Planck units (geometrical)

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.

Geometrical objects
The mathematical electron is a Planck unit model where mass $$M$$, length $$L$$, time $$T$$, and ampere $$A$$ are each assigned discrete geometrical objects from the geometry of 2 dimensionless physical constants, the (inverse)  fine structure constant α and  Omega Ω. Embedded into each object is the object function (attribute).

As the geometries of dimensionless constants, these objects are also dimensionless and so are independent of any system of units, and of any numerical system, and so could qualify as "natural units" (naturally occuring units);

As geometrical objects, they may combine 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). This requires a mathematical (unit number) relationship that defines how the objects interact with each other.

As alpha (α = 137.035 999 084) and Omega (Ω = 2.007 134 949 636) both have numerical solutions, we can assign to MLTA numerical values, i.e.: V = 2πΩ2 = 25.3123819 and use to solve geometrical physical constant equivalents.

We then find that where the unit numbers cancel, the numerical solutions agree (see Table 8).

Scalars
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*kSI = $$m_P$$), then kSI = 0.2176728e-7kg (SI units)


 * Using 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)

Scalar relationships
Because the scalars also include the SI unit, v = 11843707.905m/s ... they follow the unit number relationship uθ. 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 = a3l3), and from l and t we know the value for k.


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

This means that once any 2 scalars have been assigned values, the other scalars are then defined by default, consequently the CODATA 2014 values are used here as only 2 constants (c, 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.


 * $$v = \frac{c}{2 \pi \Omega^2}= 11 843 707.905 ...,\; units = \frac{m}{s}$$


 * $$r^7 = \frac{2^{11} \pi^5 \Omega^4 \mu_0}{\alpha};\; r = 0.712 562 514 304 ...,\; units = (\frac{kg.m}{s})^{1/4}$$

Fine structure constant
The fine structure constant can be derived from this formula (units and scalars cancel).


 * $$\frac{2 (h^*)}{(\mu_0^*) (e^*)^2 (c^*)} = 2({2^3 \pi^4 \Omega^4})/(\frac{\alpha}{2^{11} \pi^5 \Omega^4})(\frac{2^{7} \pi^4 \Omega^3}{\alpha})^2(2 \pi \Omega^2) = \color{red}\alpha \color{black}$$


 * $$units \;\frac{u^{19}}{u^{56} (u^{-27})^2 u^{17}} = 1$$


 * $$scalars \;(\frac{r^{13}}{v^5})(\frac{1}{r^7})(\frac{v^6}{r^6})(\frac{1}{v}) = 1$$

Electron formula
The electron object (formula fe) is a mathematical particle (units and scalars cancel).


 * $$f_e = 4\pi^2(2^6 3 \pi^2 \alpha \Omega^5)^3 = .23895453...x10^{23}$$ units = 1

In this example, embedded within the electron are the objects for charge, length and time ALT. AL as an ampere-meter (ampere-length) are the units for a magnetic monopole.


 * $$T = \pi \frac{r^9}{v^6},\; u^{-30}$$


 * $$\sigma_{e} = \frac{3 \alpha^2 A L}{2\pi^2} = {2^7 3 \pi^3 \alpha \Omega^5}\frac{r^3}{v^2},\; u^{-10}$$


 * $$f_e = \frac{\sigma_{e}^3}{2 T} = \frac{(2^7 3 \pi^3 \alpha \Omega^5)^3}{2\pi},\; units = \frac{(u^{-10})^3}{u^{-30}} = 1, scalars = (\frac{r^3}{v^2})^3 \frac{v^6}{r^9} = 1$$

Associated with the electron are dimensioned parameters, these parameters however are a function of the MLTA units, the formula fe dictating the frequency of these units. By setting MLTA to their SI Planck unit equivalents (Table 6.);

electron mass    $$m_e^* = \frac{M}{f_e}$$ (M =  Planck mass = $$\frac{r^4}{v})$$ = 0.910 938 232 11 e-30

electron wavelength $$\lambda_e^* = 2\pi L f_e$$ (L = Planck length = $$2\pi\Omega^2\frac{r^9}{v^5})$$ = 0.242 631 023 86 e-11

elementary charge   $$e^* = A\;T$$ (T =  Planck time) = $$\frac{2^7 \pi^4 \Omega^3}{\alpha}\frac{r^3}{v^3}$$ = 0.160 217 651 30 e-18

Rydberg constant $$R^* = (\frac{m_e}{4 \pi L \alpha^2 M}) = \frac{1}{2^{23} 3^3 \pi^{11} \alpha^5 \Omega^{17}}\frac{v^5}{r^9}\;u^{13}$$ = 10 973 731.568 508

