Good luck on that its common knowledge for those that understand GR.

Why else do you think Ive mentioned distribution throughout this thread ?

By the way your more than welcome to apply GR to prove me wrong.

1 hour ago, Mordred said:

Good luck on that its common knowledge for those that understand GR.

Why else do you think Ive mentioned distribution throughout this thread ?

By the way your more than welcome to apply GR to prove me wrong.

Since you explicitly invited me to apply GR to prove you wrong, I will do so.

Your statement: "Under SR/GR all particles that are following a geodesic are in freefall. So mass is irrelevant" demonstrates a fundamental confusion between the test mass m and the source mass M.

In GR, the trajectory of a free-falling test particle is governed by the geodesic equation. You are correct that the mass of the falling particle m does not appear here.

However, the Christoffel symbols are strictly defined by the metric tensor.

In standard General Relativity, the geometry of spacetime is dependent on M (the mass of the central body). Without M, the metric reduces to flat Minkowski space. You cannot determine the absolute scale of the system (R_s) in GR without knowing M and G.

My Chrono-Spectroscopic Theorem demonstrated exactly the opposite: the absolute system scale (R_s = 2953.3 m for the Sun) can be derived strictly from chronometry and spectroscopy, without ever invoking the central mass M_{sun} or the constant G.

You confused the mass of the observer with the mass of the system. I suggest before you commenting read what you actually attempt to critic.

Throughout this long dialog you keep showing your absolute disrespect to me by ignoring the materials presented. This remains the main barrier in communication between us.

nice try keep trying.

Why don't you start with basic classical freefall and notice that the mass term becomes irrelevant. (all objects regardless of mass fall at the same rate...)

Then look at the how proper acceleration=0 in geodesic motion ( free fall no external forces)

Tell me does the expression "Under GR gravity is not described as a force but the result of spacetime curvature men anything to you ?

https://en.wikipedia.org/wiki/Geodesics_in_general_relativity#:~:text=For%20broader%20coverage%20of%20this,(3%2DD)%20space.

In general relativity, a geodesic generalizes the notion of a "straight line" to curved spacetime. Importantly, the world line of a particle free from all external, non-gravitational forces is a particular type of geodesic. In other words, a freely moving or falling particle always moves along a geodesic.

In general relativity, gravity can be regarded as not a force but a consequence of a curved spacetime geometry where the source of curvature is the stress–energy tensor (representing matter, for instance). Thus, for example, the path of a planet orbiting a star is the projection of a geodesic of the curved four-dimensional (4-D) spacetime geometry around the star onto three-dimensional (3-D) space.

As you like AI

https://www.google.com/search?q=are+particles+in+freefall+on+a+geodesic+path&rlz=1C1VDKB_enCA1142CA1142&oq=are+particles+in+freefall+on+a+geodesic+path&gs_lcrp=EgZjaHJvbWUyBggAEEUYOdIBCjQxOTM0ajBqMTWoAgCwAgA&sourceid=chrome&ie=UTF-8

To be honest I had hoped you would have picked onto this when I had previously mentioned Newtons shell theorem. with regards to mass energy distribution.

Edited March 10Mar 10 by Mordred

3 hours ago, Mordred said:

Under SR/GR all particles that are following a geodesic are in freefall. So mass is irrelevant so is proper acceleration.

Where mass/energy density distribution becomes relevant is determining the geodesic itself.

1 hour ago, Anton Rize said:

However, the Christoffel symbols are strictly defined by the metric tensor.

.

incorrect

the stress energy tensor tells spacetime how to curve which affects the metric tensor That stress energy tensor includes energy density and pressure terms and directly affects the metric tensor

5 hours ago, KJW said:

Unfortunately, what you are asking me about is galaxy rotation curves, of which I have little knowledge or interest.

I think there is a slight, but very important, misunderstanding here.

The [math]M \sin(i)[/math] degeneracy has absolutely nothing to do with galaxy rotation curves, Modified Gravity (MOND), or Dark Matter. It is a strictly local, classical problem in orbital mechanics and stellar kinematics (specifically, radial velocity measurements of binary stars and exoplanets).

