SR Issue

Discussion in 'Alternative Theories' started by chinglu, Jun 11, 2014.

  1. paddoboy Valued Senior Member

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    27,543


    Pretty simple even for this layman chinglu1998...
    It's enforcing time dilation and length contraction as predicted by SR, from two different FoR's.
    Nothing strange about it at all. We see it everyday.
    In other words, time and space are not absolute.
    There is no universal NOW!!!!!
     
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  3. chinglu Valued Senior Member

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    OK, so explain this in the context of arfa brane's post. Don't take too long!!!
     
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  5. paddoboy Valued Senior Member

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    27,543
    and it goes on to say.....

    """""""""""""""""""""""""
    According to the special theory of relativity, it is impossible to say in an absolute sense that two distinct events occur at the same time if those events are separated in space. For example, a car crash in London and another in New York, which appear to happen at the same time to an observer on the earth, will appear to have occurred at slightly different times to an observer on an airplane flying between London and New York. The question of whether the events are simultaneous is relative: in the stationary earth reference frame the two accidents may happen at the same time but in other frames (in a different state of motion relative to the events) the crash in London may occur first, and in still other frames the New York crash may occur first. However, if the two events could be causally connected (i.e. the time between event A and event B is greater than the distance between them divided by the speed of light), the order is preserved (i.e., "event A precedes event B") in all frames of reference.

    http://en.wikipedia.org/wiki/Relativity_of_simultaneity
    """"""""""""""""""""""""""""""""""""""


    Isn't SR wonderful?

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  7. paddoboy Valued Senior Member

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    I already have.

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  8. paddoboy Valued Senior Member

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    Hi arfabrane and rpenner
    Would either of you comment on my interpretation re arfabrane's post, to help chinglu out please?
     
  9. chinglu Valued Senior Member

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    Prove one event and 2 observers satisfies the relativity of simultaneity. That is arfa brane's claim.
     
  10. paddoboy Valued Senior Member

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    I want you to prove that
    Whether one event and two observers in different FoR's, or two events and one observer, time dilation and length contraction will see claims of times of when an event happened, and where it happened as non absolute and depended on that FoR, speed and gravity well.
    Now our pure of heart SR specialists may whince at the way I have put that, but it does support SR I'm sure.

    Your claims that SR is not valid and false is obvious to all.
    In that respect, as usual, you are wrong.
     
  11. arfa brane call me arf Valued Senior Member

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    7,832
    Ok, when two observers are co-located, they are not spatially separated. (Duh!)
    If they're in relative motion, events (for either observer) which are spatially separated will have different times (as well as, you know, different locations, (duh! again)).

    The relative motion of the observers means their wordlines intersect at an angle; the point of intersection can be considered an "event" but there is no spatial separation.

    What that means is, ... well, you can work it out (you don't need to be a genius either!).
     
  12. chinglu Valued Senior Member

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    1,637
    Well, our disagreement is from this post of yours.

    You were claiming ROS can be applied to one event and I corrected you. I am glad you see the error in this.

    Now, you mentioned that you cannot say when C' and M are co-located since each has a different time on their clocks at co-location.

    In fact, the OP showed you could use the term "when". "When" in that case meant the time on the clock at C' is \(t'=\frac{d'}{c}\) and M is \(t=\frac{d'}{c\gamma}\).

    So, "when" in this case has a precise meaning for both frames and they both agree on this definition of "when". Note this definition does not require the existence of an absolute clock.

    OK, now I will ask you, where is the lightning on the positive x-axis in C' frame coordinates when C' and M are co-located? RPenner refuses to answer this simple question.
     
  13. krash661 [MK6] transitioning scifi to reality Valued Senior Member

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    prove it.
     
  14. chinglu Valued Senior Member

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    One event between two frames does not constitute an example of ROS. Otherwise, prove it.
     
  15. chinglu Valued Senior Member

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    1,637
    Let's keep it simple. The following is in the OP.

    When C' and M are co-located, SR claims the lightning is located at M' frame space-time coordinates of \((d',0,0,d'/c)\) based on the M' frame light postulate. So, this cannot be disputed.

