A Purely Hypothetical Question regarding Special Relativity Theory.

geistkiesel

01-30-05, 06:01 PM

geistkiesel:

You see. This is my problem. I have always understood that any experiment performed in any valid frame should always yeild the same results to the observer in that frame, no matter what the frame's state of motion. Especially that the speed of light is always measured to be c.

If the two clocks read differently in the spaceship, then I would calculate a different "speed of light" for each photon, given that I measure a 1m length for each leg of the apparatus. In the ship, nothing appears odd to me.

This makes no sense. Help me.

I will try. If you take the moving case and have the L and R clocks be mirrors the photons would reflect back to the observer at th eoriginal midpoint of the frame simultaneously. There will be a slight time difference in the moving and stationary cases however. The moving opbserver seeing the simultaneou arival; of the emitted photons might conclude he is at rest.

One explanation is that the "same results" expected is an imposed theoretical condition. Under the limits of the experiment here, obviosly the "same result" assumption appears nt to work, which is the natural reult expected from my pespective. In case 1 the frames are stationary wrt the embankment. In case 2 the frames are moving. One would expect a different measured result using the same laws of physics in both cases, especially of the motion is uniform. In case 2 he energy f he frame wrt the stationary case is greater. Using the C + v and c - v terms is a direct violation of the tenet iof SRT that holds all measurements of the speed of light will result in a measurment of c. This is of vcoursse, obviously correct when we all agree on the constancy of he speed f light. This dioes not, however, negate the use of relative velocity includong light in terms of c + v and c - v.

No one is asserting that light or any massive object is moving faster than the spoeed of light. C + v means only a counted measure of the closing speed of a beam of light and a massive object.

In short what you have been told and understood, like the rest of us at one time or another, seems to be erroneous. I suggest you walk through the experiment with L and R as mirrors and use the left moving photon as a measure of the distances light has traveled (without regard to observers or clocks) assuimg only the constancy of the speed of light. Detect where the "extra time" comes from in the moving case. See why the photons do not reflect at the same time and why they do combine simultaneously at the physical midpoint of the inertial frame. You can do this without any higher maths other than addition and substraction. To start: the left photon moves a distance ct before reaching L (which is moving toward l). The r photons moves the same distance to the right, but R has also moved right.

Where is l after it has moved ct + ct? How far away from the midpoint is l then? How far away from R is r after initially moving ct to the right?

Geistkiesel

superluminal

01-30-05, 06:08 PM

geistkiesel

01-30-05, 06:13 PM

Geistkiesel, and others,
Could you please give an answer to the following:

A simple question, presented with the utmost humbleness:

If I do the experiment above, sitting here in my house sipping an espresso, I measure the same travel times for each photon. Both clocks read the same.
Please Check YES or NO

If I do the experiment above, sitting in my spaceship, sipping tea this time, travelling at 0.8c toward Rigel, I measure the same travel times for each photon. Both clocks read the same.
Please Check YES or NO

Thanks.
Check Yes for the stationary case.

Check No for the moving case.

Remember the clocks are synchronized on your space ship. Remember also that the motion of light is independent of the motion of the source of the light. Also, remember you are privy only to the printed arrival time of the photons after the fact that were emitted simultaneously from the midpoint. Even if the frame is contracting and even if there is time dilation in the case 2, the clocks will still record a time difference of photon arrivals for the l and r moving photons.

This is the only point I intended to demonstrate with this thread.

Geistkiesel

geistkiesel

01-30-05, 06:32 PM

Is not the invariance in the measurement of c within a given unaccelerated frame fundamental? Is it not a direct consequence of Einstein's equivalence principle? Does this not state that you cannot determine your state of motion by performing any experiment within your isolated frame?
If this is what the invarinace principle m,eans then so be it. The question is not the invarinace principle per se, it is the correctness and accuracy of the principle. If one were to retsrict himself/herself to the invarinace principle and walk away from the thread without a physical reasoning then we are left with what? Blind faith acceptance in principles is the answer, while we ignore the physics of the situation. Who will be the first to assert the eternal truth of any physical principle?

Superlunimal,
Apply the invarinace principle here and analyze this with the experimental results? What must have occured that would manifestly have the two clocks printout the same time of flight of the two photons in the moving case? Does not motion affect both equivalently? Aren't the clocks located such that one cannot change its ticking rate without the other also changing in the same degree? If these questions aren't anwered in the positive must not one then explain the invariance principle in terms of mystery and voodoo? The observers on spaceships aren't mental numb robots, they are trained physicists and engineers who know very well the results of experiments on stationary and moving frames re the speed of light.

As an after thought, it appears to me that the invariance principle as stated by yourself is a correct statement of the principle, which only restricts the measurnment of the motion. The principle does not deny the fact of motionj itself, only the measurment. This however, as we have seen by the experiment here points to an apparent flaw in the principle as understood, or so I humbly assert.

Geistkiesel

geistkiesel

01-30-05, 06:43 PM

I started an entire thread conserning this question called Absolute Space. I had even propossed a device based on exactly the same scheme that you had draw, called PASD.

The answer is we cant know. If that is everything we know about the light source then I suppose that what Yuriu had calculated is correct:

But if you realize this here on Earth you cant be sure because except velocity v there would come into play and Earth velocity around Sun, The Sun velocity around galaxy center and so on....

So in the above equations v is not just some kind of velocity but the absolute velocity of the light source WRT the abosolute space.

If the observer is placed at the center of the light-source he will observe that both clocks are running exactly in the same rate in the case 2). That is because the photons should go back to the source and both times will becomes equal. If you had observer somewhere else he will notice the same - time is running in the same rate according to all clocks and that is noyt in contradiction to the above formulaes. But if we have two observers at the two clocks, they will observe different photon arrival times.

As I had sayed and before 'observe' is a very slicky word. You should be very carefull with it. Whenever be have movement of light back and forth observation times are equal. The PASD device however is designed so that only uni-directional movement of light gets into acount.
What is so puzzling or slicky about a printout of a time from a clock on a moving platform? Observations aren't voodo they are, in this experiment, simply looking at numbers printed on two pieces of paper.

Xgen, To begin, focus only on the arrival times opf the photons at L and R. Are the arrival times the same for the two cases discussed?

If the clocks are synchronized the light, if reflected back to the midpoint observer, will arrive their simultaneously, but if you read the printout of the arrival times at L and R the left moving photon arrives at L before the right moving photon arrives at R, which is unambiguous physical evidence that the frame is moving to the right.

How can you explain this differenltly?

Geistkiesel

geistkiesel

01-30-05, 06:45 PM

Geistkiesel,

I see and understand what you are saying, however, I cling to the principles of Einsteinian relativity. I agree that the intuitive result seems obvious, but I'm not sure it is. I'm working on it.
Go for it, just keep Ms. Objectivity forever vigilant in your explorations. Thank you for the intelligent, cogent and professionally submitted comments and questions.

Geistkiesel

superluminal

01-30-05, 06:51 PM

Geistkiesel,

If the clocks are synchronized the light, if reflected back to the midpoint observer, will arrive their simultaneously, but if you read the printout of the arrival times at L and R the left moving photon arrives at L before the right moving photon arrives at R, which is unambiguous physical evidence that the frame is moving to the right.

This seems correct at first inspection. My issue is that I have a long held belief in the Einsteinian view of things, and have had my physical intuition about this sort of experiment turned around and explained nicely in terms of relativity. I admit that I'm having difficulty with this one. Getting "unambiguous physical evidence that the frame is moving to the right" upsets my apple cart.

Yuriy

01-30-05, 07:23 PM

Superluminal,

1. To whom was addressed post with “YES or NO”?
If to me (or to me also) my answers are the following:
If L_M_R device is rested in respect to you – in both cases “YES”
If in any case L_M_R device is moving in regard to you – in that case answer is “NO”.
(BTW, look how many words in your post, that were used, are absolutely useless for the right answer: all what matters is your and device relative velocity in respect to each other)

2. I have always understood that any experiment performed in any valid frame should always yeild the same results to the observer in that frame, no matter what the frame's state of motion. Especially that the speed of light is always measured to be c.
The Principle of Relativity summarizes our knowledge in simple statement that the events are the same, not the descriptive characteristics of events. Simultaneity of two events is not an event! “Light reached receptor” – is an event. “This two events happened simultaneously” – is not an event. Same as “this line has length L” is not an event, but “my length-meter read a length L” is an event! I hope you understand what I am saying here – it is very, very important to apprehend SRT!
And now about c = constant. Here you are mixing two very important issue too. The velocity of any body is purely relative notion and strongly depends on the dynamical state of observers in respect each other. But speed of light is not! Speed of light (in vacuum!) is absolute World-constant and is absolutely independent on motion of observers and/or of source of light! This fact of Nature is a fundamental base of understanding of the natural features of Space and Time. Why it happens so and what is responsible for that – those are the gnoseological questions and we can know the answers or not (yet). But it does not change the fact that it is the fact of Nature.

