Planck units (geometrical)

Planck units as geometrical objects

Planck unit models use a set of base units for mass, length, time and charge and these are characterized at the Planck scale. In a geometrical Planck model, these 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 will encode the function of length such that, unlike numerical models, a dimensioned descriptive (km, mile ... ) will not be required.

Geometrical objects

A set of base units for mass $M$, length $L$, time $T$, and ampere $A$ can be constructed from the geometry of 2 dimensionless physical constants, the fine structure constant α and Omega Ω . 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

$M=(1)$

$T=(2\pi )$

$P=(\Omega )$

$V=(2\pi \Omega ^{2})$

$L=(2\pi ^{2}\Omega ^{2})$

$A=({\frac {2^{6}\pi ^{3}\Omega ^{3}}{\alpha }})$

Mathematical relationship

A relationship between the objects is defined using uⁿ whereby;

$(A)\;u^{3}\;$

$(L)\;u^{-13}\;$

$(M)\;u^{15}\;$

$(P)\;u^{16}\;$

$(V)\;u^{17}\;$

$(T)\;u^{-30}\;$

Attribute

Each object is assigned a unit (a dimension function);

$(A)\;$ ampere

$(L)\;$ length

$(M)\;$ mass

$(P)\;$ sqrt of momentum

$(V)\;$ velocity

$(T)\;$ time

Scalars

To translate from geometrical objects to a numerical system of units such as the SI units requires scalars (kltpva) that can be assigned numerical values with associated dimensioned units. For example, if we use k to convert M to the SI Planck mass $m_{P}$, then $u^{15}$ will equate to the unit kg.

Geometrical units

Attribute	Geometrical object	Scalar
mass	$M=1$	$k,\;unit=u^{15}$
time	$T=2\pi$	$t,\;unit=u^{-30}$
momentum (sqrt of)	$P=\Omega$	$p,\;unit=u^{16}$
velocity	$V=2\pi \Omega ^{2}$	$v,\;unit=u^{17}$
length	$L=2\pi ^{2}\Omega ^{2}$	$l,\;unit=u^{-13}$
ampere	$A={\frac {2^{6}\pi ^{3}\Omega ^{3}}{\alpha }}$	$a,\;unit=u^{3}$

Scalar relationships

The following uⁿ groups cancel, as such only 2 numerical scalars are required, for example, if we know a and l then we know k and t (as in the following examples).

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

SI Planck unit scalars

$M=m_{P}=(1)k;\;k=m_{P}=.21767281758...\;10^{-7},\;u^{15}\;(kg)$

$T=t_{p}={2\pi }t;\;t={\frac {t_{p}}{2\pi }}=.17158551284...10^{-43},\;u^{-30}\;(s)$

$L=l_{p}={2\pi ^{2}\Omega ^{2}}l;\;l={\frac {l_{p}}{2\pi ^{2}\Omega ^{2}}}=.20322086948...10^{-36},\;u^{-13}\;(m)$

$V=c={2\pi \Omega ^{2}}v;\;v={\frac {c}{2\pi \Omega ^{2}}}=11843707.9...,\;u^{17}\;(m/s)$

$A=A=({\frac {2^{6}\pi ^{3}\Omega ^{3}}{\alpha }})a;\;a={\frac {A\alpha }{64\pi ^{3}\Omega ^{3}}}=.12691858859...10^{23},\;u^{3}\;(A)$

Example MLT;

${\frac {L^{15}}{M^{9}T^{11}}}={\frac {l_{p}^{15}}{m_{P}^{9}t_{p}^{11}}}={\frac {(2\pi ^{2}\Omega ^{2}l)^{15}}{(1k)^{9}(2\pi t)^{11}}}=2^{4}\pi ^{19}\Omega ^{30}$

${\frac {l^{15}}{k^{9}t^{11}}}={\frac {(.203...x10^{-36})^{15}}{(.217...x10^{-7})^{9}(.171...x10^{-43})^{11}}}{\frac {u^{-13*15}}{u^{15*9}u^{-30*11}}}=1$

Example ALT;

${\frac {A^{3}L^{3}}{T}}={\frac {A_{p}^{3}l_{p}^{3}}{t_{p}}}={\frac {(2^{6}\pi ^{3}\Omega ^{3}a)^{3}(2\pi ^{2}\Omega ^{2}l)^{3}}{(\alpha )^{3}(2\pi t)}}={\frac {2^{20}\pi ^{14}\Omega ^{15}}{\alpha ^{3}}}$

${\frac {a^{3}l^{3}}{t}}={\frac {(.126...x10^{23})^{3}(.203...x10^{-36})^{3}}{(.171...x10^{-43})}}{\frac {u^{3*3}u^{-13*3}}{u^{-30}}}=1$

Note: the geometry Ω¹⁵ is common to unit-less ratios.

