# Planck units (geometrical)

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].

### Geometrical objects

Base units for mass ${\displaystyle M}$, length ${\displaystyle L}$, time ${\displaystyle T}$, and ampere ${\displaystyle A}$ can be constructed from the geometry of 2 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]

#### Objects

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 ${\displaystyle M=(1)}$ ${\displaystyle 15}$
time ${\displaystyle T=(\pi )}$ ${\displaystyle -30}$
sqrt(momentum) ${\displaystyle P=(\Omega )}$ ${\displaystyle 16}$
velocity ${\displaystyle V=(2\pi \Omega ^{2})}$ ${\displaystyle 17}$
length ${\displaystyle L=(2\pi ^{2}\Omega ^{2})}$ ${\displaystyle -13}$
ampere ${\displaystyle A={\frac {16V^{3}}{\alpha P^{3}}}=({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }})}$ ${\displaystyle 3}$

Comparison table of dimensioned physical constants
CODATA 2014 [5] SI unit Natural constant* Unit number
c = 299 792 458 (exact) ${\displaystyle {\frac {m}{s}}}$ c* = V = 25.3123819329 17
h = 6.626 070 040(81) e-34 ${\displaystyle {\frac {kg\;m^{2}}{s}}}$ h* = ${\displaystyle 2\pi MVL}$ = 12647.240312 15+17-13 = 19
G = 6.674 08(31) e-11 ${\displaystyle {\frac {m^{3}}{kg\;s^{2}}}}$ G* = ${\displaystyle {\frac {V^{2}L}{M}}}$ = 50950.554778 34-13-15 = 6
e = 1.602 176 620 8(98) e-19 ${\displaystyle C=As}$ e* = ${\displaystyle AT}$ = 735.706358485 3-30 = -27
kB = 1.380 648 52(79) e-23 ${\displaystyle {\frac {kg\;m^{2}}{s^{2}\;K}}}$ kB* = ${\displaystyle {\frac {2\pi VM}{A}}}$ = 0.679138336 17+15-3 = 29

#### Scalars

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 ${\displaystyle m_{P}}$ (M = 1kSI = ${\displaystyle m_{P}}$), then 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 ${\displaystyle M=(1)}$ k ${\displaystyle u^{15}}$
time ${\displaystyle T=(\pi )}$ t ${\displaystyle u^{-30}}$
sqrt(momentum) ${\displaystyle P=(\Omega )}$ p ${\displaystyle u^{16}}$
velocity ${\displaystyle V=(2\pi \Omega ^{2})}$ v ${\displaystyle u^{17}}$
length ${\displaystyle L=(2\pi ^{2}\Omega ^{2})}$ l ${\displaystyle u^{-13}}$
ampere ${\displaystyle A=({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }})}$ a ${\displaystyle u^{3}}$

##### Scalar relationships

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 = a3l3), and from l and t we know k.

${\displaystyle {\frac {u^{3*3}u^{-13*3}}{u^{-30}}}\;({\frac {a^{3}l^{3}}{t}})={\frac {u^{-13*15}}{u^{15*9}u^{-30*11}}}\;({\frac {l^{15}}{k^{9}t^{11}}})=\;...\;=1}$

This also means that by changing the 2 scalars accordingly, we can solve (G, h, c, e, me, kB) 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].

Scalars 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...

${\displaystyle v={\frac {c}{2\pi \Omega ^{2}}}=11843707.905...,\;units=m/s}$
${\displaystyle r^{7}={\frac {2^{11}\pi ^{5}\Omega ^{4}\mu _{0}}{\alpha }};\;r=0.712562514304...,\;units=({\frac {kg.m}{s}})^{1/4}}$

We can now define the relation between the constants. Setting

${\displaystyle i=\Omega ^{15}}$, units = ${\displaystyle {\sqrt {\frac {L^{15}}{M^{9}T^{11}}}}}$ = 1 (unit number = -13*15 -15*9 +11*30 = 0, no scalars)
${\displaystyle x=\Omega {\frac {v}{r^{2}}}}$ , units = ${\displaystyle {\sqrt {\frac {L}{MT}}}}$ = u1 = u (unit number = -13 -15 +30 = 2/2 = 1) ... ${\displaystyle ({\sqrt {\frac {L^{15}}{M^{9}T^{11}}}}={\sqrt {\frac {L}{MT}}}{\frac {L^{7}}{M^{4}T^{5}}}),\;{\frac {l^{7}}{k^{4}t^{5}}}={\frac {r^{2}}{v}}}$
${\displaystyle y={\frac {r^{17}}{v^{8}}}}$ , units = ${\displaystyle M^{2}T}$ = 1, (unit number = 15*2 -30 = 0) ... ${\displaystyle ({\sqrt {\frac {L^{15}}{M^{9}T^{11}}}}=M^{2}T{\sqrt {\frac {L^{15}}{M^{13}T^{13}}}}),\;{\sqrt {\frac {l^{15}}{k^{13}t^{13}}}}={\frac {v^{8}}{r^{17}}}}$

Note: The following suggests a limit (numerical boundary) to the values the SI constants can have.

