Physical constant (anomaly)

Anomalies within the dimensioned physical constants (G, h, c, e, me, kB) 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, kB) having independently assigned exact values (they cannot be derived in terms of each other), and this confirmed the independence of their associated SI units as shown in this table.

However there are anomalies which occur in certain combinations of the physical constants (G, h, c, e, me, kB) which suggest a mathematical relationship between the units (kg, m, s, A, K). In order for the dimensioned physical constants to be fundamental, the units must be independent of each other, there cannot be a unit number relationship ... however these anomalies question this fundamental assumption. Every combination predicted by the model returns an answer consistent with CODATA precision. Statistically therefore, can these anomalies be dismissed as coincidence?

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 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: $$r = (\frac{kg\;m}{s})^{1/4}$$


 * 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 relationship to the SI units kg, m, s, A, K.

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 the number of apples in a container, numbers such as 299792458 carry no unit-specific information, and so the units are treated as independent by default.

This can be resolved by assigning to each Planck unit a geometrical object (denoted MLTVA), and for which the geometry embeds the attribute (for example, the geometry of the time object T embeds the function time and so a descriptive s is not required). We may then combine these objects to form more complex objects; from electrons to galaxies, while still retaining the underlying attributes (of mass, wavelength, frequency …).

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, me, KB...).

Table 2. is an example of object orientated units; it assigns MLTA objects as the geometry of 2 dimensionless physical constants; the Sommerfeld fine structure constant alpha and Omega. As alpha and Omega are dimensionless (alpha = 137.035999084, Omega = 2.0071349496), so too are these objects.

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 fe (below) is one example of an f(X) structure

The dimensioned constants in terms of MLTA as the Planck units

As Alpha and Omega can have (dimensionless) numerical values, we can use dimensioned numerical scalars to convert from the MLTVA objects to their SI equivalents.

For example, we can use scalar v to convert from dimensionless geometrical object V to dimensioned c.


 * scalar vSI = 11843707.905 m/s gives c = V*vSI = 25.3123819 * 11843707.905 m/s = 299792458 m/s (SI units)


 * scalar vimp = 7359.3232155 miles/s gives c = V*vimp = 186282 miles/s (imperial units)

As scalar v also carries the unit designation m/s or miles/s, then it is dimensioned, and so the unit number relationship θ applies to the scalar itself, for scalar v the unit number (θ = 17). Here each attribute is assigned a scalar.

Because the scalars follow the unit number relationship (units as uθ), 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. Here the scalars are each defined in terms of the 2 scalars r, v.

Calculating the electron
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.


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


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


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

Electron parameters
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;

electron mass    $$m_e^* = \frac{M}{f_e}$$ (M =  Planck mass = $$\frac{r^4}{v}) = 0.910 938 232 11 \;x10^{-30}\;u^{15}$$

electron wavelength $$\lambda_e^* = 2\pi L f_e$$ (L = Planck length = $$2\pi\Omega^2\frac{r^9}{v^5}) = 0.242 631 023 86 \;x10^{-11}\;u^{-13}$$

elementary charge   $$e^* = A\;T$$ (T =  Planck time) = $$\frac{2^7 \pi^4 \Omega^3}{\alpha}\frac{r^3}{v^3} = 0.160 217 651 30 \;x10^{-18}\;u^{-27}$$

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

Calculating from (α, Ω)
If we can reduce the 5 SI units to 2 scalars (example; r, v in tables 5, 6), then we can find combinations of the physical constants (G, h, c, e, me, kB) where the unit numbers θ and the scalars 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 SI component has cancelled).

This should therefore apply to any set of units, even extraterrestrial and non-human ones, suggesting that these MLTVA objects could be 'natural' units, the precision of the results in following table can be used to verify this conjecture.

For example;


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


 * $$\frac{h^3}{e^{13} c^{24}} =$$ 0.228 473 639... 10-58

Here we solve physical constant combinations using only α, Ω (and the mathematical constants 2, 3, π). As the scalars (v, r) have cancelled, we do not need to know their values or the units. This means that column 1 does not equal column 2, rather column 1 (sans scalars) is column 2. The precision of the results depends on the precision of the SI constants; combinations with G and kB return the least precise values.

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

Calculating from (α, Ω, v, r)
Here the attributes are defined in terms of 2 scalars; from c (exact value) is v (θ = 17), and from μ0 (exact value) we can derive r (θ = 8), hence the rationale for choosing scalars r and v in table 1.


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


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

Calculating from (c, R, μ0, α)
By matching the unit numbers we can numerically solve the least precise dimensioned physical constants (G, h, e, me, kB ...) using the 3 most precise (CODATA 2014); speed of light c (exact value),  vacuum permeability μ0 (exact value),  Rydberg constant R (12-13 digits) and the dimensionless  fine structure constant alpha.

R = 10973731.568508 (θ=13)

c = 299792458 (θ=17)

μ0 = 4π/107 (θ=56)

α = 137.035999139 (θ=0)

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


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

Calculating from (α)
Combinations which reduce to the dimensionless (no scalars or units) fine structure constant. For example;


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


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


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

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


 * $$\color{red}i\color{black} = \pi^2 \Omega^{15}$$, units = $$\sqrt{(\frac{L^{15}}{M^9 T^{11}})}$$ = 1 (unit number θ = (-13*15) - (15*9) - (-30*11) = 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)