I What about the Big Ben Paradox?

[patdolan]

Consider a distant observer traveling at .867 c ( 𝛾=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.

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

Note: Newtonian gravity is not assumed in this paradox. Invariant spacetime curvature is assumed to be the cause of the earth's orbit.

 

[Nugatory]

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.

Note: Newtonian gravity is not assumed in this paradox. Invariant spacetime curvature is assumed to be the cause of the earth's orbit.

These two claims are mutually inconsistent.

According to the second claim, the earth is following a geodesic through curved spacetime. This curvature and its geodesics are frame-independent invariants, they're the same in all frames. The geodesic path of the earth doesn't intersect the worldline of the sun (it's a helix around the sun if we draw a 3-D diagram in which the z-axis is used for time and the earth's orbit lies in the x-y plane).

The first claim is assuming Newtonian gravity and Kepler's laws - how else would you calculate "the velocity necessary to keep the earth in stable orbit around the sun"?

So you are reasoning from two mutually inconsistent assumptions, and not surprisingly getting mutually inconsistent results. The resolution is to work out which set of assumptions are in fact applicable, and use these. Here the general relativistic solution is correct; the Newtonian orbital solution is only valid in frames in which the center of mass of the system is not moving relativistically.

 

[Dale]

Note, this is an exact duplicate of your question at physics stack exchange, so I have posted an exact duplicate of my answer.

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

No. Because this is not true:

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

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

Invariant spacetime curvature is assumed to be the cause of the earth's orbit

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

 

[Nugatory]

I notice that you've asked this question in other internet forums, and the answers you've received in some of them are correct. You aren't going to get different answers here.

 

[patdolan]

That Big Ben must run at a time dilated rate of .5 compared to the distant observer's wristwatch is an iron clad requirement of special relativity. Imagine Big Ben by itself rushing towards you at gamma = 2. Like any other clock rushing towards you, it would have to run at .5 the rate of your own clock. So in the distant observer's inertial frame of reference the earth MUST also orbit at .5 it's orbital velocity compared to the solar system's rest frame. Why? Because otherwise Big Ben would not be in sync with the orbital period of the earth around the sun--the total number of revolutions of Big Ben's little hand would indicate it was only Spring when London was in deep Winter if it were the case that Ben Ben runs slow but the earth maintained its normal orbital velocity around the sun. No Dale, If Big Ben runs slow then the earth also MUST RUN SLOW.

The above constitutes a proof sufficient to demolish your objection, Dale. Please proceed to your next issue. Or we can continue to argue this one.

 

[patdolan]

These two claims are mutually inconsistent.

According to the second claim, the earth is following a geodesic through curved spacetime. This curvature and its geodesics are frame-independent invariants, they're the same in all frames. The geodesic path of the earth doesn't intersect the worldline of the sun (it's a helix around the sun if we draw a 3-D diagram in which the z-axis is used for time and the earth's orbit lies in the x-y plane).

The first claim is assuming Newtonian gravity and Kepler's laws - how else would you calculate "the velocity necessary to keep the earth in stable orbit around the sun"?

So you are reasoning from two mutually inconsistent assumptions, and not surprisingly getting mutually inconsistent results. The resolution is to work out which set of assumptions are in fact applicable, and use these. Here the general relativistic solution is correct; the Newtonian orbital solution is only valid in frames in which the center of mass of the system is not moving relativistically.

No, Nugatory. There is no inconsistency in my assumptions. At non-relativistic velocities, such as the earth's orbital velocity around the sun, the Newtonian approximation to the GR solutions are close enough. So we can take the Newtonian calculations for the earth's geodesic to be (almost) the same as the more precise GR geodesic. Now the spacetime curvature in the vicinity of the sun is the same for all observers, as you state. So from the standpoint of the distant observer, the earth is orbiting too slowly through the sun's invariant spacetime curvature to maintain the geodesic of a stable orbit.

 

[Nugatory]

So in the distant observer's inertial frame of reference the earth MUST also orbit at .5 it's orbital velocity compared to the solar system's rest frame.

Yes, this is just another example of velocities being frame dependent and no one is disputing that. (It would be a good exercise to apply the non-parallel motion velocity addition formula here).

