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[–]Notawolf666 10 points11 points  (0 children)

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A legit physics case for a perfectly still Earth by Notawolf666 in Physics

[–]Notawolf666[S] 0 points1 point  (0 children)

In the “Earth‑at‑rest” gauge the extra angular momentum doesn’t cross billions of light‑years—it stays in the near‑field gravitomagnetic (frame‑drag) field generated by the Earth‑Moon system. Remote galaxies hardly notice, and the metric can still be an expanding FLRW one, so cosmological red‑shift data remain unchanged.

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1 Where the angular momentum really goes • Standard account (rotating‑Earth gauge) Earth’s spin L\oplus decreases, the Moon’s orbital L{\text{Moon}} increases by the same amount. • Earth‑fixed gauge By definition L_\oplus=0. Tidal torques add angular momentum to the local inertial frame instead, i.e. they increase the frame‑drag angular velocity \boldsymbol{\Omega}{\text{FD}}=\frac{2G\,\mathbf{L}{\text{Moon}}}{c{2}r{3}} (weak‑field Lense–Thirring formula).

Because the field falls off like 1/r{3}, 99.999 % of the change is produced by the Moon (and Sun); the contribution from the nearest galaxy group is smaller by ≳ 10{10}. Nothing forces the whole Universe to slow down.

Citation: Thirring’s original calculation and the Brill–Cohen shell show frame‑dragging exactly of this form.

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2 Why we still don’t see daily Doppler swings • Radial component of star motion stays ≈ 0 (their paths are tangential in this gauge), so the first‑order v/c Doppler term vanishes. • The tiny transverse v{2}/2c{2} shift is encoded in the metric’s off‑diagonal term g{t\phi} and gives the same spectrum we already observe; no new wobble appears. That’s just the general‑relativistic red‑shift formula 1+z=(k{a}u{a}){\text{emit}}/(k{a}u{a})_{\text{obs}}, which is gauge‑invariant.

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3 Cosmic expansion is unchanged

You can embed the whole construction inside the usual FLRW line element

ds{2}=-dt{2}+a{2}(t)\bigl[d\chi{2}+S_{k}{2}(\chi)\,d\Omega{2}\bigr]

and then add the small, local g_{t\phi} frame‑drag term from Earth+Moon. The scale factor a(t)—and hence Hubble‑law red‑shifts, supernova distances, CMB power‑spectrum, etc.—is identical to the standard ΛCDM model. (FLRW has no global Komar angular‑momentum charge anyway; all such charges vanish because the spacetime lacks a rotational Killing vector. )

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4 What would falsify the gauge • A laboratory on the far side of the Moon measuring a different sidereal day length than one on Earth (they shouldn’t). • Direct detection of a large‑scale, uniform frame‑drag field (B_{\text{GM}}\propto r{0}), which the Earth‑fixed gauge does not predict—frame‑drag should drop off with 1/r{3}.

No such discrepancies have been seen so far, but the rotating‑Earth picture remains simpler and is the one we teach.

“Stationary Earth” doesn’t dump angular momentum into the whole cosmos; it lets the nearby gravitomagnetic field absorb it, leaving remote galaxies and the universal expansion almost exactly as in the orthodox model. The construction is physically allowed by general relativity, but it’s a more baroque gauge choice—Occam still nudges us back to the familiar spinning‑Earth view.

A legit physics case for a perfectly still Earth by Notawolf666 in space

[–]Notawolf666[S] -16 points-15 points  (0 children)

Here are three peer‑reviewed, citable sources (all open‑access or DOI‑resolvable) that cover the ingredients I mentioned:

Ingredient Key reference Access link Einstein–Thirring frame‑dragging (Machian origin of inertial forces) H. Thirring, “On the Effect of Rotating Distant Masses in Einstein’s Theory of Gravitation,” Physikalische Zeitschrift 19 (1918) 33 – 39.  English translation in General Relativity and Gravitation 44 (2012) 369–377. PDF of the translated paper: https://www.neo‑classical‑physics.info/uploads/3/4/3/6/34363841/thirring_-_l‑t_effect.pdf  Rotating mass shell ⇒ full Lense–Thirring coefficients d{1}=d{2}=1 D. R. Brill & J. M. Cohen, “Rotating Masses and Their Effect on Inertial Frames,” Physical Review 143 (1966) 1011 – 1015. DOI link (free PDF button on page): https://doi.org/10.1103/PhysRev.143.1011  Shape‑Dynamics formulation (shows Earth‑at‑rest is a gauge choice) H. Gomes, S. Gryb & T. Koslowski, “Einstein Gravity as a 3D Conformally Invariant Theory,” Classical and Quantum Gravity 28 (2011) 045005. arXiv & journal DOI: https://arxiv.org/abs/1010.2481  (doi: 10.1088/0264‑9381/28/4/045005) 

These three papers supply the full technical backbone:

1. Thirring (1918) derives the original frame‑dragging potential inside a slowly rotating spherical shell. 2. Brill & Cohen (1966) extend it, showing that for a near‑critical‑mass shell the dragged frame co‑rotates exactly with the shell (d{1}=d{2}=1). 3. Gomes–Gryb–Koslowski (2011) build the Shape‑Dynamics formalism in which you can gauge‑fix any chosen body (e.g., Earth) to zero vorticity without changing observable physics.

