A Purely Hypothetical Question regarding Special Relativity Theory.

Discussion in 'Physics & Math' started by geistkiesel, Jan 29, 2005.

  1. everneo Re-searcher Registered Senior Member

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    You have just slipped for the second time in this thread, Yuiry. Just think over.
     
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  3. Yuriy Registered Senior Member

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    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!
     
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  5. Xgen Registered Senior Member

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    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.
     
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  7. superluminal I am MalcomR Valued Senior Member

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    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:

    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.
     
    Last edited: Jan 31, 2005
  8. superluminal I am MalcomR Valued Senior Member

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  9. superluminal I am MalcomR Valued Senior Member

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  10. superluminal I am MalcomR Valued Senior Member

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  11. fo3 acdcrocks Registered Senior Member

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    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)

    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.)
     
  12. geistkiesel Valued Senior Member

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    2,471
    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.
    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.
    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?

    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?

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

    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
     
  13. geistkiesel Valued Senior Member

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    2,471
    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
     
  14. geistkiesel Valued Senior Member

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    The photons were emitted in the moving frame.
    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
     
  15. Yuriy Registered Senior Member

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    1,080
    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 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….
     
    Last edited: Feb 1, 2005
  16. superluminal I am MalcomR Valued Senior Member

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    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.
     
  17. Yuriy Registered Senior Member

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    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!
     
    Last edited: Feb 1, 2005
  18. superluminal I am MalcomR Valued Senior Member

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    10,876
    High praise indeed! Thanks!
     
  19. geistkiesel Valued Senior Member

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    We see alike Xgen.

    Geistkiesel
     
  20. geistkiesel Valued Senior Member

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    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?

    Both rest frame clocks read the same.
    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.
    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.
    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
     
  21. MacM Registered Senior Member

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    10,104
    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 [thread=44235]Here[/thread].

    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?
     
    Last edited: Feb 1, 2005
  22. geistkiesel Valued Senior Member

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    2,471
    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.
     
  23. MacM Registered Senior Member

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    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).
     
    Last edited: Feb 2, 2005

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