The sqrt of Planck momentum
The sqrt of Planck momentum can potentially be used to link the mass constants and the charge constants [ 1] 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] 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...
u
n
i
t
=
q
{\displaystyle Q=1.019\;113\;411...\;unit=q\;}
Replacing m with q
u
n
i
t
m
=
q
2
s
k
g
{\displaystyle unit\;m={\frac {q^{2}s}{kg}}}
Speed of light :
c
,
u
n
i
t
=
q
2
k
g
{\displaystyle c,\;unit={\frac {q^{2}}{kg}}}
Planck mass :
m
P
=
2
π
Q
2
c
,
u
n
i
t
=
k
g
{\displaystyle m_{P}={\frac {2\pi Q^{2}}{c}},\;unit=kg}
Planck energy :
E
p
=
m
P
c
2
=
2
π
Q
2
c
,
u
n
i
t
s
=
k
g
.
m
2
s
2
=
q
4
k
g
{\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 :
l
p
,
u
n
i
t
=
q
2
s
k
g
{\displaystyle l_{p},\;unit={\frac {q^{2}s}{kg}}}
Planck time :
t
p
=
2
l
p
c
,
u
n
i
t
=
s
{\displaystyle t_{p}={\frac {2l_{p}}{c}},\;unit=s}
Planck force :
F
p
=
2
π
Q
2
t
p
,
u
n
i
t
s
=
q
2
s
{\displaystyle F_{p}={\frac {2\pi Q^{2}}{t_{p}}},\;units={\frac {q^{2}}{s}}}
Assigning a Planck ampere
A
Q
=
8
c
3
α
Q
3
,
u
n
i
t
A
=
m
3
q
3
s
3
=
q
3
k
g
3
{\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 :
e
=
A
Q
t
p
=
8
c
3
α
Q
3
.
2
l
p
c
=
16
l
p
c
2
α
Q
3
,
u
n
i
t
s
=
A
.
s
=
q
3
s
k
g
3
{\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 :
T
p
=
A
Q
c
π
=
8
c
3
α
Q
3
.
c
π
=
8
c
4
π
α
Q
3
,
u
n
i
t
s
=
q
5
k
g
4
{\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 :
k
B
=
E
p
T
p
=
π
2
α
Q
5
4
c
3
,
u
n
i
t
s
=
k
g
3
q
{\displaystyle k_{B}={\frac {E_{p}}{T_{p}}}={\frac {\pi ^{2}\alpha Q^{5}}{4c^{3}}},\;units={\frac {kg^{3}}{q}}}
Magnetic field :
B
=
m
P
c
2
e
α
2
l
p
=
π
Q
5
16
α
l
p
2
c
2
{\displaystyle \;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
=
1
μ
0
c
2
=
32
l
p
c
3
π
2
α
Q
8
{\displaystyle \;\epsilon _{0}={\frac {1}{\mu _{0}c^{2}}}\;={\frac {32l_{p}c^{3}}{\pi ^{2}\alpha Q^{8}}}\;}
Coulomb's constant :
k
e
=
1
4
π
ϵ
0
=
π
α
Q
8
128
l
p
c
3
{\displaystyle \;k_{e}={\frac {1}{4\pi \epsilon _{0}}}\;={\frac {\pi \alpha Q^{8}}{128l_{p}c^{3}}}\;}
2.10^{-7} newton per meter of length.
F
e
l
e
c
t
r
i
c
A
Q
2
=
F
p
α
.
1
A
Q
2
=
2
π
Q
2
α
t
p
.