Omega
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;


 * $$\mu_0^* = \frac{4 \pi V^2 M}{\alpha L A^2} = \frac{\alpha}{2^{11} \pi^5 \Omega^4} r^7,\; u^{56}$$


 * $$R^* = (\frac{m_e}{4 \pi L \alpha^2 M}) = \frac{1}{2^{23} 3^3 \pi^{11} \alpha^5 \Omega^{17}} \frac{v^5}{r^9},\;u^{13}$$


 * $$\frac{(c^*)^{35}}{(\mu_0^*)^9 (R^*)^7} = (2 \pi \Omega^2)^{35}/(\frac{\alpha}{2^{11} \pi^5 \Omega^4})^9 .(\frac{1}{2^{23} 3^3 \pi^{11} \alpha^5 \Omega^{17}})^7,\;units = \frac{(u^{17})^{35}}{(u^{56})^9 (u^{13})^7}$$


 * $$\frac{(c^*)^{35}}{(\mu_0^*)^9 (R^*)^7} = 2^{295} \pi^{157} 3^{21} \alpha^{26} (\Omega^{15})^{15}$$, units = 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).


 * $$\Omega^{225}=\frac{(c^*)^{35}}{2^{295} 3^{21} \pi^{157} (\mu_0^*)^9 (R^*)^7 \alpha^{26}}, \;units = 1$$


 * $$\Omega = 2.007\;134\;949\;636...,\; units = 1$$ (CODATA 2014 mean values)


 * $$\Omega = 2.007\;134\;949\;687...,\; units = 1$$ (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 (in the mass domain Ω occurs as Ω2 = 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).


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

Dimensionless combinations
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 kB 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).

Example


 * $$\frac{(h^*)^3}{(e^*)^{13} (c^*)^{24}} = (2^3 \pi^4 \Omega^4 \frac{r^{13}}{v^5})^3/(\frac{2^7 \pi^4 \Omega^3 r^3}{\alpha v^3})^7.(2\pi\Omega^2 v)^{24} = \frac{\alpha^{13}}{2^{106} \pi^{64} (\color{red}\Omega^{15})^5\color{black}} = $$ 0.228 473 759... 10-58

Note: the geometry $$\color{red}(\Omega^{15})^n\color{black}$$ (integer n ≥ 0) is common to all ratios where units and scalars cancel, suggesting a geometrical base-15.

Table of Constants
We can construct a table of constants using these 3 geometries. Setting


 * $$f(x)\;units = (\frac{L^{15}}{M^9 T^{11}})^n = 1$$

i.e.: unit number θ = (-13*15) - (15*9) - (-30*11) = 0


 * $$\color{red}i\color{black} = \pi^2 \Omega^{15}$$, units = $$\sqrt{f(x)}$$ = 1 (unit number = 0, no scalars)


 * $$\color{red}x\color{black} = \Omega \frac{v}{r^2}$$, units = $$\sqrt{\frac{L}{M T}}$$ = u1 = u (unit number = -13 -15 +30 = 2/2 = 1, with scalars v, r)


 * $$\color{red}y\color{black} = \pi \frac{r^{17}}{v^8}$$, units = $$M^2 T$$ = 1, (unit number = 15*2 -30 = 0, with scalars v, r)

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


 * $$\frac{v}{r^2} = a^{1/3} = \frac{1}{t^{2/15}k^{1/5}} = \frac{\sqrt{v}}{\sqrt{k}}$$ ... = 23326079.1...; unit = u


 * $$\frac{r^{17}}{v^8} = k^2 t = \frac{k^{17/4}}{v^{15/4}} = ... $$ gives a range from 0.812997... x10-59 to 0.123... x1060

Note: Influence of $$f(x)$$, units = 1


 * $$\frac{r^{17}}{v^8} \;\;units \;(\frac{M^2 L^8}{T^7}) (\frac{T}{L})^8 = M^2 T$$


 * $$r^{17} \;\;units \;(\frac{M\;L}{T})^{17/4} fx^{1/4} = \frac{M^2\;L^8}{T^7}$$


 * $$r \;\;units \;(\frac{M\;L}{T})^{1/4} fx^{1/4} = \frac{L^4}{M^2 T^3}$$

From the perspective of geometries

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

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).

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.

