Basic Special Relativity Question

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

  1. OnlyMe Valued Senior Member

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    I don't think here it is the math, that is the real issue. It is the "obvious reasons" Tach referrs to, as why his corrections are valid. On that note what is obvious to some may not be so obvious to others and since the discussion has gone on for as long as it has, those obvious reasions should be made in print, as in explained.
     
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  3. Pete It's not rocket surgery Registered Senior Member

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    Look at the platform frame animation. The floor of the train is at y''=0:

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    The events of the collision are denoted by red dots.
    As you can see, after the collision of any element of the rod with the floor, the y'' coordinate of that element is y''=0.
     
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  5. Pete It's not rocket surgery Registered Senior Member

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    The rod bending is a function of proper time in \(S''\), attributed by \(S''\) to events happening in \(S''\).
    Do you think the events don't happen in \(S''\)?
    Try to derive length contraction from the Lorentz transform without transforming the time coordinates. You'll find length extension instead.
     
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  7. Tach Banned Banned

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    Pete, stop playing games. What I said is that time labels have no physical meanings.

    You are twisting my point, again. The rod "bending" in \(S"\), the way you show it, is time dependent, it happens gradually as the points of the rod make contact with the car floor at different time labels. Please stop playing games.

    \(dx'=\gamma(dx-vdt)\)
    \(dt'=\gamma(dt-vdx/c^2)\)

    Mark both endpoints simultaneously :
    \(0=dt'=\gamma(dt-vdx/c^2)\)

    so

    \(dt=vdx/c^2\)

    resulting into:

    \(dx'=dx/\gamma\)

    No time dependent effect, it doesn't happen "sequentially".

    Now, Gron clearly stated that the rod "bending" is not physical. But length contraction IS physical, there is experimental proof of that.
     
  8. tashja Registered Senior Member

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    I don't know.. There is a weird air of uncertainty hanging over the whole thread. I, for one, would like to have closure. A simple ''Tach is right/wrong'' from Alpha would do it for me lol.
     
  9. mikelizzi Registered Senior Member

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    Your analysis is exactly right Pete. And nice animation to boot. I have a real time simulation program that shows the same thing.
     
  10. Undefined Banned Banned

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    Pete, I don't understand what point Tach is making in this bit:
    I suggested a scenario before where there are two rods and the train roof overhung the platform surface while the train floor was in sliding contact with the level surface edge of the platform. When the rods released and move (one rod directed down to the floor on the train side of the train-platform sliding contact line, and the other rod directed down to the platform surface on the platform side of the train-platform sliding contact line), they fall practically side by side with practically no "D" between the train floor-edge and platform surface-edge. So what does that practically "Zero D" mean to the above maths section invoking "invariant D" between floor of train and platform as Tach uses it? Can someone explain its relevance and implications either way for my naive understanding please?
     
  11. Undefined Banned Banned

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    Pete, I am not sure what Tach means here:
    Didn't Tach himself suggest at one point cutting a slit in train floor so rod could pass through and math done accordingly for that case to illustrate something important?
     
  12. Pete It's not rocket surgery Registered Senior Member

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    What you said is that the \(S''\) observer is just applying labels to events happening in \(S'\).
    That is wrong.
    The \(S''\) observer is using \(S''\) clocks to measure the time of events happening in \(S''\).
    For the \(S''\) observer, the \(S''\) clocks measure proper time. They are not just applying labels.

    Yes. So what? It's still exactly as physical as length contraction.
    Both the bending of the rod and length contraction of the train depend on transforming the time component of event coordinates. If you insist that the \(S''\) observer uses \(t'\) coordinates, then you'll find that they measure length extension, not length contraction.

    You are not using 'physical' is the same sense as Grøn.

    In the sense used by Grøn, length contraction is not a 'physical effect':
    They use 'physical effect' to mean absolute effects, effects that are Lorentz invariant:
    The bending of the rod is exactly as physical as length contraction:
    If you use 'physical' in such a sense that length contraction is physical, then the bending of the rod is also physical in that sense.
    If you 'physical' in such as sense that length contraction is not physical, then the bending of the rod is also not physical in that sense.

    To clarify:
    Both length contraction and the bending of the rod are real measurements in \(S''\), but neither effect the physical structure of the objects.
    Length contraction does not crush the train.
    The bending of the rod does not break the rod.
     
  13. OnlyMe Valued Senior Member

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    Tach, could you provide a reference to support the portion in bold above. To the best of my knowledge there is no direct evidence that supress length contraction. Time dilation yes. But using time dilation as a proof of length contraction is not a direct proof. It is only inferred...
     