When we observe a star orbiting a companion (like the S0-2 star orbiting the supermassive compact object at the center of our galaxy, which I used in my data), we don't know its incantation (the orbital tilt), we can only measure its 1D line-of-sight velocity via the Doppler shift. Standard methods state that we cannot disentangle the true orbital velocity from the inclination angle ([math]i[/math]) of the orbital plane. They are locked in the formula [math]K \propto v \sin(i)[/math] where [math]K=\frac{\beta}{\sqrt{1-e^{2}}}\sin(i) [/math]. So currently in order to fully determent the orbital system we have to make assumptions about its distance and relay on less accurate optical data sources. Its a concrete limitation of our current observational methods and theoretical models.

My algebraic derivation resolves this exact orbital problem strictly through the geometric asymmetry of the transverse baseline at the apsides, isolating the true velocity and the inclination angle using only 1D spectroscopic extrema.

I understand your stance on Dark Matter and the Bullet Cluster, and I am not asking you to abandon your wholehearted faith in GR or to evaluate cosmology.

I am asking you, as someone who understands pure math and standard orbital geometry, to look at the solution I provided in the previous post. It is pure kinematics and trigonometry. Since you have a keen eye for mathematical consistency, your audit of this specific orbital derivation would be highly valued.

1 hour ago, Mordred said:

In general relativity, gravity can be regarded as not a force but a consequence of a curved spacetime geometry where the source of curvature is the stress–energy tensor (representing matter, for instance).

Read the exact Wikipedia text you just posted: "...where the source of curvature is the stress–energy tensor (representing matter...)".

You are confusing Test Mass ([math]m[/math]) with Source Mass ([math]M[/math]).

1. Test Mass ([math]m[/math]): Drops out of the geodesic equation in freefall. We agree.

2. Source Mass ([math]M_{sun}[/math]): Is absolutely required in GR to generate the stress-energy tensor that curves the spacetime in the first place.

In standard GR, you cannot calculate the absolute scale of the geometry ([math]R_s[/math]) without knowing the central mass [math]M[/math] and [math]G[/math].

My derivation calculated the exact absolute scale of the Solar System ([math]R_s \approx 2953.3[/math] m) using strictly local clock ticks and light shifts -
**completely bypassing the stress-energy tensor, [math]M_{sun}[/math], and [math]G[/math]**.

You argued that the falling object's mass is irrelevant. I never said it was. I showed that the Central Star's mass is not a necessary primitive to generate the geometry.

I don't know how to communicate with you and I can't see any reasons why I should. Since you are arguing against the very text you are quoting, I will leave it at that.

No you have to think GR dont restrict yourself to the Maximally symmetric Minkowskii metric. In GR the stress energy momentum tells spacetime how to curve from the Minkowskii metric. GR describes the Field distribution of the particle ensemble not individual particle motion per se but the entire ensemble. The Minkowskii metric is what is used by SR primarily which is a specific solution of the metric tensor.

\[g_{\mu\nu}=\eta_{\mu\nu}+ h_{\mu\nu}\]

You likely already know the Einstein field equation so I won't waste time latexing that.

In the above the \{\eta\} is the Minkowskii tensor and h is the perturbation tensor the stress tensor acts upon the permutation tensor and gives the deviations from the Minkowskii tensor to result in the new metric tensor ( can often have off diagonal terms) such as a rotating star will have off diagonal terms.

However then again the Minkowskii tensor isn't always maximally symmetric take observer A rotating around observer B along the z axis

\[\eta_{\mu\nu}=\begin{pmatrix}-\frac{\omega^2 r^2}{c^2}&0&\frac{\omega r^2}{c}&0\\0&1&0&0\\\frac{\omega r^2}{c}&0&r^2&0\\0&0&0&1\end{pmatrix}\]

The Sagnac effect I mentioned previously

here is a good stepping stone to understanding geodesics, here is a quote from the following article that applies.

