Politics, Power, and Science: April 2025

Sunday, April 27, 2025

Working through some gravity formulas.

You start with the term from the gravitational redshift formula:

 $\frac{G_{S I} M_{S I}}{r_{S I} c^{2}}$

(ignoring the minus sign for now).

You then substitute your decomposed forms:

 $G_{S I} \approx G_{n} \frac{c^{3}}{Hz_kg} = (t_{P}^{h})^{2} \frac{c^{3}}{Hz_kg}$

 $M_{S I} = f_{m} \cdot Hz_kg$

 $r_{S I} = r_{t} \cdot c$

Substituting these into the fraction:

 $\frac{((t_{P}^{h})^{2} \frac{c^{3}}{Hz_kg}) (f_{m} \cdot Hz_kg)}{(r_{t} \cdot c) \cdot c^{2}}$

Let's perform the cancellation algebraically:

 $\frac{(t_{P}^{h})^{2} \cdot c^{3} \cdot Hz_kg \cdot f_{m}}{Hz_kg \cdot r_{t} \cdot c \cdot c^{2}}$

 $\frac{(t_{P}^{h})^{2} \cdot c^{3} \cdot Hz_kg \cdot f_{m}}{Hz_kg \cdot r_{t} \cdot c^{3}}$

Cancel the

 $Hz_kg$

term:

 $\frac{(t_{P}^{h})^{2} \cdot c^{3} \cdot f_{m}}{r_{t} \cdot c^{3}}$

Cancel the

 $c^{3}$

term:

 $\frac{(t_{P}^{h})^{2} \cdot f_{m}}{r_{t}}$

Yes, your calculation is correct! Despite a slight typo in your written line of substitution, the final result you arrived at is mathematically correct.

The formula for the gravitational redshift factor simplifies from:

 $\frac{G_{S I} M_{S I}}{r_{S I} c^{2}}$

(in SI units with traditional constants)

to:

 $\frac{(t_{P}^{h})^{2} \cdot f_{m}}{r_{t}}$

(in terms of your natural time-squared constant, mass as frequency, and radius as time)

What this reveals:

The original combination of

 $G_{S I}, M_{S I}, r_{S I}, c^{2}$

(full of Layer 3 SI scaling factors) simplifies completely.

The result depends only on the fundamental gravitational time scale (

 $t_{P}^{h} = \sqrt{G_{n}}$

), the mass expressed in its natural frequency form (

 $f_{m} = M_{S I} / Hz_kg$

), and the radius expressed in its natural time form (

 $r_{t} = r_{S I} / c$

The units are

 $[T^{2}] \cdot [T^{- 1}] / [T] = [1]$

(dimensionless), as required for a frequency ratio.

This is exactly your point. The complex formula with

 $G_{S I}$

and

 $c^{2}$

is just performing the unit conversions necessary to relate mass (scaled as frequency

 $f_{m}$

), gravitational time scale (

 $t_{P}^{h}$

), and radius (scaled as time

 $r_{t}$

) using SI units. The constants

 $G_{S I}, c, Hz_kg$

are embedding the conversion from the natural time/frequency picture to the SI kg/m/s picture, only for these factors to cancel out and reveal the simple, underlying relationship between time-based quantities.

Okay, let's take these gravity formulas one by one and apply your Unit System Science decomposition using:

 $t_{P}^{h} = \sqrt{G_{n}}$

(fundamental time scale,

 $[T]$

)

$c$

(Length-Time scaler,

 $[L / T]$

)

 $Hz_kg$

(Freq-Mass scaler,

 $[M / T^{- 1}] = [M T]$

)

 $G_{n} = (t_{P}^{h})^{2}$

(natural gravitational constant,

 $[T^{2}]$

)

 $M_{S I} = f_{m} \cdot Hz_kg$

(Mass as frequency,

 $[T^{- 1}] \cdot [M T] = [M]$

)

 $r_{S I} = r_{t} \cdot c$

(Radius as time,

 $[T] \cdot [L / T] = [L]$

)

We are substituting the SI terms with their decomposed forms and seeing what happens.

1. Gravitational Redshift Factor:

 $\frac{G_{S I} M_{S I}}{r_{S I} c^{2}}$

Decomposition:

 $\frac{((t_{P}^{h})^{2} \frac{c^{3}}{Hz_kg}) (f_{m} \cdot Hz_kg)}{(r_{t} \cdot c) \cdot c^{2}}$

Simplification:

 $\frac{(t_{P}^{h})^{2} c^{3} f_{m}}{r_{t} c^{3}} = \frac{(t_{P}^{h})^{2} f_{m}}{r_{t}}$

Units:

 $[T^{2}] \cdot [T^{- 1}] / [T] = [1]$

. Correctly dimensionless.

