Strong Nuclear Force in Temporal Physics

 Strong Nuclear Force

Known Values:

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

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

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

   $$F \sim \frac{g_s^2}{r_s^2}$$

   Substituting in the known values, we have:

   $$F \sim \frac{(1.2)^2}{(1 \times 10^{-15})^2} = 1.44 \times 10^{30} \, \text{N}$$
   

4. Expressing in Terms of $\mathrm{I }$: (Invariant Quantity) Assuming a relationship such as:

   $$g_s^2 \sim I \cdot \left(\frac{\hbar G}{c^3}\right)$$
   

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

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

   Substituting these values:

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

   Calculating $c^3$:

   $$c^3 = (3 \times 10^8)^3 = 2.7 \times 10^{25} \, \text{m}^3/\text{s}^3$$
   

   Now substituting this into the equation:

   $$\frac{\hbar G}{c^3} \approx (1.055 \times 10^{-34})(6.674 \times 10^{-11}) \cdot (2.7 \times 10^{25}) \approx 2.612 \times 10^{-70}$$
   

   Plugging this back into the relationship to solve for $I$:

   $$1.44 \sim I \cdot (2.612 \times 10^{-70})$$
   

   Rearranging gives:

   $$I \sim \frac{1.44}{2.612 \times 10^{-70}} \approx 5.51 \times 10^{69}$$
   

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Weak Nuclear Force

Known Values:

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

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

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

   $$F \sim \frac{g_w^2}{r_w^2}$$

   Substituting in the known values:

   $$F \sim \frac{(0.65)^2}{(1 \times 10^{-10})^2} = 4.225 \times 10^{20} \, \text{N}$$
   

4. Expressing in Terms of $I$: Using a similar relationship:

   $$g_w^2 \sim I \cdot \left(\frac{\hbar G}{c^5}\right)$$
   

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

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

   Calculating $c^5$:

   $$c^5 = (3 \times 10^8)^5 = 2.43 \times 10^{40} \, \text{m}^5/\text{s}^5$$
   

   Now substituting this into the equation:

   $$\frac{\hbar G}{c^5} \approx (1.055 \times 10^{-34})(6.674 \times 10^{-11}) \cdot (2.43 \times 10^{40}) \approx 2.37 \times 10^{-84}$$
   

   Solving for $I$:

   $$0.4225 \sim I \cdot (2.37 \times 10^{-84})$$
   

   Rearranging gives:

   $$I \sim \frac{0.4225}{2.37 \times 10^{-84}} \approx 1.78 \times 10^{83}$$