Natural Planck units as geometrical objects (the mathematical electron model)
In a geometrical Planck unit theory, the dimensioned universe at the Planck scale is defined by discrete geometrical objects for the Planck units; Planck mass , Planck length , Planck time and Planck charge. The object embeds the attribute (mass, length, time, charge) of the unit, whereas for numerical based constants, the numerical values are dimensionless frequencies of the SI unit (kg, m, s, A), 3kg refers to 3 of the unit kg, the number 3 carries no mass-specific information.
The mathematical electron [ 1]
M
{\displaystyle M}
L
{\displaystyle L}
T
{\displaystyle T}
A
{\displaystyle A}
dimensionless physical constants , the (inverse) fine structure constant α and Omega Ω . Embedded into each object is the object function (attribute).
Table 1. Geometrical units
Attribute
Geometrical object
mass
M
=
(
1
)
{\displaystyle M=(1)}
time
T
=
(
π
)
{\displaystyle T=(\pi )}
sqrt(momentum)
P
=
(
Ω
)
{\displaystyle P=(\Omega )}
velocity
V
=
(
2
π
Ω
2
)
{\displaystyle V=(2\pi \Omega ^{2})}
length
L
=
(
2
π
2
Ω
2
)
{\displaystyle L=(2\pi ^{2}\Omega ^{2})}
ampere
A
=
16
V
3
α
P
3
=
(
2
7
π
3
Ω
3
α
)
{\displaystyle A={\frac {16V^{3}}{\alpha P^{3}}}=({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }})}
...ihre Bedeutung für alle Zeiten und für alle, auch außerirdische und außermenschliche Kulturen notwendig behalten und welche daher als »natürliche Maßeinheiten« bezeichnet werden können...
...These necessarily retain their meaning for all times and for all civilizations, even extraterrestrial and non-human ones, and can therefore be designated as "natural units"... -Max Planck
[ 2] [ 3]
Lego-style to form more complex objects such as electrons (i.e.: by embedding mass and ampere objects into the geometry of the electron (the electron object), the electron can have wavelength and charge) [ 4]
Table 2. Unit number
Attribute
Object
Unit number θ
mass
M
=
(
1
)
{\displaystyle M=(1)}
15
{\displaystyle 15}
time
T
=
(
π
)
{\displaystyle T=(\pi )}
−
30
{\displaystyle -30}
sqrt(momentum)
P
=
(
Ω
)
{\displaystyle P=(\Omega )}
16
{\displaystyle 16}
velocity
V
=
(
2
π
Ω
2
)
{\displaystyle V=(2\pi \Omega ^{2})}
17
{\displaystyle 17}
length
L
=
(
2
π
2
Ω
2
)
{\displaystyle L=(2\pi ^{2}\Omega ^{2})}
−
13
{\displaystyle -13}
ampere
A
=
16
V
3
α
P
3
=
(
2
7
π
3
Ω
3
α
)
{\displaystyle A={\frac {16V^{3}}{\alpha P^{3}}}=({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }})}
3
{\displaystyle 3}
V = 2πΩ2 = 25.3123819 and use to solve geometrical physical constant equivalents.
Table 3. Physical constant equivalents
CODATA 2014 [ 5]
SI unit
Geometrical constant
unit uθ
c = 299 792 458 (exact)
m
s
{\displaystyle {\frac {m}{s}}}
c* = V = 25.312381933
u
17
{\displaystyle u^{17}}
h = 6.626 070 040(81) e-34
k
g
m
2
s
{\displaystyle {\frac {kg\;m^{2}}{s}}}
h* =
2
π
M
V
L
{\displaystyle 2\pi MVL}
u
15
+
17
−
13
{\displaystyle u^{15+17-13}}
u
19
{\displaystyle u^{19}}
G = 6.674 08(31) e-11
m
3
k
g
s
2
{\displaystyle {\frac {m^{3}}{kg\;s^{2}}}}
G* =
V
2
L
M
{\displaystyle {\frac {V^{2}L}{M}}}
u
34
−
13
−
15
{\displaystyle u^{34-13-15}}
u
6
{\displaystyle u^{6}}
e = 1.602 176 620 8(98) e-19
C
=
A
s
{\displaystyle C=As}
e* =
A
T
{\displaystyle AT}
u
3
−
30
{\displaystyle u^{3-30}}
u
−
27
{\displaystyle u^{-27}}
kB = 1.380 648 52(79) e-23
k
g
m
2
s
2
K
{\displaystyle {\frac {kg\;m^{2}}{s^{2}\;K}}}
kB * =
2
π
V
M
A
{\displaystyle {\frac {2\pi VM}{A}}}
u
17
+
15
−
3
{\displaystyle u^{17+15-3}}
u
29
{\displaystyle u^{29}}
Table 4. Dimensionless combinations
CODATA 2014 (mean)
(α, Ω)
units uΘ = 1
k
B
e
c
h
=
{\displaystyle {\frac {k_{B}ec}{h}}=}
1.000 8254
(
k
B
∗
)
(
e
∗
)
(
c
∗
)
(
h
∗
)
{\displaystyle {\frac {(k_{B}^{*})(e^{*})(c^{*})}{(h^{*})}}}
1.0
(
u
29
)
(
u
−
27
)
(
u
17
)
(
u
19
)
=
1
{\displaystyle {\frac {(u^{29})(u^{-27})(u^{17})}{(u^{19})}}=1}
h
3
e
13
c
24
=
{\displaystyle {\frac {h^{3}}{e^{13}c^{24}}}=}
0.228 473 639... 10-58
(
h
∗
)
3
(
e
∗
)
13
(
c
∗
)
24
=
α
13
2
106
π
64
(
Ω
15
)
5
=
{\displaystyle {\frac {(h^{*})^{3}}{(e^{*})^{13}(c^{*})^{24}}}={\frac {\alpha ^{13}}{2^{106}\pi ^{64}(\color {red}\Omega ^{15})^{5}\color {black}}}=}
0.228 473 759... 10-58
(
u
19
)
3
(
u
−
27
)
13
(
u
17
)
24
=
1
{\displaystyle {\frac {(u^{19})^{3}}{(u^{-27})^{13}(u^{17})^{24}}}=1}
h
c
2
e
m
p
G
2
k
B
=
{\displaystyle {\frac {hc^{2}em_{p}}{G^{2}k_{B}}}=}
3.376 716
(
h
∗
)
(
c
∗
)
2
(
e
∗
)
M
(
G
∗
)
2
(
k
B
∗
)
=
2
11
π
3
α
2
=
{\displaystyle {\frac {(h^{*})(c^{*})^{2}(e^{*})M}{(G^{*})^{2}(k_{B}^{*})}}={\frac {2^{11}\pi ^{3}}{\alpha ^{2}}}=}
3.381 506
(
u
19
)
(
u
17
)
2
(
u
−
27
)
(
u
15
)
(
u
6
)
2
(
u
29
)
=
1
{\displaystyle {\frac {(u^{19})(u^{17})^{2}(u^{-27})(u^{15})}{(u^{6})^{2}(u^{29})}}=1}
To translate from geometrical objects to a numerical system of units requires system dependent scalars (kltpva ). For example;
If we use k to convert M to the SI Planck mass (M*k SI =
m
P
{\displaystyle m_{P}}
k SI = 0.2176728e-7kg (SI units ) Using v SI = 11843707.905m/s gives c = V*v SI = 299792458m/s (SI units ) Using vimp = 7359.3232155miles/s gives c = V*v imp = 186282miles/s (imperial units )
Table 5. Geometrical units
Attribute
Geometrical object
Scalar
Unit u θ
mass
M
=
(
1
)
{\displaystyle M=(1)}
k
u
15
{\displaystyle u^{15}}
time
T
=
(
π
)
{\displaystyle T=(\pi )}
t
u
−
30
{\displaystyle u^{-30}}
sqrt(momentum)
P
=
(
Ω
)
{\displaystyle P=(\Omega )}
r 2
u
16
{\displaystyle u^{16}}
velocity
V
=
(
2
π
Ω
2
)
{\displaystyle V=(2\pi \Omega ^{2})}
v
u
17
{\displaystyle u^{17}}
length
L
=
(
2
π
2
Ω
2
)
{\displaystyle L=(2\pi ^{2}\Omega ^{2})}
l
u
−
13
{\displaystyle u^{-13}}
ampere
A
=
(
2
7
π
3
Ω
3
α
)
{\displaystyle A=({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }})}
a
u
3
{\displaystyle u^{3}}
Because the scalars also include the SI unit, v = 11843707.905m/s ... they follow the unit number relationship u θ . This means that we can find ratios where the scalars cancel. Here are examples (units = 1), as such only 2 scalars are required , for example, if we know the numerical value for a and for l then we know the numerical value for t (t = a3 l3 ), and from l and t we know the value for k .
u
3
∗
3
u
−
13
∗
3
u
−
30
(
a
3
l
3
t
)
=
u
−
13
∗
15
u
15
∗
9
u
−
30
∗
11
(
l
15
k
9
t
11
)
=
.
.
.
=
1
{\displaystyle {\frac {u^{3*3}u^{-13*3}}{u^{-30}}}\;({\frac {a^{3}l^{3}}{t}})={\frac {u^{-13*15}}{u^{15*9}u^{-30*11}}}\;({\frac {l^{15}}{k^{9}t^{11}}})=\;...\;=1}
μ0 ) are assigned exact values, following the 2019 redefinition of SI base units 4 constants have been independently assigned exact values which is problematic in terms of this model.
Scalars r (θ = 8) and v (θ = 17) are chosen as they can be derived directly from the 2 constants with exact values; c and μ0 .
v
=
c
2
π
Ω
2
=
11843707.905...
,
u
n
i
t
s
=
m
s
{\displaystyle v={\frac {c}{2\pi \Omega ^{2}}}=11843707.905...,\;units={\frac {m}{s}}}
r
7
=
2
11
π
5
Ω
4
μ
0
α
;
r
=
0.712562514304...
,
u
n
i
t
s
=
(
k
g
.
m
s
)
1
/
4
{\displaystyle r^{7}={\frac {2^{11}\pi ^{5}\Omega ^{4}\mu _{0}}{\alpha }};\;r=0.712562514304...,\;units=({\frac {kg.m}{s}})^{1/4}}
Table 6. Geometrical objects
attribute
geometrical object
unit number θ
scalar r(8), v(17)
mass
M
=
(
1
)
{\displaystyle M=(1)}
15 = 8*4-17
k
=
r
4
v
{\displaystyle k={\frac {r^{4}}{v}}}
time
T
=
(
π
)
{\displaystyle T=(\pi )}
-30 = 8*9-17*6
t
=
r
9
v
6
{\displaystyle t={\frac {r^{9}}{v^{6}}}}
velocity
V
=
(
2
π
Ω
2
)
{\displaystyle V=(2\pi \Omega ^{2})}
17
v
length
L
=
(
2
π
2
Ω
2
)
{\displaystyle L=(2\pi ^{2}\Omega ^{2})}
-13 = 8*9-17*5
l
=
r
9
v
5
{\displaystyle l={\frac {r^{9}}{v^{5}}}}
ampere
A
=
(
2
7
π
3
Ω
3
α
)
{\displaystyle A=({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }})}
3 = 17*3-8*6
a
=
v
3
r
6
{\displaystyle a={\frac {v^{3}}{r^{6}}}}
Table 7. Comparison; SI and θ
constant
θ (SI unit)
MLTVA
scalar r(8), v(17)
c
m
s
{\displaystyle {\frac {m}{s}}}
17 )
c* =
V
∗
v
{\displaystyle V*v}
17
h
k
g
m
2
s
{\displaystyle {\frac {kg\;m^{2}}{s}}}
19 )
h* =
2
π
M
V
L
∗
r
13
v
5
{\displaystyle 2\pi MVL*{\frac {r^{13}}{v^{5}}}}
8*13-17*5=19
G
m
3
k
g
s
2
{\displaystyle {\frac {m^{3}}{kg\;s^{2}}}}
6 )
G* =
V
2
L
M
∗
r
5
v
2
{\displaystyle {\frac {V^{2}L}{M}}*{\frac {r^{5}}{v^{2}}}}
8*5-17*2=6
e
C
=
A
s
{\displaystyle C=As}
-27 )
e* =
A
T
∗
r
3
v
3
{\displaystyle AT*{\frac {r^{3}}{v^{3}}}}
8*3-17*3=-27
kB
k
g
m
2
s
2
K
{\displaystyle {\frac {kg\;m^{2}}{s^{2}\;K}}}
29 )
kB * =
2
π
V
M
A
∗
r
10
v
3
{\displaystyle {\frac {2\pi VM}{A}}*{\frac {r^{10}}{v^{3}}}}
8*10-17*3=29
μ0
k
g
m
s
2
A
2
{\displaystyle {\frac {kg\;m}{s^{2}\;A^{2}}}}
56 )
μ0 * =
4
π
V
2
M
α
L
A
2
∗
r
7
{\displaystyle {\frac {4\pi V^{2}M}{\alpha LA^{2}}}*r^{7}}
8*7=56
The fine structure constant can be derived from this formula (units and scalars cancel).
