What is the Flaw of General Relativity Regarding Uniform Gravitational Fields?

[Zanket]

Numbered for reference:

1. The equivalence principle tells us that the crew of a rocket, traveling in flat spacetime and where the crew feels a constant acceleration, experiences a uniform gravitational field identical to that experienced locally by an observer on a planet.

2. Special relativity http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html that in principle the crew can traverse between any two points A and B in flat spacetime in an arbitrarily short proper time, where the two points are at rest with respect to each other and the rocket accelerates from rest at point A to the halfway point and then decelerates from the halfway point to rest at point B.

3. Then the crew can observe free-rising objects (such as an object floating stationary at the halfway point) to recede apparently arbitrarily fast—a million c is not out of the question—while causal contact is maintained since the actual velocity is always less than c.

Example: Let the rocket travel from Earth to Andromeda, two million light years away as we measure. (Assume that Earth and Andromeda are at rest with respect to each other and the spacetime between them is flat.) Let a buoy float stationary at the halfway point. Let the half of the trip from the buoy take ten proper years as the crew measures. Then during this half the buoy recedes by one million proper light years in ten proper years, an apparent (not actual) velocity of one hundred thousand c. From the crew's perspective the buoy free-rises in a uniform gravitational field.

4. Then why does general relativity not predict the same possible observation for the observer on the planet?

 

[Doc Al]

I assume you are referring to the fact that two objects a distance L_0 apart in their own frame can be passed by a rocket within an arbitrarily short time (according to the rocket) if the rocket moves fast enough with respect to the objects. But this is entirely due to the rocket's speed, not its acceleration.

 

[Zanket]

Which numbered point are you objecting to? How does what you're saying refute the point?

 

[Doc Al]

I fail to see any argument for your point (#4). #1 has to do with acceleration, while #2 & #3 have to do with speed. Your conclusion (#4) does not follow.

 

[Zanket]

You must disagree with #1, 2, or 3 for #4 to not follow. (I added some clarification. Take another look.) Free-rising objects have speed for either the crew or the observer on the planet.

 

[Doc Al]

#1 has nothing to do with #2 or #3. (Note that #2 & #3 apply regardless of acceleration.) #4 is a non sequitur.

 

[selfAdjoint]

You must disagree with #1, 2, or 3 for #4 to not follow. (I added some clarification. Take another look.) Free-rising objects have speed for either the crew or the observer on the planet.

No he doesn't. He just has to show that #4 doesn't FOLLOW from #1, #2, and #3. He did that, by pointing out that the acceleration mentioned in #1, and which you invoke in the gravity of #4, has nothing to do with the speed statements in #2 and #3. Three true statements that have nothing to do with one another do not lead to a conclusion, other than their union.

 

[Zanket]

#1 has nothing to do with #2 or #3. (Note that #2 & #3 apply regardless of acceleration.) #4 is a non sequitur.

#1 shows that whatever the crew can experience, the observer on the planet can also experience locally. #3 shows what the crew can experience, based on #2. Then if you agree with #3, #4 is a valid question.

 

[Zanket]

No he doesn't. He just has to show that #4 doesn't FOLLOW from #1, #2, and #3. He did that, by pointing out that the acceleration mentioned in #1, and which you invoke in the gravity of #4, has nothing to do with the speed statements in #2 and #3. Three true statements that have nothing to do with one another do not lead to a conclusion, other than their union.

See my reply to him above. If he agrees with #1, 2, and 3, then #4 is a valid question--it follows.

 

[ZapperZ]

#1 shows that whatever the crew can experience, the observer on the planet can also experience locally. #3 shows what the crew can experience, based on #2. Then if you agree with #3, #4 is a valid question.

You seem to think that "acceleration" implies "velocity". I think there's more a flaw in your understanding of kinematics rather than there's a flaw in GR.

I can show you something with zero velocity, yet it has an acceleration. Therefore, #2 and #3 that DEPENDS on velocity doesn't apply to #1, they are not automatically related. That is why you are being told that #4 makes no sense.

Zz.

 

[Doc Al]

#1 shows that whatever the crew can experience, the observer on the planet can also experience locally.

#1 states that the effect of the rocket's acceleration is equivalent to the planet's gravity.

#3 shows what the crew can experience, based on #2.

#3 and #2 discuss effects due to the rocket's speed not its acceleration. So #1 is irrelevent.

Then if you agree with #3, #4 is a valid question.

