<p>The equation is</p>
<p><span class="math-container">$$a = \frac{\sqrt{2} \cdot c^2}{k}$$</span></p>
<p>where a is the acceleration due to gravity, c is the speed of light and k is the radius of curvature of the geodesic. The equation is dimensionally consistent and predicts that the more highly curved is the geodesic, the greater is the acceleration due to gravity.</p>
<p>The source of the equation is as follows. If the universe is modeled as a 3-sphere expanding at the speed of light, paths of light (geodesics) always follow logarithmic (equiangular) spirals with an angle of 45 degrees. The radius of curvature of such a logarithmic spiral is 2<sup>0.5</sup> times the radius of the spiral.</p>
<p>In addition there is a cosmological acceleration due to gravity in such a universe equal to</p>
<p>a  =  c<sup>2</sup>/r</p>
<p>(I think this equation may also hold for the event horizon of a black hole)</p>
<p>Substituting for r gives the equation of the question. The 3-sphere model often seems to correctly predict features of our universe. I was wondering whether it had done so in this case, perhaps some feature of general relativity.</p>

CC BY-SA 4.0
2022-04-25T19:17:55.037Z
It might help if you summarised the "reverse engineering"  that produced this equation. What was your starting point?  How does the "root 2" factor arise? If it doesn't come from general relativity, what does k represent (since curvature of spacetime is something that is only defined in GR.)
James K
CC BY-SA 4.0
2022-04-26T11:51:37.413Z
So how did you calculate the geodesic?
Carl Witthoft
CC BY-SA 4.0
2022-04-29T16:17:39.200Z
*"The 3-sphere model often seems to correctly predict features of our universe."* Do you have a citation for that?
PM 2Ring
<gravity><cosmology>
By dint of some 'reverse engineering' I have produced the equation below. Does anyone recognise it? Does it come from general relativity?
John Hobson
# By dint of some 'reverse engineering' I have produced the equation below. Does anyone recognise it? Does it come from general relativity?

(Asked by: username_0 on 2022-04-25)

The equation is

$$a = \frac{\sqrt{2} \cdot c^2}{k}$$

where a is the acceleration due to gravity, c is the speed of light and k is the radius of curvature of the geodesic. The equation is dimensionally consistent and predicts that the more highly curved is the geodesic, the greater is the acceleration due to gravity.

The source of the equation is as follows. If the universe is modeled as a 3-sphere expanding at the speed of light, paths of light (geodesics) always follow logarithmic (equiangular) spirals with an angle of 45 degrees. The radius of curvature of such a logarithmic spiral is 20.5 times the radius of the spiral.

In addition there is a cosmological acceleration due to gravity in such a universe equal to

a = c2/r

(I think this equation may also hold for the event horizon of a black hole)

Substituting for r gives the equation of the question. The 3-sphere model often seems to correctly predict features of our universe. I was wondering whether it had done so in this case, perhaps some feature of general relativity.

**Tags:** gravity, cosmology
> **Comments:**
> **username_1**: It might help if you summarised the "reverse engineering"  that produced this equation. What was your starting point?  How does the "root 2" factor arise? If it doesn't come from general relativity, what does k represent (since curvature of spacetime is something that is only defined in GR.)  
> **username_2**: So how did you calculate the geodesic?  
> **username_3**: *"The 3-sphere model often seems to correctly predict features of our universe."* Do you have a citation for that?  

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## 0 answers

