Natural Planck units as geometrical objects
Planck unit theories use basic units for mass, length, time and charge (MLTA ), and operate at the Planck scale. In a geometrical Planck theory, these basic units are assigned geometrical objects rather than numerical values, the advantage being that the geometries themselves can encode the function of the unit, for example the object for length (L ) will encode the function of length (it is length), and so the geometrical L is 1 unit of (Planck) length, thus, unlike numerical models, a dimensioned descriptive (i.e.: kg, m, s, A , ... ) is not required.
The geometry of the objects are selected whereby they may interact with each other (the mass object for example is not independent of the length and the time objects). This permits a mathematical unit relationship linking the objects, and so a physical universe can be constructed Lego-style by combining the MLTA objects to form more complex objects such as electrons (i.e.: by embedding L and A into the geometry of the electron, the electron can have wavelength and charge) [1]
Contents
1 Geometrical objects
1.1 Objects 1.2 Scalars
1.2.1 Scalar relationships 1.2.2 Natural units MLTPA
1.2.2.1 mP , lp , tp 1.2.2.2 A, lp , tp 1.2.2.3 c, μ0 , R 1.2.2.4 c, e, kB , h 1.2.2.5 c, h, e 1.2.2.6 me , λe 1.2.2.7 c, e, me 1.2.2.8 kB , c, e, me 1.2.2.9 mP , tp , ε0 1.2.2.10 G, h, c, e, me , KB 1.2.2.11 α
1.2.3 SI Planck unit scalars
1.3 Physical constants (as geometrical formulas) 1.4 2019 SI unit revision 1.5 External links 1.6 References
Base units for mass
M
{\displaystyle M}
L
{\displaystyle L}
T
{\displaystyle T}
A
{\displaystyle A}
dimensionless physical constants , the (inverse) fine structure constant α = 137.035 999 139 and Omega Ω = 2.007 134 949 636 [2]
Being independent of any numerical system and of any system of units, these MLTA units would qualify as "natural units";
...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
[3] [4]
Each object is assigned a specific geometry, embedded in this geometry is the object function (attribute) and a numerical value (a unit number ) whereby relationships between the objects can be defined;
Geometrical units
Attribute
Geometrical 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}
Comparison table of dimensioned physical constants
CODATA 2014 [5]
SI unit
Natural constant*
Unit number
c = 299 792 458 (exact)
m
s
{\displaystyle {\frac {m}{s}}}
c* = V = 25.3123819329
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}
15+17-13 = 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}}}
34-13-15 = 6
e = 1.602 176 620 8(98) e-19
C
=
A
s
{\displaystyle C=As}
e* =
A
T
{\displaystyle AT}
3-30 = -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}}}
17+15-3 = 29
As alpha and Omega have numerical solutions, we can assign to MLTA numerical values, i.e.: V = 2πΩ2 = 25.3123819. 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
P
{\displaystyle m_{P}}
SI =
m
P
{\displaystyle m_{P}}
kSI = 0.2176728e-7kg (SI units) Using scalar vSI = 11843707.905m/s gives c = V*vSI = 299792458m/s (SI units) Using vimp = 7359.3232155miles/s gives c = V*vimp = 186282miles/s (Imperial units)
Geometrical units
Attribute
Geometrical object
Scalar
Unit uθ
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 )}
p
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 unit, whether kg , or m/s , or miles/s ... they follow the same unit number relationship. This means that we can find ratios where the scalars cancel. Here are examples (units = 1 ), as such only 2 numerical scalars are required , for example, if we know a and l then we know t (t = a3 l3 ), and from l and t we know 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}
This also means that by changing the 2 scalars accordingly, we can solve (G , h , c , e , m e , k B ) for any system of units, to quote Max Planck 'even extraterrestrial', as demonstrated by this physical constants calculator (inputs are the value for Planck mass and the speed of light) [6]
r (unit number = 8) and v were chosen as the principal scalars for the following as they can be derived directly from 2 precise constants c and μ0 .
