SR Issue

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

  1. paddoboy Valued Senior Member

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    27,543
    I would really think carefully about who you support.
    Is this just an exersise in sympathy for the underdog, or a anti establishment phobia, just for the sake of it?
    chinglu claims SR is false. He claims time dilation and length contraction do not happen. He claims a universal absolute time.
    He has been banned on other forums permanently, not for claiming such rubbish, but for refusing to accept mathematical proof that it does exist, and time dilation and length contraction are real effects of FoR's

    Clocks on orbiting satellites move slower: FACT:
    . Atomic clocks on planes move slower: FACT:
    . Michelson-Morley experiment: FACT:
    . Muon particles decay more slowly while falling:FACT:

    It's OK to sit smugly by and claim everyone has the right to think for themselves and say what they think. But people need to also be responsible for what they say.
    Science is knowledge of the whole world and Universe, and we dont, and I don't want any school age children to be mislead buy 100% false claims.
    Really, in giving chinglu free reign, which is what you are suggesting, we might as well let the Creationists enter all our schools, make claims about some divine deity of choice, and claim evolution is wrong.
    Is that what you really want?
     
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  3. Trippy ALEA IACTA EST Staff Member

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    Moderator note:

    103 posts that are off topic, trolling, or responses to tolling posts have been moved here.
     
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  5. paddoboy Valued Senior Member

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    Everyone is entitled to his own opinion, but not his own facts.
     
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  7. Neddy Bate Valued Senior Member

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    I hate to say it, but chinglu is correct that demonstrating ROS requires at least two events, not one single event. The whole idea of ROS is that two events which are simultaneous according to reference frame Σ' might not be simultaneous according to reference frame Σ. One cannot demonstrate that if one is considering only a single event.

    For the record, this does not mean that I think chinglu is correct in his OP.
     
  8. paddoboy Valued Senior Member

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    Hi Neddy.....
    That was probably my error.
    Arfa and rpenner though seem to have shown chinglu's silly conclusive results re SR.
    I was sticking my big nose in,

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  9. rpenner Fully Wired Valued Senior Member

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    Which is why I introduced line j (a line of simultaneity in frame Σ), and line k (a line of simultaneity in frame Σ'), thus providing more than one, two, or 20 million events.

    Possibly because in the OP he computes frame Σ coordinates and frame Σ' coordinates of the three events, P, Q, and R, and thus demonstrated that \(t_P = t_Q\) and \(t'_P = t'_R\) were compatible with \(t_Q \neq t_R\) in at least one situation.
     
    Last edited: Jun 26, 2014
  10. arfa brane call me arf Valued Senior Member

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    This thread appears to have propagated the question of what's needed to demonstrate that simultaneity is relative.

    Every example I've seen has two observers and at least one event in it. Simultaneous events implies that you need at least two events for each observer to compare; in examples with two observers seeing one event, the other event is the co-location of the observer's frames.

    But relativity of simultaneity doesn't depend on co-location of two observers, only that they have relative motion.

    The universe provides lots of simultaneous events, so arguably the observers won't see just one "external" event in each of their simultaneous frames, they will see, well, a universe of events, but they won't agree on which ones are simultaneous, so therefore any single event (common to both frames) "demonstrates" that simultaneity is relative.

    So what's needed is two observers in relative motion and a universe where information transfer is limited by, well, you know.
     
  11. Neddy Bate Valued Senior Member

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    2,548
    This is simply untrue. One cannot make any determination about simultaneity from a single event. It is meaningless to consider whether one event is "simultaneous" or not.

    Imagine that a firecracker is at rest at x=0.000 and it pops at time t=0.000. Both of those coordinates pertain to the Σ reference frame. Using the Lorentz transformations, we can determine that the coordinates of that one event as they pertain to the Σ' reference frame which is moving by at a velocity of v=0.866c would be:

    x' = γ(x - vt) = 0.000
    t' = γ(t - (vx / c²)) = 0.000
    Where:
    γ = 1 / √(1 - v²/c²) = 2.000

    The result is that the Σ' reference frame would measure the firecracker to be located at x'=0.000 when it pops at time t'=0.000. Note that this does not demonstrate ROS at all.

    ___________________________________


    To demonstrate ROS, we need to add another event besides the one given above. Let there be another firecracker, and this one is at rest at x=-1.000 and it pops at time t=0.000. Since the t coordinates are the same for both events, this demonstrates that those two events are simultaneous according to the Σ reference frame. Using the Lorentz transformations, we can determine that the coordinates of that other event as they pertain to the Σ' reference frame which is moving by at a velocity of v=0.866c would be:

    x' = γ(x - vt) = -2.000
    t' = γ(t - (vx / c²)) = 1.732
    Where:
    γ = 1 / √(1 - v²/c²) = 2.000

    The result is that the Σ' reference frame would measure the other firecracker to be located at x'=-2.000 when it pops at time t'=1.732. Since the t' coordinates are not the same for both events, this demonstrates that those two events are not simultaneous according to the Σ' reference frame. But we already saw that those two events were simultaneous according to the Σ reference frame, because the t coordinates are the same for both events. Thus demonstrating ROS.
     