    However, also when C' and M are co-located, by the M frame light postulate and an application of LT, the lightning is located at M' frame space-time coordinates of \((d'(1-v/c),0,0,d'(1-v/c)/c)\).

    Lightning cannot be at 2 different positive x-axis locations at the co-location event of C' and M.

    That is the issue.
     
  16. rpenner Fully Wired Valued Senior Member

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    4,833
    So in frame Σ, \(x_P = x_O\) but in frame Σ', \(x'_P < x'_O\), which demonstrates relativity of two time-like separated events which may be in the same position in one frame and in different positions in another frame.
    So in frame Σ, \(t_Q = t_P\) but in frame Σ', \(t'_Q < t'_P\), which demonstrates relativity of two space-like separated events which may be in the same time in one frame and at different times in another frame.

    Just because \(t_Q = t_P\) and \(t'_R = t'_P\) doesn't mean \(\color{red} t_Q = t_R\) or \(\color{red} t'_Q = t'_R\) because of relativity of simultaneity. The actual case is \(t_Q \lt t_R\) and \(t'_Q \lt t'_R\) which is self-consistent with both events Q and R being on the same world-path of the propagation of a beam of light that never passes through event P.
     
    Last edited: Jun 24, 2014
  17. arfa brane call me arf Valued Senior Member

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    7,832
    I didn't make an error, two observers can see the same event occur at different times, even though they agree on where it occured. So you haven't corrected anything.
    The only possible way both observers can agree on a "definition of when" is if they compare clocks, that is, they must (be able to) communicate with each other.
    No, you refuse to accept his answer. If I do the same amount of work rpenner did answering your repeated question, is it safe to assume you will refuse to accept it as well?
    If I type out an example from a textbook, will you refuse to accept that? Is there a book anywhere on earth that answers your question? It must be hard denying that such a thing could exist.

    So, tell me why I or anyone should bother with someone who is clearly delusional?
     
  18. rpenner Fully Wired Valued Senior Member

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    4,833
    So what things are not relative in Minkowski space-time? The Lorentz-invariant inner products of differences of coordinates:

    \( c^2 (t_Q - t_O) (t_Q - t_O) - (x_Q - x_O) (x_Q - x_O) = c^2 (t'_Q - t'_O) (t'_Q - t'_O) - (x'_Q - x'_O) (x'_Q - x'_O) = 0 \quad \quad \quad \textrm{"light-like"} c^2 (t_Q - t_O) (t_P - t_O) - (x_Q - x_O) (x_P - x_O) = c^2 (t'_Q - t'_O) (t'_P - t'_O) - (x'_Q - x'_O) (x'_P - x'_O) = d'^2 (1 - \beta^2) c^2 (t_Q - t_O) (t_R - t_O) - (x_Q - x_O) (x_R - x_O) = c^2 (t'_Q - t'_O) (t'_R - t'_O) - (x'_Q - x'_O) (x'_R - x'_O) = 0 \quad \quad \quad \textrm{"orthogonal"} c^2 (t_Q - t_O) (t_P - t_Q) - (x_Q - x_O) (x_P - x_Q) = c^2 (t'_Q - t'_O) (t'_P - t'_Q) - (x'_Q - x'_O) (x'_P - x'_Q) = d'^2 (1 - \beta^2) c^2 (t_Q - t_O) (t_R - t_Q) - (x_Q - x_O) (x_R - x_Q) = c^2 (t'_Q - t'_O) (t'_R - t'_Q) - (x'_Q - x'_O) (x'_R - x'_Q) = 0 \quad \quad \quad \textrm{"orthogonal"} c^2 (t_Q - t_O) (t_R - t_P) - (x_Q - x_O) (x_R - x_P) = c^2 (t'_Q - t'_O) (t'_R - t'_P) - (x'_Q - x'_O) (x'_R - x'_P) = - d'^2 (1 - \beta^2) c^2 (t_P - t_O) (t_P - t_O) - (x_P - x_O) (x_P - x_O) = c^2 (t'_P - t'_O) (t'_P - t'_O) - (x'_P - x'_O) (x'_P - x'_O) = d'^2 (1 - \beta^2) \quad \quad \quad \textrm{"time-like"} c^2 (t_P - t_O) (t_R - t_O) - (x_P - x_O) (x_R - x_O) = c^2 (t'_P - t'_O) (t'_R - t'_O) - (x'_P - x'_O) (x'_R - x'_O) = d'^2 ( 1 + \beta ) c^2 (t_P - t_O) (t_P - t_Q) - (x_P - x_O) (x_P - x_Q) = c^2 (t'_P - t'_O) (t'_P - t'_Q) - (x'_P - x'_O) (x'_P - x'_Q) = 0 \quad \quad \quad \textrm{"orthogonal"} c^2 (t_P - t_O) (t_R - t_Q) - (x_P - x_O) (x_R - x_Q) = c^2 (t'_P - t'_O) (t'_R - t'_Q) - (x'_P - x'_O) (x'_R - x'_Q) = \beta d'^2 ( 1 + \beta ) c^2 (t_P - t_O) (t_R - t_P) - (x_P - x_O) (x_R - x_P) = c^2 (t'_P - t'_O) (t'_R - t'_P) - (x'_P - x'_O) (x'_R - x'_P) = \beta d'^2 ( 1 + \beta ) c^2 (t_R - t_O) (t_R - t_O) - (x_R - x_O) (x_R - x_O) = c^2 (t'_R - t'_O) (t'_R - t'_O) - (x'_R - x'_O) (x'_R - x'_O) = 0 \quad \quad \quad \textrm{"light-like"} c^2 (t_R - t_O) (t_P - t_Q) - (x_R - x_O) (x_P - x_Q) = c^2 (t'_R - t'_O) (t'_P - t'_Q) - (x'_R - x'_O) (x'_P - x'_Q) = d'^2 ( 1 + \beta ) c^2 (t_R - t_O) (t_R - t_Q) - (x_R - x_O) (x_R - x_Q) = c^2 (t'_R - t'_O) (t'_R - t'_Q) - (x'_R - x'_O) (x'_R - x'_Q) = 0 \quad \quad \quad \textrm{"orthogonal"} c^2 (t_R - t_O) (t_R - t_P) - (x_R - x_O) (x_R - x_P) = c^2 (t'_R - t'_O) (t'_R - t'_P) - (x'_R - x'_O) (x'_R - x'_P) = - d'^2 ( 1 + \beta ) c^2 (t_P - t_Q) (t_P - t_Q) - (x_P - x_Q) (x_P - x_Q) = c^2 (t'_P - t'_Q) (t'_P - t'_Q) - (x'_P - x'_Q) (x'_P - x'_Q) = - d'^2 (1 - \beta^2) \quad \quad \quad \textrm{"space-like"} c^2 (t_P - t_Q) (t_R - t_Q) - (x_P - x_Q) (x_R - x_Q) = c^2 (t'_P - t'_Q) (t'_R - t'_Q) - (x'_P - x'_Q) (x'_R - x'_Q) = \beta d'^2 ( 1 + \beta ) c^2 (t_P - t_Q) (t_R - t_P) - (x_P - x_Q) (x_R - x_P) = c^2 (t'_P - t'_Q) (t'_R - t'_P) - (x'_P - x'_Q) (x'_R - x'_P) = d'^2 ( 1 + \beta ) c^2 (t_R - t_Q) (t_R - t_Q) - (x_R - x_Q) (x_R - x_Q) = c^2 (t'_R - t'_Q) (t'_R - t'_Q) - (x'_R - x'_Q) (x'_R - x'_Q) = 0 \quad \quad \quad \textrm{"light-like"} c^2 (t_R - t_Q) (t_R - t_P) - (x_R - x_Q) (x_R - x_P) = c^2 (t'_R - t'_Q) (t'_R - t'_P) - (x'_R - x'_Q) (x'_R - x'_P) = - \beta d'^2 ( 1 + \beta ) c^2 (t_R - t_P) (t_R - t_P) - (x_R - x_P) (x_R - x_P) = c^2 (t'_R - t'_P) (t'_R - t'_P) - (x'_R - x'_P) (x'_R - x'_P) = - d'^2 \left(1 + \beta \right)^2 \quad \quad \quad \textrm{"space-like"} \)

    (I've suppressed the y and z (and y' and z') coordinates since for every event mentioned they have the same value so their differences cannot contribute to this inner product.)
     