Using α = 137.035 999 139 (CODATA 2014), Ω = 2.007 134 949 636...
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=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}}
i
=
Ω
15
{\displaystyle i=\Omega ^{15}}
L
15
M
9
T
11
{\displaystyle {\sqrt {\frac {L^{15}}{M^{9}T^{11}}}}}
x
=
Ω
v
r
2
{\displaystyle x=\Omega {\frac {v}{r^{2}}}}
L
M
T
{\displaystyle {\sqrt {\frac {L}{MT}}}}
1 = u (unit number = -13 -15 +30 = 2/2 = 1) ...
(
L
15
M
9
T
11
=
L
M
T
L
7
M
4
T
5
)
,
l
7
k
4
t
5
=
r
2
v
{\displaystyle ({\sqrt {\frac {L^{15}}{M^{9}T^{11}}}}={\sqrt {\frac {L}{MT}}}{\frac {L^{7}}{M^{4}T^{5}}}),\;{\frac {l^{7}}{k^{4}t^{5}}}={\frac {r^{2}}{v}}}
y
=
r
17
v
8
{\displaystyle y={\frac {r^{17}}{v^{8}}}}
M
2
T
{\displaystyle M^{2}T}
(
L
15
M
9
T
11
=
M
2
T
L
15
M
13
T
13
)
,
l
15
k
13
t
13
=
v
8
r
17
{\displaystyle ({\sqrt {\frac {L^{15}}{M^{9}T^{11}}}}=M^{2}T{\sqrt {\frac {L^{15}}{M^{13}T^{13}}}}),\;{\sqrt {\frac {l^{15}}{k^{13}t^{13}}}}={\frac {v^{8}}{r^{17}}}}
Note: The following suggests a limit (numerical boundary) to the values the SI constants can have.
j
=
r
17
v
8
=
k
2
t
=
k
17
/
4
v
15
/
4
=
.
.
.
{\displaystyle j={\frac {r^{17}}{v^{8}}}=k^{2}t={\frac {k^{17/4}}{v^{15/4}}}=...}
-59 to 0.123... x1060
a
1
/
3
=
v
r
2
=
1
t
2
/
15
k
1
/
5
=
v
k
{\displaystyle a^{1/3}={\frac {v}{r^{2}}}={\frac {1}{t^{2/15}k^{1/5}}}={\frac {\sqrt {v}}{\sqrt {k}}}}
Table 3. Geometrical units
Attribute
Geometrical object
Unit
Unit number θ
(Vacuum permittivity )
ϵ
0
=
(
2
9
π
3
α
)
x
θ
i
6
y
11
=
(
2
9
π
3
α
)
1
v
2
r
7
{\displaystyle \epsilon _{0}=({\frac {2^{9}\pi ^{3}}{\alpha }})\color {red}{\frac {x^{\theta }i^{6}}{y^{11}}}\color {black}=({\frac {2^{9}\pi ^{3}}{\alpha }}){\frac {1}{v^{2}r^{7}}}}
T
M
4
{\displaystyle {\frac {T}{M^{4}}}}
−
90
{\displaystyle -90}
time
T
=
(
π
)
x
θ
i
2
y
3
=
(
π
)
r
9
v
6
{\displaystyle T=(\pi )\color {red}{\frac {x^{\theta }i^{2}}{y^{3}}}\color {black}=(\pi ){\frac {r^{9}}{v^{6}}}}
T
{\displaystyle T}
−
30
{\displaystyle -30}
(Elementary charge )
e
=
(
2
7
π
4
α
)
x
θ
i
2
y
3
=
(
2
7
π
4
Ω
3
α
)
r
3
v
3
{\displaystyle e=({\frac {2^{7}\pi ^{4}}{\alpha }})\color {red}{\frac {x^{\theta }i^{2}}{y^{3}}}\color {black}=({\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }}){\frac {r^{3}}{v^{3}}}}
L
3
/
2
T
1
/
2
M
3
/
2
{\displaystyle {\frac {L^{3/2}}{T^{1/2}M^{3/2}}}}
−
27
{\displaystyle -27}
(Planck constant )
h
−
1
=
(
1
2
3
π
4
)
x
θ
i
y
3
=
(
1
2
3
π
4
Ω
4
)
v
5
r
13
{\displaystyle h^{-1}=({\frac {1}{2^{3}\pi ^{4}}})\color {red}{\frac {x^{\theta }i}{y^{3}}}\color {black}=({\frac {1}{2^{3}\pi ^{4}\Omega ^{4}}}){\frac {v^{5}}{r^{13}}}}
T
L
2
M
{\displaystyle {\frac {T}{L^{2}M}}}
−
19
{\displaystyle -19}
length
L
=
(
2
π
2
)
x
θ
i
y
=
(
2
π
2
Ω
2