MT to LPVA

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

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

$T=(2\pi )t,\;unit=u^{-30}$

Replacing scalars pvla with kt

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

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

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

$A={\frac {8V^{3}}{\alpha P^{3}}}=\left({\frac {64\pi ^{3}\Omega ^{3}}{\alpha }}\right)\;{\frac {1}{k^{3/5}t^{2/5}}},\;unit=u^{9/15*(-15)+6/15*30=3}$

PV to MTLA

In this example units 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

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

$T^{2}=(2\pi \Omega )^{15}{\frac {P^{9}}{2\pi V^{12}}}$

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

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

$A={\frac {8V^{3}}{\alpha P^{3}}}=({\frac {2^{6}\pi ^{3}\Omega ^{3}}{\alpha }}){\frac {v^{3}}{p^{3}}},\;unit=u^{17*3-16*3=3}$

Physical constants

In this example, to maintain integer exponents, scalar p is defined in terms of a scalar r.

$r={\sqrt {p}}={\sqrt {\Omega }},\;unit\;u^{16/2=8}$

As α and Ω have fixed values, 2 scalars are also needed to solve the physical constants with numerical values. The SI Planck units are known with a low precision, conversely 2 of the CODATA 2014 physical constants have been assigned exact numerical values; c and permeability of vacuum μ₀. Thus scalars r and v were chosen as they can be derived directly from V = c and μ₀.

$v={\frac {c}{2\pi \Omega ^{2}}}=11843707.9...,\;units=m/s$

$r^{7}={\frac {2^{11}\pi ^{5}\Omega ^{4}\mu _{0}}{\alpha }};\;r=.712562514...,\;units=({\frac {kg.m}{s}})^{1/4}$

Physical constants; geometrical vs experimental (CODATA)

Constant	In Planck units	Geometrical object	SI calculated (r, v, Ω, α*)	SI CODATA 2014
Speed of light	V	$c^{*}=(2\pi \Omega ^{2})v,\;u^{17}$	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	$R^{*}=({\frac {m_{e}}{4\pi L\alpha ^{2}M}})$	$R^{*}={\frac {1}{2^{23}3^{3}\pi ^{11}\alpha ^{5}\Omega ^{17}}}{\frac {v^{5}}{r^{9}}},\;u^{13}$	R* = 10 973 731.568 508, unit = u13	R = 10 973 731.568 508(65)
Vacuum permeability	$\mu _{0}^{*}={\frac {\pi V^{2}M}{\alpha LA^{2}}}$	$\mu _{0}^{*}={\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}}r^{7},\;u^{17*2+15+13-6=7*8=56}$	μ0* = 4π/10^7, unit = u56	μ0 = 4π/10^7 (exact)
Planck constant	$h^{*}=2\pi MVL$	$h^{*}=2^{3}\pi ^{4}\Omega ^{4}{\frac {r^{13}}{v^{5}}},\;u^{15+17-13=8*13-17*5=19}$	h* = 6.626 069 134 e-34, unit = u19	h = 6.626 070 040(81) e-34
Gravitational constant	$G^{*}={\frac {V^{2}L}{M}}$	$G^{*}=2^{3}\pi ^{4}\Omega ^{6}{\frac {r^{5}}{v^{2}}},\;u^{34-13-15=8*5-17*2=6}$	G* = 6.672 497 192 29 e11, unit = u6	G = 6.674 08(31) e-11
Elementary charge	$e^{*}=AT$	$e^{*}={\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }}{\frac {r^{3}}{v^{3}}},\;u^{3-30=3*8-17*3=-27}$	e* = 1.602 176 511 30 e-19, unit = u-19	e = 1.602 176 620 8(98) e-19
Boltzmann constant	$k_{B}^{*}={\frac {\pi VM}{A}}$	$k_{B}^{*}={\frac {\alpha }{2^{5}\pi \Omega }}{\frac {r^{10}}{v^{3}}},\;u^{17+15-3=10*8-17*3=29}$	kB* = 1.379 510 147 52 e-23, unit = u29	kB = 1.380 648 52(79) e-23
Electron mass		$m_{e}^{*}={\frac {M}{f_{e}}},\;u^{15}$	me* = 9.109 382 312 56 e-31, unit = u15	me = 9.109 383 56(11) e-31
Classical electron radius		$\lambda _{e}^{*}=2\pi Lf_{e},\;u^{-13}$	λe* = 2.426 310 2366 e-12, unit = u-13	λe = 2.426 310 236 7(11) e-12
Planck temperature	$T_{p}^{*}={\frac {AV}{\pi }}$	$T_{p}^{*}={\frac {2^{7}\pi ^{3}\Omega ^{5}}{\alpha }}{\frac {v^{4}}{r^{6}}},\;u^{3+17=17*4-6*8=20}$	Tp* = 1.418 145 219 e32, unit = u20	Tp = 1.416 784(16) e32
Planck mass	M	$m_{P}^{*}=(1){\frac {r^{4}}{v}},\;u^{15}$	mP* = .217 672 817 580 e-7, unit = u15	mP = .217 647 0(51) e-7
Planck length	L	$l_{p}^{*}=(2\pi ^{2}\Omega ^{2}){\frac {r^{9}}{v^{5}}},\;u^{-13}$	lp* = .161 603 660 096 e-34, unit = u-13	lp = .161 622 9(38) e-34
Planck time	T	$t_{p}^{*}=(2\pi ){\frac {r^{9}}{v^{6}}},\;u^{-30}$	tp* = 5.390 517 866 e-44, unit = u-30	tp = 5.391 247(60) e-44
Ampere	A	$A^{*}={\frac {2^{6}\pi ^{3}\Omega ^{3}}{\alpha }}{\frac {v^{3}}{r^{6}}},\;u^{3}$	A^* = 0.148 610 6299 e25, unit = u3
Von Klitzing constant	$R_{K}^{*}=({\frac {h}{e^{2}}})^{*}$		RK* = 25812.807 455 59, unit = u73	RK = 25812.807 455 5(59)
Gyromagnetic ratio		$\gamma _{e}/2\pi ={\frac {gl_{p}^{*}m_{P}^{*}}{2k_{B}^{*}m_{e}^{*}}},\;unit=u^{-42}$	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 uⁿ 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 a strong argument for a Planck scale mathematical universe.

