Sqrt Planck momentum

The sqrt of Planck momentum

The square root of momentum is used in the mathematical electron [1], a simulation hypothesis model, to link the mass constants with the charge constants. In SI unit terms the sqrt of Planck momentum denoted Q with units q;

Planck momentum = 2 π Q2, unit = kg.m/s = q2

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

mass constants

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

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

charge constants

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

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.

${\displaystyle {\frac {F_{electric}}{A_{Q}^{2}}}={\frac {F_{p}}{\alpha }}.{\frac {1}{A_{Q}^{2}}}={\frac {2\pi Q^{2}}{\alpha t_{p}}}.({\frac {\alpha Q^{3}}{8c^{3}}})^{2}={\frac {\pi \alpha Q^{8}}{64l_{p}c^{5}}}={\frac {2}{10^{7}}}}$
${\displaystyle \mu _{0}={\frac {\pi ^{2}\alpha Q^{8}}{32l_{p}c^{5}}}={\frac {4\pi }{10^{7}}},\;units={\frac {kg.m}{s^{2}A^{2}}}={\frac {kg^{6}}{q^{4}s}}}$

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

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

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). Proposed formula for a magnetic monopole; σe = 0.13708563 x10-6;

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

The Rydberg constant R, unit = 1/m.

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

This however now gives us 2 solutions for length m, if 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:

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

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; c, μ0, R, α;

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

From the formula for Q15 the least accurate dimensioned constants can be defined in terms of the most accurate constants; 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;

${\displaystyle unit\;X={\frac {kg^{12}}{q^{15}s^{2}}}=1}$
${\displaystyle e={\frac {16l_{p}c^{2}}{\alpha Q^{3}}}={\frac {\pi ^{2}Q^{5}}{2\mu _{0}c^{3}}},\;units={\frac {q^{3}s}{kg^{3}}}}$
${\displaystyle e^{3}={\frac {\pi ^{6}Q^{15}}{8\mu _{0}^{3}c^{9}}}={\frac {4\pi ^{5}}{3^{3}c^{4}\alpha ^{8}R}},\;units={\frac {kg^{3}s}{q^{6}}}=({\frac {q^{3}s}{kg^{3}}})^{3}.X}$
${\displaystyle h=2\pi Q^{2}2\pi l_{p}={\frac {4\pi ^{4}\alpha Q^{10}}{8\mu _{0}c^{5}}},\;units={\frac {q^{4}s}{kg}}}$
${\displaystyle h^{3}=({\frac {4\pi ^{4}\alpha Q^{10}}{8\mu _{0}c^{5}}})^{3}={\frac {2\pi ^{10}\mu _{0}^{3}}{3^{6}c^{5}\alpha ^{13}R^{2}}},\;units={\frac {kg^{21}}{q^{18}s}}=({\frac {q^{4}s}{kg}})^{3}.X^{2}}$
${\displaystyle k_{B}={\frac {\pi ^{2}\alpha Q^{5}}{4c^{3}}},\;units={\frac {kg^{3}}{q}}}$
${\displaystyle k_{B}^{3}={\frac {\pi ^{5}\mu _{0}^{3}}{3^{3}2c^{4}\alpha ^{5}R}},\;units={\frac {kg^{21}}{q^{18}s^{2}}}=({\frac {kg^{3}}{q}})^{3}.X}$
${\displaystyle G={\frac {c^{2}l_{p}}{m_{P}}}={\frac {\pi \alpha Q^{6}}{64\mu _{0}c^{2}}},\;units={\frac {q^{6}s}{kg^{4}}}}$
${\displaystyle G^{5}={\frac {\pi ^{3}\mu _{0}}{2^{20}3^{6}\alpha ^{11}R^{2}}},\;units=kg^{4}s=({\frac {q^{6}s}{kg^{4}}})^{5}.X^{2}}$
${\displaystyle l_{p}^{15}={\frac {\pi ^{22}\mu _{0}^{9}}{2^{35}3^{24}c^{35}\alpha ^{49}R^{8}}},\;units={\frac {kg^{81}}{q^{90}s}}=({\frac {q^{2}s}{kg}})^{15}.X^{8}}$
${\displaystyle m_{P}^{15}={\frac {2^{25}\pi ^{13}\mu _{0}^{6}}{3^{6}c^{5}\alpha ^{16}R^{2}}},\;units=kg^{15}={\frac {kg^{39}}{q^{30}s^{4}}}.{\frac {1}{X^{2}}}}$
${\displaystyle m_{e}^{3}={\frac {16\pi ^{10}R\mu _{0}^{3}}{3^{6}c^{8}\alpha ^{7}}},\;units=kg^{3}={\frac {kg^{27}}{q^{30}s^{4}}}.{\frac {1}{X^{2}}}}$
${\displaystyle A_{Q}^{5}={\frac {2^{10}\pi 3^{3}c^{10}\alpha ^{3}R}{\mu _{0}^{3}}},\;units={\frac {q^{30}s^{2}}{kg^{27}}}=({\frac {q^{3}}{kg^{3}}})^{5}.{\frac {1}{X}}}$

