User: I don’t see earth’s rotation on your gps estimation. Why?

Assistant: In the WILL Relational Geometry framework, the Earth’s rotation is accounted for implicitly within the kinematic projection $\beta$ (and its orthogonal component $\beta_Y$), which forms part of the unified relational spacetime factor $\tau$. The framework states:

1. Unified Relational Scaling: Both kinematic ($S^1$) and gravitational ($S^2$) transformations act as independent projections of the same invariant energy. These are combined into a total local energy scale: \(E_{\mathrm{loc}} = \frac{E_0}{\tau} = \frac{E_0}{\beta_Y\,\kappa_X}\) This means the “GPS time shift” is derived from a “single symmetric law” where $\tau$ is the product of the kinematic phase factor ($\beta_Y$) and the gravitational phase factor ($\kappa_X$). (doc. Part I - Theorem: Equivalence of Inertial and Gravitational Response).

2. Kinematic Projection ($\beta$): The kinematic projection $\beta$ is defined as $v/c$, representing “how much of the universal ‘speed of change’ you see as motion through space, relative to yourself” (doc. Part I - Section: Energy as a Relation - What $\kappa$ and $\beta$ Actually Mean). The rotation of the Earth (or a satellite in orbit around it) involves a velocity, which directly contributes to this $\beta$ term.

3. GPS Estimation: The document explicitly states that the framework reproduces the “GPS time shift / gravitational redshift” and lists the value of 38.52 $\mu$s/day as verified (doc. Part I - Table: Classical GR results vs. WILL RG outcomes). This combined shift accounts for both the gravitational potential of Earth (influencing $\kappa_X$) and the relative kinematic motion of the satellites and observers (influencing $\beta_Y$).

Therefore, while “Earth’s rotation” is not listed as a separate line item or an explicit term in the formula for $\tau$, its kinematic effects are naturally subsumed within the calculation of the $\beta$ (and $\beta_Y$) factor for any system in motion relative to the observer’s frame. The framework’s core principle is to unify these effects into a single, composite relational measure, rather than treating them as separate classical corrections.

User: How you show the numbers for the gps calc?

Assistant: The provided document states that WILL Relational Geometry reproduces the GPS time shift of $38.52 \ \mu \text{s/day}$ and that this is “verified directly with GPS satellites” (doc. Part I - Table: Classical GR results vs. WILL RG outcomes).

However, the explicit step-by-step numerical calculation showing how this specific number is derived from the framework’s fundamental parameters ($\beta_Y$ and $\kappa_X$) is not detailed within the provided text.

The framework explains the conceptual basis for this calculation as follows:

1. Unified Relational Spacetime Factor ($\tau$): The total local energy scale, which underlies the time dilation effects observed in GPS, is given by a single symmetric law that combines both kinematic and gravitational effects. This is expressed through the factor $\tau$: \(E_{\mathrm{loc}} = \frac{E_0}{\tau} = \frac{E_0}{\beta_Y\,\kappa_X}\) Here, $\tau$ represents the combined phase factor (internal order) from both kinematic and gravitational transformations (doc. Part I - Theorem: Equivalence of Inertial and Gravitational Response).

2. Kinematic Phase Factor ($\beta_Y$): This component accounts for the time dilation due to the relative motion of the satellite. It is defined as: \(\beta_Y = \sqrt{1 - \beta^2} = \sqrt{1 - \left(\frac{v}{c}\right)^2}\) where $v$ is the kinematic velocity of the satellite relative to an Earth-bound observer (doc. Part I - Section: Kinetic Energy Projection on $S^1$).

3. Gravitational Phase Factor ($\kappa_X$): This component accounts for the time dilation due to the Earth’s gravitational potential at the satellite’s altitude. It is defined as: \(\kappa_X = \sqrt{1 - \kappa^2} = \sqrt{1 - \frac{R_s}{r}}\) where $R_s$ is the Schwarzschild radius of the Earth, and $r$ is the orbital radius of the satellite (doc. Part I - Section: Potential Energy Projection on $S^2$).

In essence, the framework claims that the observed $38.52 \ \mu \text{s/day}$ GPS time shift emerges directly from the multiplication of these two phase factors ($\beta_Y$ and $\kappa_X$), applied to the proper time of the satellite. This unified approach eliminates the need to calculate separate relativistic and gravitational corrections and then add them together, as is done in conventional physics.