3. If the two clocks read differently in the spaceship, then I would calculate a different "speed of light" for each photon, given that I measure a 1m length for each leg of the apparatus. In the ship, nothing appears odd to me.
I can not help you because you did not concretize relative motion of L_M_R device and observer. If you will do it my absolute answer, which always is the same “No, nobody will measure different speed of light for any leg”, could be supported by calculations why it is true and your assumption is wrong, i.e. that there is no problem, at all…

geistkiesel

1. Would not an observer always determine he is moving in case 2?
If he is a smart guy and has learn the classic Physics – never!
2. If the arrival times of the photons at L and R are different, as you concluded, then how can the observer ever conclude he is at rest when the arrival times of the photons are dofferent?
Why should he connect such a conclusion … with arriving times of two photons simultaneously emitted in some another reference frame?
Does this not tell the moving observer that he is moving wrt the stationary case?
No how! There is no way of logic to such a stupid conclusion!
All what you written below this your question (in the same post) is BS. The same concerns your next post, sorry…

3. Some days are just sadder than others Yuri.
When I read your anti-SRT posts – any day is sad in the same measure – you again did not learn anything!…

P.S. This my respons concerns only posts on page 1 of this thread. The page 2 I will comment later...

superluminal

01-30-05, 07:57 PM

This is interesting Yuiry. I agree with every thing you said and everything you said confirms my belief that SRT is the correct description of the universe. We are definitely having a difficult time agreeing that we agree.

I said:

If the two clocks read differently in the spaceship, then I would calculate a different "speed of light" for each photon, given that I measure a 1m length for each leg of the apparatus. In the ship, nothing appears odd to me.

You said:

I can not help you because you did not concretize relative motion of L_M_R device and observer. If you will do it my absolute answer, which always is the same “No, nobody will measure different speed of light for any leg”, could be supported by calculations why it is true and your assumption is wrong, i.e. that there is no problem, at all…

I will reword my statement:

If I (the guy on the ship, moving with the apparatus) found that the two clocks read differently in the spaceship, then I would calculate a different "speed of light" for each photon, given that I measure a 1m length for each leg of the apparatus. This would be in violation of everything I understand about SR. In the ship, nothing appears odd to me, therefore I will NOT see different times on my clocks.

Better?

superluminal

01-30-05, 08:02 PM

About 1/3 down the page. (Simultenaeity)

http://instruct1.cit.cornell.edu/courses/astro101/lec21.htm

geistkiesel

01-30-05, 11:56 PM

Geistkiesel,

I see and understand what you are saying, however, I cling to the principles of Einsteinian relativity. I agree that the intuitive result seems obvious, but I'm not sure it is. I'm working on it.
Superluminal, just a question that came to mind: Other than clinging to the Einsteinian principles as you sway, can you offer yourself any physical reaon to ovecome what you define as obvious? I understand you to mean you understand my input on this matter, which is trivially simple to see..

Geistkiesel

geistkiesel

01-31-05, 12:21 AM

Superluminal,

1. To whom was addressed post with “YES or NO”?
If to me (or to me also) my answers are the following:
If L_M_R device is rested in respect to you – in both cases “YES”
If in any case L_M_R device is moving in regard to you – in that case answer is “NO”.
(BTW, look how many words in your post, that were used, are absolutely useless for the right answer: all what matters is your and device relative velocity in respect to each other)

Useless to you pehaps as your res[ponse will always be the SRT answer, which has yet been proved to me.

2.
The Principle of Relativity summarizes our knowledge in simple statement that the events are the same, not the descriptive characteristics of events. Simultaneity of two events is not an event! “Light reached receptor” – is an event. “This two events happened simultaneously” – is not an event. Same as “this line has length L” is not an event, but “my length-meter read a length L” is an event! I hope you understand what I am saying here – it is very, very important to apprehend SRT!
And now about c = constant. Here you are mixing two very important issue too. The velocity of any body is purely relative notion and strongly depends on the dynamical state of observers in respect each other. But speed of light is not! Speed of light (in vacuum!) is absolute World-constant and is absolutely independent on motion of observers and/or of source of light! This fact of Nature is a fundamental base of understanding of the natural features of Space and Time. Why it happens so and what is responsible for that – those are the gnoseological questions and we can know the answers or not (yet). But it does not change the fact that it is the fact of Nature.

Yuri, take a real close look at the initial post of this thread. The question was simply what the L and R clocks would printout in the case where the frame was moving wrt the embankment, not what a stationary or moving observer would "observe". In any case the observers are not privy to any information other than what canb be observed. Here only the clock times on he moving platform are relevant. Even if there were frame contraction adn time dilation he fact remains that L would meassure an arrival time before the R clock measures a photon. The observer, if you will, strolls to L, pushes a button and gets the time of arrival of the left moving photon. Then the observer strolls to the R clock, pushes another button and reads the time of arrival of the right moving photon.

In the case of the frame at rest with respect to the embankment the tme of arrival of he photons at L and R are the same. We next put the frame in motion to the right and repeat the experiment: photons are emitted simultaneously from an emitter attached to the midpoint of the moving frame (this is the sanme emitter used previously). The clocks at L and R will printout different times of arrival of the photons.

Certainly Yuri you can have no problem with this simple physical sitution.

Geistkiesel

3.
I can not help you because you did not concretize relative motion of L_M_R device and observer. If you will do it my absolute answer, which always is the same “No, nobody will measure different speed of light for any leg”, could be supported by calculations why it is true and your assumption is wrong, i.e. that there is no problem, at all…
You are correct here. No one can calculate or measure diffeent speeds of slight for the diffeent legs. What we are measuring is the arrival time of he left two photons, only.

geistkiesel
Quote=Yuri]
1.
If he is a smart guy and has learn the classic Physics – never!
2.
Why should he connect such a conclusion … with arriving times of two photons simultaneously emitted in some another reference frame?
[/quote]

Yuri, the photons were emitted in the same physical frame from the same device. In the first case the frame was at rest wrt the embankment, in the second case the frame was moving wrt the embankment. It would have made no difference anyway what frame the photons were emitted, the results are the same.

No how! There is no way of logic to such a stupid conclusion!
All what you written below this your question (in the same post) is BS. The same concerns your next post, sorry…
What stupiid conclusion specifically are ypu referring to Yuri.

3.
When I read your anti-SRT posts – any day is sad in the same measure – you again did not learn anything!…

P.S. This my respons concerns only posts on page 1 of this thread. The page 2 I will comment later...

The "sad" statement was a joke Yuri, simply a joke.

Do the two photons arrive at the L and R clocks at the same instant in the moving frame as in the case where the frame is at rest wrt the embankment?
Geistkiesel

everneo

01-31-05, 02:15 AM

Yes.

Yuriy

01-31-05, 03:19 AM

No.
Exact quantitative answer I already gave.
Anyone who does not understand this answer never will understand SRT. Period.

geistkiesel

01-31-05, 03:21 AM

Talking to Yuri,

This is interesting Yuiry. I agree with every thing you said and everything you said confirms my belief that SRT is the correct description of the universe. We are definitely having a difficult time agreeing that we agree.

I will reword my statement:

If I (the guy on the ship, moving with the apparatus) found that the two clocks read differently in the spaceship, then I would calculate a different "speed of light" for each photon, given that I measure a 1m length for each leg of the apparatus. This would be in violation of everything I understand about SR. In the ship, nothing appears odd to me, therefore I will NOT see different times on my clocks.

Better?
Suprluminal,
You have the true grasp of the problem. Your use of the invariance principle has superceded any attempt to analyze the physics of he moving photons. Bear with me. Likewise you have gone outside the limits of the hypothetical problem.

Your use of invaiance here neglected the option that the frame (at least viewed from the embankment, which the moving observer must be aware of) was moving and that different times of arrival of the photons was due to the frame motion. Do you see what gets neglected here using SRT? Anwser: Physical data.

That you conclude differently than I is of no consequence. Think of the physical condition together with your understanding of SRT. You need not integrate the thoughts simulabneously when constructing a hypothetical situation. Put all the rlevant ideas together in a rational form that are necessary as determiend by your own comfort zone.

Beginning with the photons only, and forgetting about observers for this brief instant we see that
each photon is emitted at a unique point in space.
Each photon moves independently of the frame, moving or otherwise.
In whatever time base one uses, each photon will move the same distance in that time span, relative to the point in space from which they emitted..
neither photon knows that there are L and R clocks ahead.
in the 'frame moving to the right case', the L clock is closing (colliding), with the left moving photon while,
the right moving photon is chasing the R clock.
When the left moving photon strikes L after travelling ct, the right moving photon has also travelled ct, in the opposite direction.
when the L clock records the left moving photon arrival the right moving photon has not yet reached R.
at this point the right moving photon is a distance 2vt from R.

The clocks merely 'stamp' out the current moving frame time when the photons arrive. The data is then open for scrutiny, perusal. What do the topics of "observer at rest" or "moving observer" have to do with the results?
What t do we use? Answer: Any convenient time. The SR theorists use SRT time, the non-SRTists use embankment time. These are of no consequence as the different arrival times is the only important parameter here.

What does the "invarinace principle" have to do with the results?

Even Yuri used the terms, C + V and C - V in an earlier post to this thread. Is not this use of velocity addition assume a violation of SRT where the relative motion of the frame and light is always and axiomatically supposed to be C?

Geistkiesel

Yuriy

01-31-05, 03:31 AM

superluminal,
I will reword my statement:
If I (the guy on the ship, moving with the apparatus) found that the two clocks read differently in the spaceship, then I would calculate a different "speed of light" for each photon, given that I measure a 1m length for each leg of the apparatus. This would be in violation of everything I understand about SR. In the ship, nothing appears odd to me, therefore I will NOT see different times on my clocks.