${\displaystyle j={\frac {r^{17}}{v^{8}}}=k^{2}t={\frac {k^{17/4}}{v^{15/4}}}=...}$ gives a range from 0.812997... x10-59 to 0.123... x1060
${\displaystyle a^{1/3}={\frac {v}{r^{2}}}={\frac {1}{t^{2/15}k^{1/5}}}={\frac {\sqrt {v}}{\sqrt {k}}}}$ ... = 23326079.1...; unit = u

Table 3. Geometrical units
Attribute Geometrical object Unit Unit number θ
(Vacuum permittivity) ${\displaystyle \epsilon _{0}=({\frac {2^{9}\pi ^{3}}{\alpha }})\color {red}{\frac {x^{\theta }i^{6}}{y^{11}}}\color {black}=({\frac {2^{9}\pi ^{3}}{\alpha }}){\frac {1}{v^{2}r^{7}}}}$ ${\displaystyle {\frac {T}{M^{4}}}}$ ${\displaystyle -90}$
time ${\displaystyle T=(\pi )\color {red}{\frac {x^{\theta }i^{2}}{y^{3}}}\color {black}=(\pi ){\frac {r^{9}}{v^{6}}}}$ ${\displaystyle T}$ ${\displaystyle -30}$
(Elementary charge) ${\displaystyle e=({\frac {2^{7}\pi ^{4}}{\alpha }})\color {red}{\frac {x^{\theta }i^{2}}{y^{3}}}\color {black}=({\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }}){\frac {r^{3}}{v^{3}}}}$ ${\displaystyle {\frac {L^{3/2}}{T^{1/2}M^{3/2}}}}$ ${\displaystyle -27}$
(Planck constant) ${\displaystyle h^{-1}=({\frac {1}{2^{3}\pi ^{4}}})\color {red}{\frac {x^{\theta }i}{y^{3}}}\color {black}=({\frac {1}{2^{3}\pi ^{4}\Omega ^{4}}}){\frac {v^{5}}{r^{13}}}}$ ${\displaystyle {\frac {T}{L^{2}M}}}$ ${\displaystyle -19}$
length ${\displaystyle L=(2\pi ^{2})\color {red}{\frac {x^{\theta }i}{y}}\color {black}=(2\pi ^{2}\Omega ^{2}){\frac {r^{9}}{v^{5}}}}$ ${\displaystyle L}$ ${\displaystyle -13}$
(Gravitational constant) ${\displaystyle G^{-1}=({\frac {1}{2^{3}\pi ^{4}}})\color {red}{\frac {x^{\theta }}{y}}\color {black}=({\frac {1}{2^{3}\pi ^{4}\Omega ^{6}}}){\frac {v^{2}}{r^{5}}}}$ ${\displaystyle {\frac {MT^{2}}{L^{3}}}}$ ${\displaystyle -6}$
ampere ${\displaystyle A=({\frac {2^{7}\pi ^{3}}{\alpha }})\color {red}x^{\theta }\color {black}=({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }}){\frac {v^{3}}{r^{6}}}}$ ${\displaystyle {\frac {L^{3/2}}{M^{3/2}T^{3/2}}}=A}$ ${\displaystyle 3}$
mass ${\displaystyle M=(1)\color {red}{\frac {x^{\theta }y^{2}}{i}}\color {black}=(1){\frac {r^{4}}{v}}}$ ${\displaystyle M}$ ${\displaystyle 15}$
sqrt(momentum) ${\displaystyle P=\color {red}{\frac {x^{\theta }y^{2}}{i}}\color {black}=(\Omega )r^{2}}$ ${\displaystyle {\frac {M^{1/2}L^{1/2}}{T^{1/2}}}}$ ${\displaystyle 16}$
velocity ${\displaystyle V=(2\pi )\color {red}{\frac {x^{\theta }y^{2}}{i}}\color {black}=(2\pi \Omega ^{2})v}$ ${\displaystyle {\frac {L}{T}}}$ ${\displaystyle 17}$
(Planck temperature) ${\displaystyle T_{p}=({\frac {2^{7}\pi ^{3}}{\alpha }})\color {red}{\frac {x^{\theta }y^{2}}{i}}\color {black}=({\frac {2^{7}\pi ^{3}\Omega ^{5}}{\alpha }}){\frac {v^{4}}{r^{6}}}}$ ${\displaystyle {\frac {L^{5/2}}{M^{3/2}T^{5/2}}}=AV}$ ${\displaystyle 20}$
(Boltzmann constant) ${\displaystyle k_{B}=({\frac {\alpha }{2^{5}\pi }})\color {red}{\frac {x^{\theta }y^{4}}{i^{2}}}\color {black}=({\frac {\alpha }{2^{5}\pi \Omega }}){\frac {r^{10}}{v^{3}}}}$ ${\displaystyle {\frac {M^{5/2}T^{1/2}}{L^{1/2}}}}$ ${\displaystyle 29}$

##### Natural units MLTPA

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]. Note: the geometry ${\displaystyle \color {red}(\Omega ^{15})^{n}\color {black}}$ (integer n ≥ 0) is common to all (units = 1) ratios that include an Omega term.