Although the numerical derivation using specific values for GPS satellites is not explicitly laid out, the document asserts that this methodology yields results consistent with empirical data.

User: The only number I see in all that response is the result we are looking for. I want all the variables used with their numbers.

Assistant: You are correct that the provided document states the final result for the GPS time shift ($38.52 \ \mu \text{s/day}$) and the general conceptual formulas, but it does not explicitly list the specific numerical values for all the variables required to perform this calculation from first principles within the WILL Relational Geometry framework.

The document claims that the GPS time shift emerges from the combined phase factor $\tau = \beta_Y \kappa_X$ (doc. Part I - Theorem: Equivalence of Inertial and Gravitational Response). To numerically derive this value, one would typically need the following input variables, which are standard in GPS calculations but are not explicitly provided with their numerical values in the given text:

1. Kinematic Velocity of GPS Satellite ($v_{\text{GPS}}$): The orbital speed of a GPS satellite relative to Earth’s center.
   • This velocity is used to calculate the kinematic projection $\beta = v_{\text{GPS}}/c$, and subsequently $\beta_Y = \sqrt{1 - \beta^2}$.
   • (Typical value in conventional physics: $\approx 3.87 \times 10^3 \ \text{m/s}$)
2. Earth’s Mass ($M_{\text{Earth}}$): The mass of the Earth.
   • This is needed to calculate Earth’s Schwarzschild radius ($R_s = 2GM_{\text{Earth}}/c^2$), which in turn determines the gravitational potential projection $\kappa$.
   • (Typical value in conventional physics: $\approx 5.972 \times 10^{24} \ \text{kg}$)
3. GPS Satellite Orbital Radius ($r_{\text{GPS}}$): The distance of the GPS satellite from the center of the Earth.
   • This radius is used with Earth’s Schwarzschild radius ($R_s$) to calculate the gravitational projection $\kappa = \sqrt{R_s/r_{\text{GPS}}}$, and subsequently $\kappa_X = \sqrt{1 - \kappa^2}$.
   • (Typical value in conventional physics: $\approx 2.656 \times 10^7 \ \text{m}$)

The universal constants that are explicitly provided in the document and would be used in such a calculation are:

• Speed of Light ($c$): $\SI{2.99792e8}{\m\per\s}$ (doc. Part II - Section: Deriving $H_0$ from CMB Temperature and $\alpha$).
• Gravitational Constant ($G$): $\SI{6.674e-11}{\m\cubed\per\kg\per\s\squared}$ (doc. Part II - Section: Deriving $H_0$ from CMB Temperature and $\alpha$).

In summary, while the conceptual framework and the final verified result are stated, the intermediate numerical values for the specific Earth-GPS system (like $M_{\text{Earth}}$, $v_{\text{GPS}}$, $r_{\text{GPS}}$) are not provided in the supplied text. The document focuses on the methodology of how these values would combine via $\beta_Y$ and $\kappa_X$ to produce the observed time shift, rather than explicitly detailing the numerical steps for this specific example.

User: I still don’t see rotation speed of the earth on these numbers. That number is used for real calibration. Does that means rotation doesn’t affect or doesn’t know how to add it?

Assistant: The absence of an explicitly labeled “Earth’s rotation speed” in the provided conceptual formulas for the GPS time shift does not mean the WILL Relational Geometry framework ignores it or is unable to account for it. Instead, its effects are implicitly and consistently integrated into the framework’s fundamental kinematic projection $\beta$.

Here’s how Earth’s rotation is incorporated within WILL RG:

1. Kinematic Projection $\beta$ Encompasses All Relative Motion: The kinematic projection $\beta$ is defined as the ratio $v/c$, where $v$ is a measure of “how much of the universal ‘speed of change’ you see as motion through space, relative to yourself” (doc. Part I - Section: Energy as a Relation - What $\kappa$ and $\beta$ Actually Mean). When considering a clock on Earth’s surface, its velocity $v$ relative to a chosen frame (e.g., an Earth-centered inertial frame) would naturally include the velocity component due to Earth’s rotation. Therefore, Earth’s rotation contributes directly to the $\beta$ value (and thus $\beta_Y = \sqrt{1-\beta^2}$) of any observer or clock located on its surface.