AFTER SUCH REWORD OF YOUR INITIAL STATEMENT you are talking about Case v=0. The unswer is trivial: clocks rested in regard to you and point M will read the same times. And nothing will be "in violation of everything I understand about SR."
But we are talking about comparison of cases 1. and 2., remamber?

geistkiesel

01-31-05, 03:38 AM

The Principle of Relativity summarizes our knowledge in simple statement that the events are the same, not the descriptive characteristics of events. Simultaneity of two events is not an event! “Light reached receptor” – is an event. “This two events happened simultaneously” – is not an event. Same as “this line has length L” is not an event, but “my length-meter read a length L” is an event! I hope you understand what I am saying here – it is very, very important to apprehend SRT!

Note to the reader: I am takinhg Yuri's statement somewhat out of context here and focus only on these worsds.

Yuri, This is not a discussion of simultaneity or events as described by yourself above. When I say the light/photons were emitted simultaneously from a point source at the midpoint of the moving frame this is not a staenment discussing "simultaneity of events." When I day the photons were recorded at different times on themoving frame, I make no simultabneoty assertion. It is just a physical fact: The left moving phoptn was detected, then later the right moving photon was detected. The lengths of both legs are identical throughout from symmetry considerations an the time of flight for equal distances is identical.

A. The instant L records a photon arrival is an event.
B. The instant R records a photon arrival is an event.

A occured before B as determined by moving frame clocks.
Get in focus.
Geistkiesel

Yuriy

01-31-05, 03:56 AM

Yuri, This is not a discussion of simultaneity or events as described by yourself above. When I say the light/photons were emitted simultaneously from a point source at the midpoint of the moving frame this is not a staenment discussing "simultaneity of events." When I day the photons were recorded at different times on themoving frame, I make no simultabneoty assertion. It is just a physical fact: The left moving phoptn was detected, then later the right moving photon was detected. The lengths of both legs are identical throughout from symmetry considerations an the time of flight for equal distances is identical.
A. The instant L records a photon arrival is an event.
B. The instant R records a photon arrival is an event.
A occured before B as determined by moving frame clocks.
Get in focus.

There is analysis of your post.
1. When you say the one beam of light (or a photon) was emitted left and another one beam of ligt (or a photon) was emitted from a point source at the midpoint of the (moving or rested - it does not matter) reference frame THIS IS
a. discussion about two events
b. discussion about two events that happened simultaneously
do you like it or not;

2. When you say the one beam of light (or a photon) was arrived on the end of the left leg, and another one beam of ligt (or a photon) was arrived on the end of the right leg THIS IS
a. discussion about two events
b. discussion about simultaneouty or none-simultaneoty of two events
do you like it or not.

For your general information; SRT is only about events, nothing else!, do you like it or not...

So, get in focus.

everneo

01-31-05, 04:06 AM

No.
Exact quantitative answer I already gave.
Anyone who does not understand this answer never will understand SRT. Period.
You have just slipped for the second time in this thread, Yuiry. Just think over.

Yuriy

01-31-05, 04:26 AM

Beginning with the photons only, and forgetting about observers for this brief instant we see that each photon is emitted at a unique point in space.
Each photon moves independently of the frame, moving or otherwise.
In whatever time base one uses, each photon will move the same distance in that time span, relative to the point in space from which they emitted..
neither photon knows that there are L and R clocks ahead.
in the 'frame moving to the right case', the L clock is closing (colliding), with the left moving photon while, the right moving photon is chasing the R clock.
When the left moving photon strikes L after travelling ct, the right moving photon has also travelled ct, in the opposite direction.
when the L clock records the left moving photon arrival the right moving photon has not yet reached R.
at this point the right moving photon is a distance 2vt from R.
The clocks merely 'stamp' out the current moving frame time when the photons arrive. The data is then open for scrutiny, perusal. What do the topics of "observer at rest" or "moving observer" have to do with the results?
What t do we use? Answer: Any convenient time. The SR theorists use SRT time, the non-SRTists use embankment time. These are of no consequence as the different arrival times is the only important parameter here.
What does the "invarinace principle" have to do with the results?
Even Yuri used the terms, C + V and C - V in an earlier post to this thread. Is not this use of velocity addition assume a violation of SRT where the relative motion of the frame and light is always and axiomatically supposed to be C?

Let us analyse this your post too...

1. "Each photon moves independently of the frame, moving or otherwise."
Right.
2. "In whatever time base one uses, each photon will move the same distance in that time span, relative to the point in space from which they emitted.."
Wrong. Do not agree? Then prove your assertion!
3. "neither photon knows that there are L and R clocks ahead."
Right.
4. "in the 'frame moving to the right case', the L clock is closing (colliding), with the left moving photon while, the right moving photon is chasing the R clock."
For what observer? That one who sits on the moving point M sees nothing such a picture...
5. "When the left moving photon strikes L after travelling ct, the right moving photon has also travelled ct, in the opposite direction."
This is repetition of previous statement...
6. "when the L clock records the left moving photon arrival the right moving photon has not yet reached R."
Again, repetition of the same picture...
7. "at this point the right moving photon is a distance 2vt from R."
Wrong. Do not agree? Then prove your assertion!
8. "What do the topics of "observer at rest" or "moving observer" have to do with the results?"
What results? Yours, I just comment? Nothing! With real description of real situation? A huge 'what'!
9. "What t do we use? Answer: Any convenient time. The SR theorists use SRT time, the non-SRTists use embankment time. These are of no consequence as the different arrival times is the only important parameter here."
BS.
10. "What does the "invarinace principle" have to do with the results?"
Direct, at right description...
11. "Even Yuri used the terms, C + V and C - V in an earlier post to this thread. Is not this use of velocity addition assume a violation of SRT where the relative motion of the frame and light is always and axiomatically supposed to be C?"
c+v and c-v in my solution comes not from SRT and "velocities addition formula, but because of simple fact that ends of legs are moving to and from propagating beams, correspondingly. And namely that has nothing to do with SRT!

Xgen

01-31-05, 06:23 AM

What is so puzzling or slicky about a printout of a time from a clock on a moving platform? Observations aren't voodo they are, in this experiment, simply looking at numbers printed on two pieces of paper.

Xgen, To begin, focus only on the arrival times opf the photons at L and R. Are the arrival times the same for the two cases discussed?

If the clocks are synchronized the light, if reflected back to the midpoint observer, will arrive their simultaneously, but if you read the printout of the arrival times at L and R the left moving photon arrives at L before the right moving photon arrives at R, which is unambiguous physical evidence that the frame is moving to the right.

How can you explain this differenltly?.

Are the arrival times the same for the two cases discussed? -

Answer: No, they arrive at differen times. I cant imagine how they can arrive at the same time in a frame moving with velocity V. Thus can happen only if light is moving with c+v and c-v velocity WRT rest frame (embankment), this means that if light arives at the same times in L and R then either it velocity in embankment had been bigger then c, or its velocity in the moving frame had been biggest then c. I am ready to bet that if such experiment is maded light will arive in both frames at different times. This is the only way for c to remain constant in all frames of reference.

Accualy light is moving always with c and it do not know that there is a observer moving with v. But , yes, light velocity can appear bigger then c. If we, wrongly, conclude that light photons had passed distance of 1 meter before to arrive at L and R, we will measure that light velocity had been c-v wrt L and v+c wrt R. But that is because we had maded the wrong conclusion that paths are equal.

Anyway if we measure time difference in arrival times at L and R, this would mean that we are on a non-stationary wrt vacuum or absolute space frame. We can find velocity magnitude and direction by this time shift. Isn't that simple!

SRT had maded the unclear postulate that light is moving with c WRT alll frame of reference. Is that a reality or observation (measurement)? Why Einstein had postulated that there is no absolute space when that is the only way invariance principle to be valid? I think that they had not been clear about it and that is why in SRT there is only about what will be observed and not what is the reallity. Brilliant, in this way SRT is valid in both cases, and useless.

Because all the natural conclusion from invariance principle is the existence of the absolute space. The relativity theory had gone on the wrong way from this point and had mislead science for almost a century! One day it will be appreciated as the biggest Illusion in the whole history of science.

I am sure that if L-R experiment is realized correctly, taking in acount only uni-directional movement of light, it will show that light will arrive at different times at the L and R observer. This is an imediate prove for the existence of Absolute Space. Introduction of this concept will make SRT relevant and self-consistent.

superluminal

01-31-05, 08:40 AM

Yuiry:

I said:

I will reword my statement:
If I (the guy on the ship, moving with the apparatus) found that the two clocks read differently in the spaceship, then I would calculate a different "speed of light" for each photon, given that I measure a 1m length for each leg of the apparatus. This would be in violation of everything I understand about SR. In the ship, nothing appears odd to me, therefore I will NOT see different times on my clocks.

You said:

AFTER SUCH REWORD OF YOUR INITIAL STATEMENT you are talking about Case v=0. The unswer is trivial: clocks rested in regard to you and point M will read the same times. And nothing will be "in violation of everything I understand about SR."
But we are talking about comparison of cases 1. and 2., remamber?

Jesus H. Christ on a stick! Yuiry, you simply don't understand english well enough to see that I am agreeing with you! In english we commonly make a statement "this result would indicate something-or-other..." (which is clearly false) and then state "therefore, the result would be clearly untrue..."
Do you not have the same language tools in russian?

Geistkiesel:

I understand your frustration with me. I am working on a diagram in an attempt to show physically how these results come about and can be consistent. I am certain however that what Yuiry and every mainstream SRT researcher knows to be true is indeed true based on experimentally verified theory. Please bear with me.