If the planet moves at the same speed with respect to those objects as does the rocket, then the planet observer will see the same speed-dependent effects. (Nothing to do with the equivalence principle.)

 

[Zanket]

I can show you something with zero velocity, yet it has an acceleration. Therefore, #2 and #3 that DEPENDS on velocity doesn't apply to #1, they are not automatically related. That is why you are being told that #4 makes no sense.

Zz.

Regardless of what #3 depends on, the observer on the planet should be able to experience the same observation as described in #3, according to #1. A free-rising object in the local frame of the observer on the planet has velocity relative to the observer.

Let's try this:

A. #1 shows that whatever the crew can experience, the observer on the planet can also experience locally.

B. #3 shows what the crew can experience, based on #2.

C. Then if you agree with #3, #4 is a valid question.

Which of these statements do you object to?

 

[Zanket]

#1 states that the effect of the rocket's acceleration is equivalent to the planet's gravity.

Given that, whatever the crew can experience, the observer on the planet can also experience locally.

#3 and #2 discuss effects due to the rocket's speed not its acceleration. So #1 is irrelevent.

#3 is based on #2. Neither are based on #1. According to #1, #3 is an effect that applies equally well to a free-rising object in the local frame of an observer on a planet. A free-rising object has speed relative to either the crew or the planetary observer.

If the planet moves at the same speed with respect to those objects as does the rocket, then the planet observer will see the same speed-dependent effects.

Then #4 is a valid question, since the free-rising objects in either case can move at the same speed.

 

[ZapperZ]

Regardless of what #3 depends on,

No, you can't sweep this under the carpet. It DOES depends on velocity. This is not negotiable because if you ignore this, you are ignoring SR. So then why are we even discussing this if you wish to make up your own laws?

Acceleration is not velocity.

Zz.

 

[George Jones]

1. The equivalence principle tells us that the crew of a rocket, traveling in flat spacetime and where the crew feels a constant acceleration, experiences a uniform gravitational field identical to that experienced locally by an observer on a planet.

This is roughly true.

2. Special relativity http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html that in principle the crew can traverse between any two points A and B in flat spacetime in an arbitrarily short proper time, where the two points are at rest with respect to each other and the rocket accelerates from rest at point A to the halfway point and then decelerates from the halfway point to rest at point B.

I agree completely.

3. Then they can observe free-rising objects (such as an object floating stationary at the halfway point) to recede apparently arbitrarily fast—a million c is not out of the question—while causal contact is maintained since the actual velocity is always less than c.

Here, I'm a little lost.

What does "recede apparently arbitrarily fast—a million c is not out of the question" mean?

Regards,
George

 

[Zanket]

No, you can't sweep this under the carpet. It DOES depends on velocity. This is not negotiable because if you ignore this, you are ignoring SR. So then why are we even discussing this if you wish to make up your own laws?

Acceleration is not velocity.

Zz.

I'm not disagreeing with you on what #3 depends on. If you object by saying that #3 depends on velocity, then I say that velocity applies as well to the observer on the planet, because a free-rising object in the local frame of the observer on the planet has velocity relative to the observer. Whatever the crew can observe (velocity, acceleration, whatever), so can the observer on the planet in a local frame, according to #1.

Then what is your objection?

 

[ZapperZ]

I'm not disagreeing with you on what #3 depends on. If you object by saying that #3 depends on velocity, then I say that velocity applies as well to the observer on the planet, because a free-rising object in the local frame of the observer on the planet has velocity relative to the observer. Whatever the crew can observe (velocity, acceleration, whatever), so can the observer on the planet in a local frame, according to #1.

Then what is your objection?

Show me an example of a "free-rising" object on a "planet" and show me how another inertial frame would observe this very same object in the idential way.

Zz.

 

[Doc Al]

I'm not disagreeing with you on what #3 depends on. If you object by saying that #3 depends on velocity, then I say that velocity applies as well to the observer on the planet, because a free-rising object in the local frame of the observer on the planet has velocity relative to the observer. Whatever the crew can observe (velocity, acceleration, whatever), so can the observer on the planet in a local frame, according to #1.

Then what is your objection?

If all you are saying is that an observer on a planet moving with the same speed as the rocket with respect to those objects will see the same speed-dependent effects as an observer on the rocket, then why all the mumbo jumbo with the equivalence principle? It's your argument that doesn't make sense, not that last statement (if that's what you were trying to say).

And what does this have to do with some supposed "flaw" in GR? What flaw?

