# Physical constant (anomaly)

Anomalies within the dimensioned physical constants (G, h, c, e, m$_e$, k$_B$) suggest a mathematical relationship linking the units (kg, m, s, A, K).
A dimensioned physical constant, sometimes denoted a fundamental physical constant, is a physical quantity that is generally believed to be both universal in nature and have constant value in time. Common examples being the speed of light c, the gravitational constant G, the Planck constant h and the elementary charge e. These constants are usually measured in terms of SI units mass (kilogram), length (meter), time (second), charge (ampere), temperature (Kelvin) ... (kg, m, s, A, K ...).
These constants form the scaffolding around which the theories of physics are erected, and they define the fabric of our universe, but science has no idea why they take the special numerical values that they do, for these constants follow no discernible pattern. The desire to explain the constants has been one of the driving forces behind efforts to develop a complete unified description of nature, or "theory of everything". Physicists have hoped that such a theory would show that each of the constants of nature could have only one logically possible value. It would reveal an underlying order to the seeming arbitrariness of nature.
Notably a physical universe, as opposed to a mathematical universe (a computer simulation), has as a fundamental premise the concept that the universe scaffolding (of mass, space and time) exists, that somehow mass is, space is, time is ... these dimensions are real, and independent of each other ... we cannot measure distance in kilograms and amperes, or mass using length and temperature. The 2019 redefinition of SI base units resulted in 4 physical constants (h, c, e, k$_B$) having assigned exact values, and this confirmed the independence of their associated SI units as shown in this table.
2019 redefinition of SI base units :
| constant | | SI units |
| Speed of light | c | ${\frac {m}{s}}$ |
| Planck constant | h | ${\frac {kg\;m^{2}}{s}}$ |
| Elementary charge | e | $C=As$ |
| Boltzmann constant | k$_B$ | ${\frac {kg\;m^{2}}{s^{2}\;K}}$ |
However there are anomalies which occur in certain combinations of the fundamental (dimensioned) physical constants (G, h, c, e, m$_e$, k$_B$) which suggest a mathematical relationship between the units (kg ⇔ 15, m ⇔ -13, s ⇔ -30, A ⇔ 3, K ⇔ 20).
In order for these physical constants to be fundamental, the units must be independent of each other, there cannot be such a unit number relationship ... however these anomalies question this fundamental assumption. Physics has a set of constants defined in terms of the units (kg, m, s, A, K), these are called Planck units (Planck mass, Planck length, Planck time ...), and these Planck units are interchangeable with the physical constants.
If we include this unit number relationship (kg ⇔ 15, m ⇔ -13, s ⇔ -30, A ⇔ 3, K ⇔ 20), then we find that we may need only 3 Planck units (labelled MTP) and the fine structure constant alpha to derive and solve these 6 physical constants (G, h, c, e, m$_e$, k$_B$). This would then question their status as being fundamental. Note: (α, Ω) are dimensionless constants, (r, v) are system dependent dimensioned scalar variables.
    Planck mass $M=(1){\frac {r^{4}}{v}}$
    Planck time $T=(\pi ){\frac {r^{9}}{v^{6}}}$
    sqrt(Planck momentum) $P=(\Omega )r^{2}$
Every test listed in the following examples using this relationship returns answers consistent with the premise. Statistically therefore, can these anomalies be dismissed as coincidence?

## Theory
### Unit number

We can demonstrate a curious geometrical relationship between the units (kg, m, s, A) by selecting 2 dimensioned quantities, here are chosen r, v (we can choose others) such that
    $kg={\frac {r^{4}}{v}},\;m={\frac {r^{9}}{v^{5}}},\;s={\frac {r^{9}}{v^{6}}},\;A={\frac {v^{3}}{r^{6}}}$
The units (kg, m, s, A) remain independent of each (i.e.: the kg cannot be replaced by the m or the s ...), and so we still have 4 independent units, however if 3 (or more) units are combined together, in a specific ratio, they can cancel.
    $f_{X}={\frac {kg^{9}s^{11}}{m^{15}}}={\frac {({\frac {r^{4}}{v}})^{9}({\frac {r^{9}}{v^{6}}})^{11}}{({\frac {r^{9}}{v^{5}}})^{15}}}=1$
This f(X) embeds the units kg, m, s but itself is dimensionless, units = 1 (i.e.: it is a mathematical structure).
If we assign these SI units to the dimensioned quantities r, v;
    units: $r=({\frac {kg\;m}{s}})^{1/4}$ (r$^4$ are the units for momentum)
    units: $v={\frac {m}{s}}$
    units: $f_{X}={\frac {kg^{9}s^{11}}{m^{15}}}$ units = 1
Mass
    ${\frac {r^{4}}{v}}=({\frac {kg\;m}{s}})\;({\frac {s}{m}})=kg$
Length ($f_{X}$ = 1)
    $m={\frac {r^{9}}{v^{5}}}$
    $(r^{9})^{4}={\frac {kg^{9}\;m^{9}}{s^{9}}}$
    $({\frac {1}{v^{5}}})^{4}={\frac {s^{20}}{m^{20}}}$
    $({\frac {r^{9}}{v^{5}}})^{4}={\frac {kg^{9}s^{11}}{m^{11}}}=m^{4}{\frac {kg^{9}s^{11}}{m^{15}}}=m^{4}f_{X}=m^{4}$
Time
    $s={\frac {r^{9}}{v^{6}}}$
    $(r^{9})^{4}={\frac {kg^{9}\;m^{9}}{s^{9}}}$
    $({\frac {1}{v^{6}}})^{4}={\frac {s^{24}}{m^{24}}}$
    $({\frac {r^{9}}{v^{6}}})^{4}={\frac {kg^{9}s^{15}}{m^{15}}}=s^{4}{\frac {kg^{9}s^{11}}{m^{15}}}=s^{4}f_{X}=s^{4}$
We can also construct a unit-less structure using the ampere with length and time
    $f_{X}={\frac {A^{3}m^{3}}{s}}={\frac {({\frac {v^{3}}{r^{6}}})^{3}({\frac {r^{9}}{v^{5}}})^{3}}{\frac {r^{9}}{v^{6}}}}=1$
If we assign a numerical value θ to r (θ = 8) and to v (θ = 17), then we can assign a unit number θ to the SI units kg, m, s, A, K.
Table 1. unit relationship :
| attribute | SI equivalent | unit number θ | scalars |
| M (mass) | kg | 8*4-17 = 15 | ${\frac {r^{4}}{v}}$ |
| T (time) | s | 8*9-17*6 = -30 | ${\frac {r^{9}}{v^{6}}}$ |
| V (velocity) | m/s | 17 | $v$ |
| L (length) | m | 8*9-17*5 = -13 | ${\frac {r^{9}}{v^{5}}}$ |
| A (ampere) | A | 17*3-8*6 = 3 | ${\frac {v^{3}}{r^{6}}}$ |
| K (temperature) | K | 17*4-8*6 = 20 | ${\frac {v^{4}}{r^{6}}}$ |