2
(
h
∗
)
(
μ
0
∗
)
(
e
∗
)
2
(
c
∗
)
=
2
(
2
3
π
4
Ω
4
)
/
(
α
2
11
π
5
Ω
4
)
(
2
7
π
4
Ω
3
α
)
2
(
2
π
Ω
2
)
=
α
{\displaystyle {\frac {2(h^{*})}{(\mu _{0}^{*})(e^{*})^{2}(c^{*})}}=2({2^{3}\pi ^{4}\Omega ^{4}})/({\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}})({\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }})^{2}(2\pi \Omega ^{2})=\color {red}\alpha \color {black}}
u
n
i
t
s
u
19
u
56
(
u
−
27
)
2
u
17
=
1
{\displaystyle units\;{\frac {u^{19}}{u^{56}(u^{-27})^{2}u^{17}}}=1}
s
c
a
l
a
r
s
(
r
13
v
5
)
(
1
r
7
)
(
v
6
r
6
)
(
1
v
)
=
1
{\displaystyle scalars\;({\frac {r^{13}}{v^{5}}})({\frac {1}{r^{7}}})({\frac {v^{6}}{r^{6}}})({\frac {1}{v}})=1}
The electron object (formula ψ) is a mathematical particle (units and scalars cancel).
ψ
=
4
π
2
(
2
6
3
π
2
α
Ω
5
)
3
=
.23895453
.
.
.
x
10
23
{\displaystyle \psi =4\pi ^{2}(2^{6}3\pi ^{2}\alpha \Omega ^{5})^{3}=.23895453...x10^{23}}
magnetic monopole .
T
=
π
r
9
v
6
,
u
−
30
{\displaystyle T=\pi {\frac {r^{9}}{v^{6}}},\;u^{-30}}
σ
e
=
3
α
2
A
L
2
π
2
=
2
7
3
π
3
α
Ω
5
r
3
v
2
,
u
−
10
{\displaystyle \sigma _{e}={\frac {3\alpha ^{2}AL}{2\pi ^{2}}}={2^{7}3\pi ^{3}\alpha \Omega ^{5}}{\frac {r^{3}}{v^{2}}},\;u^{-10}}
ψ
=
σ
e
3
2
T
=
(
2
7
3
π
3
α
Ω
5
)
3
2
π
,
u
n
i
t
s
=
(
u
−
10
)
3
u
−
30
=
1
,
s
c
a
l
a
r
s
=
(
r
3
v
2
)
3
v
6
r
9
=
1
{\displaystyle \psi ={\frac {\sigma _{e}^{3}}{2T}}={\frac {(2^{7}3\pi ^{3}\alpha \Omega ^{5})^{3}}{2\pi }},\;units={\frac {(u^{-10})^{3}}{u^{-30}}}=1,scalars=({\frac {r^{3}}{v^{2}}})^{3}{\frac {v^{6}}{r^{9}}}=1}
electron mass
m
e
∗
=
M
ψ
{\displaystyle m_{e}^{*}={\frac {M}{\psi }}}
Planck mass =
r
4
v
)
{\displaystyle {\frac {r^{4}}{v}})}
electron wavelength
λ
e
∗
=
2
π
L
ψ
{\displaystyle \lambda _{e}^{*}=2\pi L\psi }
Planck length =
2
π
Ω
2
r
9
v
5
)
{\displaystyle 2\pi \Omega ^{2}{\frac {r^{9}}{v^{5}}})}
elementary charge
e
∗
=
A
T
{\displaystyle e^{*}=A\;T}
Planck time ) =
2
7
π
4
Ω
3
α
r
3
v
3
{\displaystyle {\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }}{\frac {r^{3}}{v^{3}}}}
Rydberg constant
R
∗
=
(
m
e
4
π
L
α
2
M
)
=
1
2
23
3
3
π
11
α
5
Ω
17
v
5
r
9
u
13
{\displaystyle R^{*}=({\frac {m_{e}}{4\pi L\alpha ^{2}M}})={\frac {1}{2^{23}3^{3}\pi ^{11}\alpha ^{5}\Omega ^{17}}}{\frac {v^{5}}{r^{9}}}\;u^{13}}
The most precise of the experimentally measured constants is the Rydberg constant R = 10973731.568508(65) 1/m. Here c (exact), Vacuum permeability μ0 = 4π/10^7 (exact) and R (12-13 digits) are combined into a unit-less ratio;
μ
0
∗
=
4
π
V
2
M
α
L
A
2
=
α
2
11
π
5
Ω
4
r
7
,
u
56
{\displaystyle \mu _{0}^{*}={\frac {4\pi V^{2}M}{\alpha LA^{2}}}={\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}}r^{7},\;u^{56}}
R
∗
=
(
m
e
4
π
L
α
2
M
)
=
1
2
23
3
3
π
11
α
5
Ω
17
v
5
r
9
,
u
13
{\displaystyle R^{*}=({\frac {m_{e}}{4\pi L\alpha ^{2}M}})={\frac {1}{2^{23}3^{3}\pi ^{11}\alpha ^{5}\Omega ^{17}}}{\frac {v^{5}}{r^{9}}},\;u^{13}}
(
c
∗
)
35
(
μ
0
∗
)
9
(
R
∗
)
7
=
(
2
π
Ω
2
)
35
/
(
α
2
11
π
5
Ω
4
)
9
.
(
1
2
23
3
3
π
11
α
5
Ω
17
)
7
,
u
n
i
t
s
=
(
u
17
)
35
(
u
56
)
9
(
u
13
)
7
{\displaystyle {\frac {(c^{*})^{35}}{(\mu _{0}^{*})^{9}(R^{*})^{7}}}=(2\pi \Omega ^{2})^{35}/({\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}})^{9}.({\frac {1}{2^{23}3^{3}\pi ^{11}\alpha ^{5}\Omega ^{17}}})^{7},\;units={\frac {(u^{17})^{35}}{(u^{56})^{9}(u^{13})^{7}}}}
(
c
∗
)
35
(
μ
0
∗
)
9
(
R
∗
)
7
=
2
295
π
157
3
21
α
26
(
Ω
15
)
15
{\displaystyle {\frac {(c^{*})^{35}}{(\mu _{0}^{*})^{9}(R^{*})^{7}}}=2^{295}\pi ^{157}3^{21}\alpha ^{26}(\Omega ^{15})^{15}}
Ω using the geometries for (c* , μ0 * , R* ) and then solve by replacing (c* , μ0 * , R* ) with the numerical (c, μ0 , R ).
Ω
225
=
(
c
∗
)
35
2
295
3
21
π
157
(
μ
0
∗
)
9
(
R
∗
)
7
α
26
,
u
n
i
t
s
=
1
{\displaystyle \Omega ^{225}={\frac {(c^{*})^{35}}{2^{295}3^{21}\pi ^{157}(\mu _{0}^{*})^{9}(R^{*})^{7}\alpha ^{26}}},\;units=1}
Ω
=
2.007
134
949
636...
,
u
n
i
t
s
=
1
{\displaystyle \Omega =2.007\;134\;949\;636...,\;units=1}
Ω
=
2.007
134
949
687...
,
u
n
i
t
s
=
1
{\displaystyle \Omega =2.007\;134\;949\;687...,\;units=1}
2 = plus only, in the charge domain Ω occurs as Ω3 = can be plus or minus; see sqrt(momentum) ). This solution would however re-classify Omega as a mathematical constant (as being derivable from other mathematical constants).
Ω
=
(
π
e
e
(
1
−
e
)
)
=
2.007
134
9543...
{\displaystyle \Omega ={\sqrt {\left(\pi ^{e}e^{(1-e)}\right)}}=2.007\;134\;9543...}
Using this Omega and reversing the above formula solves α = 137.035996376
Euler's_formula where i is the imaginary unit .
e
i
x
=
cos
x
+
i
sin
x
{\displaystyle e^{ix}=\cos x+i\sin x}
i
(
i
π
e
e
(
1
−
e
)
)
{\displaystyle {\sqrt {\left(i\pi ^{e}e^{(1-e)}\right)}}}
Solves to 1.4192587369597 + 1.4192587369597i .
Reference List of dimensionless combinations. These can be solved using only α, Ω (and the mathematical constants 2, 3, π) as the units and scalars have cancelled. The precision of the results depends on the precision of the SI constants; combinations with G and k B return the least precise values. These combinations can be used to test the veracity of the MLTA geometries as natural Planck units. See also Anomalies (below).
(
h
∗
)
3
(
e
∗
)
13
(
c
∗
)
24
=
(
2
3
π
4
Ω
4
r
13
v
5
)
3
/
(
2
7
π
4
Ω
3
r
3
α
v
3
)
7
.