 

[Zanket]

What does "recede apparently arbitrarily fast—a million c is not out of the question" mean?

An example: Suppose the rocket travels from Earth to Andromeda, 2 million light years away as we measure, in 20 proper years. (Assume spacetime between is flat.) Then in the crew's frame a buoy floating stationary at the halfway point, between passing the buoy and arriving at Andromeda, recedes one million proper light years in ten proper years, an apparent (not actual) velocity of one hundred thousand c. From the crew's perspective the buoy free-rises in a uniform gravitational field.

 

[Zanket]

Show me an example of a "free-rising" object on a "planet" ...

An apple thrown upwards.

... and show me how another inertial frame would observe this very same object in the idential way.

Why? The two frames discussed here, in which an object is free-rising, are non-inertial frames.

 

[George Jones]

If all you are saying is that an observer on a planet moving with the same speed as the rocket with respect to those objects will see the same speed-dependent effects as an observer on the rocket, then why all the mumbo jumbo with the equivalence principle?

Zanket is saying that the a person standing on the planet corresponds to the person accelerating in the rocket.

Maybe it would be better to consider a person using a rocket to hover above the surface of the planet. Then, I suppose the free-rising object corrresponds to a freely falling object.

But I'm not sure what the problem is.

Regards,
George

 

[George Jones]

An example: Suppose the rocket travels from Earth to Andromeda, 2 million light years away as we measure, in 20 proper years. (Assume spacetime between is flat.) Then in the crew's frame a buoy floating stationary at the halfway point, between passing the buoy and arriving at Andromeda, recedes one million proper light years in ten proper years

No, this isn't true.

Regards,
George

 

[ZapperZ]

An apple thrown upwards.



Why? The two frames discussed here, in which an object is free-rising, are non-inertial frames.

Wait a second! Did you think an apple thrown upwards in a gravitational field is IDENTICAL to a free-floating object as seen by an accelerating frame?

Zz.

 

[Zanket]

If all you are saying is that an observer on a planet moving with the same speed as the rocket with respect to those objects will see the same speed-dependent effects as an observer on the rocket, then why all the mumbo jumbo with the equivalence principle?

Because GR does not predict that, in the local frame of an observer on a planet, a free-rising object can recede apparently arbitrarily fast while causal contact is maintained, as the equivalence principle demands given that the crew can observe that. Then ...

And what does this have to do with some supposed "flaw" in GR? What flaw?

... GR is inconsistent.

 

[Zanket]

Zanket is saying that the a person standing on the planet corresponds to the person accelerating in the rocket.

Yes, in the local frame of the person on the planet.

But I'm not sure what the problem is.

The problem is that GR does not predict the same possible observation for the person on the planet as it does for the crew, even though the equivalence principle demands that it does.

 

[Zanket]

No, this isn't true.

Please be specific about what you think isn't true.

 

[Zanket]

Wait a second! Did you think an apple thrown upwards in a gravitational field is IDENTICAL to a free-floating object as seen by an accelerating frame?

Zz.

Not any free-floating object. But a free-rising object, yes. According to the equivalence principle, a free-rising apple in the crew's frame is equivalent to a free-rising apple in the local frame of the observer on the planet, all else being equal (like the acceleration they feel).

 

[ZapperZ]

Not any free-floating object. But a free-rising object, yes. According to the equivalence principle, a free-rising apple in the crew's frame is equivalent to a free-rising apple in the local frame of the observer on the planet, all else being equal (like the acceleration they feel).

Where does it say that? And besides, you never define what is a "free rising object" in an accelerated frame.

If this is where in both cases someone throws a ball "upwards", then where exactly do these two differ?

And if it is what I think it is, which is where you are equating a free falling ball in a gravitational field with a free object being observed in an accelerated frame, then I can immediately tell you that those two are NOT identical.

Zz.

 

[derz]

The equivalence principle says nothing more than that a constantly accelerated frame is equivalent to a homogenous gravitational field, i.e objects move the same way in both conditions.

This is apparently true. Keep in mind that "real" gravitational fields are never homogenous.

 

[George Jones]

Then in the crew's frame a buoy floating stationary at the halfway point, between passing the buoy and arriving at Andromeda, recedes one million proper light years in ten proper years

The distance that the buoy recedes in the crew's frame is not one million lightyears. For distances in an acclerated frame, see my posts #4 and #10 in https://www.physicsforums.com/showthread.php?t=110742&highlight=acclerated".