I like to restate the basic issue occasionally to make sure we're all still addressing the same problem:

I (and Yuiry, whether he understands my phrasing or not) claim:

1) In the moving frame, an observer in that frame will be at rest wrt to the light source and will see both photons reach the clocks simultaneously.

2) An outside observer (on an embankment) will see the rear photon hit first, and the forward photon hit later.

3) The question is: If the clocks in the moving frame record the same one-way travel time, (a printout) how is this reconciled with what the observer on the embankment sees?

Intuition says this is BS but relativity and quantum physics is full of counterintuitive results that are nevertheless experimentally verified.

superluminal

01-31-05, 08:57 AM

superluminal

01-31-05, 02:37 PM

Please go here:

http://origins.colorado.edu/~ajsh/sr/paradox.html

Ponder and we'll discuss.

superluminal

01-31-05, 02:39 PM

Please read:

http://www.phatnav.com/wiki/index.php?title=Emitter_theory

fo3

01-31-05, 04:58 PM

1) In the moving frame, an observer in that frame will be at rest wrt to the light source and will see both photons reach the clocks simultaneously.
2) An outside observer (on an embankment) will see the rear photon hit first, and the forward photon hit later.

I agree on this too. Its the first thing that any book trying to explain SRT makes clear. (everything else in this thread has just been confusing me)

3) The question is: If the clocks in the moving frame record the same one-way travel time, (a printout) how is this reconciled with what the observer on the embankment sees?
Yuriy, any explanation on this? I'm counting on you, since so far you've succeeded in explaining all questions regarding SRT. (for me, at least.)

geistkiesel

01-31-05, 08:40 PM

Let us analyse this your post too...

1. "Each photon moves independently of the frame, moving or otherwise."
Right.

2. "In whatever time base one uses, each photon will move the same distance in that time span, relative to the point in space from which they emitted.."
Wrong. Do not agree? Then prove your assertion!

If we use the moving frame clocks each photon, being emitted simultaneously, from the moving frame, will travel the same distance in the same time. The speed of light is constant and the photons are symmetrically arranged. What else do you need?

More specifically, explain your disagreement.

3. "neither photon knows that there are L and R clocks ahead."
Right.

4. "in the 'frame moving to the right case', the L clock is closing (colliding), with the left moving photon while, the right moving photon is chasing the R clock."
For what observer? That one who sits on the moving point M sees nothing such a picture...

No observer is privy to the motion and arrival times of he photons until they reach the clocks. Therefore, the experiment is "observer free". Considering just the emission of the photons from a point in space, which remains invariant, the photons are moving left and right. Each knows nothing of the other or of the frames, moving or stationary. From the emitted point when the left photon moves ct then so does the right photon. The location of the frame is arbitrary and as it is moving, wrt the stationary frame, the photons collide and chase the L and R clock respectivley. The moving observer can examine the data located at the clocks with the stationary observer after the frame stops, or the moving observer can forward the data. or the data can be tansmitted to all observers, but only after the clocks have data to broadcast.

The dynamics of the photon motion is independent of observers and frames, correct? The experiement can be duplicated exactly should the clocks, L and R be placed where expected conditioned on the velocity of he frame. Even assuming time dilation the fact of the different arrival times cannot be eaqsed by application of any theory.

5. "When the left moving photon strikes L after travelling ct, the right moving photon has also travelled ct, in the opposite direction."
This is repetition of previous statement...
6. "when the L clock records the left moving photon arrival the right moving photon has not yet reached R."
Again, repetition of the same picture...
7. "at this point the right moving photon is a distance 2vt from R."
Wrong. Do not agree? Then prove your assertion!

The photons have moved ct each, the frame has moved vt to the right. Draw it out on a piece of paper. you can prove the statement yourself.

Yuriy,
When the left photon strikes L where is the right moving photon wrt R?

8. "What do the topics of "observer at rest" or "moving observer" have to do with the results?"
What results? Yours, I just comment? Nothing! With real description of real situation? A huge 'what'!

The results being the photons arrive at different times at L and R, otherwise the photons will not arrive simultaneously back at the frame midpoint should the clocks also be mirror reflectors.

So if it is so huge spell it out.

I claim the observers have nothing to do with the measurments and that only the clocks and the arrival times of the photons are significant. You aren't going to state that the SRT affect as you see it is purely a generation of the mental condition of observers are you?

9. "What t do we use? Answer: Any convenient time. Frame, embankment whatever. The SR theorists use SRT time, the non-SRTists use embankment time. These are of no consequence as the fact of different arrival times is the only important parameter here."
BS.

Explain your rationalization for the "BS" response.

The fact of different arrival times undermines any time dilation construct as an explanation for the affect.

(Note I edit my phrase quoted by Y immediately above).

10. "What does the "invariance principle" have to do with the results?"

Direct, at right description...
Please explain "Direct, at right description.'

11. "Even Yuri used the terms, C + V and C - V in an earlier post to this thread. Is not this use of velocity addition assume a violation of SRT where the relative motion of the frame and light is always and axiomatically supposed to be C?"
c+v and c-v in my solution comes not from SRT and "velocities addition formula", but because of simple fact that ends of legs are moving to and from propagating beams, correspondingly. And namely that has nothing to do with SRT!

This what I havwe been saying, the experiment has nothoing to do with SRT.

OK. If the ends of the legs are moving to and from propagating photons that were emitted simultaneoulsy in the frame moving to the right from a common emission point, which end of the frame will the photons reach first, L or R?

Geistkiesel

geistkiesel

01-31-05, 08:49 PM

About 1/3 down the page. (Simultenaeity)

http://instruct1.cit.cornell.edu/courses/astro101/lec21.htm

Superluminal, This is the phrase you are referring to ?
"An observer on the train sees the light reach the front and back simultaneously. "

What is ambiguouis and incorrect is the fact that the observers see only the arrival of the light at L and R and only from the anaysis or viewing of data. Neither observer can possibly see the arrival of both photons simultaneously because the observers cannot be at both ends of the frame at the same time.

Also Superluminal your reference is only a statement of theory and is void of any physical analysis. I am beginning to lose you here for this reason.

Geistkiesel

geistkiesel

01-31-05, 08:56 PM

geistkiesel

1.
If he is a smart guy and has learn the classic Physics – never!
2.
Why should he connect such a conclusion … with arriving times of two photons simultaneously emitted in some another reference frame?

The photons were emitted in the moving frame.

No how! There is no way of logic to such a stupid conclusion!
All what you written below this your question (in the same post) is BS. The same concerns your next post, sorry…

3.
When I read your anti-SRT posts – any day is sad in the same measure – you again did not learn anything!…

P.S. This my respons concerns only posts on page 1 of this thread. The page 2 I will comment later...
The sad statement was a joke Yuriy.

Do they have jokes on your planet? Have you concluded investigation and research, or have you terminated future learning of the physical world?

Geistkiesel

Yuriy

01-31-05, 09:03 PM

Dear fo3,
Quoting superluminal’s statement
: If the clocks in the moving frame record the same one-way travel time, (a printout) how is this reconciled with what the observer on the embankment sees?
you are asking me:
Yuriy, any explanation on this? I'm counting on you, since so far you've succeeded in explaining all questions regarding SRT. (for me, at least)
Honestly speaking, I have already posted (and not once) everything, what one needs to explain such type problems in frames of SRT. And we have discussed almost the same problems with geistkiesel not once. Why all those my efforts do not “work”? By very simple reason: people, and especially those that do not accept SRT … because of luck of learning do not follow the basic rules of discussion about Relativity (not only SRT, but any Relativity). And the most important rule is the following one. As far we have recognized that the cozy “Phantom of Instant Action on the Distance” is gone (vanished, get "kaput", excluded of Nature), as each time we are speaking on relative notions we have to refer to the observer (i. e. reference frame, RF) in regard to which this notion is determined.
I told that to geistkiesel several times personally, and with what result? He now proclaims that … there is no need to mention observer at all! I am not surprised that he will come to any, as crazy as possible…’conclusions’!

But right now I will speak with you.

What actually is the problem we are talking about? Let me set it up in right way.
Let us consider two identical devices: the base of length 2S with equal legs L and R has the source of light in middle point M and two receivers of light on the ends of each leg, L and R.
Both devices are synchronized at the beginning: their clocks are put on zero readings at moment when their middle points M are coincided.
Let us consider one device is rested in Laboratory RF, and another one is moving in respect Laboratory with a constant velocity along direction of their bases.
(I hope, everybody understands what experimental situation I have described).
Let us now to do our experiment.
When both devices were at coincidence of their middle points, M-s, (the moment of synchronization of all clocks in both devices), the emitters in both middle points sent the light’s beams to the receivers on both ends of legs of the corresponding devices. So, for each device we have left beam that is rushing to the left end of device, and the right beam that is rushing to the right end of device, respectively in both devices.

