Reference: https://doi.org/10.5281/zenodo.17594186
Standard cosmology tells us the Universe is about 13.8 billion years old. In Quantum Traction Theory (QTT), there is a deeper absolute time, and in that clock the age comes out closer to 15.4 billion years.
In this post, we’ll do something very concrete: starting from observed baryons and the QTT White Void (WV) creation law, we will derive an absolute age of about 15.4 Gyr. No dark energy parameter, no fitted “drift” fudge factor – just baryon density, Newton’s constant, and the QTT micro-creation rule.
1. The Observed Baryon Density Today
We begin with two pieces of observational input:
- The present-day Hubble rate \(H_0\) (from e.g. Planck CMB data).
- The present baryon density parameter \(\Omega_b\).
1.1. Critical density and baryon fraction
The critical density today is
[
\rho_{c,0} = \frac{3H_0^2}{8\pi G}.
]
Take (Planck-like values for definiteness):
- \(H_0 \simeq 67.4\ \text{km s}^{-1}\text{Mpc}^{-1}\),
- \(\Omega_b \simeq 0.049\) (about 4.9% of the critical density).
Convert \(H_0\) to SI units. One megaparsec is \(1\ \text{Mpc} \approx 3.0857\times 10^{22}\ \text{m}\), so
[
H_0
= 67.4\ \frac{\text{km}}{\text{s}\,\text{Mpc}}
= 67.4\ \frac{10^3\,\text{m}}{\text{s}} \cdot \frac{1}{3.0857\times 10^{22}\,\text{m}}
\approx 2.19\times 10^{-18}\ \text{s}^{-1}.
]
Now square it:
[
H_0^2 \approx (2.19\times 10^{-18})^2
\approx 4.80\times 10^{-36}\ \text{s}^{-2}.
]
Newton’s constant is \(G = 6.6743\times 10^{-11}\ \text{m}^3\text{kg}^{-1}\text{s}^{-2}\). Compute \(8\pi G\):
[
8\pi \approx 25.133,
\qquad
8\pi G \approx 25.133\times 6.6743\times 10^{-11}
\approx 1.678\times 10^{-9}\ \text{m}^3\text{kg}^{-1}\text{s}^{-2}.
]
Then
[
\rho_{c,0}
= \frac{3H_0^2}{8\pi G}
\approx \frac{1.44\times 10^{-35}}{1.678\times 10^{-9}}
\approx 8.6\times 10^{-27}\ \text{kg m}^{-3}.
]
This is the familiar critical density. Now the baryon density is simply
[
\rho_{b,0} = \Omega_b\,\rho_{c,0}
\approx 0.049\times 8.6\times 10^{-27}
\approx 4.2\times 10^{-28}\ \text{kg m}^{-3}.
]
Cross-check: this corresponds to roughly \(0.25\) protons per cubic metre (since \(m_p\approx 1.67\times 10^{-27}\ \text{kg}\) and \(0.25\times 1.67\times 10^{-27}\approx 4.2\times 10^{-28}\)). Good.
We will now take
\(\rho_{b,0} \approx 4.2\times 10^{-28}\ \text{kg m}^{-3}\)
as our single observational input about “how many baryons per cubic metre” the Universe has today.
2. The QTT White Void Ledger: How Baryons Create Space
Quantum Traction Theory adds a microscopic law of creation: each Planck bundle of baryons seeds a fixed number of White Voids, and each White Void mints a fixed quantum of space per Planck tick.
2.1. From Planck bundles to White Voids
In the QTT ledger:
- Each Planck mass of baryons seeds exactly 24 White Voids over cosmic history.
- Each White Void (WV), once born, produces one space quantum every Planck tick \(t_P\).
- Each space quantum has volume \(V_{\rm SQ} = 4\pi \ell_P^3\), with \(\ell_P\) the Planck length.
If the domain has baryon mass \(M_b\), then the number of Planck bundles is
[
N_{\rm bundles} = \frac{M_b}{m_P},
]
and as QTT counts through all the WV births and SQ ticks, one finds that the total 3-volume minted by baryons by absolute time \(T\) is
[
V_{\rm WV}^{(b)}(T) \simeq 48\pi\,G\,M_b\,T^2.
]
This is a QTT result: the \(T^2\) comes from counting ticks, and the coefficient \(48\pi G\) comes from identifying the creation units with Planck geometry and using \(G = \ell_P^2 c^3/\hbar\).