Exercise: What is the distance in the crew's frame?

Note also what ZapperZ and derz say, i.e., the gravitational field of a planet is not homogeneous, so that the metric can only be put into its special relativistic form at a single event.

Regards,
George

 

[Zanket]

Where does it say that?

Doesn't say it; it's given by it. If "the forces produced by gravity are equivalent to the forces produced by acceleration" (Encarta's definition), then the motion of a free-rising apple will be observed identically in either the crew's frame or the local frame of the observer on the planet, all else being equal.

And besides, you never define what is a "free rising object" in an accelerated frame.

You really need that? A rising object on which no forces act except gravity (in either the crew's frame or the local frame of the observer on the planet, who both experience a uniform gravitational field).

If this is where in both cases someone throws a ball "upwards", then where exactly do these two differ?

#1 suggests that the cases are equivalent according to the equivalence principle. #4 asks why GR does not predict the same possible observation in both cases.

And if it is what I think it is, which is where you are equating a free falling ball in a gravitational field with a free object being observed in an accelerated frame, then I can immediately tell you that those two are NOT identical.

I'm equating a free-rising object in the crew's frame with a free-rising object in the local frame of the observer on the planet, all else being equal.

 

[Zanket]

The equivalence principle says nothing more than that a constantly accelerated frame is equivalent to a homogenous gravitational field, i.e objects move the same way in both conditions.

Then do you have an answer for #4 in the original post? GR does not predict that objects move the same way as in #3 in both conditions, even though the equivalence principle demands that.

Keep in mind that "real" gravitational fields are never homogenous.

That's why I'm careful to specify a local frame of the observer on the planet. See more info in my reply to George Jones below.

 

[Zanket]

The distance that the buoy recedes in the crew's frame is not one million lightyears.

What do you disagree to:

A. When they pass the buoy, distance = zero.

B. Ten proper years later they are at rest with respect to the buoy, which is one million proper light years away.

For distances in an acclerated frame, see my posts #4 and #10 in https://www.physicsforums.com/showthread.php?t=110742&highlight=acclerated".

At the exact moment they come to rest with respect to the buoy, their trip is done. They could shut off their engines and would remain at rest with respect to the buoy. So acceleration is not an issue to the final measurement of distance.

Exercise: What is the distance in the crew's frame?

The final distance is one million light years.

Note also what ZapperZ and derz say, i.e., the gravitational field of a planet is not homogeneous, so that the metric can only be put into its special relativistic form at a single event.

That's why I'm careful to specify a local frame of the observer on the planet. It doesn't matter if the local frame is infinitesimal, question #4 in the original post still applies. But it helps to imagine a local frame that is larger, such as the local frames that extend from the ground to the upper atmosphere in which special relativity has been experimentally tested (muon experiment). The gravitational field therein negligibly differs from homogenous for that experiment; i.e., the results are not significantly skewed. A local frame (throughout which the tidal force is negligible but not necessarily zero) can in principle be as large as one can imagine.

 

[pervect]

Here is my $.02 on the issue.

If we adopt the metric

ds^2 = (1+gz)^2*dt^2 - dx^2 - dy^2 - dz^2

for 'the' coordinate system of an accelerated observer, Zanket's remarks that an unaccelerated trajectory exists so that the z coordinate is zero at a time coordinate of zero, and that the z coordinate is roughly 1 million at a time coordinate of 10 is correct, in my opinion.

If we use radar coordinates, as George suggests, the metric becomes something like

ds^2 = exp(g*z)(dt^2 - dz^2) - dx^2 - dy^2

Both are reasonable choices for coordinate systems, though I'll admit a personal preference for Zanket's choice. George's different choice was very useful in the thread above, howewver and was in fact the key to cracking the problem in that instance - my own preferences as to coordinate choices was one of the reasons I had a hard time solving the problem, while George came up with the solution relatively quickly.

Note that MTW's "Gravitation" uses Zanket's choice, which has the nice property that \Gamma^z{}_{zz}=0. Why this is a nice property is another topic, but it implies things about the uniformity of the tic marcs of the coordinate system in the z direction. (Readers familiar with parallel transport might ask the question about what happens when we parallel transport a 'z' vector along the 'z' axis).

So much of the argument is about the choice of coordinates, something that is not really "physical". Hopefully we can come to an agreement that Zanket's remarks are reasonable given his choice of coordinates, while also pointing out that some confusion was generated because Zanket assumed that his choice of coordinates was unique.

