Consider a distant observer traveling at 0.867 c ( $$\gamma=2$$ ) relative to the solar system along the line that is collinear with the sun's axis of rotation. As the clockwork solar system spins beneath him, the distant observer peers through his powerful telescope at Big Ben in London. After taking relativistic doppler into account, the distant observer measures Big Ben's little hand to make one revolution for every two revolutions of his own wristwatch's little hand, in accordance with relativistic time dilation. He also observes that Big Ben's little hand still makes 730.5 revolutions for every revolution that the earth makes around the sun. From these two observations the distant observer concludes that in his inertial frame of reference the earth's orbital velocity is only half the velocity necessary to keep the earth in stable orbit around the sun, given the invariant spacetime curvature in the vicinity of the sun through which the earth's geodesic passes.

Will the earth spiral into the sun? If not, why not?

• You've asked a variant of this question a few times. Does my explanation here do anything to help you? physics.stackexchange.com/a/745531/313823. Laws like Kepler's law only hold in the object's own frame of reference. Jun 13, 2023 at 21:15
• The crux of your difficulty is when you say "So in his inertial frame of reference, the distant observer measures the earth's orbital velocity to be only half the velocity necessary to keep the earth in stable orbit around the sun." He would need to convert to the Earth's frame of reference to apply Kepler's laws. Jun 13, 2023 at 21:16
• In the same way, Earth when viewed from a moving perspective appears to be an ellipsoid, contracted in the direction of travel, despite all laws of gravity and chemical bonding predicting that a rocky planet of the mass and composition of the Earth form a spherical shape. It doesn't mean the laws of gravity and chemistry are wrong, but those laws, like all physical laws, are only valid in the planets own rest frame. To get the correct laws in a different moving frame, you need to Lorentz Transform them. Jun 13, 2023 at 21:20
• Those quotations are imprecisely stated, because they are propounding a conceptual viewpoint, which the author later follows with mathematically rigorous statements. You are not going to get any value out of this site if your approach is to attempt to teach the community novel physics, rather than attempt to learn established physics from the community. Throughout your repeated questioning, you've appeared unwilling to accept the universal opinion of the more experienced users here on the topic you raise. With that attitude, I doubt this site can help you. Jun 14, 2023 at 5:18
• @Pat Dolan - In general relativity your quote from special relativity is useless. A moving mass has different gravity than a mass at rest, similar to a moving charge that also has different electromagnetic properties than one at rest, see here. Jun 15, 2023 at 0:35

Will the earth spiral into the sun? If not, why not?

No. Because this is not true:

the distant observer measures the earth's orbital velocity to be only half the velocity necessary to keep the earth in stable orbit around the sun

No derivation of this claim is provided. It is simply claimed without any justification. The claim is incorrect.

General relativity is based on tensors and pseudo Riemannian geometry. They are explicitly designed from first principles so that all measurable results are independent of the coordinate system. If the orbit is stable in one frame then it is necessarily stable in all frames. The foundations of the math do not permit any other outcome.

Invariant spacetime curvature is assumed to be the cause of orbit.

As is stated here, curvature is invariant. So the results do not depend on the coordinates.

Edit: I wanted to add the actual GR math. The outcome of this is exactly as expected by anyone with GR experience. Indeed, from first principles it could be no other way. All equations are using geometrized units where $$c=G=1$$.

We start with the weak field metric in cylindrical coordinates: $$ds^2 = (1 - 2 U) dr^2 + (-1 - 2 U) dt^2 + (1 - 2 U) dz^2 + (r^2 - 2 r^2 U) d\phi^2$$ with the standard gravitational potential in cylindrical coordinates $$U=-\frac{M}{\sqrt{r^2+z^2}}$$