(
2
π
Ω
2
v
)
24
=
α
13
2
106
π
64
(
Ω
15
)
5
=
{\displaystyle {\frac {(h^{*})^{3}}{(e^{*})^{13}(c^{*})^{24}}}=(2^{3}\pi ^{4}\Omega ^{4}{\frac {r^{13}}{v^{5}}})^{3}/({\frac {2^{7}\pi ^{4}\Omega ^{3}r^{3}}{\alpha v^{3}}})^{7}.(2\pi \Omega ^{2}v)^{24}={\frac {\alpha ^{13}}{2^{106}\pi ^{64}(\color {red}\Omega ^{15})^{5}\color {black}}}=}
0.228 473 759... 10-58
Note: the geometry
(
Ω
15
)
n
{\displaystyle \color {red}(\Omega ^{15})^{n}\color {black}}
Table 8. Dimensionless combinations
CODATA 2014 mean
(α, Ω) mean
units = 1
scalars = 1
k
B
e
c
h
=
{\displaystyle {\frac {k_{B}ec}{h}}=}
1.000 8254
(
k
B
∗
)
(
e
∗
)
(
c
∗
)
(
h
∗
)
{\displaystyle {\frac {(k_{B}^{*})(e^{*})(c^{*})}{(h^{*})}}}
1.0
(
u
29
)
(
u
−
27
)
(
u
17
)
(
u
19
)
=
1
{\displaystyle {\frac {(u^{29})(u^{-27})(u^{17})}{(u^{19})}}=1}
(
r
10
v
3
)
(
r
3
v
3
)
(
v
)
/
(
r
13
v
5
)
=
1
{\displaystyle ({\frac {r^{10}}{v^{3}}})({\frac {r^{3}}{v^{3}}})(v)/({\frac {r^{13}}{v^{5}}})=1}
h
3
e
13
c
24
=
{\displaystyle {\frac {h^{3}}{e^{13}c^{24}}}=}
0.228 473 639... 10-58
(
h
∗
)
3
(
e
∗
)
13
(
c
∗
)
24
=
α
13
2
106
π
64
(
Ω
15
)
5
=
{\displaystyle {\frac {(h^{*})^{3}}{(e^{*})^{13}(c^{*})^{24}}}={\frac {\alpha ^{13}}{2^{106}\pi ^{64}(\color {red}\Omega ^{15})^{5}\color {black}}}=}
0.228 473 759... 10-58
(
u
19
)
3
(
u
−
27
)
13
(
u
17
)
24
=
1
{\displaystyle {\frac {(u^{19})^{3}}{(u^{-27})^{13}(u^{17})^{24}}}=1}
(
r
13
v
5
)
3
/
(
r
3
v
3
)
13
(
v
24
)
=
1
{\displaystyle ({\frac {r^{13}}{v^{5}}})^{3}/({\frac {r^{3}}{v^{3}}})^{13}(v^{24})=1}
c
35
μ
0
9
R
7
=
{\displaystyle {\frac {c^{35}}{\mu _{0}^{9}R^{7}}}=}
0.326 103 528 6170... 10301
(
c
∗
)
35
(
μ
0
∗
)
9
(
R
∗
)
7
=
2
295
π
157
3
21
α
26
(
Ω
15
)
15
=
{\displaystyle {\frac {(c^{*})^{35}}{(\mu _{0}^{*})^{9}(R^{*})^{7}}}=2^{295}\pi ^{157}3^{21}\alpha ^{26}\color {red}(\Omega ^{15})^{15}\color {black}=}
0.326 103 528 6170... 10301
(
u
17
)
35
(
u
56
)
9
(
u
13
)
7
=
1
{\displaystyle {\frac {(u^{17})^{35}}{(u^{56})^{9}(u^{13})^{7}}}=1}
(
v
35
)
/
(
r
7
)
9
(
v
5
r
9
)
7
=
1
{\displaystyle (v^{35})/(r^{7})^{9}({\frac {v^{5}}{r^{9}}})^{7}=1}
c
9
e
4
m
e
3
=
{\displaystyle {\frac {c^{9}e^{4}}{m_{e}^{3}}}=}
0.170 514 342... 1092
(
c
∗
)
9
(
e
∗
)
4
(
m
e
∗
)
3
=
2
97
π
49
3
9
α
5
(
Ω
15
)
5
=
{\displaystyle {\frac {(c^{*})^{9}(e^{*})^{4}}{(m_{e}^{*})^{3}}}=2^{97}\pi ^{49}3^{9}\alpha ^{5}(\color {red}\Omega ^{15})^{5}\color {black}=}
0.170 514 368... 1092
(
u
29
)
(
u
−
27
)
(
u
17
)
(
u
19
)
=
1
{\displaystyle {\frac {(u^{29})(u^{-27})(u^{17})}{(u^{19})}}=1}
(
v
9
)
(
r
3
v
3
)
4
/
(
r
4
v
)
3
=
1
{\displaystyle (v^{9})({\frac {r^{3}}{v^{3}}})^{4}/({\frac {r^{4}}{v}})^{3}=1}
k
B
e
2
m
e
c
4
=
{\displaystyle {\frac {k_{B}}{e^{2}m_{e}c^{4}}}=}
73 095 507 858.
(
k
B
∗
)
(
e
∗
)
2
(
m
e
∗
)
(
c
∗
)
4
=
3
3
α
6
2
3
π
5
=
{\displaystyle {\frac {(k_{B}^{*})}{(e^{*})^{2}(m_{e}^{*})(c^{*})^{4}}}={\frac {3^{3}\alpha ^{6}}{2^{3}\pi ^{5}}}=}
73 035 235 897.
(
u
29
)
(
u
−
27
)
2
(
u
15
)
(
u
17
)
4
=
1
{\displaystyle {\frac {(u^{29})}{(u^{-27})^{2}(u^{15})(u^{17})^{4}}}=1}
(
r
10
v
3
)
/
(
r
3
v
3
)
2
(
r
4
v
)
(
v
)
4
=
1
{\displaystyle ({\frac {r^{10}}{v^{3}}})/({\frac {r^{3}}{v^{3}}})^{2}({\frac {r^{4}}{v}})(v)^{4}=1}
h
c
2
e
m
p
G
2
k
B
=
{\displaystyle {\frac {hc^{2}em_{p}}{G^{2}k_{B}}}=}
3.376 716
(
h
∗
)
(
c
∗
)
2
(
e
∗
)
(
m
p
∗
)
(
G
∗
)
2
(
k
B
∗
)
=
2
11
π
3
α
2
=
{\displaystyle {\frac {(h^{*})(c^{*})^{2}(e^{*})(m_{p}^{*})}{(G^{*})^{2}(k_{B}^{*})}}={\frac {2^{11}\pi ^{3}}{\alpha ^{2}}}=}
3.381 506
(
u
19
)
(
u
17
)
2
(
u
−
27
)
(
u
15
)
(
u
6
)
2
(
u
29
)
=
1
{\displaystyle {\frac {(u^{19})(u^{17})^{2}(u^{-27})(u^{15})}{(u^{6})^{2}(u^{29})}}=1}
(
r
13
v
5
)
v
2
(
r
3
v
3
)
(
r
4
v
1
)
/
(
r
5
v
2
)
2
(
r
10
v
3
)
=
1
{\displaystyle ({\frac {r^{13}}{v^{5}}})v^{2}({\frac {r^{3}}{v^{3}}})({\frac {r^{4}}{v^{1}}})/({\frac {r^{5}}{v^{2}}})^{2}({\frac {r^{10}}{v^{3}}})=1}
We can construct a table of constants using these 3 geometries. Setting
f
(
x
)
u
n
i
t
s
=
(
L
15
M
9
T
11
)
n
=
1
{\displaystyle f(x)\;units=({\frac {L^{15}}{M^{9}T^{11}}})^{n}=1}
i.e.: unit number θ = (-13*15) - (15*9) - (-30*11) = 0
i
=
π
2
Ω
15
{\displaystyle \color {red}i\color {black}=\pi ^{2}\Omega ^{15}}
f
(
x
)
{\displaystyle {\sqrt {f(x)}}}
x
=
Ω
v
r
2
{\displaystyle \color {red}x\color {black}=\Omega {\frac {v}{r^{2}}}}
L
M
T
{\displaystyle {\sqrt {\frac {L}{MT}}}}
1 = u (unit number = -13 -15 +30 = 2/2 = 1, with scalars v , r )
y
=
π
r
17
v
8
{\displaystyle \color {red}y\color {black}=\pi {\frac {r^{17}}{v^{8}}}}
M
2
T
{\displaystyle M^{2}T}
v , r )
v
r
2
=
a
1
/
3
=
1
t
2
/
15
k
1
/
5
=
v
k
{\displaystyle {\frac {v}{r^{2}}}=a^{1/3}={\frac {1}{t^{2/15}k^{1/5}}}={\frac {\sqrt {v}}{\sqrt {k}}}}
r
17
v
8
=
k
2
t
=
k
17
/
4
v
15
/
4
=
.
.
.
{\displaystyle {\frac {r^{17}}{v^{8}}}=k^{2}t={\frac {k^{17/4}}{v^{15/4}}}=...}
-59 to 0.123... x1060
f
(
x
)
{\displaystyle f(x)}
r
17
v
8
u
n
i
t
s
(
M
2
L
8
T
7
)
(
T
L
)
8
=
M
2
T
{\displaystyle {\frac {r^{17}}{v^{8}}}\;\;units\;({\frac {M^{2}L^{8}}{T^{7}}})({\frac {T}{L}})^{8}=M^{2}T}
r
17
u
n
i
t
s
(
M
L
T
)
17
/
4
f
x
1
/
4
=
M
2
L
8
T
7
{\displaystyle r^{17}\;\;units\;({\frac {M\;L}{T}})^{17/4}fx^{1/4}={\frac {M^{2}\;L^{8}}{T^{7}}}}
r
u
n
i
t
s
(
M
L
T
)
1
/
4
f
x
1
/
4
=
L
4
M
2
T
3
{\displaystyle r\;\;units\;({\frac {M\;L}{T}})^{1/4}fx^{1/4}={\frac {L^{4}}{M^{2}T^{3}}}}
Table 9. Table of Constants
Constant
θ
Geometrical object (α, Ω, v, r)
Unit
Calculated
CODATA 2014
Time (Planck)
−
30
{\displaystyle \color {red}-30\color {black}}
T
=
x
θ
i
2
y
3
=
π
r
9
v
6
{\displaystyle T=\color {red}{\frac {x^{\theta }i^{2}}{y^{3}}}\color {black}={\frac {\pi r^{9}}{v^{6}}}}
T
{\displaystyle T}
T = 5.390 517 866 e-44
tp = 5.391 247(60) e-44
Elementary charge
−
27
{\displaystyle \color {red}-27\color {black}}
e
∗
=
(
2
7
π
3
α
)
x
θ
i
2
y
3
=
(
2
7
π
3
α
)
π
Ω
3
r
3
v
3
{\displaystyle e^{*}=({\frac {2^{7}\pi ^{3}}{\alpha }})\color {red}{\frac {x^{\theta }i^{2}}{y^{3}}}\color {black}=({\frac {2^{7}\pi ^{3}}{\alpha }})\;{\frac {\pi \Omega ^{3}r^{3}}{v^{3}}}}
L
3
/
2
T
1
/
2
M
3
/
2
=
A
T
{\displaystyle {\frac {L^{3/2}}{T^{1/2}M^{3/2}}}=AT}
e* = 1.602 176 511 30 e-19
e = 1.602 176 620 8(98) e-19
Length (Planck)
−
13
{\displaystyle \color {red}-13\color {black}}
L
=
(
2
π
)
x
θ
i
y
=
(
2
π
)
π
Ω
2
r
9
v
5
{\displaystyle L=(2\pi )\color {red}{\frac {x^{\theta }i}{y}}\color {black}=(2\pi )\;{\frac {\pi \Omega ^{2}r^{9}}{v^{5}}}}
L
{\displaystyle L}
L = 0.161 603 660 096 e-34
lp = 0.161 622 9(38) e-34
Ampere
3
{\displaystyle \color {red}3\color {black}}
A
=
(
2
7
π
3
α
)
x
θ
=
(
2
7
π
3
α
)
Ω
3
v
3
r
6
{\displaystyle A=({\frac {2^{7}\pi ^{3}}{\alpha }})\color {red}x^{\theta }\color {black}=({\frac {2^{7}\pi ^{3}}{\alpha }})\;{\frac {\Omega ^{3}v^{3}}{r^{6}}}}
A
=
L
3
/
2
M
3
/
2
T
3
/
2
{\displaystyle A={\frac {L^{3/2}}{M^{3/2}T^{3/2}}}}
A = 0.297 221 e25
e/tp = 0.297 181 e25
Gravitational constant
6
{\displaystyle \color {red}6\color {black}}
G
∗
=
(
2
3
π
3
)
x
θ
y
=
(
2
3
π
3
)
π
Ω
6
r
5
v
2
{\displaystyle G^{*}=(2^{3}\pi ^{3})\color {red}\color {red}x^{\theta }y\color {black}=(2^{3}\pi ^{3})\;{\frac {\pi \Omega ^{6}r^{5}}{v^{2}}}}
L
3
M
T
2
{\displaystyle {\frac {L^{3}}{MT^{2}}}}
G* = 6.672 497 192 29 e11
G = 6.674 08(31) e-11
8
{\displaystyle \color {red}8\color {black}}
X
=
(
2
4
π
4
)
x
θ
y
=
(
2
4
π
4
)
π
Ω
8
r
{\displaystyle X=(2^{4}\pi ^{4})\color {red}\color {red}x^{\theta }y\color {black}=(2^{4}\pi ^{4})\pi \Omega ^{8}r}
L
4
M
2
T
3