###### mP, lp, tp

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 10-7kg
l = 0.203 220 869 487... x 10-36m
t = 0.171 585 512 841... x 10-43s
${\displaystyle {\frac {L^{15}}{M^{9}T^{11}}}={\frac {(2\pi ^{2}\Omega ^{2})^{15}}{(1)^{9}(\pi )^{11}}}({\frac {l^{15}}{k^{9}t^{11}}})={\frac {l_{p}^{15}}{m_{P}^{9}t_{p}^{11}}}}$ (CODATA 2018 mean)

The klt scalars cancel, leaving;

${\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 , ${\displaystyle ({\frac {l^{15}}{k^{9}t^{11}}})=1,\;{\frac {u^{-13*15}}{u^{15*9}u^{-30*11}}}=1}$

Solving for the SI units;

${\displaystyle {\frac {l_{p}^{15}}{m_{P}^{9}t_{p}^{11}}}={\frac {(1.616255e-35)^{15}}{(2.176434e-8)^{9}(5.391247e-44)^{11}}}=}$ 0.109 485... 1024

###### A, lp, tp
a = 0.126 918 588 592... x 1023A
${\displaystyle {\frac {A^{3}L^{3}}{T}}=({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }})^{3}{\frac {(2\pi ^{2}\Omega ^{2})^{3}}{(\pi )}}({\frac {a^{3}l^{3}}{t}})={\frac {2^{24}\pi ^{14}\color {red}(\Omega ^{15})^{1}\color {black}}{\alpha ^{3}}}=}$ 0.205 571... 1013, ${\displaystyle ({\frac {a^{3}l^{3}}{t}})=1,\;{\frac {u^{3*3}u^{-13*3}}{u^{-30}}}=1}$
${\displaystyle {\frac {(e/t_{p})^{3}l_{p}^{3}}{t_{p}}}={\frac {(1.602176634e-19/5.391247e-44)^{3}(1.616255e-35)^{3}}{(5.391247e-44)}}=}$ 0.205 543... 1013, ${\displaystyle units={\frac {(C/s)^{3}m^{3}}{s}}}$

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, ${\displaystyle \mu _{0}}$; CODATA 2014 mean values), we can test to greater precision;

###### c, μ0, R
${\displaystyle {\frac {(c^{*})^{35}}{(\mu _{0}^{*})^{9}(R^{*})^{7}}}=(2\pi \Omega ^{2}v)^{35}/({\frac {\alpha r^{7}}{2^{11}\pi ^{5}\Omega ^{4}}})^{9}.({\frac {v^{5}}{2^{23}3^{3}\pi ^{11}\alpha ^{5}\Omega ^{17}r^{9}}})^{7}=2^{295}\pi ^{157}3^{21}\alpha ^{26}\color {red}(\Omega ^{15})^{15}\color {black}=}$ 0.326 103 528 6170... 10301, ${\displaystyle {\frac {(u^{17})^{35}}{(u^{56})^{9}(u^{13})^{7}}}=1,\;(v^{35})/(r^{7})^{9}({\frac {v^{5}}{r^{9}}})^{7}=1}$
${\displaystyle {\frac {c^{35}}{\mu _{0}^{9}R^{7}}}={\frac {(299792458)^{35}}{(4\pi /10^{7})^{9}(10973731.568160)^{7}}}=}$ 0.326 103 528 6170... 10301, ${\displaystyle units={\frac {m^{33}A^{18}}{s^{17}kg^{9}}}=={\frac {(u^{-13})^{33}(u^{3})^{18}}{(u^{-30})^{17}(u^{15})^{9}}}=1}$

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;

${\displaystyle h^{*}=2\pi MVL=2^{3}\pi ^{4}\Omega ^{4}{\frac {r^{13}}{v^{5}}},\;u^{15+17-13=19}}$
${\displaystyle e^{*}=AT={\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }}{\frac {r^{3}}{v^{3}}},\;u^{3-30=-27}}$
${\displaystyle k_{B}^{*}=2\pi MV/A={\frac {\alpha }{2^{5}\pi \Omega }}{\frac {r^{10}}{v^{3}}},\;u^{17+15-3=29}}$

###### c, e, kB, h
${\displaystyle {\frac {(k_{B}^{*})(e^{*})(c^{*})}{(h^{*})}}=({\frac {\alpha }{2^{5}\pi \Omega }}{\frac {r^{10}}{v^{3}}})({\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }}{\frac {r^{3}}{v^{3}}})(2\pi \Omega ^{2}v)/(2^{3}\pi ^{4}\Omega ^{4}{\frac {r^{13}}{v^{5}}})}$ = 1.0, ${\displaystyle {\frac {(u^{29})(u^{-27})(u^{17})}{(u^{19})}}=1,\;({\frac {r^{10}}{v^{3}}})({\frac {r^{3}}{v^{3}}})(v)/({\frac {r^{13}}{v^{5}}})=1}$
${\displaystyle {\frac {k_{B}ec}{h}}=}$ 1.000 8254, ${\displaystyle units={\frac {mC}{s^{2}K}}=={\frac {(u^{-13})(u^{-27})}{(u^{-30})^{2}(u^{20})}}=1}$

