A possible proof that negative ageing doesn't occur in special relativity?

Discussion in 'Alternative Theories' started by Mike_Fontenot, Jun 13, 2021.

  1. Neddy Bate Valued Senior Member

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    2,548
    What is that supposed to mean? At time t=40 she knows what?

    You have At=40 and At=20 in the same inexplicable stream of thought, and then you are back to x=17.3 again. Why don't you write x=34.6 instead?

    Please try to be more clear.

    Okay, but notice that x=17.3 is not in there anyplace, because t=40 is not t=20, and x=17.3 is not x=34.6.
     
    Last edited: Jun 30, 2021
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  3. Neddy Bate Valued Senior Member

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    Phyti,

    This is the post where you lost me:

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    Do you see where I am talking about x=34.6 and you are saying he thinks he is at x=17.3? Why?

    Do you also see where you are talking about him receiving a signal "t=5.4" which happens at x=34.6, and yet you are saying he thinks he is at 17.3? Why?

    You keep talking about x=17.3 but that does not have anything to do with x=34.6. I am trying to figure out what you are saying, but you are making it impossible.
     
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  5. Neddy Bate Valued Senior Member

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    phyti,

    In the part above where you quote me saying, "Knowing he is 34.6 light years away from her," I was talking about if he stops moving relative to her when he is at x=34.6.

    The full quote is, "...the traveling twin stops when he is 20 years old, and using the CMIF method, he calculates the home twin is 20*2=40 years old. At that same moment, he looks through a powerful telescope, and sees her and her clock displaying 5.359 years. Knowing that he is 34.641 light years away from her..."

    Maybe that is part of the confusion, I guess you think he is still moving or something. But when he is at x=34.6 he does not think he is at x=17.3! Jeez.
     
    Last edited: Jun 30, 2021
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  7. TonyYuan Gravitational Fields and Gravitational Waves Registered Senior Member

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    The assumption that the speed of light is constant is not true. SR is already a mathematical logic game.

    https://tonyyuan7.academia.edu/research
    Why Morley Experiment Could Not Observe the Movement of Interference Fringe
    1. The influence of water on the speed of a car
    2. The influence of air on light propagation
    3. The influence of gravitational field on light propagation
    4. Eddington Observation
    5. Mass-energy equation
    6. Fiber optic gyroscope
    7. Explanation of the Sagnac effect
    If the weakest gravitational wave could rise the fluctuation of the LIGO interference fringe of light, how should we ignore the gravitational field influences on light? We raise the possibility that the light is affected by the gravitational field of the earth. In this way, the Morley Experiment would not find the movement of interference fringe either in the air or in the vacuum environment.
     
  8. arfa brane call me arf Valued Senior Member

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    7,832
    And yet, as I've pointed out elsewhere, if the speed of light isn't constant and the assumption is false, modern electronics and optonics wouldn't exist.

    The fact that radar "works as assumed" strongly supports Einstein's assumption. If you suppose the assumption isn't supported by evidence you have a lot of explaining to do; you could start with explaining why the inconstant speed of light makes it useless as a signalling option.

    Which is to say, explain why modern communications tech is unsuccessful at transmitting light signals.
    That is what's known as a direct contradiction of evidence. Go for it, my man.
     
  9. TonyYuan Gravitational Fields and Gravitational Waves Registered Senior Member

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    852
    What you said has almost nothing to do with the constant speed of light required by SR theory.
    Without SR, human technology will make greater progress.
     
  10. TonyYuan Gravitational Fields and Gravitational Waves Registered Senior Member

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    852
    modern electronics and optonics What do they have to do with the constant speed of light? Do you know what is the constant speed of light?
    Radar, I have been engaged in radio communication for more than 10 years. Radio communication has nothing to do with SR, and has nothing to do with the speed of light.
    Are you talking about fiber optic communication? This has nothing to do with "constant speed of light".
    SR is a mathematical game. If you assume that the speed of sound does not change, you can also create a theory and you will be the contemporary Einstein.

    Do you know Fiber optic gyroscope? Do you know why it can't measure the rotation of the earth?

    Light is held by the gravitational field. Why don't you understand such a simple truth? Why believe that the speed of light is constant? Fiber optic gyroscope has been enough to prove that "the speed of light does not change" is a lie, and SR is a fallacy.
     
    Last edited: Jul 2, 2021
  11. arfa brane call me arf Valued Senior Member

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    But radio signals propagate at the speed of light. So your statement is stunningly ignorant at least.
    The connection between SR and solid state physics has been known for a while (like about 100 years).
     
