SR Issue

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

  1. paddoboy Valued Senior Member

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    27,543
    When you answer my many previous questions.
    You have nothing chingy. If you did have anything, you would not be here.
    You would be presenting a proper peer reviewed scientific paper.
    Time and space are not absolute. While you continue to fight that, you'll continue to be wrong.
    Live with it.
     
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  3. chinglu Valued Senior Member

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    1,637
    OK, can you prove the math is flawed?
     
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  5. paddoboy Valued Senior Member

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    27,543
    I've answered that question.
    Why do you not accept that time and space are not absolute, despite observational evidence to 100% support that?
    While you so stubbornly refuse to accept that, you will continue to always be in error, and applying misinterpretations as to what is really happening.
    This has been shown to you, many times in many threads, that obviously you do not want to discuss.
    That's OK, you shall run away again.
     
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  7. Neddy Bate Valued Senior Member

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    2,548
    Oh good, so we agree on this. The coordinates you gave in the OP say the light is located at two different places at two different times. But you are claiming the light should be located at the same place at two different times. Who should we believe, you or SR?
     
  8. arfa brane call me arf Valued Senior Member

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    7,832
    No it isn't; the correct answer includes which observer sees the event at \((d',0,0,d'/c)\), you have failed to specify this, so you aren't describing special relativity but something else.
    Again, the correct answer includes an observer, there are two of these, one for each frame. Which observer are you (intentionally not) referring to?
    No, chinglu got the answer wrong, or came to a completely false conclusion. You have, in fact, only demonstrated that, given a vague enough description of simultaneity, you can get all kinds of wrong answers.
    You haven't defined what you mean by "unique", and it is not something you find described in special relativity. There are no unique events, or times, or distances; what the hell are you doing introducing a term that has nothing to do with the physical reality? Wait, don't answer that.
    What you seem to be incapable of comprehending is that each observer sees what they "deem to be true", and truth is not unique (how can it be?).
    SR doesn't "claim" anything, observers do.

    It doesn't matter what a Lorentz transform "says" about where another observer moving relative to a stationary frame will see an event; the stationary observer sees the event in their own system of coordinates, the moving observer does too. It doesn't matter what they calculate about their relatively moving counterpart or how many transforms they try, they cannot, and will never, see what the other sees.
    That's why your attempt at discrediting Einstein is so pathetic; you think the transform can "do something" or something . . .

    You are a complete idiot.
     
  9. chinglu Valued Senior Member

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    1,637
    This is not what the OP says.

    It says, given M and C' are co-located, the primed frame puts the light at \((d',0,0,d'/c)\).

    Given M and C' are co-located, the unprimed frame puts the light at \((d'(1-v/c),0,0,d'(1-v/c)/c)\).

    Einstein wrote,
    https://www.fourmilab.ch/etexts/einstein/specrel/www/

    So, the OP is talking about one time and only one time, that being \(d'/c\). If M and C' are co-located, then all clocks in the primed frame are \(d'/c\) according to Einstein.

    Now, given M and C' are co-located, where does LT claim the light is located? The answer is \((d'(1-v/c),0,0,d'(1-v/c)/c)\). Well, that is false. First, given M and C' are co-located, there is only one time in the primed frame and that is \(d'/c\). So, LT gets the time wrong.

    LT is required to provide what is true in the primed frame, but it gets it wrong.

    Further, the light is located at \((d',0,0)\). Yet LT says it is at \((d'(1-v/c),0,0)\). LT got that wrong also.

    So, there is only one time in the primed frame with M and C' co-located and that time is \(d'/c\) for all clocks in the primed frame.

    LT is required to match the truth in the primed frame, but as we can all see it fails.
     
  10. chinglu Valued Senior Member

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    1,637
    Nope that is not the way SR works. All observers in the primed frame agree the light pulse is at \((d',0,0,d'/c)\) given M and C' and all observers in the primed frame agree their clocks read \(d'/c\) with M and C' co-located.

    Again you are confused. The OP with the primed frame is only concerned with primed frame observers. All agree on events in the primed frame and all have the same time on their clocks.