What is disputed is the claim that this frame-dependent lower orbital velocity means that the earth will not maintain its stable orbit. That claim is incorrect and comes from naively applying Newton’s law under conditions where it does not apply.

 

[PeterDonis]

The above constitutes a proof sufficient to demolish your objection, Dale.

There is no inconsistency in my assumptions.

@patdolan you are way too confident in the correctness of your arguments. Both of your statements quoted above are false. I strongly suggest that you take a step back. The path you are on now is very likely to lead to you receiving a warning and this thread being closed.

 

[Nugatory]

So from the standpoint of the distant observer, the earth is orbiting too slowly through the sun's invariant spacetime curvature to maintain the geodesic of a stable orbit.

The earth is following the same invariant and geodesic path through spacetime either way. The frame-dependent difference in speed (note that this a three-velocity not a four-velocity) is just a the result of projecting this geodesic path onto different spacelike hypersurfaces of constant coordinate time.

 

[patdolan]

@patdolan you are way too confident in the correctness of your arguments. Both of your statements quoted above are false. I strongly suggest that you take a step back. The path you are on now is very likely to lead to you receiving a warning and this thread being closed.

Nugatory, the Big Ben Paradox thread is always closed at some point on every moderated physics platform. Usually in short order. What other argument can be made against it, other than closure? Before you close the Big Ben Paradox thread, could you kindly provide us some links to other threads which were closed due to overconfidence? Thank you.

 

[Dale]

That Big Ben must run at a time dilated rate of .5 compared to the distant observer's wristwatch is an iron clad requirement of special relativity. Imagine Big Ben by itself rushing towards you at gamma = 2. Like any other clock rushing towards you, it would have to run at .5 the rate of your own clock. So in the distant observer's inertial frame of reference the earth MUST also orbit at .5 it's orbital velocity compared to the solar system's rest frame. Why? Because otherwise Big Ben would not be in sync with the orbital period of the earth around the sun--the total number of revolutions of Big Ben's little hand would indicate it was only Spring when London was in deep Winter if it were the case that Ben Ben runs slow but the earth maintained its normal orbital velocity around the sun. No Dale, If Big Ben runs slow then the earth also MUST RUN SLOW.

I have no objection to that except for a minor quibble that this is a GR problem and not an SR problem. So I am not sure why you thought “No Dale” belongs with the rest.

The above constitutes a proof sufficient to demolish your objection, Dale.

No it doesn’t. It doesn’t have anything to do with my objection. My objection is that your claim “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” is wrong and unfounded.

At non-relativistic velocities, such as the earth's orbital velocity around the sun, the Newtonian approximation to the GR solutions are close enough

But the velocities are not non-relativistic. They are approximately $0.867 c$

 

[patdolan]

And why is that?

 

[Nugatory]

At non-relativistic velocities, such as the earth's orbital velocity around the sun, the Newtonian approximation to the GR solutions are close enough.

Of course, but the relevant velocity here is the .867c that the solar system is moving at in the frame in which the spaceship is at rest and which you are using to calculate the half-speed earth. That’s a relativistic velocity, and the Newtonian approximation won’t work.

I’ve already suggested that doing the velocity addition calculation (non-parallel case) would be a good exercise. Try it, and you’ll see more clearly why the Newtonian approximation doesn’t work here. (And do remember that the Newtonian approximation is a special case of Schwarzschild coordinates in which the sun is at rest).

 

[Dale]

And why is that?

Because you didn’t calculate the required velocity using GR

 

[PeterDonis]

Nugatory

In your post #10, quoted here, you quoted a post from me, @PeterDonis, not @Nugatory.

the Big Ben Paradox thread is always closed at some point on every moderated physics platform. Usually in short order.

I'm glad to hear it. It's good to know that other moderated physics platforms are doing their jobs.

What other argument can be made against it, other than closure?

I can't say what arguments were or were not made on other platforms, but you already have good arguments against it here on PF, in this thread, which hasn't (yet) been closed. The fact that you are sticking your fingers in your ears and refusing to listen to them does not mean they aren't there.

Before you close the Big Ben Paradox thread, could you kindly provide us some links to other threads which were closed due to overconfidence?

You must be joking. This happens often enough here that the list of links would go on for pages.