Now we have the following series of questions:
#1. What will be the readings of clocks in Laboratory when beams of the rested device will reach the receivers of the rested device?
#2. What will be the readings of clocks in Laboratory when beams of the moving device will reach the receivers of the moving device?
#3. What will be the readings of clocks in the moving device when beams of the rested device will reach the receivers of the rested device?
#4. What will be the readings of clocks in the moving device when beams of the moving device will reach the receivers of the moving device?
(I hope, everybody will understand these simple and clear questions)

The SRT gives the simple and clear answers on each of these questions and they are the following.
Answer on the question #1:
1. Laboratory clocks will read time
T1 = S/c
when the left beam in the rested device will reach the left receiver of the rested device;
2. Laboratory clocks will read time
T2 = S/c
when the right beam in the rested device will reach the right receiver of the rested device;
Conclusion: according to Laboratory clocks both beams of the rested device are reaching the receivers of the rested device simultaneously.
Answer on the question #2:
3. Laboratory clocks will read time
t1 = γS/(c+v)
when the left beam in the moving device will reach the left receiver of the moving device;
4. Laboratory clocks will read time
t2 = γS/(c-v)
when the right beam in the moving device will reach the right receiver of the moving device;
Here γ = (1- v²/c²)^½. It appears in this formula because Laboratory observer sees the legs of the moving device shrunk in γ times!

Conclusion: according to Laboratory clocks both beams of the moving device are reaching the receivers of the moving device not simultaneously: first the left beam reaches the left receiver, and then right beam reaches the right receiver.
Answer on the question #3:
5. The moving clocks will read time
T1’ = γS/(c-v)
when the left beam in the rested device will reach the left receiver of the rested device;
6. Laboratory clocks will read time
T2’ = γS/(c+v)
when the right beam in the rested device will reach the right receiver of the rested device;
Here γ again is (1- v²/c²)^½. It appears in this formula because moving observer sees the legs of the laboratory device shrunk in γ times!
Conclusion: according to moving clocks both beams of the rested device are reaching the receivers of the rested device not simultaneously: first the right beam reaches the right receiver, and then left beam reaches the left receiver.
Answer on the question #4:
7. Laboratory clocks will read time
t1’ = S/c
when the left beam in the moving device will reach the left receiver of the moving device;
8. Laboratory clocks will read time
t2’ = S/c
when the right beam in the moving device will reach the right receiver of the moving device;
Conclusion: according to moving clocks both beams of the moving device are reaching the receivers of the moving device simultaneously.
All what we need to accomplish our answers is the notice that any observer (in any inertial reference frame) will give the same answers on our main four questions!
For instance, if we ask the Laboratory observer to answer on question #4, he will tell as the following:
“I see the end of the right leg of the moving device as a point that is moving by the law
XR= vt + γS
and the end of the left leg of the moving device as a point that is moving by the law
XL= vt - γS
Therefore if I will use the Lorentz transformation formulas
T’ = (t – vx/c²) / γ
I will get
TR’ = (γ²tR – γS v/c²)/ γ = γtR – S v/c²
for the readings of the moving clocks when the right beam is reaching the right receiver of the moving device, and
TL’ = (γ²tL + γS v/c²)/ γ = γtL + S v/c²
for the readings of the moving clocks when the left beam is reaching the left receiver of the moving device.
But
tR = t1 = γS/(c-v) and tL = t2 = γS/(c+v) (see my answer on question #2).
Therefore, I predict that
TR’ = γt1 – S v/c² = γ²S/(c-v) – S v / c² = (S/c)* (1 + v/c – v/c) = S/c
and
TL’ = γt2 – S v/c² = γ²S/(c+v) + S v / c² = (S/c)* (1 – v/c + v/c) = S/c
I.e. I predict that the moving clocks will read equal times for both events”.

As one can see, the Laboratory observer confirms that the clocks of moving device will read the same times for events “each beam reached its receiver”, i.e. confirms that the moving observer sees its beams reaching the receivers simultaneously. Just as our answer #4 is saying! And it will be so for any third observer. Because the Lorentz transformations form the group!

Now my dear friend, knowing and understanding what SRT says, try to find any sense in all accusations of SRT you have read in the posts of geistkiesel and his supporters….

superluminal

01-31-05, 10:01 PM

Geistkiesel wrote:
Also Superluminal your reference is only a statement of theory and is void of any physical analysis. I am beginning to lose you here for this reason.

The physical analysis of the solution to this whole issue is here. I think the key concept is that the "hypersurfaces of simultaneity" for each observer are different. This is just the way it is.

http://origins.colorado.edu/~ajsh/sr/centre.html

The results for each observer within his/her own frame are completely valid and make sense to them. In our example, the "printouts" from the ship will read that the photons arrived simultaneously, because the measurement was made in a self-contained frame. We, on the "embankment" might violently disagree based on what we observe, unless we understood the effects of relativity on simultenaeity.

If this does not reprersent enough of a physical analysis, then I am sorry. I know no better. Certain things are inevitable consequences of observed phenomena. Light is in fact measured to be c for any observer, independent of their motion. All valid experiments I know of show this.

The nature of light itself along with the classic double-slit experiment is another example of something that exhibits counterintuitive behavior, and has no good "physical analysis" that will show a person it is valid - like handing someone a rock and saying "here's a rock, understand?" "Yep, no problem." Things are just the way they are in the realms far outside our normal experience. Should we expect to be able to have an intuitive grasp of every phenomenon we encounter? I don't believe so.

Yuriy

01-31-05, 10:08 PM

Wouh, superluminal.
When you want you can be so clear and straight speaking that even I ... can understand what you just said...
And you just said wonderful speach. Bravo!

superluminal

01-31-05, 10:13 PM

High praise indeed! Thanks!

geistkiesel

01-31-05, 10:33 PM

geistkiesel

01-31-05, 11:31 PM

Dear fo3,
Quoting superluminal’s statement

you are asking me:

Honestly speaking, I have already posted (and not once) everything, what one needs to explain such type problems in frames of SRT. And we have discussed almost the same problems with geistkiesel not once. Why all those my efforts do not “work”? By very simple reason: people, and especially those that do not accept SRT … because of luck of learning do not follow the basic rules of discussion about Relativity (not only SRT, but any Relativity). And the most important rule is the following one. As far we have recognized that the cozy “Phantom of Instant Action on the Distance” is gone (vanished, get "kaput", excluded of Nature), as each time we are speaking on relative notions we have to refer to the observer (i. e. reference frame, RF) in regard to which this notion is determined.
I told that to geistkiesel several times personally, and with what result? He now proclaims that … there is no need to mention observer at all! I am not surprised that he will come to any, as crazy as possible…’conclusions’!

But right now I will speak with you.

What actually is the problem we are talking about? Let me set it up in right way.
Let us consider two identical devices: the base of length 2S with equal legs L and R has the source of light in middle point M and two receivers of light on the ends of each leg, L and R.
Both devices are synchronized at the beginning: their clocks are put on zero readings at moment when their middle points M are coincided.
Let us consider one device is rested in Laboratory RF, and another one is moving in respect Laboratory with a constant velocity along direction of their bases.
(I hope, everybody understands what experimental situation I have described).
Let us now to do our experiment.
When both devices were at coincidence of their middle points, M-s, (the moment of synchronization of all clocks in both devices), the emitters in both middle points sent the light’s beams to the receivers on both ends of legs of the corresponding devises. So, for each device we have left beam that is rushing to the left end of devise, and the right beam that is rushing to the right end of device, respectively in both devices.

Now we have the following series of questions:
1. What will be the readings of clocks in Laboratory when beams of the rested device will reach the receivers of the rested device?
2. What will be the readings of clocks in Laboratory when beams of the moving device will reach the receivers of the moving device?
3. What will be the readings of clocks in the moving device when beams of the rested device will reach the receivers of the rested device?
4. What will be the readings of clocks in the moving device when beams of the moving device will reach the receivers of the moving device?
(I hope, everybody will understand these simple and clear questions)

The SRT gives the simple and clear answers on each of these questions and they are the following.
Answer on the question #1:
1. Laboratory clocks will read time
T1 = S/c
when the left beam in the rested device will reach the left receiver of the rested device;
2. Laboratory clocks will read time
T2 = S/c
when the right beam in the rested device will reach the right receiver of the rested device;
Conclusion: according to Laboratory clocks both beams of the rested device are reaching the receivers of the rested device simultaneously.
Answer on the question #2:
3. Laboratory clocks will read time
t1 = γS/(c+v)
when the left beam in the moving device will reach the left receiver of the moving device;
4. Laboratory clocks will read time
t2 = γS/(c-v)
when the right beam in the moving device will reach the right receiver of the moving device;
Here γ = (1- v²/c²)^½. It appears in this formula because Laboratory observer sees the legs of the moving device shrunk in γ times!

