Speed of light for different observers

In summary, the conversation discusses the principles of Special Relativity in relation to the speed of light. It explains how the idea of luminiferous aether was refuted by the Michelson-Morley experiment and how Einstein proposed that the speed of light in a vacuum is constant for all observers. The conversation goes on to discuss a thought experiment involving a moving train-car and three observers, demonstrating how the speed of light can appear different for different observers. It concludes by discussing the implications of this understanding for our observations of the Universe and the limitations of measuring the speed of light.
  • #1
sisoev
117
0
The following is not a theory but only attempt to understand the principles of Special Relativity in connection with the speed of light.

Introduction

The idea that light waves require luminiferous aether to move through was refuted by Michelson-Morley experiment and that permitted Einstein to postulate that the speed of light in a vacuum is the same for all observers, regardless of their relative motion or of the motion of the source of the light.
No one ever considered that the space itself appears as aether for the light waves.
Space as distance between two objects is not any different for the light than a fiber cable.
Imagine that you accelerate fiber cable through which travels light. The same way we can accelerate space.
The only difference is the different speed in these two mediums.
Can we accelerate space thus accelerating the light which travels through it?
The answer “Yes” may sound stupid on a first though but here is a simple thought experiment, which can be actually performed with the predicted results.

The Experiment

speed-of-light1.gif


The above graphic shows the principle of the experiment.
A train-car with certain length is moving in right direction with certain constant speed.
There is a light source attached at the back of the train-car which sticks out of it.
In the front of the train-car there is a mirror, which for the purpose of the experiment we call “observer A”.
That will be the frame of reference which we will call “frame A”.
Outside this frame is located frame of reference B, which we will call “frame B”.
While in motion (with constant speed) the light source on the train-car produces flash toward observer A. The light from the flash needs certain time to reach observer A, during which time the train-car advances in right direction.
Knowing the speed of the train-car and the distance between the light source and observer A, we calculate at which place in frame B, observer A will see the light and on the same line in frame B we place observer B, so both A and B will see the light simultaneously.

speed-of-light2.gif


The above graphic is the position of the train-car at the moment of the simultaneously observed light by A and B.
Since A is stationary to the source, the distance is measured from the source to the observer.
However the distance covered by the light to observer B has to be measured from the place of the emission.
In that case the result is two different distances, covered by the light from one emission, and perceived simultaneously by two different observers.
Special Relativity would argue the above result with different running clocks in the two frames of reference.
The different running clocks though, will not be favorable arguments since B will see the light shifted in the blue spectrum, due to its velocity in respect to the source at the moment of the emission.
Let’s extend the experiment and put third observer (observer C) behind the train-car in frame B.
Observer C will not see the flash, but only the reflection of the light beam for A and B.
These reflections for C will be simultaneous, but also seen with different light frequencies.

Conclusion
The above thought experiment shows two different distances covered by the light from one emission in simultaneous observation from three different observers.
Since the light emission is one, the change of the frequency is due to the different speed of the light in respect to the observer.
In other words, observer B measures the speed of light as higher than the measurements of observer A and the result will be seen in the different patterns left by the light.
.
A question comes to mind after this experiment; can we consider the different light frequency as different speed of the light.
The answer is very simple; if two observers perceive the light from one emission with two different frequencies, then their velocity is different not only in respect to the source, but to the light as well.

And here is the my thought for the right understanding of the expression “speed of light for different observers”.
Light is electromagnetic emission with certain frequency.
One electromagnetic wave is not light. Even if we know the precise speed of one photon, that wouldn’t be the speed of light.

The speed of light is characterized by the speed, with which every next wave due to the velocity at the moment of the emission approaches the observer.
Imagine that the source and the observer are in approaching motion at the moment of the emission. The light waves will approach the observer with speed greater than c and that will shorten the time between the waves.
In this case the light frequency is characterized not by wave length, which by my opinion is a wrong understanding about light frequency, but by the time between the light waves.
Respectively, different time between the light waves is perceived as different speed of the light and will leave different light patterns when measured.

Follows the question, “does the different pattern left by light from stationary sources with different frequencies shows different speed of the light”?
No. Two stationary sources with two different frequencies will emit light with the same speed.
All sources emit light with the same speed, but the velocity of the light should be measured as any other velocities.