### Planck units

The Planck units are direct measures of the SI units; Planck mass in kg, Planck length in m, Planck time in s ... and so they are analogues to the attributes in the above table. The SI Planck units have numerical values, however to derive a mathematical relation between these SI units we cannot use numerical values, this is because numerical values are simply dimensionless frequencies of the SI unit itself, 299792458 could refer to the speed of light 299792458m/s or equally to the number of apples in a container (299792458 apples), numbers such as 299792458 carry no unit-specific information, and so the units are treated as independent by default. This therefore requires that to the number 299792458 is added a descriptive (the unit), which could be m/s or apples.
This inherent restriction can be resolved by assigning to each unit a geometrical object for which the geometry embeds the attribute (for example, the geometry of the time object T embeds the function time and so a descriptive unit s = seconds is not required). We may then combine these objects Lego-style to form more complex objects; from electrons to galaxies, while still retaining the underlying attributes (of mass M, wavelength L, frequency T ...). An apple has mass because its 'geometry' includes the geometrical object for mass.
Table 2. lists a set of attributes as the geometry of 2 dimensionless physical constants; the inverse fine structure constant alpha and Omega. As alpha and Omega are dimensionless (alpha = 137.035999139, Omega = 2.00713495...), so too are these geometrical objects.
The dimensioned variables (in this example we are using r and v) are unit specific. They are used to translate between the geometrical objects (being dimensionless they are universal) and local unit systems - the numerical value for v will depend on whether we are using m/s or miles/hour or ...
Table 2. MLTVA Geometrical objects :
| attribute | geometrical object | numerical | unit number | dimensioned component |
| mass | $M=(1)$ | 1 | 15 | ${\frac {r^{4}}{v}}$ |
| time | $T=(\pi )$ | 3.1415926535... | -30 | ${\frac {r^{9}}{v^{6}}}$ |
| sqrt(momentum) | $P=(\Omega )$ | 2.00713495... | 16 | $r^{2}$ |
| velocity | $V={\frac {2\pi P^{2}}{M}}=(2\pi \Omega ^{2})$ | 25.3123819... | 17 | $v$ |
| length | $L=VT=(2\pi ^{2}\Omega ^{2})$ | 79.5211931... | -13 | ${\frac {r^{9}}{v^{5}}}$ |
| ampere | $A={\frac {2^{4}V^{3}}{\alpha P^{3}}}=({\frac {2^{7}\pi ^{3}\Omega ^{3}}{\alpha }})$ | 234.182607... | 3 | ${\frac {v^{3}}{r^{6}}}$ |
| temperature | $K={\frac {AV}{2\pi }}=({\frac {2^{7}\pi ^{3}\Omega ^{5}}{\alpha }})$ | 943.425875... | 20 | ${\frac {v^{4}}{r^{6}}}$ |
We can now solve those f(X) structures as dimensionless geometrical forms.
$f_{X}={\frac {kg^{9}s^{11}}{m^{15}}}={\frac {(1)^{9}(\pi )^{11}}{(2\pi ^{2}\Omega ^{2})^{15}}}$
The electron f$_e$ (below) is one example of an f(X) structure
As this particular geometrical approach requires that the objects be interrelated (they are not independent of each other), this unit number relationship hypothesis can be easily tested. This is because, if these MLTVPA objects are natural Planck units, then they will be embedded within our dimensioned SI physical constants (G, h, c, e, m$_e$, K$_B$...).
We can use these text-book formulas to convert between Planck units and the common physical constants.
Table 3. Physical constant unit numbers :
| SI constant | geometrical analogue | unit number θ |
| Speed of light | c* = V | 17 |
| Planck constant | $h^{*}=2\pi MVL$ | 15+17-13=19 |
| Gravitational constant | $G^{*}={\frac {V^{2}L}{M}}$ | 34-13-15=6 |
| Elementary charge | $e^{*}=AT$ | 3-30=-27 |
| Boltzmann constant | $k_{B}^{*}={\frac {2\pi VM}{A}}$ | 17+15-3=29 |
| Vacuum permeability | $\mu _{0}^{*}={\frac {4\pi V^{2}M}{\alpha LA^{2}}}$ | 34+15+13-6=56 |

### Scalars

We can assign numerical values to alpha (the inverse fine structure constant) and Omega, and then use the dimensioned scalars (r, v) to convert from the MLTVA objects to their SI equivalents.
alpha α = 137.035999139
Omega Ω = 2.0071349496
For example, we can use scalar v to convert from dimensionless geometrical object V to dimensioned c.
    scalar v = 11843707.905 m/s gives c = V*v = 25.3123819 * 11843707.905 m/s = 299792458 m/s (SI units)
    scalar v = 7359.3232155 miles/s gives c = V*v = 186282 miles/s (imperial units)
    r = 0.712562514304 ... (SI units)
As the scalars also carry the unit designation m/s or miles/hour ... (as well as an associated numerical value), they are dimensioned, and so we can apply the unit number relationship θ to them; scalar v (θ = 17), for r (θ = 8). We find that the unit number relationship is consistent regardless of the constants and the system of units used.
Table 4. Comparison θ; SI units and scalars :
| constant | θ from SI units | MLTVA | θ from r(8), v(17) |
| c | ${\frac {m}{s}}$ (-13+30 = 17) | c* = $V*v$ | 17 |
| h | ${\frac {kg\;m^{2}}{s}}$ (15-26+30=19) | h* = $2\pi MVL*{\frac {r^{13}}{v^{5}}}$ | 8*13-17*5=19 |
| G | ${\frac {m^{3}}{kg\;s^{2}}}$ (-39-15+60=6) | G* = ${\frac {V^{2}L}{M}}*{\frac {r^{5}}{v^{2}}}$ | 8*5-17*2=6 |
| e | $C=As$ (3-30=-27) | e* = $AT*{\frac {r^{3}}{v^{3}}}$ | 8*3-17*3=-27 |
| k$_B$ | ${\frac {kg\;m^{2}}{s^{2}\;K}}$ (15-26+60-20=29) | k$_B$* = ${\frac {2\pi VM}{A}}*{\frac {r^{10}}{v^{3}}}$ | 8*10-17*3=29 |
| μ$_0$ | ${\frac {kg\;m}{s^{2}\;A^{2}}}$ (15-13+60-6=56) | μ$_0$* = ${\frac {4\pi V^{2}M}{\alpha LA^{2}}}*r^{7}$ | 8*7=56 |

## Anomalies

Dimensioned physical constants (G, h, c, e, m$_e$, k$_B$ ...) "form the scaffolding around which the theories of physics are erected, and they define the fabric of our universe, but science has no idea why they take the special numerical values that they do, for these constants follow no discernible pattern".
They are referred to as fundamental constants, for they cannot be derived from a more fundamental group of constants. However, if there is a unit number relationship, then it may be possible to relate (and so define) these constants in terms of each other.