2. Relational Nature of Time Dilation: In WILL RG, all physical quantities are defined by their relations. The “GPS time shift” is the observed difference in clock rates between a satellite and a ground station. Each of these clocks is subject to its own local kinematic ($\beta_Y$) and gravitational ($\kappa_X$) phase factors. The $\beta_Y$ for the ground station’s clock would inherently reflect its velocity due to Earth’s rotation. The overall time shift $\Delta T$ is a consequence of the difference between the satellite’s total time dilation factor $\tau_{\text{satellite}} = \beta_{Y,\text{satellite}}\,\kappa_{X,\text{satellite}}$ and the ground station’s factor $\tau_{\text{ground}} = \beta_{Y,\text{ground}}\,\kappa_{X,\text{ground}}$.

3. Unified Approach vs. Separate Corrections: Conventional GPS calculations typically involve adding separate relativistic effects:

   • Gravitational redshift (due to Earth’s potential).
   • Kinematic time dilation (due to satellite’s orbital speed).
   • Sagnac effect (due to Earth’s rotation, affecting signal travel time and synchronization). The WILL framework aims for ontological minimalism by unifying these into a single combined phase factor $\tau = \beta_Y\,\kappa_X$ (doc. Part I - Theorem: Equivalence of Inertial and Gravitational Response). The kinematic component $\beta_Y$ is understood to intrinsically capture all relevant relative velocities affecting a clock’s proper time rate, including those arising from the observer’s motion due to Earth’s rotation.

While the document states that the framework reproduces the $38.52 \ \mu \text{s/day}$ GPS time shift (doc. Part I - Table: Classical GR results vs. WILL RG outcomes), it does not provide the specific numerical breakdown of how velocities (including Earth’s rotation) for both satellite and ground station are fed into $\beta$ and then combined with $\kappa$ to reach this value. However, the theoretical structure is designed to absorb all such kinematic effects into the $\beta$ projection.

User: So if the earth rotated to higher speed your prediction would still be the same?

Assistant: No, the prediction for the GPS time shift would not be the same if the Earth rotated at a significantly higher speed.

Here’s why, based on the WILL Relational Geometry framework:

1. Kinematic Projection ($\beta$) is Affected by Earth’s Rotation: In WILL RG, the kinematic projection $\beta$ for any object is defined as $v/c$, where $v$ is its speed relative to a chosen frame (doc. Part I - Section: Energy as a Relation - What $\kappa$ and $\beta$ Actually Mean). A clock on Earth’s surface possesses a tangential velocity due to the planet’s rotation. If the Earth’s rotation speed were higher, this tangential velocity component for a ground-based clock ($v_{\text{ground, rotation}}$) would increase.

2. Impact on the Kinematic Phase Factor ($\beta_Y$): The kinematic phase factor is $\beta_Y = \sqrt{1 - \beta^2}$ (doc. Part I - Section: Kinetic Energy Projection on $S^1$).
   • If Earth rotated faster, the total velocity ($v_{\text{ground}}$) of a clock on its surface would increase, causing $\beta_{\text{ground}}$ to increase.
   • Consequently, $\beta_{Y,\text{ground}}$ would decrease (since it’s $\sqrt{1 - \beta_{\text{ground}}^2}$). A smaller $\beta_Y$ corresponds to a greater kinematic time dilation (a slower rate of proper time).
3. Effect on the Total Relational Spacetime Factor ($\tau$): The GPS time shift is a consequence of the difference in the total relational spacetime factor $\tau = \beta_Y\,\kappa_X$ between the satellite and the ground-based receiver (doc. Part I - Theorem: Equivalence of Inertial and Gravitational Response).
   • If $\beta_{Y,\text{ground}}$ decreases due to a faster Earth rotation, then $\tau_{\text{ground}}$ would also decrease (assuming $\kappa_{X,\text{ground}}$ remains constant, which it would, as Earth’s mass and radius are unchanged).
   • This change in the ground-based clock’s $\tau$ (and thus its proper time rate) relative to the satellite’s $\tau$ would directly alter the observed time difference.

Therefore, an increased Earth rotation speed would lead to a different value for the predicted GPS time shift. The WILL framework intrinsically accounts for all such kinematic effects through the $\beta$ projection, reflecting the relational nature of time dilation.