For example, if we solve using the above formulas;

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

$${(m_e^*)}^3 = \frac{2^4 \pi^{10} R \mu_0^3}{3^6 c^8 \alpha^7},\;m_e^* = 9.109\;382\;3259 \;10^{-31}$$

$${(\mu_0^*)}^3 = \frac{3^6 h^3 c^5 \alpha^{13} R^2}{2 \pi^{10}},\;\mu_0^* = 1.256\;637\;251\;88\;10^{-6}$$

$${(h^*)}^3 = \frac{2 \pi^{10} \mu_0^3}{3^6 c^5 \alpha^{13} R^2},\;h^* = 6.626\;069\;149\;10^{-34}$$

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

Anomalies
The following are notes on the anomalies as evidence of a simulation universe source code.

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


 * $$\frac{L^{15}}{M^{9} T^{11}} = \frac{(2\pi^2\Omega^2)^{15}}{(1)^{9} (\pi)^{11}} (\frac{l^{15}}{k^9 t^{11}}) = \frac{l_p^{15}}{m_P^{9} t_p^{11}} $$ (CODATA 2018 mean)

The klt scalars cancel, leaving;


 * $$\frac{L^{15}}{M^{9} T^{11}} = \frac{(2\pi^2\Omega^2)^{15}}{(1)^{9} (\pi)^{11}} (\frac{l^{15}}{k^9 t^{11}}) = 2^{15} \pi^{19} \color{red}(\Omega^{15})^2\color{black} = $$0.109 293... 1024, $$(\frac{l^{15}}{k^9 t^{11}}) = 1, \;\frac{u^{-13*15}}{u^{15*9} u^{-30*11}} = 1$$

Solving for the SI units;


 * $$\frac{l_p^{15}}{m_P^{9} t_p^{11}} = \frac{(1.616255e-35)^{15}}{(2.176434e-8)^{9} (5.391247e-44)^{11}} = $$ 0.109 485... 1024

A, lp, tp

 * a = 0.126 918 588 592... x 1023A


 * $$\frac{A^3 L^3}{T} = (\frac{2^7 \pi^3 \Omega^3}{\alpha})^3 \frac{(2\pi^2\Omega^2)^3}{(\pi)} (\frac{a^3 l^3}{t}) = \frac{2^{24} \pi^{14} \color{red}(\Omega^{15})^1\color{black}}{\alpha^3} = $$ 0.205 571... 1013, $$(\frac{a^3 l^3}{t}) = 1,\; \frac{u^{3*3} u^{-13*3}}{u^{-30}} = 1$$


 * $$\frac{(e / t_p)^3 l_p^3}{t_p} = \frac{(1.602176634e-19/5.391247e-44)^3 (1.616255e-35)^3}{(5.391247e-44)} = $$ 0.205 543... 1013, $$units = \frac{(C/s)^3 m^3}{s} $$

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, $$\mu_0$$; CODATA 2014 mean values), we can test to greater precision;

c, μ0, R

 * $$\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} = $$ 0.326 103 528 6170... 10301, $$\frac{(u^{17})^{35}}{(u^{56})^9 (u^{13})^7} = 1, \;(v^{35})/(r^7)^9 (\frac{v^5}{r^9})^7 = 1$$


 * $$\frac{c^{35}}{\mu_0^9 R^7} = \frac{(299792458)^{35}}{(4 \pi/10^7)^9 (10973731.568160)^7} = $$ 0.326 103 528 6170... 10301, $$units = \frac{m^{33}A^{18}}{s^{17}kg^9} == \frac{(u^{-13})^{33} (u^{3})^{18}}{(u^{-30})^{17} (u^{15})^9} = 1$$

c, e, kB, h

 * $$\frac{(k_B^*) (e^*) (c^*)}{(h^*)} = (\frac{\alpha}{2^5 \pi \Omega} \frac{r^{10}}{v^3}) (\frac{2^7 \pi^4 \Omega^3}{\alpha} \frac{r^3}{v^3}) (2 \pi \Omega^2 v) / (2^3 \pi^4 \Omega^4 \frac{r^{13}}{v^5}) $$ = 1.0, $$\frac{ (u^{29}) (u^{-27}) (u^{17}) }{ (u^{19}) } = 1,\; (\frac{r^{10}}{v^3}) (\frac{r^3}{v^3}) (v) / (\frac{r^{13}}{v^5}) = 1$$