###### c, h, e
${\displaystyle {\frac {(h^{*})^{3}}{(e^{*})^{13}(c^{*})^{24}}}=(2^{3}\pi ^{4}\Omega ^{4}{\frac {r^{13}}{v^{5}}})^{3}/({\frac {2^{7}\pi ^{4}\Omega ^{3}r^{3}}{\alpha v^{3}}})^{7}.(2\pi \Omega ^{2}v)^{24}={\frac {\alpha ^{13}}{2^{106}\pi ^{64}(\color {red}\Omega ^{15})^{5}\color {black}}}=}$ 0.228 473 759... 10-58, ${\displaystyle {\frac {(u^{19})^{3}}{(u^{-27})^{13}(u^{17})^{24}}}=1,\;({\frac {r^{13}}{v^{5}}})^{3}/({\frac {r^{3}}{v^{3}}})^{13}(v^{24})=1}$
${\displaystyle {\frac {h^{3}}{e^{13}c^{24}}}=}$ 0.228 473 639... 10-58, ${\displaystyle units={\frac {kg^{3}s^{21}}{m^{18}C^{13}}}=={\frac {(u^{15})^{3}(u^{-30})^{21}}{(u^{-13})^{18}(u^{-27})^{13}}}=1}$

###### me, λe
${\displaystyle \sigma _{e}={\frac {3\alpha ^{2}AL}{2\pi ^{2}}}={2^{7}3\pi ^{3}\alpha \Omega ^{5}}{\frac {r^{3}}{v^{2}}},\;u^{-10}}$
${\displaystyle f_{e}={\frac {\sigma _{e}^{3}}{2T}}=2^{20}3^{3}\pi ^{8}\alpha ^{3}(\color {red}\Omega ^{15})\color {black},\;{\frac {(u^{-10})^{3}}{u^{-30}}}=1,\;({\frac {r^{3}}{v^{2}}})^{3}{\frac {v^{6}}{r^{9}}}=1}$
${\displaystyle (m_{e}^{*})={\frac {M}{f_{e}}}=\color {blue}9.109\;382\;3227\;10^{-31}\color {black}\;u^{15}}$
${\displaystyle (m_{e}^{*})={\frac {2^{3}\pi ^{5}(h^{*})}{3^{3}\alpha ^{6}(e^{*})^{3}(c^{*})^{5}}}={\frac {1}{2^{20}\pi ^{8}3^{3}\alpha ^{3}(\color {red}\Omega ^{15})\color {black}}}{\frac {r^{4}u^{15}}{v}}=\color {blue}9.109\;382\;3227\;10^{-31}\color {black}\;u^{15}}$
${\displaystyle m_{e}=\color {blue}9.109\;383\;7015...\;10^{-31}\color {black}\;kg}$
${\displaystyle (\lambda _{e}^{*})=2\pi Lf_{e}=\color {purple}2.426\;310\;238\;667\;10^{-12}\color {black}\;u^{-13}}$
${\displaystyle \lambda _{e}={\frac {h}{m_{e}c}}=\color {purple}2.426\;310\;238\;67\;10^{-12}\color {black}\;m}$

###### c, e, me
${\displaystyle (m_{e}^{*})={\frac {M}{f_{e}}},\;f_{e}=2^{20}3^{3}\pi ^{8}\alpha ^{3}(\color {red}\Omega ^{15})^{1}\color {black}}$, units = scalars = 1 (me formula)
${\displaystyle {\frac {(c^{*})^{9}(e^{*})^{4}}{(m_{e}^{*})^{3}}}=2^{97}\pi ^{49}3^{9}\alpha ^{5}(\color {red}\Omega ^{15})^{5}\color {black}=}$ 0.170 514 368... 1092, ${\displaystyle {\frac {(u^{17})^{9}(u^{-27})^{4}}{(u^{15})^{3}}}=1,\;(v^{9})({\frac {r^{3}}{v^{3}}})^{4}/({\frac {r^{4}}{v}})^{3}=1}$
${\displaystyle {\frac {c^{9}e^{4}}{m_{e}^{3}}}=}$ 0.170 514 342... 1092, ${\displaystyle units={\frac {m^{9}C^{4}}{s^{9}kg^{3}}}=={\frac {(u^{-13})^{9}(u^{-27})^{4}}{(u^{-30})^{9}(u^{15})^{3}}}=1}$

###### kB, c, e, me
${\displaystyle {\frac {(k_{B}^{*})}{(e^{*})^{2}(m_{e}^{*})(c^{*})^{4}}}={\frac {3^{3}\alpha ^{6}}{2^{3}\pi ^{5}}}=}$ 73 035 235 897., ${\displaystyle {\frac {(u^{29})}{(u^{-27})^{2}(u^{15})(u^{17})^{4}}}=1,\;({\frac {r^{10}}{v^{3}}})/({\frac {r^{3}}{v^{3}}})^{2}({\frac {r^{4}}{v}})(v)^{4}=1}$
${\displaystyle {\frac {k_{B}}{e^{2}m_{e}c^{4}}}=}$ 73 095 507 858., ${\displaystyle units={\frac {s^{2}}{m^{2}KC^{2}}}=={\frac {(u^{-30})^{2}}{(u^{-13})^{2}(u^{20})(u^{-27})^{2}}}=1}$

###### mP, tp, ε0

These 3 constants, Planck mass, Planck time and the vacuum permittivity have no Omega term.

