Basic Special Relativity Question

Discussion in 'Physics & Math' started by Fednis48, Apr 22, 2013.

  1. Tach Banned Banned

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    No, it doesn't , it simply means exactly what I said, that you are unable to measure \(t"=-\frac{Vk\gamma'}{c^2}\). I challenged you to do that and you are still fumbling and bumbling but you have no ability to answer.


    Nice try, I can demonstrate length contraction using the plane of simultaneity in \(S"\) , \(dt"=0\) (as I have already done) , this doesn't mean that the timestamps can be measured, as you amply demonstrate in you inability to do so.
     
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  3. Tach Banned Banned

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    No, I am not saying that, you should stop lying. I have told you repeatedly that this is not what I am saying, I challenged you with finding out what I am saying and you managed to find the exact quote, it doesn't support your lying assertion.
    If you do not understand what I am saying, then you should stop posting because it just makes you look like you are either ignorant or a liar (or both).
     
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  5. Neddy Bate Valued Senior Member

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    Let's see...

    It seems to me that you were saying that the rod does not hit sequentially, according to the platform. Maybe you should clarify your position.
     
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  7. Tach Banned Banned

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    Do you understand the sentence "the contact is the same in all frames"? I.e. the physics of the contact is frame invariant?
     
  8. OnlyMe Valued Senior Member

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    What you are saying then, Tach.., is that if the rod is hits the floor of the train parallel to the floor in any frame it hits the floor parallel, in all frames?
     
  9. Tach Banned Banned

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    Yes, and I have proved it, mathematically. What brings you back to this thread, last you were here, you were desperately denying length contraction.
     
  10. Pete It's not rocket surgery Moderator

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    Liar.
    Here are the values for the collision, measured by \(S''\) clocks lined up along the \(x''\) axis:
    (k, t'')
    (0.00, 0)
    (0.75, -1)
    (1.50, -2)
    (2.25, -3)
    (3.00, -4)
    (3.75, -5)
    (4.50, -6)
    (5.25, -7)
    (6.00, -8)

    I can demonstrate that the rod is bent using the plane of simultaneity in \(S''\), \(dt"=0\) (as I have already done).
    The rod bending is exactly as physical as length contraction.

    Tach, we agreed that the Lorentz transform tells us what the \(S''\) observer would measure.
    Why don't you just face up to the facts? Why do you keep wriggling further into ludicrous arguments?
    Do you really want your argument to rely on the \(S''\) not being able to measure the proper time of events in \(S''\)?
    Are you hoping that those events are somehow not real if you can stop them from being measured?
     
  11. Tach Banned Banned

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    How did you measure these values , Pete? What instrument(s) did you use, Pete?


    Nope, length contraction is a "measurable" effect as per the links you asked for, "rod bending" , not so much.
     
  12. Pete It's not rocket surgery Moderator

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    What Tach is saying is that the events of the collision occur parallel to the floor of the train in all reference frames, which is trivially true because the events o the collision obviously occur on the floor.

    He knows that these events happen at different times in the platform frame.
    He knows that while the rod is moving across the train it is not parallel to the floor in the platform frame.
     
  13. Neddy Bate Valued Senior Member

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    1,430
    Let's see it in context:


    Look, you clearly did not think the rod hit the floor sequentially according to the platform. You went to great lengths to explain that the different times shown on the platform clocks were only meaningless labels, and that they did not make the rod hit sequentially, according to the platform. If I am misunderstanding you, then simply come out and say that the rod does in fact hit sequentially in the platform frame. Simply come out and say that the different times on the platform clocks represent the different times of impact of the rod, according to the platform frame. Simply come out and say that regardless of all that, the rod does not become dented on one end, because the physics of the problem remain as they are in the rod's own reference frame. That is what Pete has been saying, and you have been arguing with him about it. So, just say it and you two can finally agree.
     
  14. Tach Banned Banned

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    Good, you are learning.

    ....which times, aren't measurable through any experimental setup.
    ....which times, are simply timestamps assigned by the observer in \(S"\), devoid of any physical meaning

    Sure , that is a trivial case of Thomas precession, I taught you about it.
    The issue in discussion is not the motion through the car frame but what happens at collision. The motion is frame-variant, the physics of the collision, not so much.
     