 * $$\frac{k_B e c}{h} = $$ 1.000 8254, $$units = \frac{m C}{s^2 K} == \frac{(u^{-13}) (u^{-27})}{(u^{-30})^2 (u^{20})} = 1$$

c, h, e

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


 * $$\frac{h^3}{e^{13} c^{24}} = $$ 0.228 473 639... 10-58, $$units = \frac{kg^3 s^{21}}{m^{18} C^{13}} == \frac{(u^{15})^3 (u^{-30})^{21}}{(u^{-13})^{18} (u^{-27})^{13}} = 1$$

me, λe

 * $$\sigma_{e} = \frac{3 \alpha^2 A L}{2\pi^2} = {2^7 3 \pi^3 \alpha \Omega^5}\frac{r^3}{v^2},\; u^{-10}$$


 * $$f_e = \frac{\sigma_{e}^3}{2 T} = 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$$


 * $$(m_e^*) = \frac{M}{f_e} = \color{blue}9.109\;382\;3227 \;10^{-31}\color{black}\;u^{15}$$


 * $$(m_e^*) = \frac{2^3 \pi^5 (h^*)}{3^3 \alpha^6 (e^*)^3 (c^*)^5} = \frac{1}{2^{20} \pi^8 3^3 \alpha^3 (\color{red}\Omega^{15})\color{black}} \frac{r^4 u^{15}}{v} = \color{blue}9.109\;382\;3227 \;10^{-31}\color{black}\;u^{15}$$


 * $$m_e = \color{blue}9.109\;383\;7015... \;10^{-31}\color{black}\;kg$$


 * $$(\lambda_e^*) = 2 \pi L f_e = \color{purple}2.426\;310\;238\;667 \;10^{-12}\color{black}\;u^{-13}$$


 * $$\lambda_e = \frac{h}{m_e c} = \color{purple}2.426 \;310 \;238 \;67 \;10^{-12}\color{black}\;m$$

c, e, me

 * $$(m_e^*)= \frac{M}{f_e}, \;f_e = 2^{20} 3^3 \pi^8 \alpha^3 (\color{red}\Omega^{15})^1\color{black} $$, units = scalars = 1 (me formula)


 * $$\frac{(c^*)^9 (e^*)^4}{(m_e^*)^3} = 2^{97} \pi^{49} 3^9 \alpha^5 (\color{red}\Omega^{15})^5\color{black} = $$ 0.170 514 368... 1092, $$\frac{(u^{17})^9 (u^{-27})^4}{(u^{15})^3} = 1,\; (v^9) (\frac{r^3}{v^3})^4 / (\frac{r^4}{v})^3 = 1$$


 * $$\frac{c^9 e^4}{m_e^3} = $$ 0.170 514 342... 1092, $$units = \frac{m^9 C^4}{s^9 kg^3} == \frac{(u^{-13})^9 (u^{-27})^4}{(u^{-30})^9 (u^{15})^3} = 1$$

kB, c, e, me

 * $$\frac{(k_B^*)}{(e^*)^2 (m_e^*) (c^*)^4} = \frac{3^3 \alpha^6}{2^3 \pi^5} = $$ 73 035 235 897., $$\frac{(u^{29})}{(u^{-27})^2 (u^{15}) (u^{17})^4} = 1,\; (\frac{r^{10}}{v^3}) / (\frac{r^3}{v^3})^2 (\frac{r^4}{v}) (v)^4 = 1$$


 * $$\frac{k_B}{e^2 m_e c^4} = $$ 73 095 507 858., $$units = \frac{s^2}{m^2 K C^2} == \frac{(u^{-30})^2}{(u^{-13})^2 (u^{20}) (u^{-27})^2} = 1$$

mP, tp, ε0
These 3 constants, Planck mass, Planck time and the vacuum permittivity have no Omega term.
 * $$\frac{M^4 (\epsilon_0^*)}{T} = (1) (\frac{2^9 \pi^3}{\alpha}) / (\pi) = \frac{2^9 \pi^2}{\alpha} = $$ 36.875, $$\frac{(u^{15})^4 (u^{-90})}{(u^{-30})} = 1,\; (\frac{r^4}{v})^4 (\frac{1}{r^7 v^2}) / (\frac{r^9}{v^6}) = 1$$


 * $$\frac{m_p^4 (\epsilon_0)}{t_p} = $$ 36.850, $$units = \frac{kg^4}{s} \frac{s^4 A^2}{m^3 kg} = \frac{kg^3 A^2 s^3}{m^3} == \frac{(u^{15})^3 (u^{3})^2 (u^{-30})^3}{(u^{-13})^3}  = 1$$

