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Effects of Gravity on pi

  1. May 19, 2007 #1
    I have heard that while pi is known to many millions, even billions of decimal places, gravity makes it both fundamentally and practically impossible to measure pi to more than 10 significant figures.

    I find this to be a very strage theory, and have never heard of it before. I am also finding it difficult to find any information on the area.

    Does anyone know of any good websites or books that could help answer this mystery? Or if you know anything on the matter yourself? The question is currently haunting me lol
  2. jcsd
  3. May 19, 2007 #2
    I think there is a common misunderstanding here. [itex]\pi[/itex] is the ratio of the circumference of a circle to the diameter. [itex]\pi[/itex] is a constant, it does not change.

    Now it is true that on a deformed surface this ratio is no longer [itex]\pi[/itex] but that does not mean that [itex]\pi[/itex] is changed.

    It is easy to see that the ratio changes if you deform a flat surface. Take a circular piece of clay and stick it on the top of a football, make sure there is no space between the clay and the football. The ratio between the diameter and the circumference is no longer [itex]\pi[/itex].

    Only curvature that deforms changes the ratio. For instance just wrapping the clay around a cylinder does not deform the clay so it does not change the ratio. But sticking it onto a Vietnamese hat deforms it very much, so that means that the ratio changes.

    Gravitation curves spacetime, which is like deforming a surface, but then in four dimensions.
    Last edited: May 19, 2007
  4. May 19, 2007 #3


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    Pi isn't pi because we measure it, it is calculated from theoretically perfect geometric shapes.
  5. May 19, 2007 #4


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    I'm also finding this very strange, I think something important has gotten lost or misrepresented somewhere.

    As far as general curvature goes, the ratio of the circumference of a "circle" (a set of points a constant distance away from a central point to it's radius depends on the value of the radius. The error is quadratic in the radius, see for instance


    or Misner, Thorne, Wheeler, "Gravitation", pg 336.

    This is also discussed somewhat in http://www.eftaylor.com/pub/chapter2.pdf

    this is a chapter from one of the coauthors of "Gravitation" that has some of the same material, parts of which are online, and which is much closer to a popular level.

    The significance of this is not major though - even on the Earth, one can get 10 digits of accuracy for the ratio of circumference /diameter as long as the radius of the circle is less than about 10^-5 of the radius of the Earth - that's about 60 meters.

    The extrinsic curvature of [correction]space due to gravity is much less than the extrinsic curvature of the surface of Earth. It depends on how you "slice" space-time into space+time, i.e. what defintion of simultaneity one adopts, as well as one's distance away from a large mass.

    It's not quite what you are looking for, but http://www.pitt.edu/~jdnorton/teaching/HPS_0410_2007/Assignments/09_Non_Euc_GR/index.html

    may also be helpful about the concept of "curvature" in general relativity,
    Last edited: May 20, 2007
  6. May 19, 2007 #5


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    As has been pointed out, [itex]\pi[/itex], at least the [itex]\pi[/itex] that is "known to many millions, even billions of decimal places" is a specifically defined mathematical constant. It is true that, according to general relativity, in the geometry around a massive object, the "ratio of the circumference of a circle to its diameter" is not even a constant, much less "equal to [itex]\pi[/itex]".

    I'm not at all clear what you mean by "makes it both fundamentally and practically impossible to measure pi to more than 10 significant figures." I'm sure there are difficulties with that that have nothing to do with gravity!
  7. May 22, 2007 #6
    One does not measure pi, it is a defined value: the ratio of a circle's circumference to it's diameter in *Euclidean space*.
  8. May 22, 2007 #7

    Chris Hillman

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    I've seen this nonsense claim at some math crank sites. (Others have already explained why it is nonsense.) I hope you didn't hear it in school or anyplace like that.
  9. May 23, 2007 #8


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    [itex] \pi [/itex] , [itex] \alpha [/itex] , it's all the same.

    maybe at higher temperatures or energies, [itex] \pi [/itex] changes to a different number.
  10. May 23, 2007 #9
    haha, actually I heard it from my university lecturer...

    that online link that pervect gave for the gravitation book gave me an idea as 2 what my lecturer was on about... theres a bit in it that has to do with space being warped by gravity, and it uses an example of a black hole to make the point really obvious... section 4, titled the r-coordinate: reduced circumference, the 7th page down...

    basically, coz the black hole warps spacetime, if u have a circle/sphere around the black hole, the actual radius is not going 2 b the ratio we'r used to, the radius will b greater... but yeh, that site explains it beta than i do haha
  11. May 23, 2007 #10

    George Jones

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    Last week I met my niece's boyfriend, who just finished his first year of university, and he said that his physics prof told the class that dark energy is responsible for the slowing down of the expansion of the universe.
  12. May 23, 2007 #11


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    No, no, no, no, no, no. (one for each post where someone corrected this misconception already....)
  13. May 23, 2007 #12

    Chris Hillman

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    Trying again to explain the point

    Robert, I really hope you were joking. If not, please move this to "Skepticism and debunking".

    Teresa: I think you are missing the point. Pi is defined to be the circumference of a euclidean circle of unit diameter. What the book you read (and probably what our lecturer) actually said--- or if not, what they should have said--- is that while the notion of "circles" still makes sense in noneuclidean geometries, the relation to circumference and area to diameter of these circles is not given by the euclidean formulae
    C=\pi \, d, \; A= \pi \, d^2/2
    but rather by more complicated relations [itex]C=C(d), \, A=A(d)[/itex]. In some cases, e.g. hyperbolic or spherical geometry, it is possible to write down these relations precisely. In general, all one can say is that for very small circles, the euclidean relations hold almost exactly in any curved manifold. That is one reason by euclidean geometry holds a special place in Riemannian geometry (the study of curved manifolds).
    Last edited: May 23, 2007
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