${\displaystyle {\frac {M^{4}(\epsilon _{0}^{*})}{T}}=(1)({\frac {2^{9}\pi ^{3}}{\alpha }})/(\pi )={\frac {2^{9}\pi ^{2}}{\alpha }}=}$ 36.875, ${\displaystyle {\frac {(u^{15})^{4}(u^{-90})}{(u^{-30})}}=1,\;({\frac {r^{4}}{v}})^{4}({\frac {1}{r^{7}v^{2}}})/({\frac {r^{9}}{v^{6}}})=1}$
${\displaystyle {\frac {m_{p}^{4}(\epsilon _{0})}{t_{p}}}=}$ 36.850, ${\displaystyle units={\frac {kg^{4}}{s}}{\frac {s^{4}A^{2}}{m^{3}kg}}={\frac {kg^{3}A^{2}s^{3}}{m^{3}}}=={\frac {(u^{15})^{3}(u^{3})^{2}(u^{-30})^{3}}{(u^{-13})^{3}}}=1}$

###### G, h, c, e, me, KB
${\displaystyle {\frac {(h^{*})(c^{*})^{2}(e^{*})(m_{e}^{*})}{(G^{*})^{2}(k_{B}^{*})}}=(m_{e}^{*})({\frac {2^{11}\pi ^{3}}{\alpha ^{2}}})=}$ 0.1415... 10-21, ${\displaystyle {\frac {(u^{19})(u^{17})^{2}(u^{-27})(u^{15})}{(u^{6})^{2}(u^{29})}}=1,\;({\frac {r^{13}}{v^{5}}})v^{2}({\frac {r^{3}}{v^{3}}})({\frac {r^{4}}{v^{1}}})/({\frac {r^{5}}{v^{2}}})^{2}({\frac {r^{10}}{v^{3}}})=1}$
${\displaystyle {\frac {hc^{2}em_{e}}{G^{2}k_{B}}}=}$ 0.1413... 10-21, ${\displaystyle units={\frac {kg^{3}s^{3}CK}{m^{4}}}=={\frac {(u^{15})^{3}(u^{-30})^{3}(u^{-27})(u^{20})}{(u^{-13})^{4}}}=1}$

###### α
${\displaystyle {\frac {2(h^{*})}{(\mu _{0}^{*})(e^{*})^{2}(c^{*})}}=2({2^{3}\pi ^{4}\Omega ^{4}})/({\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}})({\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }})^{2}(2\pi \Omega ^{2})=\color {blue}\alpha \color {black},\;{\frac {u^{19}}{u^{56}(u^{-27})^{2}u^{17}}}=1,\;({\frac {r^{13}}{v^{5}}})({\frac {1}{r^{7}}})({\frac {v^{6}}{r^{6}}})({\frac {1}{v}})=1}$

Note: The above will apply to any combinations of constants (alien or terrestrial) where scalars = 1.

##### SI Planck unit scalars
${\displaystyle M=m_{P}=(1)k;\;k=m_{P}=.217\;672\;817\;58...\;10^{-7},\;u^{15}\;(kg)}$
${\displaystyle T=t_{p}={\pi }t;\;t={\frac {t_{p}}{\pi }}=.171\;585\;512\;84...10^{-43},\;u^{-30}\;(s)}$
${\displaystyle L=l_{p}={2\pi ^{2}\Omega ^{2}}l;\;l={\frac {l_{p}}{2\pi ^{2}\Omega ^{2}}}=.203\;220\;869\;48...10^{-36},\;u^{-13}\;(m)}$
${\displaystyle V=c={2\pi \Omega ^{2}}v;\;v={\frac {c}{2\pi \Omega ^{2}}}=11\;843\;707.905...,\;u^{17}\;(m/s)}$
${\displaystyle A=e/t_{p}=({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }})a=.126\;918\;588\;59...10^{23},\;u^{3}\;(A)}$

###### MT to LPVA

In this example LPVA are derived from MT. The formulas for MT;

${\displaystyle M=(1)k,\;unit=u^{15}}$
${\displaystyle T=(\pi )t,\;unit=u^{-30}}$

Replacing scalars pvla with kt

${\displaystyle P=(\Omega )\;{\frac {k^{12/15}}{t^{2/15}}},\;unit=u^{12/15*15-2/15*(-30)=16}}$
${\displaystyle V={\frac {2\pi P^{2}}{M}}=(2\pi \Omega ^{2})\;{\frac {k^{9/15}}{t^{4/15}}},\;unit=u^{9/15*15-4/15*(-30)=17}}$
${\displaystyle L=TV=(2\pi ^{2}\Omega ^{2})\;k^{9/15}t^{11/15},\;unit=u^{9/15*15+11/15*(-30)=-13}}$
${\displaystyle A={\frac {2^{4}V^{3}}{\alpha P^{3}}}=\left({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }}\right)\;{\frac {1}{k^{3/5}t^{2/5}}},\;unit=u^{9/15*(-15)+6/15*30=3}}$

###### PV to MTLA

In this example MLTA are derived from PV. The formulas for PV;

${\displaystyle P=(\Omega )p,\;unit=u^{16}}$
${\displaystyle V=(2\pi \Omega ^{2})v,\;unit=u^{17}}$

Replacing scalars klta with pv

${\displaystyle M={\frac {2\pi P^{2}}{V}}=(1){\frac {p^{2}}{v}},\;unit=u^{16*2-17=15}}$
${\displaystyle T=(\pi ){\frac {p^{9/2}}{v^{6}}},\;unit=u^{16*9/2-17*6=-30}}$
${\displaystyle L=TV=(2\pi ^{2}\Omega ^{2}){\frac {p^{9/2}}{v^{5}}},\;unit=u^{16*9/2-17*5=-13}}$
${\displaystyle A={\frac {2^{4}V^{3}}{\alpha P^{3}}}=({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }}){\frac {v^{3}}{p^{3}}},\;unit=u^{17*3-16*3=3}}$

#### Physical constants (as geometrical formulas)

note: ${\displaystyle \color {red}(u^{15})^{n}\color {black}}$ constants have no Omega term.

