Relativistic parallel rods

Discussion in 'Physics & Math' started by Pete, Apr 30, 2013.

  1. Tach Banned Banned

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    Because it was interesting to see how many different ways I could disprove Pete's claims. This is not our first rodeo, this is the third time he's modified the scenario, so I'm used to it.



    But Pete was not giving the correct interpretation of the result, his interpretation is exactly the opposite to the one given by the paper authors.

    The above is downright hilarious, I see your persistent confusion. The fact that I make \(\Delta t'=0\) in order to get the coordinates \(\Delta r'\) in S' as a function of the coordinates \(\Delta r\) in S does not imply that I am forcing \(\Delta t=0\). Nowhere in the derivation there is any such implication. Nor does that mean that the relationship derived is incorrect. It is just basic algebra.....Fednis48 has been harping on the same thing.....
     
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  3. Fednis48 Registered Senior Member

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    I did find that bit interesting. Since the "extended present" necessarily includes other reference frames, it's a bit of a cop-out when we want to know how the laws of physics play out in a particular reference frame. Janus' argument was that any stress signal propagates more slowly than simultaneity can catch up with itself, which is just a less formal way of saying that the two ends of the rod are out of causal contact between the two collisions (ie. the collisions are in each other's extended presents). I'd be intrigued to hear if anyone can come up with a non-stress-based mechanism underlying the "extended present" in this situation.

    Regardless, though, I find it cute that in response to three quotes from the paper that call your math into question, you cherry pick the one that's easiest to answer and pretend the other two didn't happen. Even if Janus got some details wrong, your insistence that the rods hit each other parallel in all frames is flat-out contradicted on page 3, and your method of "instances" is discredited shortly thereafter. I don't suppose there's any chance you'd actually stoop to admitting an error, though.
     
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  5. Tach Banned Banned

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    Look,

    You can't do any math
    You do not understand what I am claiming (hint: my claims do not have anything to do with RoS, so you can stop beating this old and tired strwaman)
    You do not understand the paper I linked

    So why do you keep posting is a mystery.
     
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  7. przyk squishy Valued Senior Member

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    Posting a calculation for a different problem than the one defined in the OP does not prove that the OP is wrong. It only serves to make you look like you haven't understood anything. Is that the impression you want to give? Then think about how you respond more carefully in future.


    What paper are you talking about? arXiv:0809.1721 [physics.class-ph]? Have you actually read the paper? Because it comes to pretty much the same conclusion as Pete and everyone else here did. For instance, as Fednis48 already pointed out, near the end of the introduction they say:

    Isn't this exactly what's illustrated on Pete's pretty pictures? Isn't this exactly what you called "unphysical"? The only way I see the authors agreeing with you is in that they seem to be under the impression that this is a difficult paradox.
     
  8. Tach Banned Banned

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    I didn't need any calculation, post 2 explains the flaws in Pete's post 1.



    So, you think that you can devise an experiment that can execute measurements in the "extended present" since the so called "collisions" take place in the "extended present". Nice!
     
  9. Fednis48 Registered Senior Member

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    Oh, no you don't. I had to jump back to the previous thread to find a mathematical formulation of your "instances" method, but since you've continued to insist on instances in this thread, I think the quote is still relevant.

    (These formulas were going from \(S'\) to \(S''\), whereas Tach was just talking about going from \(S\) to \(S'\), but the math doesn't depend on how many primes we use to label each frame.) A cursory glance at these equations will show that you fixed \(t'=t'_0\) in the original frame, then transformed only the spatial coordinates into the new frame. Setting \(\Delta t''=0\) in the boosted frame is the correct method, it's the method I've been arguing in favor of, and it's the method you've been arguing bitterly against. To claim otherwise makes you a sleazeball and/or a pathological liar.
     
  10. Tach Banned Banned

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    You need to jump back to the basics, you know the part that you are unable to follow. Here is the link.
     
  11. przyk squishy Valued Senior Member

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    Post #2 amounts to denial of the relativity of simultaneity effect.


    Did I say I could? Did I even express any interest whatsoever in this "extended present" idea, that I hadn't even heard of before I'd seen this paper? No. So what are you talking about? You said the paper agreed with you. Actually, interpretation aside, they derive the same raw result that Pete did in the OP, which you have specifically insisted was wrong. The only odd thing about this paper is that the authors seem to think this is hard and actually worth discussing.
     
  12. Tach Banned Banned

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    Good, you learned this.

    Correction: they show the same derivation but they arrive to contradictory conclusions. The 'angle' is not measurable. Last I checked, non measurable means non-existent.
     