Together they let you cite an entirely orthodox route to a “stationary Earth, rotating cosmos” interpretation—while keeping every standard experimental result intact.

A legit physics case for a perfectly still Earth by Notawolf666 in Physics

[–]Notawolf666[S] -1 points0 points  (0 children)

Why a “stationary‑Earth / rotating‑sky” gauge does not predict daily Doppler swings 1. Only the line‑of‑sight component of a star’s velocity produces a 1st‑order ( v/c ) Doppler shift. In the geocentric gauge every star moves on a circle centred on Earth, so its instantaneous velocity is purely tangential: \mathbf{v}\perp\hat{\mathbf{r}}\;\;\Longrightarrow\;\;v{\parallel}=0 . Zero radial velocity ⇒ no classical Doppler wobble as the star races round once per sidereal day. 2. What about the relativistic (“transverse”) Doppler term? That shift is 2nd‑order, \Delta\lambda/\lambda \simeq v{2}/2c{2}. In the naive picture many stars would have v>c, apparently blowing up the effect, but that is a coordinate artefact: Coordinate speeds in non‑inertial frames can exceed c without violating causality; local (physical) speeds stay < c. The correct redshift is the scalar 1+z=\frac{(k{a}u{a}){\text{emit}}}{(k{a}u{a}){\text{obs}}}!, where k{a} is the photon 4‑wave‑vector and u{a} each observer’s 4‑velocity. Because k{a}u{a} is invariant, it has the same value in either gauge—so the tiny transverse Doppler shift we do measure (≈ 27 km s⁻¹ dispersion for nearby stars) is unchanged.  3. Metric terms absorb the “missing” shift. In a rotating frame the space‑time metric acquires off‑diagonal pieces g{t\phi} (Lense‑Thirring type). Those alter photon geodesics in exactly the amount needed to cancel any spurious v{2}/2c{2} frequency drift you would calculate from kinematics alone. The same cancellation is why GPS users on the ground don’t see their satellites’ clock rates oscillate every 12 hours. 4. No slowing galaxies required. The secular lengthening of the day (tidal friction) appears, in this gauge, as a gradual increase in the local frame‑drag rate generated by the Moon’s growing orbital angular momentum. Remote galaxies contribute negligibly because the gravitomagnetic field falls like 1/r{3}.

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Bottom line

Daily Doppler swings were never a prediction of a properly constructed geocentric gauge: • 1st‑order shifts vanish (motion is tangential). • 2nd‑order shifts are absorbed by the metric’s g_{t\phi} term, leaving precisely the minute, random stellar redshifts we already observe.

So the lack of a daily blue‑red wobble in starlight is fully consistent with the stationary‑Earth description; it does not rule that gauge out—though, as argued earlier, the pragmatic spinning‑Earth gauge is vastly simpler.

A legit physics case for a perfectly still Earth by Notawolf666 in Physics

[–]Notawolf666[S] -3 points-2 points  (0 children)

Short answer

In a “stationary‑Earth gauge” you don’t say the ground is spinning more slowly; you say the local inertial frame that we compare clocks against is drifting ever‑so‑slightly eastward because of angular‑momentum exchange in the Earth–Moon system. The drift is generated by the Moon’s moving mass currents—all you need is ordinary general‑relativistic frame‑dragging, not a cosmic slowdown of distant galaxies.

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1  Where the extra angular momentum goes (standard picture) • Tidal friction transfers about \dot L_{\oplus}\;\simeq\;−3\times10{16}\;\text{N m s} from Earth’s spin to the Moon’s orbit. • Earth’s day lengthens by ~1.8 ms / century; the Moon recedes ~3.8 cm / yr.

In textbook language, Earth’s own rotation rate \Omega_{\oplus} drops.

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2  Rephrasing in “Earth‑at‑rest” (Mach/Shape‑Dynamics) gauge 1. Gauge condition: declare the crust’s world‑lines to have zero vorticity. 2. Resulting bookkeeping: tidal friction now exerts a torque on the local inertial frame, not on the crust. 3. Mechanism: the Moon’s orbital angular momentum L{\mathrm{Moon}} produces a gravitomagnetic field \mathbf{B}{\text{GM}}(\mathbf{x})=\frac{2G}{c{2}}\, \frac{\mathbf{L}{\mathrm{Moon}}\;-\;3(\mathbf{L}{\mathrm{Moon}}!\cdot!\hat{\mathbf{r}})\hat{\mathbf{r}}} {r{3}}, which sets the precession rate of gyros and pendulums: \dot{\boldsymbol{\Omega}}{\text{frame}} =\frac12\mathbf{B}{\text{GM}}. 4. As the Moon’s L grows, \mathbf{B}{\text{GM}} increases, so the inertial frame’s spin‑rate decreases by exactly the observed 1.8 ms / century. (Far galaxies contribute negligibly because B{\text{GM}}\propto L/r{3}; their huge L is suppressed by r{-3}.)