### Planck units

This section uses the unit number relationship to formulate the constants (G, h, c, e, k$_B$) in terms of the 3 Planck units MTP and the fine structure constant alpha.
Initial constants: Note: for convenience α is assigned the value 137.035999139 (which is the CODATA 2014 inverse alpha).
    α = 137.035999139, (dimensionless)
    Ω = 2.007134949638 (dimensionless)
    v = 11843707.9052487, unit u = 17
    r = 0.71256251430467, unit u = 8
If we select MTP as the principal Planck units (they include a dimensionless geometrical component and 2 dimensioned variables; r and v);
    M = (1) * r^4/v, unit = 15 (Planck mass)
    T = (π) * r^9/v^6, unit = -30 (Planck time)
    P = (Ω) * r^2, unit = 16 (sqrt of Planck momentum)
then from MTP and alpha we can formulate VLA
    $V={\frac {2\pi P^{2}}{M}}$, unit = 17 (velocity c)
    $L=VT$, unit = -13 (Planck length)
    $A={\frac {2^{4}V^{3}}{\alpha P^{3}}}$, unit = 3 (Planck ampere)
and from MTPVLA we can formulate the dimensioned fundamental physical constants using the textbook formulas for converting from Planck units. Note: (kg ⇔ 15, m ⇔ -13, s ⇔ -30, A ⇔ 3, K ⇔ 20).
    $c=V$, unit = 17 (speed of light)
    $\mu _{0}={\frac {4\pi V^{2}M}{\alpha LA^{2}}}$, unit = 56 (permeability of vacuum)
    $h=2\pi MVL$, unit = 19 (Planck's constant)
    $e=AT$, unit = -27 (elementary charge)
    $G={\frac {V^{2}L}{M}}$, unit = 6 (gravitational constant)
    $k_{B}={\frac {2\pi VM}{A}}$, unit = 29 (Boltzmann constant)

### Calculating the electron

We can now construct the electron from magnetic monopoles AL and time T (AL units ampere-meter (ampere-length) are the units for a magnetic monopole).
    $T=\pi {\frac {r^{9}}{v^{6}}},\;u^{-30}$
    $\sigma _{e}={\frac {3\alpha ^{2}AL}{2\pi ^{2}}}={2^{7}3\pi ^{3}\alpha \Omega ^{5}}{\frac {r^{3}}{v^{2}}},\;u^{-10}$
    $\psi ={\frac {\sigma _{e}^{3}}{2T}}={\frac {(2^{7}3\pi ^{3}\alpha \Omega ^{5})^{3}}{2\pi }},\;units={\frac {(u^{-10})^{3}}{u^{-30}}}=1,scalars=({\frac {r^{3}}{v^{2}}})^{3}{\frac {v^{6}}{r^{9}}}=1$
Both units and scalars cancel (units = scalars = 1), and so the formula ψ is dimensionless. We can solve the electron parameters; electron mass, wavelength, frequency, charge ... as the frequency of the Planck units, and this frequency is ψ. Our results (calculated) agree with CODATA 2014. This means that the formula ψ not only determines the frequency of the Planck units (and so the magnitude or duration of the electron parameters), but it also embeds those Planck units.
In other words, this formula ψ contains all the information needed to make the electron, and so by definition this formula ψ is the electron. However it is dimensionless (units = 1), and this means that the electron is a mathematical particle, not a physical particle. And if the electron is not a physical particle, then it is these electron parameters (wavelength, charge, mass ...), and not the electron itself, that we are measuring. The existence of the electron is inferred, it is not observed.
Solving the electron parameters using ψ
V = (2*pi*Omega^2*v) = 299792458m/s (V ⇔ speed of light)
T = (pi*r^9/v^6) = 0.53905178661e-43s (T ⇔ Planck time)
L = V*T = 0.1616036601e-34m (L ⇔ Planck length)
ψ = 4*pi^2*(2^6*3*π^2*α*Omega^5)^3 = 0.2389545307369 e23 (dimensionless)
1. Compton wavelength
λ$_e$ = 2.4263102367e-12m (CODATA 2014)
λ$_e$ = 2*π*L*ψ = 2.4263102386e-12m (calculated)
2. Electron mass
m$_e$ = 9.10938356e-31kg (CODATA 2014)
M = (1*r^4/v) = 0.21767281758e-7kg (M ⇔ Planck mass)
ψ = 4*pi^2*(2^6*3*π^2*α*Omega^5)^3 = 0.2389545307369 e23 (dimensionless)
M/ψ = (1*r^4/v)/(4*pi^2*(2^6*3*π^2*α*Omega^5)^3) = (0.21767281758e-7/0.2389545307369e23)kg
m$_e$ = M/ψ = 0.91093823211e-30kg (calculated)
3. Elementary charge
e = 1.6021766208e-19C (CODATA 2014)
A = 2^4*V^3/(alpha*P^3) = 0.2972212598e25C
e = A*T = 0.16021765130e-18C (calculated)
4. Rydberg constant
R = 10973731.568508/m (CODATA 2014)
$R=({\frac {m_{e}}{4\pi L\alpha ^{2}M}})={\frac {1}{2^{23}3^{3}\pi ^{11}\alpha ^{5}\Omega ^{17}}}{\frac {v^{5}}{r^{9}}}\;u^{13}$ = 10973731.568508/m (calculated)
In summary, we have a dimensionless geometrical mathematical electron formula ψ that resembles the formula for the volume of a torus or surface area of a 4-axis hypersphere ($4\pi ^{2}(AL)^{3}$), and that includes the information needed to make both the electron parameters and to make the Planck units. It can also be divided into 3 magnetic monopoles $(AL)^{3}$ and these suggest a potential 'quark' model for the electron as well as a rationale for electron spin.

### Calculating from (α, Ω, v, r)

In this section we solve the constants (G, h, c, e, k$_B$) using 2 dimensionless constants; the (inverse) fine structure constant and Omega, and 2 scalars (r, v) which have been assigned (best fit) numerical values consistent with the SI units. r can be defined in terms of μ$_0$ and v can be defined in terms of c, thus we may use either (α, Ω, r, v) or (α, Ω, c, μ$_0$) as the initial constants, however the (α, Ω, c, μ$_0$) set is more complicated to formulate, conversely (α, Ω, r, v) is more intuitive.
Table 5. Dimensioned constants (α, Ω, v, r) :
| constant | geometrical object | calculated (α, Ω, r, v) | CODATA 2014 (mean) |
| Planck constant | $h^{*}=2\pi MVL=2^{3}\pi ^{4}\Omega ^{4}{\frac {r^{13}}{v^{5}}}$ | 6.626069134e-34, u$^{19}$ | 6.626070040e-34 |
| Gravitational constant | $G^{*}={\frac {V^{2}L}{M}}=2^{3}\pi ^{4}\Omega ^{6}{\frac {r^{5}}{v^{2}}}$ | 6.67249719229e11, u$^6$ | 6.67408e-11 |
| Elementary charge | $e^{*}=AT={\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }}{\frac {r^{3}}{v^{3}}}$ | 1.60217651130e-19, u$^{-27}$ | 1.6021766208e-19 |
| Boltzmann constant | $k_{B}^{*}={\frac {2\pi VM}{A}}={\frac {\alpha }{2^{5}\pi \Omega }}{\frac {r^{10}}{v^{3}}}$ | 1.37951014752e-23, u$^{29}$ | 1.38064852e-23 |
| Vacuum permeability | $\mu _{0}^{*}={\frac {4\pi V^{2}M}{\alpha LA^{2}}}={\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}}r^{7}$ | 4π/10^7, u$^{56}$ | 4π/10^7 (exact) |