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 ${\displaystyle c^{*}=(2\pi \Omega ^{2})v,\;u^{17}}$ 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 ${\displaystyle R^{*}=({\frac {m_{e}}{4\pi L\alpha ^{2}M}})}$ ${\displaystyle R^{*}={\frac {1}{2^{23}3^{3}\pi ^{11}\alpha ^{5}\Omega ^{17}}}{\frac {v^{5}}{r^{9}}},\;u^{13}}$ R* = 10 973 731.568 508, unit = u13 R = 10 973 731.568 508(65)
Vacuum permeability ${\displaystyle \mu _{0}^{*}={\frac {4\pi V^{2}M}{\alpha LA^{2}}}}$ ${\displaystyle \mu _{0}^{*}={\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}}r^{7},\;u^{17*2+15+13-6=7*8=56}}$ μ0* = 4π/10^7, unit = u56 μ0 = 4π/10^7 (exact)
Vacuum permittivity ${\displaystyle \epsilon _{0}^{*}={\frac {1}{\mu _{0}^{*}(c^{*})^{2}}}}$ ${\displaystyle \epsilon _{0}^{*}={\frac {2^{9}\pi ^{3}}{\alpha }}{\frac {1}{r^{7}v^{2}}},\;\color {red}1/(u^{15})^{6}\color {black}=u^{-90}}$
Planck constant ${\displaystyle h^{*}=2\pi MVL}$ ${\displaystyle h^{*}=2^{3}\pi ^{4}\Omega ^{4}{\frac {r^{13}}{v^{5}}},\;u^{15+17-13=8*13-17*5=19}}$ h* = 6.626 069 134 e-34, unit = u19 h = 6.626 070 040(81) e-34
Gravitational constant ${\displaystyle G^{*}={\frac {V^{2}L}{M}}}$ ${\displaystyle G^{*}=2^{3}\pi ^{4}\Omega ^{6}{\frac {r^{5}}{v^{2}}},\;u^{34-13-15=8*5-17*2=6}}$ G* = 6.672 497 192 29 e11, unit = u6 G = 6.674 08(31) e-11
Elementary charge ${\displaystyle e^{*}=AT}$ ${\displaystyle e^{*}={\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }}{\frac {r^{3}}{v^{3}}},\;u^{3-30=3*8-17*3=-27}}$ e* = 1.602 176 511 30 e-19, unit = u-27 e = 1.602 176 620 8(98) e-19
Boltzmann constant ${\displaystyle k_{B}^{*}={\frac {2\pi VM}{A}}}$ ${\displaystyle k_{B}^{*}={\frac {\alpha }{2^{5}\pi \Omega }}{\frac {r^{10}}{v^{3}}},\;u^{17+15-3=10*8-17*3=29}}$ kB* = 1.379 510 147 52 e-23, unit = u29 kB = 1.380 648 52(79) e-23
Electron mass ${\displaystyle m_{e}^{*}={\frac {M}{f_{e}}},\;u^{15}}$ me* = 9.109 382 312 56 e-31, unit = u15 me = 9.109 383 56(11) e-31
Classical electron radius ${\displaystyle \lambda _{e}^{*}=2\pi Lf_{e},\;u^{-13}}$ λe* = 2.426 310 2366 e-12, unit = u-13 λe = 2.426 310 236 7(11) e-12
Planck temperature ${\displaystyle T_{p}^{*}={\frac {AV}{\pi }}}$ ${\displaystyle T_{p}^{*}={\frac {2^{7}\pi ^{3}\Omega ^{5}}{\alpha }}{\frac {v^{4}}{r^{6}}},\;u^{3+17=17*4-6*8=20}}$ Tp* = 1.418 145 219 e32, unit = u20 Tp = 1.416 784(16) e32
Planck mass M ${\displaystyle m_{P}^{*}=(1){\frac {r^{4}}{v}},\;\color {red}(u^{15})^{1}\color {black}}$ mP* = .217 672 817 580 e-7, unit = u15 mP = .217 647 0(51) e-7
Planck length L ${\displaystyle l_{p}^{*}=(2\pi ^{2}\Omega ^{2}){\frac {r^{9}}{v^{5}}},\;u^{-13}}$ lp* = .161 603 660 096 e-34, unit = u-13 lp = .161 622 9(38) e-34
Planck time T ${\displaystyle t_{p}^{*}=(\pi ){\frac {r^{9}}{v^{6}}},\;\color {red}1/(u^{15})^{2}\color {black}}$ tp* = 5.390 517 866 e-44, unit = u-30 tp = 5.391 247(60) e-44
Ampere ${\displaystyle A={\frac {16V^{3}}{\alpha P^{3}}}}$ ${\displaystyle A^{*}={\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }}{\frac {v^{3}}{r^{6}}},\;u^{3}}$ A* = 0.297 221 e25, unit = u3 e/tp = 0.297 181 e25
Von Klitzing constant ${\displaystyle R_{K}^{*}=({\frac {h}{e^{2}}})^{*}}$ ${\displaystyle R_{K}^{*}={\frac {\alpha ^{2}}{2^{11}\pi ^{4}\Omega ^{2}}}r^{7}v,\;u^{73}}$ RK* = 25812.807 455 59, unit = u73 RK = 25812.807 455 5(59)
Gyromagnetic ratio ${\displaystyle \gamma _{e}/2\pi ={\frac {gl_{p}^{*}m_{P}^{*}}{2k_{B}^{*}m_{e}^{*}}},\;unit=u^{-42}}$ 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 uθ with the SI unit equivalents (u15 → kg, u-13 → m, u-30 → s, ...), the geometrical objects (i.e.: c* = 2πΩ2v = 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.