Now, an object in orbit is in free-fall, so the worldline of the planet is a geodesic. To calculate the orbit of the earth we therefore calculate the geodesic using the equations described here. When we do so, we get the following equations: $$0 = \left( \begin{array}{c} 0 \\ \frac{2 r^2 \left(z^2-2 M^2\right) r''+r \left(-z^2 \left(z^2-4 M^2\right) \phi '^2-M r'^2 \left(2 M+3 \sqrt{r^2+z^2}\right)+M z'^2 \left(\sqrt{r^2+z^2}-2 M\right)+M \left(\sqrt{r^2+z^2}-2 M\right)\right)+z \left(z \left(z^2-4 M^2\right) r''-4 M \sqrt{r^2+z^2} r' z'\right)+r^3 \phi '^2 \left(M \left(2 M+\sqrt{r^2+z^2}\right)-2 z^2\right)+r^4 r''+r^5 \left(-\phi '^2\right)}{\left(r^2+z^2\right) \left(-4 M^2+r^2+z^2\right)} \\ \frac{2 r' \phi ' \left(-4 M^2+r^2 \left(1-\frac{2 M}{\sqrt{r^2+z^2}}\right)+z^2\right) +r \left(\left(r^2-4 M^2\right) \phi ''-\frac{4 M z z' \phi '}{\sqrt{r^2+z^2}}+z^2 \phi ''\right)}{r \left(-4 M^2+r^2+z^2\right)} \\ \frac{r \left(r \left(r^2-4 M^2\right) z''-4 M \sqrt{r^2+z^2} r' z'\right)+2 z^2 \left(r^2-2 M^2\right) z''+M z \left(r'^2 \left(\sqrt{r^2+z^2}-2 M\right)-z'^2 \left(2 M+3 \sqrt{r^2+z^2}\right)+\left(\sqrt{r^ 2+z^2}-2 M\right) \left(r^2 \phi '^2+1\right)\right)+z^4 z''}{\left(r^2+z^2\right) \left(-4 M^2+r^2+z^2\right)} \\ \end{array} \right)$$

To specifically find a circular orbit we can set $$z=0$$ and $$r=R$$ and $$\phi = d\phi \ t$$. That simplifies the geodesic equation to: $$0=\left( \begin{array}{c} 0 \\ \frac{-\text{d\phi }^2 M R^2-\text{d\phi }^2 R^3+M}{2 M R+R^2} \\ 0 \\ 0 \\ \end{array} \right)$$ so $${d\phi}=\frac{\sqrt{M}}{\sqrt{M R^2+R^3}}$$

Solving for $$\phi = 2\pi$$ we get $$t_{2\pi}=\frac{2 \pi \sqrt{R^2 (M+R)}}{\sqrt{M}}$$

Evaluating proper time along the worldline of the earth we get $$\frac{{d\tau}}{{dt}}=\frac{\sqrt{-4 M^2-2 M R+R^2}}{\sqrt{M R+R^2}}$$ which we can integrate to get Big Ben's time over one year to get $$\tau_{2\pi} =\int_0^{t_{2\pi}} \frac{d\tau}{dt} dt = 2 \pi R \sqrt{-\frac{4 M}{R}+\frac{R}{M}-2}$$ Plugging in the mass of the sun $$M=1480$$ in geometrized units (meters) and the orbital radius of the Earth $$R=1.496 \ 10^{11}$$ we get the proper time $$\tau_{2\pi} = 9.45 \ 10^{15}$$ which is one year in geometrized units.

OK, now adding the math for the other frame that the OP was going on about. We will boost along the cylindrical axis with the Lorentz transform in that direction so $$t=\frac{T-v Z}{\sqrt{1-v^2}}$$ $$z=\frac{Z-v T}{\sqrt{1-v^2}}$$ Note that the $$r$$ and $$\phi$$ coordinates are unchanged. Transforming the line element we get $$ds^2= {d}\phi ^2 \left(r^2-2 r^2 U\right)+{{d}r}^2 (1-2 U)+{{d}T}^2 \left(\frac{2 U \left(v^2+1\right)}{v^2-1}-1\right)-\frac{8 {dT} {dZ} U v}{v^2-1}+{dZ}^2 \left(\frac{2 U \left(v^2+1\right)}{v^2-1}+1\right)$$ And transforming the potential we get $$U=-\frac{M}{\sqrt{r^2 + \left( \frac{Z-vT}{\sqrt{1-v^2}} \right)^2}}$$