"From this follows that the gravitational effects are locally indistinguishable from the physical ones experienced in an accelerated frame. Such an equivalence enables us to eliminate gravity in a sufficiently small region of space-time. This is done by introducing a suitable chart which supports a locally inertial frame. One can therefore assume that freely falling, inertial frames can be defined in a sufficiently small neighborhood UP of each space-time point P, and the laws of Special Relativity, which may strictly hold only at each point P, will also be sufficiently good approximations of the true laws inside UP for all inertial observers defined therein. A typical example of such a possibility is provided by a free-falling elevator in the gravitational field of the Earth. In fact, a test body inside the falling elevator will fluctuate freely as if the elevator would be placed in empty space, in a region free from any gravitational field."

The article later shows how the FLRW metric equations of state are involved in the Christoffels themself

further down another relevant quote

"Previously, we mentioned the Einstein’s conceptual experiment in the case of one test body inside the free-falling elevator. However, if we put two test bodies (instead of one) inside the elevator, then there is an important physical difference between the two configurations previously mentioned i.e., free-fall in a given field and real absence of field, which soon clearly emerges. Suppose, for instance, that the two bodies are initially at rest at the initial time t0. Then, for t > t0, they will keep both at rest in the absence of a real external field; on the contrary, they will start approaching each other with a relative accelerated motion if the elevator is free falling. The relative motion is unavoidable, in the second case, due to the fact that the test bodies 15 are falling along geodesic trajectories which are not parallel, but converging toward the source of the physical field. So, even if the relative velocity of the two bodies is initially vanishing, v(t0) = 0, their initial relative acceleration, a(t0), is always non-vanishing"

https://amslaurea.unibo.it/id/eprint/18755/1/Raychaudhuri.pdf

I invite you to study the contents in greater detail.

the last should highlight the distinction between local reference frames from the global geometry

5 hours ago, Mordred said:

Under SR/GR all particles that are following a geodesic are in freefall. So mass is irrelevant so is proper acceleration.

Where mass/energy density distribution becomes relevant is determining the geodesic itself.

So tell me I established above and you agree the mass of the test particle isn't relevent. The proper acceleration is zero and it is the field distribution of mass/energy that determines the geodesic equations of motion "

the stress energy momentum tensor tells spacetime how to curve"

so where am I wrong in the quoted statement that you accused me of being wrong ?

Edited March 10Mar 10 by Mordred

13 hours ago, Mordred said:

Another question one can ask is if mass is a primitive concept " why do different particles or other objects require greater force to accelerate the same distance than others "?

Depend on the extend to which they interact with Higgs field.

11 hours ago, KJW said:

Why would opposite signs unify them?

My opinion is that if space and time is equivalent their magnitude should be zero when subtracted from each other,therefore,the need for opposite signs... therefore,they are unified by their equivalency.

6 hours ago, Anton Rize said:

But what if we could derive the exact same absolute scale of a stellar system without ever knowing its "Mass" and without ever using G? If the geometry of the system can be completely solved using only clocks and light, then Mass is not a fundamental pillar of reality

The question should be why so? ... It just shows there is fundamental relationship/interconnection between Mass,Gravity,time(clocks) and light that we are not aware of...clocks might be telling us about mass while light might be telling us something a bout gravity...I think the current definition of mass and gravity don't tell us fundamentally what they are.(Just a suggestion based on my perspectives)

2 hours ago, Anton Rize said:

I don't know how to communicate with you

1 hour ago, Mordred said:

GR dont restrict yourself to the Maximally symmetric Minkowskii metric

@Anton Rize and @Mordred

😂 you people are hilarious,You remind me of construction of Tower of babel.

@Mordred , So you just incapable of admitting your own mistakes? I see...
Non of this has anything to do with my derivation or your clear misunderstanding of it:

6 hours ago, Mordred said:

Under SR/GR all particles that are following a geodesic are in freefall. So mass is irrelevant so is proper acceleration.