But now we address the original question.

The metric of a real planet is NEITHER of the above metrics. The metric for the gravity field of a planet is the Schwarzschild metric (or an equivalent formulation in other coordinates, such as isotropic coordinates).

Rather than write down the formula for the Schwarzschild or isotropic metrics, I want to explain why the metric of a planet cannot be either of the above. The technical answer is simple - the Riemann tensor evaluates to identically zero for both of the above metrics, as they represent flat space-time. The Riemann tensor is NOT zero for a planet, as space-time is curved around a planet - thus they cannot be the same.

This may be a bit heavy on the jargon, so let me try and explain the error without the technical language. The error being made here was to assume that a planet had a uniform gravitational field. As several posters have remarked, the gravitational field of a planet is not uniform (in fact it falls off as 1/r^2 in the Newtonian limit). This explains the difference in what we see from an accelerated spaceship and from a planet - the physical situation is not the same, because the planet has a 1/r^2 gravitational field, while the spaceship has a uniform (from the Newtonian POV) field.

It should therefore be utterly non-surprising that we get different results for the trajectory of a free-falling object in a planet's non-uniform gravitational field compared to the trajectory of a free-falling object in a spaceships uniform gravitational field.

In terms of the numbered arguments, the equivalence principle was misinterpreted (#1). What the equivalence principle says is that it is possible to create a laboratory small enough where one can get equivalent results from a gravitational field and an accelerating spaceship.

It is not true that it's possible to extend the results to an arbitrarily large laboratory, which is the error in Zanket's argument.

 

[ZapperZ]

I'm equating a free-rising object in the crew's frame with a free-rising object in the local frame of the observer on the planet, all else being equal.

Again, you are being very vague in describing the scenario.

Case 1:
I have a ball and I'm standing on earth. I throw it up in the air.

I have a ball and I'm in an accelerating space ship. I throw it up "vertically", which is in the same direction of my acceleration.

This the above the scenario you are asking for? Or is it the one below?

Case 2:
I have a ball and I'm standing on earth. I throw it up in the air.

I have a ball that I'm observing that is moving with a constant velocity v at that instant in the same direction as my accelerating spaceship.

Zz.

 

[Zanket]

It is not true that it's possible to extend the results to an arbitrarily large laboratory, which is the error in Zanket's argument.

I don't do that. #4 in the original post applies to an arbitrarily small laboratory. Keep in mind that special relativity has been experimentally tested in rather large "laboratories", like those extending from the ground to the upper atmosphere (muon experiment). In any given sized laboratory, the tidal force throughout can in principle be arbitrarily small so as to only negligibly skew the results of an experiment.

 

[Zanket]

Case 1:
I have a ball and I'm standing on earth. I throw it up in the air.

I have a ball and I'm in an accelerating space ship. I throw it up "vertically", which is in the same direction of my acceleration.

This the above the scenario you are asking for? Or is it the one below?

It's the case above. In #3 in the original post, a ball thrown upward outside the rocket at the exact moment the crew passes the buoy (assume they've just begun their deceleration phase), and thrown upward at the same relative velocity as the buoy, remains at rest with respect to the buoy. The ball's movement would track with the buoy's. And the buoy can recede apparently arbitrarily fast in the crew's frame.

In #4, I ask why GR does not predict such possible observations for the local frame of the observer on the planet.

 

[RandallB]

The flaw in “A Flaw of General Relativity?”

1. The equivalence principle tells us that the crew of a rocket, traveling in flat spacetime and where the crew feels a constant acceleration, experiences a uniform gravitational field identical to that experienced locally by an observer on a planet.

Planets do not have a uniform gravitational field, but we can imagine one.

The real flaw in your statements comes here:

3. Then the crew can observe free-rising objects (such as an object floating stationary at the halfway point) ……………

I assume you’re taking about an object as if it were a helium balloon released on the planets surface rising away from the stationary observer.
Your flaw is, where exactly do you see a “free-rising” object while traveling in the ship you describe.
In detail:
Earth, mid point buoy & Andromeda all in a common reference frame.
While traveling from Earth you observe the buoy coming towards you at an ever-increasing speed just as you’d expect someone dropping something from high above while on the planet. No problem.