2.2. Baryon density as a function of absolute time
From that volume, the baryon density at absolute age \(T\) is simply mass over volume:
[
\rho_b(T) = \frac{M_b}{V_{\rm WV}^{(b)}(T)}
= \frac{M_b}{48\pi G M_b T^2}
= \frac{1}{48\pi G T^2}.
]
QTT baryon density law: \[ \rho_b(T) = \frac{1}{48\pi G T^2}. \]
Notice something crucial: the baryon mass \(M_b\) cancels out. Once you accept the WV ledger, the relation between baryon density and absolute age is completely independent of how big a chunk of the Universe you choose. It’s a pure law of the form \(\rho_b \propto 1/T^2\) with a fixed prefactor.
3. Equating QTT and Observations: Solve for the Absolute Age
We now impose that the QTT baryon density at “today” equals the observed baryon density:
[
\rho_b(T_0) = \rho_{b,0}.
]
Using the QTT law,
[
\frac{1}{48\pi G T_0^2} = \rho_{b,0}.
]
Solve this for the absolute age \(T_0\):
[
T_0^2 = \frac{1}{48\pi G\rho_{b,0}},
\qquad
T_0 = \frac{1}{\sqrt{48\pi G\rho_{b,0}}}.
]
3.1. Plug in the numbers
We already have:
- \(\rho_{b,0} \approx 4.2\times 10^{-28}\ \text{kg m}^{-3}\),
- \(G = 6.6743\times 10^{-11}\ \text{m}^3\text{kg}^{-1}\text{s}^{-2}\),
- \(48\pi \approx 48\times 3.14159265 \approx 150.80.\)
First compute \(G\,\rho_{b,0}\):
[
G\rho_{b,0}
\approx (6.6743\times 10^{-11})\times(4.2\times 10^{-28})
= 6.6743\times 4.2\times 10^{-39}.
]
Multiply the mantissas:
- \(6.6743\times 4 \approx 26.6972\),
- \(0.2\times 6.6743 \approx 1.3349\),
- sum \(\approx 28.032\).
So
[
G\rho_{b,0} \approx 2.803\times 10^{-38}\ \text{s}^{-2}.
]
Now multiply by \(48\pi\):
[
48\pi\,G\rho_{b,0}
\approx 150.80\times 2.803\times 10^{-38}.
]
Compute the mantissa:
- \(150.8\times 2.8 \approx 422.2\),
- \(150.8\times 0.003 \approx 0.45\),
- total \(\approx 422.7\).
Thus
[
48\pi\,G\rho_{b,0}
\approx 4.23\times 10^{2}\times 10^{-38}
= 4.23\times 10^{-36}\ \text{s}^{-2}.
]
Therefore
[
T_0^2 \approx \frac{1}{4.23\times 10^{-36}}\ \text{s}^2
= \frac{1}{4.23}\times 10^{36}\ \text{s}^2.
]
Since \(1/4.23 \approx 0.2364\),
[
T_0^2 \approx 2.364\times 10^{35}\ \text{s}^2.
]
Take the square root:
- \(\sqrt{2.364} \approx 1.538\) (because \(1.5^2=2.25\) and \(1.54^2\approx 2.37\)),
- \(\sqrt{10^{35}} = 10^{17.5} = 10^{17}\sqrt{10} \approx 3.1623\times 10^{17}.\)
So
[
T_0 \approx 1.538\times 3.1623\times 10^{17}\ \text{s}
\approx 4.86\times 10^{17}\ \text{s}.
]
3.2. Convert seconds to billions of years
Convert to years using \(1\ \text{yr} \approx 3.15576\times 10^7\ \text{s}\):
[
T_0\ \text{(yr)} \approx
\frac{4.86\times 10^{17}}{3.15576\times 10^7}
\approx 1.54\times 10^{10}\ \text{yr}.
]
Divide by \(10^9\) to get gigayears:
\(T_0 \approx 15.4\ \text{Gyr}.\)
We have just derived an absolute age of about 15.4 billion years directly from
- the observed baryon density \(\rho_{b,0}\),
- Newton’s constant \(G\),
- and the QTT WV microcreation law \(\rho_b(T) = 1/(48\pi G T^2)\).
No dark energy term, no arbitrary cosmological constant, and no free “time drift” parameter entered this derivation.