  12. TonyYuan Gravitational Fields and Gravitational Waves Registered Senior Member

    Messages:
    852
    Do you know what is the constant speed of light? What is the Lorenz transformation? It will be better to express your opinion after you understand this.
     
  13. phyti Registered Senior Member

    Messages:
    732
    Neddy;

    We will keep it simple without any fluff from the original 'twin' problem.

    Imagine each observer at the apex of a light cone. They are only aware of events on and within their light cone via the signals they receive.
    An observer can only make measurements or get information at the speed of light.
    Place A and B in remote space with no significant masses nearby, and moving at a constant speed and direction.
    If A or B receive a (blue) light signal from some random direction, they have no way of knowing how far its source was.
    What is needed for measurement is a 2-way light signal. That provides a means of determining distance. If they know the emission time t1 and the return time t2, they can calculate distance x=(t2-t1)/2.
    left:
    A sends signal at 2.7, which returns at 37.3 with a timestamp from B reading Bt=10.
    She calculates her clock read At=(37.3+2.7)/2=20 when Bt=10, in agreement with a td of 1/2. She calculates their separation at At=20 as .866*20=17.3.
    That is all she knows 'now' at At=40, since the remaining events haven't happened yet for her.
    right:
    B sends signal at 1.6, which returns at 20 with a timestamp from A reading At=5.4.
    He calculates his clock read Bt=(20+1.6)/2=10.8 when At=5.4, in agreement with a td of 1/2. He calculates their separation at Bt=10.8 as .866*10.8=9.2.
    That is all he knows 'now' at Bt=20, since the remaining events haven't happened yet for him.
    Their 'nows' are simultaneous.
    Their conclusions are based on historical events.

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  14. phyti Registered Senior Member

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    732
    Tony;
    response to your revised paper
     

    Attached Files:

  15. Neddy Bate Valued Senior Member

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    2,548
    Note that your calculations conclude something factual about At=20 even though she cannot actually perform those calculations until later at 37.3 which is well after At=20. Well my post was about concluding something factual about At=40, not making sure the calculations could be completed before that time.

    So instead you should have said that A sends a signal at 5.4, which returns at 74.6 with a timestamp from B reading Bt=20. She would then calculate her clock must have read At=(5.4+74.6)/2=40 when Bt=20, in agreement with a td of 1/2. She can also calculate their separation at At=40 as .866*40=34.6.

    Or let B send a signal at 2.7, which returns at 37.3 with a timestamp from A reading At=10. He would then calculate his clock must have read Bt=(2.7+37.4)/2=20 when At=10, in agreement with a td of 1/2. He can also calculate their separation at Bt=20 as .866*20=17.3.

    No matter what times we choose, this signal method only tells us what we already should have known just from the td of 1/2: When she is At years old she says he is At/2 years old, and when he is Bt years old he says she is Bt/2 years old, for the whole time they are in relative motion.

    If we want to make some verifiable measurements and avoid the time delay in waiting for signals to travel, we can just let A have a partner called AA who is stationary with respect to her, located at x=34.6 and possessing a clock that is synchronised to hers. When B passes closely by AA, they both instantly record the respective times on their respective clocks as AAt=40 and Bt=20 which is an event that all reference frames must agree on. This confirms that when she is 40 years old and she says he is 20 years old, she is correct.

    And we can also let B have a partner BB stationary with respect to him, located at x'=-17.3 and possessing a clock that is synchronised to his. When BB passes closely by A, they both instantly record the respective times on their respective clocks as BBt=20 and At=10 which is also an event that all reference frames must agree on. This confirms that when he is 20 years old and he says she is 10 years old, he is also correct.

    No, there is no universal now while the twins are in relative motion, because each one says the other is younger by 1/2 at any given time.

    If one of them decelerates/stops so that they are stationary with respect to one another, then they can agree on now, and that is what my post was about. If he decelerates/stops at his own time Bt=20, then he will have to conclude she is At=40 instead of 10. But the reverse is also true, if, for example, if she instead accelerates to his speed at her own time At=10, then she will have to conclude that he is Bt=20 instead of 5.
     
    Last edited: Jul 8, 2021
  16. TonyYuan Gravitational Fields and Gravitational Waves Registered Senior Member

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    852
    Light travels through a long glass. (take the ground as reference)
    If the speed of the glass is 0, the speed of light is v0;
    If the speed of the glass is 100m/s, the speed of light is v1.
    What is the relationship between v0 and v1? Will it be equal or not equal?

