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Question Regarding. Gravity, Relativity and Curve Space

  1. May 1, 2004 #1
    Draw a set of 4 horizontal parallel "x" lines, and 4 vertical "y" lines that intersect to form a square grid. This should make 16 squares. Within each sqaure, draw a circle that is just large enough to fit snuggly in each sqaure. Then pick a point on the paper somewhere within the squares and mark it with a pencil or pen. Next, using a ruler, measure a straight line from the center of every circle through the point you marked on the paper and where your ruler intersects the circumference of the circle, mark a pencil dot. Do this for all 16 circles. After you have completed this step, connect all the dots on each circle within each collumn and row you made on each circle's circumference. You will notice that there exists an underlying series of varying curves by the pattern of dots you made on the circles' circumferences that get more extreme as you approach the point on the piece of paper. This curvature is pretty obvious with just sixteen circles, but if you were to fill in 1,000 overlapping circles within that square grid and repeat the steps above, you would notice a pronounced curvature, that almost seems like Einstein's curved space. If this grid were without bounds, then the degree of curvature of space would approach zero as you go further and further from the point in space you marked on the paper. It is presumable that any where within a finite distance of the point, the space that we formed by the method mentioned above, has negative curvature. It is presumable that as you approach an infinite distance from the point on the paper, the curvature of the space we made, approaches zero. The question I have is as follows: is there a solution such that the arbitrary negative curvature of the mathematical space mentioned above exactly equals the degree of negative curvature of space-time predicted for an object composed of x number of oscillators with a mass m kilograms?


    Edwin G. Schasteen
  2. jcsd
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