(
α
Q
3
8
c
3
)
2
=
π
α
Q
8
64
l
p
c
5
=
2
10
7
{\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 :
μ
0
=
π
2
α
Q
8
32
l
p
c
5
=
4
π
10
7
,
u
n
i
t
s
=
k
g
.
m
s
2
A
2
=
k
g
6
q
4
s
{\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 ;
l
p
=
π
2
α
Q
8
32
μ
0
c
5
,
u
n
i
t
=
q
2
s
k
g
=
m
{\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 σe ;
σ
e
=
3
α
2
e
c
2
π
2
=
0.13708563
x
10
−
6
,
u
n
i
t
s
=
q
5
s
k
g
4
{\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
f
e
=
σ
e
3
t
p
=
2
8
3
3
α
3
l
p
2
c
10
π
6
Q
9
=
3
3
α
5
Q
7
4
π
2
μ
0
2
,
u
n
i
t
s
=
q
15
s
2
k
g
12
{\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}}}}
R∞ , unit = 1/m (see Electron mass).
R
∞
=
m
e
e
4
μ
0
2
c
3
8
h
3
=
2
5
c
5
μ
0
3
3
3
π
α
8
Q
15
,
u
n
i
t
s
=
1
m
=
k
g
13
q
17
s
3
{\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;
m
=
q
2
s
k
g
.
q
15
s
2
k
g
12
=
q
17
s
3
k
g
13
;
t
h
u
s
q
15
s
2
k
g
12
=
1
{\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}
f
e
=
σ
e
3
t
p
,
u
n
i
t
s
=
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 ;
s
=
k
g
6
q
15
/
2
{\displaystyle s={\frac {kg^{6}}{q^{15/2}}}}
μ
0
=
k
g
6
q
4
s
=
q
7
/
2
{\displaystyle \mu _{0}={\frac {kg^{6}}{q^{4}s}}=q^{7/2}}
Replacing q with the more familiar m gives this ratio;
q
2
=
k
g
.
m
s
;
q
30
=
(
k
g
.
m
s
)
15
=
k
g
24
s
4
{\displaystyle q^{2}={\frac {kg.m}{s}};\;q^{30}=({\frac {kg.m}{s}})^{15}={\frac {kg^{24}}{s^{4}}}}
u
n
i
t
s
=
k
g
9
s
11
m
15
=
1
{\displaystyle units={\frac {kg^{9}s^{11}}{m^{15}}}=1}
Electron mass as frequency of Planck mass:
m
e
=
m
P
f
e
,
u
n
i
t
=
k
g
{\displaystyle m_{e}={\frac {m_{P}}{f_{e}}},\;unit=kg}
Electron wavelength via Planck length:
λ
e
=
2
π
l
p
f
e
,
u
n
i
t
s
=
m
=
q
2
s
k
g
{\displaystyle \lambda _{e}=2\pi l_{p}f_{e},\;units=m={\frac {q^{2}s}{kg}}}
Gravitation coupling constant:
α
G
=
(
m
e
m
P
)
2
=
1
f
e
2
,
u
n
i
t
s
=
1
{\displaystyle \alpha _{G}=({\frac {m_{e}}{m_{P}}})^{2}={\frac {1}{f_{e}^{2}}},\;units=1}
The Rydberg constant R∞ 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
=
2
5
c
5
μ
0
3
3
3
π
α
8
R
,
u
n
i
t
s
=
k
g
12
s
2
=
q
15
{\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;
u
n
i
t
s
X
=
k
g
9
s
11
m
15
=
k
g
12
q
15
s
2
=
1
{\displaystyle units\;X={\frac {kg^{9}s^{11}}{m^{15}}}={\frac {kg^{12}}{q^{15}s^{2}}}=1}
e
=
16
l
p
c
2
α
Q
3
=
π
2
Q
5
2
μ
0
c
3
,
u
n
i
t
s
=
q
3
s
k
g
3
{\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}}}}
e
3
=
π
6
Q
15
8
μ
0
3
c
9
=
4
π
5
3
3
c
4
α
8
R
,
u
n
i
t
s
=
k
g
3
s
q
6
=
(
q
3
s
k
g
3
)
3
.
X
{\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}
h
=
2
π
Q
2
2
π
l
p
=
4
π
4
α
Q
10
8
μ
0
c
5
,
u
n
i
t
s
=
q
4
s
k
g
{\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}}}
h
3
=
(
4
π
4
α
Q
10
8
μ
0
c
5
)
3
=
2
π
10
μ
0
3
3
6
c
5
α
13
R
2
,
u
n
i
t
s
=
k
g
21
q
18
s
=
(
q
4
s
k
g
)
3
.
