Apparent Violation of 1st Postulate of SRT

Discussion in 'Pseudoscience' started by HeyBert, Feb 16, 2015.

  1. HeyBert Registered Member

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    The following scenario utilizes events that are applicable to Einstein’s Special Relativity Theory. What follows requires further analysis regarding application of the first postulate WRT SRT. For information with images, see http://gsjournal.net/Science-Journals/Research Papers-Relativity Theory/Download/5916.


    - Definitions; (Beta) β = v/c, (Gamma) γ = (1-v^2/c^2)^-1/2.

    - Electromagnetic relations (second postulate); x = cΔτ, x’ = cΔτ’.

    - There are two parallel linear events (A and B) with uniform velocity along the positive x-axis.

    - Event A is an electromagnetic event (velocity = c) following the path from point x = 0 to x.

    - Event B is an inertial body event (velocity = u) following an identical and parallel path from point x = 0 to x.


    EVENTS OCCURRING IN REST FRAME (v = 0)

    - These events now occur within a reference frame considered to be at rest (rest frame).

    - The event time for A is; Δτ = x/c.

    - The event time for B is; Δt = x/u.

    - The ratio of these event times gives us the relationship between their event times;

    Δτ/Δt = (x/c)/(x/u) = u/c.


    EVENTS OCCURRING IN INERTIAL FRAME (v)

    - These events now occur within a reference frame considered to be in uniform motion (inertial frame).

    - The SRT event time for A is; Δτ = (Δτ’+vx’/c^2)γ. Applying the second postulate;

    Δτ = (Δτ’+vx’/c^2)γ, x’ = cΔτ’

    Δτ = (Δτ’+v(cΔτ’)/c^2)γ

    Δτ = (Δτ’+vΔτ’/c)γ

    Δτ = Δτ’(1+ β)γ

    - The SRT event time for B is; Δt = (Δt’+vx’/c^2)γ.

    - The ratio of these event times gives us the relationship between their event times;

    Δτ/Δt = [Δτ’(1+ β)γ]/[(Δt’+vx’/c^2)γ]

    Δτ/Δt = [Δτ’(1+ β)]/[(Δt’+vx’/c^2)]

    Δτ/Δt = [Δτ’(1+ β)]/[Δt’(1+vx’/(c^2)Δt’)]

    Δτ/Δt = [Δτ’(1+ β)]/[Δt’(1+ βx’/cΔt’)]


    In order to satisfy the first postulate, the relationship between their event times must remain unchanged in order to prevent one from ascertaining the motion of the inertial frame utilizing this scenario. As we can clearly see, the relationship between their event times can only be regained by allowing an electromagnetic relation (x’ = cΔt’) to be utilized to reduce the Δt relation to (1+ β) in order to regain equality for these events.

    Δτ/Δt = [Δτ’(1+ β)]/[Δt’(1+ βx’/cΔt’)], x’ = cΔt’

    Δτ/Δt = [Δτ’(1+ β)]/[Δt’(1+ β)]

    Δτ/Δt = Δτ/Δt

    If we were to allow this substitution, then we would have to admit that in order for the first postulate to remain valid, then the Lorentz transformations (in their current format) could only apply to electromagnetic events. It appears that the only recourse would be to reformat the Lorentz temporal transformation to;

    ***Δτ = Δτ’(1+ β)γ
     
    Last edited: Feb 16, 2015
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  3. Daecon Kiwi fruit Valued Senior Member

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    So... what exactly is it that you're trying to say?

    Einstein was wrong?
     
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  5. Russ_Watters Not a Trump supporter... Valued Senior Member

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    I don't think you understand what an "event" is.
     
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  7. HeyBert Registered Member

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    All that I show is that the current format of the Lorentz temporal transformation complies with the first postulate for round-trip events, but not one-way events. We can either choose to accept this as a limitation of STR, or choose a format for the temporal transformation that does comply with the first postulate for one-way events. Einstein is no more wrong than Newton was, but his theory is not perfect either.
     
  8. paddoboy Valued Senior Member

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    27,543


    The last anti relativist we had, that claimed SR/GR were wrong was chinglu....
    His threads have all ended up in the sewer, but he kept coming back.
    I'm not mathematically savvy enough to invalidate his apparent claim, but I'm sure rpenner will be along to straighten out, one way or the other.
    Irrespective, any poster coming here, and first post some anti SR/GR agenda, shows one up [in my opinion] that he basically has nothing.
    As I tell them all, if they have anything to refute/invalidate SR/GR, or any other postulate, or claim to have a TOE and to rewrite cosmology, would not be here!

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    They would be getting there stuff peer reviewed through the proper channels, and then standing in line for this year's Physic's Nobel.

