Apparent Violation of 1st Postulate of SRT

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

  1. rpenner Fully Wired Valued Senior Member

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    Hmm, some minor mistakes from being 11 hours driving through California. Apparently, the speed in the original coordinate system was \(u\) not \(\beta c\) and this allows \(v\) to be used as the Lorentz parameter after all. Also I found a typo.

    ---

    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:
    \(x_a - x_O = c ( t_a - t_O ) \\ x_b - x_O = u ( t_b - t_O ) \\ x_d - x_O = 0 \\ x_e - x_O = L \)
    where L is the distance between d and e as measured in this frame, which makes much more sense for a name than "x".

    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 \(x_B = x_O + L , \; t_B = t_O + \frac{L}{u}\)

    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. (We're using the language of algebra, as did Einstein in 1905, but in 1908 Minkowski demonstrated that one can also use the language of geometry which provides deeper math reasons for the basic self-consistency of Special Relativity.)

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

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

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

    Step 2 Find the equations for a, b, d and e in terms of the new coordinates:
    \(x_a - x_O = c ( t_a - t_O ) \Rightarrow \frac{x'_a + v t'_a}{\sqrt{1 - \frac{v^2}{c^2}}} - \frac{x'_O + v t'_O}{\sqrt{1 - \frac{v^2}{c^2}}} = c ( \frac{t'_a + \frac{v}{c^2} x'_a}{\sqrt{1 - \frac{v^2}{c^2}}} - \frac{t'_O + \frac{v}{c^2} x'_O}{\sqrt{1 - \frac{v^2}{c^2}}} ) \\ \Rightarrow x'_a + v t'_a - x'_O - v t'_O = c t'_a + \frac{v}{c} x'_a - c t'_O + \frac{v}{c} x'_O \\ \Rightarrow \frac{1 - \frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}} (x'_a - x'_O ) = c \frac{1 - \frac{v}{c}}{\sqrt{1 - \frac{v^2}{c^2}}} ( t'_a - t'_O ) \\ \Rightarrow x'_a - x'_O = c ( t'_a - t'_O ) \)
    Above we see that a has identical description with reference to event O in every coordinate system.
    \(x_b - x_O = u ( t_b - t_O ) \Rightarrow \frac{x'_b + v t'_b}{\sqrt{1 - \frac{v^2}{c^2}}} - \frac{x'_O + v t'_O}{\sqrt{1 - \frac{v^2}{c^2}}} = u ( \frac{t'_b + \frac{v}{c^2} x'_b}{\sqrt{1 - \frac{v^2}{c^2}}} - \frac{t'_O + \frac{v}{c^2} x'_O}{\sqrt{1 - \frac{v^2}{c^2}}} ) \\ \Rightarrow x'_b - x'_O = \frac{u - v}{1 - \frac{u v}{c^2} } ( t'_b - t'_O ) \)
    Above we see the origin of the Einstein velocity composition law. (\(u' = \frac{u + v}{1 + \frac{u v}{c^2} }\) for the case where they are parallel and \(u' = \frac{u - v}{1 - \frac{u v}{c^2} }\) (change of sign on v) for the case where u and v are oppositely oriented (antiparallel).)
    \(x_d - x_O = 0 \Rightarrow x'_d - x'_O = (- v) ( t'_d - t'_O ) \\ x_e - x_O = L \Rightarrow x'_e - x'_O = \sqrt{1 - \frac{v^2}{c^2}} L - v ( t'_e - t'_O )\)
    Now e and d don't correspond to a state of rest in these coordinates. (But since L was defined in the coordinate system where both d and e were at rest, above we see the origin of the Lorentz contraction of length, \(L' = \sqrt{1 - \frac{v^2}{c^2}} L \).)

    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:

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

    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 (respectively).
    ---

    So in the OP the ratio \(r = \frac{t_A - t_O}{t_B - t_O} = \frac{u}{c}\) is highlighted. In the transformed frame we defined \(r' = \frac{t'_A - t'_O}{t'_B - t'_O} = \frac{ c^2 u - c u v }{ c^3 - c u v } \neq \frac{u}{c}\).