    You somehow think primed frame observers disagree on events and that is completely false under SR.

    So, it is OK that M has a different space and time coordinate from C' but it is not OK that LT gets the space-time location of the light flash in the primed frame wrong with M and C' are co-located.

    I explained unique in the OP by proving each frame agrees on the same times on the clocks at C' and M given they are co-located. That means neither clock can have a different time with C' and M being co-located. Therefore, it is a unique event.

    You completely got this wrong again.

    Given C' and M are co-located, the light flash is at \((d',0,0,d'/c)\) period and no where else.

    But, given C' and M are co-located, M claims it is at \((d'(1-v/c),0,0,d'(1-v/c)/c)\), which is false.

    So LT got it wrong.
     
  11. chinglu Valued Senior Member

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    1,637
    To all. It is easy to think about the problem this way.

    Assume you are the C' observer in the primed frame. Given C' and M are co-located, you place the light pulse at \((d',0,0,d'/c)\). That is the correct answer period based on the light postulate. There are not 2 correct answers.

    Now, at your common location, the M observer in the unprimed frame says you are full of it and your light is actually located at \((d'(1-v/c),0,0,d'(1-v/c)/c)\).

    What would you say to M? You would tell that observer they are wrong. It is that simple.

    LT gets the answer wrong so LT fails and contradicts the primed frame light postulate.
     
  12. brucep Valued Senior Member

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    4,098
    You couldn't be more wrong chinglu. Why you're allowed to 'stonewall' the scientific truth over 550 posts is outside my ability to comprehend.
     
  13. arfa brane call me arf Valued Senior Member

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    7,832
    No, the unprimed frame has unprimed coordinates. You are committing the sin of mixing primed and unprimed frames, exactly what you were accused of on page 1.
    How did you manage to work out that Einstein's reference to a stationary frame with stationary clocks is about a moving frame? This is something nobody else has ever done!
    Who cares? what matters is where the primed or unprimed frames "observe" this light.
    No, you fail to see your error. You've been failing since the OP. Are you a failure, or what? Wait, don't answer that.

    Here is your mistake, once again:
    The M observer doesn't see anything at \((d'(1-v/c),0,0,d'(1-v/c)/c)\), because those coordinates are in the primed frame, not the frame M is actually in.
     
  14. chinglu Valued Senior Member

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    1,637
    Then you will state why you think I am wrong and that way I can correct you.
     
  15. chinglu Valued Senior Member

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    1,637
    Well, when you translate from the unprimed frame to the primed frame and compare that to what the primed frame see, how exactly is that frame mixing?

    You see, it is a comparison of primed to primed coordinates. That was the whole point of the OP. Perhaps you should re-read it.

    1) I am operating only in the primed frame and what is true there. This is not the moving frame. Then, the moving frame (unprimed frame) gives its incorrect answer.

    2) OK, you answer the OP. Where is the light in the primed frame if C' and M are co-located. Since you say you are right, answer this question.

    Good, at least you have that right.

    Where is the light in the primed frame if M and C' are co-located. M says it is at \((d'(1-v/c),0,0,d'(1-v/c)/c)\) and that is wrong and I have proven that. Prove it is right if you disagree.
     
  16. arfa brane call me arf Valued Senior Member

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    7,832
    Here is your original post.

    What it looks like you've tried to define is a single observer, C'. But the location of this observer is defined by a velocity, so M and M' have a relative velocity.
    This appears to be you implying that colocation of M and M' means that event is common to both frames. This is not true, however, and it can't possibly be true if observer C' has a velocity relative to M.
    No, observers located at M and M' (if they are there) will be able to calculate all of that if the relative velocity between frames is known.

    Note though, how you appear to imply that the time on the clock at M can be defined in terms of d', as if d' is a common distance for M and M'.
    You should have an equation that relates d' to units of distance in frame M; you intentionally leave this step out because it makes it more confused and that's what you really want. you want to show everyone how confusing Einstein's theory is when you leave out some critical details. You want us to believe that colocation is some kind of unique event that means an axis is shared. That idea is obviously crazy, and so, I guess, are you.
    You say this is the M frame solutions, but you've added solutions for M', why? Doesn't that contradict what we as observers actually do? You then repeat this procedure (adding the other frame's solutions) below.
    You say the M' frame has solutions for itself and for M. This is confused, misleading rubbish.