)
r
9
v
5
{\displaystyle L=(2\pi ^{2})\color {red}{\frac {x^{\theta }i}{y}}\color {black}=(2\pi ^{2}\Omega ^{2}){\frac {r^{9}}{v^{5}}}}
L
{\displaystyle L}
−
13
{\displaystyle -13}
(Gravitational constant )
G
−
1
=
(
1
2
3
π
4
)
x
θ
y
=
(
1
2
3
π
4
Ω
6
)
v
2
r
5
{\displaystyle G^{-1}=({\frac {1}{2^{3}\pi ^{4}}})\color {red}{\frac {x^{\theta }}{y}}\color {black}=({\frac {1}{2^{3}\pi ^{4}\Omega ^{6}}}){\frac {v^{2}}{r^{5}}}}
M
T
2
L
3
{\displaystyle {\frac {MT^{2}}{L^{3}}}}
−
6
{\displaystyle -6}
ampere
A
=
(
2
7
π
3
α
)
x
θ
=
(
2
7
π
3
Ω
3
α
)
v
3
r
6
{\displaystyle A=({\frac {2^{7}\pi ^{3}}{\alpha }})\color {red}x^{\theta }\color {black}=({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }}){\frac {v^{3}}{r^{6}}}}
L
3
/
2
M
3
/
2
T
3
/
2
=
A
{\displaystyle {\frac {L^{3/2}}{M^{3/2}T^{3/2}}}=A}
3
{\displaystyle 3}
mass
M
=
(
1
)
x
θ
y
2
i
=
(
1
)
r
4
v
{\displaystyle M=(1)\color {red}{\frac {x^{\theta }y^{2}}{i}}\color {black}=(1){\frac {r^{4}}{v}}}
M
{\displaystyle M}
15
{\displaystyle 15}
sqrt(momentum)
P
=
x
θ
y
2
i
=
(
Ω
)
r
2
{\displaystyle P=\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}}}}
16
{\displaystyle 16}
velocity
V
=
(
2
π
)
x
θ
y
2
i
=
(
2
π
Ω
2
)
v
{\displaystyle V=(2\pi )\color {red}{\frac {x^{\theta }y^{2}}{i}}\color {black}=(2\pi \Omega ^{2})v}
L
T
{\displaystyle {\frac {L}{T}}}
17
{\displaystyle 17}
(Planck temperature )
T
p
=
(
2
7
π
3
α
)
x
θ
y
2
i
=
(
2
7
π
3
Ω
5
α
)
v
4
r
6
{\displaystyle T_{p}=({\frac {2^{7}\pi ^{3}}{\alpha }})\color {red}{\frac {x^{\theta }y^{2}}{i}}\color {black}=({\frac {2^{7}\pi ^{3}\Omega ^{5}}{\alpha }}){\frac {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}
20
{\displaystyle 20}
(Boltzmann constant )
k
B
=
(
α
2
5
π
)
x
θ
y
4
i
2
=
(
α
2
5
π
Ω
)
r
10
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 \Omega }}){\frac {r^{10}}{v^{3}}}}
M
5
/
2
T
1
/
2
L
1
/
2
{\displaystyle {\frac {M^{5/2}T^{1/2}}{L^{1/2}}}}
29
{\displaystyle 29}
Regardless of which system of units we use, alien or terrestrial, any combination of constants where scalars = 1 (i.e.: the scalars overlap and cancel) will give the same numerical result, they will default to the MLTPA objects. This implies that these objects are Planck's 'natural' units, i.e.: that all possible systems of units are based on these objects, and so, given that these are dimensionless geometrical objects, they can be construed as evidence of a mathematical universe. The following are examples of units = scalars = 1 ratios using SI units [7]
(
Ω
15
)
n
{\displaystyle \color {red}(\Omega ^{15})^{n}\color {black}}
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}
The 2019 SI unit revision assigned exact numerical values to 4 constants (c, e, kB , h).
