motion catches the lead flash before the rear flash.

Question is: what is the mathematics behind this?

Assume the train is 150,000 km in lngth which places the train observer 17000 km in the middle. It will take the light from the front flash 0.5 sec to reach the stationary observer. If the train is moving at 0.5 c, how long will it take the front falsh light to reach the moving observer on the train - DO THE MATH! Not just words - DO THE MATH

I have a feeling it is still 0.5 seconds but it shouldn't be
 
  • #34
stevmg said:
Assume the train is 150,000 km in lngth which places the train observer 17000 km in the middle. It will take the light from the front flash 0.5 sec to reach the stationary observer. If the train is moving at 0.5 c, how long will it take the front falsh light to reach the moving observer on the train - DO THE MATH! Not just words - DO THE MATH
How long will it take according to whom? The observer on the train or the observer on the platform? What's the proper length of the train? It the train is 150,000 km long, the middle would be 75,000 km from each end.

Since you use a time of 0.5 seconds, perhaps you meant the train to be 300,000 km long? If so, then the light from each flash would take 0.5 sec to reach the middle of the train according to train observers.

According to the platform observer, it would take less time since the length is shorter and the train moves towards the incoming light. The length of the train would be shorter by a factor of γ = 1.155. So if the proper distance between the middle and ends of the train is L = 0.5 light-seconds, that distance would be L/γ according to the platform observer. To find the time for the front flash to reach the middle of the train, use:
L/γ - vt = ct

Solving for time, you get:
t = L/[γ(c+v)] ≈ 0.29 sec
 

1. What is special relativity?

Special relativity is a theory developed by Albert Einstein in 1905 that explains how objects behave at high speeds and in the presence of strong gravitational fields. It is based on two main principles: the laws of physics are the same for all observers in uniform motion, and the speed of light in a vacuum is constant for all observers regardless of their relative motion.

2. What are the main concepts of special relativity?

The main concepts of special relativity include the relativity of simultaneity, time dilation, length contraction, and the equivalence of mass and energy (E=mc^2). These concepts help to explain how space and time are interconnected and how they are affected by the relative motion of objects.

3. How does special relativity differ from Newtonian mechanics?

Special relativity differs from Newtonian mechanics in several ways. Firstly, it takes into account the constant speed of light and the relativity of simultaneity, which means that time and space are not absolute. Additionally, special relativity predicts that time and space are affected by an object's motion, while Newtonian mechanics assumes that they are fixed and independent of an object's motion.

4. What are some real-world applications of special relativity?

Special relativity has many real-world applications, such as in GPS technology, particle accelerators, and nuclear energy. It also helps to explain phenomena like time dilation in space travel and the existence of black holes.

5. How has special relativity been tested and confirmed?

Special relativity has been extensively tested and confirmed through various experiments and observations. Some of the most famous experiments include the Michelson-Morley experiment, which showed the constancy of the speed of light, and the Hafele-Keating experiment, which demonstrated time dilation. Additionally, the predictions of special relativity have been consistently supported by astronomical observations and technological advancements.

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