  12. arfa brane call me arf Valued Senior Member

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    Then I guess I have to ask this: Are two observers in relative motion one event, or two events? If the two observers have intersecting worldlines, is the point of intersection one event or two events?

    I think the answer to the first question is that every point on each of the observer's worldlines is an event, and the answer to the second question is that the point of intersection is a different event for each observer, since they assign different times to it.
    So what can be said to be simultaneous for either observer when they intersect?
     
  13. rpenner Fully Wired Valued Senior Member

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    arfa brane is not wholly wrong, in that you don't need two or more concrete events to demonstrate relativity of simultaneity.

    Let Σ and Σ' be any two inertial coordinate frames that don't share a common standard of rest. Let \(\vec{v}\) describe the relative velocity of the standards of rest to each other. Let P be any event in the space-time that is parametrized by Σ and Σ'.

    Then it should be obvious that the concept of "location" is frame specific. The locus of all events in the same location (according to frame Σ) is straight, time-like line in spacetime. We can call this line \(\vec{x} = \vec{x}_P\). Similarly there is a second straight, time-like line in spacetime which is described as \(\vec{x}' = \vec{x}'_P\). Each of these lines describes a standard of rest for their respective frames, therefore the two lines are not equal to each other. Also event P is common to both lines. Therefore it follows that the two lines intersect in concrete event P.

    Minkowski-Orthogonal (as opposed to a Euclid-Orthogonal) to each time-like line at event P is space-like, 3-dimensional hyper-plane. These planes are described by \(t = t_P\) and \(t' = t'_P\), respectively. Minkowski-Orthogonal at event P means for every event A on the line and every event B on the hyper-plane the following is true in every inertial coordinate system:
    \(c^2 \left( t_A - t_P \right) \left( t_B - t_P \right) - \left( \vec{x}_A - \vec{x}_P \right) \cdot \left( \vec{x}_B - \vec{x}_P \right) = c^2 \left( t'_A - t'_P \right) \left( t'_B - t'_P \right) - \left( \vec{x}'_A - \vec{x}'_P \right) \cdot \left( \vec{x}'_B - \vec{x}'_P \right) = 0\)
    This statement is obviously true if A or B is the same event as P. It's also obviously true in the Σ frame that \(\vec{x}_A = \vec{x}_P\) and the \(t_B = t_P\) implies \(c^2 \left( t_A - t_P \right) \left( t_B - t_P \right) - \left( \vec{x}_A - \vec{x}_P \right) \cdot \left( \vec{x}_B - \vec{x}_P \right) = 0\). But because for any four events in space-time, A, B, C and D, the Lorentz transform guarantees \(c^2 \left( t_A - t_C \right) \left( t_B - t_D \right) - \left( \vec{x}_A - \vec{x}_C \right) \cdot \left( \vec{x}_B - \vec{x}_D \right) = c^2 \left( t'_A - t'_C \right) \left( t'_B - t'_D \right) - \left( \vec{x}'_A - \vec{x}'_C \right) \cdot \left( \vec{x}'_B - \vec{x}'_D \right) \) this condition of Minkowski-Orthogonality is a statement of space-time geometry, not mere coordinates.

    So relativity of simultaneity is demonstrated in that the two different time-like lines that meet at event P have two different Minkowski-Orthogonal 3-dimensional hyper-planes associated with them. So to say event B is on the hyper-plane Minkowski-Orthogonal to the line \(\vec{x} = \vec{x}_P\) at event P is to say event B is Σ-simultaneous with event P. But event B is not Σ'-simultaneous with P if \(\left( \vec{x}_B - \vec{x}_P \right) \cdot \vec{v} \neq 0\). (Here we use the usual 3-dimensional dot-product.) This follows from the Lorentz transform:

    Given: \(t_B = t_P\) and \(\left( \vec{x}_B - \vec{x}_P \right) \cdot \vec{v} \neq 0\), it follows that t' coordinate associated with event B is \(t'_B = \gamma \left( t_B - \frac{\vec{x}_B \cdot \vec{v}}{c^2} \right) = \gamma \left( t_P - \frac{\vec{x}_P \cdot \vec{v}}{c^2} - \frac{ \left( \vec{x}_B - \vec{x}_P \right) \cdot \vec{v}}{c^2} \right) = t'_P - \frac{\gamma}{c^2} \left( \vec{x}_B - \vec{x}_P \right) \cdot \vec{v} \neq t'_P\).​
    http://en.wikipedia.org/wiki/Lorentz_transformation#Boost_in_any_direction

    Since the generic event B proves the two space-like hyper-planes \(t = t_P\) and \(t' = t'_P\) are not identical, the non-equivalence of the hyper-planes is a geometrical truth about space-time, independent of coordinates. As each hyper-plane constitutes a locus of all frame-simultaneous events, non-equivalence of hyper-planes demonstrates relativity of simultaneity. And thus the only thing concretely used was the event P and the two different standards of rest of frames Σ and Σ'. ​
     
  14. Neddy Bate Valued Senior Member

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    2,548
    In relativity, the word 'event' has a very specific meaning: An event is an occurrence of zero duration which has a specific spacial location, and a specific time of occurring. Merely saying "two observers in relative motion" does not describe any event, because there is no specific occurrence of zero duration, and there is no specific location or time of occurring. Thus it would not be correct to call "two observers in relative motion" one event or two events. Neither is the case.

