Considering Counting Triangles to Unveiling Temporal Waves

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  Considering Counting Triangles to Unveiling Temporal Waves By: John Gavel For years, my work in Temporal Flow Physics (TFP) has pursued a radical idea: what if spacetime itself —with all its gravitational curves and quantum fluctuations—isn't fundamental at all? What if it emerges from a deeper reality: a network of one-dimensional temporal flows , weaving the universe together moment by moment? It’s bold, yes—but I believe this view holds the key to a truly unified theory of physics , one that roots both quantum mechanics and gravity in the same temporal fabric. From Counting Triangles to Counting Time My earliest simulations: I counted triangles. More specifically, I measured how triangular motifs in temporal flow networks dissipated under coarse-graining. The decay rate of these patterns—captured by a parameter I called A₃ —served as a stand-in for emergent gravitational effects. If motifs faded predictably with scale, it suggested that macroscopic structure (like sp...
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Strong Nuclear Force in Temporal Physics

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 Strong Nuclear Force

Known Values:

  1. Coupling Constant gsg_s: The strong coupling constant at low energy scales is approximately 1.2.

  2. Range of Strong Force rsr_s: The strong force operates at a range on the order of the nuclear scale, approximately 1 femtometer (1 fm = 1×10−15 m1 \times 10^{-15} \, \text{m}).

  3. Force Equation: The strong nuclear force can be approximated using the Yukawa potential:

    F∼gs2rs2F \sim \frac{g_s^2}{r_s^2}

    Substituting in the known values, we have:

    F∼(1.2)2(1×10−15)2=1.44×1030 NF \sim \frac{(1.2)^2}{(1 \times 10^{-15})^2} = 1.44 \times 10^{30} \, \text{N}

  4. Expressing in Terms of I : (Invariant Quantity) Assuming a relationship such as:

    gs2∼I⋅(ℏGc3)g_s^2 \sim I \cdot \left(\frac{\hbar G}{c^3}\right)

    We first need to calculate ℏGc3\frac{\hbar G}{c^3}:

    • Planck’s Constant ℏ=1.055×10−34 Js\hbar = 1.055 \times 10^{-34} \, \text{Js}
    • Gravitational Constant G=6.674×10−11 m3⋅kg−1⋅s−2G = 6.674 \times 10^{-11} \, \text{m}^3 \cdot \text{kg}^{-1} \cdot \text{s}^{-2}
    • Speed of Light c=3×108 m/sc = 3 \times 10^8 \, \text{m/s}

    Substituting these values:

    ℏGc3=(1.055×10−34)(6.674×10−11)(3×108)3\frac{\hbar G}{c^3} = \frac{(1.055 \times 10^{-34})(6.674 \times 10^{-11})}{(3 \times 10^8)^3}

    Calculating c3c^3:

    c3=(3×108)3=2.7×1025 m3/s3c^3 = (3 \times 10^8)^3 = 2.7 \times 10^{25} \, \text{m}^3/\text{s}^3

    Now substituting this into the equation:

    ℏGc3≈(1.055×10−34)(6.674×10−11)⋅(2.7×1025)≈2.612×10−70\frac{\hbar G}{c^3} \approx (1.055 \times 10^{-34})(6.674 \times 10^{-11}) \cdot (2.7 \times 10^{25}) \approx 2.612 \times 10^{-70}

    Plugging this back into the relationship to solve for II:

    1.44∼I⋅(2.612×10−70)1.44 \sim I \cdot (2.612 \times 10^{-70})

    Rearranging gives:

    I∼1.442.612×10−70≈5.51×1069I \sim \frac{1.44}{2.612 \times 10^{-70}} \approx 5.51 \times 10^{69}

Weak Nuclear Force

Known Values:

  1. Coupling Constant gwg_w: The weak coupling constant at low energy scales is approximately 0.65.

  2. Range of Weak Force rwr_w: The range of the weak force is approximately 0.1 nm, or 1×10−10 m1 \times 10^{-10} \, \text{m}.

  3. Force Equation: The weak nuclear force can be approximated as:

    F∼gw2rw2F \sim \frac{g_w^2}{r_w^2}

    Substituting in the known values:

    F∼(0.65)2(1×10−10)2=4.225×1020 NF \sim \frac{(0.65)^2}{(1 \times 10^{-10})^2} = 4.225 \times 10^{20} \, \text{N}

  4. Expressing in Terms of II: Using a similar relationship:

    gw2∼I⋅(ℏGc5)g_w^2 \sim I \cdot \left(\frac{\hbar G}{c^5}\right)

    First, we calculate ℏGc5\frac{\hbar G}{c^5}:

    ℏGc5=(1.055×10−34)(6.674×10−11)(3×108)5\frac{\hbar G}{c^5} = \frac{(1.055 \times 10^{-34})(6.674 \times 10^{-11})}{(3 \times 10^8)^5}

    Calculating c5c^5:

    c5=(3×108)5=2.43×1040 m5/s5c^5 = (3 \times 10^8)^5 = 2.43 \times 10^{40} \, \text{m}^5/\text{s}^5

    Now substituting this into the equation:

    ℏGc5≈(1.055×10−34)(6.674×10−11)⋅(2.43×1040)≈2.37×10−84\frac{\hbar G}{c^5} \approx (1.055 \times 10^{-34})(6.674 \times 10^{-11}) \cdot (2.43 \times 10^{40}) \approx 2.37 \times 10^{-84}

    Solving for II:

    0.4225∼I⋅(2.37×10−84)0.4225 \sim I \cdot (2.37 \times 10^{-84})

    Rearranging gives:

    I∼0.42252.37×10−84≈1.78×1083I \sim \frac{0.4225}{2.37 \times 10^{-84}} \approx 1.78 \times 10^{83}

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