# 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, 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 [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.

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

### Mass constants

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

### Charge constants

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 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). A proposed formula for a magnetic monopole σe;

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

### Dimensionless electron formula

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:

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

### 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);

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

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;

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

Physical constants; calculated vs experimental (CODATA)
Constant Calculated from (R*, c, μ0, α*) CODATA 2014 [3]
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)

### Fine structure constant alpha

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

### 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.

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 ${\displaystyle A={\frac {2^{6}\pi ^{3}\Omega ^{3}}{\alpha }}}$  ${\displaystyle unit=u^{3}}$

### References

1. Macleod, Malcolm; "The Sqrt of Planck momentum and the Mathematical Electron". https://papers.ssrn.com/sol3/papers.cfm?abstract_id=3585466"
2. Macleod, M.J. "Programming Planck units from a virtual electron; a Simulation Hypothesis". Eur. Phys. J. Plus 113: 278. 22 March 2018. doi:10.1103/10.1140/epjp/i2018-12094-x.
3. [1] | CODATA, The Committee on Data for Science and Technology | (2014)