G, h, c, e, me, KB

 * $$\frac{(h^*) (c^*)^2 (e^*) (m_e^*)}{(G^*)^2 (k_B^*)} = (m_e^*) (\frac{2^{11} \pi^3}{\alpha^2}) = $$ 0.1415... 10-21, $$\frac{ (u^{19}) (u^{17})^2 (u^{-27}) (u^{15}) }{ (u^{6})^2 (u^{29}) } = 1,\; (\frac{r^{13}}{v^5}) v^2 (\frac{r^{3}}{v^3})(\frac{r^{4}}{v^1}) / (\frac{r^5}{v^2})^2 (\frac{r^{10}}{v^3}) = 1$$


 * $$\frac{h c^2 e m_e}{G^2 k_B} = $$ 0.1413... 10-21, $$units = \frac{kg^3 s^3 C K}{m^4} == \frac{(u^{15})^3 (u^{-30})^3 (u^{-27}) (u^{20}) }{(u^{-13})^4}  = 1$$

α

 * $$\frac{2 (h^*)}{(\mu_0^*) (e^*)^2 (c^*)} = 2({2^3 \pi^4 \Omega^4})/(\frac{\alpha}{2^{11} \pi^5 \Omega^4})(\frac{2^{7} \pi^4 \Omega^3}{\alpha})^2(2 \pi \Omega^2) = \color{blue}\alpha \color{black},\; \frac{u^{19}}{u^{56} (u^{-27})^2 u^{17}} = 1,\; (\frac{r^{13}}{v^5})(\frac{1}{r^7})(\frac{v^6}{r^6})(\frac{1}{v}) = 1$$

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

SI Planck unit scalars

 * $$M = m_P = (1)k;\; k = m_P = .217\;672\;817\;58... \;10^{-7},\; u^{15}\; (kg)$$


 * $$T = t_p = {\pi}t;\; t = \frac{t_p}{\pi} = .171\;585\;512\;84... 10^{-43},\; u^{-30}\; (s)$$


 * $$L = l_p = {2\pi^2\Omega^2}l;\; l = \frac{l_p}{2\pi^2\Omega^2} = .203\;220\;869\;48... 10^{-36},\; u^{-13}\; (m)$$


 * $$V = c = {2\pi\Omega^2}v;\; v = \frac{c}{2\pi\Omega^2} = 11\;843\;707.905... ,\; u^{17}\; (m/s)$$


 * $$A = e/t_p = (\frac{2^7 \pi^3 \Omega^3}{\alpha})a = .126\;918\;588\;59... 10^{23},\; u^{3}\; (A)$$

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


 * $$M = (1)k,\; unit = u^{15}$$


 * $$T = (\pi) t,\; unit = u^{-30}$$

Replacing scalars pvla with kt


 * $$P = (\Omega)\;\frac{k^{12/15}}{t^{2/15}},\; unit = u^{12/15*15-2/15*(-30)=16}$$


 * $$V = \frac{2 \pi P^2}{M} = (2 \pi \Omega^2)\; \frac{k^{9/15}}{t^{4/15}},\; unit = u^{9/15*15-4/15*(-30)=17} $$


 * $$L = T V = (2 \pi^2 \Omega^2) \; k^{9/15} t^{11/15},\; unit = u^{9/15*15+11/15*(-30)=-13}$$


 * $$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;


 * $$P = (\Omega)p,\; unit = u^{16}$$


 * $$V = (2\pi\Omega^2)v,\; unit = u^{17}$$

Replacing scalars klta with pv


 * $$M = \frac{2\pi P^2}{V} = (1)\frac{p^2}{v},\; unit = u^{16*2-17=15} $$


 * $$T = (\pi) \frac{p^{9/2}}{v^6},\; unit = u^{16*9/2-17*6=-30} $$


 * $$L = T V = (2\pi^2\Omega^2)\frac{p^{9/2}}{v^5},\; unit = u^{16*9/2-17*5=-13}$$


 * $$A = \frac{2^4 V^3}{\alpha P^3} = (\frac{2^7 \pi^3 \Omega^3}{\alpha})\frac{v^3}{p^3},\; unit = u^{17*3-16*3=3}$$

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;


 * $${(h^*)}^3 = (2^3 \pi^4 \Omega^4 \frac{r^{13} u^{19}}{v^5})^3 = \frac{3^{19} \pi^{12} \Omega^{12} r^{39} u^{57}}{v^{15}},\; \theta = 57$$ ... and ...


 * $$\frac{2\pi^{10} {(\mu_0^*)}^3} {3^6 {(c^*)}^5 \alpha^{13} {(R^*)}^2} = \frac{3^{19} \pi^{12} \Omega^{12} r^{39} u^{57}}{v^{15}},\; \theta = 57$$