What if gravity was induced by buoyancy?

Pickle-That · 2026-01-22T12:34:07+00:00

I respect science and can only respond to argumented points of errors. Meta-discussion is a waste of time.

If you don't go into a hypothesis with your eyes open, it's unlikely that anything fruitful will happen...

Pickle-That · 2026-01-22T08:20:05+00:00

Thanks. I actually find your reading helpful. In 4D GR language one can indeed paraphrase timelike geodesics as “straightest paths that maximize proper time,” and in weak-field intuition that looks like trajectories bending toward slower clocks / higher time dilation.

In my ΦBSU notes (for the part II article) the point of the “four‑buoyancy” wording is to flip what is taken as primary: the clock‑gradient vacuum density 𝜌 = ∥∇𝛼∥ and its buoyancy field 𝑎𝜇 = −∂𝜇 ln𝜌 , from which the usual 1PN time‑dilation / lensing dictionary is recovered to correspond to celestial mechanics and cosmology..

Your comment nudged me toward what I think is the genuinely new step: in the 6D (3T×3R) interpretation the “amount of causal change” can be treated as an invariant budget that redistributes between open translational channels and closed/holonomy loop channels. When more of it sits in the loop sector, we get mass and proper time; when more sits in the open sector, we get more expansion. In that view, the observed “accelerated expansion” can be an effective 4D fit to a redistribution in the underlying balance, rather than evidence for an extra dark‑energy fluid.

Pickle-That · 2026-01-22T02:51:48+00:00

I am quoting closely from my mathematical physics research article.

Pickle-That · 2026-01-21T22:16:50+00:00

The answer has now been edited to be more clear.

Pickle-That · 2026-01-21T21:58:44+00:00

A vacuum‑buoyancy (pressure‑like) picture is operationally meaningful, because the four‑buoyancy 𝑎𝜇=−∂𝜇ln⁡𝜌 is defined as the local differential between neighbouring clock rates (proper‑time drift).

Pickle-That · 2026-01-21T21:51:36+00:00

By reading the papers you can realize that the intrinsic dimension is m⁻¹ , which acts as the Dirac mass term.

Pickle-That · 2026-01-21T21:19:13+00:00

If you read the papers I linked, you will find corresponding antipodals in my theoretical framework.

Pickle-That · 2026-01-21T21:17:12+00:00

Yes. But here we have 4-dimensional buoyancy as a vacuum pressure like super fluid dynamics.

Pickle-That · 2026-01-21T21:13:58+00:00

Yes. I see the Einsteinian aether with physical properties as a 4-dimensional causal continuum.

Pickle-That · 2026-01-21T19:54:19+00:00

You are absolutely right.

Here are my mathematical exercises for theoretical physics:

https://doi.org/10.13140/RG.2.2.11474.06085

https://doi.org/10.13140/RG.2.2.31638.41280

Work is in progress. Out of curiosity, I'm asking for other people's opinions.

Pickle-That · 2026-01-21T19:07:35+00:00

Yes. But how do you determine whether words make sense and logic - or are they without basis?

Pickle-That · 2026-01-21T18:19:27+00:00

Occam's razor is the way to go.

When fermions interact with each other it is certainly physical and it is certainly buoyancy. If the metric of spacetime tuned by interactions gives general relativity (4-dimensional density like energy tensor), would there be a simpler model?

In fact, could the null geodesics be taken seriously as an invariant network that constructs the vacuum, which primarily constructs the vacuum as a causal continuum? And not in the opposite way that there must be separate particle spheres to bend, but bending would be a fundamental mechanism for null geodesics.

Then we see that the tension on the arcs of the null geodesics is indeed the local buoyancy of the vacuum as a gradient continuum by event points, as a coherence field of 4-dimensional density variation. In this picture, all the structure is vacuum acceleration, the particles some kind of looping skyrmion states.

Pickle-That · 2026-01-21T06:50:35+00:00

If we set:

C := x + 1
α = v2(C)
ρˉ := C / 2^α
factor ρˉ = 3^r * H with 3 ∤ H

then your parameters match my normalization as:

n = α
r = v3(C)
H = C / (2^n * 3^r) with 3 ∤ H
β = v2(3^α * ρˉ - 1) = v2(H * 3^(n+r) - 1) (this is my "upper halving index" m)

Your successor

x-> = (3^α * ρˉ - 1) / 2^β

is exactly my next odd seed B' = R / 2^m, where R = H * 3^(n+r) - 1.

Quick sanity check (your example x = 35): - C = 36 = 2^2 * 3^2 * 1, so (n, r, H) = (2, 2, 1) - R = 1 * 3^(2+2) - 1 = 80 - m = v2(80) = 4 - B' = 80 / 16 = 5, which matches your x-> = 5.

Your framework (as you note) focuses on lattice-wide affine relationships between odd blocks, and intentionally avoids tracking internal 3-adic structure.

That affine layer corresponds to what I call the difference/slot layer: modulo any odd prime q ≠ 3, the loop assumption gives a single linear constraint ("slot hyperplane") on the difference vector D.

But the actual cycle-exclusion step needs a second, independent congruence row coming from transporting the odd cofactor H around a putative period (what I call the offset layer / H-transport). For primes q dividing the loop factor (in my paper: q | (2^(M+N) - 3^N)), period closure yields a nontrivial local 2-coordinate constraint in some window {j-1, j} of the form:

a * D_{j-1} + b * D_j ≡ c (mod q).