Electron formula

Main article: Electron (mathematical)

Although the Planck units MLTA are embedded within the electron formula f_e, this formula is both unit-less and non scalable (units = scalars = 1). Furthermore it is the geometry of 2 dimensionless physical constants and so can also be defined as a dimensionless physical constant.

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

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

$T=2\pi {\frac {r^{9}}{v^{6}}},\;u^{-30}$

$\sigma _{e}={\frac {3\alpha ^{2}AL}{\pi ^{2}}}={2^{7}3\pi ^{3}\alpha \Omega ^{5}}{\frac {r^{3}}{v^{2}}},\;u^{-10}$

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

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

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

elementary charge $e=A.T$

Fine structure constant

The Sommerfeld fine structure constant alpha is a dimensionless physical constant, the CODATA inverse alpha = 137.035999084. Alpha is associated with orbitals (orbital radius and period) and so relates orbital period (a time dimension) with orbital radius (a length dimension) . As period is involved and as Planck mass has no time component, alpha is typically associated with the charge domain.

$\alpha ={\frac {2h}{\mu _{0}e^{2}c}}$

$\alpha =2({8\pi ^{4}\Omega ^{4}})/({\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}})({\frac {128\pi ^{4}\Omega ^{3}}{\alpha }})^{2}(2\pi \Omega ^{2})=\alpha$

$scalars={\frac {r^{13}}{v^{5}}}.{\frac {1}{r^{7}}}.{\frac {v^{6}}{r^{6}}}.{\frac {1}{v}}=1$

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

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;

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

$units={\frac {(u^{17})^{35}}{(u^{56})^{9}(u^{13})^{7}}}=1$

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) CODATA 2014 values.

$\Omega ^{225}={\frac {(c^{*})^{35}}{2^{295}3^{21}\pi ^{157}(\mu _{0}^{*})^{9}(R^{*})^{7}\alpha ^{26}}},\;units=1$

$\Omega =2.007\;134\;9496...,\;units=1$

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).

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

From this Omega and the above; (inverse) alpha = 137.035 996 369

G, h, e, m_e, k_B

As geometrical objects, the physical constants (G, h, e, m_e, k_B) can be defined using the geometrical formulas for (c^*, μ₀^*, R^*) and solved using the CODATA 2014 numerical values for (c, μ₀, R, α), i.e.:.