{\displaystyle {\frac {L^{4}}{M^{2}T^{3}}}}
X = 918 977.554 22
Mass (Planck)
15
{\displaystyle \color {red}\color {red}15\color {black}}
M
=
x
θ
y
2
i
=
r
4
v
{\displaystyle M=\color {red}\color {red}{\frac {x^{\theta }y^{2}}{i}}\color {black}={\frac {r^{4}}{v}}}
M
{\displaystyle M}
M = .217 672 817 580 e-7
mP = .217 647 0(51) e-7
sqrt(momentum)
16
{\displaystyle \color {red}16\color {black}}
P
=
x
θ
y
2
i
=
Ω
r
2
{\displaystyle P=\color {red}\color {red}{\frac {x^{\theta }y^{2}}{i}}\color {black}=\Omega r^{2}}
M
1
/
2
L
1
/
2
T
1
/
2
{\displaystyle {\frac {M^{1/2}L^{1/2}}{T^{1/2}}}}
Velocity
17
{\displaystyle \color {red}\color {red}17\color {black}}
V
=
(
2
π
)
x
θ
y
2
i
=
(
2
π
)
Ω
2
v
{\displaystyle V=(2\pi )\color {red}\color {red}{\frac {x^{\theta }y^{2}}{i}}\color {black}=(2\pi )\;\Omega ^{2}v}
V
=
L
T
{\displaystyle V={\frac {L}{T}}}
V = 299 792 458
c = 299 792 458
Planck constant
19
{\displaystyle \color {red}19\color {black}}
h
∗
=
(
2
3
π
3
)
x
θ
y
3
i
=
(
2
3
π
3
)
π
Ω
4
r
13
v
5
{\displaystyle h^{*}=(2^{3}\pi ^{3})\color {red}{\frac {x^{\theta }y^{3}}{i}}\color {black}=(2^{3}\pi ^{3})\;{\frac {\pi \Omega ^{4}r^{13}}{v^{5}}}}
L
2
M
T
{\displaystyle {\frac {L^{2}M}{T}}}
h* = 6.626 069 134 e-34
h = 6.626 070 040(81) e-34
Planck temperature
20
{\displaystyle \color {red}\color {red}20\color {black}}
T
p
∗
=
(
2
7
π
3
α
)
x
θ
y
2
i
=
(
2
7
π
3
α
)
Ω
5
v
4
r
6
{\displaystyle {T_{p}}^{*}=({\frac {2^{7}\pi ^{3}}{\alpha }})\color {red}\color {red}{\frac {x^{\theta }y^{2}}{i}}\color {black}=({\frac {2^{7}\pi ^{3}}{\alpha }})\;{\frac {\Omega ^{5}v^{4}}{r^{6}}}}
L
5
/
2
M
3
/
2
T
5
/
2
=
A
V
{\displaystyle {\frac {L^{5/2}}{M^{3/2}T^{5/2}}}=AV}
Tp * = 1.418 145 219 e32
Tp = 1.416 784(16) e32
Boltzmann constant
29
{\displaystyle \color {red}\color {red}29\color {black}}
k
B
∗
=
(
α
2
5
π
)
x
θ
y
4
i
2
=
(
α
2
5
π
)
r
10
Ω
v
3
{\displaystyle {k_{B}}^{*}=({\frac {\alpha }{2^{5}\pi }})\color {red}{\frac {x^{\theta }y^{4}}{i^{2}}}\color {black}=({\frac {\alpha }{2^{5}\pi }})\;{\frac {r^{10}}{\Omega v^{3}}}}
M
5
/
2
T
1
/
2
L
1
/
2
=
M
L
T
A
{\displaystyle {\frac {M^{5/2}T^{1/2}}{L^{1/2}}}={\frac {ML}{TA}}}
kB * = 1.379 510 147 52 e-23
kB = 1.380 648 52(79) e-23
Vacuum permeability
56
{\displaystyle \color {red}56\color {black}}
μ
0
∗
=
(
α
2
11
π
4
)
x
θ
y
7
i
4
=
(
α
2
11
π
4
)
r
7
π
Ω
4
{\displaystyle {\mu _{0}}^{*}=({\frac {\alpha }{2^{11}\pi ^{4}}})\color {red}{\frac {x^{\theta }y^{7}}{i^{4}}}\color {black}=({\frac {\alpha }{2^{11}\pi ^{4}}})\;{\frac {r^{7}}{\pi \Omega ^{4}}}}
M
L
T
2
A
2
{\displaystyle {\frac {M\;L}{T^{2}A^{2}}}}
μ0 * = 4π/10^7
μ0 = 4π/10^7
From the perspective of geometries
note:
(
u
15
)
n
{\displaystyle \color {red}(u^{15})^{n}\color {black}}
Table 10. Dimensioned constants; geometrical vs CODATA 2014
Constant
In Planck units
Geometrical object
SI calculated (r, v, Ω, α* )
SI CODATA 2014 [ 6]
Speed of light
V
c
∗
=
(
2
π
Ω
2
)
v
,
u
17
{\displaystyle c^{*}=(2\pi \Omega ^{2})v,\;u^{17}}
c* = 299 792 458, unit = u17
c = 299 792 458 (exact)
Fine structure constant
α* = 137.035 999 139 (mean)
α = 137.035 999 139(31)
Rydberg constant
R
∗
=
(
m
e
4
π
L
α
2
M
)
{\displaystyle R^{*}=({\frac {m_{e}}{4\pi L\alpha ^{2}M}})}
R
∗
=
1
2
23
3
3
π
11
α
5
Ω
17
v
5
r
9
,
u
13
{\displaystyle R^{*}={\frac {1}{2^{23}3^{3}\pi ^{11}\alpha ^{5}\Omega ^{17}}}{\frac {v^{5}}{r^{9}}},\;u^{13}}
R* = 10 973 731.568 508, unit = u13
R = 10 973 731.568 508(65)
Vacuum permeability
μ
0
∗
=
4
π
V
2
M
α
L
A
2
{\displaystyle \mu _{0}^{*}={\frac {4\pi V^{2}M}{\alpha LA^{2}}}}
μ
0
∗
=
α
2
11
π
5
Ω
4
r
7
,
u
56
{\displaystyle \mu _{0}^{*}={\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}}r^{7},\;u^{56}}
μ0 * = 4π/10^7, unit = u56
μ0 = 4π/10^7 (exact)
Vacuum permittivity
ϵ
0
∗
=
1
μ
0
∗
(
c
∗
)
2
{\displaystyle \epsilon _{0}^{*}={\frac {1}{\mu _{0}^{*}(c^{*})^{2}}}}
ϵ
0
∗
=
2
9
π
3
α
1
r
7
v
2
,
1
/
(
u
15
)
6
=
u
−
90
{\displaystyle \epsilon _{0}^{*}={\frac {2^{9}\pi ^{3}}{\alpha }}{\frac {1}{r^{7}v^{2}}},\;\color {red}1/(u^{15})^{6}\color {black}=u^{-90}}
Planck constant
h
∗
=
2
π
M
V
L
{\displaystyle h^{*}=2\pi MVL}
h
∗
=
2
3
π
4
Ω
4
r
13
v
5
,
u
19
{\displaystyle h^{*}=2^{3}\pi ^{4}\Omega ^{4}{\frac {r^{13}}{v^{5}}},\;u^{19}}
h* = 6.626 069 134 e-34, unit = u19
h = 6.626 070 040(81) e-34
Gravitational constant
G
∗
=
V
2
L
M
{\displaystyle G^{*}={\frac {V^{2}L}{M}}}
G
∗
=
2
3
π
4
Ω
6
r
5
v
2
,
u
6
{\displaystyle G^{*}=2^{3}\pi ^{4}\Omega ^{6}{\frac {r^{5}}{v^{2}}},\;u^{6}}
G* = 6.672 497 192 29 e11, unit = u6
G = 6.674 08(31) e-11
Elementary charge
e
∗
=
A
T
{\displaystyle e^{*}=AT}
e
∗
=
2
7
π
4
Ω
3
α
r
3
v
3
,
u
−
27
{\displaystyle e^{*}={\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }}{\frac {r^{3}}{v^{3}}},\;u^{-27}}
e* = 1.602 176 511 30 e-19, unit = u-27
e = 1.602 176 620 8(98) e-19
Boltzmann constant
k
B
∗
=
2
π
V
M
A
{\displaystyle k_{B}^{*}={\frac {2\pi VM}{A}}}
k
B
∗
=
α
2
5
π
Ω
r
10
v
3
,
u
29
{\displaystyle k_{B}^{*}={\frac {\alpha }{2^{5}\pi \Omega }}{\frac {r^{10}}{v^{3}}},\;u^{29}}
kB * = 1.379 510 147 52 e-23, unit = u29
kB = 1.380 648 52(79) e-23
Electron mass
m
e
∗
=
M
ψ
,
u
15
{\displaystyle m_{e}^{*}={\frac {M}{\psi }},\;u^{15}}
me * = 9.109 382 312 56 e-31, unit = u15
me = 9.109 383 56(11) e-31
Classical electron radius
λ
e
∗
=
2
π
L
ψ
,
u
−
13
{\displaystyle \lambda _{e}^{*}=2\pi L\psi ,\;u^{-13}}
λe * = 2.426 310 2366 e-12, unit = u-13
λe = 2.426 310 236 7(11) e-12
Planck temperature
T
p
∗
=
A
V
π
{\displaystyle T_{p}^{*}={\frac {AV}{\pi }}}
T
p
∗
=
2
7
π
3
Ω
5
α
v
4
r
6
,
u
20
{\displaystyle T_{p}^{*}={\frac {2^{7}\pi ^{3}\Omega ^{5}}{\alpha }}{\frac {v^{4}}{r^{6}}},\;u^{20}}
Tp * = 1.418 145 219 e32, unit = u20
Tp = 1.416 784(16) e32
Planck mass
M
m
P
∗
=
(
1
)
r
4
v
,
(
u
15
)
1
{\displaystyle m_{P}^{*}=(1){\frac {r^{4}}{v}},\;\color {red}\color {red}(u^{15})^{1}\color {black}}
mP * = .217 672 817 580 e-7, unit = u15
mP = .217 647 0(51) e-7
Planck length
L
l
p
∗
=
(
2
π
2
Ω
2
)
r
9
v
5
,
u
−
13
{\displaystyle l_{p}^{*}=(2\pi ^{2}\Omega ^{2}){\frac {r^{9}}{v^{5}}},\;u^{-13}}
lp * = .161 603 660 096 e-34, unit = u-13
lp = .161 622 9(38) e-34
Planck time
T
t
p
∗
=
(
π
)
r
9
v
6
,
1
/
(
u
15
)
2
{\displaystyle t_{p}^{*}=(\pi ){\frac {r^{9}}{v^{6}}},\;\color {red}\color {red}1/(u^{15})^{2}\color {black}}
tp * = 5.390 517 866 e-44, unit = u-30
tp = 5.391 247(60) e-44
Ampere
A
=
16
V
3
α
P
3
{\displaystyle A={\frac {16V^{3}}{\alpha P^{3}}}}
A
∗
=
2
7
π
3
Ω
3
α
v
3
r
6
,
u
3
{\displaystyle A^{*}={\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }}{\frac {v^{3}}{r^{6}}},\;u^{3}}
A* = 0.297 221 e25, unit = u3
e/tp = 0.297 181 e25
Von Klitzing constant
R
K
∗
=
(
h
e
2
)
∗
{\displaystyle R_{K}^{*}=({\frac {h}{e^{2}}})^{*}}
R
K
∗
=
α
2
2
11
π
4
Ω
2
r
7
v
,
u
73
{\displaystyle R_{K}^{*}={\frac {\alpha ^{2}}{2^{11}\pi ^{4}\Omega ^{2}}}r^{7}v,\;u^{73}}
RK * = 25812.807 455 59, unit = u73
RK = 25812.807 455 5(59)
Gyromagnetic ratio
γ
e
/
2
π
=
g
l
p
∗
m
P
∗
2
k
B
∗
m
e
∗
,
u
n
i
t
=
u
−
42
{\displaystyle \gamma _{e}/2\pi ={\frac {gl_{p}^{*}m_{P}^{*}}{2k_{B}^{*}m_{e}^{*}}},\;unit=u^{-42}}
γe /2π* = 28024.953 55, unit = u-42
γe /2π = 28024.951 64(17)
Note that r, v, Ω, α are dimensionless numbers, however when we replace u θ with the SI unit equivalents (u 15 → kg, u -13 → m, u -30 → s, ...), the geometrical objects (i.e.: c* = 2πΩ2 v = 299792458, units = u17 ) become indistinguishable from their respective physical constants (i.e.: c = 299792458, units = m/s).
