Lorentz contraction

  1. I think I am correct in saying that Lorentz relativity and standard SR are experimentally indistinguishable. I think it is also the case that in the first, the space between objects does not contract while in the second evrything inclding space contracts pro rata. As a personal choice I prefer standard SR. Its simpler.

    Reading up on Bell's standard spaceship paradox in wiki and other places, I see quite a heated argument and some physicists being accused of some basic misunderstandings. To cut a long story short, one of the arguments is that in one formulation the string breaks and in the other it does not, because space does not contract in one but it does in the other. Now if this argument was valid would it not be possible experimentally to decide between the two formulations.

    I have always understood that the formulations are not distinguishable and see no reason to change my mind. So someones argument is wrong. Can anyone correct my suppositions or enlighten me in any other way.

    Matheinste.
     
  2. jcsd
  3. Doc Al

    Staff: Mentor

    I don't quite understand the controversy. In Bell's paper, if I recall correctly, he shows that the string breaks using the Lorentz approach. It certainly breaks according to standard SR.
     
  4. Yes, they are indistinguishable, they both use the Lorentz transform to make their predictions.
     
  5. DocAl

    Thanks for you quick response. The arguments were mostly on the discussion page of the Wiki entry. I will take a longer, closer look and perhaps I can learn even from the incorrect ones. While I am hardly conversant with the subject, I have always found it surprsing that there have been differing opinions. If you stick by the postulates and physical laws and use logic, surely there is no room for controversy. I must say that some of the discussions seem quite heated and almost personal.

    DaleSpam. Thanks also.

    Matheinste.
     
  6. atyy

    atyy 10,859
    Science Advisor

  7. clem

    clem 1,276
    Science Advisor

    I think Bell claims that the distance between the ships does not change, but a rope connecting them gets shorter. That is wrong.
     
  8. So it would break. So do you think Bell is correct for the wrong reasons?

    My reading in the last couple of days leads me to conclude that the string breaks. Whether it breaks or not was initially of no interest to me. What was of interest was how such a proposed scenario could result in opposite outcomes when analysed by physicists who knew their subject. Perhaps I made the mistake of including some of the contributors to the discussion page of the Wiki entry on Bell's paradox in that category. However, I have learnt a lot from it.

    Matheinste.
     
  9. This is the standard on the paradox.
    http://www.mathpages.com/home/kmath422/kmath422.htm

    If I read it correctly though, the equations assume the string is pulled by the forward ship. What I do not see in the equations is the corresponding push by the accelerating ship from behind.



    Also, here are some excellent papers on calculating the integral for constant acceleration.

    http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html
    http://arxiv.org/PS_cache/physics/pdf/0411/0411233v1.pdf
    http://www.ejournal.unam.mx/rmf/no521/RMF52110.pdf


    I simply cannot see why the string will break though that is what most folks think.

    If the perspective of the launch frame is utilized and the accelerations are perfectly the same, the launch frame applies the acceleration equations and sees the distance between the ships maintained perfectly.
     
    Last edited: Nov 14, 2009
  10. JesseM

    JesseM 8,491
    Science Advisor

    But the electromagnetic forces between atoms in the string will be constantly changing, increasing the tension in the string until it reaches the breaking point. This should be true even if you analyze things wholly in the launch frame, although the details of such a calculation would be beyond me. Still, it's obvious they must change--just think of the case of two identical springs, one at rest in the launch frame and one moving at relativistic velocity, the equilibrium length of the fast-moving spring must be shorter than the equilibrium length of the spring that's at rest in this frame, due to Lorentz contraction. If there were no change in electromagnetic force between atoms as a function of distance in the launch frame, then the two springs would have the same equilibrium spacing between atoms in this frame and therefore the same equilibrium length.
     
  11. I agree, the front part of the string will be pulled and the back part will be pushed.

    Equilibrium will occur in the middle.

    If it breaks, it would be on the front ship side between the middle and the front.

    But, the launch frame in and of itself presents a problem.

    For all t, the equations predict if d is the initial distance between the ships, then d wil be the distance for the ships during the acceleration.

    You can see this in the links to the papers I presented by calculating x after any burntime in the proper time of the launch frame given a constant acceleration.
     
  12. JesseM

    JesseM 8,491
    Science Advisor

    But why do you think this presents a problem? Presumably a detailed calculation of the inter-atomic forces inside the string done from the perspective of the launch frame would show that these forces depend on the velocity of the atoms as well as their distance, so that the tension in the string can be continually increasing even if the length (and the average spacing between atoms) remains constant.
     
  13. Yea, I think of it another way and only from the front ship side.

    Given any segment of the string, there exists a "gravity" potential difference between the side toward the front and the side toward the back.

    Hopefully you agree the back side is not in play because of the reverse gravity potential difference.

    Therefore, there exists a constant distance for the string to operate but a difference in gravity potentials given any segment of the string.

    Now, the original solution claims the string stretches and then breaks without considering the back side and thus the distance between the ships apparently increases.

    I would therefore say, it needs to be decided with this stretching in the front part, because of the gravity differential, what the maximum stress on the string will be.

    Therefore, unless this string is of perfect rigidity, it is not decidable if it will break.
     
  14. JesseM

    JesseM 8,491
    Science Advisor

    Gravity? This is an SR problem...are you talking about a fictitious force in an accelerating frame? I thought you wanted to analyze things from the inertial launch frame.
    Since I don't understand what you mean by "gravity" none of this makes any sense to me.
    What do you mean "without considering the back side"? The distance only increases if you analyze things from the perspective of either ship's instantaneous inertial rest frame at different moments, the distance in the instantaneous rest frame will be increasing regardless if you are looking at the instantaneous rest frame of the back or the front (or of some segment of string in the middle).
    Since your analysis seems to be based on some hard-to-follow conceptual picture and not on any math, and physicists who have done the math all agree the string will break, this suggests that there must be some error in your thinking. If you want help trying to understand where your conceptual picture goes wrong we can discuss that, but this forum is not the place to advance original ideas which contradict mainstream thinking.
     
