Sqrt Planck momentum

The sqrt of Planck momentum

The sqrt of Planck momentum can potentially be used to link the mass constants and the charge constants and so can be used to reduce the required number of SI units, this permits the least accurate physical constants, (G, h, e, me, kB ...) to be defined and solved using the 4 most precise constants (c, μ0, R, α). The electron is reduced to a construct of magnetic monopoles. In more general terms the sqrt of momentum is used to reference the dimensionless mathematical electron, a  simulation hypothesis model.

As it has no assigned SI unit, it is denoted here as Q with units q whereby Planck momentum = 2 π Q2, unit = kg.m/s = q2. It can be argued that Q qualifies as an independent Planck unit.


 * $$Q = 1.019\; 113\; 411...\; unit = q \;$$

Mass constants
Replacing m with q


 * $$unit\; m = \frac{q^2s}{kg}$$

Speed of light :$$c,\; unit = \frac{q^2}{kg}$$

Planck mass :$$m_P = \frac{2 \pi Q^2}{c},\; unit = kg$$

Planck energy :$$E_p = m_P c^2 = 2 \pi Q^2 c,\; units = \frac{kg.m^2}{s^2} = \frac{q^4}{kg}$$

Planck length :$$l_p,\; unit = \frac{q^2s}{kg}$$

Planck time :$$t_p = \frac{2 l_p}{c},\; unit = s$$

Planck force :$$F_p = \frac{2 \pi Q^2}{t_p},\; units = \frac{q^2}{s}$$

Charge constants
Assigning a Planck ampere


 * $$A_Q = \frac{8 c^3}{\alpha Q^3},\; unit\;A = \frac{m^3}{q^3s^3} = \frac{q^3}{kg^3}$$

gives;

elementary charge :$$e = A_Q t_p =  \frac{8 c^3}{\alpha Q^3}. \frac{2 l_p}{c}\; = \frac{16 l_p c^2}{\alpha Q^3},\; units = A.s = \frac{q^3 s}{kg^3}$$

Planck temperature :$$T_p = \frac{A_Q c}{\pi} = \frac{8 c^3}{\alpha Q^3}. \frac{c}{\pi}\; = \frac{8 c^4}{\pi \alpha Q^3},\; units = \frac{q^5}{kg^4}$$

Boltzmann constant :$$k_B = \frac{E_p}{T_p} = \frac{\pi^2 \alpha Q^5}{4 c^3},\; units = \frac{kg^3}{q}$$

Magnetic field :$$\; B = \frac{m_P c}{2 e \alpha^2 l_p}\; = \frac{\pi Q^5}{16 \alpha l_p^2 c^2} \;$$

Vacuum permittivity :$$\; \epsilon_0 = \frac{1}{\mu_0 c^2} \; = \frac{32 l_p c^3}{\pi^2 \alpha Q^8} \;$$

Coulomb's constant :$$\; k_e = \frac{1}{4 \pi \epsilon_0}\; = \frac{\pi \alpha Q^8}{128 l_p c^3} \;$$

The ampere is that constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross section, and placed 1 meter apart in vacuum, would produce between these conductors a force equal to exactly 2.10^{-7} newton per meter of length.


 * $$\frac{F_{electric}}{A_Q^2} = \frac{F_p}{\alpha}.\frac{1}{A_Q^2}  = \frac{2 \pi Q^2}{\alpha t_p}.(\frac{\alpha Q^3}{8 c^3})^2 = \frac{\pi \alpha Q^8}{64 l_p c^5} =  \frac{2}{10^7}$$

Vacuum permeability :$$\mu_0 = \frac{\pi^2 \alpha Q^8}{32 l_p c^5} = \frac{4 \pi}{10^7},\; units = \frac{kg.m}{s^2 A^2} = \frac{kg^6}{q^4 s}$$

Rewriting Planck length lp in terms of Q, c, α, μ0;


 * $$l_p = \frac{\pi^2 \alpha Q^8}{32 \mu_0 c^5},\; unit = \frac{q^2 s}{kg} = m $$

