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 ^[1] and so can be used to reduce the required number of SI units, this permits the least accurate physical constants, (G, h, e, m_e, k_B ...) 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 ^[2], 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.

${\displaystyle Q=1.019\;113\;411...\;unit=q\;}$

Replacing m with q

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

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

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

Planck energy :${\displaystyle E_{p}=m_{P}c^{2}=2\pi Q^{2}c,\;units={\frac {kg.m^{2}}{s^{2}}}={\frac {q^{4}}{kg}}}$

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

Planck time :${\displaystyle t_{p}={\frac {2l_{p}}{c}},\;unit=s}$

Planck force :${\displaystyle F_{p}={\frac {2\pi Q^{2}}{t_{p}}},\;units={\frac {q^{2}}{s}}}$

Assigning a Planck ampere

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

gives;

elementary charge :${\displaystyle e=A_{Q}t_{p}={\frac {8c^{3}}{\alpha Q^{3}}}.{\frac {2l_{p}}{c}}\;={\frac {16l_{p}c^{2}}{\alpha Q^{3}}},\;units=A.s={\frac {q^{3}s}{kg^{3}}}}$

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

Boltzmann constant :${\displaystyle k_{B}={\frac {E_{p}}{T_{p}}}={\frac {\pi ^{2}\alpha Q^{5}}{4c^{3}}},\;units={\frac {kg^{3}}{q}}}$

Magnetic field :${\displaystyle \;B={\frac {m_{P}c}{2e\alpha ^{2}l_{p}}}\;={\frac {\pi Q^{5}}{16\alpha l_{p}^{2}c^{2}}}\;}$

Vacuum permittivity :${\displaystyle \;\epsilon _{0}={\frac {1}{\mu _{0}c^{2}}}\;={\frac {32l_{p}c^{3}}{\pi ^{2}\alpha Q^{8}}}\;}$

Coulomb's constant :${\displaystyle \;k_{e}={\frac {1}{4\pi \epsilon _{0}}}\;={\frac {\pi \alpha Q^{8}}{128l_{p}c^{3}}}\;}$

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.

${\displaystyle {\frac {F_{electric}}{A_{Q}^{2}}}={\frac {F_{p}}{\alpha }}.{\frac {1}{A_{Q}^{2}}}={\frac {2\pi Q^{2}}{\alpha t_{p}}}.({\frac {\alpha Q^{3}}{8c^{3}}})^{2}={\frac {\pi \alpha Q^{8}}{64l_{p}c^{5}}}={\frac {2}{10^{7}}}}$

Vacuum permeability :${\displaystyle \mu _{0}={\frac {\pi ^{2}\alpha Q^{8}}{32l_{p}c^{5}}}={\frac {4\pi }{10^{7}}},\;units={\frac {kg.m}{s^{2}A^{2}}}={\frac {kg^{6}}{q^{4}s}}}$

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

${\displaystyle l_{p}={\frac {\pi ^{2}\alpha Q^{8}}{32\mu _{0}c^{5}}},\;unit={\frac {q^{2}s}{kg}}=m}$

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 σ_e;

${\displaystyle \sigma _{e}={\frac {3\alpha ^{2}ec}{2\pi ^{2}}}=0.13708563x10^{-6},\;units={\frac {q^{5}s}{kg^{4}}}}$

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

${\displaystyle f_{e}={\frac {\sigma _{e}^{3}}{t_{p}}}={\frac {2^{8}3^{3}\alpha ^{3}l_{p}^{2}c^{10}}{\pi ^{6}Q^{9}}}={\frac {3^{3}\alpha ^{5}Q^{7}}{4\pi ^{2}\mu _{0}^{2}}},\;units={\frac {q^{15}s^{2}}{kg^{12}}}}$

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

${\displaystyle R_{\infty }={\frac {m_{e}e^{4}\mu _{0}^{2}c^{3}}{8h^{3}}}={\frac {2^{5}c^{5}\mu _{0}^{3}}{3^{3}\pi \alpha ^{8}Q^{15}}},\;units={\frac {1}{m}}={\frac {kg^{13}}{q^{17}s^{3}}}}$

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;

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

${\displaystyle f_{e}={\frac {\sigma _{e}^{3}}{t_{p}}},\;units=1}$

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

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

${\displaystyle \mu _{0}={\frac {kg^{6}}{q^{4}s}}=q^{7/2}}$

Replacing q with the more familiar m gives this ratio;

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

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

Electron mass as frequency of Planck mass:

${\displaystyle m_{e}={\frac {m_{P}}{f_{e}}},\;unit=kg}$

Electron wavelength via Planck length:

${\displaystyle \lambda _{e}=2\pi l_{p}f_{e},\;units=m={\frac {q^{2}s}{kg}}}$

Gravitation coupling constant:

${\displaystyle \alpha _{G}=({\frac {m_{e}}{m_{P}}})^{2}={\frac {1}{f_{e}^{2}}},\;units=1}$

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);

${\displaystyle Q^{15}={\frac {2^{5}c^{5}\mu _{0}^{3}}{3^{3}\pi \alpha ^{8}R}},\;units={\frac {kg^{12}}{s^{2}}}=q^{15}}$

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;

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

${\displaystyle e={\frac {16l_{p}c^{2}}{\alpha Q^{3}}}={\frac {\pi ^{2}Q^{5}}{2\mu _{0}c^{3}}},\;units={\frac {q^{3}s}{kg^{3}}}}$

