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Pi and Gravity

  1. Sep 14, 2010 #1
    Hi, this thought came to my mind..not sure how correct this is. In extreme gravity, the space curves. Will the value of Pi change in this case, for a circle near such gravity ? pi = circumference/diameter.
     
  2. jcsd
  3. Sep 14, 2010 #2

    Mentz114

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    If you draw a circle on the surface of a ball, is pi = circumference/diameter ?

    Spatial curvature does change the ratio circumference/diameter and also the total angle in polygons.
     
  4. Sep 14, 2010 #3
    In extreme gravity you may even have problems with distinguishing space from time.
     
  5. Sep 14, 2010 #4

    D H

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    In general the ratio of circumference to diameter for a circle varies with the size of the circle and the nature of the distance metric (which depends on geometry). In Euclidean space and using the Euclidean norm the ratio is constant and is equal to pi by definition. In any other space, or any other metric, the ratio of circumference to diameter may or may not be equal to pi. This does not mean that pi has a different value in that space / with that metric.
     
  6. Sep 14, 2010 #5

    bcrowell

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    Except in Indiana, where it equals 3.
     
  7. Sep 14, 2010 #6
    pi is a component of the fine structure constant and what i was thinking was....light follows curved path near high gravity. will the fine structure constant be different near such gravity because of pi ?
     
  8. Sep 14, 2010 #7

    Mentz114

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    It is after all, very flat in Indiana.:smile:

    Isn't the angle all the way around a circle always 2 pi ? And similarly with solid angles ?
     
  9. Sep 15, 2010 #8

    bcrowell

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    If you study D H's #4, you'll see that there are two possible cases. (a) A pi appears in a particular formula because it's the ratio of the circumference of a circle to its diameter. (b) A pi appears there for some other reason.

    In case a, it may be reasonable to speculate that the formula would be different in curved spacetime. In case b, not so much.

    Do you think your example is case a, or b?
     
  10. Sep 15, 2010 #9

    bcrowell

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    Yes. An angle is a local thing, and the local geometry of spacetime is always flat.
     
  11. Sep 15, 2010 #10

    pervect

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    If you make sure you keep away from any singularities inside your circle, you'll find that for a small enough circle, the flatness of space-time will give you a value of pi for the ratio of circumference to diameter for a small enough circle as well.

    If you draw a circle around a cosmic string, though, I'd expect that you'd find the ratio of circumference/diameter being non-pi even for a small circle - as long as the circle encloses the singularity (the cosmic string in this case).
     
  12. Sep 23, 2010 #11
    An interesting, but perhaps no so trivial, question would be:

    What is the circumference and area of a unit circle of radius (ruler distance) = 1 with its center at Schwarzschild coordinate r and a Schwarzschild radius of R0 (we assume theta and phi is 0, so the circle is drawn 'flat')?

    If we have a formula we can plot the diversion from pi in terms of the r coordinate and R0.

    Anyone?

    Seems to me that to find the area we need to integrate 'slices' of the circle with constant r, the 'slices' are arcs of a circle of radius r.
     
    Last edited: Sep 23, 2010
  13. Sep 23, 2010 #12

    pervect

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    By definition, we know the circumference of a circle at Schwarzschild coordinate R is 2*pi*R, and the area is 4*pi*R^2 - that's how the schwarzschild radial coordinate R is defined. What we don't know is the "radial distance to the center of the black hole". The question is probably basically meaningless. Among other issues, inside the black hole, r is not a spatial coordinate - i.e. if we consider two nearby points (r,t) and (r+dr,t), there is a timelike separation between these points, not a spacelike separation.
     
  14. Sep 23, 2010 #13
    Completely agree. I added a graph to show this. If we measure the ruler distance between R and R+1 for decreasing values of R we observe that the ruler distance increases. Also, observe the behavior of the radar distance (total roundtrip time) in an additional graph.

    Just in case you wrote this in answer to my question then you completely misunderstood what I was asking, let me try it again but with slightly different words:

    For a given Schwarzschild coordinate value Rc draw a unit circle of ruler length 1 around this point (we assume theta and phi is 0, so the circle is drawn 'flat'). What is the circumference and area given a Schwarzschild radius R0. And obviously Rc > R0.

    But I assume you simply responded to the whole topic at hand.
     

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    Last edited: Sep 23, 2010
  15. Sep 23, 2010 #14

    DrGreg

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    I've no idea what you mean by this. If [itex]\theta[/itex] and [itex]\phi[/itex] are both zero, you must be talking about a radial straight line, not a circle.
     
  16. Sep 23, 2010 #15
    Aarg...I see my mistake.

    I meant [itex]\theta=0[/itex], of course [itex]\phi[/itex] is not!

    Ok, perhaps a picture would help.
     

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    Last edited: Sep 23, 2010
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