The above helps to understand better the events in the Universe and our place in it. We know that a star moves toward us if we observe its light in the blue spectrum, but we can never know the place of the star, since we don’t know the initial frequency of the light and its speed which depends on the velocity at the moment of the emission.
Not knowing all this we can never determine the present place of the star whose light we observe.
Considering the above, we cannot predict what happens in the Universe right now if we measure all events with same speed for the light.
We see the past without knowing the present. In that regard it won’t help much if we do our measurements with different speed for the light, but it will definitely make a change.
 

Attachments

  • speed-of-light1.gif
    speed-of-light1.gif
    27.3 KB · Views: 697
Physics news on Phys.org
  • #3
Aaron_Shaw said:
Aaron, in that forum I presented the experiment like for kids ;)
Here the presentation is like for students and teachers :)
And by the way, try to be more respectful and decent, please.
 
Last edited by a moderator:
  • #4
Very few question marks for someone seeking understanding.

Ok, I'll give you the benefit of the doubt and help you with your attempt to understand.

Draw two spacetime diagrams for your Gedankenexperiment. One in A's frame, the other in B's frame. (Better not use one single diagram for both points of view, I think)
If you don't know how to do that: That's the first thing you must learn. Look it up, if you get stuck, ask here.

Having done that, look at the time coordinate of the emission event in both frames. Different or not?
Look at the space coordinate of said event. Calculate the speed of light in each frame. Different or not?
If different: search for the error in your calculations. If you don't find it, ask. We'll help.

That's how SR handles the situation. You have to understand it before you jump to conclusions.
 
  • #5
Ich said:
Very few question marks for someone seeking understanding.
The question is actually one: is the speed of light the same for all observers.
I built a thought experiment on the question and I expect someone to comment on it.
Your patronizing attitude doesn't help much, Ich.
Would you please tell me why do you think that the experiment is wrong or draws wrong conclusions?
The experiment is as simple as Michelson-Morley experiment and if their experiment doesn't need your recommendations I wonder why my experiment would need them.
 
  • #6
The experiment is as simple as Michelson-Morley experiment and if their experiment doesn't need your recommendations I wonder why my experiment would need them.
If you don't need recommendations, I wonder what you're doing here.

Would you please tell me why do you think that the experiment is wrong or draws wrong conclusions?
Answer the questions, and you'll see what's wrong (Hint: has to do with the time coordinates of the emission event).
Your only chance to understand is IMHO that you draw the diagrams yourself. If you try, we'll help. If not, that would be proof that you don't seek understanding, but want to promote fringe views. I don't think that's your intention.
 
  • #7
Ich said:
If you don't need recommendations, I wonder what you're doing here.


Answer the questions, and you'll see what's wrong (Hint: has to do with the time coordinates of the emission event).
Your only chance to understand is IMHO that you draw the diagrams yourself. If you try, we'll help. If not, that would be proof that you don't seek understanding, but want to promote fringe views. I don't think that's your intention.
Not that I don't need recommendations but I wonder why such a simple experiment need to be complicated in such way :)

There is one only light emission for both A and B and I don't see your point.
I guess you are seeing the experiment from SR perspective, and you try to argue it from that point of view.
Wouldn't it be correct to say that if this experiment is correct Special Relativity would stand incorrect and your argument would be invalid?
Shouldn't you argue the experiment without using arguments drawn from SR?

ONE QUESTION: do you agree that B will see the light as a blue-shifted (different light pattern from A if measured with photographic plate)?
If yes, how would you explain it?

Please don't mind my "aggressive" manner of discussion.
I found out that this is the best way for me to understand something ;)
 
  • #8
I would suggest to look at the experiment as a proof that Space is actually medium through which the light propagates and if we accelerate the space, the speed of the light changes with the speed of the accelerated space.

Thinking that way, we wouldn't need Michelson-Morley experiment because it would appear pointless.

If you find the above not appealing to your logic, forget it and concentrate on the experiment.
 