    In my previous post, I used a firecracker popping to represent an event. A firecracker is a little bit of gunpowder wrapped in a small paper tube. One lights a fuse at one end, and when it ignites it pops. In the second half of my previous post, the specific spacial location of this event was x=-1.000 (according to reference frame Σ), and the specific time it occurred was t=0.000 (according to reference frame Σ). However, according to reference frame Σ', the specific spacial location of that same event was x'=-2.000 and the specific time it occurred was t'=1.732. The two different reference frames used two different sets of coordinates to describe the same event.


    Yes, that sounds correct, because every point on a worldline represents an occurrence of zero duration, which has a specific spacial location, and a specific time of occurring.


    That one single firecracker popped at t=0.000 (according to reference frame Σ) and popped at t'=1.732 (according to reference frame Σ'). According to a thousand other reference frames, that one event can have a thousand different time coordinates. Does that mean the one single firecracker pop is actually 1000 different events? No, because all 1000 reference frames would agree that the firecracker only popped once, not 1000 times.


    I'm not sure what you are asking here. Regardless of where the two observers are located on their own worldlines, you still need a minimum of two events to even consider whether those events are simultaneous or not. Events which are simultaneous according to one observer's reference frame, may or may not be simultaneous according to the other observer's reference frame, because ROS only applies to events which are spacially separated along the axis of motion between the two observers.
     
  15. Neddy Bate Valued Senior Member

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    Yes, I agree that the non-equivalence of hyper-planes demonstrates ROS, and that multiple concrete events do not have to be considered. However, I interpreted arfa brane's argument to be that ROS can be demonstrated by just the fact that different reference frames can assign different time coordinates to a single event. I just think that, by itself, is an unsatisfactory explanation for ROS. He further confuses the issue by saying that multiple different time coordinates means multiple different events. I just think taking the word "simultaneous" at face value makes more sense. One cannot say, "This single event is simultaneous," and still expect to make sense to anyone. The appropriate follow up question must be, "Simultaneous with what?" and the correct answer, whatever it is, must be at least one other event.
     
  16. arfa brane call me arf Valued Senior Member

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    Read rpenner's latest post.
    This part pretty much sums it up:
    He appears to be agreeing with what Roger Penrose says about relativity of simultaneity--it's evident when you have two worldlines intersecting (at the point P in rpenner's post).

    Each worldline has a different simultaneous space--the 3-dimensional hyperplane labelled "the present" in Minkowski spacetime (and in spacetime diagrams!).
    Uniform relative motion translates to each observer having different velocities.

    Now look at Penrose's example: two observers moving past each other at low velocities (walking past each other on the earth's surface) have different views of "the present".
    This is more noticeable the further you extend their respective hyperplanes, so that an event in the Andromeda galaxy (Penrose's example) will "occur" at widely separated times for this pair of observers. The remote event, of course, can't be observed for a long time so the separation between these observer's hyperplanes (an angle in Minkowski spacetime) will remain quite small, they will most likely not be alive to confirm any separation relative to the event when worldines intersected.

    So, how many events is that altogether?
     
  17. Neddy Bate Valued Senior Member

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    ------------------------------

    The time and place where the car passes close by the pedestrian is an event.

    The space admirals in the Andromeda galaxy asking the question, "Shall we invade??" on Monday is another event.

    The fleet of space ships heading for earth and screaming, "Kill all humans!!!" on Tuesday is another event.

    I agreed with rpenner's point that the difference in the hyperplanes is enough to demonstrate ROS, and that the hypothetical "Monday" and "Tuesday" events do not have to be considered as concrete events. What I don't agree with is the idea that one can say, "This single event is simultaneous," and still make sense. The appropriate follow up question would be, "Simultaneous with what?" and whatever the answer is, it must be at least one other event,
     
  18. krash661 [MK6] transitioning scifi to reality Valued Senior Member

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    2,973
    and the result is...
    starting over..
    it's obvious what's going here.
    (shakes head)
     
  19. krash661 [MK6] transitioning scifi to reality Valued Senior Member

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    yes
     
  20. krash661 [MK6] transitioning scifi to reality Valued Senior Member

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    why do you continue to say such things.
    it's massively obvious to those, that do not need meds, this is no where in the vicinity of accurate.
     
  21. krash661 [MK6] transitioning scifi to reality Valued Senior Member

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    another attempt to insert the obvious " trolling "
     
  22. krash661 [MK6] transitioning scifi to reality Valued Senior Member

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    do you realize there's a contradiction on your part ?
     
  23. paddoboy Valued Senior Member

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    27,543
    At the very least, this thread should be moved to Alternative section, based on past records of the claimant, and the quite obvious fact that this maths illusion is just a ploy to continue to deride and falsify SR.
    It's that simple.
     

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