Where mass/energy density distribution becomes relevant is determining the geodesic itself.

At this point you only making it worse. You basically forcing me to ignore you. Is that what you trying to achieve?

So you believe yet you haven't actually stated any argument to counter my quoted statement. However that's a typical response I often see when I describe something that can be found in any textbook on GR.

Feel free to go right on ahead on what you feel is right. I know GR is a field treatment and why it is a field treatment. Boils down to that key word distribution and a word I haven't mentioned is flow...

If you want to test that try tracking 20000 test particles.

Edited March 10Mar 10 by Mordred

7 minutes ago, MJ kihara said:

It just shows there is fundamental relationship/interconnection between Mass,Gravity,time(clocks) and light that we are not aware of...clocks might be telling us about mass while light might be telling us something a bout gravity...

Yes I see where you pointing at. The problem is that you putting light, time and mass on epistemologically equal footing - they not.
Light, time - directly measurable.
mass - model output.

They are on vastly different epistemological levels. If we can derive all phenomena of a the system from only directly measurables - introducing any extra unmeasurable entities is just speculation.

9 minutes ago, Mordred said:

So you believe yet you haven't actually stated any argument to counter my quoted statement.

What was my derivation about and how do you understand it? In the end its about this derivation your comments supposed to be. But with you its hard to tell...

Im

6 minutes ago, Anton Rize said:

Yes I see where you pointing at. The problem is that you putting light, time and mass on epistemologically equal footing - they not.
Light, time - directly measurable.
mass - model output.

What was my derivatiod how do you understand it? In the end its about this derivation your comments supposed to be. But with you its hard to tell...

How many times has the question of mass come up this thread ?

Would not distinguishing when mass becomes relevant be useful to answer those questions ?

We cross posted

On 3/10/2026 at 1:37 PM, Anton Rize said:

In standard GR, you cannot calculate the absolute scale of the geometry ([math]R_s[/math]) without knowing the central mass [math]M[/math] and [math]G[/math].

That's not correct.

From the formula for escape velocity (or alternatively from the Schwarzschild metric):

[math]r_s = \dfrac{2GM}{c^2}[/math]

From Kepler's third law of planetary motion:

[math]\dfrac{a^3}{T^2} = \dfrac{GM}{4\pi^2}[/math]

Eliminating [math]GM[/math] from both formulae:

[math]r_s = \dfrac{8\pi^2}{c^2} \dfrac{a^3}{T^2}[/math]

Thus, we have an expression for [math]r_s[/math] in terms an orbiting object without [math]G[/math] or [math]M[/math].

I recently said that using [math]r_s[/math] as the arbitrary constant in the Schwarzschild solution is purely mathematical, not requiring any connection to physical mass. Even the original equation:

[math]R_{\mu \nu} = 0[/math]

does not contain any reference to physical energy-momentum. However, there is a natural connect to distance within the mathematics.

Not only can GR work without [math]G[/math] and [math]M[/math], but it takes extra effort to work with [math]G[/math] and [math]M[/math].

Edited March 11Mar 11 by KJW

7 hours ago, KJW said:

That's not correct.

From the formula for escape velocity (or alternatively from the Schwarzschild metric):

rs=2GMc2

From Kepler's third law of planetary motion:

a3T2=GM4π2

Eliminating GM from both formulae:

rs=8π2c2a3T2

Thus, we have an expression for rs in terms an orbiting object without G or M.

I recently said that using rs as the arbitrary constant in the Schwarzschild solution is purely mathematical, not requiring any connection to physical mass. Even the original equation:

Rμν=0

does not contain any reference to physical energy-momentum. However, there is a natural connect to distance within the mathematics.

Not only can GR work without G and M, but it takes extra effort to work with G and M.

That brings to mind an interesting scenario

If you go from \[R_{\mu\nu}=0\] of the Schwartzchild metric to the FLRW metric which has the time component as

\[R_{00}=-\frac{3}{c^2}\frac{\ddot{a}}{a}\]

With space components

\[R_{ij}= (a\ddot{a}+2a^2+2c^2)\frac{g_{ij}}{a^2}\]

The question of the transition distance comes to mind between the 2 scenarios. I would surmise the latter would take effect when the transition between gravitational bound spacetime regions to spacetime regions not gravitationally bound.