Conclusion: according to Laboratory clocks both beams of the moving device are reaching the receivers of the moving device not simultaneously: first the left beam reaches the left receiver, and then right beam reaches the right receiver.
Answer on the question #3:5. The moving clocks will read time
T1’ = γS/(c-v)
when the left beam in the rested device will reach the left receiver of the rested device;
6. Laboratory clocks will read time
T2’ = γS/(c+v)
when the right beam in the rested device will reach the right receiver of the rested device;
Here γ again is (1- v²/c²)^½. It appears in this formula because moving observer sees the legs of the laboratory device shrunk in γ times!
Conclusion: according to moving clocks both beams of the rested device are reaching the receivers of the rested device not simultaneously: first the right beam reaches the right receiver, and then left beam reaches the left receiver.
Answer on the question #4:7. Laboratory clocks will read time
t1’ = S/c
when the left beam in the moving device will reach the left receiver of the moving device;
8. Laboratory clocks will read time
t2’ = S/c
when the right beam in the moving device will reach the right receiver of the moving device;
Conclusion: according to moving clocks both beams of the moving device are reaching the receivers of the moving device simultaneously.
All what we need to accomplish our answers is the notice that any observer (in any inertial reference frame) will give the same answers on our main four questions!
For instance, if we ask the Laboratory observer to answer on question #4, he will tell as the following:
“I see the end of the right leg of the moving device as a point that is moving by the law
XR= vt + γS
and the end of the left leg of the moving device as a point that is moving by the law
XL= vt - γS
Therefore if I will use the Lorentz transformation formulas
T’ = (t – vx/c²) / γ
I will get
TR’ = (γ²tR – γS v/c²)/ γ = γtR – S v/c²
for the readings of the moving clocks when the right beam is reaching the right receiver of the moving device, and
TL’ = (γ²tL + γS v/c²)/ γ = γtL + S v/c²
for the readings of the moving clocks when the left beam is reaching the left receiver of the moving device.
But
tR = t1 = γS/(c-v) and tL = t2 = γS/(c+v) (see my answer on question #2).
Therefore, I predict that
TR’ = γt1 – S v/c² = γ²S/(c-v) – S v / c² = (S/c)* (1 + v/c – v/c) = S/c
and
TL’ = γt2 – S v/c² = γ²S/(c+v) + S v / c² = (S/c)* (1 – v/c + v/c) = S/c
I.e. I predict that the moving clocks will read equal times for both events”.

As one can see, the Laboratory observer confirms that the clocks of moving device will read the same times for events “each beam reached its receiver”, i.e. confirms that the moving observer sees its beams reaching the receivers simultaneously. Just as our answer #4 is saying! And it will be so for any third observer. Because the Lorentz transformations form the group!

Now my dear friend, knowing and understanding what SRT says, try to find any sense in all accusations of SRT you have read in the posts of geistkiesel and his supporters….

This is all very interesting Yuriy. It is too sad that you cannot answer without using your off the shelf formulae. Can you describe the situation from 1st principles?

Now put the clocks in the rest frame at the exact point the light reaches L and R in the moving frame. this can be done from experiment. There is no physical way the lights can reach the L and R clocks simultaneously in the moving frame. If this were so then the lights would not arrive at the midpoint simultaneously were the clocks on the moving frame also mirrors. Also the ligts will have taken different path lengths in order for the lights to arrive simultaneously at L and R in the moving frame.

There is no need to run the experiments together. Run the stationary frame experiment first with the observer of the moving frame a passive observer in the rest frame. This observer then boards the frame and the moving experiment is conducted. Therefore the moving observer has prior information that the photons arrive at L and R simultaneously in the stationary frame. This observer also knows the stationary frame actual times. To be otherwise is to impose unnatural attributes on the motion of light which is completely independent of the motion of the frames. This is a true statement is it not Yuriy?

Now we have the following series of questions:
1. What will be the readings of clocks in Laboratory when beams of the rested device will reach the receivers of the rested device?
Both rest frame clocks read the same.

2. What will be the readings of clocks in Laboratory when beams of the moving device will reach the receivers of the moving device?
The rest frame clocks, if stationed where the moving frame clocks are located as the moving frame passes, will be in sequence, 1st the L clock is triggered, then the R clock.

3. What will be the readings of clocks in the moving device when beams of the rested device will reach the receivers of the rested device?
The beams used are he same for both frames of reference.
If the clock data in the rest frame are made available to the moving frame, the light arrives at L and R as: The photons reach L in the rest fframe after the photos reach L in the moving frame. The photons reach R in the rest frame before the photons reach R in the moving frame.

4. What will be the readings of clocks in the moving device when beams of the moving device will reach the receivers of the moving device?
(I hope, everybody will understand these simple and clear questions)
See 3 above. The beams are the same in both cases. The beams wwill register arrival at L before the beam arrives at R.

The times do not cancel. The total time for the beams to reac L and R in the rest frame is t = S/c. The time when both photons have reached their especive clocks in the moving frame is t = S/c + t'

When the photons arrive at L the right photon is 2vt from R. In order for the the photon to reach R it must pass through 2vt plus the small distance the frame moves during this time, or

ct' = 2vt + vt', or

t' = t(2v)/(C - V),

As we can see this t' is 0 when V = 0 and t' > 0 when V >0.

This time "discrepency" accounts for all the SRT time differences without unnecessary SRT calculations. It does take longer for he moving frame to complete he activities that are completed in the rest frame, hence, absoluite motion, absolute time are alive and well..

Observes aren't necessary or relevant to the solution of this problem.

Geistkiesel

MacM

01-31-05, 11:48 PM

Please read:

http://www.phatnav.com/wiki/index.php?title=Emitter_theory

Interesting discussion. Something came to mind having read this in your referenced link regarding Emitter Theory and Pion testing.

In the 1960s, it became practical to test this theory. Particles called neutral pions were accelerated to near the speed of light in a particle accelerator, and the speed of the photons emitted by decay of those particles was measured. The speed was found to be exactly the same as that of light emitted by the decay of stationary particles.

I have read of that test before. Giving it some thought one must conclude that the measurement of the speed of light is being made relative to the lab observer. That would mean we should have also seen that the forward moving photon had a velocity of (c-v) relative to the pion and the trailing photon would have been (c+v) to the pion, from our perspective.

That is because if you argue that the photon had a velocity of "c" relative to its source, the pion, then it would have had a velocity of almost 2c (forward) and rest (trailing) relative to the lab observer.

The other issue which really stood out was that setting aside all the fiat or text book descriptions of light, the fact that the light appeared to have a velocity of 'c' in all directions, even for a relavistically moving source, really matches the observation of no lateral motion of the source raised Here.

That is one could seem to state: "Photons are the consequence of some instantaneous stimulation or pertabation of a "OMG" absolute background linked to the observer".

That is it carries no momentum from its source's velocity, lateral or parallel.

Thoughts or comments?

geistkiesel

01-31-05, 11:55 PM

Michelson and Morley were looking for the effect, on light, of the postulated "Aether". In the process, they found not only was there no Aether, but that the speed of light is completely independent of the motion of the source and observer. Everybody gets the same result.

Here is a cool flash app:
http://galileoandeinstein.physics.virginia.edu/more_stuff/flashlets/mmexpt6.htm
Superluminal,

The followiwng was posted by MacM and shows that lateral motion of emitted photns does not occur. This is a physical condition assumed by MM, that is thta light has a lateral momentum component.
http://www.physicsnews1.com/article_12.html

Therefore, MM as an expefriment must be properly conducted redone.

Geistkiesel.

MacM

02-01-05, 12:41 AM

As far we have recognized that the cozy “Phantom of Instant Action on the Distance” is gone (vanished, get "kaput", excluded of Nature),

I can't believe you said this or meant this.

Are you saying you reject "Particle Entanglement"? The testing and data seem rather strong in favor of that unique feature of nature. What is your basis for such rejection? (Other than gut belief).

geistkiesel

02-01-05, 05:43 AM

The physical analysis of the solution to this whole issue is here. I think the key concept is that the "hypersurfaces of simultaneity" for each observer are different. This is just the way it is.

http://origins.colorado.edu/~ajsh/sr/centre.html

The results for each observer within his/her own frame are completely valid and make sense to them. In our example, the "printouts" from the ship will read that the photons arrived simultaneously, because the measurement was made in a self-contained frame. We, on the "embankment" might violently disagree based on what we observe, unless we understood the effects of relativity on simultenaeity.
You copied the wordes "hypersurface of simultaneity" as if you understood the words,. Earlier you professed basic ignorance of all the SRT stuff. What are you Superluminal, a propagandist?

"Because the measurements were made in a self contained frame" You used this as justification for the special relativity conclusions you made. What does the self contained frame have to do with photon velocity and clocks?

Why do you use he phrase that "the results ...make sense to them"? what does this have to do with photons of light moving through space and the clocks that measure the time of arrival of the photons? The photons know nothing of the frame or the frame motion. The clocks are synchronized. When the clocks and the point of emission of the photons is invariant the clocks measure simultaneous arrival of the photons only in thje rest frame.

Now Superluminal consider for an onstant that the point of emission of the photons is invariant in the sense that when a photon is emitted from a small photon emitter in deep space the point of emission does not move, because the point is not a thing it is just a point where the photon was emited. Let us spray a light vapor in space and emit a dozen photon simultaneously in a dozen directions. Th trajectory of all the photons are straight linea and all the lines meet at the point the photons were emitted, correct? The answer is of course the trajectory lines meet at the point of emission.

Read this for your information that the point of the emitted photons is invariant. (http://www.physicsnews1.com/article_12.html)

Back on earth when the photons are emitted that point of emission which coincidentally is at the midpoint of the clocks defines an invarinat point in space. The photons that move can be referenced to that point and ergo the clocks can be referenced to that point as can the frame in its entirety. The light and photons will meet as we expect and as Yuriy properly acknowledged when he gave his C -V and C + V terms earlier in the thread. At what point does the observers have abn affect on the clocks and the photons?

You are saying that the frame motion has no affect on the arrival sequence and is effectively identical to the stationary frame experiment where "every one agrees". You are concluding this are you not? If the observer on the moving frame knows the frame is moving will the recorded arrival times vary?

You iuse a simultaneity argument in your referenced link that has only the opinion of that author who created the link's opinion. All matters in that link are phrased in what the observers see and do not address the issue of the photons moving with respect to a point in space that can be located in our scenario where photons and frame are both referenced.