Then you reach the midway point while the buoy has been “falling” down towards you all this time and has reached a very high speed as you said. You turn the ship around and prepare to restart engines to decelerate. The buoy would have “dropped” by you had you not turned around but as you start to decelerate after turning around you can clearly see the buoy has an initial velocity UP from your view as you begin to re-experience gravity in the deceleration – and so does the buoy as you observe it. You can see its speed slowing down in accord with the “gravity” you feel, just as if you had tossed a ball straight up while on a planet. Again – no problem for GR.
By the time you reach Andromeda, the buoy has slowed enough because of the “gravity” of your acceleration to bring it to a complete stop, just as your stopping at Andromeda.
If you don’t turn your engines off, it will start to fall towards you from that stopped position as you depart Andromeda.

Do you see there is no “free-rising object” observed from the ship?
GR does predicts the same thing in both places.
Your arbitrarily large laboratory is fine, you just need to take your measurements correctly.

 

[ZapperZ]

It's the case above. In #3 in the original post, a ball thrown upward outside the rocket at the exact moment the crew passes the buoy (assume they've just begun their deceleration phase), and thrown upward at the same relative velocity as the buoy, remains at rest with respect to the buoy. The ball's movement would track with the buoy's. And the buoy can recede apparently arbitrarily fast in the crew's frame.

In #4, I ask why GR does not predict such possible observations for the local frame of the observer on the planet.

And you do not see why the two situations are NOT identical?

In the accelerating ship, once the ball leaves the "hand" it is moving with a constant velocity. I can always transform myself to an inertial frame where that ball is at rest.

In the second situation, I can't do that. A ball being thrown upwards in a gravitational field never maintain a constant velocity. There are NO inertial frame to which I can transform to for that ball to be at rest.

So your insistance on applying the equivalence principle here is faulty. These two are not "equivalent" and GR does not insist that they are.

Zz.

 

[pervect]

I don't do that. #4 in the original post applies to an arbitrarily small laboratory. Keep in mind that special relativity has been experimentally tested in rather large "laboratories", like those extending from the ground to the upper atmosphere (muon experiment). In any given sized laboratory, the tidal force throughout can in principle be arbitrarily small so as to only negligibly skew the results of an experiment.

I'm afraid I'm not following your argument then - possibly I'm not even aware of what you think the problem is.

It appeared to me that you were claiming that the trajectory of a free-falling object on a planet would be the same as the trajectory of a free-falling object on a spaceship. This is obviously not a true statement.

Looking closer at what you actually ask, you make a bunch of statements, you introduce an example of a rocket traveling from here to Andromeda (does the rocket break to a stop? I'm not quite sure), then you finally ask.

Then why does general relativity not predict the same possible observation for the observer on the planet?

Unfortunately this is extremely vague.

To which "possible observation" do you refer? Why such vague language? ("possible" observation, not "observation"). You apparently had something specific in mind, but it's not clear as to what it was from just reading your post :-(.

Why do you think that GR does not predict the "same possible observation"? You make a bunch of statements as assumptions (that is good), but you do not present a chain of logic that starts with these assumptions and then reaches your conclusion. (This chain of logic would help disambiguate your unfortunately vague question).

Basically, I'm not following you (apparently nobody else is either).

Are you perhaps concerned about the fact that the rate of change of a position coordinate with respect to a time coordinate can be greater than 'c'? If that's your quesiton, I can answer it, but I don't think it's worthwhile to answer it at this point until I'm sure that that's what your question is.

 

[Zanket]

Planets do not have a uniform gravitational field, but we can imagine one.

That’s what the qualifier “locally” is for in #1, where “local” means “of a range throughout which the tidal force is negligible.” While planets have a nonuniform gravitational field, the uniformity of an arbitrary planet's gravitational field can in principle negligibly differ from perfect throughout any range desired for the purposes of a given experiment.

Your flaw is, where exactly do you see a “free-rising” object while traveling in the ship you describe.

Here:

You can see its speed slowing down in accord with the “gravity” you feel, just as if you had tossed a ball straight up while on a planet. Again – no problem for GR.
By the time you reach Andromeda, the buoy has slowed enough because of the “gravity” of your acceleration to bring it to a complete stop, just as your stopping at Andromeda.

You described a free-rising object (maybe there's a better term; by "free-rising" I mean that it is in free fall, rising upward). According to the equivalence principle, this case negligibly differs from that of projecting a buoy upward within the local frame of an observer on a planet, all else being equal.

GR does predicts the same thing in both places.