What are your thoughts on that @KJW ?

Not sure why that thought came to mind but to me it does pose an interesting question

Edited March 11Mar 11 by Mordred

5 hours ago, Mordred said:

That brings to mind an interesting scenario

If you go from \[R_{\mu\nu}=0\] of the Schwartzchild metric to the FLRW metric which has the time component as

\[R_{00}=-\frac{3}{c^2}\frac{\ddot{a}}{a}\]

With space components

\[R_{ij}= (a\ddot{a}+2a^2+2c^2)\frac{g_{ij}}{a^2}\]

The question of the transition distance comes to mind between the 2 scenarios. I would surmise the latter would take effect when the transition between gravitational bound spacetime regions to spacetime regions not gravitationally bound.

What are your thoughts on that @KJW ?

Not sure why that thought came to mind but to me it does pose an interesting question

I'm not sure, but the only black hole solution I'm aware of that is not in an otherwise flat background is the De Sitter-Schwarzschild metric.

For the Schwarzschild metric:

[math](ds)^2 = -(1 - a r^{-1}) (c\ dt)^2 + (1 - a r^{-1})^{-1} (dr)^2 + r^2 ((d\theta)^2 + (\sin\theta\ d\phi)^2)[/math]

and the De Sitter metric:

[math](ds)^2 = -(1 - b r^{2}) (c\ dt)^2 + (1 - b r^{2})^{-1} (dr)^2 + r^2 ((d\theta)^2 + (\sin\theta\ d\phi)^2)[/math] for [math]b > 0[/math]

the De Sitter-Schwarzschild metric is:

[math](ds)^2 = -(1 - a r^{-1} - b r^{2}) (c\ dt)^2 + (1 - a r^{-1} - b r^{2})^{-1} (dr)^2 + r^2 ((d\theta)^2 + (\sin\theta\ d\phi)^2)[/math] for [math]b > 0[/math]

The thing I found to be surprising is that one can simply superimpose the Schwarzschild metric directly onto the De Sitter metric. However, given that the Schwarzschild and De Sitter metrics have the same form, I suppose it is to be expected that they can be superimposed to produce another metric of the same form. By contrast, the Schwarzschild metric and FLRW metric do not have the same form, and so are probably not amenable to being directly superimposed.

Solid point its food for thought when I get a chance I will likely look into it in greater detail. Thanks for your opinion on an interesting problem to examine.

The Desittter and anti-Desitter spacetime route may be a decent approach to the scenario to start with.

Did some digging and found 3 classes of solutions all three examined in this paper.

https://arxiv.org/pdf/2308.07374

With Mc Vittie being more commonly referenced (Schwartzchild FLRW metric). There are a few others but the three in this paper are seemingly examined more

Anyways though there is some relevance (static vs commoving coordinate embeddings) it runs the risk of derailing the thread but thought I'd share what I found.

9 hours ago, KJW said:

I'm not sure, but the only black hole solution I'm aware of that is not in an otherwise flat background

How about the Vaidya class of black holes? These spacetimes are not asymptotically flat.

On 3/11/2026 at 2:46 PM, KJW said:

That's not correct.

From the formula for escape velocity (or alternatively from the Schwarzschild metric):

rs=2GMc2

From Kepler's third law of planetary motion:

a3T2=GM4π2

Eliminating GM from both formulae:

rs=8π2c2a3T2

Thus, we have an expression for rs in terms an orbiting object without G or M.

I recently said that using rs as the arbitrary constant in the Schwarzschild solution is purely mathematical, not requiring any connection to physical mass. Even the original equation:

Rμν=0

does not contain any reference to physical energy-momentum. However, there is a natural connect to distance within the mathematics.

Not only can GR work without G and M, but it takes extra effort to work with G and M.