  • #9
There is one only light emission for both A and B and I don't see your point.
I know that you don't see my point. That's why you must draw the diagrams.
I guess you are seeing the experiment from SR perspective, and you try to argue it from that point of view.
Exactly.
Shouldn't you argue the experiment without using arguments drawn from SR?
You analyze the thought experiment with SR. Either you find an internal inconsistency in SR, or not.
So, will you, or won't you?
And, please, make sure you've read the forum guidelines, especially https://www.physicsforums.com/showthread.php?t=17355".

ONE QUESTION: do you agree that B will see the light as a blue-shifted (different light pattern from A if measured with photographic plate)?
If yes, how would you explain it?
Yes.
Doppler effect.
 
Last edited by a moderator:
  • #10
Ich said:
I know that you don't see my point. That's why you must draw the diagrams.

Exactly.

You analyze the thought experiment with SR. Either you find an internal inconsistency in SR, or not.
So, will you, or won't you?
And, please, make sure you've read the forum guidelines, especially https://www.physicsforums.com/showthread.php?t=17355".


Yes.
Doppler effect.

I read the forum rules and as I state in my opening post, I'm here to gain better understanding of SR.
I guess that I have the right to say what is the obstacle for me to understand the theory.
The obstacle for me is presented in the experiment.

Pardon me, but there is no rule in your forums that say how fast must I agree with you or understand your explanation for that matter :smile:

Now, about the Doppler effect: obviously it is, but why two different observers see the light from one emission, simultaneously, at one and the same point with two different frequencies?
Would you help me to understand this?
 
Last edited by a moderator:
  • #11
The obstacle for me is presented in the experiment.
Then learn how to analyze the experiment.
Pardon me, but there is no rule in your forums that say how fast must I agree with you or understand your explanation for that matter
I didn't give an explanation. I offer to guide you in finding it. You may accept or refuse.

Would you help me to understand this?
Yes. Right after you've learned to draw and read said diagrams.


If you don't want me to help you, just say so. No problem.
 
  • #12
Ich said:
Then learn how to analyze the experiment.

I didn't give an explanation. I offer to guide you in finding it. You may accept or refuse.

I expected direct answer as a help :smile:
Sorry, it is my misunderstanding.

Yes. Right after you've learned to draw and read said diagrams.

I'm afraid I don't understand what do you want me to do. Would you elaborate on that, please? Tell me your idea and I'll draw it.


If you don't want me to help you, just say so. No problem.
No, no!
I would be happy to be helped by you.

I'd like to ask something more on the Doppler effect issue in my experiment and I hope this time you'll answer me. That would help me great deal.
Perhaps I am deceived by Michelson-Morley experiment, but please correct me:
Michelson expected to register different speed of the light on the two light beams, because one of them was parallel to the flow of ether.
That difference in the speed would be seen in the different interference.

Why the different interference in my experiment wouldn't be seen as different speed of the light?
 
  • #13
sisoev said:
The above thought experiment shows two different distances covered by the light from one emission in simultaneous observation from three different observers.
Different observers measure different distances between emission at one end of the train (event 1) and arrival at the other end (event 2). They also measure different times.
Since the light emission is one, the change of the frequency is due to the different speed of the light in respect to the observer.
No. The change in frequency is due to the different speed of the source in the different frames. The speed of the light is the same.
In other words, observer B measures the speed of light as higher than the measurements of observer A and the result will be seen in the different patterns left by the light.
No. To measure the speed of the light, you'd need to measure the travel time. Do that and you'll find that everyone measures the same speed for the light.
 
  • #14
Doc Al said:
Different observers measure different distances between emission at one end of the train (event 1) and arrival at the other end (event 2). They also measure different times.

For observer C (from the extended part of the experiment) the distance between the emission and the two mirrors is the same, but C still sees the light with different frequency, which is my problem here - it would mean that the light for observer A was traveling slower.

No. The change in frequency is due to the different speed of the source in the different frames. The speed of the light is the same.

Hm...
I don't understand how the source would change the frequency of the light which was already emitted before it change its position.
And I tried to explain my view on the way the light changes it frequency when the source is in motion, but obviously (at least according to you) I am mistaken.
Would you comment please the part in my post which is around this line:
In this case the light frequency is characterized not by wave length, which by my opinion is a wrong understanding about light frequency, but by the time between the light waves.

Doc Al said:
No. To measure the speed of the light, you'd need to measure the travel time. Do that and you'll find that everyone measures the same speed for the light.

