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, 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 , 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.

$Q=1.019113411...unit=q$

Mass constants

Replacing m with q

$unitm=\frac{q^{2}s}{kg}$

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 :$l_p,unit=\frac{q^{2}s}{kg}$

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

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

Charge constants

Assigning a Planck ampere

$A_Q=\frac{8c^3}{\alpha Q^3},unitA=\frac{m^3}{q^3{s}^3}=\frac{q^3}{k{g}^3}$

gives;

elementary charge :$e=A_Q{t}_p=\frac{8c^3}{\alpha Q^3}.\frac{2l_p}{c}=\frac{16l_p{c}^2}{\alpha Q^3},units=A.s=\frac{q^{3}s}{k{g}^3}$

Planck temperature :$T_p=\frac{A_{Q}c}{\pi }=\frac{8c^3}{\alpha Q^3}.\frac{c}{\pi }=\frac{8c^4}{\pi \alpha Q^3},units=\frac{q^5}{k{g}^4}$

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

Magnetic field :$B=\frac{m_{P}c}{2e{\alpha }^2{l}_p}=\frac{\pi Q^5}{16\alpha l_p^2{c}^2}$

Vacuum permittivity :${ϵ}_0=\frac{1}{{\mu }_0{c}^2}=\frac{32l_p{c}^3}{{\pi }^2\alpha Q^8}$

Coulomb's constant :$k_e=\frac{1}{4\pi {ϵ}_0}=\frac{\pi \alpha Q^8}{128l_p{c}^3}$

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.

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

Vacuum permeability :${\mu }_0=\frac{{\pi }^2\alpha Q^8}{32l_p{c}^5}=\frac{4\pi }{10^7},units=\frac{kg.m}{s^2{A}^2}=\frac{k{g}^6}{q^{4}s}$

Rewriting Planck length l_p 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 σ_e;

${\sigma }_e=\frac{3{\alpha }^{2}ec}{2{\pi }^2}=0.13708563x10^{-6},units=\frac{q^{5}s}{k{g}^4}$

Dimensionless electron formula

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

$f_e=\frac{{\sigma }_e^3}{t_p}=\frac{2^8{3}^3{\alpha }^3{l}_p^2{c}^{10}}{{\pi }^6{Q}^9}=\frac{3^3{\alpha }^5{Q}^7}{4{\pi }^2{\mu }_0^2},units=\frac{q^{15}{s}^2}{k{g}^{12}}$

The Rydberg constant R_{Template:Math}, unit = 1/m (see Electron mass).

$R_{\mathrm{\infty }}=\frac{m_e{e}^4{\mu }_0^2{c}^3}{8h^3}=\frac{2^5{c}^5{\mu }_0^3}{3^3\pi {\alpha }^8{Q}^{15}},units=\frac{1}{m}=\frac{k{g}^{13}}{q^{17}{s}^3}$

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;

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

$f_e=\frac{{\sigma }_e^3}{t_p},units=1$

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

$s=\frac{k{g}^6}{q^{15/2}}$

${\mu }_0=\frac{k{g}^6}{q^{4}s}=q^{7/2}$

Replacing q with the more familiar m gives this ratio;

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

$units=\frac{k{g}^9{s}^{11}}{m^{15}}=1$

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_{Template:Math} = 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);

$Q^{15}=\frac{2^5{c}^5{\mu }_0^3}{3^3\pi {\alpha }^{8}R},units=\frac{k{g}^{12}}{s^2}=q^{15}$

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;

$unitsX=\frac{k{g}^9{s}^{11}}{m^{15}}=\frac{k{g}^{12}}{q^{15}{s}^2}=1$

$e=\frac{16l_p{c}^2}{\alpha Q^3}=\frac{{\pi }^2{Q}^5}{2{\mu }_0{c}^3},units=\frac{q^{3}s}{k{g}^3}$

$e^3=\frac{{\pi }^6{Q}^{15}}{8{\mu }_0^3{c}^9}=\frac{4{\pi }^5}{3^3{c}^4{\alpha }^{8}R},units=\frac{k{g}^{3}s}{q^6}=(\frac{q^{3}s}{k{g}^3}{)}^3.X$

