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Penrose stairs on the torus

  1. Jan 2, 2016 #1


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    Nobody mentions that Penrose stairs is possible on the torus. I wonder Why. isn't this obvious?
  2. jcsd
  3. Jan 3, 2016 #2


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    Maybe you can explain to us what a Penrose stair is?
  4. Jan 3, 2016 #3


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    Penrose stairs is this impossible object:


    Such an object can exist locally but not globally. The point is, that walking in one direction on the stairs, we are always rising, still we can make a closed walk on it. These stairs are impossible because we cannot draw a closed curve on the surface of a cylinder having the property that both of its coordinates are always growing.

    I say that we can obviously draw such a curve on the torus, but, unfortunately, nobody tells this while dealing with these strairs.

    I mean a description something like this:

    The torus is [itex]S^1\times S^1[/itex]. [itex]S^1[/itex] can be coordinatised on two overlapping charts [itex](U,x)[/itex] and [itex](V,y)[/itex] by [itex]x: U\to (0,1)[/itex] and [itex]y:V\to (0,1)[/itex] such, that there is a [itex]f:\mathbb R\to S^1[/itex] differentiable surjetion, that [itex]\frac {d (x\circ f)}{dt}>0[/itex] and [itex]\frac {d (y\circ f)}{dt}>0[/itex] where they are defined. The curve [itex] \mathbb R\to S^1\times S^1: t\mapsto (f(t),f(t))[/itex] is a walk on the torus that is always rising and returns to a point.

    I am not a professional mathematician, so perhaps my formulation isn't the best, that's why I am looking for a professional formulation of this.
    Last edited: Jan 3, 2016
  5. Jan 3, 2016 #4
    @mma I've wondered about this and this is my drawing. It may be helpful because I was after a torus-like structure.
  6. Jan 3, 2016 #5


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    My stairs on the torus look like this:

    where the the points of the opposite sites of the square are identified, so it is is a topological torus:

    Last edited: Jan 3, 2016
  7. Jan 5, 2016 #6


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    Perhaps this is in relation somehow with the curvature.

    We live in a real line bundle [itex]L[/itex]. Its base space is [itex]S^2[/itex] (the surface of the Earth), and the fibers are the vertical lines. The gravitation defines a connection on this bundle, i.e. determines the horizontal subspaces of the tangent bundle [itex]TL[/itex]. of [itex]L[/itex]. This connection must be curvature-free, because all real line bundles over [itex]S^2[/itex] are trivial. This means, that the horizontal lift of a closed curve from the base space to [itex]L[/itex] is also a closed curve, The horizontal lift of a curve has by definition everywhere horizontal tangent vectors. An unidirectional walk on a Pentose stair has nowhere horizontal tangent vectors. I suspect that curvature-free property rules out the possibility for such curves being closed. That's why Penrose stairs are impossible in our world.

    In contrast of this, torus is a circle bundle.over the circle. It may have a curved connection, so it is possible, that a curve in it has nowhere horizontal tangent vectors, still it is closed.
  8. Jan 6, 2016 #7

    It's not obvious to me. The idea is that stairs go up and down, to overcome gravity. Under this condition a Penrose stairs would be a free energy device. Just roll a ball down the stairs.

    If you are talking about the Penrose stairs as an optical illusion, then this illusion may be drawn on a wide variety of objects. They needn't be topologically a torus.
  9. Jan 6, 2016 #8

    jim mcnamara

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    Hmm. http://mathworld.wolfram.com/PenroseStairway.html

    This says that Penrose Stairs are an impossible object. There is no math in the article describing it - or the links. The links refer to:
    Hofstadter, D. R. Gödel, Escher, Bach: An Eternal Golden Braid. New York: Vintage Books, p. 15, 1989
    And books about optical illusions, M C Escher, and articles in Psychology. So, all are: Art, Philosophy, and Psychology links.

    I would seriously consider this to indicate that this construct is an Escher-illusion, not a mathematically definable object. I am also guessing: it is in mathworld to diffuse discussions like what we have in this thread.
  10. Jan 6, 2016 #9


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    @jim mcnamara
    Still matematicians sometimes take it seriously, see for example here. I would regard these two properties mentioned there by anon:

  11. Jan 7, 2016 #10
    This was what I thought of when I read the thread title... saw it on wikipedia under torus.

    Attached Files:

  12. Jan 7, 2016 #11
    Seriously in the sense that Penrose feels the illusion is worth studying. Possibly new and compelling illusions would be discovered. Or perhaps a real-world 3D object could be made that gives the illusion of being a Penrose stair when viewed from a certain angle under certain conditions.

    Or perhaps there is some imaginary mathematical world within which Penrose stairs are consistent.
  13. Jan 8, 2016 #12


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    I meant originnaly exactly this.
  14. Jan 9, 2016 #13


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    Or an energy losing device. If we neglect the energy dissipation, then in our world we can call height on equal right the vertial distance measured from a fixed point and the mechanical work done since started from this point (because then [itex] E = mgh[/itex]). But if we take into account also the dissipation (e.g. energy loss by air resistance), then the "heigth" on the energy axis will always increase during a walk on closed loop. So, if we take into account the dissipation, then in our world we cannot use this energy axis as a spatial coordinate line, while in the torus-world we can.
  15. Jan 21, 2016 #14
    There are (at least) two ways in which "impossible objects" are possible.

    One is that, because these are based on 2D drawings, there is almost always a 3D object which, when viewed from the appropriate angle, will appear exactly as an "impossible object". A number of science museums have actual rooms designed to appear impossible when viewed from some viewpoint.

    Perhaps more interesting is that, as purely geometric objects, many impossible objects have a totally rigorous existence, even if such things are not subsets of 3-dimensional Euclidean space.

    For instance, consider the helix given parametrically by

    C(t) = (cos(2πt), sin(2πt), t), -∞ t < ∞.

    Now define the set B to be those points in space whose distance to the helix {C(t)} is exactly 1/2.

    Finally, define a new set B' by identifying any point (cos(2πt), sin(2πt), t) of the helix with (cos(2πt), sin(2πt), t+1), which is also on the helix.

    B' is a forever-ascending helical cylinder that is topologically a torus.

    This is similar to the circle of sounds that results from identifying each pure frequency with double that frequency.
  16. Jan 25, 2016 #15
    What do you mean "always rising"? How do you define "up"?
  17. Jan 25, 2016 #16
    Any such geometry would not have a consistent measure of distance in the "up" direction.

    Consider the condition for moving from one stair to the next in a "real" Penrose staircase: you must increase your distance in the vertical direction from the origin. You can never decrease this distance (if you allowed this it would be easy to construct in our world - you simply ramp down the tread of each step). Consider now the condition for completing a circuit of the staircase: you must return to a position which has the same vertical distance from the origin as your start point (because it is your start point). A measure that is always increasing can never return to its starting value.
  18. Jan 25, 2016 #17
    Oh, maybe the forces aren't conservative. Maybe the measure decreases when traveling the stairs in the opposite direction. I feel sure it is possible to come up with some fantasy world where it works after a fashion, but I can't say I'm interested in pursuing that.
  19. Jan 25, 2016 #18
    My argument has nothing to do with forces or direction of travel.
    There is no point in making that statement. Either find a flaw in my argument or a counter-example.
  20. Jan 25, 2016 #19
    I have a third option. I can do something more to my liking.
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