The problem for me is not the exact measurement of the speed, but the difference shown in the different light patterns.
Because the way you explain it, we have to know the time measured in the frames.
But in order to know the time we use the speed of light which is in argument :smile:
Seems that SR does not allow gap to avoid this circle ;)
That is why I am interested in the difference visible in the patterns.
If you can help me with that I'll be very grateful.

I think that it would be easy to start with observer C.
Sorry, I am a bit slow but it has its advantages :wink:
 
  • #15
sisoev said:
For observer C (from the extended part of the experiment) the distance between the emission and the two mirrors is the same,
Since observers C and B are in the same frame, they agree on the distance between the two events. Observers in the train car frame disagree.
but C still sees the light with different frequency,
You mean that C will see the light reflected from the moving mirror (at A) and the light reflected from the stationary mirror (at B) to have different frequencies? So?
which is my problem here - it would mean that the light for observer A was traveling slower.
No it wouldn't. Why would you think that?


Hm...
I don't understand how the source would change the frequency of the light which was already emitted before it change its position.
You might want to learn about the Doppler effect.
And I tried to explain my view on the way the light changes it frequency when the source is in motion, but obviously (at least according to you) I am mistaken.
Would you comment please the part in my post which is around this line:
In this case the light frequency is characterized not by wave length, which by my opinion is a wrong understanding about light frequency, but by the time between the light waves.​
Not quite sure of the point you are trying to make, but frequency and wavelength are related by: speed = frequency X wavelength. The "time between the light waves" is the period, which is 1/frequency.
 
  • #16
sisoev said:
Imagine that the source and the observer are in approaching motion at the moment of the emission. The light waves will approach the observer with speed greater than c and that will shorten the time between the waves.

This is incorrect, there are already multiple experiments that prove your above statement to be false.

In addition to that, your gedank suffers from the following major flaws:

1. It is virtually impossible to understand.

2. The lack of mathematical support (there are no calculations) makes it unfalsifiable.

3. You present ZERO experimental data, yet you conclude that light speed is dependent on the source speed.
 
  • #17
I'm afraid I don't understand what do you want me to do.
Google spacetime diagram.
http://physics.syr.edu/courses/modules/LIGHTCONE/events.html" a site that seems to explain the concept.
Make a diagram with the time axis upwards, the x-axis (=direction of motion) to the right. Forget about y and z.
For each interesting point (e.g. back of train, observer B...) draw its position as a function of time. These are all straight lines.

This forces you to set up your experiment unambiguously, with concrete numbers.
Ask if you get stuck.


Why the different interference in my experiment wouldn't be seen as different speed of the light?
Because the source is moving. Moving sources must exhibit Doppler shift, relativity or not. This could be easily explained with a spacetime diagram.
 
Last edited by a moderator:
  • #18
I am fairly certain that consideration of such trains is what lead to SR.

Lorentz certainly considered an eather, refining his eather until Einstein developed SR and it was shown the two were equivelent, without the need for the eather's preferred reference frame.

Then Einstein went on to explain gravity in this context as a curvature of space-time. Curvature and flow are both mathematiaclly modeled as the differential of one variable with respect to another variable.

The better question might be why no one ever thought to consider gravity as a flow of SR with time. (Eddington does talk about the equivelent medium representation.)
 
  • #19
Doc Al said:
Since observers C and B are in the same frame, they agree on the distance between the two events. Observers in the train car frame disagree.
That's the point with observer C - he sees the reflection of the light traveled in frame A, which is not the same length as in frame B.
You would probably argue with the fact that the emission for B and C was at the same place, but it was at the same place for A as well :smile:
Now if we measure the distance in frame A as different, the reflection from mirror A seen by C will be from the light traveled that different distance.
Yet, it will still be seen simultaneously with the reflection from mirror B.


Doc Al said:
You mean that C will see the light reflected from the moving mirror (at A) and the light reflected from the stationary mirror (at B) to have different frequencies? So?
That would mean different speed of light.

Doc Al said:
No it wouldn't. Why would you think that?
Why would Michelson and Morley think it?

Doc Al said:
You might want to learn about the Doppler effect.
I did.