Magnetic monopole
A magnetic monopole is a hypothesized particle that is a magnet with only 1 pole. The unit for the magnetic monopole is the ampere-meter, the SI unit for pole strength (the product of charge and velocity) in a magnet (A m = e c). A proposed formula for a magnetic monopole σe;


 * $$\sigma_{e} = \frac{3 \alpha^2 e c}{2 \pi^2} = 0.13708563 x 10^{-6},\; units = \frac{q^5 s}{kg^4}

$$

Dimensionless electron formula
The formula for an electron in terms of magnetic monopoles and Planck time


 * $$f_e = \frac{\sigma_e^3}{t_p} = \frac{2^8 3^3 \alpha^3 l_p^2 c^{10}}{\pi^6 Q^9} = \frac{3^3 \alpha^5 Q^7}{4 \pi^2 \mu_0^2},\; units = \frac{q^{15} s^2}{kg^{12}}$$

The Rydberg constant R$∞$, unit = 1/m (see Electron mass).


 * $$R_\infty = \frac{m_e e^4 \mu_0^2 c^3}{8 h^3} = \frac{2^5 c^5 \mu_0^3}{3^3 \pi \alpha^8 Q^{15}},\; units = \frac{1}{m} = \frac{kg^{13}}{q^{17} s^3}$$

This however now gives us 2 solutions for length m, if we conjecture that they are both valid then there must be a ratio whereby the units q, s, kg overlap and cancel;


 * $$m = \frac{q^2 s}{kg}.\frac{q^{15} s^2}{kg^{12}} = \frac{q^{17} s^3}{kg^{13}};\; thus\; \frac{q^{15} s^2}{kg^{12}} = 1$$


 * $$f_e = \frac{\sigma_e^3}{t_p},\; units = 1$$

and so we can further reduce the number of units required, for example we can define s in terms of kg, q;


 * $$s = \frac{kg^6}{q^{15/2}}$$


 * $$\mu_0 = \frac{kg^6}{q^4 s} = q^{7/2}$$

Replacing q with the more familiar m gives this ratio;


 * $$q^2 = \frac{kg.m}{s};\; q^{30} = (\frac{kg.m}{s})^{15} = \frac{kg^{24}}{s^4}$$


 * $$units = \frac{kg^{9} s^{11}}{m^{15}} = 1$$

Electron mass as frequency of Planck mass:


 * $$m_e = \frac{m_P}{f_e},\; unit = kg$$

Electron wavelength via Planck length:


 * $$\lambda_e = 2 \pi l_p f_e, \;units = m = \frac{q^2 s}{kg} $$

Gravitation coupling constant:


 * $$\alpha_G = (\frac{m_e}{m_P})^2 = \frac{1}{f_e^2},\; units = 1$$

Q15
The Rydberg constant R$∞$ = 10973731.568508(65) has been measured to a 12 digit precision. The known precision of Planck momentum and so Q is low, however with the solution for the Rydberg we may re-write Q as Q15 in terms of the 4 most precise constants; c (exact), μ0 (CODATA 2014 exact), R (12 digits), α (11 digits);


 * $$Q^{15} = \frac{2^5 c^5 \mu_0^3}{3^3 \pi \alpha^8 R},\;units = \frac{kg^{12}}{s^2} = q^{15}$$

From the above formula for Q15, the least accurate dimension-ed constants can now be defined in terms of c, μ0, R, α. The constants are first arranged until they include a Q15 term which can then be replaced by the above formula. Setting unit X as;


 * $$units\; X = \frac{kg^{9} s^{11}}{m^{15}} = \frac{kg^{12}}{q^{15}s^2} = 1$$


 * $$e = \frac{16 l_p c^2}{\alpha Q^3} = \frac{\pi^2 Q^5}{2\mu_0 c^3}, \;units = \frac{q^3 s}{kg^3}$$