  19. chinglu Valued Senior Member

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    1,637
    So, how is the co-location event translated to two space-like separated events which may be in the same time in one frame and at different times in another frame?

    You claimed the co-location event was an implementation of ROS. Now prove it.
     
  20. chinglu Valued Senior Member

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    1,637
    I don't know why you waste your time.

    Where is the lightning along the positive x-axis in the M' frame when C' and M are co-located.

    Can you answer this or no.
     
  21. rpenner Fully Wired Valued Senior Member

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    You misunderstand. P is the co-location event. Events Q and R are both space-like separated from event P. But events Q and R are not space-like separated from each other.
    So there exists coordinate frames where P and Q are simultaneous (for example, frame Σ). And there exist coordinate frames where P and R are simultaneous (for example, frame Σ').

    Simple: Line j is defined by the equation \(t = t_P\). Line k is defined by the equation \(t' = t'_P\). Lines j and k meet at only one event. Calculate the (x,t) or (x', t') of that event.* That event is the same event P defined as when C' is co-located with M. Thus there is only one place and time where both \(t = t_P\) and \(t' = t'_P\) are true.

    Events Q and R don't happen in the right place for that to be true. Q.E.D. Relativity of Simultaneity is demonstrated.

    [HR][/HR]


    * Assuming chinglu is too lazy to calculate, because we have seen no new calculations since the OP:
    • Start with \(t_P = \gamma^{-1} \frac{d'}{c}, \quad t'_P = \frac{d'}{c}, \quad t = t_P = \gamma^{-1} \frac{d'}{c}, \quad t' = t'_P = \frac{d'}{c}, \quad t' = \gamma (t - \beta c^{-1} x)\) The only way the last three equations can be true is if \( \frac{d'}{c} = \gamma( \gamma^{-1} \frac{d'}{c} - \beta c^{-1} x) \) or \(x = 0\). So the event where \( t = t_P \) and \( t' = t'_P \) are both true is \( \left( x = 0, \; t = \gamma^{-1} \frac{d'}{c} \right)\) which are the frame Σ coordinates of event P as calculated in the OP.
    • Alternately, start with \(t = \gamma^{-1} \frac{d'}{c}, \quad t' = \frac{d'}{c}, \quad t = \gamma (t' + \beta c^{-1} x')\) and solve \(\gamma^{-1} \frac{d'}{c} = \gamma ( \frac{d'}{c} + \beta c^{-1} x') \) with solution \(x' = - \beta d'\). So the event where \( t = t_P \) and \( t' = t'_P \) are both true is \( \left( x' = - \beta d', \; t' = \frac{d'}{c} \right)\) which are the frame Σ' coordinates of event P as calculated in the OP.
     
  22. chinglu Valued Senior Member

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    OK, where is the lightning along the positive x-axis in the M' frame when C' and M are co-located? Answer the question.
     
  23. rpenner Fully Wired Valued Senior Member

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    OK means all my calculations are 100% correct and Line j is defined by the equation \(t = t_P\), line k is defined by the equation \(t' = t'_P\) and you agree with me that the only event on both lines j and k is event P.
    Because you agree with me that the only event on both lines j and k is event P, you therefore agree with me that the phrase "when C' and M are co-located" can only be properly interpreted with a a frame-dependent convention of simultaneity. As you have selected frame Σ', then you are asking for the x' coordinate of the intersection of line ℓ with the line where \(t'=t'_P\). The only intersection between lines ℓ and k is event R. So you are asking for \(x'_R\) which was calculated in both the OP and [POST=3198606]Post #2[/POST] as \(x'_R = d'\).

    Also, I remind you that in [POST=3198939]Post #15[/POST], I wrote: “It's not "lightning" (闪电)-- it's a flash of light (点样光的颗粒).” Otherwise it would not move at the speed of light as you said in the OP.

    To the contrary, I don't know why you waste your time asking questions that have been answered long ago.
     

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