Doc Al said:
The "time between the light waves" is the period, which is 1/frequency.
Exactly.
So, if the velocity is different then the frequency changes.
My point was that the velocity between the light waves and the observer changes.
The velocity between the source and the observer is important only as "speed giver" to the light waves. After the emission the speed of the source is unimportant. It may as well stop and move backwards but we will still see it at the place of the emission and the frequency will still be changed.
I suppose that you will disagree with this, and if you do, please explain why.
Explain please how do you see the change of the frequency due to different velocity.
May be I'm missing something and I'll be thankful to you to correct me.
 
  • #20
Ich said:
Google spacetime diagram.
http://physics.syr.edu/courses/modules/LIGHTCONE/events.html" a site that seems to explain the concept.
Make a diagram with the time axis upwards, the x-axis (=direction of motion) to the right. Forget about y and z.
For each interesting point (e.g. back of train, observer B...) draw its position as a function of time. These are all straight lines.

This forces you to set up your experiment unambiguously, with concrete numbers.
Ask if you get stuck.

Thank You very much.
I greatly appreciate your help, and I promise that I'll check it and I'll see what I can do.

Because the source is moving. Moving sources must exhibit Doppler shift, relativity or not. This could be easily explained with a spacetime diagram.

I already said in my previous post that the speed of the source, by my opinion, doesn't matter after the emission.
Please correct me if I am wrong but if I am right your answer from the above will be wrong.

I am afraid that I may look double-fool giving the same wrong answers two times, so you may want to wait for the answer of Doc Al on the above :smile:
 
Last edited by a moderator:
  • #21
sisoev said:
That's the point with observer C - he sees the reflection of the light traveled in frame A, which is not the same length as in frame B.
Not true. For one thing, the light does not belong to one frame. There's no such thing as 'light that traveled in frame A'. There is just one beam of light that is emitted and happens to hit mirrors at A and B at the same moment (because those mirrors are in the same place at that moment). The distance that the light travels depends on whose frame you are viewing things from. Observers in the C-B frame measure one distance; Observers in the train frame measure a different distance.
You would probably argue with the fact that the emission for B and C was at the same place, but it was at the same place for A as well :smile:
There's only one emission, so I'm not sure what you're talking about here.
Now if we measure the distance in frame A as different, the reflection from mirror A seen by C will be from the light traveled that different distance.
Nope. See my comment above.
Yet, it will still be seen simultaneously with the reflection from mirror B.
Of course. The light reflects from A and B at the same place at the same time, so of course it reaches C at the same time.
 
  • #22
utesfan100 said:
The better question might be why no one ever thought to consider gravity as a flow of SR with time. (Eddington does talk about the equivelent medium representation.)

I see that flaw but the rules of the forum do not permit me to present private theories :biggrin:
 
  • #23
sisoev said:
So, if the velocity is different then the frequency changes.
But it isn't. Next?
My point was that the velocity between the light waves and the observer changes.
I've tried to point out your error.
The velocity between the source and the observer is important only as "speed giver" to the light waves.
The relative speed of source and observer gives the light its observed frequency, not its speed.
After the emission the speed of the source is unimportant. It may as well stop and move backwards but we will still see it at the place of the emission and the frequency will still be changed.
Sure. If a pulse of light is sent out, it will have some observed frequency. Once that pulse has been sent, the source could blow up for all we care.
I suppose that you will disagree with this, and if you do, please explain why.
Explain please how do you see the change of the frequency due to different velocity.
Again I refer you to the Doppler effect.
 
  • #24
Doc Al said:
Not true. For one thing, the light does not belong to one frame. There's no such thing as 'light that traveled in frame A'. There is just one beam of light that is emitted and happens to hit mirrors at A and B at the same moment (because those mirrors are in the same place at that moment). The distance that the light travels depends on whose frame you are viewing things from. Observers in the C-B frame measure one distance; Observers in the train frame measure a different distance.

I am sorry "traveled in frame A" was used as a figure of speech.
We measure the distance for A, not from the place of emission but from the place of the source, because that is the right way to do it.
Lets think the same way about observer C.
At the moment of the emission, observer C sees mirror A closer and this is the distance for the light to cover. For C mirror A moves while the light hits it, the same way as A moves to position B while the light hits it. So observe C sees the length of frame A as observer A sees it.
Mirror B is stationary and that distance is seen longer by C.