    Hohum

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  9. HeyBert Registered Member

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    Russ,
    If it is not an "event", then what would you call it? Perhaps a "happening", or an "occurring"? Please do not get wrapped around the words and lose sight of the message.
     
  10. HeyBert Registered Member

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    No value added...but thanks for stopping by!
     
  11. paddoboy Valued Senior Member

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    Must agree with that. Each has there parameters of applicability.
     
  12. Daecon Kiwi fruit Valued Senior Member

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    How come you're the only person to have ever noticed this and not any of the hundreds of people who study this kind of thing as part of their jobs?
     
  13. paddoboy Valued Senior Member

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    Just telling it like it is Bert.
    We have had anti relativists around before. That can be verified.
    And don't you agree that if any of them had anything of substance, they would be elsewhere?
     
  14. Russ_Watters Not a Trump supporter... Valued Senior Member

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    Probably a "process", but you didn't describe them well enough to know for sure. Is your "electromagnetic event" supposed to be something like a moving photon and your "inertial body event" a moving massive object?

    An event is something that happens at a single moment in time, at a specific location.
    The words are what convey the message. If they aren't right, the message can only be wrong.
     
  15. Russ_Watters Not a Trump supporter... Valued Senior Member

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    Assuming I understood what you really meant, then this is wrong:
    Two observers in inertial motion with respect to each other do not need to agree on the travel time of a photon from one place to another. All that will tell is their motion with respect to each other and doesn't imply absolute motion.
     
  16. rpenner Fully Wired Valued Senior Member

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    Sorry, currently in Lost Hills, California. Not a great place to critique science claims.

    It would be nice if the option existed to force people who make crazy claims to support them or face permanent bans for raising related crazy claims.
     
  17. HeyBert Registered Member

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    Russ,
    I hope you take the time to read http://gsjournal.net/Science-Journals/Research Papers-Relativity Theory/Download/5916 in hopes of better understanding where my argument originates from.

    To the rest,
    In the end, this was meant to inspire healthy debate and perhaps enlightenment, not unwarranted insults from the "Science Gang" that trolls these posts and flexes their collective ignorance. If that is all you have to offer, feel free to ignore this post and present your own original content elsewhere.
     
  18. paddoboy Valued Senior Member

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    27,543


    Just telling it like it is Bert.
    We have had anti relativists around before...just the plain old anti science commoner with his usual angst, and then the God Botherer kind, that would like to show science as submissive to their mythical deity. That can be verified in many threads.
    And don't you agree that if any of them had anything of substance, any real science, they would be elsewhere?

    Now maybe you are an exception, I don't know. But like I said, first post, an anti SR flavour, makes people rather wary of another agenda laden diatribe.
    I'll certainly apologise if someone can verify what you are claiming, let's wait and see shall we?
    Oh, and please note, I'm not a part of any gang, science or otherwise.
    I may only be a lay person, and just know the bare essentials, as I always say, the number of claims that people claim behind the safety of a computer screen and Internet to impress is staggering.
    I know for certain, that if I did stumble across some previously hidden scientific model that explained all, a forum would be the last place for me to announce it on.

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  19. Russ_Watters Not a Trump supporter... Valued Senior Member

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  20. Daecon Kiwi fruit Valued Senior Member

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    Ignoring the "Science Gang" for the moment, how would you explain your ideas to laypeople who don't have any formal training in the sciences? Y'know, laypeople and such?
     
  21. rpenner Fully Wired Valued Senior Member

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    Hello, from Pea Soup Andersen's, a quaint California institution/motel/gift shop/restaurant.

    A thought experiment is a rigorous argument from a physical theory of what would be the consequences if a certain situation arises and the theory is applicable. It is useful in three cases: a counter-intuitive result used to learn the limits of reliance on common sense when exploring the consequences of the theory, a physically never seen result which suggests the situation never arises or the theory is basically wrong in some situations, and a demonstration that the theory is self-contradictory -- which means its successes are shadows of some not yet known theory and its logical pathology suggests that it is in some respects worse than common sense.

    The OP seeks to present the third but fails for not working within the framework of a formal logical argument and not strictly relying on the theory at issue. In fact, as someone already observed, the poster misuses the very basic term "event." In addition, there appears to be a hidden assumption of absolute space and clock synchronization, which ruins any point the OP wishes to attempt.

    An "event" is a 0-dimensional part of space-time. A "trajectory" is a 1-dimensional part of space-time which is not superluminal (space-like). A "place" is an inertial (straight) slower-than-light trajectory.

