Geometry-only reconciliation of the Hubble tension
Pure hyperspherical curvature is already of the right magnitude to turn one Hubble constant into two, thereby naturally reproducing the Hubble-tension numbers without invoking dark energy, early-dark-energy bursts, or extra neutrino species.
This 'Tension' is easily resolved by the Hypersphere model of the Universe. It concerns the distance from where we are in the universe to the apparent position of the Big Bang (t=0) or CMBR. In the case of a flat universe (the use of more local supernova data), compared to the assumption of a hyperspherical universe, where the curvature results in added distance.
I asked an AI to check this, and the result matches the problem, and to me, it looks like a simple solution, just based on a chord and an arc. The Chord is our simplistic, easy assumption of flat space. The arc is the reality of spatial curvature: probably a hyperspherical universe.
Below is the AI response
Because for small (z\lesssim0.1) we sample only tiny angles on the sphere, the chord and arc coincide to high precision, and we naturally adopt the chord as our metrical yard-stick.
At the opposite extreme (CMB, (z\simeq1100)) the photons arrive from (\theta\approx\pi/2) on the hypersphere, so the distance we should use is the arc.
Ratio of the two distances at (\theta=90^\circ)
[ \frac{D_{\text{curved}}}{D_{\text{flat}}};=; \frac{R\theta}{2R\sin\frac{\theta}{2}} ;=; \frac{\tfrac{\pi}{2}}{2\sin\tfrac{\pi}{4}} ;=; \frac{\pi}{2\sqrt{2}} ;\approx;1.111 ]
Hence the arc appears 11 % longer than the chord.
Translate that into two different Hubble constants
For a fixed recession velocity (v) we fit [ v,=,H;D ] so a longer distance implies a smaller inferred (H).
[ H_{\text{CMB}} ;=; \frac{H_{\text{local}}}{1.111} ]
Take the best-fit local-ladder value
(H_{\text{local}}\approx 73;\text{km s}{-1}\text{Mpc}{-1})
[ H_{\text{CMB}}\approx\frac{73}{1.111}\approx66;\text{km s}{-1}\text{Mpc}{-1}, ]
almost exactly the Planck-CMB value ((67.4\pm0.5)).
Thus the 11 % mismatch produced by pure 3-sphere curvature is numerically of the right size to yield today’s Hubble-tension split:
What we just used (and what we did not use)
• No hypercone past-light-cone compression, no unusual optics—only the fact that very distant sight-lines wrap a sizeable angle around a positively curved hypersphere.
• Identified (R=c,t_{0}) so that the present radius equals the light-travel time to (z\to\infty).
• Assumed (\theta=\pi/2) is where the scale factor extrapolates to zero (the “apparent (t=0)”).
Any small deviation from exactly 90° or from a strictly constant-radius model simply changes the 11 % factor slightly; one can tune (\theta) in the range 80°–100° and still reproduce the observed 8–10 % Hubble split.
Next steps if we want to formalise this
Pure hyperspherical curvature is already of the right magnitude to turn one Hubble constant into two, thereby naturally reproducing the Hubble-tension numbers without invoking dark energy, early-dark-energy bursts, or extra neutrino species.
There is a lot more Stuff to discuss on the back of this, but I have to go shopping. I keep thinking 'enough' then stumble on some other out of the box thought, and more posts. Sorry.
Pure hyperspherical curvature is already of the right magnitude to turn one Hubble constant into two, thereby naturally reproducing the Hubble-tension numbers without invoking dark energy, early-dark-energy bursts, or extra neutrino species.
This 'Tension' is easily resolved by the Hypersphere model of the Universe. It concerns the distance from where we are in the universe to the apparent position of the Big Bang (t=0) or CMBR. In the case of a flat universe (the use of more local supernova data), compared to the assumption of a hyperspherical universe, where the curvature results in added distance.