    Where is the logic in your extremely vague and misleading definition of a "problem"? When are you going to stop pushing this piece of shit up the hill?
     
  17. chinglu Valued Senior Member

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    1,637
    Prove this with the Winstein paper.

    You are refuting Einstein.

    https://www.fourmilab.ch/etexts/einstein/specrel/www/

    I simply used LT for my calculations. Is LT wrong?
    I can't understand why it threatens you to ask where is the light in the primed frame. Can you explain that?

    Further, can you explain given M and C' are co-located why the unprimed frame gets the position of the light in the primed frame wrong?
     
  18. arfa brane call me arf Valued Senior Member

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    7,832
    How do you reconcile this:
    with your calculations:
    where you have something other than what Einstein specifies in his paper? You have nonzero values for t and t', but they're supposed to be zero at colocation.

    You can see why you've done this and why you use d' exclusively: it's confusing, and it has an immediate problem, which is that t is not defined in the stationary system, but rather in the moving system.
     
  19. Neddy Bate Valued Senior Member

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    2,548
    Your mistake is in thinking that M does not understand relativity. If M understands relativity, then he will know that the location where he places the light at the time when C' and M are co-located is not the place where C' would place the light. All M would have to do is look at the co-located C' clock and see that it displays \(t'=d'/c\) and he will know that C' would place the light at \(x'=ct'=cd'/c=d'\).


    And the situation is reciprocal. If C' understands relativity then he will know that the location where he places the light at the time when C' and M are co-located is not the place where M would place the light. All C' would have to do is look at the co-located M clock and see that it displays \(t=d'/c\gamma\) and he will know that M would place the light at \(x=ct=cd'/c\gamma=d'/\gamma\). Now it is true that the event \((x,y,z,t)=(d'/\gamma,0,0,d'/c\gamma)\) transforms to the primed frame as \((x',y',z',t')=(d'(1-v/c),0,0,d'(1-v/c)/c)\) but the time coordinate \(t'=d'(1-v/c)/c\) does not match the time when C' and M are co-located, which is \(t'=d'/c\). So relativity is telling you straight to your face that event is not simultaneous with the co-location of C' and M according to the primed frame. Thus, the only way to make relativity get the 'wrong' answer is to totally misunderstand what relativity is telling you, as you, chinglu, have been doing all along. Next!
     
  20. arfa brane call me arf Valued Senior Member

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    7,832
    With this definition:
    chinglu can apply a bit of mathematical trickery: without ever defining d', or which observer measures d', and then defining t and t' in terms of d', he can produce something that appears to contradict "nature".

    Of course, it doesn't matter too much what d' is (as Neddy just showed, and others have shown this), as long as you stick to the "rules" of frame dependence, something chinglu wants us to believe is unnecessary because, according to him, two frames can share a common x-axis (we are supposed to interpret "share" and "common" freely, since neither terms are defined by him anywhere).

    He is just trying to pull the fast one.
     
  21. brucep Valued Senior Member

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    4,098
    Pull a 'slow one' is more appropriate to his scholarship on this subject. I enjoyed reading the posts which demonstrate chinglus complete lack of scholarship and intellectual honesty during his presentation. Otherwise ?
     
  22. chinglu Valued Senior Member

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    1,637
    Well, if you have x and t, you can use :T acquire x' and t' of the corresponding observer and clock time at that event.

    That is how I reconcile it. It is simple SR.
     
  23. chinglu Valued Senior Member

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    1,637
    Now, what you are saying here is that LT is completely useless and fails to give the correct answer. Try to keep in mind, LT is part of SR.

    Keep in mind, we are only talking about primed frame coordinates. So you are confessing that M will not place the light pulse at the same location in the primed frame that the primed frame light postulate places it given M and C' are co-located.

    Therefore, LT contradicts the light postulate in the primed frame.
     

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