    Using correct language, lower-case letters for 1- and 3-dimensional flat entities and upper-case letters for events, the given situation is:
     
  22. rpenner Fully Wired Valued Senior Member

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    4,833
    a: a light-like inertial trajectory
    b: a time-like inertial trajectory

    a and b meet in one event, O

    d: a time-like inertial trajectory which passes through O but is distinct from b
    e: a time-like inertial trajectory which represents the same state of motion as d and which intersects both a and b at a time later than the time of O

    A: the intersection of a and e
    B: the intersection of b and e

    Once written in the proper language, we can talk about three concepts of time, the elapsed coordinate time in a coordinate system where d and e are considered at rest, the elaspsed time in a coordinate system where b is at rest (b's proper time) and a third example where neither b nor d are considered at rest.

    In proper language, no self-contradiction arises because we don't introduce bad assumptions about absolute time and space through sloppiness.
     
  23. rpenner Fully Wired Valued Senior Member

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    4,833
    Also in proper language, we can describe these events and trajectories in coordinates. In the inertial coordinates (x,t) where d and e are the standard of rest:
    \(x_a - x_O = c ( t_a - t_O ) \\ x_b - x_O = \beta c ( t_b - t_O ) \\ x_d - x_O = 0 \\ x_e - x_O = L \)
    where L is the distance between d and e as measured in this frame, which makes much more sense for a name than "x". Similarly, it is peculiar and unnecessarily confusing to use v both as the coordinate-dependent speed of world line b and the Lorentz transform parameter, so I will use "u" for that.

    To find the coordinates of \(A = a \cap e\), we solve:
    \(x_a - x_O = c ( t_a - t_O ) \\ x_e - x_O = L \\ x_A = x_a = x_e \\ t_A = t_a = t_e\)
    with solution \(x_A = x_O + L , \; t_A = t_O + \frac{L}{c}\)
    Likewise, the coordinates of \(B = b \cap e\) are \(x_B = x_O + L , \; t_B = t_O + \frac{L}{\beta c}\)

    Now to check if the Lorentz equations are self-consistent, we need to transform a, b, d,and e into new coordinates and see if they give the same solution as directly transforming A and B.

    \(t = \frac{t' + \frac{u}{c^2} x'}{\sqrt{1 - \frac{u^2}{c^2}}} \\ x = \frac{x' + u t'}{\sqrt{1 - \frac{u^2}{c^2}}}\)

    Step 1: Find the coordinates for O, A and B. I will illustrate the general solution with O:
    \(t_O = \frac{t'_O + \frac{u}{c^2} x'_O}{\sqrt{1 - \frac{u^2}{c^2}}} \\ x_O = \frac{x'_O + u t'_O}{\sqrt{1 - \frac{u^2}{c^2}}}\)
    Then
    \(t_O - \frac{u}{c^2} x_O = \frac{t'_O + \frac{u}{c^2} x'_O - \frac{u}{c^2} x'_O - \frac{u}{c^2} u t'_O}{\sqrt{1 - \frac{u^2}{c^2}}} = t'_O \frac{1 - \frac{u^2}{c^2}}{\sqrt{1 - \frac{u^2}{c^2}}} = t'_O \sqrt{1 - \frac{u^2}{c^2}}\)
    Thus \(t'_O = \frac{t_O - \frac{u}{c^2} x_O}{\sqrt{1 - \frac{u^2}{c^2}}}\)
    Similarly: \(x'_O = \frac{x_O - u t_O}{\sqrt{1 - \frac{u^2}{c^2}}}\)
    Likewise,
    \(t'_A = \frac{t_A - \frac{u}{c^2} x_A}{\sqrt{1 - \frac{u^2}{c^2}}} = \frac{t_O + \frac{L}{c} - \frac{u}{c^2} x_O - \frac{u}{c^2} L}{\sqrt{1 - \frac{u^2}{c^2}}} = t'_O + \frac{L}{c} \frac{ 1 - \frac{u}{c} }{\sqrt{1 - \frac{u^2}{c^2}}} \\ x'_A = \frac{x_A - u t_A}{\sqrt{1 - \frac{u^2}{c^2}}} = x'_O + L \frac{1 - \frac{u}{c}}{\sqrt{1 - \frac{u^2}{c^2}}} \\ t'_B = \frac{t_B - \frac{u}{c^2} x_B}{\sqrt{1 - \frac{u^2}{c^2}}} = t'_O + \frac{\frac{L}{\beta c} - \frac{u}{c^2} L}{\sqrt{1 - \frac{u^2}{c^2}}} = t'_O + \frac{L}{\beta c} \frac{1 - \beta \frac{u}{c} }{\sqrt{1 - \frac{u^2}{c^2}}} \\ x'_B = \frac{x_B - u t_B}{\sqrt{1 - \frac{u^2}{c^2}}} = x'_O + \frac{ L - u \frac{L}{\beta c}}{\sqrt{1 - \frac{u^2}{c^2}}} = x'_O + L \frac{ 1 - \frac{u}{\beta c}}{\sqrt{1 - \frac{u^2}{c^2}}} \)

