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Posers and Puzzles

Posers and Puzzles

  1. 09 Jan '10 15:39
    Would we be able of seeing through a sphere where for every point (x,y,z) , x,y,z are rational numbers?
  2. Subscriber AThousandYoung
    Proud Boys Beware
    09 Jan '10 17:16
    Originally posted by smaia
    Would we be able of seeing through a sphere where for every point (x,y,z) , x,y,z are rational numbers?
    As long as the sphere isn't larger than our ability to see, yes, because that's a mathematically defined sphere, not a real material sphere.
  3. Standard member wolfgang59
    Infidel
    09 Jan '10 17:43
    Originally posted by smaia
    Would we be able of seeing through a sphere where for every point (x,y,z) , x,y,z are rational numbers?
    What????????????
  4. Standard member TheMaster37
    Kupikupopo!
    09 Jan '10 18:32
    Originally posted by smaia
    Would we be able of seeing through a sphere where for every point (x,y,z) , x,y,z are rational numbers?
    I don't think so. There might be infinitely many sight-lines but all of them are too small to let any photons through. I don't have the proof, but it results in something like this;

    In every part of the sphere there are infinitely many rational points. So a photon would always hit infinitely many points on the sphere.
  5. 09 Jan '10 20:17
    Originally posted by smaia
    Would we be able of seeing through a sphere where for every point (x,y,z) , x,y,z are rational numbers?
    This is not a sphere, and spheres don't exist in nature.
  6. Standard member Palynka
    Upward Spiral
    09 Jan '10 20:37
    I think what smaia is getting at is more related to the continuity of a sphere (i.e. is the mesh fine enough). Like wondering if the rational numbers are not complete, for example (which they are not).

    If that is the question, then the answer should be "no" as a sphere is a compact.
  7. 09 Jan '10 21:34
    Originally posted by Palynka
    I think what smaia is getting at is more related to the continuity of a sphere (i.e. is the mesh fine enough). Like wondering if the rational numbers are not complete, for example (which they are not).

    If that is the question, then the answer should be "no" as a sphere is a compact.
    That's precisely the case.
    All that needs to be done is to prove such "sphere" is compact.
  8. Standard member TheMaster37
    Kupikupopo!
    10 Jan '10 10:35
    Originally posted by smaia
    That's precisely the case.
    All that needs to be done is to prove such "sphere" is compact.
    The term I was thinking of was 'dense'.

    "A space is compact", crudely said; "you can cover your space with finitely many 'tiles' of given size."

    "A subset is dense", equally crudely said; "you can approximate every point in your space arbritarily well with points of the subset."

    That you can cover the rational sphere with finitely many tiles of given size doesn't imply that you don't leave a hole somewhere. I think you need that the rational sphere is dense to be able to say that.
  9. Standard member wolfgang59
    Infidel
    10 Jan '10 11:04
    Originally posted by smaia
    Would we be able of seeing through a sphere where for every point (x,y,z) , x,y,z are rational numbers?
    Consider the sphere created by all the points (x,y,z)
    such that x^2 + y^2 + z^2 < 1
    and x,y and z are rational
    (ie a sphere of radius 1)

    If we we only consider points (Ap, Bp, Cp) where A,B,C are integers and p is half the size of a photon then we would have a mesh too fine to see through.

    If however we assume the OP used 'see' to indicate the possibility of a 'line of sight' through the sphere then there are an infinite number of such lines of sight e.g any line on the plane x=pi/4
  10. Standard member Palynka
    Upward Spiral
    10 Jan '10 13:40 / 1 edit
    Originally posted by Palynka
    I think what smaia is getting at is more related to the continuity of a sphere (i.e. is the mesh fine enough). Like wondering if the rational numbers are not complete, for example (which they are not).

    If that is the question, then the answer should be "no" as a sphere is a compact.
    Oops, I don't know why I read real, where smaia wrote rational.

    As wolfgang says, then the answer is yes and there are many such lines of sight as the sphere will not be locally compact.
  11. Standard member Palynka
    Upward Spiral
    10 Jan '10 13:49 / 1 edit
    Originally posted by TheMaster37
    The term I was thinking of was 'dense'.

    "A space is compact", crudely said; "you can cover your space with finitely many 'tiles' of given size."

    "A subset is dense", equally crudely said; "you can approximate every point in your space arbritarily well with points of the subset."

    That you can cover the rational sphere with finitely many tiles of ewhere. I think you need that the rational sphere is dense to be able to say that.
    In topology, dense sets are subspaces of a topological space and so are said to be dense in that space. It's a characteristic of subsets, not spaces. In that sense, the rationals are actually a dense subset of the real numbers!

    However, the space generated by a real sphere with a point removed will not be a compact because you can construct a sequence of points within that space that in the limit approaches the removed point (outside the space).
  12. 10 Jan '10 16:56
    Originally posted by TheMaster37
    There might be infinitely many sight-lines but all of them are too small to let any photons through.
    Can I throw a spanner in the works? It seems likely to me that firing a light source at this object would lead to scattering, using the wave-like behaviour of light. Thinking in terms of single protons probably doesn't work.

    Not that I feel like trying to calculate the scattering pattern .
  13. 10 Jan '10 18:42
    Originally posted by mtthw
    Can I throw a spanner in the works? It seems likely to me that firing a light source at this object would lead to scattering, using the wave-like behaviour of light. Thinking in terms of single protons probably doesn't work.

    Not that I feel like trying to calculate the scattering pattern .
    I think there is a caveat - It is believed, but not yet proved, in nature, the distance between two points cannot be smaller than the planck lenght-> 10^^-32 meter. In this case, any finite length taken in the sphere has to be multiple of the planck length (Assuming the conjecture is correct). What happens then if a single photon hits the sphere?
    Some commenters mentioned the "size"of the photon. What is the size of a photon? Assume the conjectiure is false and nature allows any distance to be as small as you want. Then the question is: what is the minimum distance that allows a photon to pass through?
  14. Standard member wolfgang59
    Infidel
    10 Jan '10 20:12
    Originally posted by smaia
    I think there is a caveat - It is believed, but not yet proved, in nature, the distance between two points cannot be smaller than the planck lenght-> 10^^-32 meter. In this case, any finite length taken in the sphere has to be multiple of the planck length (Assuming the conjecture is correct). What happens then if a single photon hits the sphere?
    Some commen ...[text shortened]... want. Then the question is: what is the minimum distance that allows a photon to pass through?
    Yes ... you have caught me out!
    What is the size of a photon?

    I assumed that the question was a mathematical one, even though I gave two answers! (A 'reality' one and a mathematical one)
  15. 12 Jan '10 10:12
    Originally posted by wolfgang59
    Yes ... you have caught me out!
    What is the size of a photon?
    The size of a photon is of no interest. The transparancy of a material is.

    A photon can go miles through glass (think fibre optics).
    But a photon cannot pass a micron of gold.