    Who can give the answer?
     
  17. Write4U Valued Senior Member

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    20,092
    I don't know the answer but this scenario reminded me of a slow motion filming of a photon by the fasted photographic method to slow down high speed motion.

    I hope this maybe of interest in context of the question
     
  18. James R Just this guy, you know? Staff Member

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    \(v_1=\frac{v_0+100}{1+\frac{100v_0}{c^2}}\)
     
  19. TonyYuan Gravitational Fields and Gravitational Waves Registered Senior Member

    Messages:
    852
    Perfect formula, let's substitute the speed of the earth's revolution to calculate the speed of light in different directions.
    Initial conditions:
    The earth's revolution speed is 29.783km/s
    The speed of light in the air is 300000 km / s

    The speed of light in the direction of the earth's Revolution:
    v1=(300000+29.783)/[1+300000*29.783/(300000*300000)]

    The speed of light in the opposite direction of the earth's Revolution:
    v2=(300000-29.783)/[1-300000*29.783/(300000*300000)]

    Speed of light in vertical direction:
    v3=(300000+0)/[1-300000*0(300000*300000)]

    Obviously v1, v2 and v3 are not equal, so Morley should be able to observe the obvious movement of interference fringes, but why didn't he observe it?

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  20. Ssssssss Registered Senior Member

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    v3 is irrelevant if you are working in the Earth centered inertial frame which I guess from the mess above is what you are trying and failing to do. Why would you think different speeds in different directions are a problem? You need to work out the travel times not the speeds and you need to use the correct velocities when you do it.
     
  21. TonyYuan Gravitational Fields and Gravitational Waves Registered Senior Member

    Messages:
    852
    ok,
    1. The time required on a back and forth path parallel to the direction of revolution velocity is
    time1 = len/(v1-29.783)+len/(v2+29.783)
    time1 = len/{(300000+29.783)/[1+300000*29.783/(300000*300000)] - 29.783}+ len/{(300000-29.783)/[1-300000*29.783/(300000*300000)] + 29.783}

    2.The time required on the back and forth path perpendicular to the direction of revolution velocity is
    time2 = len/v3 + len/v3
    time2 = len/{(300000+0)/[1-300000*0(300000*300000)]} + len/{(300000+0)/[1-300000*0(300000*300000)]}


    time1 != time2

    3. In other directions, the required time is also different.

    You should be able to observe the obvious movement of interference fringes!





     
    Last edited: Jul 8, 2021
  22. Ssssssss Registered Senior Member

    Messages:
    302
    Why are you adding or subtracting 29.783? Your v1 and v2 are already the velocity relative to the Earth centered frame you seem to be trying to use so why would you do another transform and why would you do a Galilean transform? That would definitely give you inconsistent results even if there were a point to using a transform.
    As I already said v3 is irrelevant if you are using the Earth centered inertial frame because the light is not travelling perpendicular to the rotation of the Earth in this frame. Try again with the correct velocity transform or calculate v1 and v2 correctly in the Earth's surface rest frame.
    Your calculation of time1 and possibly time2 are wrong so this conclusion does not follow.

    By the way are you trying to work in the rest frame of the Earth's surface or the Earth centered inertial frame? You seem to be trying to do both and mixing relativistic and Newtonian formulas into the bargain which is why you get nonsense.
     
  23. TonyYuan Gravitational Fields and Gravitational Waves Registered Senior Member

    Messages:
    852
    The light is drawn by the medium, you can look at James' formula.

    You can look at my paper, I just use classic physics for analysis. Now James gives a formula for calculating speed. I just apply this formula to the calculation of classical physics. Isn't that okay? Can't James' speed be used for things on earth?
    It seems that it is not me who get nonsense. If the speed formula given by James is not suitable for simple classical physics calculations, please explain.

    You can ignore everything I said earlier.
    Please use your SR theory to calculate the elapsed time of light in different directions in Morley's experiment.
    Initial conditions:
    1. len = 10m
    2. The earth's revolution speed is 29.783km/s.
    3. The speed of light in the air is 300000 km/s.
    (If you think the len and these speeds are derived from classical physics, please give the len and the speed of light and the speed of the earth under SR.)

    1. Parallel to the direction of the earth's revolution speed
    Your ans:


    2. Perpendicular to the direction of the earth's revolution speed
    Your ans:

     
    Last edited: Jul 8, 2021

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