From the table geometrical physical constants , we get geometrical formulas and scalars for;
h
∗
=
2
π
M
V
L
=
2
3
π
4
Ω
4
r
13
v
5
,
u
15
+
17
−
13
=
19
{\displaystyle h^{*}=2\pi MVL=2^{3}\pi ^{4}\Omega ^{4}{\frac {r^{13}}{v^{5}}},\;u^{15+17-13=19}}
e
∗
=
A
T
=
2
7
π
4
Ω
3
α
r
3
v
3
,
u
3
−
30
=
−
27
{\displaystyle e^{*}=AT={\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }}{\frac {r^{3}}{v^{3}}},\;u^{3-30=-27}}
k
B
∗
=
2
π
M
V
/
A
=
α
2
5
π
Ω
r
10
v
3
,
u
17
+
15
−
3
=
29
{\displaystyle k_{B}^{*}=2\pi MV/A={\frac {\alpha }{2^{5}\pi \Omega }}{\frac {r^{10}}{v^{3}}},\;u^{17+15-3=29}}
(
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}}
f
e
=
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 f_{e}={\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
f
e
=
9.109
382
3227
10
−
31
u
15
{\displaystyle (m_{e}^{*})={\frac {M}{f_{e}}}=\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
f
e
=
2.426
310
238
667
10
−
12
u
−
13
{\displaystyle (\lambda _{e}^{*})=2\pi Lf_{e}=\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
f
e
,
f
e
=
2
20
3
3
π
8
α
3
(
Ω
15
)
1
{\displaystyle (m_{e}^{*})={\frac {M}{f_{e}}},\;f_{e}=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}}
Physical constants (as geometrical formulas) [ edit | edit source ] note:
(
u
15
)
n
{\displaystyle \color {red}(u^{15})^{n}\color {black}}
Dimensioned constants; geometrical vs CODATA 2014
Constant
In Planck units
Geometrical object
SI calculated (r, v, Ω, α* )
SI CODATA 2014 [8]
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
17
∗
2
+
15
+
13
−
6
=
7
∗
8
=
56
{\displaystyle \mu _{0}^{*}={\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}}r^{7},\;u^{17*2+15+13-6=7*8=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
15
+
17
−
13
=
8
∗
13
−
17
∗
5
=
19
{\displaystyle h^{*}=2^{3}\pi ^{4}\Omega ^{4}{\frac {r^{13}}{v^{5}}},\;u^{15+17-13=8*13-17*5=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
34
−
13
−
15
=
8
∗
5
−
17
∗
2
=
6
{\displaystyle G^{*}=2^{3}\pi ^{4}\Omega ^{6}{\frac {r^{5}}{v^{2}}},\;u^{34-13-15=8*5-17*2=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
3
−
30
=
3
∗
8
−
17
∗
3
=
−
27
{\displaystyle e^{*}={\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }}{\frac {r^{3}}{v^{3}}},\;u^{3-30=3*8-17*3=-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
17
+
15
−
3
=
10
∗
8
−
17
∗
3
=
29
{\displaystyle k_{B}^{*}={\frac {\alpha }{2^{5}\pi \Omega }}{\frac {r^{10}}{v^{3}}},\;u^{17+15-3=10*8-17*3=29}}
kB * = 1.379 510 147 52 e-23, unit = u29
kB = 1.380 648 52(79) e-23
Electron mass
m
e
∗
=
M
f
e
,
u
15
{\displaystyle m_{e}^{*}={\frac {M}{f_{e}}},\;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
f
e
,
u
−
13
{\displaystyle \lambda _{e}^{*}=2\pi Lf_{e},\;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
3
+
17
=
17
∗
4
−
6
∗
8
=
20
{\displaystyle T_{p}^{*}={\frac {2^{7}\pi ^{3}\Omega ^{5}}{\alpha }}{\frac {v^{4}}{r^{6}}},\;u^{3+17=17*4-6*8=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}(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}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). If this mathematical relationship can therefore be identified within the SI units themselves, then we have an argument for a Planck scale mathematical universe.