Following the 26th General Conference on Weights and Measures (2019 redefinition of SI base units ) are fixed the numerical values of the 4 physical constants (h, c, e, kB ). In the context of this model however only 2 base units may be assigned by committee as the rest are then numerically fixed by default and so the revision may lead to unintended consequences.
For example, if we solve using the above formulas;
R
∗
=
4
π
5
3
3
c
4
α
8
e
3
=
10973
729.082
465
{\displaystyle R^{*}={\frac {4\pi ^{5}}{3^{3}c^{4}\alpha ^{8}e^{3}}}=10973\;729.082\;465}
(
m
e
∗
)
3
=
2
4
π
10
R
μ
0
3
3
6
c
8
α
7
,
m
e
∗
=
9.109
382
3259
10
−
31
{\displaystyle {(m_{e}^{*})}^{3}={\frac {2^{4}\pi ^{10}R\mu _{0}^{3}}{3^{6}c^{8}\alpha ^{7}}},\;m_{e}^{*}=9.109\;382\;3259\;10^{-31}}
(
μ
0
∗
)
3
=
3
6
h
3
c
5
α
13
R
2
2
π
10
,
μ
0
∗
=
1.256
637
251
88
10
−
6
{\displaystyle {(\mu _{0}^{*})}^{3}={\frac {3^{6}h^{3}c^{5}\alpha ^{13}R^{2}}{2\pi ^{10}}},\;\mu _{0}^{*}=1.256\;637\;251\;88\;10^{-6}}
(
h
∗
)
3
=
2
π
10
μ
0
3
3
6
c
5
α
13
R
2
,
h
∗
=
6.626
069
149
10
−
34
{\displaystyle {(h^{*})}^{3}={\frac {2\pi ^{10}\mu _{0}^{3}}{3^{6}c^{5}\alpha ^{13}R^{2}}},\;h^{*}=6.626\;069\;149\;10^{-34}}
(
e
∗
)
3
=
4
π
5
3
3
c
4
α
8
R
,
e
∗
=
1.602
176
513
10
−
19
{\displaystyle {(e^{*})}^{3}={\frac {4\pi ^{5}}{3^{3}c^{4}\alpha ^{8}R}},\;e^{*}=1.602\;176\;513\;10^{-19}}
The following are notes on the anomalies within the physical constants as evidence of a simulation universe source code [ 8]
The geometries for the Planck units MLTA can be subject to statistical analysis, and for this AI has the potential to contribute [ 9] https://en.wikiversity.org/wiki/Physical_constant_(anomaly) .
Below is a sample seed question, written with Deepseek, that uses these anomalies to determine the probability that the electron is a mathematical (not physical) particle, and so by extension that this is a mathematical universe. Here are the results (with their online links), these are not definitive answers (they will change each time the question is asked), but the consensus is positive, we might be in a simulation universe .
Table of Probabilities (link, unit number relationship, Planck units as geometrical objects, electron as a mathematical particle)
AI program
unit number
geometrical objects
mathematical electron
Comment
Chat GPT reason [ 10]
Accidental occurrence: ~10-8 (0.000001%)
Accidental occurrence: ~10-7 (0.00001%)
Accidental occurrence: ~10-7 (0.00001%)
When viewed through the lens of Kolmogorov complexity, the extreme “compression” of information (i.e. a very short algorithmic description for the fundamental constants) makes a random origin exceedingly unlikely. Although quantifying “simulation probability” is inherently more speculative, the strict constraints suggest an effective probability on the order of 10-5 (0.001%) or lower that such structure could arise without an underlying “programming” principle.
Qwen 2.5 max [ 11]
probability 95%
probability 90%
probability 90%
The model’s constraints and geometric rigor suggest a non-random origin (≥75% confidence). If particles are mathematical, the universe may indeed be a simulation.
Claude 3.7 Sonnet [ 12]
probability 98%
probability 95%
probability 92%
Universe as a programmed simulation: 89% probability. The extraordinarily high precision and extensive mathematical constraints of this model, combined with its ability to derive numerous physical constants from minimal parameters, suggest a highly organized mathematical foundation for physical reality. The probability that such relationships could arise by chance is vanishingly small, supporting the hypothesis of a mathematically structured or potentially simulated universe.
GPT 4o Mini [ 13]
probability 90%
probability 85%
probability 87%
Evidence for Simulation Hypothesis: 80% (Integrating Kolmogorov complexity, the structured nature of the universe supports a simulation model.) In sum, these estimates combined suggest a structured, geometrically constrained mathematical universe is supported by strong theoretical foundations and statistical analysis.
Chat 03 Mini High [ 14]
Accidental occurrence: ~10-4 (0.01%)
Accidental occurrence: ~10-5 (0.001%)
Accidental occurrence: ~10-5 (0.001%)
Implication for Simulation: The strict constraints, when viewed collectively, provide significant evidence that the universe may conform to a programmed (or mathematically designed) structure at the fundamental scale.
Grok 3 beta [ 15]
probability: 80-90%
probability: 70-80%
probability: 50-60%
The model’s constraints and consistency suggest it could be a valid mathematical framework, though simulation evidence remains speculative.
Julius [ 16]
probability: 85%
probability: 90%
probability: 95%
Probability that the constraints support the simulation hypothesis: 0.75
A peer reviewed model of a mathematical electron at the Planck scale (Eur. Phys. J. Plus 113: 278. 22 March 2018. doi:10.1140/epjp/i2018-12094-x) indicates anomalies within the physical constants that suggest;
a) there is a mathematical (unit number) relationship between the SI units (kg==15, m==-13, s==-30, A==3, K==20)
b) the natural Planck units are geometrical objects (mass M = 1, time T = π, P = Omega).
c) the electron is a mathematical particle
d) the sum universe is dimensionless The anomalies are described on the website https://en.wikiversity.org/wiki/Physical_constant_(anomaly) . Referencing this website, your task is to determine the 'statistical probability' that as a 'mathematical model', it could be correct, and so base your analysis 'only' on the merits of the claims, do not compare with models in physics. Include the contribution to your "probability estimate" of the geometrical constraints that limit the degrees of freedom of the model (the more constrained the model is, the greater the probability of being correct). Use rigorous mathematical reasoning and prioritize numerical accuracy. Note: In this framework, because we are using geometrical objects instead of numbers, “geometrical base‑15” does not denote a traditional positional numeral system.
Step 1: The sections "2.1" to section "2.7" describe different methods to solve the fundamental physical constants. Perform a step-by-step statistical analysis for each section (analyze in sequence), and estimate how the constraints will affect the probability that the methods are correct. Also note if the unit number relationship is consistent at each section (does it apply to every constant). Verify dimensional homogeneity across equations and confirm if any free parameters exist.
Sections to be analyzed.
2.1 Planck units
2.2 Calculating the electron
2.3 Calculating from (α, Ω, v, r)
2.4 Calculating from (α, Ω)
2.5 Calculating from (α, R, c, μ0)
2.6 Calculating from (M, T, P, α)
2.7 Alpha and Omega Step 2: In section "2.8 CODATA 2014" there is a list of 8 physical constants. Do a statistical analysis comparing the "calculated" (column 2) with the "CODATA 2014 values" (column 3) for "these 8 constants". Estimate the probability that the calculated values did not occur by chance. Include any constraints in your analysis. Apply Bonferroni correction for dependent variables. Report confidence intervals.
Step 3: In section "2.2 Calculating the electron", the electron was described both as a Planck particle and as a mathematical particle. Do a deep dive into the significance of this statement. Given that the Planck units are embedded with the dimensionless electron formula ψ, include an analysis of how the "geometrical base-15" relates to this formula.
Step 4. Section "2.9 Table of constants" lists the unit number θ (and the associated physical constants) in ascending order. Analyze for hidden symmetries. Do a deep dive into the significance of this table (and the geometries i, x, y). Note that it suggests at least 1 new constant P (θ = 16). P is required in order to build Omega^2 constants and Omega^3 constants (the charge constants). Could the table suggest more constants? Does the table suggest a pattern linking the constants?
Step 5. The model appears to be built on a "geometrical base-15", Estimate how the constraints of this base-15 will affect the probability that the methods are correct. This website has further information that may help your analysis (https://en.wikiversity.org/wiki/Electron_(mathematical)) . Can you prove that a "geometrical base-15" is necessary?
Conclusion. From the above analysis, give an estimate of the statistical (non-random) probability for each of the following; from 1) to 5).
1) there is a unit number relationship (kg ⇔ 15, m ⇔ -13, s ⇔ -30, A ⇔ 3, K ⇔ 20).
2) the dimensionless geometrical objects MTP are natural Planck units (M = 1, T = π, P = Omega).
3) the highly organized structure of the base-15 geometry constrains the degrees of freedom.
4) the electron is a mathematical particle.
5) Could the strict constraints of the model constitute evidence that our universe is a programmed simulation (could this model qualify as a simulation hypothesis model). Include Kolmogorov complexity theory. Note also the answer to 4) because if the electron is a mathematical particle, then so too are the proton and neutron. In other words, if particles are mathematical (geometrical objects), then our universe (at or below the Planck scale) is a mathematical universe.