A slight variation.
If the photons are emitted (or reflected) simultaneously from L and R in the moving frame will the photons meet at the midpoint simultaneously also? If a point on the embankment is colocated with the moving frame midpoint just as the photons are reflected what will be the result?
Will he photons arrive simultaneously at the two midpoints?

The nature of light itself along with the classic double-slit experiment is another example of something that exhibits counterintuitive behavior, and has no good "physical analysis" that will show a person it is valid - like handing someone a rock and saying "here's a rock, understand?" "Yep, no problem." Things are just the way they are in the realms far outside our normal experience. Should we expect to be able to have an intuitive grasp of every phenomenon we encounter? I don't believe so.

What does your belief have to do with photon motion and clocks?

What is counterintuitive about the double slit experiment? Just becaue you say it is counterintuitive to you then you are saying it must be counterintuitive to everyone else on the planet correct? And you are using this time worn and hardly original statement as if you just thought it up. You say you aren't a physicist, well I believe you, just what are you then? And why are you here? Simple personal interest?

I will say that you are one slick dude, with a wide eyed innocence of just wanting to know the truth, confused, but rescued by the logic of Yuriy ,which nevertheless frustrates you when he is unable to see that you support his position as if this matter were for a jury to decide. But you aren't so slick as to remain undetectable.

Geistkiesel

geistkiesel

02-01-05, 05:49 AM

MacM, The thoughts are starting to generate. I will post something soon. Thaks for the reference and for E. Skyler's work also.
Geistkiesel.

geistkiesel

02-01-05, 06:24 AM

geistkiesel,
I have to emphasize for the record:
in my answers on your questions v meant exactly what you have defined in your initial post - WE see once v =0, and in another case the same WE see v>0. So v is velocity of your device in respect to WE (I hope, you meant WE as a simple Laboratory Reference Frame.) So in my post there was no trace of any mysterious Absolute Space, Absolute Time or Absolute Motion...
I am not quite sure of your point but yes t v is wrt the stationary frame.

geistkiesel

02-01-05, 06:49 AM

Yes.
Everneo, your yes as lonely as it sits might be more useful with some explanation. What of the condition where the photons are reflected simultaneously from L and R back to the midpoint. Will the photons arrive at the midpoint simultaneously in the moving frame? This would follow from, your conclusion that the lights arrive at L aqnd R at the same time, woul it not?

If the photons arrive at the clocks simultabneousl then the light would reflect back tot he midpoint simultaneously?

Aren't you saying that all moving frame clocks will always record simultaneous arrival times of photons emitted simultaneously from the midpoint of the frame? If as Yuriy noted, L and the left moving photon are in a collison course and the right moving photon is chasing R, then how do the photons arrive simultaneously at L and R?

You must conclude from your answer that the photons emitted simultaneously must necessarily travel the same distance in the same time if they are to arrive at the L and R clocks simultaneously. The clocks are sycnchronized and the distances between the clocks and the emission point are identical, yet the photons still arrive simultaneously in the moving frame?

Is this your final answer?
Geistkiesel

geistkiesel

02-01-05, 07:03 AM

No.
Exact quantitative answer I already gave.
Anyone who does not understand this answer never will understand SRT. Period.
Yuriy, Please make up your mind. Either the photons arrive simultaneously in the moving frame experiment or they do not, which is it?
If I have misread your posts then please forgive this note with my apologies.
Geistkiesel

geistkiesel

02-01-05, 07:23 AM

Are the arrival times the same for the two cases discussed? -

Answer: No, they arrive at differen times. I cant imagine how they can arrive at the same time in a frame moving with velocity V. Thus can happen only if light is moving with c+v and c-v velocity WRT rest frame (embankment), this means that if light arives at the same times in L and R then either it velocity in embankment had been bigger then c, or its velocity in the moving frame had been biggest then c. I am ready to bet that if such experiment is maded light will arive in both frames at different times. This is the only way for c to remain constant in all frames of reference.

Accualy light is moving always with c and it do not know that there is a observer moving with v. But , yes, light velocity can appear bigger then c. If we, wrongly, conclude that light photons had passed distance of 1 meter before to arrive at L and R, we will measure that light velocity had been c-v wrt L and v+c wrt R. But that is because we had maded the wrong conclusion that paths are equal.

Anyway if we measure time difference in arrival times at L and R, this would mean that we are on a non-stationary wrt vacuum or absolute space frame. We can find velocity magnitude and direction by this time shift. Isn't that simple!

SRT had maded the unclear postulate that light is moving with c WRT alll frame of reference. Is that a reality or observation (measurement)? Why Einstein had postulated that there is no absolute space when that is the only way invariance principle to be valid? I think that they had not been clear about it and that is why in SRT there is only about what will be observed and not what is the reallity. Brilliant, in this way SRT is valid in both cases, and useless.

Because all the natural conclusion from invariance principle is the existence of the absolute space. The relativity theory had gone on the wrong way from this point and had mislead science for almost a century! One day it will be appreciated as the biggest Illusion in the whole history of science.

I am sure that if L-R experiment is realized correctly, taking in acount only uni-directional movement of light, it will show that light will arrive at different times at the L and R observer. This is an imediate prove for the existence of Absolute Space. Introduction of this concept will make SRT relevant and self-consistent.
Xgen,
I think I misread your post , but am now back on track with you. I t appears that SRT denies the reality of motion, such that in space ships that necessarily accelerated to get where they are the ships must be moving so there is nothing special about detecting different arrival times and thereopfre motion, absolute as it is, for observers on space ships. The only rational way in which an observer can assert an "at rest: consifiton is if he measured a deceleration of his space ship.

Geistkiesel.

everneo

02-01-05, 08:06 AM

Everneo, your yes as lonely as it sits might be more useful with some explanation. What of the condition where the photons are reflected simultaneously from L and R back to the midpoint. Will the photons arrive at the midpoint simultaneously in the moving frame? This would follow from, your conclusion that the lights arrive at L aqnd R at the same time, woul it not?
Yes.
If the photons arrive at the clocks simultabneousl then the light would reflect back tot he midpoint simultaneously?
Yes.

Aren't you saying that all moving frame clocks will always record simultaneous arrival times of photons emitted simultaneously from the midpoint of the frame? If as Yuriy noted, L and the left moving photon are in a collison course and the right moving photon is chasing R, then how do the photons arrive simultaneously at L and R?
The total distance travelled by both the photons are same when they meet at the mid point again.

You must conclude from your answer that the photons emitted simultaneously must necessarily travel the same distance in the same time if they are to arrive at the L and R clocks simultaneously. The clocks are sycnchronized and the distances between the clocks and the emission point are identical, yet the photons still arrive simultaneously in the moving frame?
In an intertial frame, whether the whole setup is moving with velocity v1 wrt another reference frame RF1 (say earth) or its moving with velocity v2 wrt yet another reference frame RF2 (say venus) etc, it does not matter IF the distance between ML & MR remains same. Accepting that there will be a difference in the time recorded in clocks L & R upon the arrival of photons in L & R clocks is akin to accept measuring 'absolute velocity'. Choosing a specific velocity v to add/subtract with c itself is BS under this setup. Why not v2 (relative velocity with venus as mentioned above) ? Would the time recorded in the clocks L & R change accordingly ?

Sorry Geistkiesel, there is no absolute velocity to be calculated from the time stamps (because they would be same) on the clocks in your setup whether it is moving with v1 or v2 or v3 or at lab.

geistkiesel

02-01-05, 08:10 AM

Yuiry:

I said:

You said:

Jesus H. Christ on a stick! Yuiry, you simply don't understand english well enough to see that I am agreeing with you! In english we commonly make a statement "this result would indicate something-or-other..." (which is clearly false) and then state "therefore, the result would be clearly untrue..."
Do you not have the same language tools in russian?

Geistkiesel:

I understand your frustration with me. I am working on a diagram in an attempt to show physically how these results come about and can be consistent. I am certain however that what Yuiry and every mainstream SRT researcher knows to be true is indeed true based on experimentally verified theory. Please bear with me.

I like to restate the basic issue occasionally to make sure we're all still addressing the same problem:

[quote=Superluminal]I (and Yuiry, whether he understands my phrasing or not) claim:

1) In the moving frame, an observer in that frame will be at rest wrt to the light source and will see both photons reach the clocks simultaneously.
Tyes the observer is at rest wrt his own frame of reference, but the frame is moving and ergo so o=is th eobw=server, he just doesn't know it or so all SRTist wowuld aver.

2) An outside observer (on an embankment) will see the rear photon hit first, and the forward photon hit later.
Correct and so will the moving observer who is moving with his frame which is moving with respect to the stationary frame of reference

3) The question is: If the clocks in the moving frame record the same one-way travel time, (a printout) how is this reconciled with what the observer on the embankment sees?
It is reconciled with an error in the clocks synchronization, the distances between the "midpoint" of the clocks are different, the physical environment in both legs is differnet and the SOL is ergo different, the photons in one of the legs was diverted to a longer optical path, the cfrew rfeading the clocks sbatoged the readings in order to maintain the fallacy of SRT, the times were simply misread, the ink on the oprintourts was smudged whatever.

Intuition says this is BS but relativity and quantum physics is full of counterintuitive results that are nevertheless experimentally verified.

Why not present one of your experiemental proofs verifying SRT?