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

Physical constants; calculated vs experimental (CODATA)

Constant	Geometry	Calculated from (R, c, μ0, α)	CODATA 2014
Planck constant	${(h^{*})}^{3}={\frac {2\pi ^{10}{\mu _{0}}^{3}}{3^{6}{c}^{5}\alpha ^{13}{R_{\infty }}^{2}}},\;unit=u^{57}$	h* = 6.626 069 134 e-34, unit = u19	h = 6.626 070 040(81) e-34
Gravitational constant	${(G^{*})}^{5}={\frac {\pi ^{3}{\mu _{0}}}{2^{20}3^{6}\alpha ^{11}{R_{\infty }}^{2}}},\;unit=u^{30}$	G* = 6.672 497 192 29 e11, unit = u6	G = 6.674 08(31) e-11
Elementary charge	${(e^{*})}^{3}={\frac {4\pi ^{5}}{3^{3}{c}^{4}\alpha ^{8}{R_{\infty }}}},\;unit=u^{-81}$	e* = 1.602 176 511 30 e-19, unit = u-19	e = 1.602 176 620 8(98) e-19
Boltzmann constant	${(k_{B}^{*})}^{3}={\frac {\pi ^{5}{\mu _{0}}^{3}}{3^{3}2{c}^{4}\alpha ^{5}{R_{\infty }}}},\;unit=u^{87}$	kB* = 1.379 510 147 52 e-23, unit = u29	kB = 1.380 648 52(79) e-23
Electron mass	${(m_{e}^{*})}^{3}={\frac {16\pi ^{10}{R_{\infty }}{\mu _{0}}^{3}}{3^{6}{c}^{8}\alpha ^{7}}},\;unit=u^{45}$	me* = 9.109 382 312 56 e-31, unit = u15	me = 9.109 383 56(11) e-31
Planck mass	$(m_{P}^{*})^{15}={\frac {2^{25}\pi ^{13}\mu _{0}^{6}}{3^{6}c^{5}\alpha ^{16}R_{\infty }^{2}}},\;unit=(u^{15})^{15}$	mP* = .217 672 817 580 e-7, unit = u15	mP = .217 647 0(51) e-7
Planck length	$(l_{p}^{*})^{15}={\frac {\pi ^{22}\mu _{0}^{9}}{2^{35}3^{24}\alpha ^{49}c^{35}R_{\infty }^{8}}},\;unit=(u^{-13})^{15}$	lp* = .161 603 660 096 e-34, unit = u-13	lp = .161 622 9(38) e-34
Gyromagnetic ratio	$(\gamma _{e}/2\pi )^{3}={\frac {g^{3}3^{3}c^{4}}{2^{8}\pi ^{8}\alpha \mu _{0}^{3}R_{\infty }^{2}}},\;unit=u^{-126}$	γe/2π* = 28024.953 55, unit = u-42	γe/2π = 28024.951 64(17)

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, k_B). 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
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

For example;

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

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

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

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

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

u as √{length/mass.time}

We find there is a single base unit u from which the other units and numerical values can be derived. This base unit incorporates MLT as square roots.

u = √{L/M.T}

$u,\;units={\sqrt {\frac {L}{MT}}}={\sqrt {u^{-13-15+30=2}}}=u^{1}$

$x,\;units={\sqrt {\frac {M^{9}T^{11}}{L^{15}}}}=u^{0}=1$

$y,\;units=M^{2}T=u^{0}=1$

Gives;

$u^{3}={\frac {L^{3/2}}{M^{3/2}T^{3/2}}}=A,\;(ampere)$

$u^{6}(y)=L^{3}/T^{2}M,\;(G)$

$u^{13}(xy)=1/L,\;(1/l_{p})$

$u^{15}(xy^{2})=M,\;(m_{P})$

$u^{17}(xy^{2})=V,\;(c)$

$u^{19}(xy^{3})=ML^{2}/T,\;(h)$

$u^{20}(xy^{2})={\frac {L^{5/2}}{M^{3/2}T^{5/2}}}=AV,\;(T_{P})$

$u^{27}(x^{2}y^{3})={\frac {M^{3/2}{\sqrt {T}}}{L^{3/2}}}=1/AT,\;(1/e)$

$u^{29}(x^{2}y^{4})={\frac {M^{5/2}{\sqrt {T}}}{\sqrt {L}}}=ML/AT,\;(k_{B})$

$u^{30}(x^{2}y^{3})=1/T,\;(1/t_{p})$

$u^{56}(x^{4}y^{7})={\frac {M^{4}T}{L^{2}}}={\frac {ML}{T^{2}A^{2}}},\;(\mu _{0})$

β (unit = u)

i (from x) and j (from y).