Although the Planck units MLTA are embedded within the electron formula fe , this formula is both unit-less and non scalable (units = 1, scalars = 1). Furthermore it is the geometry of 2 dimensionless physical constants and so can also be defined as a dimensionless physical constant (if units = scalars = 1, then that constant will be independent of any numerical system and of any system of units, and so would qualify as a "natural unit").
f
e
=
4
π
2
(
2
6
3
π
2
α
Ω
5
)
3
=
.23895453
.
.
.
x
10
23
{\displaystyle f_{e}=4\pi ^{2}(2^{6}3\pi ^{2}\alpha \Omega ^{5})^{3}=.23895453...x10^{23}}
AL as an ampere-meter (ampere-length) are the units for a 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}}
f
e
=
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 f_{e}={\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}
fe is a mathematical function that dictates how these Planck objects are applied, it does not have dimension units of its own, consequently there is no physical electron.
electron mass
m
e
=
M
f
e
{\displaystyle m_{e}={\frac {M}{f_{e}}}}
Planck mass )
electron wavelength
λ
e
=
2
π
L
f
e
{\displaystyle \lambda _{e}=2\pi Lf_{e}}
Planck length )
elementary charge
e
=
A
.
T
{\displaystyle e=A.T}
The Sommerfeld fine structure constant alpha is a dimensionless physical constant, the CODATA 2018 inverse alpha = 137.035999084.
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}
The most precise of the experimentally measured constants is the Rydberg R = 10973731.568508(65) 1/m . Here c, μ0 , R are combined into a unit-less ratio;
(
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
=
1
{\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}}}=1}
We can now define 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}
There is a close natural number for Ω that is a square root implying that Ω can have a plus or a minus solution and this agrees with theory. This solution would however re-classify Omega as a mathematical constant (as being derivable from other mathematical constants).
Ω
=
(
π
e
e
(
e
−
1
)
)
=
2.007
134
9543...
{\displaystyle \Omega ={\sqrt {\left({\frac {\pi ^{e}}{e^{(e-1)}}}\right)}}=2.007\;134\;9543...}
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}
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 [9]
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}
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}}
↑ A Planck scale mathematical universe model ↑ 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.
↑ [1] | CODATA, The Committee on Data for Science and Technology | (2014)↑ *Macleod, M. J, 2018, "Physical constants calculator "
↑ Macleod, Malcolm J. "Do the physical constants embed evidence of a simulation universe at the Planck scale?". RG . doi:10.13140/RG.2.2.15874.15041/4 .
↑ [2] | CODATA, The Committee on Data for Science and Technology | (2014)↑ [3] | CODATA, The Committee on Data for Science and Technology | (2014)↑ [4] | CODATA, The Committee on Data for Science and Technology | (2018)