 

[Nugatory]

OP has been temporarily blocked from this thread because they’re repeating the same misunderstanding. The thread remains open for anyone who has more to say about the correct treatment of this genuinely interesting problem.

 

[Dale]

@patdolan let me clarify the key issue that both @Nugatory and I have touched on.

First, what is not objectionable: the year and Big Ben have a fixed proportionality. So in a reference frame where Big Ben is time dilated the year is also time dilated by the same amount. That is not disputed.

What is objectionable: your claim that therefore “the earth's orbital velocity is only half the velocity necessary to keep the earth in stable orbit around the sun”.

To determine what velocity is necessary to keep the earth in stable orbit around the sun you have to use general relativity to do the calculations. You have not done so, so your claim is unfounded. Because the basis of GR is invariant tensors, we are guaranteed a priori that if the orbit is stable in one frame then it is stable in all frames. It is built into the math from the beginning. So your claim is also wrong.

This is what you need to address. You need to actually do the general relativistic calculations.

Your justification for not doing those calculations is

At non-relativistic velocities, such as the earth's orbital velocity around the sun, the Newtonian approximation to the GR solutions are close enough

This is a glaringly wrong justification since the velocities in this problem are around $0.867c$ which are highly relativistic. The Newtonian approximations are off by roughly a factor of 2. The full GR calculations are necessary.

 

[Orodruin]

But only if someone will actually do the calculations and show the work proving that the earth will maintain a stable geodesic around the sun in view of it's orbital velocity being halved (to which Dale agrees?). Will anyone step forward and do this?

You seem to be under the faulty impression that someone else must do these computations for you. You are the one making the extraordinary claim so you are the one that needs to provide the computation showing what you claim: that the Earth cannot maintain orbit. Of course, everyone in this thread already know what the result will be because we have actually studied GR and kniw that the outcome is invariant regardless of the coordinates used by construction.

 

[PeroK]

There's a simpler way to show the incompatibility of Newton's law of gravitation and special relativity.

Imagine an object dropped on Earth from a height of $5m$, say. In the local reference frame of the Earth's surface, the object takes $1s$ to hit the ground - as measured by a local clock.

Now, imagine the Earth receding at approximately $0.867c$ at a right angle to the direction the object falls. In this reference frame the local clock runs slow by a factor of 2, so the object takes $2s$ to fall, as measured by a distant clock. Yet there is no length contraction of the distance the object falls, as it falls at a right angle to the motion. Hence, the basic law of Newtonian gravitation does not apply.

Note also that if we try to invoke relativistic mass, where the Earth and the objet have twice their rest mass, then that only makes the calculation worse. In that case, in the frame in which the Earth is moving, the object should fall four times faster.

In any case, $F = \dfrac{GMm}{r^2}$ cannot apply in the case of relativistic velocities of the masses.

PS it was this incompatibility and the impossibility of resolving it that led to the General Theory in the first place.

 

[Orodruin]

There's a simpler way to show the incompatibility of Newton's law of gravitation and special relativity.

I don’t think this is the OP’s problem. I think the problem is that he thinks the GR prediction won’t change from the Newtonian one because orbital velocity is slow and gravity weak. Of course, neither assumption is correct once you boost the system to relativistic speeds.

 

[PeroK]

I don’t think this is the OP’s problem. I think the problem is that he thinks the GR prediction won’t change from the Newtonian one because orbital velocity is slow and gravity weak. Of course, neither assumption is correct once you boost the system to relativistic speeds.

That applies in my thought experiment as well. The orbital velocity in that case is zero. You could throw the object and the time to fall through its locally parabolic path would be the same.

 

[Demystifier]

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.

That's wrong, because the speed needed for that is not invariant. The needed speed depends on the velocity of the observer.

 

[Dale]

You are the one making the extraordinary claim so you are the one that needs to provide the computation

This is really the issue. @patdolan is making two extraordinary claims.

The first claim is the direct technical claim, that their “Big Ben paradox” shows that GR is self contradictory. That it predicts that a stable orbit in one frame is an unstable orbit in another frame.

The second claim is the implicit claim that the past 108 years of physicists missed this issue. Physicists who know this math inside and out. He is implicitly claiming that he is smarter than Hawking, Susskind, or even Einstein.