  15. http://articles.adsabs.harvard.edu//full/1992ASPC...30....1L/0000008.000.html

    Pages 8 and 9 indicate a well known equivalence between acceleration and gravity.

    Not true, CERN's theory division asserted the string would not break.

    Objections and counter-objections have been published to the above analysis. For example, Paul Nawrocki suggests that the string should not break,[3] while Edmond Dewan defends his original analysis from these objections in a reply.[4] Bell reported that he encountered much skepticism from "a distinguished experimentalist" when he presented the paradox. To attempt to resolve the dispute, an informal and non-systematic canvas was made of the CERN theory division. According to Bell, a "clear consensus" of the CERN theory division arrived at the answer that the string would not break

    http://en.wikipedia.org/wiki/Bell_spaceship_paradox
     
  16. All this emphasises my point. With such levels of disagreement, quote and counter quote, reference and counter reference do not help or give any learners confidence in the answers. A clear and carefully presented analysis should speak for itself. The answer cannot depend on who we care to believe.

    Break or not break is a discusson I would not get involved in. With such "eminent" names being cited as references with different conclusions no-one would, or should, take notice of anything someone at my level has to say.

    Matheinste.
     
  17. JesseM

    JesseM 8,491
    Science Advisor

    But the pseudo-gravitational force only appears in an accelerating coordinate system, there is no such force observed in an inertial coordinate system like the launch frame. In any case, you really need to explain your ideas in more detail, I still have no idea what you mean by phrases like "without considering the back side" or "reverse gravity potential".
    This sounds like it was just an informal canvas in which physicists were asked for their initial reaction to hearing the paradox, much like the different reactions of physicists to hearing Feynman's underwater sprinkler puzzle. You cut out the part where Bell added "Of course, many people who get the wrong answer at first get the right answer on further reflection", and this paper says on p. 11 that "Though many of Bell's CERN colleagues originally thought the thread would not break, it is now universally agreed that it, indeed, will." And this paper on the paradox notes on p.4 that "stress is an absolute (frame-independent) physical quantity, which is represented by a tensor". So I'm pretty confident you wouldn't find any peer-reviewed papers disagreeing with the claim that the stress in the string will continually increase as the ships accelerate--the references to the papers by Dewan and Nawrocki are impossible to judge without seeing the details of these papers, but this abstract of another paper refers to the existence of a special case of the Dewan's thought-experiment where the string never breaks, suggesting Dewan might agree that in other cases where the distance is constant in the launch frame it would break (perhaps the special case could be one where the stress in the string increases but in a way that approaches a finite limit instead of increasing without bound, so that if the string was strong enough to withstand the stress of that limit it would never break?)
     
    Last edited: Nov 14, 2009
  18. Assuming that the distance between two objects would be equal to the length of a rope stretched between them, what is the difference between "length contraction" and "space contraction"? It seems to me that if the "space between objects does not contract" then a rope stretched between them would not be contracted either.

    It's not like a rope would stretch between two objects in their rest frame, but only reach part way as viewed from a different frame, because the rope contracted while the distance between the objects didn't.
     
    Last edited: Nov 15, 2009
  19. Obviously, I was talking about the instantaneous frame of the rockets. And, these are not my ideas. The description of the rope I as talking about was from this link.

    Consider a uniform distribution of particles at rest along some segment of the x axis of an inertial coordinate system x,t at the time t = 0. Each particle is subjected to a constant proper acceleration (hyperbolic motion) such that, with respect to its instantaneously co-moving inertial rest frames, the distances to each of the other particles remain constant
    http://www.mathpages.com/home/kmath422/kmath422.htm


    Well, the article did not state that CERN changed its mind. Further, I am not in a position to claim that CERN would look at a problem casually.

    Finally, there is no literature that I am aware of that says CERN backed away from its decision on the rope.

    Here is the x for constant acceleration.
    It is in the links I posted.
    x(t) = c^2/a ( sqrt( 1 + (a t/c)^2 ) -1 )

    Now, by adding 0, ie choosing the back back to initialize at 0 and x0 for the front rocket, you can readily see the x(t) is based only on a and t from the launch frame. Thus, the launch frame will not see a deviation with the distance between the rockets.
     
    Last edited: Nov 15, 2009
  20. I was aware of the fracture for solving this problem.

    I gave both sides to the problem, one by doing the integral in the instantaneous frame of the rockets and rope here
    http://www.mathpages.com/home/kmath422/kmath422.htm

    And, the other by viewing the problem from the launch frame using the constant acceleration equations of SR here
    http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html
    http://arxiv.org/PS_cache/physics/pdf/0411/0411233v1.pdf
    http://www.ejournal.unam.mx/rmf/no521/RMF52110.pdf


    Then, I began analyzing the problem but Jesse does not want me to so I am going to stop.
     
  21. Hello cfrogue,

    I can asure you that my remarks were not aimed at you personally. I was commenting generally on the confusion amomg people, many of which I assume, are well versed in relativity.

    My position is that from what I have read I believe that the correct answer is that the string breaks. However, I am as yet not confident in my ability to explain the reasons to anyone else, which of course is the acid test of understanding. I am also aware that there are variations on the scenario and so we may not all be singing from the same songbook.

    Matheinste.
     
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