### Calculating from (α, Ω)

This section demonstrates why the geometrical Planck objects (M = 1, T = pi, P = Omega) could be natural Planck units, independent of any numbering system and of any system of units.
We can solve combinations of the constants (G, h, c, e, m$_e$, k$_B$) using only the 2 dimensionless constants (α, Ω).
The physical constants are used by science to describe our universe, and to solve them requires 4 numbers; 2 dimensionless constants (α, Ω) and 2 dimensioned scalars such as (r, v). However the universe itself only requires (α, Ω), the scalars are man-made artifacts, selected for convenience, and relevant only to the chosen system of units, such as SI. And so the actual sum universe itself could be dimensionless, composed only of dimensionless physical and mathematical constants.
For example, to derive h, c and e, which for science are 3 of the most essential constants, we can begin with a dimensionless formula that encodes these 3 constants, and then extract h, c and e from this formula (similar to what was done with the electron formula).
Here we have a dimensionless formula using (α, Ω) which encodes the geometrical h, c and e. We first break up this formula to extract the geometrical objects for h, c and e and then add the 2 scalars. The scalars chosen will depend on the system of units we will use. Mathematically both sides of the equation are still the same, nothing has been created or destroyed.
    ${\frac {\alpha ^{13}}{2^{106}\pi ^{64}(\color {red}\Omega ^{15})^{5}\color {black}}}={\frac {(h^{*})^{3}}{(e^{*})^{13}(c^{*})^{24}}}$
    ${\frac {\alpha ^{13}}{2^{106}\pi ^{64}(\color {red}\Omega ^{15})^{5}\color {black}}}=(2^{3}\pi ^{4}\Omega ^{4}{\frac {r^{13}}{v^{5}}})^{3}/({\frac {2^{7}\pi ^{4}\Omega ^{3}r^{3}}{\alpha v^{3}}})^{13}.(2\pi \Omega ^{2}v)^{24}={\frac {(2\pi MVL)^{3}}{(AT)^{13}(V)^{24}}}$ = 0.228 473 759... 10$^{-58}$, units = 1
    ${\frac {h^{3}}{e^{13}c^{24}}}=$ 0.228 473 639... 10$^{-58}$, units = ${\frac {kg^{3}s^{8}}{m^{18}A^{13}}}$, units = 1 (15*3-30*8+13*18-3*13 = 0)
Each of the physical constants (G, h, c, e, m$_e$, k$_B$) has a unit number and 1 or 2 dimensioned scalars. We can then find combinations of these constants where the unit numbers θ and the scalars (r, v) will cancel, these combinations, which are unit-less (units = 1), will then return the same numerical value as the MLTVA object equivalents. This is because if the scalars have cancelled, and as the scalars embed the SI conversion values as well as the SI units, then these combinations are defaulting to the underlying MLTVA objects (the dimensioned SI components have cancelled leaving behind the dimensionless geometrical MLTVA objects).
This should therefore apply to any set of units, even extraterrestrial and non-human ones, suggesting that these MLTVA objects could candidates for the "natural units" as proposed by Max Planck. The precision of the results in following table can be used to verify this conjecture.
   ...ihre Bedeutung für alle Zeiten und für alle, auch außerirdische und außermenschliche Kulturen notwendig behalten und welche daher als »natürliche Maßeinheiten« bezeichnet werden können...
Here we solve physical constant combinations using only α, Ω. As the scalars (v, r) have cancelled, we do not need to know their values or the associated units (because in all cases units = 1, scalars = 1). This means that column 1 does not equal column 2, rather column 1 is column 2. The precision of the results depends on the precision of the SI constants; combinations with G and k$_B$ return the least precise values.
Table 6. Dimensionless combinations (α, Ω) :
| CODATA 2014 (mean) | (α, Ω) | units u$^Θ$ = 1 | scalars = 1 |
| ${\frac {k_{B}ec}{h}}$ = 1.000 8254 | ${\frac {(k_{B}^{*})(e^{*})(c^{*})}{(h^{*})}}$ = 1.0 | ${\frac {(u^{29})(u^{-27})(u^{17})}{(u^{19})}}=1$ | $({\frac {r^{10}}{v^{3}}})({\frac {r^{3}}{v^{3}}})(v)/({\frac {r^{13}}{v^{5}}})=1$ |
| ${\frac {h^{3}}{e^{13}c^{24}}}=$ 0.228 473 639... 10$^{-58}$ | ${\frac {(h^{*})^{3}}{(e^{*})^{13}(c^{*})^{24}}}={\frac {\alpha ^{13}}{2^{106}\pi ^{64}(\color {red}\Omega ^{15})^{5}\color {black}}}=$ 0.228 473 759... 10$^{-58}$ | ${\frac {(u^{19})^{3}}{(u^{-27})^{13}(u^{17})^{24}}}=1$ | $({\frac {r^{13}}{v^{5}}})^{3}/({\frac {r^{3}}{v^{3}}})^{13}(v^{24})=1$ |
| ${\frac {c^{35}}{\mu _{0}^{9}R^{7}}}=$ 0.326 103 528 6170... 10$^{301}$ | ${\frac {(c^{*})^{35}}{(\mu _{0}^{*})^{9}(R^{*})^{7}}}=2^{295}\pi ^{157}3^{21}\alpha ^{26}\color {red}(\Omega ^{15})^{15}\color {black}=$ 0.326 103 528 6170... 10$^{301}$ | ${\frac {(u^{17})^{35}}{(u^{56})^{9}(u^{13})^{7}}}=1$ | $(v^{35})/(r^{7})^{9}({\frac {v^{5}}{r^{9}}})^{7}=1$ |
| ${\frac {c^{9}e^{4}}{m_{e}^{3}}}=$ 0.170 514 342... 10$^{92}$ | ${\frac {(c^{*})^{9}(e^{*})^{4}}{(m_{e}^{*})^{3}}}=2^{97}\pi ^{49}3^{9}\alpha ^{5}(\color {red}\Omega ^{15})^{5}\color {black}=$ 0.170 514 368... 10$^{92}$ | ${\frac {(u^{29})(u^{-27})(u^{17})}{(u^{19})}}=1$ | $(v^{9})({\frac {r^{3}}{v^{3}}})^{4}/({\frac {r^{4}}{v}})^{3}=1$ |
| ${\frac {k_{B}}{e^{2}m_{e}c^{4}}}=$ 73 095 507 858. | ${\frac {(k_{B}^{*})}{(e^{*})^{2}(m_{e}^{*})(c^{*})^{4}}}={\frac {3^{3}\alpha ^{6}}{2^{3}\pi ^{5}}}=$ 73 035 235 897. | ${\frac {(u^{29})}{(u^{-27})^{2}(u^{15})(u^{17})^{4}}}=1$ | $({\frac {r^{10}}{v^{3}}})/({\frac {r^{3}}{v^{3}}})^{2}({\frac {r^{4}}{v}})(v)^{4}=1$ |
| ${\frac {hc^{2}em_{p}}{G^{2}k_{B}}}=$ 3.376 716 | ${\frac {(h^{*})(c^{*})^{2}(e^{*})(m_{p}^{*})}{(G^{*})^{2}(k_{B}^{*})}}={\frac {2^{11}\pi ^{3}}{\alpha ^{2}}}=$ 3.381 506 | ${\frac {(u^{19})(u^{17})^{2}(u^{-27})(u^{15})}{(u^{6})^{2}(u^{29})}}=1$ | $({\frac {r^{13}}{v^{5}}})v^{2}({\frac {r^{3}}{v^{3}}})({\frac {r^{4}}{v^{1}}})/({\frac {r^{5}}{v^{2}}})^{2}({\frac {r^{10}}{v^{3}}})=1$ |