GR does not predict the same thing in both places. In the local frame of an observer on a planet, GR does not predict that objects rising freely (e.g. thrown upward) can recede to any given proper distance in an arbitrarily short proper time while causal contact is maintained, as the crew can observe in #3.

 

[Zanket]

In the accelerating ship, once the ball leaves the "hand" it is moving with a constant velocity.

Not in the frame of the crew, the frame in #3. In that frame a ball (or buoy) thrown upward decelerates.

A ball being thrown upwards in a gravitational field [in the local frame of an observer on a planet] never maintain a constant velocity.

Yep, just like in the crew’s frame.

So your insistance on applying the equivalence principle here is faulty. These two are not "equivalent" and GR does not insist that they are.

According to the equivalence principle, the two situations above (in the frames I specify) negligibly differ, all else being equal. (By "local" in "local frame" I mean "of a range throughout which the tidal force is negligible".)

Going back to something you said before:

Again, you are being very vague in describing the scenario.

Please, what shorter terminology if any means "an object in free fall, rising upward in the given frame" to you, if not a "free-rising object"? I wish to be less confusing.

 

[Zanket]

It appeared to me that you were claiming that the trajectory of a free-falling object on a planet would be the same as the trajectory of a free-falling object on a spaceship. This is obviously not a true statement.

According to the equivalence principle, in the local frame of an observer on a planet the trajectory would negligibly differ, all else being equal. “Local” here means “of a range throughout which the tidal force is negligible”.

Looking closer at what you actually ask, you make a bunch of statements, you introduce an example of a rocket traveling from here to Andromeda (does the rocket break to a stop? I'm not quite sure), ...

#2 says “the rocket ... decelerates from the halfway point to rest at point B”. So it brakes to a stop.

To which "possible observation" do you refer?

#4 refers to the observation described in #3.

Why such vague language? ("possible" observation, not "observation").

In #4 I’m saying, why doesn’t general relativity predict that the observer on the planet can (i.e. possibly) observe the same as the crew does?

Why do you think that GR does not predict the "same possible observation"?

In the local frame of an observer on a planet, GR does not predict that objects rising freely (e.g. thrown upward) can recede to any given proper distance in an arbitrarily short proper time while causal contact is maintained, as the crew can observe.

 

[derz]

2. Special relativity http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html that in principle the crew can traverse between any two points A and B in flat spacetime in an arbitrarily short proper time --

Not in "arbitrarily short proper time", since the crew will always observe their velocity being less than c relative to every other inertial frame.

3. Then the crew can observe free-rising objects (such as an object floating stationary at the halfway point) to recede apparently arbitrarily fast—a million c is not out of the question-

"A million times c" is out of the question. The crew will always observe the objects velocity lower than c.

Then during this half the buoy recedes by one million proper light years in ten proper years, an apparent (not actual) velocity of one hundred thousand c.

According to the crew the buoy is only a little over 10 lightyears away; Andromeda 20 lightyears away. Assuming that the spaceship's speed is nearly c relative to Andromeda.

 

[ZapperZ]

Not in the frame of the crew, the frame in #3. In that frame a ball (or buoy) thrown upward decelerates.



Yep, just like in the crew’s frame.



According to the equivalence principle, the two situations above (in the frames I specify) negligibly differ, all else being equal. (By "local" in "local frame" I mean "of a range throughout which the tidal force is negligible".)

And this is where you make your fatal error. You never once addressed the fact that I can always transform myself to an inertial frame in which the ball thrown in the accelerating frame is at rest, whereas I can't in the second. These are NOT identical situations dispite your misinterpretation of GR.

Zz.

 

[Zanket]

Not in "arbitrarily short proper time", since the crew will always observe their velocity being less than c relative to every other inertial frame.

Yes in an arbitrarily short proper time, according to the equations http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html , even though the velocity is always less than c.

"A million times c" is out of the question. The crew will always observe the objects velocity lower than c.

In #3 I say “apparently arbitrarily fast—a million c is not out of the question” and “the actual velocity is always less than c”. I say “the buoy recedes by one million proper light years in ten proper years, an apparent (not actual) velocity of one hundred thousand c”.

Then during this half the buoy recedes by one million proper light years in ten proper years, an apparent (not actual) velocity of one hundred thousand c.

According to the crew the buoy is only a little over 10 lightyears away; Andromeda 20 lightyears away. Assuming that the spaceship's speed is nearly c relative to Andromeda.