${\displaystyle e^{3}={\frac {\pi ^{6}Q^{15}}{8\mu _{0}^{3}c^{9}}}={\frac {4\pi ^{5}}{3^{3}c^{4}\alpha ^{8}R}},\;units={\frac {kg^{3}s}{q^{6}}}=({\frac {q^{3}s}{kg^{3}}})^{3}.X}$

${\displaystyle h=2\pi Q^{2}2\pi l_{p}={\frac {4\pi ^{4}\alpha Q^{10}}{8\mu _{0}c^{5}}},\;units={\frac {q^{4}s}{kg}}}$

${\displaystyle h^{3}=({\frac {4\pi ^{4}\alpha Q^{10}}{8\mu _{0}c^{5}}})^{3}={\frac {2\pi ^{10}\mu _{0}^{3}}{3^{6}c^{5}\alpha ^{13}R^{2}}},\;units={\frac {kg^{21}}{q^{18}s}}=({\frac {q^{4}s}{kg}})^{3}.X^{2}}$

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

${\displaystyle k_{B}^{3}={\frac {\pi ^{5}\mu _{0}^{3}}{3^{3}2c^{4}\alpha ^{5}R}},\;units={\frac {kg^{21}}{q^{18}s^{2}}}=({\frac {kg^{3}}{q}})^{3}.X}$

${\displaystyle G={\frac {c^{2}l_{p}}{m_{P}}}={\frac {\pi \alpha Q^{6}}{64\mu _{0}c^{2}}},\;units={\frac {q^{6}s}{kg^{4}}}}$

${\displaystyle G^{5}={\frac {\pi ^{3}\mu _{0}}{2^{20}3^{6}\alpha ^{11}R^{2}}},\;units=kg^{4}s=({\frac {q^{6}s}{kg^{4}}})^{5}.X^{2}}$

${\displaystyle l_{p}^{15}={\frac {\pi ^{22}\mu _{0}^{9}}{2^{35}3^{24}c^{35}\alpha ^{49}R^{8}}},\;units={\frac {kg^{81}}{q^{90}s}}=({\frac {q^{2}s}{kg}})^{15}.X^{8}}$

${\displaystyle m_{P}^{15}={\frac {2^{25}\pi ^{13}\mu _{0}^{6}}{3^{6}c^{5}\alpha ^{16}R^{2}}},\;units=kg^{15}={\frac {kg^{39}}{q^{30}s^{4}}}.{\frac {1}{X^{2}}}}$

${\displaystyle m_{e}^{3}={\frac {16\pi ^{10}R\mu _{0}^{3}}{3^{6}c^{8}\alpha ^{7}}},\;units=kg^{3}={\frac {kg^{27}}{q^{30}s^{4}}}.{\frac {1}{X^{2}}}}$

${\displaystyle A_{Q}^{5}={\frac {2^{10}\pi 3^{3}c^{10}\alpha ^{3}R}{\mu _{0}^{3}}},\;units={\frac {q^{30}s^{2}}{kg^{27}}}=({\frac {q^{3}}{kg^{3}}})^{5}.{\frac {1}{X}}}$

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

${\displaystyle \sigma ={\frac {2\pi ^{5}k_{B}^{4}}{15h^{3}c^{2}}},\;r_{d}={\frac {4\sigma }{c}},\;units=r}$

${\displaystyle r_{d}^{3}={\frac {3^{3}4\pi ^{5}\mu _{0}^{3}\alpha ^{19}R^{2}}{5^{3}c^{10}}},\;units={\frac {kg^{30}}{q^{36}s^{5}}}.{\frac {1}{X^{2}}}={\frac {kg^{6}}{q^{6}s}}=r^{3}}$

${\displaystyle \alpha ={\frac {2h}{\mu _{0}e^{2}c}}={2}\;{2\pi Q^{2}2\pi l_{p}}\;{\frac {32l_{p}c^{5}}{\pi ^{2}\alpha Q^{8}}}\;{\frac {\alpha ^{2}Q^{6}}{256l_{p}^{2}c^{4}}}\;{\frac {1}{c}}\;=\alpha ,\;units={\frac {q^{4}s}{kg}}{\frac {q^{4}s}{kg^{6}}}{\frac {kg^{6}}{q^{6}s^{2}}}{\frac {kg}{q^{2}}}=1}$

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.

Geometrical units

Designation	Geometrical object	Unit
mass	${\displaystyle M=1}$	${\displaystyle unit=u^{15}}$
time	${\displaystyle T=2\pi }$	${\displaystyle unit=u^{-30}}$
momentum (sqrt of)	${\displaystyle P=\Omega }$	${\displaystyle unit=u^{16}}$
velocity	${\displaystyle V=2\pi \Omega ^{2}}$	${\displaystyle unit=u^{17}}$
length	${\displaystyle L=2\pi ^{2}\Omega ^{2}}$	${\displaystyle unit=u^{-13}}$
ampere	${\displaystyle A={\frac {2^{6}\pi ^{3}\Omega ^{3}}{\alpha }}}$	${\displaystyle unit=u^{3}}$

• Programming Planck units as geometrical objects
• Mathematical electron
• Programming relativity at the Planck scale
• Programming gravity at the Planck scale
• Programming the cosmic microwave background at the Planck scale
• The Programmer God
• The Simulation hypothesis
• Programming at the Planck scale using geometrical objects -Malcolm Macleod's website
• Simulation Argument -Nick Bostrom's website
• Our Mathematical Universe: My Quest for the Ultimate Nature of Reality -Max Tegmark
• The mathematical electron model in a Planck scale universe -(article)
• Dirac-Kerr-Newman black-hole electron -Alexander Burinskii (article)