    But that's not a violation of the first postulate because the law r = r' is not a law of physics. It's just a made-up ratio of coordinate times among three events which are related only pair-wise. You might as well be complaining that triangles have different heights when the coordinate system is rotated.

    Here are quantities that SR does preserve:

    \( <A - O, A - O> = (c t_A - c t_O)(c t_A - c t_O) - (x_A - x_O)(x_A - x_O) = L^2 - L^2 = 0 \\ <A - O, A - O> = (c t'_A - c t'_O)(c t'_A - c t'_O) - (x'_A - x'_O)(x'_A - x'_O) = L^2 \frac{\left(1 - \frac{v}{c} \right)^2}{1 - \frac{v^2}{c^2}} - L^2 \frac{ \left( 1 - \frac{v}{c} \right)^2}{1 - \frac{v^2}{c^2}} = 0\)
    Both coordinate systems agree, the trajectory from O to A is light-like.
    \( <B - O, B - O> = (c t_B - c t_O)(c t_B - c t_O) - (x_B - x_O)(x_B - x_O) = \frac{c^2 L^2}{u^2} - L^2 = \frac{c^2 - u^2}{u^2} L^2 \\ <B - O, B - O> = (c t'_B - c t'_O)(c t'_B - c t'_O) - (x'_B - x'_O)(x'_B - x'_O) = \frac{\left(1 - \frac{u v}{c^2} \right)^2 }{1 - \frac{v^2}{c^2}} \frac{c^2 L^2}{u^2} - \frac{\left(u - v\right)^2}{1 - \frac{v^2}{c^2} } \frac{L^2}{u^2} = \frac{L^2}{u^2} \frac{c^2 - 2 u v + \frac{u^2 v^2}{c^2} - u^2 + 2 u v - v^2 }{1 - \frac{v^2}{c^2}} = \frac{c^2 - u^2}{u^2} L^2 \)
    Both coordinate systems agree, the trajectory from O to B is time-like. The value is proportional to the square of the proper elapsed time.

    \( <A - O, B - O> = (c t_A - c t_O)(c t_B - c t_O) - (x_A - x_O)(x_B - x_O) = \frac{c^2 L^2}{c u} - L^2 = \frac{c - u}{u} L^2 \\ <A - O, B - O> = (c t'_A - c t'_O)(c t'_B - c t'_O) - (x'_A - x'_O)(x'_B - x'_O) = L^2 \frac{c}{u} \frac{\left(1 - \frac{v}{c} \right)\left(1 - \frac{u v}{c^2} \right)}{1 - \frac{v^2}{c^2}} - L^2 \frac{ \left( 1 - \frac{v}{c} \right)\left(1 - \frac{v}{u}\right)}{1 - \frac{v^2}{c^2}} = L^2 \frac{ \frac{c}{u} - \frac{v}{u} - \frac{v}{c} + \frac{v^2}{c^2} - 1 + \frac{v}{c} + \frac{v}{u} - \frac{v^2}{c u} }{1 - \frac{v^2}{c^2}} = L^2 \left( \frac{c}{u} - 1 \right) = \frac{c - u}{u} L^2 \)
    This is analogous to the dot product of Euclidean vectors.
     
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  3. Daecon Kiwi fruit Valued Senior Member

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    It looks like you've opened an HTML file in notepad.
     
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  5. rpenner Fully Wired Valued Senior Member

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    The forum uses JavaScript to render LaTeX.
     
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  7. Daecon Kiwi fruit Valued Senior Member

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    Ah, my apologies. I disable Javascript when browsing on my PS Vita.

    Why do you think nobody else has discovered this discrepancy?
     
  8. rpenner Fully Wired Valued Senior Member

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    Unclear antecedent. To what, specifically, do you refer?
     
  9. Daecon Kiwi fruit Valued Senior Member

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    The thread topic is titled "Apparent Violation of 1st Postulate of SRT".

    Why isn't this violation common knowledge?
     
  10. Russ_Watters Not a Trump supporter... Valued Senior Member

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    Are you begging the question? Clearly, the answer is that the supposed discrepanacy doesn't exist.
     
  11. Daecon Kiwi fruit Valued Senior Member

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    I had a brainfart somewhere and somehow got rpenner confused with the OP.