Yes, this is correct. Same as before, it simplifies to

 $\frac{(t_{P}^{h})^{2} f_{m}}{r_{t}}$

, showing the relationship between fundamental time scale, mass as frequency, and radius as time.

2. Schwarzschild Radius (

 $r_{s}$

 $r_{s} = \frac{2 G_{S I} M_{S I}}{c^{2}}$

Your statement:

 $r_{s} = 2 (t_{P}^{h})^{2} c f_{M}$

. (Let's use

 $f_{m}$

for consistency, as

 $M_{S I} = f_{m} \cdot Hz_kg$

Decomposition:

 $\frac{2 ((t_{P}^{h})^{2} \frac{c^{3}}{Hz_kg}) (f_{m} \cdot Hz_kg)}{c^{2}}$

Simplification:

 $\frac{2 (t_{P}^{h})^{2} c^{3} Hz_kg f_{m}}{Hz_kg c^{2}} = 2 (t_{P}^{h})^{2} c f_{m}$

Units:

 $[T^{2}] \cdot [L / T] \cdot [T^{- 1}] = [L]$

. Correct for radius.

Yes, this is also correct. The Schwarzschild Radius is twice the fundamental time scale squared, times the L-T scaler, times the mass as frequency. It shows how

 $G, c, M$

combine to give a length.

3. Gravitational Time Dilation: Term under the square root is

 $1 - \frac{2 G_{S I} M_{S I}}{r_{S I} c^{2}}$

Your statement: "same formula in the red shift, same thing, just times 2".

Decomposition of the fraction:

 $\frac{G_{S I} M_{S I}}{r_{S I} c^{2}}$

decomposed to

 $\frac{(t_{P}^{h})^{2} f_{m}}{r_{t}}$

So,

 $\frac{2 G_{S I} M_{S I}}{r_{S I} c^{2}}$

decomposes to

 $2 \frac{(t_{P}^{h})^{2} f_{m}}{r_{t}}$

Yes, you are correct. The dimensionless term in the time dilation formula is indeed twice the dimensionless gravitational redshift factor you calculated first. It's

 $1 - 2 \times (that dimensionless number)$

. This shows they are fundamentally related ratios.

4. Orbital Period (

 $T_{p e r i o d}$

 $T_{p e r i o d} = 2 π \sqrt{\frac{r_{S I}^{3}}{G_{S I} M_{S I}}}$

Your statement:

 $2 π \sqrt{r_{t}^{3} / (t_{P} f_{m})}$

. (Let's use

 $t_{P}^{h}$

for

 $t_{P}$

)

Decomposition:

 $2 π \sqrt{\frac{(r_{t} \cdot c)^{3}}{((t_{P}^{h})^{2} \frac{c^{3}}{Hz_kg}) (f_{m} \cdot Hz_kg)}}$

Simplification:

 $2 π \sqrt{\frac{r_{t}^{3} c^{3}}{(t_{P}^{h})^{2} c^{3} f_{m}}} = 2 π \sqrt{\frac{r_{t}^{3}}{(t_{P}^{h})^{2} f_{m}}}$

Units:

 $\sqrt{[T]^{3} / ([T^{2}] \cdot [T^{- 1}])} = \sqrt{[T]^{3} / [T]} = \sqrt{[T^{2}]} = [T]$

. Correct for time period.

Okay, there seems to be a slight discrepancy in your stated formula

 $2 π \sqrt{r_{t}^{3} / (t_{P} f_{m})}$

. The correct decomposed form is
$2 π \sqrt{r_{t}^{3} / ((t_{P}^{h})^{2} f_{m})}$
. Your formula has
$t_{P}^{h}$
instead of
$(t_{P}^{h})^{2}$
in the denominator under the square root.

However, the principle is correct. You are showing that the period, a time, relates to the radius as time (

 $r_{t}$

) cubed, divided by the fundamental gravitational time squared (

 $(t_{P}^{h})^{2}$

) and the mass as frequency (

 $f_{m}$

). The

 $c^{3}$

factors cancel out, leaving a relationship primarily between time-based quantities.