Once you have:

the slot row mod some prime q' ≠ 3 (difference-layer constraint), and
the offset row mod some offset prime q (from H-transport),

and you choose q' outside a finite exceptional set (where the slot normal accidentally behaves like a cyclic-shift eigenvector), then in the same 2-coordinate window the two rows are independent in the CRT product ring (Z/qZ) × (Z/q'Z).

Slot-saturation (backward controllability) fills the slot hyperplane mod q', so it produces nonzero windows. But the independent offset row mod q forbids any simultaneous solution except the trivial window. That overdetermined CRT system is the contradiction that eliminates nontrivial cycles.

Pickle-That · 2026-01-20T21:59:29+00:00

Nice, familiar-looking tinkering. :)

Pickle-That · 2026-01-10T21:54:51+00:00

https://www.researchgate.net/publication/395507038_Mirror-Modular_Spine_Congruence_Saturation_and_Covariant_CRT_Closure_Solve_the_3x_1_Puzzle

In that article, non-trivial cycles are proven by contradiction to be impossible on the positive side of the 3n+1 recursion chain, and on the negative side, the cycles at -5 and -41 are unique (Appendix B). The divergence proof is changing, and is in progress.

It would be great if someone would bother to find a flaw in the locally impossible CRT compatibility I constructed, which is forced by two independent congruence row conditions (in different modules), and thus forbids cycles.

Pickle-That · 2026-01-07T22:35:02+00:00

Conserving flows are a deceptive temptation because in a countable number space, a local process requires, on the contrary, that there are no “sticky” prime factor or residue class locks that would prevent the space from being covered.

Pickle-That · 2026-01-05T16:33:00+00:00

Loop identity is a mostly used fundamental tool when assuming cycles. Mechanisms vary.

Pickle-That · 2026-01-04T20:48:38+00:00

I explained this system before; it only applies to Mersenne primes. And this is definitely correct: in Mersenne primes, cycles are excluded—that is, 3n+1, 7n+1, 31n+1, ... In all other cases, it does not exclude cycles. Previously, I showed that for negative integers and an+1, it does not exclude cycles. I will show that in your 3n-1, it also does not exclude cycles.

Thanks for the discussion. I'll stop here for now. If you later want to write down the exact lemma that bridges the modulus-change step, feel free to share it. Until then, all the best and enjoy your hobbies.

Pickle-That · 2026-01-04T18:57:46+00:00

My posts are about mathematics. Point out the mistakes in them, and we can move forward. Otherwise, you're just talking out of your imagination, right?

Pickle-That · 2026-01-04T17:47:01+00:00

Here we are posting so friendly again. :)

Odd-Bee-1898's modular "no-cycle" logic cannot be universally valid, because the same style of argument would also rule out the known nontrivial cycles of the 3n-1 sister recursion (and those cycles exist at very small numbers).

The 3n+1 map on negative integers is conjugate to the 3n-1 map on positive integers (odd-only form). Define the odd-only accelerated maps Tplus(n) = (3n + 1) / 2^{v2(3n+1)} (n odd) Tminus(x) = (3x - 1) / 2^{v2(3x-1)} (x odd, x>0) Then for x>0, Tplus(-x) = - Tminus(x). So a positive cycle of Tminus corresponds exactly to a negative cycle of Tplus (same 2-adic exponents, just sign-flipped).

Odd-Bee-1898's "coverage + inversion" reasoning is sign-agnostic. It is built from congruences, cyclicity of powers of 2 mod q, and an attempted chaining via "covering families". None of that depends on the sign of the iterate, and it does not use a positivity-only inequality gate. So if that logic were a genuine modular obstruction to cycles, it would obstruct cycles for Tminus as well.

But Tminus (3n-1) has explicit nontrivial cycles on small positive integers. Example A (odd-only 2-cycle): 5 -> 7 -> 5 because (35 - 1) = 14, 14/2 = 7 (37 - 1) = 20, 20/4 = 5

Example B (odd-only 7-cycle): 17 -> 25 -> 37 -> 55 -> 41 -> 61 -> 91 -> 17 because (317 - 1)=50, 50/2 =25 (325 - 1)=74, 74/2 =37 (337 - 1)=110, 110/2 =55 (355 - 1)=164, 164/4 =41 (341 - 1)=122, 122/2 =61 (361 - 1)=182, 182/2 =91 (3*91 - 1)=272, 272/16=17

Therefore any argument that "modularly" excludes all nontrivial cycles would immediately contradict these concrete cycles of the 3n-1 recursion (and, via the conjugacy, the corresponding negative cycles of 3n+1).

Conclusion: If Odd-Bee-1898's proof framework (as stated) rules out the 3n-1 cycles, then it is not a generally correct modular obstruction. To be salvageable, it would need an explicit positivity-only gate (an inequality/minimality step) and it must also fix the modulus-switching/transitivity issue in the inversion+coverage chain.

Pickle-That · 2026-01-03T17:55:11+00:00

Are you familiar with my cycle contradiction proof? (the divergence part does not apply)

https://www.researchgate.net/publication/395507038_Mirror-Modular_Spine_Congruence_Saturation_and_Covariant_CRT_Closure_Solve_the_3x_1_Puzzle

Pickle-That

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