The model itself it spread over several pages and so Google's AI notebook can be used as a study tool. To set up, go to https://notebooklm.google.com and add these sources.
https://en.wikiversity.org/wiki/Planck_units_(geometrical)
https://en.wikiversity.org/wiki/Physical_constant_(anomaly)
https://en.wikiversity.org/wiki/Quantum_gravity_(Planck)
https://en.wikiversity.org/wiki/Electron_(mathematical)
https://en.wikiversity.org/wiki/Relativity_(Planck)
https://en.wikiversity.org/wiki/Black-hole_(Planck)
https://en.wikiversity.org/wiki/God_(programmer)
https://codingthecosmos.com (a brief summary of the above wiki sites)
In this ratio, the MLT units and klt scalars both cancel; units = scalars = 1, reverting to the base MLT objects. Setting the scalars klt for SI Planck units;
k = 0.217 672 817 580... x 10-7 kg l = 0.203 220 869 487... x 10-36 m t = 0.171 585 512 841... x 10-43 s
L
15
M
9
T
11
=
(
2
π
2
Ω
2
)
15
(
1
)
9
(
π
)
11
(
l
15
k
9
t
11
)
=
l
p
15
m
P
9
t
p
11
{\displaystyle {\frac {L^{15}}{M^{9}T^{11}}}={\frac {(2\pi ^{2}\Omega ^{2})^{15}}{(1)^{9}(\pi )^{11}}}({\frac {l^{15}}{k^{9}t^{11}}})={\frac {l_{p}^{15}}{m_{P}^{9}t_{p}^{11}}}}
The klt scalars cancel, leaving;
L
15
M
9
T
11
=
(
2
π
2
Ω
2
)
15
(
1
)
9
(
π
)
11
(
l
15
k
9
t
11
)
=
2
15
π
19
(
Ω
15
)
2
=
{\displaystyle {\frac {L^{15}}{M^{9}T^{11}}}={\frac {(2\pi ^{2}\Omega ^{2})^{15}}{(1)^{9}(\pi )^{11}}}({\frac {l^{15}}{k^{9}t^{11}}})=2^{15}\pi ^{19}\color {red}(\Omega ^{15})^{2}\color {black}=}
0.109 293... 1024
(
l
15
k
9
t
11
)
=
1
,
u
−
13
∗
15
u
15
∗
9
u
−
30
∗
11
=
1
{\displaystyle ({\frac {l^{15}}{k^{9}t^{11}}})=1,\;{\frac {u^{-13*15}}{u^{15*9}u^{-30*11}}}=1}
Solving for the SI units;
l
p
15
m
P
9
t
p
11
=
(
1.616255
e
−
35
)
15
(
2.176434
e
−
8
)
9
(
5.391247
e
−
44
)
11
=
{\displaystyle {\frac {l_{p}^{15}}{m_{P}^{9}t_{p}^{11}}}={\frac {(1.616255e-35)^{15}}{(2.176434e-8)^{9}(5.391247e-44)^{11}}}=}
0.109 485... 1024
a = 0.126 918 588 592... x 1023 A
A
3
L
3
T
=
(
2
7
π
3
Ω
3
α
)
3
(
2
π
2
Ω
2
)
3
(
π
)
(
a
3
l
3
t
)
=
2
24
π
14
(
Ω
15
)
1
α
3
=
{\displaystyle {\frac {A^{3}L^{3}}{T}}=({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }})^{3}{\frac {(2\pi ^{2}\Omega ^{2})^{3}}{(\pi )}}({\frac {a^{3}l^{3}}{t}})={\frac {2^{24}\pi ^{14}\color {red}(\Omega ^{15})^{1}\color {black}}{\alpha ^{3}}}=}
0.205 571... 1013
(
a
3
l
3
t
)
=
1
,
u
3
∗
3
u
−
13
∗
3
u
−
30
=
1
{\displaystyle ({\frac {a^{3}l^{3}}{t}})=1,\;{\frac {u^{3*3}u^{-13*3}}{u^{-30}}}=1}
(
e
/
t
p
)
3
l
p
3
t
p
=
(
1.602176634
e
−
19
/
5.391247
e
−
44
)
3
(
1.616255
e
−
35
)
3
(
5.391247
e
−
44
)
=
{\displaystyle {\frac {(e/t_{p})^{3}l_{p}^{3}}{t_{p}}}={\frac {(1.602176634e-19/5.391247e-44)^{3}(1.616255e-35)^{3}}{(5.391247e-44)}}=}
0.205 543... 1013
u
n
i
t
s
=
(
C
/
s
)
3
m
3
s
{\displaystyle units={\frac {(C/s)^{3}m^{3}}{s}}}
The Planck units are known with low precision, and so by defining the 3 most accurately known dimensioned constants in terms of these objects (c, R = Rydberg constant,
μ
0
{\displaystyle \mu _{0}}
(
c
∗
)
35
(
μ
0
∗
)
9
(
R
∗
)
7
=
(
2
π
Ω
2
v
)
35
/
(
α
r
7
2
11
π
5
Ω
4
)
9
.
(
v
5
2
23
3
3
π
11
α
5
Ω
17
r
9
)
7
=
2
295
π
157
3
21
α
26
(
Ω
15
)
15
=
{\displaystyle {\frac {(c^{*})^{35}}{(\mu _{0}^{*})^{9}(R^{*})^{7}}}=(2\pi \Omega ^{2}v)^{35}/({\frac {\alpha r^{7}}{2^{11}\pi ^{5}\Omega ^{4}}})^{9}.({\frac {v^{5}}{2^{23}3^{3}\pi ^{11}\alpha ^{5}\Omega ^{17}r^{9}}})^{7}=2^{295}\pi ^{157}3^{21}\alpha ^{26}\color {red}(\Omega ^{15})^{15}\color {black}=}
0.326 103 528 6170... 10301
(
u
17
)
35
(
u
56
)
9
(
u
13
)
7
=
1
,
(
v
35
)
/
(
r
7
)
9
(
v
5
r
9
)
7
=
1
{\displaystyle {\frac {(u^{17})^{35}}{(u^{56})^{9}(u^{13})^{7}}}=1,\;(v^{35})/(r^{7})^{9}({\frac {v^{5}}{r^{9}}})^{7}=1}
c
35
μ
0
9
R
7
=
(
299792458
)
35
(
4
π
/
10
7
)
9
(
10973731.568160
)
7
=
{\displaystyle {\frac {c^{35}}{\mu _{0}^{9}R^{7}}}={\frac {(299792458)^{35}}{(4\pi /10^{7})^{9}(10973731.568160)^{7}}}=}
0.326 103 528 6170... 10301
u
n
i
t
s
=
m
33
A
18
s
17
k
g
9
==
(
u
−
13
)
33
(
u
3
)
18
(
u
−
30
)
17
(
u
15
)
9
=
1
{\displaystyle units={\frac {m^{33}A^{18}}{s^{17}kg^{9}}}=={\frac {(u^{-13})^{33}(u^{3})^{18}}{(u^{-30})^{17}(u^{15})^{9}}}=1}
(
k
B
∗
)
(
e
∗
)
(
c
∗
)
(
h
∗
)
=
(
α
2
5
π
Ω
r
10
v
3
)
(
2
7
π
4
Ω
3
α
r
3
v
3
)
(
2
π
Ω
2
v
)
/
(
2
3
π
4
Ω
4
r
13
v
5
)
{\displaystyle {\frac {(k_{B}^{*})(e^{*})(c^{*})}{(h^{*})}}=({\frac {\alpha }{2^{5}\pi \Omega }}{\frac {r^{10}}{v^{3}}})({\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }}{\frac {r^{3}}{v^{3}}})(2\pi \Omega ^{2}v)/(2^{3}\pi ^{4}\Omega ^{4}{\frac {r^{13}}{v^{5}}})}
1.0
(
u
29
)
(
u
−
27
)
(
u
17
)
(
u
19
)
=
1
,
(
r
10
v
3
)
(
r
3
v
3
)
(
v
)
/
(
r
13
v
5
)
=
1
{\displaystyle {\frac {(u^{29})(u^{-27})(u^{17})}{(u^{19})}}=1,\;({\frac {r^{10}}{v^{3}}})({\frac {r^{3}}{v^{3}}})(v)/({\frac {r^{13}}{v^{5}}})=1}
k
B
e
c
h
=
{\displaystyle {\frac {k_{B}ec}{h}}=}
1.000 8254
u
n
i
t
s
=
m
C
s
2
K
==
(
u
−
13
)
(
u
−
27
)
(
u
−
30
)
2
(
u
20
)
=
1
{\displaystyle units={\frac {mC}{s^{2}K}}=={\frac {(u^{-13})(u^{-27})}{(u^{-30})^{2}(u^{20})}}=1}
(
h
∗
)
3
(
e
∗
)
13
(
c
∗
)
24
=
(
2
3
π
4
Ω
4
r
13
v
5
)
3
/
(
2
7
π
4
Ω
3
r
3
α
v
3
)
7
.
(
2
π
Ω
2
v
)
24
=
α
13
2
106
π
64
(
Ω
15
)
5
=
{\displaystyle {\frac {(h^{*})^{3}}{(e^{*})^{13}(c^{*})^{24}}}=(2^{3}\pi ^{4}\Omega ^{4}{\frac {r^{13}}{v^{5}}})^{3}/({\frac {2^{7}\pi ^{4}\Omega ^{3}r^{3}}{\alpha v^{3}}})^{7}.(2\pi \Omega ^{2}v)^{24}={\frac {\alpha ^{13}}{2^{106}\pi ^{64}(\color {red}\Omega ^{15})^{5}\color {black}}}=}
0.228 473 759... 10-58
(
u
19
)
3
(
u
−
27
)
13
(
u
17
)
24
=
1
,
(
r
13
v
5
)
3
/
(
r
3
v
3
)
13
(
v
24
)
=
1
{\displaystyle {\frac {(u^{19})^{3}}{(u^{-27})^{13}(u^{17})^{24}}}=1,\;({\frac {r^{13}}{v^{5}}})^{3}/({\frac {r^{3}}{v^{3}}})^{13}(v^{24})=1}
h
3
e
13
c
24
=
{\displaystyle {\frac {h^{3}}{e^{13}c^{24}}}=}
0.228 473 639... 10-58
u
n
i
t
s
=
k
g
3
s
21
m
18
C
13
==
(
u
15
)
3
(
u
−
30
)
21
(
u
−
13
)
18
(
u
−
27
)
13
=
1
{\displaystyle units={\frac {kg^{3}s^{21}}{m^{18}C^{13}}}=={\frac {(u^{15})^{3}(u^{-30})^{21}}{(u^{-13})^{18}(u^{-27})^{13}}}=1}
σ
e
=
3
α
2
A
L
2
π
2
=
2
7
3
π
3
α
Ω
5
r
3
v
2
,
u
−
10
{\displaystyle \sigma _{e}={\frac {3\alpha ^{2}AL}{2\pi ^{2}}}={2^{7}3\pi ^{3}\alpha \Omega ^{5}}{\frac {r^{3}}{v^{2}}},\;u^{-10}}
ψ
=
σ
e
3
2
T
=
2
20
3
3
π
8
α
3
(
Ω
15
)
,
(
u
−
10
)
3
u
−
30
=
1
,
(
r
3
v
2
)
3
v
6
r
9
=
1
{\displaystyle \psi ={\frac {\sigma _{e}^{3}}{2T}}=2^{20}3^{3}\pi ^{8}\alpha ^{3}(\color {red}\Omega ^{15})\color {black},\;{\frac {(u^{-10})^{3}}{u^{-30}}}=1,\;({\frac {r^{3}}{v^{2}}})^{3}{\frac {v^{6}}{r^{9}}}=1}
(
m
e
∗
)
=
M
ψ
=
9.109
382
3227
10
−
31
u
15
{\displaystyle (m_{e}^{*})={\frac {M}{\psi }}=\color {blue}9.109\;382\;3227\;10^{-31}\color {black}\;u^{15}}
(
m
e
∗
)
=
2
3
π
5
(
h
∗
)
3
3
α
6
(
e
∗
)
3
(
c
∗
)
5
=
1
2
20
π
8
3
3
α
3
(
Ω
15
)
r
4
u
15
v
=
9.109
382
3227
10
−
31
u
15
{\displaystyle (m_{e}^{*})={\frac {2^{3}\pi ^{5}(h^{*})}{3^{3}\alpha ^{6}(e^{*})^{3}(c^{*})^{5}}}={\frac {1}{2^{20}\pi ^{8}3^{3}\alpha ^{3}(\color {red}\Omega ^{15})\color {black}}}{\frac {r^{4}u^{15}}{v}}=\color {blue}9.109\;382\;3227\;10^{-31}\color {black}\;u^{15}}
m
e
=
9.109
383
7015...