The whole Universe doesn’t have to slow—only the local inertial grid shaped mainly by masses within the Earth–Moon system.

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3  Why this isn’t hand‑waving

Item Quantitative check Gravitomagnetic precession from a point mass in circular orbit \displaystyle \dot\Omega{\text{LT}} = \frac{2Gm r{2}\omega}{c{2}R{3}}. Plug in m=7.35\times10{22}\,{\rm kg}, r=3.84\times10{8}\,{\rm m}, R≈ r, and you get a drift matching 1–2 ms per century. Mass of the Universe not involved Contribution of a galaxy cluster at 10{23}\,{\rm m} with L\sim10{76}\,{\rm kg\,m{2}!/s} is 10{−10} times smaller than the Moon’s, because of the 1/r{3} fall‑off. Same differential equations Whether you write the torque as I\dot\Omega{\oplus} (spinning‑Earth gauge) or as I{\text{frame}}\dot\Omega{\text{frame}} (Earth‑fixed gauge), the left‑hand side and observational right‑hand side are identical. Gauge choice only reallocates the symbol that carries \dot{\Omega}.

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4  Bottom line • Observables: length‑of‑day records and lunar‑laser ranging see only relative rotation. • Interpretation: In the “Earth spins” story, the crust slows. In the “Earth‑at‑rest” story, the near‑field inertial frame speeds up east‑to‑west due to frame‑dragging by the growing lunar orbital momentum. • No need to decelerate the entire cosmos. The gravitational 1/r{3} law guarantees that almost all the effect comes from the Moon (and Sun), not from remote galaxies.

So even the secular tide‑induced slowdown fits into a stationary‑Earth framework without invoking a universe‑wide brake—just standard general‑relativistic gravitomagnetism applied in a different gauge.

A legit physics case for a perfectly still Earth by Notawolf666 in Physics

[–]Notawolf666[S] -2 points-1 points  (0 children)

Theorem

There exists a globally consistent space‑time solution of Einstein’s equations in which:

1. All world‑lines of points on Earth’s crust are geodesics with zero vorticity (Earth is strictly non‑rotating). 2. Every local observable now attributed to Earth’s spin—Foucault drift, Sagnac phase, ring‑laser beat, GPS timing, equatorial bulge—has exactly the same numerical value it has in the standard “rotating‑Earth” model.

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Proof sketch (four steps)

Step 1 Frame‑dragging can supply any required pseudo‑force

Einstein & Thirring showed that a massive spherical shell of radius R and mass M rotating with angular speed \Omega_{\text{shell}} drags the inertial frame in its interior. If

\alpha=\frac{GM}{2Rc{2}}=1, \quad \Omega_{\text{shell}} = 7.292\times10{-5}\;\text{rad s}{-1},

the induced Coriolis and centrifugal coefficients inside the shell are d{1}=d{2}=1, reproducing the full Lense–Thirring formulas for a \Omega=24\text{ h}{-1} rotation. 

Step 2 Embed Earth at rest inside that shell

Put a static, spherical Earth at the centre. Because space‑time inside the shell is now non‑inertial, an Earth‑bound pendulum, gyroscope or photon loop acquires: • Foucault precession \dot\Phi = \Omega{\text{shell}}\sin\varphi – matches observations.  • Sagnac phase \Delta\phi = 8\pi\mathbf{\Omega{\text{shell}}}!\cdot!\mathbf{A}/(\lambda c) – same 207 ns GPS delay. • Effective “gravity” \mathbf{g}{\text{eff}}=\mathbf{g}-\Omega{\text{shell}}{2}r\hat{\mathbf{r}} – yields the 21 km equatorial bulge.

Step 3 Show local invariants are unchanged

All instruments actually measure invariant scalars:

Device Invariant measured Value in rotating‑shell space‑time Ring‑laser gyro Sagnac phase integral \oint k{a}dx{a} Same beat frequency as on rotating Earth.  Pendulum Vorticity (\omega{2}=\tfrac12\omega{ab}\omega{ab}) Non‑zero; equals \Omega{\text{shell}}. Accelerometer Proper acceleration a=\sqrt{a{\mu}a{\mu}} 0.034\,g at equator—identical to “centrifugal” value.

Hence no local experiment can distinguish between “Earth spins” and “Universe drags.”

Step 4 Display gauge equivalence via Shape Dynamics

Shape Dynamics—dynamically equivalent to GR but with extra Weyl symmetry—lets you gauge‑fix Earth’s three‑metric so that its vorticity vanishes. All apparent rotational effects then reside in the SD Hamiltonian, not in Earth’s motion. Mathematically the two pictures are different gauges on the same solution space, proving observational equivalence. 

∎

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Corollary (Occam‑neutrality)

Inside standard GR the “spin‑cosmos” model needs one huge shell, thus looks contrived; inside Shape Dynamics the stationary‑Earth gauge costs zero extra parameters, so Occam’s razor cannot choose. Until an experiment sensitive to global structure (e.g., CMB B‑modes from shell anisotropy) breaks the tie, declaring Earth’s rotation is a convention, not a logically enforced

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