$R={\sqrt {P}}={\sqrt {\Omega }}r,\;units=u^{8}$

$\beta ={\frac {V}{R^{2}}}={\frac {2\pi R^{2}}{M}}={\frac {A^{1/3}\alpha ^{1/3}}{2}}\;...,\;unit=u$

$i={\frac {1}{2\pi {(2\pi \Omega )}^{15}}},\;unit=1$

$j={\frac {r^{17}}{v^{8}}}=k^{2}t={\frac {k^{8}}{r^{15}}}...,\;unit={\frac {u^{17*8}}{u^{8*17}}}=u^{15*2}u^{-30}...=1$

For example; the constants solved in terms of (r, v)

$\beta ={\frac {V}{R^{2}}}={\frac {2\pi \Omega ^{2}v}{\Omega r^{2}}},\;u$

$A=\beta ^{3}({\frac {2^{3}}{\alpha }})={\frac {2^{6}\pi ^{3}\Omega ^{3}}{\alpha }}{\frac {v^{3}}{r^{6}}},\;u^{3}$

$G={\frac {\beta ^{6}}{2^{3}\pi ^{2}}}(j)=2^{3}\pi ^{4}\Omega ^{6}{\frac {r^{5}}{v^{2}}},\;u^{6}$

$L^{-1}=4\pi \beta ^{13}(ij)={\frac {1}{2\pi ^{2}\Omega ^{2}}}{\frac {v^{5}}{r^{9}}},\;u^{13}$

$M=2\pi \beta ^{15}(ij^{2})={\frac {r^{4}}{v}},\;u^{15}$

$P=\beta ^{16}(ij^{2})=\Omega r^{2},\;u^{16}$

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

$h=\pi \beta ^{19}(ij^{3})=8\pi ^{4}\Omega ^{4}{\frac {r^{13}}{v^{5}}},\;u^{19}$

$T_{P}^{*}={\frac {2^{3}\beta ^{20}}{\pi \alpha }}(ij^{2})={\frac {2^{7}\pi ^{3}\Omega ^{5}}{\alpha }}{\frac {v^{4}}{r^{6}}},\;u^{20}$

$e^{-1}={\frac {\alpha \pi \beta ^{27}(i^{2}j^{3})}{4}}={\frac {\alpha }{128\pi ^{4}\Omega ^{3}}}{\frac {v^{3}}{r^{3}}},\;u^{27}$

$k_{B}={\frac {\alpha \pi ^{2}\beta ^{29}(i^{2}j^{4})}{4}}={\frac {\alpha }{32\pi \Omega }}{\frac {r^{10}}{v^{3}}},\;u^{29}$

$T^{-1}=2\pi \beta ^{30}(i^{2}j^{3})={\frac {1}{2\pi }}{\frac {v^{6}}{r^{9}}},\;u^{30}$

$\mu _{0}^{*}={\frac {\pi ^{3}\alpha \beta ^{56}}{2^{3}}}(i^{4}j^{7})={\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}}r^{7},\;u^{56}$

$\epsilon _{0}^{*-1}={\frac {\pi ^{3}\alpha \beta ^{90}}{2^{3}}}(i^{6}j^{11})={\frac {\alpha }{2^{9}\pi ^{3}}}v^{2}r^{7},\;u^{90}$

limit j

The SI values for j suggest a limit (numerical boundary) to the values the SI constants can have.

$j={\frac {r^{17}}{v^{8}}}=k^{2}t={\frac {k^{17/4}}{v^{15/4}}}=...=.812997...x10^{-59},\;units=1$

In SI terms unit β has this value;

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

The unit-less ratios;

$(AL)^{3}/T=A^{3}T^{-1}/(L^{-1})^{3};\;units={\frac {u^{3}(u^{30}x^{2}y^{3})}{(u^{13}xy)^{3}}}=1/x$

$T^{2}T_{P}^{3}={\frac {T_{P}^{3}}{(T^{-1})^{2}}};\;units={\frac {(u^{20}xy^{2})^{3}}{(u^{30}x^{2}y^{3})^{2}}}=1/x$

${M^{9}(L^{-1})^{15}}/{(T^{-1})^{11}};\;units={\frac {(u^{15}xy^{2})^{9}(u^{13}xy)^{15}}{(u^{30}x^{2}y^{3})^{11}}}=x^{2}$

Rydberg formula

The Rydberg formula can now be re-written in terms of amperes $A^{2}$

${\frac {hc}{2\pi \alpha ^{2}}}={\frac {j^{2}A^{2}}{2^{8}2\pi t_{p}}}$

External links

• Relativity at the Planck scale
• Quantized gravity at the Planck scale
• Cosmic microwave background at the Planck scale
• Mathematical electron
• Dirac-Kerr-Newman black-hole electron -Alexander Burinskii
• Simulation Hypothesis modelling at the Planck scale
• The Planck universe, a programming approach -online resource site