The first claim would be evidenced by correctly doing the rigorous GR-based calculation and showing the mathematical result. The second claim is already refuted by the OP demanding that others do the calculation for them.

 

[Vanadium 50]

genuinely interesting problem.

Is it?

When all the cruft is swept away, the question is about the orbit of a moving pair of bodies under mutual gravitational interaction. This has been known to require GR since before there was GR. (i.e. GR was created partially to solve this problem)

The OPs argument boils down to "I won't post the actually calculation myself, but I am sure everybody else is doing it wrong", which is hard to take seriously.

One can consider the related electromagnetic case, and in that case - which SR handles just fine - you don't get the right answer without considering magnetism as well. So there is no reason to think that the quasi-Newtonian perspective is anything but nonsense. This has been known since 1905.

 

[Ibix]

My tuppence ha'penny:

The weak field approximation involves an assumption that, if you write the metric in Cartesian coordinates, the metric only differs from Minkowski in the $tt$ term, and only by a small amount:$$g_{ab}=\left(
\begin{array}{cccc}
1+h_{tt}&0&0&0\\
0&-1&0&0\\
0&0&-1&0\\
0&0&0&-1
\end{array}\right)$$where $|h_{tt}|\ll 1$.

Anything that looks like a "rest frame of a ship approaching at speed $v$" is going to have a metric that has to look approximately like a Lorentz boosted version of this, so observers at rest with respect to the Sun are doing $-v$ relative to such a coordinate system's notion of rest. That means that the metric must look like$$\begin{eqnarray*}
g'_{ab}&=&\Lambda_a{}^c\Lambda_b{}^dg_{cd}\\
&=&\left( \begin{array}{cccc}
1+\gamma^2h_{tt}&0&0&v\gamma^2 h_{tt}\\
0&-1&0&0\\
0&0&-1&0\\
v\gamma^2h_{tt}&0&0&-1+v^2\gamma^2h_{tt}
\end{array}\right)
\end{eqnarray*}$$You see we have three extra components with non-zero perturbations which are of similar size to the $tt$ perturbation if $v$ is anywhere near 1 (working in units where $c=1$ here). Those extra perturbations mean that the observer approaching the solar system at high speed is wrong to try to apply naive Newtonian reasoning to estimate the orbit of the Earth. OP can either switch to the rest frame of the Sun, calculate the Earth's period using the Newtonian approximation, and transform the result, or can solve the geodesic equations in this more complex transformed metric. OP isn't doing either, and it's that neglect that leads into self-contradiction.

This maths is not really rigorous, since the Lorentz transforms aren't the correct global transforms here. But they can't be far off because $h_{tt}$ is small, and hence corrections to the result ought to be of order $h_{tt}^2$ - i.e., much smaller than the problem you have with those off-diagonal components. If OP wants to do everything rigorously, be my guest.

 

[Nugatory]

Is it?

“Interesting” is a bit subjective, but I find the followups interesting - when I wrote that bit about the question being interesting I was thinking along the lines of @Ibix's #25 above.

 

[Orodruin]

My tuppence ha'penny:

The weak field approximation involves an assumption that, if you write the metric in Cartesian coordinates, the metric only differs from Minkowski in the $tt$ term, and only by a small amount:$$g_{ab}=\left(
\begin{array}{cccc}
1+h_{tt}&0&0&0\\
0&-1&0&0\\
0&0&-1&0\\
0&0&0&-1
\end{array}\right)$$where $|h_{tt}|\ll 1$.

That’s not the general form of the metric in the weak field limit …

In harmonic gauge and a quasi-stationary setting, the trace reversed metric perturbation $\bar h$ can be argued to be dominated by the 00 component, but as far as I am aware this is not the case for the metric perturbation itself. Indeed, using harmonic gauge, the weak field approximation would be
$$
ds^2 = (1+2\phi)dt^2 - (1-2\phi)(dx^2 + dy^2 + dz^2)
$$
where $\phi$ is the Newtonian gravitational potential.