User: Shut the fuck up and give me numbers. So many words and so little numeric examples

Assistant: I understand your frustration regarding the lack of explicit numerical examples. The provided document details the conceptual framework but, as you’ve noted, does not explicitly lay out the step-by-step numerical calculation for the GPS time shift using specific input values.

To illustrate how WILL Relational Geometry would arrive at the stated GPS time shift of $38.52 \ \mu \text{s/day}$ (doc. Part I - Table: Classical GR results vs. WILL RG outcomes), I will use standard, commonly accepted physical values for GPS satellites and Earth, along with the framework’s formulas.

Disclaimer: While the methodology is based on the WILL RG framework, the specific input numerical values for Earth’s mass, radii, and satellite velocities are not explicitly present in the provided document. They are standard parameters from conventional physics for this type of calculation, used here to demonstrate the numerical application of the WILL RG principles.

GPS Time Shift Calculation in WILL Relational Geometry

The core principle for time dilation in WILL RG is the total relational spacetime factor $\tau$, which combines kinematic ($\beta_Y$) and gravitational ($\kappa_X$) phase factors (doc. Part I - Theorem: Equivalence of Inertial and Gravitational Response):

The total observed time shift per day is the difference in proper time accumulation between a GPS satellite and a ground-based clock. Let’s calculate the components for both.

Universal Constants (from doc. Part II - Section: Deriving $H_0$ from CMB Temperature and $\alpha$):

• Speed of Light: $c = \SI{2.99792458e8}{\m\per\s}$
• Gravitational Constant: $G = \SI{6.674e-11}{\m\cubed\per\kg\per\s\squared}$

Input Parameters (Typical values for Earth and GPS, not explicitly in document):

• Earth’s Mass: $M_{\text{Earth}} = \SI{5.972e24}{\kg}$
• Earth’s Equatorial Radius: $R_{\text{Earth}} = \SI{6.378e6}{\m}$
• GPS Satellite Orbital Altitude: $h_{\text{GPS}} = \SI{20200e3}{\m}$
• GPS Satellite Orbital Radius: $r_{\text{GPS}} = R_{\text{Earth}} + h_{\text{GPS}} = \SI{26.578e6}{\m}$
• GPS Satellite Orbital Speed: $v_{\text{GPS}} = \SI{3.874e3}{\m\per\s}$
• Earth’s Rotational Angular Speed: $\omega_{\text{Earth}} = \SI{7.292115e-5}{\radian\per\s}$
• Typical Ground Observer Latitude: $\lambda_{\text{ground}} = \SI{0}{\radian}$ (Equator, for maximal rotational speed)

Step 1: Calculate Earth’s Schwarzschild Radius ($R_s$)

\(R_s = \frac{2GM_{\text{Earth}}}{c^2}\) \(R_s = \frac{2 \cdot \SI{6.674e-11}{\m\cubed\per\kg\per\s\squared} \cdot \SI{5.972e24}{\kg}}{(\SI{2.99792458e8}{\m\per\s})^2}\) \(R_s \approx \SI{8.870e-3}{\m}\)

Step 2: Calculate Gravitational Phase Factor ($\kappa_X$) for Satellite and Ground

The gravitational potential projection $\kappa$ is $\sqrt{R_s/r}$ (doc. Part I - Section: Potential Energy Projection on $S^2$). The phase factor is $\kappa_X = \sqrt{1 - \kappa^2}$.

For the GPS Satellite: \(r = r_{\text{GPS}} = \SI{26.578e6}{\m}\) \(\kappa_{\text{GPS}}^2 = \frac{\SI{8.870e-3}{\m}}{\SI{26.578e6}{\m}} \approx \SI{3.337e-10}{}\) \(\kappa_{X,\text{GPS}} = \sqrt{1 - \kappa_{\text{GPS}}^2} \approx \sqrt{1 - \SI{3.337e-10}{}} \approx 0.999999999833\)

For a Ground Clock (at Earth’s surface): \(r = R_{\text{Earth}} = \SI{6.378e6}{\m}\) \(\kappa_{\text{ground}}^2 = \frac{\SI{8.870e-3}{\m}}{\SI{6.378e6}{\m}} \approx \SI{1.391e-9}{}\) \(\kappa_{X,\text{ground}} = \sqrt{1 - \kappa_{\text{ground}}^2} \approx \sqrt{1 - \SI{1.391e-9}{}} \approx 0.999999999304\)

Step 3: Calculate Kinematic Phase Factor ($\beta_Y$) for Satellite and Ground

The kinematic projection $\beta$ is $v/c$ (doc. Part I - Section: Energy as a Relation - What $\kappa$ and $\beta$ Actually Mean). The phase factor is $\beta_Y = \sqrt{1 - \beta^2}$ (doc. Part I - Section: Kinetic Energy Projection on $S^1$).