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

${\displaystyle \sigma ={\frac {2\pi ^{5}k_{B}^{4}}{15h^{3}c^{2}}},\;r_{d}={\frac {4\sigma }{c}},\;units=r}$
${\displaystyle r_{d}^{3}={\frac {3^{3}4\pi ^{5}\mu _{0}^{3}\alpha ^{19}R^{2}}{5^{3}c^{10}}},\;units={\frac {kg^{30}}{q^{36}s^{5}}}.{\frac {1}{X^{2}}}={\frac {kg^{6}}{q^{6}s}}=r^{3}}$

Physical constants; calculated vs experimental (CODATA)
Constant Calculated from (R*, c, μ0, α*) CODATA 2014 [2]
Speed of light c* = 299 792 458, units = u17 c = 299 792 458 (exact)
Fine structure constant α* = 137.035 999 139 (mean) α = 137.035 999 139(31)
Rydberg constant R* = 10 973 731.568 508, units = u13 (mean) R = 10 973 731.568 508(65)
Vacuum permeability μ0* = 4π/10^7, units = u56 μ0 = 4π/10^7 (exact)
Planck constant h* = 6.626 069 134 e-34, units = u19 h = 6.626 070 040(81) e-34
Gravitational constant G* = 6.672 497 192 29 e11, units = u6 G = 6.674 08(31) e-11
Elementary charge e* = 1.602 176 511 30 e-19, units = u-19 e = 1.602 176 620 8(98) e-19
Boltzmann constant kB* = 1.379 510 147 52 e-23, units = u29 kB = 1.380 648 52(79) e-23
Electron mass me* = 9.109 382 312 56 e-31, units = u15 me = 9.109 383 56(11) e-31
Classical electron radius λe* = 2.426 310 2366 e-12, units = u-13 λe = 2.426 310 236 7(11) e-12
Planck mass mP* = .217 672 817 580 e-7, units = u15 mP = .217 647 0(51) e-7
Planck length lp* = .161 603 660 096 e-34, units = u-13 lp = .161 622 9(38) e-34
Von Klitzing constant RK* = 25812.807 455 59, units = u73 RK = 25812.807 455 5(59)
Gyromagnetic ratio γe/2π* = 28024.953 55, units = u-42 γe/2π = 28024.951 64(17)

Analysis

Sample standard formulas in terms of Q

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

${\displaystyle E_{n}={-}{\frac {2\pi ^{2}k_{e}^{2}m_{e}e^{4}}{h^{2}n^{2}}}={2\pi ^{2}}\;{\frac {\pi ^{2}\alpha ^{2}Q^{16}}{16384l_{p}^{2}c^{6}}}\;{m_{e}}\;{\frac {65536l_{p}^{4}c^{8}}{\alpha ^{4}Q^{12}}}\;{\frac {1}{4\pi ^{2}Q^{4}4\pi ^{2}l_{p}^{2}}}\;={-}{\frac {m_{e}c^{2}}{2\alpha ^{2}n^{2}}}\;}$

${\displaystyle q_{p}={\sqrt {4\pi \epsilon _{0}\hbar c}}={\sqrt {4\pi \;{\frac {32l_{p}c^{3}}{\pi ^{2}\alpha Q^{8}}}\;{2\pi Q^{2}l_{p}}\;c}}\;={\sqrt {\alpha }}{e}\;}$

${\displaystyle r_{e}={\frac {e^{2}}{4\pi \epsilon _{0}m_{e}c^{2}}}={\frac {256l_{p}^{2}c^{4}}{\alpha ^{2}Q^{6}}}\;{\frac {1}{4\pi }}\;{\frac {\pi ^{2}\alpha Q^{8}}{32l_{p}c^{3}}}\;{\frac {1}{m_{e}c^{2}}}\;={\frac {l_{p}m_{P}}{\alpha m_{e}}}\;}$

${\displaystyle {\frac {4\pi \epsilon _{0}Gm_{e}m_{p}}{e^{2}}}\;={4\pi }\;{\frac {32l_{p}c^{3}}{\pi ^{2}\alpha Q^{8}}}\;{\frac {l_{p}c^{3}}{2\pi Q^{2}}}\;{m_{e}m_{p}}\;{\frac {\alpha ^{2}Q^{6}}{256l_{p}^{2}c^{4}}}\;={\frac {\alpha m_{e}m_{p}}{m_{P}^{2}}}\;}$

${\displaystyle m_{P}={\frac {B^{2}r^{2}e}{2V_{p}}}\;={\frac {B^{2}r^{2}e^{2}}{2E_{p}}}\;={\frac {\pi ^{2}\alpha ^{2}Q^{10}}{64l_{p}^{4}c^{4}}}\;{l_{p}^{2}}\;{\frac {256l_{p}^{2}c^{4}}{\alpha ^{2}Q^{6}}}\;{\frac {1}{2\pi Q^{2}c}}}$

where:

• ${\displaystyle E_{n}}$  refers to the Bohr model energy levels,
• ${\displaystyle q_{p}}$  is Planck charge,
• ${\displaystyle r_{e}}$  is the Classical electron radius.