I asked an AI to check this, and the result matches the problem, and to me, it looks like a simple solution, just based on a chord and an arc. The Chord is our simplistic, easy assumption of flat space. The arc is the reality of spatial curvature: probably a hyperspherical universe.
Below is the AI response
Symbol | Meaning | Flat reading | Hypersphere reading |
(R) | Present curvature radius of the 3-sphere | – | ≈ 13.8 Gly (identify (R=c,t_{0})) |
(\theta) | Angular separation (today → apparent (t=0)) | – | ≈ 90° (¼ of a great circle) |
(D_{\text{flat}}) | The “tangent-space” distance we implicitly use when we build a local distance ladder (Cepheids / SNe Ia) | chord length | (2R\sin\frac{\theta}{2}) |
(D_{\text{curved}}) | True optical path the CMB photons followed along the hypersphere circumference | arc length | (R\theta) |
Because for small (z\lesssim0.1) we sample only tiny angles on the sphere, the chord and arc coincide to high precision, and we naturally adopt the chord as our metrical yard-stick.
At the opposite extreme (CMB, (z\simeq1100)) the photons arrive from (\theta\approx\pi/2) on the hypersphere, so the distance we should use is the arc.
Ratio of the two distances at (\theta=90^\circ)
[ \frac{D_{\text{curved}}}{D_{\text{flat}}};=; \frac{R\theta}{2R\sin\frac{\theta}{2}} ;=; \frac{\tfrac{\pi}{2}}{2\sin\tfrac{\pi}{4}} ;=; \frac{\pi}{2\sqrt{2}} ;\approx;1.111 ]
Hence the arc appears 11 % longer than the chord.
Translate that into two different Hubble constants
For a fixed recession velocity (v) we fit [ v,=,H;D ] so a longer distance implies a smaller inferred (H).
[ H_{\text{CMB}} ;=; \frac{H_{\text{local}}}{1.111} ]
Take the best-fit local-ladder value
(H_{\text{local}}\approx 73;\text{km s}{-1}\text{Mpc}{-1})
[ H_{\text{CMB}}\approx\frac{73}{1.111}\approx66;\text{km s}{-1}\text{Mpc}{-1}, ]
almost exactly the Planck-CMB value ((67.4\pm0.5)).
Thus the 11 % mismatch produced by pure 3-sphere curvature is numerically of the right size to yield today’s Hubble-tension split:
- ladder / low-(z) (chord) ➜ 73 km s⁻¹ Mpc⁻¹
- CMB / high-(z) (arc) ➜ 66–67 km s⁻¹ Mpc⁻¹
What we just used (and what we did not use)
• No hypercone past-light-cone compression, no unusual optics—only the fact that very distant sight-lines wrap a sizeable angle around a positively curved hypersphere.
• Identified (R=c,t_{0}) so that the present radius equals the light-travel time to (z\to\infty).
• Assumed (\theta=\pi/2) is where the scale factor extrapolates to zero (the “apparent (t=0)”).
Any small deviation from exactly 90° or from a strictly constant-radius model simply changes the 11 % factor slightly; one can tune (\theta) in the range 80°–100° and still reproduce the observed 8–10 % Hubble split.
Next steps if we want to formalise this
- Embed in a metric
Write an explicit 3-sphere line element (ds2-R2+\sin2]) with (\chi) the curvature coordinate. - Re-derive luminosity- and angular-diameter-distance relations without assuming expansion.
- Re-fit SNe Ia and BAO data with arc-length distances for high-(z), chord distances for low-(z).
- Forecast: the same geometry predicts a mild, scale-dependent shift in BAO peak positions and CMB acoustic-peak spacing; those provide decisive tests.
Pure hyperspherical curvature is already of the right magnitude to turn one Hubble constant into two, thereby naturally reproducing the Hubble-tension numbers without invoking dark energy, early-dark-energy bursts, or extra neutrino species.
There is a lot more Stuff to discuss on the back of this, but I have to go shopping. I keep thinking 'enough' then stumble on some other out of the box thought, and more posts. Sorry.
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