You should take a close look at your posts, and Yuriy's. You are reconsiled to clinging to SRT to the extent you cannot justify the conclusions you offer without some other effort , but hey we all do this. The conclusions you are rejecting you have admitted are obvious, even intuitive. It seems to me that rejecting the moving frame clocks from recording different arrival times you are accepting a form irrational thought processes as a substitute. It appears that the observers on your ships have no clue to the reality of motion. The ships accelerated from their home planets, calculations for the journey were made, motions and velocitities were calculated, yet when the photons emitted simultaneously from the frame midpoint do not arrive at the L and R clocks at the same time, you conclude that the clocks were in error instead of contemplating the possibility that the ship was moving.
What would you do Superluminal if you were the observer, and just about the time the experimet was conducted on you ship you determined that the clocks showed different arrival times and that you did the saame mental gymnastic trick you stated earlier: that the clocks were in error. Then you received a message from a stationary observer that the lights arrived at L and R at different times as observed by the stationary observer. Would you still "cling" to SRT?

Do ships's observers ever contemplate the fact that the ships acclerated at one time and have not decelerated? If you detrect another ship and make a meauremnt of the relative velocity of you and the ohter ship to mbe 10000 speed units (you aren't supposed to be able top measure the absolute velocity of either ship, just he relative velocity) then what do you do?
If you are a fail]hful SRTist clinging to theory you assume yourself at rest anf theother ship moving right, which is what h eother ship does, right?

What if you adopted the train of thought that it was statistically rare to the extent of near impossibility for either ship to be completely at rest. Would it not be a closer assumption to physical reality to take the average of the relative velocties instead of arbitrarilly assuming an extreme case of either at rest or moving with all the relative velocity that the two frames generated?

What if the two frames came from mearth and each had its velocity calibrated with respect to Ve = 0, the eadrth frame. Your ship was measured at 6000 units, tyhe ther ship at 4000 units and both of you measured 10000 units iof relative velocity in aan apparent collision course. Neither ship had decelerated. Would you still insist on using SRT and assume yourself at v = 0 wrt the oher frame of reference? What if you both exchanged your velocity data as given in the earth frame calibrations? Would you still cling to SRT?

What of Yuriy's errors here? He could have just been distracted when writing his posts so no real damage to this position could be honestly charged, or he could have been confused by the truth, correct?

Observers on space ships do accept the concept of velocity do they not?

All thiose ivariance data you refer to, I'll bet a bundle you haven't seriously looked at the experiment by reading the original papers, have you? Is it blind clibnging Supelunminal?

Geistkiesel

Geistkiesel

geistkiesel

02-01-05, 08:59 AM

Interesting discussion. Something came to mind having read this in your referenced link regarding Emitter Theory and Pion testing.

I have read of that test before. Giving it some thought one must conclude that the measurement of the speed of light is being made relative to the lab observer. That would mean we should have also seen that the forward moving photon had a velocity of (c-v) relative to the pion and the trailing photon would have been (c+v) to the pion, from our perspective.

That is because if you argue that the photon had a velocity of "c" relative to its source, the pion, then it would have had a velocity of almost 2c (forward) and rest (trailing) relative to the lab observer.

The other issue which really stood out was that setting aside all the fiat or text book descriptions of light, the fact that the light appeared to have a velocity of 'c' in all directions, even for a relavistically moving source, really matches the observation of no lateral motion of the source raised Here.

That is one could seem to state: "Photons are the consequence of some instantaneous stimulation or pertabation of a "OMG" absolute background linked to the observer".

That is it carries no momentum from its source's velocity, lateral or parallel.

Thoughts or comments?

So much for the SOL being measured as C from all inertial frames. If as you stated the SOL was determined from the lab frame of reference does it not appear that the measurement of C in all reference frames is not necessarily a C that is SRT imposed?
I mean if SRT has us measure C from all frames of reference, then C as a limiting velocity must be reevaluated, which would require some seriou mind manipulations for of all of us. Just as I smugly conclude I know it all something like your link arrives and I have to humble myself back to a level of human being, what drag that is. Giving up my othewise exalted state of divinity is sometimes a difficult task and very inconvenient to say the least (WHAT DO I TELL MY PUBLIC?). I ask all of you to bear with me from time to time.

Another thought that is aslo sad (for some) , is the loud clanging silence of th SRT industry in integrating the experimental results of these kinds of informaion with their precious theory, even to trash it, but then these experiments that are so close to perturbing SRT don't get brought up in discussison by SRTists, do they? Nope, that's our job and someone has to do the grunt work don't they? So here I go again, late for another dinner. Oh well. I dare not let on how much fun this stuff can be, "they" would make it illegal, sure as hell.

Hey we need the SRT industry. They offer us ample room to exert our o own mental energies and thinking power in the scientific disciplines. We get to go to places we wouldn't otherwise visit because we hadn't been pushed by the dynamics of discourse in the rhetorical exchanges.

There just might be massive space travel available to the common traveler yet. When the speed of sound was overcome theadvancement of aircraft development accelerated. When the "four minute mile" was broken the times shortened considerably and is still going down. I will bet a bundle though that we wont see a 2 minute mile for a long time, but I also predict that human speed will definitely arrive. Thirty miles an hour is nothing. All one needs is a more efficient way in which to intake and process enrgy that is intended for velocity increases instead of providing unintended inner tubes in the form of stored energy, i.e. fat, for dubious uses in emergencies when energy i s scarce.

Thnx MacM, every once in a while I find mysel making scientific progress in learning the nuances of what else is there under Mother Nature's skirt that tempts us to keep peeking other than he well turned ankles that is. The story get deeper as the story unfolds, but the cobwebs of past dogma getting purged seems very relieving like a weight lifted from the shoulders, like smoking marijuana for the first time as was described to me by a pot head friend of mine. He told me that after his first high his thoughts were, "This stuff shouldn't be illegal".
Geistkiesel

Yuriy

02-01-05, 09:17 AM

everneo

02-01-05, 09:45 AM

What I can say, reading the last page of this thread?
Is our Forum a some kind of a confessional ... in Madhouse? What all this has to do with any "frontier physics", education, or even "a discussion"?
Oh, yes, I forgot "the Freedom of Speech"...
In this 4th page only geistkiesel & myself posted so far apart from you. If you are not specific then i take offence to your comments. Clarify.

Yuriy

02-01-05, 10:55 AM

everneo,
check this link and you will find out who should worry about my post...
http://www.crank.net/day.html
Little by little our forum becomes ... a lectorium of members of crank.net!

geistkiesel

02-01-05, 11:03 AM

In this 4th page only geistkiesel & myself posted so far apart from you. If you are not specific then I take offence to your comments. Clarify.
Everneo, I think Yuriy is referring to my comments regarding the MacM post where he referred to an experiment where photons were measured emitted from pions travelling a v ~ c. There they foind, apparently c+v and c -v values measured from the lab. Read the MacM post a must for every serious SRTist and sissident alike. MacMasked for comments and thoughts and I to drifted off the road a tad. It was 4 or 5 AM my time. I didn't hog anything excessively were not so lengthy , as I view it, that the progress of the thread or the forum was jeoprodized.

And by teh way Evereno, hypothetically speaking where would you prefer youraself in the case where SR was unabkbiguously "terminated" with proof? A die hard embittered with defeat or a recognized part opf a truly momentous event? When SR goes, as I predict the ultimate fate, probably different from yours, are you gracious enough not to admit defeat, but to recognized the error in your understanding of the matter. We are assumed to be mature adults after all. There aren't victors and losers here, only science.?
Go for it, your post I mean, and look out for Y who sneers at "Freedomo of Speech" freaks.
geistkiesel.

geistkiesel

02-01-05, 06:21 PM

You answered yes to the question of whether the photons would arrive at L and R . You answered "Yes".
You anwered 'Yes' to the question whether the photons would arrive simultaneously back at M.

Then the following exchange took place:

Geistkiesel asks:
“ Aren't you saying that all moving frame clocks will always record simultaneous arrival times of photons emitted simultaneously from the midpoint of the frame? If as Yuriy noted, L and the left moving photon are in a collison course and the right moving photon is chasing R, then how do the photons arrive simultaneously at L and R?”

The total distance travelled by both the photons are same when they meet at the mid point again.

IYou did not explain how the photons arrive simultabneously at L and R, even though the LM and RM distances are equivalent, the frame observers assume a state of rest and you arsserted the final trajectories if the photons is the same, which I agree with. You stated at the beginning of this thread that the photons would arrive simultaneously at L and R in the moving frame.

Is this still your answer?

Very specifically: Are you asserting here that any reference to how much faster or slower the frame is moving wrt C is irelavant, or of no physical significance?

Here are Yuriy's figures for the same question which he posted at the beginning of the thread.

1.Case v=0.
Tlm = S/c = Tmr
2.Case v>0.
Tmr = S/(c-v)
Tlr = S/(c+v)

Do you have any comment?

The moving frame is distinguished by Yuriy and in fact he makes two velocity additions in doing so (However, Y did deny his "additions" were of any SRT significance).

In an intertial frame, whether the whole setup is moving with velocity v1 wrt another reference frame RF1 (say earth) or its moving with velocity v2 wrt yet another reference frame RF2 (say venus) etc, it does not matter IF the distance between ML & MR remains same. Accepting that there will be a difference in the time recorded in clocks L & R upon the arrival of photons in L & R clocks is akin to accept measuring 'absolute velocity'. Choosing a specific velocity v to add/subtract with c itself is BS under this setup. Why not v2 (relative velocity with venus as mentioned above) ? Would the time recorded in the clocks L & R change accordingly ?

Everneo,
If I never as