I didn't get answer on my last comment which was:

So, if the velocity is different then the frequency changes.
My point was that the velocity between the light waves and the observer changes.
The velocity between the source and the observer is important only as "speed giver" to the light waves. After the emission the speed of the source is unimportant. It may as well stop and move backwards but we will still see it at the place of the emission and the frequency will still be changed.
I suppose that you will disagree with this, and if you do, please explain why.
Explain please how do you see the change of the frequency due to different velocity.
May be I'm missing something and I'll be thankful to you to correct me.
 
  • #25
sisoev said:
Hm...
I don't understand how the source would change the frequency of the light which was already emitted before it change its position.
And I tried to explain my view on the way the light changes it frequency when the source is in motion, but obviously (at least according to you) I am mistaken.

Consider the following animations.
The first shows a source emitting light and two observers (the red and blue dots) the light is represented by the expanding circles with the bright green as the peaks of each successive wave. In this situation the source is at rest with respect to the observers. Both observers see the same frequency as is emitted at the source.

[URL]http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/doppler1.gif[/URL]

The second shows the same source and observers with the source moving with respect to the observers. Note that each successive wave expands at the same speed as before in all directions and as a circle from the point of emission.

[URL]http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/doppler2.gif[/URL]

Consider two successive peaks. Between the time that the fist peak is emitted and the second peak is emitted, the source has moved closer to the blue observer. This means that the second peak has to travel less distance than the first peak and takes less time to reach the observer.
For example: Assume a source frequency of 1 Mhz and a initial distance of 1 light sec between source and observer. Also assume relative speed of 1/2c for the source.

The initial peak is emitted and 1 millisecond later, the second peak is emitted. Between the emission of the peaks, the source has moved 150 meters (0.00000005 light sec) closer to the observer. 1 sec after the first peak was emitted, it reaches the observer. 0.99999995 sec after it was emitted, the second peak reaches the observer. The time difference between peaks as seen by the observer is (0.99999995+.0000001)-1 = 0.00000005 sec. This works out to be equal to a frequency of 2 Mhz. He sees a blue-shift frequency, and the speed of the light passing is no different than it was in the first animation.
 
Last edited by a moderator:
  • #26
sisoev said:
We measure the distance for A, not from the place of emission but from the place of the source, because that is the right way to do it.
Huh? The place of emission is the source! If you're interested in the distance that the light travels, you must measure from the place where the light started, which is where the source was at the time of emission.
Lets think the same way about observer C.
At the moment of the emission, observer C sees mirror A closer and this is the distance for the light to cover.
Nope. Mirror A moves. You need to measure to the point where mirror A is when the light gets there, not where mirror A was at the moment of emission.
For C mirror A moves while the light hits it, the same way as A moves to position B while the light hits it.
During the time that the light is traveling, mirror A moves with respect to frame C-B.
So observe C sees the length of frame A as observer A sees it.
Nope. For A, the distance that the light travels is just the proper length of the train. For C (and B) the distance that the light travels is different because: (1) C measures the length of the train as shorter, and (2) the train moves.
Mirror B is stationary and that distance is seen longer by C.
As seen by C, the light travels the same distance to both mirrors.

I didn't get answer on my last comment which was:
Sorry, that ended up in a separate post. Check out the cool animations posted by Janus! I think they will help your understanding of the Doppler effect.
 
  • #27
Janus said:
The second shows the same source and observers with the source moving with respect to the observers. Note that each successive wave expands at the same speed as before in all directions and as a circle from the point of emission.

[URL]http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/doppler2.gif[/URL]

Consider two successive peaks. Between the time that the fist peak is emitted and the second peak is emitted, the source has moved closer to the blue observer. This means that the second peak has to travel less distance than the first peak and takes less time to reach the observer.
For example: Assume a source frequency of 1 Mhz and a initial distance of 1 light sec between source and observer. Also assume relative speed of 1/2c for the source.