  14. Tach Banned Banned

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    Sure they are labels, they are the transformed of the car floor impact times \(t'\) into the platform frame, they aren't direct measurements. They cannot be since the car is moving wrt. the platform.



    Nowhere does Gron write such nonsense, there is experimental confirmation of length contraction, you keep trying to conflate is with "rod bending". Length contraction is a measurable effect via experiment, the rod "bending" is not.





    SR is a theory that concerns itself with the measurement of invariants, hence, the "rod bending" is not physical , no matter how you twist and turn. If the gradual bending in your pictures were real, the rod would not end up as a wet noodle, it would end up with a permanent deformation (or breaking). Since the whole effect is simply the result of difference in time labeling between two different frames, the effect is non-measurable, therefore, non-physical.



    No, it isn't , there is experimental confirmation of length contraction. The "rod bending" is non-measurable, therefore there will never be an experimental confirmation, hence it isn't equivalent to length contraction.
     
  15. Tach Banned Banned

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    "supress"? what are you talking about?
    Anyways, Pete knows what I am talking about, let me finish with him.
     
  16. Pete It's not rocket surgery Registered Senior Member

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    You seem to be saying that the \(S''\) observer can't make direct measurements of events in the \(S''\) frame, which is obviously wrong.
    The \(S''\) observer uses \(S''\) clocks and rulers to directly measure the events involved in the rod's collision with the floor.
    The Lorentz transforms of the events from \(S'\) tell us what the \(S''\) observer would directly measure.

    Read the quote.

    You keep saying this, I keep asking for proof, and you never deliver.
    Prove it. Find a relativistic theory of materials, and show us the maths.
     
  17. Undefined Banned Banned

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    Pete, why does Tach think that the very same kind of clock "labels" which are involved in measuring "length contraction" from remote frames in "spacetime terms" are not valid for measuring rod properties in this exercise? This is where he says one is experimentally proven but the other can never be?
    My naive question is: IF the very same kind of clock labels are valid in one case, they are valid in the other case. Tach should explain himself when saying it's "just unphysical labels" in your case and not just the same "unphysical labels" in the case of spacetime relativity calculated length contraction interpretation of the physics and properties he is saying are real?
     
  18. Tach Banned Banned

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    The train car is closed. Have an observer on the platform do the measurements for the rod falling inside the car. I dare you.



    Don't have to, materials bend permanently, regardless of speed regime. So, the rod should have some permanent deformation in \(S"\) yet none in \(S'\).
     
  19. OnlyMe Valued Senior Member

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    OK "supports"
     
  20. Tach Banned Banned

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    There is experimental confirmation of length contraction. This experimental confirmation has direct application in particle accelerators (otherwise, particle accelerators would not work). Pete knows about it and he also knows that there could not be any experimental measurement of the rod "bending" effect , this is why he now avoids the subject. I will get back to you with more details once I am done with Pete.
     
  21. Pete It's not rocket surgery Registered Senior Member

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    So you are saying that the platform observer can't make direct measurements of the events?!
    Is that seriously what you want to base your argument on?
    Would your argument fall apart if the train car was open?
    We agreed that this scenario is in the domain of SR, therefore The Lorentz transform from \(S'\) tells us what the \(S''\) observer measures.
    This is what they measure, in agreement with your own equations:

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    Handwaving. Show us the maths.
    Also, Grøn and Johannesen disagree with you:
    ...as observed in \(\Sigma'\) the rod is severely bent, and should surely be broken-at least according to conventional elasticity theory. The fact that the rod does not break even if it is bent very much, tells us that one needs a relativistic theory of elasticity.
    The rod is bent.
    It does not break.
     
  22. Motor Daddy Valued Senior Member

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    Is this physical or not, Tach?

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  23. Tach Banned Banned

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    I dared you to make the measurement when the train car is closed. Do it.

    Not at all. Measure the impact times \(t"(k)\) of each point of the rod while the train car is zipping by at , say, 0.8c. I dare you.

    True but Lorentz transforms do not replace measuring devices.

    If it is bent, why doesn't it stay bent , how come there is no "memory" of any bending in either frame. Do you think that the rigid rod is a wet noodle because your pictures "prove" that?

    Why not? If it is brittle, according to the "proof" in your pictures , it should break in the platform frame, while not breaking in the car frame. Or it would comminute different ways in the two frames.
     

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