With the line element determined in the boosted frame we simply apply the same math as before to obtain the geodesic equations in this frame. We get the following monstrosity (thank you Mathematica) $$0= \left( \begin{array}{c} 0 \\ \frac{\left(v^2-1\right)^2 r \phi '^2 \left(-\frac{4 M^2 \left(v^2-1\right)}{(Z-T v)^2-\left(v^2-1\right) r^2}-1\right)+\frac{M \left(1-v^2\right)^3 r r'^2 \left(2 M \left(v^2-1\right)+4 v Z' \sqrt{r^2-\frac{(Z-T v)^2}{v^2-1}}-\left(v^2+3\right) \sqrt{r^2-\frac{(Z-T v)^2}{v^2-1}}\right)}{\left((Z-T v)^2-\left(v^2-1\right) r^2\right)^2}-\frac{M r' (Z-T v) \left(-\frac{2 M v \left(\left(v^2+1\right) Z'^2-4 v Z'+v^2+1\right)}{\sqrt{r^2-\frac{(Z- T v)^2}{v^2-1}}}+Z' \left(v \left(v^2-3\right) Z'+4\right)+v \left(v^2-3\right)\right)}{\left(r^2-\frac {(Z-T v)^2}{v^2-1}\right)^{3/2}}+\frac{M \left(v^2-1\right)^3 r^2 \phi '^2 \left(\sqrt{r^2-\frac{(Z-T v)^2}{v^2-1}}-2 M\right) \left(v r' (Z-T v)+\left(v^2-1\right) r\right)}{\left((Z-T v)^2-\left(v^2-1\right) r^2\right)^2}-\frac{M v \left(v^2-1\right)^3 r'^3 (T v-Z) \left(\sqrt{r^2-\frac{(Z-T v)^2}{v^2-1}}-2 M\right)}{\left((Z-T v)^2-\left(v^2-1\right) r^2\right)^2}+\frac{M \left(1-v^2\right)^3 r \left(\left(v^2+1\right) Z'^2-4 v Z'+v^2+1\right) \left(\sqrt{r^2-\frac{(Z-T v)^2}{v^2-1}}-2 M\right)}{\left((Z-T v)^2-\left(v^2-1\right) r^2\right)^2}}{\left(v^2-1\right)^2 \left(\frac{4 M^2 \left(v^2-1\right)}{(Z-T v)^2-\left(v^2-1\right) r^2}+1\right)}+r'' \\ \frac{\phi ' \left(\frac{4 M \left(v^2-1\right)^2 r r' \left(v Z'-1\right)}{\left((Z-T v)^2-\left(v^2-1\right) r^2\right)^{3/2}}+\frac{M (Z-T v) \left(\frac{2 M v \left(-\left(v^2-1\right) r'^2+\left(v^2+1\right) Z'^2-4 v Z'+v^2+1\right)}{\sqrt{r^2-\frac{(Z- T v)^2}{v^2-1}}}+v \left(\left(v^2-1\right) r'^2-v^2+3\right)-v \left(v^2-3\right) Z'^2-4 Z'\right)}{\sqrt{1-v^2} \left(r^2-\frac{(Z-T v)^2}{v^2-1}\right)^{3/2}}\right)-\left(1-v^2 \right)^{3/2} \phi '' \left(-\frac{4 M^2 \left(v^2-1\right)}{(Z-T v)^2-\left(v^2-1\right) r^2}-1\right)}{\left(1-v^2\right)^{3/2} \left(\frac{4 M^2 \left(v^2-1\right)}{(Z-T v)^2-\left(v^2-1\right) r^2}+1\right)}+\frac{M v r^2 (T v-Z) \phi '^3}{\left((Z-T v)^2-\left(v^2-1\right) r^2\right) \left(2 M+\sqrt{r^2-\frac{(Z-T v)^2}{v^2-1}}\right)}+\frac{2 r' \phi '}{r} \\ \frac{\left(v Z'-1\right) \left(\frac{M (Z-T v) \left(\frac{2 M \left(\left(v^2-1\right) r'^2+\left(v^2-1\right) r^2 \phi '^2-\left(v^2+1\right) Z'^2+4 v Z'-v^2-1\right)}{\sqrt{r^2-\frac{(Z- T v)^2}{v^2-1}}}-\left(v^2-1\right) r'^2-\left(v^2-1\right) r^2 \phi '^2+v^2 Z'^2+4 v Z'-3 Z'^2-3 v^2+1\right)}{\sqrt{1-v^2} \left(r^2-\frac{(Z-T v)^2}{v^2-1}\right)^{3/2}}+\frac{4 M \left(v^2-1\right)^2 r r' \left(v-Z'\right)}{\left((Z-T v)^2-\left(v^2-1\right) r^2\right)^{3/2}}\right)}{\left(1-v^2\right)^{3/2} \left(-\frac{4 M^2 \left(v^2-1\right)}{(Z-T v)^2-\left(v^2-1\right) r^2}-1\right)}+Z'' \\ \end{array} \right)$$

