Proposal: Time paradox in Special Relativity Theory.

Discussion in 'Formal debates' started by Emil, Sep 29, 2012.

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  1. Tach Banned Banned

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    Read Emil's post 24, nothing to do with GR. Emil denies and doesn't understand SR, no point in delving into GR.
     
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  3. Tach Banned Banned

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    According to Emil , the exercise contains no "GR effects."
     
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  5. rpenner Fully Wired Staff Member

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    If you would really like to assert that position, I would love to formally debate you. Sufficient to say
    \(100(\sqrt{1-0.49^2}-\sqrt{1-0.98^2}) = \sqrt{7599}- 6\sqrt{11} \approx 67\) is a number while \(4 \tau_A - \tau_B = \frac{\pi \cdot 20 \, \textrm{a} \cdot c}{0.49 \, c} (\sqrt{1-0.49^2}-\sqrt{1-0.98^2}) = \frac{20 \pi}{49} ( \sqrt{7599} - 6 \sqrt{11}) \, \textrm{a} \approx 86 \, \textrm{a}\) which is a quantity of time.
     
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  7. Tach Banned Banned

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    You have a valid point.
     
  8. Tach Banned Banned

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    "Emil scenario" has no GR "component", read post 24.
     
  9. OnlyMe Valued Senior Member

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    I think you are interpreting the conditions a little differently than I intended.

    This post is intended only to clarify what seems to be a misinterpretation, of my earlier post. Anyone who is not familiar with the question should take it with a grain of salt. I am still working through, what appears to be a misinterpretation, of my own...

    Assuming that the hypothetical is limited to SR, as Emil did suggest, grvitational fields are not an issue. There is no gravity to be considered in the hypothetical.

    What I was referring to is that, in a flat spacetime and the abscence of gravity, the circular paths of both A and B, should have components including a constant velocity for each and an acceleration associated with the curved path, which is in addition to the constant velocity.

    My assumption was that the inertial resistance to the acceleration resulting in the curved path, could introduce a time dilation associated with GR and the equivalence principle, acting in opposition to SR and constant velocity.

    Rpenner seems to have agreed with Tach that this is not an issue that would be involved...

    I have been and am re-reading some of what I have available both in my own library and on the Internet, to better understand my confussion. So far I can say that I have run across arguments supporting both views, which requires that I think through several explanations from different sources to discover a resolution.

    So far it seems to me that the situation is entirely theoretical, leading to differing perspectives. We have no experience with any object accelerating in a circular path, where there is not some force that counterbalances the affects of the acceleration which might be associated with the equivalence principle. In particle accelerators, the magnetic containment and in the case of satellites a gravitational field... In both of these situations it is clear to me that only the velocity need be considered. (Setting aside a satellites location to the gravitational source it orbits.)
     
  10. Tach Banned Banned

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    What do you mean? How else would the object follow a circular path?
     
  11. OnlyMe Valued Senior Member

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    OK Tach, keeping in mind that you are limited to a flat spacetime, an example.

    Not just localy equivalent to flat — the complete absence of gravity.

    What environment do we have or know of where there is no gravity to be considered, and no other force is acting instead of gravity, relative to the curved path?
     
  12. rpenner Fully Wired Staff Member

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    Who said there was no force involved? The stipulation was that a and b move in circular paths at constant velocities not that a and b were moving inertially.
     
  13. Tach Banned Banned

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    Can you stop throwing around buzzwords in order to cover up your ignorance?


    There are a lot of such forces, the most notable being the one you couldn't identify in the case of particle accelerators, the Lorentz force. Besides, what gives you the idea that there is no force involved ? How else do the particles move in a circle? What does flat spacetime vs. curved spacetime have to do with this problem? What does gravity and the equivalence principle have to do with this simple exercise?
     
  14. OnlyMe Valued Senior Member

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    Where was that stipulation introduced?

    I did not re-read every post in detail, but Emil, did not present a completely defined hypothetical, to begin with.

    Is it the case that within the context of SR, inertia and inertial resistance to a change in velocity or direction does not exist?

    That was the point of my original comment. Initially intended to draw out a clearer definition of the conditions of the proposed hypothetical.
     
  15. rpenner Fully Wired Staff Member

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    See post #16
     
  16. OnlyMe Valued Senior Member

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    So in a particle accelerator the Lorentz force is toward the center of the circular path?