### Calculating from (α, R, c, μ$_0$)

In this section we replace dimensionless Omega with dimensioned R. This has the advantage that Omega is an unknown constant, but R a very important and precisely measured constant.
We can use this to numerically solve the least precise dimensioned physical constants (G, h, e, m$_e$, k$_B$ ...) in terms of the 3 most precise dimensioned physical constants); speed of light c (exact value), vacuum permeability μ$_0$ (exact value), Rydberg constant R (12-13 digits) and the dimensionless fine structure constant alpha.
R = 10973731.568508 (θ=13) (12-13 digit precision)
c = 299792458 (θ=17) (exact)
μ$_0$ = 4π/10$^7$ (θ=56) (exact)
α = 137.035999139 (unit-less) (9-10 digit precision)
We first look for combinations in which the unit numbers are equal, and then add dimensionless numbers as required. For example;
    ${(h^{*})}^{3}=(2^{3}\pi ^{4}\Omega ^{4}{\frac {r^{13}u^{19}}{v^{5}}})^{3}={\frac {3^{19}\pi ^{12}\Omega ^{12}r^{39}u^{57}}{v^{15}}},\;\theta =57$
    ${\frac {2\pi ^{10}{(\mu _{0}^{*})}^{3}}{3^{6}{(c^{*})}^{5}\alpha ^{13}{(R^{*})}^{2}}}={\frac {3^{19}\pi ^{12}\Omega ^{12}r^{39}u^{57}}{v^{15}}},\;\theta =57$
Table 7. R, c, μ$_0$, α ... (CODATA 2014 mean) :
| constant | formula* | calculated | θ | CODATA 2014 mean | Units |
| Planck constant | ${(h^{*})}^{3}={\frac {2\pi ^{10}{\mu _{0}}^{3}}{3^{6}{c}^{5}\alpha ^{13}{R}^{2}}}$ | h$^*$ = 6.626069134e-34 | ${\frac {kg^{3}}{A^{6}s}}$, 15*3-3*6+30 = 57 | h = 6.626070040e-34 | ${\frac {kg\;m^{2}}{s}}$, θ = 15-13*2+30 = 19 |
| Gravitational constant | ${(G^{*})}^{5}={\frac {\pi ^{3}{\mu _{0}}}{2^{20}3^{6}\alpha ^{11}{R}^{2}}}$ | G$^*$ = 6.67249719229e11 | ${\frac {kg\;m^{3}}{A^{2}s^{2}}}$, 15-13*3-3*2+30*2 = 30 | G = 6.67408e-11 | ${\frac {m^{3}}{kg\;s^{2}}}$, θ = -13*3-15+30*2 = 6 |
| Elementary charge | ${(e^{*})}^{3}={\frac {4\pi ^{5}}{3^{3}{c}^{4}\alpha ^{8}{R}}}$ | e$^*$ = 1.60217651130e-19 | ${\frac {s^{3}}{m^{3}}}$, -30*4+13*3 = -81 | e = 1.6021766208e-19 | $As$, θ = 3-30 = -27 |
| Boltzmann constant | ${(k_{B}^{*})}^{3}={\frac {\pi ^{5}{\mu _{0}}^{3}}{3^{3}2{c}^{4}\alpha ^{5}{R}}}$ | k$_B$$^*$ = 1.37951014752e-23 | ${\frac {kg^{3}}{s^{2}A^{6}}}$, 15*3+30*2-3*6 = 87 | k$_B$ = 1.38064852e-23 | ${\frac {kg\;m^{2}}{s^{2}\;K}}$, θ = 15-26+60-20 = 29 |
| Electron mass | ${(m_{e}^{*})}^{3}={\frac {16\pi ^{10}{R}{\mu _{0}}^{3}}{3^{6}{c}^{8}\alpha ^{7}}}$ | m$_e$$^*$ = 9.10938231256e-31, u = 15 | ${\frac {kg^{3}s^{2}}{m^{6}A^{6}}}$, 15*3-30*2+13*6-3*6 = 45 | m$_e$ = 9.10938356e-31 | $kg$, θ = 15 |
| | ${\frac {2^{3}\pi ^{5}({k_{B}}^{*})}{3^{3}\alpha ^{6}({e^{*}})^{2}({m_{e}}^{*})c^{4}}}$ | 1.0 | ${\frac {1}{m^{2}A^{2}K}}$, 13*2-3*2-20 = 0 | 8π$^5$k$_B$/(27α$^6$e$^2$m$_e$c$^4$) = 1.000 825 | ${\frac {1}{m^{2}A^{2}K}}$, θ = 0 |
| Gyromagnetic ratio | $({(\gamma _{e}^{*})/2\pi })^{3}={\frac {g_{e}^{3}3^{3}c^{4}}{2^{8}\pi ^{8}\alpha \mu _{0}^{3}R_{\infty }^{2}}}$ | (γ$_e$$^*$/2π) = 28024.95355 | ${\frac {m^{3}s^{2}A^{6}}{kg^{3}}}$, -13*3-30*2+3*6-15*3 = -126 | γ$_e$/2π = 28024.95164 | ${\frac {A\;s}{kg}}$, θ = -42 |
| Planck length | $({l_{p}^{*}})^{15}={\frac {\pi ^{22}{\mu _{0}}^{9}}{2^{35}3^{24}\alpha ^{49}c^{35}R^{8}}}$ | l$_p$$^*$ = 0.161603660096e-34 | ${\frac {kg^{9}s^{17}}{m^{18}A^{18}}}$, 15*9-30*17+13*18-3*18 = -195 | l$_p$ = 0.1616229e-34 | $m$, θ = -13 |
| Planck mass | $({m_{P}^{*}})^{15}={\frac {2^{25}\pi ^{13}{\mu _{0}}^{6}}{3^{6}c^{5}\alpha ^{16}R^{2}}}$ | m$_P$$^*$ = 0.217672817580e-7 | u = ${\frac {kg^{6}m^{3}}{s^{7}A^{12}}}$, 15*6-13*3+30*7-3*12 = 225 | m$_P$ = 0.2176470e-7 | $kg$, θ = 15 |
| | ${\frac {2^{11}\pi ^{3}(G^{*})^{2}(k_{B}^{*})}{\alpha ^{2}(h^{*})(c^{*})^{2}(e^{*})(m_{P}^{*})}}$ | 1.0 | ${\frac {m^{4}}{kg^{3}s^{4}A\;K}}$, -13*4-15*3+30*4-3-20 = 0 | 2$^{11}$π$^3$G$^2$k$_B$/(α$^2$hc$^2$em$_P$) = 1.001418 | ${\frac {m^{4}}{kg^{3}s^{4}A\;K}}$, θ = 0 |