##### Electron formula

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").

${\displaystyle f_{e}=4\pi ^{2}(2^{6}3\pi ^{2}\alpha \Omega ^{5})^{3}=.23895453...x10^{23}}$

AL as an ampere-meter (ampere-length) are the units for a magnetic monopole.

${\displaystyle T=\pi {\frac {r^{9}}{v^{6}}},\;u^{-30}}$
${\displaystyle \sigma _{e}={\frac {3\alpha ^{2}AL}{2\pi ^{2}}}={2^{7}3\pi ^{3}\alpha \Omega ^{5}}{\frac {r^{3}}{v^{2}}},\;u^{-10}}$
${\displaystyle f_{e}={\frac {\sigma _{e}^{3}}{2T}}={\frac {(2^{7}3\pi ^{3}\alpha \Omega ^{5})^{3}}{2\pi }},\;units={\frac {(u^{-10})^{3}}{u^{-30}}}=1,scalars=({\frac {r^{3}}{v^{2}}})^{3}{\frac {v^{6}}{r^{9}}}=1}$

The electron has dimensioned parameters, however the dimensions derive from the Planck units, 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 ${\displaystyle m_{e}={\frac {M}{f_{e}}}}$ (M = Planck mass)

electron wavelength ${\displaystyle \lambda _{e}=2\pi Lf_{e}}$ (L = Planck length)

elementary charge ${\displaystyle e=A.T}$

##### Fine structure constant

The Sommerfeld fine structure constant alpha is a dimensionless physical constant, the CODATA 2018 inverse alpha = 137.035999084.

${\displaystyle {\frac {2(h^{*})}{(\mu _{0}^{*})(e^{*})^{2}(c^{*})}}=2({2^{3}\pi ^{4}\Omega ^{4}})/({\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}})({\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }})^{2}(2\pi \Omega ^{2})=\color {blue}\alpha \color {black},\;{\frac {u^{19}}{u^{56}(u^{-27})^{2}u^{17}}}=1,\;({\frac {r^{13}}{v^{5}}})({\frac {1}{r^{7}}})({\frac {v^{6}}{r^{6}}})({\frac {1}{v}})=1}$

##### Omega

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;

${\displaystyle {\frac {(c^{*})^{35}}{(\mu _{0}^{*})^{9}(R^{*})^{7}}}=(2\pi \Omega ^{2})^{35}/({\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}})^{9}.({\frac {1}{2^{23}3^{3}\pi ^{11}\alpha ^{5}\Omega ^{17}}})^{7},\;units={\frac {(u^{17})^{35}}{(u^{56})^{9}(u^{13})^{7}}}=1}$

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).

${\displaystyle \Omega ^{225}={\frac {(c^{*})^{35}}{2^{295}3^{21}\pi ^{157}(\mu _{0}^{*})^{9}(R^{*})^{7}\alpha ^{26}}},\;units=1}$
${\displaystyle \Omega =2.007\;134\;949\;636...,\;units=1}$ (CODATA 2014 mean values)
${\displaystyle \Omega =2.007\;134\;949\;687...,\;units=1}$ (CODATA 2018 mean values)

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).

${\displaystyle \Omega ={\sqrt {\left({\frac {\pi ^{e}}{e^{(e-1)}}}\right)}}=2.007\;134\;9543...}$

##### G, h, e, me, kB

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;

${\displaystyle {(h^{*})}^{3}=(2^{3}\pi ^{4}\Omega ^{4}{\frac {r^{13}u^{19}}{v^{5}}})^{3}={\frac {3^{19}\pi ^{12}\Omega ^{12}r^{39}u^{57}}{v^{15}}},\;\theta =57}$ ... and ... ${\displaystyle {\frac {2\pi ^{10}{(\mu _{0}^{*})}^{3}}{3^{6}{(c^{*})}^{5}\alpha ^{13}{(R^{*})}^{2}}}={\frac {3^{19}\pi ^{12}\Omega ^{12}r^{39}u^{57}}{v^{15}}},\;\theta =57}$