    *embarrassed*
     
  12. HeyBert Registered Member

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    Fair enough. My linguistics is definitely not in line with the mainstream physics community, but the improper use (in the physics sense) of the word "event" should never have been a first line of defense if you know what someone is talking about. IAW the Merriam Webster dictionary;

    Full Definition of EVENT
    1
    a archaic : outcome

    b : the final outcome or determination of a legal action

    c : a postulated outcome, condition, or eventuality <in the event that I am not there, call the house>
    2
    a : something that happens : occurrence

    b : a noteworthy happening

    c : a social occasion or activity

    d : an adverse or damaging medical occurrence <a heart attack or other cardiac event>
    3
    : any of the contests in a program of sports
    4
    : the fundamental entity of observed physical reality represented by a point designated by three coordinates of place and one of time in the space-time continuum postulated by the theory of relativity
    5
    : a subset of the possible outcomes of an experiment
     
  13. HeyBert Registered Member

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    Since your reply is in a completely different mathematical style that I am used to working with, I'll have to look over your response in more detail at a later time as I have a day job. Thank you for taking the time to respond in a professional non-demeaning manner, and I'll try to understand what you are trying to explain in your post.
     
  14. paddoboy Valued Senior Member

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    You were also asked a question further back.
    Why isn't this supposed violation common knowledge?
    And why if you are fair dinkum, do you not get it properly peer reviewed?

    I mean surely, you don't really expect this forum and its members, that is open to any Tom, Dick and Harry, to just accept what you say?
     
  15. PhysBang Valued Senior Member

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    I don't like saying it, but you are coming off like a piece of shit.

    Like so many cranks, you know absolutely nothing about the physics you talk about and your first impulse is to publicly claim that you have found a serious conceptual problem with the physics that you did not study.
     
  16. HeyBert Registered Member

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    Are you serious? What is wrong with you?
     
  17. paddoboy Valued Senior Member

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    PhysBang is a hard task master....
    I'm far easier to deal with.[tic mode on

    Please Register or Log in to view the hidden image!

    ]

    I asked you a couple of simple questions, that should be easy to answer if you are fair dinkum.
    Here they are again....
    Why isn't this supposed violation common knowledge?
    And why if you are fair dinkum, do you not get it properly peer reviewed?

    I mean surely, you don't really expect this forum and its members, that is open to any Tom, Dick and Harry, to just accept what you say?
     
  18. HeyBert Registered Member

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    Sorry, a lot of posts so hard to keep up and I am new to forums in general (just learned tonight how to get the quotes to show up on replies).

    Cannot say why this is not common knowledge. I'm just a guy who loves physics and enjoys putting the pieces together in interesting ways.

    As for the peer review, don't even know where to start other than the little bit of googling I did. I don't know LATEX, so would have to learn that whole process before even thinking about it.

    As I am learning on these forums, I need to alter my vocabulary to use physics specific words (like not using the offensive word 'event') to describe my ideas. Doing math is really easy for me, but relating these ideas to others in a "common accepted" vocabulary is proving very difficult indeed. The forum makes it even harder as I am trying to describe pictures, but everyone keeps getting hung up on the words I am using.
     
  19. HeyBert Registered Member

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    BTW...how do you get equations to show up in the text?
     
  20. origin Heading towards oblivion Valued Senior Member

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    Like this.

    And this.
     
  21. Daecon Kiwi fruit Valued Senior Member

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    "Interesting" is not a synonym for "correct".
     
  22. HeyBert Registered Member

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    Trolling on responses to others is your thing, huh?
     
  23. rpenner Fully Wired Valued Senior Member

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    I fail to see how your citation of the dictionary supports your use of event as the history of the object's or pulse of lights trajectory through space-time. It does however, in the sole sense qualified for theoretical physics, support out usage as a zero-dimensional element of space-time geometry:
    I am reminded of Dingle's famous problem with learning Special Relativity -- he never got that special relativity was geometry of space and time and tried to talk about ratios of time and space separately which has no physical justification for being a fundamental quantity respected by the universe. It's a conceptual problem as crippling as thinking there is something fundamentally important in SI units or the numbers produced when physical quantities are measured in SI units.
     

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