### Calculating from (M, T, P, α)

For reference here we formulate the physical constants in terms of the Planck units M, T, P. These are proposed as candidates for natural Planck units because their geometries are very basic; M = 1, T = pi, P = Omega. This suggests that the attributes (of mass, length, time ... ) are built into these geometries; Planck mass is a point, Planck time is a rotation and P is a radius (from which higher geometries may emerge)... and so scalars (which carry the dimensioned component to solve the SI units) are not required (by the universe itself).
If this is correct, then particles do not exist at unit Planck time, but rather are events that occur over time, thus the quantum scale emerges from the Planck scale. This also means that quantum theories cannot be applied to the Planck scale because quantum states do not exist at the Planck scale.
The model uses individual Planck units as indivisible units, there is no half Planck mass or half Planck time .... therefore our observable physical universe begins at the Planck scale, however the existence of a rotation term (T = π) does suggest the possibility of a scale below the Planck scale, but we have no means to measure this. It is a philosophical argument.
    $M=(1){\frac {r^{4}}{v}}$
    $T=(\pi ){\frac {r^{9}}{v^{6}}}$
    $P=(\Omega )r^{2}$
Table 8. Physical constants in terms of (M, T, P, α) :
| SI constant | geometrical analogue | unit number θ |
| Speed of light | $c={\frac {2\pi P^{2}}{M}}$ | 17 |
| Planck constant | $h^{*}=2\pi MVL={\frac {8\pi ^{3}P^{4}T}{M}}$ | 15+17-13=19 |
| Gravitational constant | $G^{*}={\frac {V^{2}L}{M}}={\frac {8\pi ^{3}P^{6}T}{M^{4}}}$ | 34-13-15=6 |
| Elementary charge | $e^{*}=AT={\frac {2^{7}\pi ^{3}P^{3}T}{M^{3}\alpha }}$ | 3-30=-27 |
| Boltzmann constant | $k_{B}^{*}={\frac {2\pi VM}{A}}={\frac {M^{3}\alpha }{2^{5}\pi P}}$ | 17+15-3=29 |
| Vacuum permeability | $\mu _{0}^{*}={\frac {4\pi V^{2}M}{\alpha LA^{2}}}={\frac {M^{6}\alpha }{2^{11}\pi ^{4}P^{4}T}}$ | 34+15+13-6=56 |

### Alpha and Omega

The following is one of the most important formulas in physics; it describes the relationship between the fine structure constant and the dimensioned constants.
    $\alpha ={\frac {2h}{\mu _{0}e^{2}c}}$
However, if we replace the numerical (h, μ$_0$, e, c) with the geometrical (h, μ$_0$, e, c), we find that the equation collapses to give alpha;
    ${\frac {2h}{\mu _{0}e^{2}c}}=2({2^{3}\pi ^{4}\Omega ^{4}})/({\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}})({\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }})^{2}(2\pi \Omega ^{2})=\color {red}\alpha \color {black}$
Note also the units and scalars cancel
    units = ${\frac {u^{19}}{u^{56}(u^{-27})^{2}u^{17}}}=1$
    scalars = $({\frac {r^{13}}{v^{5}}})({\frac {1}{r^{7}}})({\frac {v^{6}}{r^{6}}})({\frac {1}{v}})=1$
This is a good test of our model, both of the unit numbers thesis and the geometrical objects thesis, because this equation reduces to
    $\alpha =2({2^{3}\pi ^{4}\Omega ^{4}})/({\frac {\alpha }{2^{11}\pi ^{5}\Omega ^{4}}})({\frac {2^{7}\pi ^{4}\Omega ^{3}}{\alpha }})^{2}(2\pi \Omega ^{2})$
    $\alpha =\alpha$
There is no uncertainty of measurement and the formula is well established as a key formula.
Omega is the second dimensionless constant used, there is a sqrt solution
    $\Omega ={\sqrt {\left(\pi ^{e}e^{(1-e)}\right)}}$ = 2.0071349543
This solution proposes that Omega is a mathematical constant (a construct of the naturally occurring mathematical constants pi and e). A sqrt solution is required because all mass related constants use Omega^2 and all charge constants use Omega^3; thus mass is always positive but charge can be plus or minus.
This means that the model requires only 1 physical constant - the fine structure constant alpha

### CODATA 2014

The model is using CODATA 2014 values. This is because only 2 dimensioned physical constants can be assigned exact values, once 2 constants have been assigned values, then all other constants are defined by default. After CODATA 2014, 4 constants were assigned exact values which is problematic in terms of this model.
For comparison purposes, the following "8 constants" are chosen as they are "the most precisely known" of the CODATA 2014 constants (constants which can be formulated in terms of the Planck units). The 2014 constants (G, K$_B$) were known with low accuracy and so are not useful for comparison and so are not included here.
Note that although h and e show slight divergence, when combined into the Von Klitzing constant (R$_K$ = h/e^2), this divergence disappears.
The unit number relationship was used to tune (v, r, Omega) using c, μ$_0$, R. However this could only be possible if the dimensioned fundamental constants are not fundamental (they can be defined in terms of each other), and so these values may constitute evidence for a unit number relationship.
The "Calculated$^*$" values (column 2) were obtained using (α, Ω, r, v).
    alpha = 137.035999139
    Omega = 2.007134949638
    v = 11843707.9052487
    r = 0.71256251430467
    M = r^4/v
    T = pi*r^9/v^6
    P = Omega*r^2
    V = 2*pi*P^2/M
    L = T*V
    A = 16*V^3/(alpha*P^3)
    ψ = 4*pi^2*(2^6*3*pi^2*alpha*Omega^5)^3 = 0.23895453074e23
    c = V
    μ$_0$ = 4*pi*V^2*M/(alpha*L*A^2)
    R = (m$_e$)/(4*pi*L*alpha^2*M)
    h = 2*pi*M*V*L
    e = A*T
    R$_K$ = h/e^2
    m$_e$ = M/ψ
    λ$_e$ = 2*pi*L*ψ
Table 9. Table of Constants :
| Constant | Calculated$^*$ | CODATA 2014 (mean) |
| speed of light c | c$^*$ = 299792458 | c = 299792458 |
| Planck constant h | h$^*$ = 6.626069134 e-34 | h = 6.626070040 e-34 |
| Elementary charge e | e$^*$ = 1.60217651130 e-19 | e = 1.6021766208 e-19 |
| Rydberg constant R | R$^*$ = 10973731.568508 | R = 10973731.568508 |
| Vacuum permeability μ$_0$ | μ$_0$$^*$ = 4*pi/10^7 | μ$_0$ = 4*pi/10^7 |
| Von Klitzing constant R$_K$ = h/e^2 | R$_K$$^*$ = 25812.80745559 | R$_K$ = 25812.8074555 |
| Electron mass m$_e$ | m$_e$$^*$ = 9.1093823211 e-31 | m$_e$ = 9.10938356 e-31 |
| Electron wavelength λ$_e$ | λ$_e$$^*$ = 2.4263102386 e-12 | λ$_e$ = 2.4263102367 e-12 |