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 ${\displaystyle {(h^{*})}^{3}={\frac {2\pi ^{10}{\mu _{0}}^{3}}{3^{6}{c}^{5}\alpha ^{13}{R}^{2}}}}$ ${\displaystyle {\frac {kg^{3}}{A^{6}s}}}$, θ = 57 h* = 6.626 069 134 e-34, θ = 19 h = 6.626 070 040(81) e-34 ${\displaystyle {\frac {kg\;m^{2}}{s}}}$, θ = 19
Gravitational constant ${\displaystyle {(G^{*})}^{5}={\frac {\pi ^{3}{\mu _{0}}}{2^{20}3^{6}\alpha ^{11}{R}^{2}}}}$ ${\displaystyle {\frac {kg\;m^{3}}{A^{2}s^{2}}}}$, θ = 30 G* = 6.672 497 192 29 e11, θ = 6 G = 6.674 08(31) e-11 ${\displaystyle {\frac {m^{3}}{kg\;s^{2}}}}$, θ = 6
Elementary charge ${\displaystyle {(e^{*})}^{3}={\frac {4\pi ^{5}}{3^{3}{c}^{4}\alpha ^{8}{R}}}}$ ${\displaystyle {\frac {s^{4}}{A^{3}}}}$, θ = -81 e* = 1.602 176 511 30 e-19, θ = -27 e = 1.602 176 620 8(98) e-19 ${\displaystyle As}$, θ = -27
Boltzmann constant ${\displaystyle {(k_{B}^{*})}^{3}={\frac {\pi ^{5}{\mu _{0}}^{3}}{3^{3}2{c}^{4}\alpha ^{5}{R}}}}$ ${\displaystyle {\frac {kg^{3}}{s^{2}A^{6}}}}$, θ = 87 kB* = 1.379 510 147 52 e-23, θ = 29 kB = 1.380 648 52(79) e-23 ${\displaystyle {\frac {kg\;m^{2}}{s^{2}\;K}}}$, θ = 29
Electron mass ${\displaystyle {(m_{e}^{*})}^{3}={\frac {16\pi ^{10}{R}{\mu _{0}}^{3}}{3^{6}{c}^{8}\alpha ^{7}}}}$ ${\displaystyle {\frac {kg^{3}s^{2}}{m^{6}A^{6}}}}$, θ = 45 me* = 9.109 382 312 56 e-31, θ = 15 me = 9.109 383 56(11) e-31 ${\displaystyle kg}$, θ = 15
Gyromagnetic ratio ${\displaystyle ({(\gamma _{e}^{*})/2\pi })^{3}={\frac {g_{e}^{3}3^{3}c^{4}}{2^{8}\pi ^{8}\alpha \mu _{0}^{3}R_{\infty }^{2}}}}$ ${\displaystyle {\frac {m^{3}s^{2}A^{6}}{kg^{3}}}}$, θ = -126 e*/2π) = 28024.953 55, θ = -42 γe/2π = 28024.951 64(17) ${\displaystyle {\frac {A\;s}{kg}}}$, θ = -42
Planck mass ${\displaystyle ({m_{P}^{*}})^{15}={\frac {2^{25}\pi ^{13}{\mu _{0}}^{6}}{3^{6}c^{5}\alpha ^{16}R^{2}}}}$ ${\displaystyle {\frac {kg^{6}m^{3}}{s^{7}A^{12}}}}$, θ = 225 mP* = 0.217 672 817 580 e-7, θ = 15 mP = 0.217 647 0(51) e-7 ${\displaystyle kg}$, θ = 15
Planck length ${\displaystyle ({l_{p}^{*}})^{15}={\frac {\pi ^{22}{\mu _{0}}^{9}}{2^{35}3^{24}\alpha ^{49}c^{35}R^{8}}}}$ ${\displaystyle {\frac {kg^{9}s^{17}}{m^{18}A^{18}}}}$, θ = -195 lp* = 0.161 603 660 096 e-34, θ = -13 lp = 0.161 622 9(38) e-34 ${\displaystyle m}$, θ = -13

#### 2019 SI unit revision

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.

Physical constants
Constant CODATA 2018 [10]
Speed of light c = 299 792 458 (exact)
Planck constant h = 6.626 070 15 e-34 (exact)
Elementary charge e = 1.602 176 634 e-19 (exact)
Boltzmann constant kB = 1.380 649 e-23 (exact)
Fine structure constant α = 137.035 999 084(21)
Rydberg constant R = 10973 731.568 160(21)
Electron mass me = 9.109 383 7015(28) e-31
Vacuum permeability μ0 = 1.256 637 062 12(19) e-6
Von Klitzing constant RK = 25812.807 45 (exact)

For example, if we solve using the above formulas;

${\displaystyle R^{*}={\frac {4\pi ^{5}}{3^{3}c^{4}\alpha ^{8}e^{3}}}=10973\;729.082\;465}$

${\displaystyle {(m_{e}^{*})}^{3}={\frac {2^{4}\pi ^{10}R\mu _{0}^{3}}{3^{6}c^{8}\alpha ^{7}}},\;m_{e}^{*}=9.109\;382\;3259\;10^{-31}}$

${\displaystyle {(\mu _{0}^{*})}^{3}={\frac {3^{6}h^{3}c^{5}\alpha ^{13}R^{2}}{2\pi ^{10}}},\;\mu _{0}^{*}=1.256\;637\;251\;88\;10^{-6}}$

${\displaystyle {(h^{*})}^{3}={\frac {2\pi ^{10}\mu _{0}^{3}}{3^{6}c^{5}\alpha ^{13}R^{2}}},\;h^{*}=6.626\;069\;149\;10^{-34}}$

${\displaystyle {(e^{*})}^{3}={\frac {4\pi ^{5}}{3^{3}c^{4}\alpha ^{8}R}},\;e^{*}=1.602\;176\;513\;10^{-19}}$