4. Coasting Gauge Check: The Absolute Hubble Rate
QTT’s coasting gauge says the absolute Hubble rate is simply
[
H_\tau(T) = \frac{1}{T}.
]
So at \(T_0 \approx 4.86\times 10^{17}\ \text{s}\),
[
H_{\tau 0} = \frac{1}{T_0}
\approx 2.06\times 10^{-18}\ \text{s}^{-1}.
]
Convert this to the usual km s\(^{-1}\) Mpc\(^{-1}\):
- 1 Mpc \(\approx 3.0857\times 10^{22}\ \text{m}\),
- and 1 km = 1000 m.
Thus
[
H_{\tau 0}
\approx 2.06\times 10^{-18}\ \text{s}^{-1}
\times 3.0857\times 10^{22}\ \frac{\text{m}}{\text{Mpc}}
\times \frac{1\ \text{km}}{10^3\ \text{m}}.
]
Combine the powers of ten:
[
2.06\times 3.0857 \approx 6.36,
\qquad
10^{-18}\times 10^{22}\times 10^{-3} = 10^{1}.
]
So
\(H_{\tau 0} \approx 6.36\times 10^1\ \text{km s}^{-1}\text{Mpc}^{-1} \approx 63.6\ \text{km s}^{-1}\text{Mpc}^{-1}.\)
This matches the QTT “ledger values”:
- \(\tau_0 \approx 15.4\ \text{Gyr}\),
- \(H_{\tau 0} = 1/\tau_0 \approx 63.5\ \text{km s}^{-1}\text{Mpc}^{-1}\).
5. Where Does the 13.8 Gyr Lab Age Enter?
The derivation above never used the familiar 13.8 Gyr. That number appears when we project absolute time onto our tilted laboratory time axis.
QTT says our lab time axis is not aligned with absolute time. There is:
- a fixed Time Tilt from an eightfold symmetry in the time plane, \(\theta_\star = \pi/8\), giving a baseline factor \[ I_{\rm clk} = \cos\left(\frac{\pi}{8}\right) \approx 0.92388, \qquad \frac{1}{I_{\rm clk}} \approx 1.0824, \] i.e. roughly an 8.2 % age boost;
- and a small extra Time Drift \(\delta_{\rm eff}\) from creation that adds a few degrees more tilt.
If \(t_0\) is the age you infer assuming a single lab clock with no tilt/drift, while \(\tau_0\) is the QTT absolute age, then
[
\tau_0 = \frac{t_0}{\cos(\theta_\star + \delta_{\rm eff})}.
]
We already saw that:
- tilt alone (no drift) would give \[ \tau_0^{(\star)} = \frac{t_0}{\cos\left(\frac{\pi}{8}\right)} \approx 1.0824\,t_0; \]
- with the drift we just implicitly used, you need \[ \frac{\tau_0}{\tau_0^{(\star)}} = \frac{\cos\theta_\star}{\cos(\theta_\star+\delta_{\rm eff})} \approx 1.03 \] to go from \(\tau_0^{(\star)} \approx 14.94\) Gyr to \(\tau_0\approx 15.4\) Gyr.
That extra 3 % corresponds to a small drift angle \(\delta_{\rm eff}\approx 0.067\) rad (about \(3.8^\circ\)) in the time plane.
What we have shown here is the hard part: starting from the observed baryon density alone, the QTT White Void law fixes the absolute age at about 15.4 Gyr and, via coasting, the absolute Hubble scale. The 13.8 Gyr then appears as a projection effect of that absolute history onto our slightly tilted and drifted lab clocks.
6. Summary: Baryons, WVs, and a 15.4-Gyr Universe
- Observed baryon density today: \(\rho_{b,0} \approx 4.2\times 10^{-28}\ \text{kg m}^{-3}\).
- QTT White Void creation law: \(\rho_b(T) = 1/(48\pi G T^2)\).
- Equating them and solving for \(T\) gives: \(T_0 \approx 15.4\ \text{Gyr}\).
- In coasting gauge, \(H_{\tau 0} = 1/T_0 \approx 63.5\ \text{km s}^{-1}\text{Mpc}^{-1}\).
- The familiar 13.8 Gyr lab age is a tilted, slightly drifted projection of this 15.4 Gyr absolute history onto our local clocks.
In other words, the Universe tells you how old it is in absolute time just by how many baryons it has per cubic metre – once you include the QTT White Void creation ledger.
QTT - Quantum Traction Theory 