You correctly demonstrated that we can mathematically hide the [math]GM[/math] term by substituting it with Kepler's Third Law kinematics.

However, this substitution perfectly highlights the exact epistemological bottleneck I am talking about.

Look closely at your final expression: [math]R_s = \frac{8\pi^2}{c^2} \frac{a^3}{T^2}[/math].

1. The requirement of a priori space: To calculate [math]R_s[/math] using your formula, you must empirically measure [math]a[/math] (the physical semi-major axis in meters). How do you measure [math]a[/math]? You must rely on the cosmic distance ladder (parallax, radar ranging). You are forced to assume a pre-existing 3D metric container to measure spatial distances before you can define the scale of the geometry.

2. The hidden mass: Furthermore, Kepler's third law ([math]\frac{a^3}{T^2} = \frac{GM}{4\pi^2}[/math]) is inherently Newtonian; it explicitly assumes the source mass [math]M[/math] drives the orbit. You didn't eliminate Mass from the fundamental ontology of the system; you just substituted the explicit [math]GM[/math] variable with its Newtonian equivalent. The GR stress-energy tensor still fundamentally requires the physical mass to curve the spacetime.

In stark contrast, look at the WILL RG Chrono-Spectroscopic equation:

[math]R_s = T c \frac{\kappa^2\beta}{2\pi}[/math]

There is no spatial distance [math]a[/math] here. There are no meters. The inputs are strictly local chronometry ([math]T[/math]) and dimensionless spectroscopic projections
([math]\kappa, \beta[/math] derived purely from redshift and optical angles).

WILL RG doesn't need to measure spatial distances to find the scale. It generates the absolute spatial scale purely from time and light relationships, without ever needing a "meter stick" or a central mass assumption.

This is the exact difference between descriptive physics (algebraically substituting variables within a pre-existing 3D background) and generative physics (creating the physical scale from pure relational tension).

Here's the desmos project: https://www.desmos.com/calculator/iymnd3tw3z

(P.S. I see the thread is venturing into exotic embeddings like McVittie and Vaidya metrics, but I would like to keep the focus strictly on this fundamental ontological difference regarding local scale generation before we jump to cosmological scale models).

Edited March 14Mar 14 by Anton Rize

How will that methodology work if you are given two stars of equal mass orbiting each ?

How would you determine the common barycenter for other mass variations of 2 stellar objects.

12 hours ago, Anton Rize said:

You correctly demonstrated that we can mathematically hide the [math]GM[/math] term by substituting it with Kepler's Third Law kinematics.

However, this substitution perfectly highlights the exact epistemological bottleneck I am talking about.

Look closely at your final expression: [math]R_s = \frac{8\pi^2}{c^2} \frac{a^3}{T^2}[/math].

1. The requirement of a priori space: To calculate [math]R_s[/math] using your formula, you must empirically measure [math]a[/math] (the physical semi-major axis in meters). How do you measure [math]a[/math]? You must rely on the cosmic distance ladder (parallax, radar ranging). You are forced to assume a pre-existing 3D metric container to measure spatial distances before you can define the scale of the geometry.

Thanks for the clarification. However, the specific statement I said was incorrect was:

In standard GR, you cannot calculate the absolute scale of the geometry ([math]R_s[/math]) without knowing the central mass [math]M[/math] and [math]G[/math].

The formula I provided was an expression for [math]R_s[/math] that did not require knowing the central mass [math]M[/math] and [math]G[/math]. Now you are saying that [math]a[/math], the semi-major axis of the orbit, is also unacceptable. That's a case of moving the goalposts. However, I did anticipate this. There are other parameters that one can choose to specify an orbit than [math]a[/math]. For example, on page 6 of this thread, I derived the time dilation for an object in a circular orbit:

[math]\dfrac{\Delta t_R}{\Delta t_\infty} = \sqrt{1 - \dfrac{3GM}{c^2 R}}[/math]

Substituting [math]r_s = \dfrac{2GM}{c^2}[/math] gives:

