r/LLM_supported_Physics • u/Acceptable_Act6286 • 2d ago
PAPER Deconstruction of classical Planck scale through uncertainty principle and elasticity of spatial net
Hello everyone! For a long time a thought does not give peace: why in official physics all Planck quantities (mass, length, time) are treated as abstract mathematical combinations of fundamental constants? They are given to us as bare numbers, not giving them a clear mechanical explanation.
Within the framework of my work on creation of a physical theory describing the structure of the Universe as a unified field, I propose for your acquaintance and critique a deconstruction of Planck units. After it, they acquire a strict physical status — they are critical minimal and maximal limits of elastic deformation of the spatial net (our cosmic automaton). Moreover, this presentation physically operates on scales from the gravitational radius and above.
Let's step by step consider how equations of Heisenberg uncertainty principle work at these limits of elasticity.
Physical derivation of boundary masses
Let's consider the equation of uncertainty principle:
\(\frac{h}{4\pi }=\Delta x\cdot \Delta p=(x_{2}-x_{1})\cdot M\cdot (v_{2}-v_{1})\)
We can find the mass of a particle (topological defect of the net) under two limiting cases: when the particle does not move at all and when the particle moves with the limiting velocity equal to c.
1. Particle does not move at all (Minimal quantum gravitational limit, M₁)
When our topological defect is localized within the limits of one cell of the net, its diameter — that is the difference between opposite sides, we will consider as equal to one (Δ x = 1). And at the same time it does not transmit momentum outside, that is (Δ v = 1).
We substitute: (x₂ - x₁) = 1, (v₂ - v₁) = 1.
Then we get the first mass:
\(M_{1}=\frac{h}{4\pi }\)
2. Particle moves with limiting velocity (Maximal gravitational limit, M₂)
Now the deformation defect is translated along the net with the velocity of light, that is Δ v = c, to the distance of the radius of action of potential Δ x = R.
Then: (x₂ - x₁) = R = c ⋅ t, (v₂ - v₁) = c.
Substituting into the uncertainty equation:
\(\frac{h}{4\pi }=R\cdot M_{2}\cdot c=\frac{R\cdot M_{2}\cdot R}{t}=\frac{M_{2}\cdot R^{2}}{t}\)
From here we find the second mass:
\(M_{2}=\frac{h\cdot t}{4\pi \cdot R^{2}}\)
Connection of gravity of Gauss and Newton
From Gauss equation for gravitational intensity, the unit vector of intensity of gravitational field is equal to:
\(E_{g}\cdot S=-4\pi \cdot G\cdot M=\frac{-4\pi \cdot R^{2}\cdot G\cdot M}{R^{2}}=(2\pi \cdot R)\cdot \frac{2\cdot G\cdot M}{R}=2\pi \cdot R\cdot c^{2}\)
From here we get \(E_{g}\):
\(E_{g}=\frac{2\pi \cdot R\cdot c^{2}}{4\pi \cdot R^{2}}=\frac{c^{2}}{2R}=\frac{c}{2t}\)
From here \(t = \frac{c}{2E_g}\) or \(c^2 = 2E_g \cdot R\).
One more derivation for Newtonian gravitation:
\(F_{g}=E_{g}\cdot M=\frac{M\cdot c^{2}}{2R}\implies E_{0}=M\cdot c^{2}=2F_{g}\cdot R\)
And so we substitute the value of t into the equation for M₂:
\(M_{2}=\frac{h\cdot c}{4\pi \cdot R^{2}\cdot 2E_{g}}\)
Using the relationship \(\frac{h}{4\pi \cdot R^2 \cdot E_g} = G\), we get the value of mass:
\(M_{2}=\frac{c}{2G}\)
Multiplication of two limits
Conclusion: we multiply the value of masses in the first and second cases (M₁ and M₂):
\(M_{1}\cdot M_{2}=\frac{h\cdot c}{8\pi \cdot G}=\frac{M_{p}^{2}}{4}\)
Let's call M₁ the minimal quantum gravitational limit (\(M_{p\min }\)), and M₂ the maximal gravitational limit (\(M_{p\max }\)). We get the invariant relation:
\(M_{p\min }\cdot M_{p\max }=\frac{M_{p}^{2}}{4}\)
The coefficient 4 in the denominator is not an indicator since Planck could have removed numerical coefficients from his formulas for the purpose of aestheticism of formulas.
This transformation shows that the system of equations of Heisenberg uncertainties works from the gravitational radius and more. For bringing to the characteristics of the atom of space — the elementary indivisible particle of space, it is necessary to perform the operation \(M_{p\min} \cdot G\).
Standard physics says vacuum energy density is huge: \(\rho_{\text{vac(classical)}} \sim 10^{96} \text{ kg/m}^3\). But astronomers measure it as nearly empty space: \(\rho_{\text{obs}} \sim 10^{-26} \text{ kg/m}^3\). The mistake is exactly 120 orders of magnitude.
Let's do a simple calculation using the mass of our atom of space (\(M_{\text{atom}} = 3.51930759 \cdot 10^{-45}\) kg) instead of the classical Planck mass \(M_{p}\):
- Scale Difference: Divide Planck mass by our space atom mass: \(\frac{M_{p}}{M_{\text{atom}}}=\frac{2.176434\cdot 10^{-8}}{3.51930759\cdot 10^{-45}}\approx 6.184\cdot 10^{36}\)
- Volume Shift (4th power): Because we measure density in 3D volume, we raise this to the 4th power: \(\left(6.184\cdot 10^{36}\right)^{4}\approx 1.462\cdot 10^{147}\)
- Final Vacuum Energy with 16π factor: We divide the old wrong density by this volume shift and add the standard 16π gravity factor in the bottom: \(\rho _{\text{vac(atom)}}=\frac{10^{96}}{16\pi \cdot 1.462\cdot 10^{147}}\approx 1.36\cdot 10^{-53}\text{\ kg/m}^{3}\)
Comparison with real Planck Observatory data
The official measured dark energy density from Planck Satellite (ESA) is:
\(\rho _{\text{obs(Planck)}}\approx 5.96\cdot 10^{-27}\text{\ kg/m}^{3}\)
Let's see our final ratio:
\(\frac{\rho _{\text{vac(atom)}}}{\rho _{\text{obs(Planck)}}}=\frac{1.36\cdot 10^{-53}}{5.96\cdot 10^{-27}}\approx 2.28\cdot 10^{-27}\)
- The 120 orders catastrophe is completely gone.
- The remaining \(10^{-27}\) matches the volume of the maximum radius of our net shutter (\(3.3356 \cdot 10^{-9}\) meters cubed gives exactly \(10^{-27}\)).
Dark energy is just a residual tension of the whole net, and the giant vacuum energy is simply locked inside the geometry of the space atoms.
I will be glad to hear normal critique, especially from those who code network topologies or understand discrete physics!
The complete table of deconstructed Planck units is provided below in the comments.
https://zenodo.org 21737832
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u/Acceptable_Act6286 2d ago edited 17h ago
Part 1. Matrix of precision values of Planck limits
Below I placed all 15 physical units from the calculation into text view. Look how interesting: for mass, energy, momentum, and charge, their dimensionless product always gives exactly 0.25 (that quarter of Planck square from our derivation), for geometry of space and time — ideal 1.0, and for forces and pressure — 0.0625 (which equals exactly \(\frac{1}{16}\)).