$h=2\pi Q^{2}2\pi l_p=\frac{4{\pi }^4\alpha Q^{10}}{8{\mu }_0{c}^5},units=\frac{q^{4}s}{kg}$

$h^3=(\frac{4{\pi }^4\alpha Q^{10}}{8{\mu }_0{c}^5}{)}^3=\frac{2{\pi }^{10}{\mu }_0^3}{3^6{c}^5{\alpha }^{13}{R}^2},units=\frac{k{g}^{21}}{q^{18}s}=(\frac{q^{4}s}{kg}{)}^3.X^2$

$k_B=\frac{{\pi }^2\alpha Q^5}{4c^3},units=\frac{k{g}^3}{q}$

$k_B^3=\frac{{\pi }^5{\mu }_0^3}{3^{3}2c^4{\alpha }^{5}R},units=\frac{k{g}^{21}}{q^{18}{s}^2}=(\frac{k{g}^3}{q}{)}^3.X$

$G=\frac{c^2{l}_p}{m_P}=\frac{\pi \alpha Q^6}{64{\mu }_0{c}^2},units=\frac{q^{6}s}{k{g}^4}$

$G^5=\frac{{\pi }^3{\mu }_0}{2^{20}{3}^6{\alpha }^{11}{R}^2},units=k{g}^{4}s=(\frac{q^{6}s}{k{g}^4}{)}^5.X^2$

$l_p^{15}=\frac{{\pi }^{22}{\mu }_0^9}{2^{35}{3}^{24}{c}^{35}{\alpha }^{49}{R}^8},units=\frac{k{g}^{81}}{q^{90}s}=(\frac{q^{2}s}{kg}{)}^{15}.X^8$

$m_P^{15}=\frac{2^{25}{\pi }^{13}{\mu }_0^6}{3^6{c}^5{\alpha }^{16}{R}^2},units=k{g}^{15}=\frac{k{g}^{39}}{q^{30}{s}^4}.\frac{1}{X^2}$

$m_e^3=\frac{16{\pi }^{10}R{\mu }_0^3}{3^6{c}^8{\alpha }^7},units=k{g}^3=\frac{k{g}^{27}}{q^{30}{s}^4}.\frac{1}{X^2}$

$A_Q^5=\frac{2^{10}\pi 3^3{c}^{10}{\alpha }^{3}R}{{\mu }_0^3},units=\frac{q^{30}{s}^2}{k{g}^{27}}=(\frac{q^3}{k{g}^3}{)}^5.\frac{1}{X}$

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

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

$r_d^3=\frac{3^{3}4{\pi }^5{\mu }_0^3{\alpha }^{19}{R}^2}{5^3{c}^{10}},units=\frac{k{g}^{30}}{q^{36}{s}^5}.\frac{1}{X^2}=\frac{k{g}^6}{q^{6}s}=r^3$

Physical constants; calculated vs experimental (CODATA)

Constant	Calculated from (R*, c, μ0, α*)	CODATA 2014
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

$\alpha =\frac{2h}{{\mu }_0{e}^{2}c}=22\pi Q^{2}2\pi l_p\frac{32l_p{c}^5}{{\pi }^2\alpha Q^8}\frac{{\alpha }^2{Q}^6}{256l_p^2{c}^4}\frac{1}{c}=\alpha ,units=\frac{q^{4}s}{kg}\frac{q^{4}s}{k{g}^6}\frac{k{g}^6}{q^6{s}^2}\frac{kg}{q^2}=1$

Mathematical electron

Template:Main Template:Main 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	$M=1$	$unit=u^{15}$
time	$T=2\pi$	$unit=u^{-30}$
momentum (sqrt of)	$P=\mathrm{\Omega }$	$unit=u^{16}$
velocity	$V=2\pi {\mathrm{\Omega }}^2$	$unit=u^{17}$
length	$L=2{\pi }^2{\mathrm{\Omega }}^2$	$unit=u^{-13}$
ampere	$A=\frac{2^6{\pi }^3{\mathrm{\Omega }}^3}{\alpha }$	$unit=u^3$

External links

• 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)