Anything that looks like a "rest frame of a ship approaching at speed $v$" is going to have a metric that has to look approximately like a Lorentz boosted version of this, so observers at rest with respect to the Sun are doing $-v$ relative to such a coordinate system's notion of rest. That means that the metric must look like$$\begin{eqnarray*}
g'_{ab}&=&\Lambda_a{}^c\Lambda_b{}^dg_{cd}\\
&=&\left( \begin{array}{cccc}
1+\gamma^2h_{tt}&0&0&v\gamma^2 h_{tt}\\
0&-1&0&0\\
0&0&-1&0\\
v\gamma^2h_{tt}&0&0&-1+v^2\gamma^2h_{tt}
\end{array}\right)
\end{eqnarray*}$$You see we have three extra components with non-zero perturbations which are of similar size to the $tt$ perturbation if $v$ is anywhere near 1 (working in units where $c=1$ here). Those extra perturbations mean that the observer approaching the solar system at high speed is wrong to try to apply naive Newtonian reasoning to estimate the orbit of the Earth. OP can either switch to the rest frame of the Sun, calculate the Earth's period using the Newtonian approximation, and transform the result, or can solve the geodesic equations in this more complex transformed metric. OP isn't doing either, and it's that neglect that leads into self-contradiction.

This maths is not really rigorous, since the Lorentz transforms aren't the correct global transforms here. But they can't be far off because $h_{tt}$ is small, and hence corrections to the result ought to be of order $h_{tt}^2$ - i.e., much smaller than the problem you have with those off-diagonal components. If OP wants to do everything rigorously, be my guest.

 

[Ibix]

That’s not the general form of the metric in the weak field limit …

In harmonic gauge and a quasi-stationary setting, the trace reversed metric perturbation $\bar h$ can be argued to be dominated by the 00 component, but as far as I am aware this is not the case for the metric perturbation itself. Indeed, using harmonic gauge, the weak field approximation would be
$$
ds^2 = (1+2\phi)dt^2 - (1-2\phi)(dx^2 + dy^2 + dz^2)
$$
where $\phi$ is the Newtonian gravitational potential.

<Hastily checks notes>

Drat, you're right. I don't think it changes the argument, though - we still get significant off-diagonal terms that we can't ignore.

 

[Orodruin]

<Hastily checks notes>

Drat, you're right. I don't think it changes the argument, though - we still get significant off-diagonal terms that we can't ignore.

Indeed, but I would be surprised if anything but an actual derivation of the correct orbital rate will satisfy the OP … or that OP has enough GR background to understand it …

 

[Orodruin]

This maths is not really rigorous, since the Lorentz transforms aren't the correct global transforms here. But they can't be far off because htt is small, and hence corrections to the result ought to be of order htt2 - i.e., much smaller than the problem you have with those off-diagonal components.

Also, regarding this: The Lorentz transformations are generally fine. As they define a global coordinate transformation there is absolutely nothing stopping you from using them. The worst that can happen is that you might get into cases where a coordinate gets lightlike or similar.

 

[Ibix]

Indeed, but I would be surprised if anything but an actual derivation of the correct orbital rate will satisfy the OP

Well, he can do that for himself. In the full GR calculation the orbital period is likely to be a function of time anyway since (looking at it in Schwarzschild coordinates) the gravitational time dilation between the Earth and the infaller is changing. Unless it's possible to construct a coordinate system that cancels that out somehow.

The Lorentz transformations are generally fine. As they define a global coordinate transformation there is absolutely nothing stopping you from using them. The worst that can happen is that you might get into cases where a coordinate gets lightlike or similar.

Fair enough - what I'm doing is a valid global coordinate change that has local properties I want, but there are no guarantees that the global properties are friendly.

 

[Vanadium 50]

Unless it's possible to construct a coordinate system that cancels that out somehow.

This sounds unlikely.

Remember, in GR, the Earth's orbit isn't even an ellipse. It's a rosette. What a "period" is is at least partially determined by convention. Hard to see how a coordinate change will hit on the right convention.

 

[morrobay]

If I may ask a B level question here: What is the correlation with gamma equals 2 and the 730.5 Big Bens little hand revolutions for every revolution the earth makes around the sun?

 

[PeroK]

If I may ask a B level question here: What is the correlation with gamma equals 2 and the 730.5 Big Bens little hand revolutions for every revolution the earth makes around the sun?

The little hand of a clock goes round twice per day, which is 730.5 times per year.

 

[morrobay]

The little hand of a clock goes round twice per day, which is 730.5 times per year.