[math]\dfrac{\Delta t_R}{\Delta t_\infty} = \sqrt{1 - \dfrac{3}{2} \dfrac{r_s}{R}}[/math]

Rearranging leads to:

[math]R = \dfrac{\dfrac{3}{2} r_s}{1 - \left(\dfrac{\Delta t_R}{\Delta t_\infty}\right)^2}[/math]

Substituting into [math]r_s = \dfrac{8\pi^2}{c^2} \dfrac{R^3}{T^2}[/math] (where [math]R = a[/math]) gives after rearranging:

[math]r_s = \dfrac{c\ T}{\sqrt{27} \pi} \left(1 - \left(\dfrac{\Delta t_R}{\Delta t_\infty}\right)^2\right)^{3/2}[/math]

Thus, we have an expression for [math]r_s[/math] in terms of the orbit period and time dilation of a circular orbit. Obviously, it would be better to have an expression for [math]r_s[/math] in terms of a general orbit rather than just a circular orbit, but I don't have the appropriate formulae for general orbits. That does not mean that such formulae do not exist in principle. The point is that it is quite incorrect to suggest that GR is not sufficiently powerful as a theory. GR is more difficult than Newtonian theory because it needs to provide corrections to Newtonian theory due to spacetime curvature that make equations more difficult to solve. But not having solutions because the equations are difficult to solve is not the same as such solutions not existing in principle.

12 hours ago, Anton Rize said:

2. The hidden mass: Furthermore, Kepler's third law ([math]\frac{a^3}{T^2} = \frac{GM}{4\pi^2}[/math]) is inherently Newtonian; it explicitly assumes the source mass [math]M[/math] drives the orbit. You didn't eliminate Mass from the fundamental ontology of the system; you just substituted the explicit [math]GM[/math] variable with its Newtonian equivalent. The GR stress-energy tensor still fundamentally requires the physical mass to curve the spacetime.

There has been discussion on this forum about whether energy-momentum is spacetime curvature or energy-momentum causes spacetime curvature. But either way, spacetime curvature is physically real and associated with energy-momentum. You are going to have a very difficult time trying to convince me that the ontology of physical reality is different.

One detail I would like to add with regards to energy momentum. Time translations conserve energy while space translations conserve the momentum. Combining the two gives energy momentum conservation.

I honestly hope you pick up on my last post orbiting bodies aren't orbiting each other. They orbit their common barycenter. With the Sun and Mercury system that's negligible that isn't true if both mass terms are similar in value. Good example Jupiter and the Sun they orbit the common barycenter just outside the Suns radius.

Just a little trivial sidenot. Ever wonder how gravity wave detectors get calibrated ? They literally place a 1 ton or greater known mass beside the beam. No orbit involved yet the beam will change in angle

Sorry guys for late reply. Im back now and I decided to commit to the open research concept fully.

I set a challenge for myself: To produce and upload on YouTube 2 videos per week.

So Im happy to share with you the first video from this challenge:

https://www.youtube.com/watch?v=6YkDZGLLxnY

Also this is probably my best desmos project so far

https://www.desmos.com/calculator/mjen4ms452

Would you agree with my Corollary?:

Corollary (Epistemic Mandate and Ontological Redundancy):

In information theory and formal logic, if a parameter is strictly absent from the complete algebraic generative chain of a system, its reintroduction constitutes an epistemic violation. Because the full structural and dynamical parameterization ([math]e[/math], [math]\Delta\varphi[/math], [math]R_s[/math]) is algebraically closed using only directly measurable parameters ([math]e[/math], [math]\theta_{\odot}[/math], [math]T_M/T_{\oplus}[/math], [math]z_{sun}[/math]) and derived relational projections ([math]\kappa[/math], [math]\beta[/math]), the variables [math]G[/math] and [math]M[/math] possess zero independent predictive power. They are not fundamental primitives. Their retention required only for conversion of pure relational geometry into legacy units of kilograms.

Edited Thursday at 12:22 PM1 day by Anton Rize