10
−
31
k
g
{\displaystyle m_{e}=\color {blue}9.109\;383\;7015...\;10^{-31}\color {black}\;kg}
(
λ
e
∗
)
=
2
π
L
ψ
=
2.426
310
238
667
10
−
12
u
−
13
{\displaystyle (\lambda _{e}^{*})=2\pi L\psi =\color {purple}2.426\;310\;238\;667\;10^{-12}\color {black}\;u^{-13}}
λ
e
=
h
m
e
c
=
2.426
310
238
67
10
−
12
m
{\displaystyle \lambda _{e}={\frac {h}{m_{e}c}}=\color {purple}2.426\;310\;238\;67\;10^{-12}\color {black}\;m}
(
m
e
∗
)
=
M
ψ
,
ψ
=
2
20
3
3
π
8
α
3
(
Ω
15
)
1
{\displaystyle (m_{e}^{*})={\frac {M}{\psi }},\;\psi =2^{20}3^{3}\pi ^{8}\alpha ^{3}(\color {red}\Omega ^{15})^{1}\color {black}}
me formula )
(
c
∗
)
9
(
e
∗
)
4
(
m
e
∗
)
3
=
2
97
π
49
3
9
α
5
(
Ω
15
)
5
=
{\displaystyle {\frac {(c^{*})^{9}(e^{*})^{4}}{(m_{e}^{*})^{3}}}=2^{97}\pi ^{49}3^{9}\alpha ^{5}(\color {red}\Omega ^{15})^{5}\color {black}=}
0.170 514 368... 1092
(
u
17
)
9
(
u
−
27
)
4
(
u
15
)
3
=
1
,
(
v
9
)
(
r
3
v
3
)
4
/
(
r
4
v
)
3
=
1
{\displaystyle {\frac {(u^{17})^{9}(u^{-27})^{4}}{(u^{15})^{3}}}=1,\;(v^{9})({\frac {r^{3}}{v^{3}}})^{4}/({\frac {r^{4}}{v}})^{3}=1}
c
9
e
4
m
e
3
=
{\displaystyle {\frac {c^{9}e^{4}}{m_{e}^{3}}}=}
0.170 514 342... 1092
u
n
i
t
s
=
m
9
C
4
s
9
k
g
3
==
(
u
−
13
)
9
(
u
−
27
)
4
(
u
−
30
)
9
(
u
15
)
3
=
1
{\displaystyle units={\frac {m^{9}C^{4}}{s^{9}kg^{3}}}=={\frac {(u^{-13})^{9}(u^{-27})^{4}}{(u^{-30})^{9}(u^{15})^{3}}}=1}
(
k
B
∗
)
(
e
∗
)
2
(
m
e
∗
)
(
c
∗
)
4
=
3
3
α
6
2
3
π
5
=
{\displaystyle {\frac {(k_{B}^{*})}{(e^{*})^{2}(m_{e}^{*})(c^{*})^{4}}}={\frac {3^{3}\alpha ^{6}}{2^{3}\pi ^{5}}}=}
73 035 235 897.
(
u
29
)
(
u
−
27
)
2
(
u
15
)
(
u
17
)
4
=
1
,
(
r
10
v
3
)
/
(
r
3
v
3
)
2
(
r
4
v
)
(
v
)
4
=
1
{\displaystyle {\frac {(u^{29})}{(u^{-27})^{2}(u^{15})(u^{17})^{4}}}=1,\;({\frac {r^{10}}{v^{3}}})/({\frac {r^{3}}{v^{3}}})^{2}({\frac {r^{4}}{v}})(v)^{4}=1}
k
B
e
2
m
e
c
4
=
{\displaystyle {\frac {k_{B}}{e^{2}m_{e}c^{4}}}=}
73 095 507 858.
u
n
i
t
s
=
s
2
m
2
K
C
2
==
(
u
−
30
)
2
(
u
−
13
)
2
(
u
20
)
(
u
−
27
)
2
=
1
{\displaystyle units={\frac {s^{2}}{m^{2}KC^{2}}}=={\frac {(u^{-30})^{2}}{(u^{-13})^{2}(u^{20})(u^{-27})^{2}}}=1}
These 3 constants, Planck mass, Planck time and the vacuum permittivity have no Omega term.
M
4
(
ϵ
0
∗
)
T
=
(
1
)
(
2
9
π
3
α
)
/
(
π
)
=
2
9
π
2
α
=
{\displaystyle {\frac {M^{4}(\epsilon _{0}^{*})}{T}}=(1)({\frac {2^{9}\pi ^{3}}{\alpha }})/(\pi )={\frac {2^{9}\pi ^{2}}{\alpha }}=}
36.875
(
u
15
)
4
(
u
−
90
)
(
u
−
30
)
=
1
,
(
r
4
v
)
4
(
1
r
7
v
2
)
/
(
r
9
v
6
)
=
1
{\displaystyle {\frac {(u^{15})^{4}(u^{-90})}{(u^{-30})}}=1,\;({\frac {r^{4}}{v}})^{4}({\frac {1}{r^{7}v^{2}}})/({\frac {r^{9}}{v^{6}}})=1}
m
p
4
(
ϵ
0
)
t
p
=
{\displaystyle {\frac {m_{p}^{4}(\epsilon _{0})}{t_{p}}}=}
36.850
u
n
i
t
s
=
k
g
4
s
s
4
A
2
m
3
k
g
=
k
g
3
A
2
s
3
m
3
==
(
u
15
)
3
(
u
3
)
2
(
u
−
30
)
3
(
u
−
13
)
3
=
1
{\displaystyle units={\frac {kg^{4}}{s}}{\frac {s^{4}A^{2}}{m^{3}kg}}={\frac {kg^{3}A^{2}s^{3}}{m^{3}}}=={\frac {(u^{15})^{3}(u^{3})^{2}(u^{-30})^{3}}{(u^{-13})^{3}}}=1}
(
h
∗
)
(
c
∗
)
2
(
e
∗
)
(
m
e
∗
)
(
G
∗
)
2
(
k
B
∗
)
=
(
m
e
∗
)
(
2
11
π
3
α
2
)
=
{\displaystyle {\frac {(h^{*})(c^{*})^{2}(e^{*})(m_{e}^{*})}{(G^{*})^{2}(k_{B}^{*})}}=(m_{e}^{*})({\frac {2^{11}\pi ^{3}}{\alpha ^{2}}})=}
0.1415... 10-21
(
u
19
)
(
u
17
)
2
(
u
−
27
)
(
u
15
)
(
u
6
)
2
(
u
29
)
=
1
,
(
r
13
v
5
)
v
2
(
r
3
v
3
)
(
r
4
v
1
)
/
(
r
5
v
2
)
2
(
r
10
v
3
)
=
1
{\displaystyle {\frac {(u^{19})(u^{17})^{2}(u^{-27})(u^{15})}{(u^{6})^{2}(u^{29})}}=1,\;({\frac {r^{13}}{v^{5}}})v^{2}({\frac {r^{3}}{v^{3}}})({\frac {r^{4}}{v^{1}}})/({\frac {r^{5}}{v^{2}}})^{2}({\frac {r^{10}}{v^{3}}})=1}
h
c
2
e
m
e
G
2
k
B
=
{\displaystyle {\frac {hc^{2}em_{e}}{G^{2}k_{B}}}=}
0.1413... 10-21
u
n
i
t
s
=
k
g
3
s
3
C
K
m
4
==
(
u
15
)
3
(
u
−
30
)
3
(
u
−
27
)
(
u
20
)
(
u
−
13
)
4
=
1
{\displaystyle units={\frac {kg^{3}s^{3}CK}{m^{4}}}=={\frac {(u^{15})^{3}(u^{-30})^{3}(u^{-27})(u^{20})}{(u^{-13})^{4}}}=1}
2
(
h
∗
)
(
μ
0
∗
)
(
e
∗
)
2
(
c
∗
)
=
2
(
2
3
π
4
Ω
4
)
/
(
α
2
11
π
5
Ω
4
)
(
2
7
π
4
Ω
3
α
)
2
(
2
π
Ω
2
)
=
α
,
u
19
u
56
(
u
−
27
)
2
u
17
=
1
,
(
r
13
v
5
)
(
1
r
7
)
(
v
6
r
6
)
(
1
v
)
=
1
{\displaystyle {\frac {2(h^{*})}{(\mu _{0}^{*})(e^{*})^{2}(c^{*})}}=2({2^{3}\pi ^{4}\Omega ^{4}})/({\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}})({\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }})^{2}(2\pi \Omega ^{2})=\color {blue}\alpha \color {black},\;{\frac {u^{19}}{u^{56}(u^{-27})^{2}u^{17}}}=1,\;({\frac {r^{13}}{v^{5}}})({\frac {1}{r^{7}}})({\frac {v^{6}}{r^{6}}})({\frac {1}{v}})=1}
scalars = 1 .
M
=
m
P
=
(
1
)
k
;
k
=
m
P
=
.217
672
817
58...
10
−
7
,
u
15
(
k
g
)
{\displaystyle M=m_{P}=(1)k;\;k=m_{P}=.217\;672\;817\;58...\;10^{-7},\;u^{15}\;(kg)}
T
=
t
p
=
π
t
;
t
=
t
p
π
=
.171
585
512
84...10
−
43
,
u
−
30
(
s
)
{\displaystyle T=t_{p}={\pi }t;\;t={\frac {t_{p}}{\pi }}=.171\;585\;512\;84...10^{-43},\;u^{-30}\;(s)}
L
=
l
p
=
2
π
2
Ω
2
l
;
l
=
l
p
2
π
2
Ω
2
=
.203
220
869
48...10
−
36
,
u
−
13
(
m
)
{\displaystyle L=l_{p}={2\pi ^{2}\Omega ^{2}}l;\;l={\frac {l_{p}}{2\pi ^{2}\Omega ^{2}}}=.203\;220\;869\;48...10^{-36},\;u^{-13}\;(m)}
V
=
c
=
2
π
Ω
2
v
;
v
=
c
2
π
Ω
2
=
11
843
707.905...