The “half” referenced is the latter half of the crew’s whole trip, their trip from the buoy to Andromeda. During this half the crew’s distance to the buoy is initially zero and, ten proper years later when the rocket arrives at Andromeda at rest with respect to both it and the buoy, the crew’s distance to the buoy is one million proper light years.

 

[Zanket]

And this is where you make your fatal error. You never once addressed the fact that I can always transform myself to an inertial frame in which the ball thrown in the accelerating frame is at rest, whereas I can't in the second. These are NOT identical situations dispite your misinterpretation of GR.

It’s not a fatal error, because the two situations can negligibly differ, as I pointed out to you. In the second situation you can be an ant sitting on the ball, with the ball at rest with respect to you. When you said that you can’t do that I assumed that your point was that there are no perfectly uniform gravitational fields for a planet, and I addressed that point. Did I assume right, or were you making a different point?

 

[ZapperZ]

It’s not a fatal error, because the two situations can negligibly differ, as I pointed out to you. In the second situation you can be an ant sitting on the ball, with the ball at rest with respect to you. When you said that you can’t do that I assumed that your point was that there are no perfectly uniform gravitational fields for a planet, and I addressed that point. Did I assume right, or were you making a different point?

Think again. If the ball was thrown in a gravitational field, at NO POINT in time does it ever have a constant velocity, be it in a uniform or none uniform field. You cannot transform to any inertial frame in which the ball is at rest! On the other hand, the ball that was thrown in an accelerating frame can! There's nothing "negligibly" different here. It's A LOT different.

Zz.

 

[RandallB]

That’s what the qualifier “locally” is for in #1, where “local” means “of a range throughout which the tidal force is negligible.”

As I said we can imagine...

You described a free-rising object (maybe there's a better term; by "free-rising" I mean that it is in free fall, rising upward). According to the equivalence principle, this case negligibly differs from that of projecting a buoy upward within the local frame of an observer on a planet, all else being equal.

No I didn't - Rising upward would mean acelerating upward. The object has upward speed BUT that speed is slowing down due to the 'gravity' the travelers feel. That slowing down is the same as FALLING.

GR does not predict the same thing in both places. In the local frame of an observer on a planet, GR does not predict that objects rising freely (e.g. thrown upward) can recede to any given proper distance in an arbitrarily short proper time while causal contact is maintained, as the crew can observe in #3.

Of course that's the same as seen on a planet - you just keep calling an intial velocity up as "rising" even when it's slowing down - that does cut it, your just WRONG. You need to define all of what the object is really doing - like stopping its "rise" when you reach Andromeda and then your “free rising” object with no change in gravity (that is you don’t turn your engines off at Andromeda and your artificial gravity is taking you back to earth) is now falling on you!

I notice you didn’t address that small detail in my comments.
“If you don’t turn your engines off, it will start to fall towards you from that stopped position as you depart Andromeda.”

Care to explain just how your free rising object decides to start falling with no change in gravity?? Seems exactly like gravity (uniform tidal negligible if you want) on a planet too me.

 

[pervect]

In the local frame of an observer on a planet, GR does not predict that objects rising freely (e.g. thrown upward) can recede to any given proper distance in an arbitrarily short proper time while causal contact is maintained, as the crew can observe.

GR does predict this, it just takes a "planet" that is on the verge of becoming a black hole, so that one can get a large enough time dilation factor.

Lets's suppose we have an actual black hole, and a sufficiently powerful rocket and a sufficiently powerful engine so that we can hover arbitrarily close to the event horizon. This hovering rocket is equivalent (in a GR sense) to living on a "large enough" planet.

By hovering close enough to the event horizon of the black hole, one can obtain a very large time dilation factor. This allows one to watch the universe (far away from the black hole) age quickly, like a film being played "fast-forwards".

This quick aging process would allow objects following a geodesic (what you call free-rising) to reach an arbitrarily large distance in an arbitrarily short amount of your time, while maintaining causal contact, the exact features you want to duplicate as nearly as I can tell from your text. (You will have to move arbitrarily close to the event horizon to achieve this).

Note that the time dilation factor is NOT related to the value of acceleration required to hold station, but to the total potential energy of the hovering observer. This may be a subtle point that you are missing. In terms of your numerically listed principles, two trajectories can be identical as required by the equivalence principle, but appear to be different because different coordinate systems have been adopted. One needs to develop more structure on how coordinates behave in order to correctly compare trajectories at different locations in space - something equivalent to GR's notion of "parallel transport".