 * $$e^3 = \frac{\pi^6 Q^{15}}{8 \mu_0^3 c^9} = \frac{4 \pi^5}{3^3 c^4 \alpha^8 R}, \; units = \frac{kg^3 s}{q^6} = (\frac{q^3 s}{kg^3})^3.X$$


 * $$h = 2\pi Q^2 2\pi l_p = \frac{4 \pi^4 \alpha Q^{10}}{8 \mu_0 c^5}, \;units = \frac{q^4 s}{kg}$$


 * $$h^3 = (\frac{4 \pi^4 \alpha Q^{10}}{8 \mu_0 c^5})^3 = \frac{2\pi^{10} \mu_0^3}{3^6 c^5 \alpha^{13} R^2}, \; units = \frac{kg^{21}}{q^{18}s} = (\frac{q^4 s}{kg})^3.X^2$$


 * $$k_B = \frac{\pi^2 \alpha Q^5}{4 c^3}, \;units = \frac{kg^3}{q} $$


 * $$k_B^3 = \frac{\pi^5 \mu_0^3}{3^3 2 c^4 \alpha^5 R}, \; units = \frac{kg^{21}}{q^{18}s^2} = (\frac{kg^3}{q})^3.X $$


 * $$G = \frac{c^2 l_p}{m_P} = \frac{\pi \alpha Q^6}{64 \mu_0 c^2},\;units = \frac{q^6 s}{kg^4}$$


 * $$G^5 = \frac{\pi^3 \mu_0}{2^{20} 3^6 \alpha^{11} R^2},\;units = kg^4 s = (\frac{q^6 s}{kg^4})^5.X^2 $$


 * $$l_p^{15} = \frac{\pi^{22} \mu_0^9}{2^{35} 3^{24} c^{35} \alpha^{49} R^8}   ,\; units =  \frac{kg^{81}}{q^{90}s} = (\frac{q^2 s}{kg})^{15}.X^8$$


 * $$m_P^{15} = \frac{2^{25} \pi^{13} \mu_0^6 }{ 3^6 c^5 \alpha^{16} R^2},\; units = kg^{15} = \frac{kg^{39}}{q^{30}s^4}.\frac{1}{X^2}$$


 * $$m_e^3 = \frac{16 \pi^{10} R \mu_0^3}{3^6 c^8 \alpha^7},\; units = kg^3 = \frac{kg^{27}}{q^{30}s^4}.\frac{1}{X^2}$$


 * $$A_Q^5 = \frac{2^{10} \pi 3^3 c^{10} \alpha^3 R}{\mu_0^3},\; units = \frac{q^{30}s^2}{kg^{27}}  = (\frac{q^3}{kg^3})^5.\frac{1}{X}$$

There is a solution for an r2 = q, it is the radiation density constant from the Stefan Boltzmann constant σ;


 * $$\sigma = \frac{2\pi^5 k_B^4}{15h^3c^2},\; r_d = \frac{4\sigma}{c},\;units = r$$


 * $$r_d^3 = \frac{3^3 4 \pi^5 \mu_0^3 \alpha^{19} R^2}{5^3 c^{10}},\; units = \frac{kg^{30}}{q^{36}s^5}.\frac{1}{X^2} = \frac{kg^6}{q^6 s} = r^3$$

Fine structure constant alpha
$$\alpha = \frac{2 h}{\mu_0 e^2 c} = {2}\; {2 \pi Q^2 2 \pi l_p}\; \frac{32 l_p c^5}{\pi^2 \alpha Q^8}\; \frac{\alpha^2 Q^6}{256 l_p^2 c^4}\; \frac{1}{c} \; = \alpha,\; units = \frac{q^4 s}{kg} \frac{q^4 s}{kg^6} \frac{kg^6}{q^6 s^2} \frac{kg}{q^2} = 1$$

Mathematical electron
Q is used in the context of SI units and so is related to the SI Planck momentum. The mathematical electron model uses geometrical objects for the Planck units and defines P as the sqrt of momentum with the unit u16. Although different sets of geometrical objects may be used in the mathematical electron model, so far only the following set can also translate to the Q related formulas and so Q places a limit on this model.