X
2
{\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}}
k
B
=
π
2
α
Q
5
4
c
3
,
u
n
i
t
s
=
k
g
3
q
{\displaystyle k_{B}={\frac {\pi ^{2}\alpha Q^{5}}{4c^{3}}},\;units={\frac {kg^{3}}{q}}}
k
B
3
=
π
5
μ
0
3
3
3
2
c
4
α
5
R
,
u
n
i
t
s
=
k
g
21
q
18
s
2
=
(
k
g
3
q
)
3
.
X
{\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}
G
=
c
2
l
p
m
P
=
π
α
Q
6
64
μ
0
c
2
,
u
n
i
t
s
=
q
6
s
k
g
4
{\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}}}}
G
5
=
π
3
μ
0
2
20
3
6
α
11
R
2
,
u
n
i
t
s
=
k
g
4
s
=
(
q
6
s
k
g
4
)
5
.
X
2
{\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}}
l
p
15
=
π
22
μ
0
9
2
35
3
24
c
35
α
49
R
8
,
u
n
i
t
s
=
k
g
81
q
90
s
=
(
q
2
s
k
g
)
15
.
X
8
{\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}}
m
P
15
=
2
25
π
13
μ
0
6
3
6
c
5
α
16
R
2
,
u
n
i
t
s
=
k
g
15
=
k
g
39
q
30
s
4
.
1
X
2
{\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}}}}
m
e
3
=
16
π
10
R
μ
0
3
3
6
c
8
α
7
,
u
n
i
t
s
=
k
g
3
=
k
g
27
q
30
s
4
.
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}}}}
A
Q
5
=
2
10
π
3
3
c
10
α
3
R
μ
0
3
,
u
n
i
t
s
=
q
30
s
2
k
g
27
=
(
q
3
k
g
3
)
5
.
1
X
{\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 ;
σ
=
2
π
5
k
B
4
15
h
3
c
2
,
r
d
=
4
σ
c
,
u
n
i
t
s
=
r
{\displaystyle \sigma ={\frac {2\pi ^{5}k_{B}^{4}}{15h^{3}c^{2}}},\;r_{d}={\frac {4\sigma }{c}},\;units=r}
r
d
3
=
3
3
4
π
5
μ
0
3
α
19
R
2
5
3
c
10
,
u
n
i
t
s
=
k
g
30
q
36
s
5
.
1
X
2
=
k
g
6
q
6
s
=
r
3
{\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)
α
=
2
h
μ
0
e
2
c
=
2
2
π
Q
2
2
π
l
p
32
l
p
c
5
π
2
α
Q
8
α
2
Q
6
256
l
p
2
c
4
1
c
=
α
,
u
n
i
t
s
=
q
4
s
k
g
q
4
s
k
g
6
k
g
6
q
6
s
2
k
g
q
2
=
1
{\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
M
=
1
{\displaystyle M=1}
u
n
i
t
=
u
15
{\displaystyle unit=u^{15}}
time
T
=
2
π
{\displaystyle T=2\pi }
u
n
i
t
=
u
−
30
{\displaystyle unit=u^{-30}}
momentum (sqrt of)
P
=
Ω
{\displaystyle P=\Omega }
u
n
i
t
=
u
16
{\displaystyle unit=u^{16}}
velocity
V
=
2
π
Ω
2
{\displaystyle V=2\pi \Omega ^{2}}
u
n
i
t
=
u
17
{\displaystyle unit=u^{17}}
length
L
=
2
π
2
Ω
2
{\displaystyle L=2\pi ^{2}\Omega ^{2}}
u
n
i
t
=
u
−
13
{\displaystyle unit=u^{-13}}
ampere
A
=
2
6
π
3
Ω
3
α
{\displaystyle A={\frac {2^{6}\pi ^{3}\Omega ^{3}}{\alpha }}}
u
n
i
t
=
u
3
{\displaystyle unit=u^{3}}