The initial peak is emitted and 1 millisecond later, the second peak is emitted. Between the emission of the peaks, the source has moved 150 meters (0.00000005 light sec) closer to the observer. 1 sec after the first peak was emitted, it reaches the observer. 0.99999995 sec after it was emitted, the second peak reaches the observer. The time difference between peaks as seen by the observer is (0.99999995+.0000001)-1 = 0.00000005 sec. This works out to be equal to a frequency of 2 Mhz. He sees a blue-shift frequency, and the speed of the light passing is no different than it was in the first animation.

This is very well done. May I add that the math used for generating the above is very simple.
Let f_blue_dot be the frequency measured in the blue dot and f_red_dot be the frequency measured in the red dot. Let f be the frequency emitted by the source, +v the speed between the source and the blue dot, -v the speed between the source and the red dot and, most importantly, c, the invariant speed of light. Then:

f_blue_dot=f*sqrt((1+v/c)/(1-v/c)), so f_blue_dot>f

Replacing +v with -v we get:

f_red_dot=f*sqrt((1-v/c)/(1+v/c)), so f_red_dot<f

That's all the math we need.
 
Last edited by a moderator:
  • #28
Janus said:
Consider the following animations.
The first shows a source emitting light and two observers (the red and blue dots) the light is represented by the expanding circles with the bright green as the peaks of each successive wave. In this situation the source is at rest with respect to the observers. Both observers see the same frequency as is emitted at the source.

[URL]http://home.earthlink.net/~parvey/sitebuildercontent/sitebuilderpictures/doppler1.gif[/URL]

You got me wrong here, Doc Al :smile:
I said:
I don't understand how the source would change the frequency of the light which was already emitted before it change its position.
You agreed that we can blow up the source after the light is emited.
Now let's take a look at the animation.
Lets imagine that this is emitted flash of light and the source is not there anymore.
Now imagine that the blue observer moves toward the center.
For him every next wave will come faster, due to the different velocity and that will change the way he observes the light, shifting it in the blue spectrum.

In my experiment we observe it the other way round - the observer is stationary and the waves move toward him with speed c+the speed of the train, and that is why the frequency is different for B. Well, this is how I see it. Where am I wrong?

Did I just stupidly broke some law of the physics :smile:
 
Last edited by a moderator:
  • #29
sisoev said:
You agreed that we can blow up the source after the light is emited.
Sure. If the sun disappeared 'right now', we wouldn't know it for about 8 minutes. Nonetheless, to understand how the frequency shifts it's best to imagine the source still there.
Now let's take a look at the animation.
Lets imagine that this is emitted flash of light and the source is not there anymore.
Now imagine that the blue observer moves toward the center.
That's illustrated in the second animation.
For him every next wave will come faster, due to the different velocity and that will change the way he observes the light, shifting it in the blue spectrum.
It's not any difference in the speed of the light--the speed of the outgoing waves doesn't change. It's the fact that the source moved while the light was being emitted. Thus the wave crests are closer together.
 
  • #30
sisoev said:
In my experiment we observe it the other way round - the observer is stationary and the waves move toward him with speed c+the speed of the train, and that is why the frequency is different for B. Well, this is how I see it. Where am I wrong?

Did I just stupidly broke some law of the physics :smile:

The speed of light is not C+V(train). The length ot the train would be contracted just enough to make the light travel at C.

You are confusing Doppler redshift with Lorentian redshift. In fact, both are present.
 
  • #31
utesfan100 said:
You are confusing Doppler redshift with Lorentian redshift. In fact, both are present.
What are you referring to by the term "Lorentian redshift"?
 
  • #32
sisoev said:
In my experiment we observe it the other way round - the observer is stationary and the waves move toward him with speed c+the speed of the train, and that is why the frequency is different for B. Well, this is how I see it. Where am I wrong?

Did I just stupidly broke some law of the physics :smile:
As has been mentioned a couple of times the part where you are wrong is the idea that the waves move at c+v instead of c. The fact that light always moves at c regardless of the motion of the emitter is called the second postulate of relativity and has been experimentally confirmed many times to very high precision:

http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html#moving-source_tests
 
  • #34
My apologies, you certainly did, even to linking right to the same sub-section.
 
  • #35
DaleSpam said:
My apologies, you certainly did, even to linking right to the same sub-section.

Nothing to apologise about, it is easy to overlook. :-)
 
Back
Top