    This inverse Lorentz transform is just as general as the forward direction, so we may summarize by writing:
    \(t' = \frac{t - \frac{u}{c^2} x}{\sqrt{1 - \frac{u^2}{c^2}}} \\ x' = \frac{x - u t}{\sqrt{1 - \frac{u^2}{c^2}}}\)

    Step 2 Find the equations for a, b, d and e in terms of the new coordinates:
    \(x_a - x_O = c ( t_a - t_O ) \Rightarrow \frac{x'_a + u t'_a}{\sqrt{1 - \frac{u^2}{c^2}}} - \frac{x'_O + u t'_O}{\sqrt{1 - \frac{u^2}{c^2}}} = c ( \frac{t'_a + \frac{u}{c^2} x'_a}{\sqrt{1 - \frac{u^2}{c^2}}} - \frac{t'_O + \frac{u}{c^2} x'_O}{\sqrt{1 - \frac{u^2}{c^2}}} ) \\ \Rightarrow x'_a + u t'_a - x'_O - u t'_O = c t'_a + \frac{u}{c} x'_a - c t'_O + \frac{u}{c} x'_O \\ \Rightarrow \frac{1 - \frac{u}{c}}{\sqrt{1 - \frac{u^2}{c^2}}} (x'_a - x'_O ) = c \frac{1 - \frac{u}{c}}{\sqrt{1 - \frac{u^2}{c^2}}} ( t'_a - t'_O ) \\ \Rightarrow x'_a - x'_O = c ( t'_a - t'_O ) \)
    Above we see that a has identical description with reference to event O in every coordinate system.
    \(x_b - x_O = \beta c ( t_b - t_O ) \Rightarrow \frac{x'_b + u t'_b}{\sqrt{1 - \frac{u^2}{c^2}}} - \frac{x'_O + u t'_O}{\sqrt{1 - \frac{u^2}{c^2}}} = \beta c ( \frac{t'_b + \frac{u}{c^2} x'_b}{\sqrt{1 - \frac{u^2}{c^2}}} - \frac{t'_O + \frac{u}{c^2} x'_O}{\sqrt{1 - \frac{u^2}{c^2}}} ) \\ \Rightarrow x'_b - x'_O = \frac{\beta c - u}{1 - \beta \frac{u}{c} } ( t'_b - t'_O ) \)
    Above we see the origin of the Einstein velocity composition law.
    \(x_d - x_O = 0 \Rightarrow x'_d - x'_O = (- u) ( t'_d - t'_O ) \\ x_e - x_O = L \Rightarrow x'_e - x'_O = \sqrt{1 - \frac{u^2}{c^2}} L - u ( t'_e - t'_O )\)
    Now e and d don't correspond to a state of rest in these coordinates.

    All four of these show that Newton's law of inertia is obeyed the same in both frames, so that tends to support Einstein's first postulate.

    Step 3 Calculate \(C = a \cap b, D = a \cap e, E = b \cap e \) in these new coordinates:

    \(x'_C - x'_O = c ( t'_C - t'_O ), \quad x'_C - x'_O = \frac{\beta c - u}{1 - \beta \frac{u}{c} } ( t'_C - t'_O )\) with solution \(x'_C = x'_O, t'_C = t'_O\).
    \(x'_D - x'_O = c ( t'_D - t'_O ), \quad x'_D - x'_O = \sqrt{1 - \frac{u^2}{c^2}} L - u ( t'_D - t'_O )\) with solution:
    \( t'_D = t'_O + \frac{\sqrt{1 - \frac{u^2}{c^2}}}{c + u} L = t'_O + \frac{L}{c} \frac{1 - \frac{u}{c}}{\sqrt{1 - \frac{u^2}{c^2}}} = t'_A \\ x'_D = x'_O + L\frac{1 - \frac{u}{c}}{\sqrt{1 - \frac{u^2}{c^2}}} = t'_A\)
    \(x'_E - x'_O = \frac{\beta c - u}{1 - \beta \frac{u}{c} } ( t'_E - t'_O ), \quad x'_E - x'_O = \sqrt{1 - \frac{u^2}{c^2}} L - u ( t'_E - t'_O )\) with solution:
    \( t'_E = t'_O + \frac{1 - \beta \frac{u}{c} }{\sqrt{1 - \frac{u^2}{c^2}} } \frac{L}{\beta c} = t'_B \\ x'_E = x'_O + \frac{\beta c - u}{\sqrt{1 - \frac{u^2}{c^2}} } \frac{L}{\beta c} = x'_B\)

    Step 4 Highlight the logical self-contradiction.

    But there is none since O and C, A and D, and B and E all identify the same event coordinates.
     

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