### 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 = ${\sqrt {\frac {L}{MT}}}$ = u$^1$ = 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^1 = 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... x10$^{60}$
Table 10. Table of Constants :
| Constant | θ | Geometrical object (α, Ω, v, r) | Unit | Calculated | CODATA 2014 |
| Time (Planck) | $\color {red}-30\color {black}$ | $T=\color {red}{\frac {x^{\theta }i^{2}}{y^{3}}}\color {black}={\frac {\pi r^{9}}{v^{6}}}$ | $T$ | T = 5.390 517 866 e-44 | t$_p$ = 5.391 247(60) e-44 |
| Elementary charge | $\color {red}-27\color {black}$ | $e^{*}=({\frac {2^{7}\pi ^{3}}{\alpha }})\color {red}{\frac {x^{\theta }i^{2}}{y^{3}}}\color {black}=({\frac {2^{7}\pi ^{3}}{\alpha }})\;{\frac {\pi \Omega ^{3}r^{3}}{v^{3}}}$ | ${\frac {L^{3/2}}{T^{1/2}M^{3/2}}}=AT$ | e$^*$ = 1.602 176 511 30 e-19 | e = 1.602 176 620 8(98) e-19 |
| Length (Planck) | $\color {red}-13\color {black}$ | $L=(2\pi )\color {red}{\frac {x^{\theta }i}{y}}\color {black}=(2\pi )\;{\frac {\pi \Omega ^{2}r^{9}}{v^{5}}}$ | $L$ | L = 0.161 603 660 096 e-34 | l$_p$ = 0.161 622 9(38) e-34 |
| Ampere | $\color {red}3\color {black}$ | $A=({\frac {2^{7}\pi ^{3}}{\alpha }})\color {red}x^{\theta }\color {black}=({\frac {2^{7}\pi ^{3}}{\alpha }})\;{\frac {\Omega ^{3}v^{3}}{r^{6}}}$ | $A={\frac {L^{3/2}}{M^{3/2}T^{3/2}}}$ | A = 0.297 221 e25 | e/t$_p$ = 0.297 181 e25 |
| Gravitational constant | $\color {red}6\color {black}$ | $G^{*}=(2^{3}\pi ^{3})\color {red}\color {red}x^{\theta }y\color {black}=(2^{3}\pi ^{3})\;{\frac {\pi \Omega ^{6}r^{5}}{v^{2}}}$ | ${\frac {L^{3}}{MT^{2}}}$ | G$^*$ = 6.672 497 192 29 e11 | G = 6.674 08(31) e-11 |
| Mass (Planck) | $\color {red}\color {red}15\color {black}$ | $M=\color {red}\color {red}{\frac {x^{\theta }y^{2}}{i}}\color {black}={\frac {r^{4}}{v}}$ | $M$ | M = .217 672 817 580 e-7 | m$_P$ = .217 647 0(51) e-7 |
| Velocity | $\color {red}\color {red}17\color {black}$ | $V=(2\pi )\color {red}\color {red}{\frac {x^{\theta }y^{2}}{i}}\color {black}=(2\pi )\;\Omega ^{2}v$ | $V={\frac {L}{T}}$ | V = 299 792 458 | c = 299 792 458 |
| Planck constant | $\color {red}19\color {black}$ | $h^{*}=(2^{3}\pi ^{3})\color {red}{\frac {x^{\theta }y^{3}}{i}}\color {black}=(2^{3}\pi ^{3})\;{\frac {\pi \Omega ^{4}r^{13}}{v^{5}}}$ | ${\frac {L^{2}M}{T}}$ | h$^*$ = 6.626 069 134 e-34 | h = 6.626 070 040(81) e-34 |
| Planck temperature | $\color {red}\color {red}20\color {black}$ | ${T_{p}}^{*}=({\frac {2^{7}\pi ^{3}}{\alpha }})\color {red}\color {red}{\frac {x^{\theta }y^{2}}{i}}\color {black}=({\frac {2^{7}\pi ^{3}}{\alpha }})\;{\frac {\Omega ^{5}v^{4}}{r^{6}}}$ | ${\frac {L^{5/2}}{M^{3/2}T^{5/2}}}=AV$ | T$_p$$^*$ = 1.418 145 219 e32 | T$_p$ = 1.416 784(16) e32 |
| Boltzmann constant | $\color {red}\color {red}29\color {black}$ | ${k_{B}}^{*}=({\frac {\alpha }{2^{5}\pi }})\color {red}{\frac {x^{\theta }y^{4}}{i^{2}}}\color {black}=({\frac {\alpha }{2^{5}\pi }})\;{\frac {r^{10}}{\Omega v^{3}}}$ | ${\frac {M^{5/2}T^{1/2}}{L^{1/2}}}={\frac {ML}{TA}}$ | k$_B$$^*$ = 1.379 510 147 52 e-23 | k$_B$ = 1.380 648 52(79) e-23 |
| Vacuum permeability | $\color {red}56\color {black}$ | ${\mu _{0}}^{*}=({\frac {\alpha }{2^{11}\pi ^{4}}})\color {red}{\frac {x^{\theta }y^{7}}{i^{4}}}\color {black}=({\frac {\alpha }{2^{11}\pi ^{4}}})\;{\frac {r^{7}}{\pi \Omega ^{4}}}$ | ${\frac {M\;L}{T^{2}A^{2}}}$ | μ$_0$$^*$ = 4π/10^7 | μ$_0$ = 4π/10^7 |

