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• patdolan
patdolan
Consider a distant observer traveling at .867 c ( đť›ľ=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.

bhobba and weirdoguy
patdolan said:
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.

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

patdolan said:
Will the earth spiral into the sun? If not, why not?

No. Because this is not true:

patdolan said:
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.

patdolan said:
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.

Klystron, bhobba, russ_watters and 2 others
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.

pinball1970, russ_watters, Vanadium 50 and 4 others
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.

bhobba, weirdoguy, PeroK and 1 other person
Nugatory said:
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.

patdolan said:
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 Newtonâ€™s law under conditions where it does not apply.

bhobba, russ_watters and Dale
patdolan said:
The above constitutes a proof sufficient to demolish your objection, Dale.
patdolan said:
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.

bhobba, russ_watters and Dale
patdolan said:
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.

bhobba and PeterDonis
PeterDonis said:
@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.

bhobba, weirdoguy and PeroK
patdolan said:
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 â€śNo Daleâ€ť belongs with the rest.
patdolan said:
The above constitutes a proof sufficient to demolish your objection, Dale.
No it doesnâ€™t. It doesnâ€™t have anything to do with my objection. My objection is that your claim â€śthe 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 sunâ€ť is wrong and unfounded.

patdolan said:
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##

bhobba, Orodruin and PeterDonis
And why is that?

patdolan said:
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. Thatâ€™s a relativistic velocity, and the Newtonian approximation wonâ€™t work.

Iâ€™ve already suggested that doing the velocity addition calculation (non-parallel case) would be a good exercise. Try it, and youâ€™ll see more clearly why the Newtonian approximation doesnâ€™t work here. (And do remember that the Newtonian approximation is a special case of Schwarzschild coordinates in which the sun is at rest).

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bhobba, PeterDonis and Dale
patdolan said:
And why is that?
Because you didnâ€™t calculate the required velocity using GR

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

patdolan said:
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.

patdolan said:
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.

patdolan said:
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.

russ_watters and Dale
OP has been temporarily blocked from this thread because theyâ€™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.

bhobba and russ_watters
@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 â€śthe earth's orbital velocity is only half the velocity necessary to keep the earth in stable orbit around the sunâ€ť.

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
patdolan said:
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.

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Klystron, bhobba, russ_watters and 2 others
patdolan said:
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.

Dale and bhobba
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.

SiennaTheGr8, Hornbein and hutchphd
PeroK said:
There's a simpler way to show the incompatibility of Newton's law of gravitation and special relativity.
I donâ€™t think this is the OPâ€™s problem. I think the problem is that he thinks the GR prediction wonâ€™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.

Orodruin said:
I donâ€™t think this is the OPâ€™s problem. I think the problem is that he thinks the GR prediction wonâ€™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.

russ_watters and Dale
patdolan said:
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.

Orodruin said:
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 â€śBig Ben paradoxâ€ť 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.

Klystron, Gleb1964, hutchphd and 3 others
Nugatory said:
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.

ersmith, russ_watters, PeterDonis and 1 other person
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.

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PeterDonis, Dale and Nugatory
Is it?
â€śInterestingâ€ť 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.

Ibix
Ibix said:
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##.
Thatâ€™s not the general form of the metric in the weak field limit â€¦

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.

Ibix said:
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.

Dale
Orodruin said:
Thatâ€™s not the general form of the metric in the weak field limit â€¦

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.

Nugatory and Dale
Ibix said:
<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 â€¦ or that OP has enough GR background to understand it â€¦

Ibix and Motore
Ibix said:
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
Orodruin said:
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.
Orodruin said:
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.

Ibix said:
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.

bhobba
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?

morrobay said:
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
PeroK said:
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 ?

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