Now, as before we will simplify this substantially by considering only circular orbits which will wind up as helical orbits in this frame. We will use $$r=R$$ and $$\phi = d\phi \ T$$ as before, but this time we will have $$Z=v T$$. These are the transforms of the original circular orbit. With these, the geodesic equation simplifies to $$0=\left( \begin{array}{c} 0 \\ -\frac{M \left(\text{d\phi }^2 R^2+v^2-1\right)+\text{d\phi }^2 R^3}{R (2 M+R)} \\ 0 \\ 0 \\ \end{array} \right)$$

Solving for $$d\phi$$ we get $$\text{d\phi }=\frac{\sqrt{M-M v^2}}{\sqrt{M R^2+R^3}}$$ Note that this is slower than $$d\phi$$ in the other frame by a factor of $$1/\gamma=\sqrt{1-v^2}$$. So in this frame the angular speed required to maintain a stable orbit (geodesic) is "dilated". This is the key fact that everyone with any experience in GR already knew.

We continue with the rest of the calculations. In this frame the time required to get to $$\phi=2\pi$$ is $$T_{2\pi}=\frac{2 \pi \sqrt{R^2 (M+R)}}{\sqrt{-M \left(v^2-1\right)}}$$ which we see is also "dilated". Meaning that the year is longer relative to coordinate time $$T$$ in the boosted frame. And finally, we plug this back into the line element to find the proper time $$\frac{\text{d\tau }}{\text{dT}}=\sqrt{\frac{4 M^2 v^2}{R (M+R)}-\frac{4 M^2}{R (M+R)}+\frac{2 M v^2}{M+R}-\frac{R v^2}{M+R}-\frac{2 M}{M+R}+\frac{R}{M+R}}$$ and integrate it over the year to obtain $$\tau_{2\pi}= \int_0^{T_{2\pi}}\frac{d\tau}{dT} dT =2 \pi \sqrt{R \left(\frac{R^2}{M}-4 M-2 R\right)}$$ Which is the exact same expression for proper time as before, and substituting numbers gets the same numbers as before. So Big Ben measures the same amount of time in a year. Both the year and Big Ben are "dilated" the same, and the angular velocity required to maintain a stable orbit matches the actual angular velocity.

So the conclusion was exactly as expected. Big Ben and the year both time dilate the same and the boosted velocity is the correct orbital velocity. Indeed, from first principles it could not be any other way. Any invariants must be the same in all frames. So the conclusion was guaranteed. What was actually tested by the above math was whether or not I can program Mathematica for GR calculations.

• Newtonian gravity is not assumed, nor is it mentioned in this paradox. Invariant spacetime curvature is assumed. Can you please amend your answer to reflect this? I will amend my question to make sure others do not repeat your false assumption. Jun 13, 2023 at 18:23
• @PatDolan done. Unfortunately, there is not much more to say since you did not derive or explain your reasoning for this claim
– Dale
Jun 13, 2023 at 18:33
• @PatDolan that would be an appropriate thing to post on the Meta site. None of the voters are required to respond to that, but they may choose to. Or you may get some useful feedback from the broader community
– Dale
Jun 14, 2023 at 15:15