# Question/Puzzle Regarding Space-Curvature Model

1. May 2, 2004

### Edwin

I've been working on the concept below, but have little experience with standard calculas notation and methods.
I was wondering if someone might be able to quantify the model below using equations, you will be credited for your work, of course, in an upcoming 'invited' research paper to be presented at the 6th WSEAS International Conference on Mathematics as a co-author and co-researcher of the concept. The completed paper will be a series of overlapping spaces formed by synching multiple radii in a space of circles to different points in space that are moving. The paper, tentatively, will concentrate on describing the pattern that arrises within each spaces' contour as one point, through which a radius from each circle is said to point to, is taken to infinity along some path that is tangent to some circle(s). The first question is listed below, and is somewhat "off topic" in terms of what I mentioned above, but if the results are positive, then it will be featured as the primary theme of the paper, instead of the contour variance, and the researcher here on this forum that discovers the underlying equations will be featured in the paper as the discoverer of the relation of the model below to the Gravitational "space-time curvature" proposed by Einstein in his "General Theory of Relativity." The purpose of the question posed in the model below is to determine whether this model is compatible with Einstein's Gravitational Model: "space-time curvature."

Best Regards,

Edwin G. Schasteen

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?

Inquistively,

Edwin G. Schasteen

Last edited: May 2, 2004
2. May 2, 2004

### Edwin

In the model above, I forgot to mention, there are two cases where the curvature approaches zero: At an infinite distance from the point marked in space, and any path that passes through the very point itself. The lines that pass through the point that a radius from all given circles are pointing to, are straight lines extending as radii in all directions from the point. If you are using an ordinary Cartesian Graph; then, depending on your "Cartesian plane's relative phase orientation" the curvature of the given radius will become negative if you move up toward the top of your graph, and positive if you move down toward the bottom of your graph. In real space, the overall curvature of the space is negative for any line that does not include the point through which all the circles' radii are pointing to as part of their set. As one moves to an infinite distance from a point, the angle between all points in a finite region of space that is the vertex at the point that is approaching infinity, approaches zero. Thus, at infinity, it might be that the angle between the vertex and all finite points is zero, such that the collapsed "cone" becomes a single infinitely long straight line. This one dimensional straight line, originating from infinity, culminates as an infinite number of parallel lines at infinity to the left, which is our mathematical universe. Thus, in this particular case, the line starting at infinity and approaching our finite universe, if it should make the infinite transition to our universe, would arrive at our universe not as a single straight line, but an infinite number of parallel lines. Thus it would include the point through which all the radii of each circle is pointing to within our finite space as part of it's set, and thus at infinity, the curvature of space would be 0 as consistent with the axiom implied above which reads as follows: If a path is defined by the collection of points in space created by the collection of intersections of a series of overlapping circles' radii with those circles' circumference, and the radii of all those circles are pointing to a single point, then the degree of curvature of the path formed by those collection of points on the circles' circumferences is zero for any path that includes that point as part of their set. If, at an infinite distance from our universe, the angle between the vertex at infinity and all points in our finite universe is 0, then the vertex collapses to a one dimensional line that is one dimensional for any section of the line within a finite distance from the vertex point, that is at infinity, and >1 dimensional for any section of the line that is within an infinite distance from the vertex point that is at infinity. Then a single straight line starting at infinity and approaching our finite n-dimensional universe will include all points in our finite region of n-space as part of their set, and thus the degree of curvature of the space at infinity is 0. I appologize if this is a little confusing, just start with the first post and reconstruct the model, and over time you will see what it is I am refering too, if you do not immediately envision it, which I can't imagine that anyone could without doing the work on paper.

What do you think?

Inquisitively,

Edwin G. Schasteen

Last edited: May 2, 2004
3. May 2, 2004

### Edwin

Also, if the ideas mentioned above are provable via some proof, it may be possible to formulate a proof that any correspondent space that is at an infinite distance from our curve space must have a layer of Euclidean-like space.

Inquisitively,

Edwin G. Schasteen

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?