Concerning the 3d surface of a Hypersphere

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JR Gibson
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Asked Ai for a Cosmology forum
Although 'undefined' it may be reasonable to assume that the hypersphere radius and circumference will respond to the normal relationship of a sphere i.e. 2 times Pi times r. Reasonable I suggest because the 3d surface of a hypersphere is a sphere.
To get to my point, time and distance can be exchanged as 1 second = 299792458 meters* then when you use the age of the universe (approx
13.8 billion years as a radius): -
1. The circumference increase per second gives a Hubble Constant of 71.
2. The 4 dimensions of our spacetime have time relegated to a process occurring in 4 spatial dimensions and not 'the' 4th dimension
3. It is hardly surprising that we cannot perceive all 4 spatial dimensions directly (often depicted as a flatlander watching a sphere penetrating his flatland; he only sees a circle expanding then growing smaller and finally disappearing)
4. As a result a) what is thought of as viewing the early universe is in error it is just viewing 90 per cent around the hypersphere where curvature dilates time to look like t=0. Think of a black hole near the event horizon judged from far away ( to get this you need to envisage the whole of time as a spacetime interval of one) b) The actual distance to t=0 is - if I remember, something in the '20s of billions of years. c) the assumption of a flat universe suggests the error of a tangent from the Hypersphere rather than the curvature (a greater distance in time and space is the real situation, maybe)
* Spacetime Physics, Taylor Wheeler

Just a thought....
 
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oops looks like I mis posted
 
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