Then there seems a disconnect: With gamma of 2 and Big Ben running slow ,1/2 of the proper time of the traveling observer. Then how is the clock stated to be still running normally during the revolution around the sun ?

 

[Dale]

Then there seems a disconnect: With gamma of 2 and Big Ben running slow ,1/2 of the proper time of the traveling observer. Then how is the clock stated to be still running normally during the revolution around the sun ?

Big Ben is a clock. The orbit of the earth around the sun is a clock. All clocks are time dilated equally.

 

[PeroK]

Then there seems a disconnect: With gamma of 2 and Big Ben running slow ,1/2 of the proper time of the traveling observer. Then how is the clock stated to be still running normally during the revolution around the sun ?

That's the definition of a year in the local reference frame of the solar system.

 

[Nugatory]

This thread is closed.
As with all such thread closures, if there is something else to add you can PM any mentor to ask that it reopened for a new contribution.

 

[Dale]

I wanted to add the actual GR math. The outcome of this is exactly as everyone who has any experience in GR said. Indeed, from first principles it could be no other way. But I had time yesterday to play around with this. 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: $$0 = \left(
\begin{array}{c}
0 \\
\frac{2 r^2 \left(z^2-2 M^2\right)
\ddot r+r \left(-z^2 \left(z^2-4
M^2\right) \ \dot \phi ^2-M \dot r^2 \left(2 M+3
\sqrt{r^2+z^2}\right)+M \dot z^2
\left(\sqrt{r^2+z^2}-2 M\right)+M
\left(\sqrt{r^2+z^2}-2
M\right)\right)+z \left(z
\left(z^2-4 M^2\right) \ddot r-4 M
\sqrt{r^2+z^2} \dot r
\dot z\right)+r^3 \ \dot phi^2 \left(M
\left(2 M+\sqrt{r^2+z^2}\right)-2
z^2\right)+r^4 \ddot r+r^5
\left(-\dot \phi
^2\right)}{\left(r^2+z^2\right)
\left(-4 M^2+r^2+z^2\right)} \\
\frac{2 \dot r \dot \phi \left(-4 M^2+r^2
\left(1-\frac{2
M}{\sqrt{r^2+z^2}}\right)+z^2\right)
+r \left(\left(r^2-4 M^2\right) \ddot \phi
-\frac{4 M z \dot z \dot \phi
}{\sqrt{r^2+z^2}}+z^2 \ddot \phi
\right)}{r \left(-4
M^2+r^2+z^2\right)} \\
\frac{r \left(r \left(r^2-4 M^2\right)
\ddot z-4 M \sqrt{r^2+z^2} \dot r
\dot z\right)+2 z^2 \left(r^2-2
M^2\right) \ddot z+M z \left(\dot r^2
\left(\sqrt{r^2+z^2}-2 M\right)-\dot z^2
\left(2 M+3
\sqrt{r^2+z^2}\right)+\left(\sqrt{r^
2+z^2}-2 M\right) \left(r^2 \dot \phi
^2+1\right)\right)+z^4
\dot z}{\left(r^2+z^2\right) \left(-4
M^2+r^2+z^2\right)} \\
\end{array}
\right) $$

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.

I will post the analysis in a moving frame in a separate post.

 

[Dale]

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 \dot \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 \dot r^2 \left(2 M \left(v^2-1\right)+4 v
\dot 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 \dot r (Z-T v)
\left(-\frac{2 M v \left(\left(v^2+1\right)
\dot Z^2-4 v
\dot Z+v^2+1\right)}{\sqrt{r^2-\frac{(Z-
T v)^2}{v^2-1}}}+\dot Z \left(v
\left(v^2-3\right) \dot 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\dot \phi^2
\left(\sqrt{r^2-\frac{(Z-T
v)^2}{v^2-1}}-2 M\right) \left(v \dot 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 \dot 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) \dot Z^2-4 v
\dot 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)}+\ddot r \\
\frac{\dot \phi \left(\frac{4 M
\left(v^2-1\right)^2 r \dot r \left(v
\dot 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)
\dot r^2+\left(v^2+1\right) \dot Z^2-4 v
\dot Z+v^2+1\right)}{\sqrt{r^2-\frac{(Z-
T v)^2}{v^2-1}}}+v \left(\left(v^2-1\right)
\dot r^2-v^2+3\right)-v \left(v^2-3\right)
\dot Z^2-4 \dot 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} \ddot \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)
\dot \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 \dot r \dot \phi
}{r} \\
\frac{\left(v \dot Z-1\right) \left(\frac{M
(Z-T v) \left(\frac{2 M
\left(\left(v^2-1\right)
\dot r^2+\left(v^2-1\right) r^2 \dot \phi
^2-\left(v^2+1\right) \dot Z^2+4 v
\dot Z-v^2-1\right)}{\sqrt{r^2-\frac{(Z-
T v)^2}{v^2-1}}}-\left(v^2-1\right)
\dot r^2-\left(v^2-1\right) r^2 \dot \phi
^2+v^2 \dot Z^2+4 v \dot Z-3 \dot 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 \dot r
\left(v-\dot 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)}+\ddot 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 $Z=v T$. 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.