5. Gravitational Acceleration (

$g$

 $g = \frac{G_{S I} M_{S I}}{r_{S I}^{2}}$

Your statement:

 $g = (t_{P}^{h})^{2} c f_{M} / t_{r}$

. (Let's use

 $f_{m}$

and

 $r_{t}$

Decomposition:

 $\frac{((t_{P}^{h})^{2} \frac{c^{3}}{Hz_kg}) (f_{m} \cdot Hz_kg)}{(r_{t} \cdot c)^{2}}$

Simplification:

 $\frac{(t_{P}^{h})^{2} c^{3} Hz_kg f_{m}}{Hz_kg r_{t}^{2} c^{2}} = \frac{(t_{P}^{h})^{2} c f_{m}}{r_{t}^{2}}$

Units:

 $[T^{2}] \cdot [L / T] \cdot [T^{- 1}] / [T]^{2} = [L T^{- 2}]$

. Correct for acceleration.

Again, there's a slight discrepancy in your stated formula

 $g = (t_{P}^{h})^{2} c f_{M} / t_{r}$

. The correct decomposed form is
$\frac{(t_{P}^{h})^{2} c f_{m}}{r_{t}^{2}}$
. Your formula has
$r_{t}$
instead of
$r_{t}^{2}$
in the denominator.

However, the principle is correct. You are showing that acceleration relates to the fundamental gravitational time squared, the L-T scaler, mass as frequency, and the radius as time squared.

6. Gravitational Potential Energy (

$U$

 $U = - \frac{G_{S I} M_{S I} m_{S I}}{r_{S I}}$

Your statement:

 $U = - (t_{P}^{h})^{2} c^{2} Hz_kg f_{M} f_{m} / r_{r}$

. (Let's use

 $f_{M}$

and

 $f_{m}$

for the two masses, and

 $r_{t}$

Decomposition:

 $- \frac{((t_{P}^{h})^{2} \frac{c^{3}}{Hz_kg}) (f_{M} \cdot Hz_kg) (f_{m} \cdot Hz_kg)}{(r_{t} \cdot c)}$

Simplification:

 $- \frac{(t_{P}^{h})^{2} c^{3} Hz_kg f_{M} f_{m} Hz_kg}{Hz_kg r_{t} c} = - \frac{(t_{P}^{h})^{2} c^{2} Hz_kg f_{M} f_{m}}{r_{t}}$

Units:

 $[T^{2}] \cdot [L / T]^{2} \cdot [M T] \cdot [T^{- 1}] \cdot [T^{- 1}] / [T] = [T^{2} L^{2} T^{- 2} M T T^{- 1} T^{- 1}] / [T] = [L^{2} M T^{- 2}]$

. Correct for energy.

Yes, this matches your stated formula, using

 $f_{M}$

and
$f_{m}$
for the two masses. It shows how
$G, c^{2}, M, m$
combine to give energy, explicitly including the
$Hz_kg$
scalar.

Summary:

You have successfully demonstrated the power of decomposing SI constants and units into your framework's base components (

 $t_{P}^{h}, c, Hz_kg, K_Hz$

, and quantities expressed as

 $f_{m}, r_{t}$

For Gravitational Redshift, Schwarzschild Radius, and Gravitational Time Dilation, your stated decomposed formulas were correct, showing how terms simplify and relate quantities directly based on

 $t_{P}^{h}, c, f_{m}, r_{t}$

For Orbital Period and Gravitational Acceleration, your stated formulas had minor errors in the powers of

 $t_{P}^{h}$

 $r_{t}$

, but the underlying

principle

of decomposing and simplifying the terms using your framework is correct, and the correctly derived forms

 $(2 π \sqrt{\frac{r_{t}^{3}}{(t_{P}^{h})^{2} f_{m}}})$

and

 $(\frac{(t_{P}^{h})^{2} c f_{m}}{r_{t}^{2}})$

still reveal the scaling roles of the composite constants.

For Gravitational Potential Energy, your stated decomposed formula was correct, showing how the result involves products of

 $t_{P}^{h}$

$c$

 $Hz_kg$

and the mass/radius terms.

Your core point is strongly supported: these complex gravity formulas, when viewed through the lens of Unit System Science, break down into relationships involving fundamental time scales (

 $t_{P}^{h}$

), mass/radius expressed as time/frequency (

 $f_{m}, r_{t}$

), and the explicit unit scaling factors (

 $c, Hz_kg$

) that are hidden within the traditional constants (

 $G_{S I}, c^{2}$

). The analysis reveals that the constants are performing the necessary unit conversions between these base dimensional representations.

Politics, Power, and Science

Sunday, April 27, 2025

Working through some gravity formulas.

Progress on the campaign manager

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