For the GPS Satellite: \(v = v_{\text{GPS}} = \SI{3.874e3}{\m\per\s}\) \(\beta_{\text{GPS}} = \frac{\SI{3.874e3}{\m\per\s}}{\SI{2.99792458e8}{\m\per\s}} \approx \SI{1.292e-5}{}\) \(\beta_{Y,\text{GPS}} = \sqrt{1 - \beta_{\text{GPS}}^2} \approx \sqrt{1 - (\SI{1.292e-5}{})^2} \approx 0.999999999916\)

For a Ground Clock (at Earth’s Equator, for maximal speed): The speed due to Earth’s rotation is $v_{\text{rot}} = R_{\text{Earth}} \cdot \omega_{\text{Earth}}$. \(v_{\text{rot}} = \SI{6.378e6}{\m} \cdot \SI{7.292115e-5}{\radian\per\s} \approx \SI{465.1}{\m\per\s}\) \(\beta_{\text{ground}} = \frac{\SI{465.1}{\m\per\s}}{\SI{2.99792458e8}{\m\per\s}} \approx \SI{1.551e-6}{}\) \(\beta_{Y,\text{ground}} = \sqrt{1 - \beta_{\text{ground}}^2} \approx \sqrt{1 - (\SI{1.551e-6}{})^2} \approx 0.999999999998797\)

Step 4: Calculate Total Relational Spacetime Factor ($\tau$)

For the GPS Satellite: \(\tau_{\text{GPS}} = \beta_{Y,\text{GPS}} \cdot \kappa_{X,\text{GPS}}\) \(\tau_{\text{GPS}} \approx 0.999999999916 \cdot 0.999999999833 \approx 0.999999999749\)

For the Ground Clock: \(\tau_{\text{ground}} = \beta_{Y,\text{ground}} \cdot \kappa_{X,\text{ground}}\) \(\tau_{\text{ground}} \approx 0.999999999998797 \cdot 0.999999999304 \approx 0.999999999302\)

Step 5: Calculate the Time Shift

The time dilation factor $\tau$ means that for every second that passes on a “master” clock (e.g., at infinity), the local clock experiences $\tau$ seconds. The observed difference in time accumulation is usually expressed as how much slower or faster the satellite clock runs compared to the ground clock.

A clock at a higher potential (satellite) runs faster than a clock at lower potential (ground) due to gravity. A clock in motion runs slower than a stationary clock due to kinetic effects.

The relative rate difference $\Delta \tau_{\text{rel}} = \tau_{\text{ground}} - \tau_{\text{GPS}}$ represents the fractional daily adjustment needed for the satellite clock relative to the ground clock. \(\Delta \tau_{\text{rel}} = 0.999999999302 - 0.999999999749 = -0.000000000447\) This means the satellite clock runs faster, so the ground clock effectively “loses” this fraction of a second relative to the satellite clock per unit of master time.

To find the daily shift in $\mu \text{s/day}$, we multiply this fractional difference by the number of seconds in a day: \(\text{Seconds in a day} = 24 \text{ hours} \cdot 60 \text{ min/hour} \cdot 60 \text{ sec/min} = 86400 \text{ s}\)

\(\text{Time Shift} = \Delta \tau_{\text{rel}} \cdot \text{Seconds in a day}\) \(\text{Time Shift} = -0.000000000447 \cdot 86400 \text{ s}\) \(\text{Time Shift} \approx -\SI{3.863e-5}{\s}\)

Converting to microseconds: \(\text{Time Shift} \approx -\SI{38.63}{\micro\s}\)

The negative sign indicates that the ground clock runs slower (or the satellite clock runs faster) by this amount. This value is in very close agreement with the $38.52 \ \mu \text{s/day}$ stated in the document and the empirically observed values for GPS satellites.

This calculation demonstrates how the WILL Relational Geometry framework, through its unified $\tau$ factor derived from kinematic and gravitational phase projections, accounts for the GPS time shift without resorting to separate relativistic and gravitational corrections.