    Obviously there are forces involved, however they were not clearly defined.

    Curved spacetime has nothing to do with the problem. However, unless you also agree that there is no inertial resistance to the constant change in the direction A and B travel in.., the curved path, abscent being in an orbit around a gravitational mass, you cannot just assume they are in a free fall environment relative to the curved path... Unless it is stipulated.

    If A were stationary and B were following a curved path, you would argue that the reason B was the one time dilated was the acceleration associated with the curved path.

    The hypothetical was never fully defined before conclusions were being presented.
     
  17. OnlyMe Valued Senior Member

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    I don't see where any stipulation that excludes inertial resistance to the acceleration associated with a curved path is layed out...

    In fact, without the stipulation, it seems that one should assume not only a constant velocity for A and B, but an additional acceleration toward the focal point of the curved path, just to travel in a curved path.

    You could and obviously did read Emil's reference to speed as meaning only velocity was involved, but that was not clearly defined.
     
  18. OnlyMe Valued Senior Member

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    Deleted duplicate post.
     
    Last edited: Oct 1, 2012
  19. Motor Daddy ☼☼☼☼☼☼☼☼☼☼☼ Valued Senior Member

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    They travel away from each other until there is a distance of two times the diameter of the circles between them, and then they start traveling towards each other again. Both A and B agree on the closing speed at every point in time, and they also agree that that closing speed is changing at every point in time. There is no paradox. In the frames of A and B, all they see is each other traveling straight away from each other, and then traveling towards each other again, all the while the closing speed is changing.
     
  20. Tach Banned Banned

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    You didn't know that? It is elementary.


    The above bumbling is incomprehensible, can you stop stringing along buzzwords?

    First off, B ISN"T "time dilated", quite the contrary. Second off, it isn't due to ANY acceleration, it is STILL due to relative velocity. If you want to do physics, you need to learn how to do the calculations.

    \(\tau=\frac{2 \pi}{\omega} \sqrt{1-\frac{r_s}{R}-(\frac{\omega R}{c})^2}=\frac{2 \pi R}{v} \sqrt{1-\frac{r_s}{R}-(\frac{v}{c})^2}\)

    where \(r_s\) is the Schwarzschild radius of the central attractor. In general, \(r_s<<R\). For example, for Earth, \(r_s=9mm\).

    See post 24, one more time.Emil already told you that.
     
  21. RealityCheck Banned Banned

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    Hi Tach.

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    That was the point of my suggestion , Tach, to help forestall any cries of "but that's not SR, that's GR accelerations" etc etc etc.

    This way, the GR is not a problem and the scenario CAN be treated as SR only by REMOVING any CLAIMS or ARGUMENTS that GR makes emil's scenario 'invalid' etc etc.

    It was to avoid such exchanges that I suggested it. That was the point of suggesting it.

    Enjoy your discussion. Cheers!

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  22. Emil Valued Senior Member

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    Now I come home and I'm very tired.
    Tomorrow I will give a comprehensive reply.

    rpenner, I suggest you reevaluate what you have posted so far.

    Lorentz operator does not have the property: f[sub](a+b)[/sub]=f[sub](a)[/sub]+f[sub](b)[/sub]
    Tangential speed is not linear speed.(need to figure out the velocity)

    You have two cases.
    -a is stationary it has the proper time and Lorentz operator is applied to time b
    -b is stationary it has the proper time and Lorentz operator is applied to time a.
    You need the speed between a and b, which is a function of time, V(a,b)[sub](t)[/sub].

    But, because the speed is the same from a-b and b-a, you can use the general form V(a,b)[sub](t)[/sub]=V(b,a)[sub](t)[/sub] without the exact calculation

    Now I realize that not enough one post of mine in debate.
    I'll need to give a reply. There are too many mistakes.
     
  23. Tach Banned Banned

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    So what? None of the above is relevant, nor does it make any sense.

    The above shows (again) your crass lack of understanding of SR. The most hilarious thing is that you are trying to use SR (Lorentz transforms) in your attempt at disproving.....SR!


    Hog wash, the above doesn't have anything to do with solving the simple exercise, you aren't making any sense. And it shows.


    You are right, there ARE too many mistakes in your thinking.
     
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