  15. Pete It's not rocket surgery Moderator

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    I'm sure you can figure something out, but here are some possibilities.
    The time values are measured with clocks.
    The k values are barcoded on the rod, and detected by barcode readers attached to the clocks.
    I disagree.
    Both length contraction and rod-bending are demonstrated in \(S''\) in the same way - by examining events on a plane of simultaneity in \(S''\).
     
  16. Pete It's not rocket surgery Moderator

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    10,166
    :bravo:
    Liar
    The \(S''\) observer strongly disagrees. \(S''\) clocks measure proper \(S''\) times.

    :bravo:
     
  17. Tach Banned Banned

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    5,265
    Good, looks like you finally got it.


    One last time, the sequence of numbers \(t^"_k=-\frac{kV\gamma'}{c^}\) has no physical meaning, it is not measurable through any experimental setup, it is just a bunch of decreasing numbers assigned by the \(S"\) observer to the contact points \(k\).



    Trouble is, this is NOT what Pete is saying, Pete insists that the "angle" between the rod and the platform is real, measurable and, worst of all, non-zero.
    At this point, you seem to have a better grasp of the underlying physics than Pete. It is good that he's studying to be a doctor, he makes a terrible physicist. The sad part is that being so obstinate and dishonest will make him into a terrible physician as well.
     
  18. Pete It's not rocket surgery Moderator

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    While the rod is moving across the train. And you agree.

    :roflmao:
     
  19. OnlyMe Valued Senior Member

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    First I never denied length contraction. Re-read slowly and carefully. I was attempting to make a distinction, between length contraction, as a theoretical component of SR and a proven measured fact. Time dilation was once theory. It has been proven experimentally for both SR and GR and is now accepted as fact.., but length contraction has never been directly measured, and so it remains theory. A component of a theory that has a good track record of predictions. Still not all predictions have yet been raised above the level of theory. But that is a different issue and I was trying to let it rest until some resolution of the OP was had... So ...


    The problem, in the discussion of the OP, as I see.., and I have been inclined to agree with your conclusion if not your methods, is that on one hand you have the rod is parallel with the floor of the train, at all times, in the train's frame.., and the floor of the train is parallel with the platform, at all times, in the train's frame.., so the rod is parallel with the platform, at all times, in the train's frame. Which given your above agreement with my last post, should make the rod parallel to the platform at all times in all frames, without respect to measurement variations between frames... Simple algebra postulate of Equality, if a=b and b=c, the a=c.

    But Pete has a transformation from (or rather to) the platform frame backed up by a published paper (and from comments rpenner has made I think backed up by rpenner), that says the rod is not at all times parallel to the floor of the train, when measured from the platform frame...
     
  20. OnlyMe Valued Senior Member

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    Your last sentence above is what I have trouble with. I can easily get that measurements from the platform frame would not measure the rod to be parallel, in a direct sense, but that seems like it has to be a measurement issue rather than a representation of the physical conditions. There should be no real difference between the physical relationship between the rod and the floor as it falls and when it hits. If they are parallel, then they are parallel. Measurements from another frame that do not reflect that immediately must be either a RoS issue or some difficulty in measurement... Or so it would seem.

    When you say is not parallel to the floor in the platform frame, are you intending that as an absolute not parallel or as not measured to be parallel?

    I should really take the time to go over the paper you referenced. Things have just been a bit, busy lately.
     
  21. Pete It's not rocket surgery Moderator

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    10,166
    It's not absolute, it's relative. In general, parallelism is relative, just like length is relative.
    Did you see the other thread?
    [thread=134529]Relativistic parallel rods[/thread]
     
  22. Tach Banned Banned

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    Sure, I even corrected your calculations, remember?
    The point is that the collision results into a zero angle in all frames, you need to come to grips with that.
     
  23. Pete It's not rocket surgery Moderator

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    Yes, of course it does. That was always obvious to everyone.

    Look, you agree the rod is not parallel to the floor in the platform frame as it moves across the train.
    So of course it will bend as it meets the floor. I don't know why this is so difficult for you to accept.
    What do you think the rod does in the platform frame in the time between the two ends of the rod meeting the floor?
    Will it not do this?

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    What else could it possibly do?
     

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