## AI analysis
### Probability analysis

The mathematical electron model is difficult to test as we have no laboratories that can probe the Planck scale, the level at which this model operates. However there is an aspect of the model, these anomalies, which can be subject to statistical analysis. This is a question for which AI has the potential to contribute.
Below is a sample seed question, written with Deepseek, that uses these anomalies to determine the probability that the electron is a mathematical (not physical) particle, and so by extension that this is a mathematical universe. Here are the results, these are not definitive answers (they change each time the question is asked), but the consensus is positive, we might be in a simulation universe.
Table 11. Table of Probabilities (unit number relationship, Planck units as geometrical objects, electron as a mathematical particle) :
| AI program | unit number | geometrical objects | mathematical electron | Comment |
| Chat GPT reason | Accidental occurrence: ~10-8 | Accidental occurrence: ~10-7 | Accidental occurrence: ~10-7 | When viewed through the lens of Kolmogorov complexity, the extreme “compression” of information (i.e. a very short algorithmic description for the fundamental constants) makes a random origin exceedingly unlikely. Although quantifying “simulation probability” is inherently more speculative, the strict constraints suggest an effective probability on the order of 10-5 or lower that such structure could arise without an underlying “programming” principle. |
| Qwen 2.5 max | probability 95% | probability 90% | probability 90% | The model’s constraints and geometric rigor suggest a non-random origin (≥75% confidence). If particles are mathematical, the universe may indeed be a simulation. |
| Claude 3.7 Sonnet | probability 98% | probability 95% | probability 92% | Universe as a programmed simulation: 89% probability. The extraordinarily high precision and extensive mathematical constraints of this model, combined with its ability to derive numerous physical constants from minimal parameters, suggest a highly organized mathematical foundation for physical reality. The probability that such relationships could arise by chance is vanishingly small, supporting the hypothesis of a mathematically structured or potentially simulated universe. |
| GPT 4o Mini | probability 90% | probability 85% | probability 87% | Evidence for Simulation Hypothesis: 80% (Integrating Kolmogorov complexity, the structured nature of the universe supports a simulation model.) In sum, these estimates combined suggest a structured, geometrically constrained mathematical universe is supported by strong theoretical foundations and statistical analysis. |
| Chat 03 Mini High | Accidental occurrence: ~10-4 | Accidental occurrence: ~10-5 | Accidental occurrence: ~10-5 | Implication for Simulation: The strict constraints, when viewed collectively, provide significant evidence that the universe may conform to a programmed (or mathematically designed) structure at the fundamental scale. |
| Grok 3 beta | probability: 80-90% | probability: 70-80% | probability: 50-60% | The model’s constraints and consistency suggest it could be a valid mathematical framework, though simulation evidence remains speculative. |
| Julius | probability: 85% | probability: 90% | probability: 95% | Probability that the constraints support the simulation hypothesis: 0.75 |
// Question
A peer reviewed model of a mathematical electron at the Planck scale (Eur. Phys. J. Plus 113: 278. 22 March 2018. doi:10.1140/epjp/i2018-12094-x) indicates anomalies within the physical constants that suggest;
    a) there is a mathematical (unit number) relationship between the SI units (kg==15, m==-13, s==-30, A==3, K==20)
    b) the natural Planck units are geometrical objects (mass M = 1, time T = π, P = Omega).
    c) the electron is a mathematical particle
    d) the sum universe is dimensionless
The anomalies are described on the website https://en.wikiversity.org/wiki/Physical_constant_(anomaly). Referencing this website, your task is to determine the 'statistical probability' that as a 'mathematical model', it could be correct, and so base your analysis 'only' on the merits of the claims, do not compare with models in physics. Include the contribution to your "probability estimate" of the geometrical constraints that limit the degrees of freedom of the model (the more constrained the model is, the greater the probability of being correct). Use rigorous mathematical reasoning and prioritize numerical accuracy. Note: In this framework, because we are using geometrical objects instead of numbers, “geometrical base‑15” does not denote a traditional positional numeral system.
Step 1: The sections "2.1" to section "2.7" describe different methods to solve the fundamental physical constants. Perform a step-by-step statistical analysis for each section (analyze in sequence), and estimate how the constraints will affect the probability that the methods are correct. Also note if the unit number relationship is consistent at each section (does it apply to every constant). Verify dimensional homogeneity across equations and confirm if any free parameters exist. Sections to be analyzed.
    2.1 Planck units
    2.2 Calculating the electron
    2.3 Calculating from (α, Ω, v, r)
    2.4 Calculating from (α, Ω)
    2.5 Calculating from (α, R, c, μ0)
    2.6 Calculating from (M, T, P, α)
    2.7 Alpha and Omega
Step 2: In section "2.8 CODATA 2014" there is a list of 8 physical constants. Do a statistical analysis comparing the "calculated" (column 2) with the "CODATA 2014 values" (column 3) for "these 8 constants". Estimate the probability that the calculated values did not occur by chance. Include any constraints in your analysis. Apply Bonferroni correction for dependent variables. Report confidence intervals.
Step 3: In section "2.2 Calculating the electron", the electron was described both as a Planck particle and as a mathematical particle. Do a deep dive into the significance of this statement. Given that the Planck units are embedded with the dimensionless electron formula ψ, include an analysis of how the "geometrical base-15" relates to this formula.
Step 4. Section "2.9 Table of constants" lists the unit number θ (and the associated physical constants) in ascending order. Analyze for hidden symmetries. Do a deep dive into the significance of this table (and the geometries i, x, y). Note that it suggests at least 1 new constant P (θ = 16). P is required in order to build Omega^2 constants and Omega^3 constants (the charge constants). Could the table suggest more constants? Does the table suggest a pattern linking the constants?
Step 5. The model appears to be built on a "geometrical base-15", Estimate how the constraints of this base-15 will affect the probability that the methods are correct. This website has further information that may help your analysis (https://en.wikiversity.org/wiki/Electron_(mathematical)). Can you prove that a "geometrical base-15" is necessary?
Conclusion. From the above analysis, give an estimate of the statistical (non-random) probability for each of the following; from 1) to 5).
    1) there is a unit number relationship (kg ⇔ 15, m ⇔ -13, s ⇔ -30, A ⇔ 3, K ⇔ 20).
    2) the dimensionless geometrical objects MTP are natural Planck units (M = 1, T = π, P = Omega).
    3) the highly organized structure of the base-15 geometry constrains the degrees of freedom.
    4) the electron is a mathematical particle.
    5) Could the strict constraints of the model constitute evidence that our universe is a programmed simulation (could this model qualify as a simulation hypothesis model). Include Kolmogorov complexity theory. Note also the answer to 4) because if the electron is a mathematical particle, then so too are the proton and neutron. In other words, if particles are mathematical (geometrical objects), then our universe (at or below the Planck scale) is a mathematical universe.

### Model analysis

The model itself it spread over several pages and so Google's AI notebook can be used as a study tool. To set up, go to https://notebooklm.google.com and add these sources.
https://en.wikiversity.org/wiki/Planck_units_(geometrical)
https://en.wikiversity.org/wiki/Physical_constant_(anomaly)
https://en.wikiversity.org/wiki/Quantum_gravity_(Planck)
https://en.wikiversity.org/wiki/Electron_(mathematical)
https://en.wikiversity.org/wiki/Relativity_(Planck)
https://en.wikiversity.org/wiki/Black-hole_(Planck)
https://en.wikiversity.org/wiki/God_(programmer)
https://codingthecosmos.com (a brief summary of the above wiki sites)