Edit: so what did we actually learn from this exercise? The conclusion was exactly as everyone but the OP said. 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.

 

[Dale]

Based on discussions with the OP elsewhere, it seems that even after the math above the OP still has the misunderstanding expressed here:

If the earth does not spiral into the sun then the spacetime curvature in the vicinity of the sun has decreased by the factor 1/γ. From this it follows that spacetime curvature is not invariant, general relativity is wrong and starlight can't bend.

As shown above, the earth does not spiral into the sun in the boosted frame, and the orbit is time-dilated the same as Big Ben. However, the claim that this implies that spacetime curvature has decreased is false. This can be refuted by calculating a complete set of spacetime curvature invariants, such as the Carminati–McLenaghan invariants.

Here I evaluated all of the CM curvature invariants at $R=1.496 \ 10^{11}$ and $M=1480$ corresponding to the orbital radius of the earth and the mass of the sun respectively in geometrized units, as above. The boosted frame was boosted to $v=0.6$ and the curvature was evaluated at $z=Z=T=0$. The curvature is not a function of either $t$ or $\phi$.

$$\begin{array}{ c c c }
\text{Invariant} & \text{Sun's frame} & \text{Ship's frame} \\
\hline
R & 4.37 \ 10^{-38} & 4.37 \ 10^{-38} \\
R_1 & 6.26 \ 10^{-76} & 6.26 \ 10^{-76} \\
R_2 & -7.06 \ 10^{-114} & -7.06 \ 10^{-114} \\
R_3 & 2.05 \ 10^{-151} & 2.05 \ 10^{-151} \\
M_3 & 3.49 \ 10^{-136} & 3.49 \ 10^{-136} \\
M_4 & 1.38 \ 10^{-174} & 1.38 \ 10^{-174} \\
W_1 & 1.17 \ 10^{-60} + 1.17 \ 10^{-60} \ i & 1.17 \ 10^{-60} + 1.17 \ 10^{-60} \ i \\
W_2 & -5.18 \ 10^{-91} -5.18 \ 10^{-91} \ i & -5.18 \ 10^{-91} -5.18 \ 10^{-91} \ i \\
M_1 & -2.37 \ 10^{-106} - 2.37 \ 10^{-106} \ i & -2.37 \ 10^{-106} - 2.37 \ 10^{-106} \ i \\
M_2 & 3.49 \ 10^{-136} + 0 \ i & 3.49 \ 10^{-136} + 0 \ i\\
M_5 & -2.47 \ 10^{-166} + 0 \ i & -2.47 \ 10^{-166} + 0 \ i \\
\hline
\end{array}$$

This shows that the curvature is indeed invariant. Despite the changes to the components of the various tensors (including the curvature tensors), none of the invariants are changed (including the curvature invariants). This is a complete set of curvature invariants, so it completely characterizes the curvature tensor at the specified event.

That it must come out this way is a foregone conclusion from first principles, but it is good to see it demonstrated.

There are two additional facts that are worth noting with respect to this specific problem. First, the fact that all of these curvature invariants are very small supports the validity of the weak field approximation in the sun's frame. Second, the fact that they are the same in the ship's boosted frame shows that the boost is not an additional approximation. So, the weak field metric is a valid approximation in this problem, and the boost is not an additional approximation.