  13. Fednis48 Registered Senior Member

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    Aaaand the ad hominems come out again. Even ignoring the bit where Janus was right, I quoted two parts of the paper that say you're wrong. You can't get out of that by insulting my math skills.

    Sure I can. Make a vertical stack of clocks, synchronized in their rest frame. Whenever a segment of the rod passes by a clock, stop that clock. Then measure the velocity of the rod. Given its velocity and the clock measurements, you can easily determine what its angle was as it zoomed by. Do this for two rods going in opposite directions, and you can deduce whether the rods were parallel. Woah!:xctd:
     
  14. przyk squishy Valued Senior Member

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    Correction: they reach the same result as Pete and then claim it is a contradiction. In reality, the result is not a contradiction and should not be surprising to anyone who is actually familiar with relativity, and in particular the relativity of simultaneity effect.
     
  15. Tach Banned Banned

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    ...meaning that it cannot physically happen.
     
  16. Fednis48 Registered Senior Member

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    Good notes, but utterly nonresponsive. The paper flat-out says you are wrong, for at least two reasons I quoted. Calling me names won't change that. Reminding me how length contraction works won't, either.

    ^This. I'm glad Tach posted the paper, if only because I can now quote a source he can't disagree with when I call out his mistakes, but it seems like a pretty unremarkable result.
     
  17. przyk squishy Valued Senior Member

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    So, you're an anti-relativity crackpot now? The authors derived that result because it is specifically what relativity predicts. They didn't present another solution. The only way to disagree with it is to disagree with relativity.
     
  18. Tach Banned Banned

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    "Thus, as explained, this paradox presents a real contradiction" (top of page 4).
    You obviously don't understand what the authors mean by that.




    When you do that, you get the "real contradiction" that the authors were referring top of page 4. Performing math does not mean that you are going to get physically meaningful results:

    Blue observer says: the rods got dented at the "A" (bottom) end.
    Red observer says: the rods got dented at the "B" (top) end.




    You got this totally backwards, the paper contradicts Janus' approach.
     
  19. Tach Banned Banned

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    No, it is just that your skills are too limited in following the argument.

    I don't disagree with the conclusion of the paper, the problem is that you don't understand the conclusion of the paper <shrug>
     
  20. Fednis48 Registered Senior Member

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    Oh, ok. When you posted the paper, I thought you were saying that the authors' resolution of the apparent contradiction was relevant to our discussion. I didn't realize you fixated on "this paradox presents a real contradiction" and concluded from there that the presented scenario is unphysical. Having a relevant source that you and I agreed on had my hopes up, and finding out now that you don't understand your own source's conclusions is discouraging. I'll leave the thread with one last summary of your viewpoints, in light of this new discovery.

    Let the record show that, according to Tach, the following scenario is unphysical:
    "Two inclined rods, identical in their respective rest frames, are moving toward each other with constant velocity."
     
  21. przyk squishy Valued Senior Member

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    One of the conclusions the paper arrives at is that relativity predicts the result depicted in Pete's diagrams in the OP. The only way you can disagree with that is to disagree with relativity or disagree with the author's derivation of that result. You cannot maintain you agree with the paper on this point and yet maintain Pete is wrong.

    Another conclusion the paper arrives at is that this result presents some sort of difficult paradox. This, I don't agree with, and have no reason to agree with since it seems like a simple consequence of relativity of simultaneity.
     
  22. Tach Banned Banned

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    You are being dishonest again (or downright incompetent). The unphysical part is:

    Blue observer says: the rods got dented at the "A" (bottom) end.
    Red observer says: the rods got dented at the "B" (top) end.
     
  23. rpenner Fully Wired Valued Senior Member

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    Just for the record, let's translate this PDF from 2013-04-27 into a real post:

    The problem with this purported proof is that the assumption is that the Lorentz transform transforms vector \( \mathbf{a} = \mathbf{a}_{\parallel} + \mathbf{a}_{\perp}\) to \( \mathbf{a}' = \frac{1}{\gamma} \mathbf{a}_{\parallel} + \mathbf{a}_{\perp}\) which is not what the Lorentz transform says because the Lorentz transform does not operate on isolated coordinate values, or even triplets of coordinate values. -- The Lorentz transform works on all four coordinates. // Added: Or rather it is a space-time transformation so you can't worry about the temporal parts and neglect the spatial parts or vice-versa.

    If the Lorentz transform did operate on spatial vectors, then you could use the fact that it is a linear transform \( \Lambda k \mathbf{x} = k \Lambda \mathbf{x}\) to prove parallel vectors stay parallel without all that awkward explanation of the dot product. But it doesn't and therefore you can't.
     

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