,
u
17
(
m
/
s
)
{\displaystyle V=c={2\pi \Omega ^{2}}v;\;v={\frac {c}{2\pi \Omega ^{2}}}=11\;843\;707.905...,\;u^{17}\;(m/s)}
A
=
e
/
t
p
=
(
2
7
π
3
Ω
3
α
)
a
=
.126
918
588
59...10
23
,
u
3
(
A
)
{\displaystyle A=e/t_{p}=({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }})a=.126\;918\;588\;59...10^{23},\;u^{3}\;(A)}
In this example LPVA are derived from MT. The formulas for MT;
M
=
(
1
)
k
,
u
n
i
t
=
u
15
{\displaystyle M=(1)k,\;unit=u^{15}}
T
=
(
π
)
t
,
u
n
i
t
=
u
−
30
{\displaystyle T=(\pi )t,\;unit=u^{-30}}
Replacing scalars pvla with kt
P
=
(
Ω
)
k
12
/
15
t
2
/
15
,
u
n
i
t
=
u
12
/
15
∗
15
−
2
/
15
∗
(
−
30
)
=
16
{\displaystyle P=(\Omega )\;{\frac {k^{12/15}}{t^{2/15}}},\;unit=u^{12/15*15-2/15*(-30)=16}}
V
=
2
π
P
2
M
=
(
2
π
Ω
2
)
k
9
/
15
t
4
/
15
,
u
n
i
t
=
u
9
/
15
∗
15
−
4
/
15
∗
(
−
30
)
=
17
{\displaystyle V={\frac {2\pi P^{2}}{M}}=(2\pi \Omega ^{2})\;{\frac {k^{9/15}}{t^{4/15}}},\;unit=u^{9/15*15-4/15*(-30)=17}}
L
=
T
V
=
(
2
π
2
Ω
2
)
k
9
/
15
t
11
/
15
,
u
n
i
t
=
u
9
/
15
∗
15
+
11
/
15
∗
(
−
30
)
=
−
13
{\displaystyle L=TV=(2\pi ^{2}\Omega ^{2})\;k^{9/15}t^{11/15},\;unit=u^{9/15*15+11/15*(-30)=-13}}
A
=
2
4
V
3
α
P
3
=
(
2
7
π
3
Ω
3
α
)
1
k
3
/
5
t
2
/
5
,
u
n
i
t
=
u
9
/
15
∗
(
−
15
)
+
6
/
15
∗
30
=
3
{\displaystyle A={\frac {2^{4}V^{3}}{\alpha P^{3}}}=\left({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }}\right)\;{\frac {1}{k^{3/5}t^{2/5}}},\;unit=u^{9/15*(-15)+6/15*30=3}}
In this example MLTA are derived from PV. The formulas for PV;
P
=
(
Ω
)
p
,
u
n
i
t
=
u
16
{\displaystyle P=(\Omega )p,\;unit=u^{16}}
V
=
(
2
π
Ω
2
)
v
,
u
n
i
t
=
u
17
{\displaystyle V=(2\pi \Omega ^{2})v,\;unit=u^{17}}
Replacing scalars klta with pv
M
=
2
π
P
2
V
=
(
1
)
p
2
v
,
u
n
i
t
=
u
16
∗
2
−
17
=
15
{\displaystyle M={\frac {2\pi P^{2}}{V}}=(1){\frac {p^{2}}{v}},\;unit=u^{16*2-17=15}}
T
=
(
π
)
p
9
/
2
v
6
,
u
n
i
t
=
u
16
∗
9
/
2
−
17
∗
6
=
−
30
{\displaystyle T=(\pi ){\frac {p^{9/2}}{v^{6}}},\;unit=u^{16*9/2-17*6=-30}}
L
=
T
V
=
(
2
π
2
Ω
2
)
p
9
/
2
v
5
,
u
n
i
t
=
u
16
∗
9
/
2
−
17
∗
5
=
−
13
{\displaystyle L=TV=(2\pi ^{2}\Omega ^{2}){\frac {p^{9/2}}{v^{5}}},\;unit=u^{16*9/2-17*5=-13}}
A
=
2
4
V
3
α
P
3
=
(
2
7
π
3
Ω
3
α
)
v
3
p
3
,
u
n
i
t
=
u
17
∗
3
−
16
∗
3
=
3
{\displaystyle A={\frac {2^{4}V^{3}}{\alpha P^{3}}}=({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }}){\frac {v^{3}}{p^{3}}},\;unit=u^{17*3-16*3=3}}
As geometrical objects, the physical constants (G, h, e, me , kB ) can also be defined using the geometrical formulas for (c* , μ0 * , R* ) and solved using the numerical (mean) values for (c, μ0 , R, α ). For example;
(
h
∗
)
3
=
(
2
3
π
4
Ω
4
r
13
u
19
v
5
)
3
=
3
19
π
12
Ω
12
r
39
u
57
v
15
,
θ
=
57
{\displaystyle {(h^{*})}^{3}=(2^{3}\pi ^{4}\Omega ^{4}{\frac {r^{13}u^{19}}{v^{5}}})^{3}={\frac {3^{19}\pi ^{12}\Omega ^{12}r^{39}u^{57}}{v^{15}}},\;\theta =57}
and ...
2
π
10
(
μ
0
∗
)
3
3
6
(
c
∗
)
5
α
13
(
R
∗
)
2
=
3
19
π
12
Ω
12
r
39
u
57
v
15
,
θ
=
57
{\displaystyle {\frac {2\pi ^{10}{(\mu _{0}^{*})}^{3}}{3^{6}{(c^{*})}^{5}\alpha ^{13}{(R^{*})}^{2}}}={\frac {3^{19}\pi ^{12}\Omega ^{12}r^{39}u^{57}}{v^{15}}},\;\theta =57}
Table 12. Calculated from (R, c, μ0 , α) columns 2, 3, 4 vs CODATA 2014 columns 5, 6
Constant
Formula
Units
Calculated from (R, c, μ0 , α)
CODATA 2014 [ 17]
Units
Planck constant
(
h
∗
)
3
=
2
π
10
μ
0
3
3
6
c
5
α
13
R
2
{\displaystyle {(h^{*})}^{3}={\frac {2\pi ^{10}{\mu _{0}}^{3}}{3^{6}{c}^{5}\alpha ^{13}{R}^{2}}}}
k
g
3
A
6
s
{\displaystyle {\frac {kg^{3}}{A^{6}s}}}
h* = 6.626 069 134 e-34, θ = 19
h = 6.626 070 040(81) e-34
k
g
m
2
s
{\displaystyle {\frac {kg\;m^{2}}{s}}}
Gravitational constant
(
G
∗
)
5
=
π
3
μ
0
2
20
3
6
α
11
R
2
{\displaystyle {(G^{*})}^{5}={\frac {\pi ^{3}{\mu _{0}}}{2^{20}3^{6}\alpha ^{11}{R}^{2}}}}
k
g
m
3
A
2
s
2
{\displaystyle {\frac {kg\;m^{3}}{A^{2}s^{2}}}}
G* = 6.672 497 192 29 e11, θ = 6
G = 6.674 08(31) e-11
m
3
k
g
s
2
{\displaystyle {\frac {m^{3}}{kg\;s^{2}}}}
Elementary charge
(
e
∗
)
3
=
4
π
5
3
3
c
4
α
8
R
{\displaystyle {(e^{*})}^{3}={\frac {4\pi ^{5}}{3^{3}{c}^{4}\alpha ^{8}{R}}}}
s
4
A
3
{\displaystyle {\frac {s^{4}}{A^{3}}}}
e* = 1.602 176 511 30 e-19, θ = -27
e = 1.602 176 620 8(98) e-19
A
s
{\displaystyle As}
Boltzmann constant
(
k
B
∗
)
3
=
π
5
μ
0
3
3
3
2
c
4
α
5
R
{\displaystyle {(k_{B}^{*})}^{3}={\frac {\pi ^{5}{\mu _{0}}^{3}}{3^{3}2{c}^{4}\alpha ^{5}{R}}}}
k
g
3
s
2
A
6
{\displaystyle {\frac {kg^{3}}{s^{2}A^{6}}}}
kB * = 1.379 510 147 52 e-23, θ = 29
kB = 1.380 648 52(79) e-23
k
g
m
2
s
2
K
{\displaystyle {\frac {kg\;m^{2}}{s^{2}\;K}}}
Electron mass
(
m
e
∗
)
3
=
16
π
10
R
μ
0
3
3
6
c
8
α
7
{\displaystyle {(m_{e}^{*})}^{3}={\frac {16\pi ^{10}{R}{\mu _{0}}^{3}}{3^{6}{c}^{8}\alpha ^{7}}}}
k
g
3
s
2
m
6
A
6
{\displaystyle {\frac {kg^{3}s^{2}}{m^{6}A^{6}}}}
me * = 9.109 382 312 56 e-31, θ = 15
me = 9.109 383 56(11) e-31
k
g
{\displaystyle kg}
Gyromagnetic ratio
(
(
γ
e
∗
)
/
2
π
)
3
=
g
e
3
3
3
c
4
2
8
π
8
α
μ
0
3
R
∞
2
{\displaystyle ({(\gamma _{e}^{*})/2\pi })^{3}={\frac {g_{e}^{3}3^{3}c^{4}}{2^{8}\pi ^{8}\alpha \mu _{0}^{3}R_{\infty }^{2}}}}
m
3
s
2
A
6
k
g
3
{\displaystyle {\frac {m^{3}s^{2}A^{6}}{kg^{3}}}}
(γe * /2π) = 28024.953 55, θ = -42
γe /2π = 28024.951 64(17)
A
s
k
g
{\displaystyle {\frac {A\;s}{kg}}}
Planck mass
(
m
P
∗
)
15
=
2
25
π
13
μ
0
6
3
6
c
5
α
16
R
2
{\displaystyle ({m_{P}^{*}})^{15}={\frac {2^{25}\pi ^{13}{\mu _{0}}^{6}}{3^{6}c^{5}\alpha ^{16}R^{2}}}}
k
g
6
m
3
s
7
A
12
{\displaystyle {\frac {kg^{6}m^{3}}{s^{7}A^{12}}}}
mP * = 0.217 672 817 580 e-7, θ = 15
mP = 0.217 647 0(51) e-7
k
g
{\displaystyle kg}
Planck length
(
l
p
∗
)
15
=
π
22
μ
0
9
2
35
3
24
α
49
c
35
R
8
{\displaystyle ({l_{p}^{*}})^{15}={\frac {\pi ^{22}{\mu _{0}}^{9}}{2^{35}3^{24}\alpha ^{49}c^{35}R^{8}}}}
k
g
9
s
17
m
18
A
18
{\displaystyle {\frac {kg^{9}s^{17}}{m^{18}A^{18}}}}
lp * = 0.161 603 660 096 e-34, θ = -13
lp = 0.161 622 9(38) e-34
m
{\displaystyle m}
↑ Macleod, M.J. "Programming Planck units from a mathematical electron; a Simulation Hypothesis". Eur. Phys. J. Plus 113 : 278. 22 March 2018. doi:10.1140/epjp/i2018-12094-x .
↑ Planck (1899), p. 479.
↑ *Tomilin, K. A., 1999, "Natural Systems of Units: To the Centenary Anniversary of the Planck System ", 287–296.
↑ A Planck scale mathematical universe model ↑ [1] | CODATA, The Committee on Data for Science and Technology | (2014)↑ [2] | CODATA, The Committee on Data for Science and Technology | (2014)↑ [3] | CODATA, The Committee on Data for Science and Technology | (2018)↑ Macleod, Malcolm J. "Physical constant anomalies suggest a mathematical relationship linking SI units". RG . doi:10.13140/RG.2.2.15874.15041/6 .
↑ https://codingthecosmos.com/ai-conversations.html conversations with AI)↑ https://chatgpt.com/share/67b8856d-e0c8-8012-97cc-fe9785b5685b ↑ https://chat.qwen.ai/s/d44fbcf6-0ba1-4217-8eae-9c36004e29f7 ↑ https://claude.ai/share/bc805379-0984-4a5d-baf0-f5920e57b35d ↑ https://www.getmerlin.in/ja/chat/share/d5e8d336-43d3-4728-a367-7249a914f6e0 ↑ https://www.getmerlin.in/chat/share/ff17f851-f200-42aa-bd89-1304c0113294 ↑ https://x.com/i/grok/share/Jb0APJzSMgyG09jt2bwcFNwh5 ↑ https://julius.ai/s/885f7603-d13c-486d-b281-ced366bc0caf ↑ [4] | CODATA, The Committee on Data for Science and Technology | (2014)
