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

[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".

 

[Zanket]

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.

Now I am completely confused as to what you’re getting at. And I really do want to understand what you’re getting at.

Let the gravitational field of the planet be uniform. In this case, according to the equivalence principle, “it is theoretically impossible to distinguish between gravitational and accelerational forces by experiment” (Encarta). Then how could the situations possibly be a LOT different? No experiment could detect a difference, all else being equal, right? In either situation you can transform to any inertial frame in which the ball is at rest—just be an ant sitting on the ball in either situation. Please elaborate as to how you think an ant sitting on the ball in the planet’s field is not in an inertial frame in which the ball is at rest.

 

[Zanket]

No I didn't - Rising upward would mean acelerating upward.

When I throw an apple upward, it is rising upward but not accelerating upward. Then they don’t mean the same.

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.

The object is in free fall, but not falling. The dictionary definition of “falling” is “moving downward”. The dictionary definition of “rising” is “ascending”. These terms are not changed in physics. The buoy ascends in their frame, it does not move downward; hence it is rising, not falling. As I said, maybe there's a better term, but by "free-rising" I mean that it is in free fall, rising upward. Do you really think that I should call an object that is ascending a “falling” object? How would I distinguish such an object from one that is “moving downward”?

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.

I call it “rising” because it meets the dictionary definition. An object is rising when it’s ascending, whether or not it is slowing down. A rocket accelerating upward, as well as an apple thrown upward, rises.

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!

What happens after they reach Andromeda is irrelevant to #3, which deals with objects in free fall that are rising, not falling. I agree that if they don’t turn their engines off at Andromeda then the buoy falls toward them, but that is irrelevant. I don’t see how that fact addresses #4.

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.”

I agree with this small detail but I didn’t address it because it’s irrelevant, as above.

Care to explain just how your free rising object decides to start falling with no change in gravity??

No, I don’t care to address anything that happens after they arrive at Andromeda, because it’s irrelevant, as above. I want to stick to the case I put forth, which ends when they arrive at Andromeda. The case ends when the object that is rising upward reaches its highest point.

Seems exactly like gravity (uniform tidal negligible if you want) on a planet too me.

The situations are not the same according to GR, even though the equivalence principle demands that they can negligibly differ. In the local frame of an observer on a planet, GR does not predict that objects rising in free fall (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. For example, in the local frame of an observer on a planet (a frame that we agree can be as large as imaginable), a buoy thrown upward cannot recede to a distance of one million proper light years in ten proper years while causal contact is maintained.

How do you think what can happen after the crew arrives at Andromeda, i.e. after the buoy reaches its highest point, resolves that? I don’t get why you even bring that up.

 

[ZapperZ]

Now I am completely confused as to what you’re getting at. And I really do want to understand what you’re getting at.

Let the gravitational field of the planet be uniform. In this case, according to the equivalence principle, “it is theoretically impossible to distinguish between gravitational and accelerational forces by experiment” (Encarta). Then how could the situations possibly be a LOT different? No experiment could detect a difference, all else being equal, right? In either situation you can transform to any inertial frame in which the ball is at rest—just be an ant sitting on the ball in either situation. Please elaborate as to how you think an ant sitting on the ball in the planet’s field is not in an inertial frame in which the ball is at rest.

Put yourself on a spaceship that is far enough away from that planet and in which it is in the same inertial frame as the planet (assuming there's no other bodies around). Now look at the ball being throw vertically up on the planet. You (the one in the "stationary" ship) will see the ball decelerating before it turns around and accelerates back to the ground.

Now tell me what inertial frame can you transform to be stationary with the ball. NONE! The ball is being accelerated by the gravitational field continuously. There is always an "F" acting on the ball!

Compare this with another spaceship which is accelerating "upwards". After the ball is thrown, there's no "F" left acting on it. It moves with a constant velocity. I can easily transform to an inertial frame in which the ball is at rest in that frame.

These two are NOT identical. There's no symmetry between the two. This is the VERY reason why we can tell that it is US that is orbiting around the sun and NOT the sun orbiting around us even when it appears that way just because the sun rises and sets every day.

Zz.

 

[RandallB]

When I throw an apple upward, it is rising upward but not accelerating upward. Then they don’t mean the same.

The object is in free fall, but not falling. The dictionary definition of “falling” is “moving downward”.

O come on Zanket .
We aren’t talking a dictionary world here, this is physics!
You need to decide what you mean by “FREE RISING” that's not in the dictionary so you tell us.
When you hit a pop-up in baseball how is it not “Free-Rising”. Just like your space buoy?
Yes or no is your object slowing down or not, does stop rising when you reach Andromeda or not.

You agree that if they don’t turn their engines off at Andromeda then the buoy falls toward them, but some how that is irrelevant?
Why?
From that point, you must agree then that it “falls” just like the baseball does for the pop-up once it reaches it high point. So, on the ship how do you make a ball do the same thing? How fast and how high would you have to “pop it up” for it to fall just like the buoy will fall if you let the “gravity” continue after reaching Andromeda?
Just how close will it be to the buoy? If you can not solve this problem you certainly cann’t work the one you’ve put forward.

 

[Zanket]

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.

This is a good post. You're getting to the heart of the matter. I'm going to think about this and answer it later, in the meantime responding to other posts.

 

[Zanket]

Put yourself on a spaceship that is far enough away from that planet and in which it is in the same inertial frame as the planet (assuming there's no other bodies around).

You mean a spaceship in free fall that is far enough away that the ship essentially stays at rest with respect to the planet? I’ll assume that for now.

Now look at the ball being throw vertically up on the planet. You (the one in the "stationary" ship) will see the ball decelerating before it turns around and accelerates back to the ground.

Agreed.

Now tell me what inertial frame can you transform to be stationary with the ball. NONE!

I did that already. I transform to the inertial frame of an ant sitting on the ball (well, clinging, since the ant is weightless). Then it’s possible and not NONE, right?

The ball is being accelerated by the gravitational field continuously. There is always an "F" acting on the ball!

That’s the Newtonian viewpoint. It is not gravity we feel, Einstein says, but simply the ground pushing up on our feet. Likewise, in an accelerating rocket, it is not gravity the crew feels, but simply the rocket pushing up on their feet. Gravity is not a force in general relativity. No F acts on the ball in either situation.

Compare this with another spaceship which is accelerating "upwards". After the ball is thrown, there's no "F" left acting on it. It moves with a constant velocity. I can easily transform to an inertial frame in which the ball is at rest in that frame.

In either situation, after the ball is thrown, there's no "F" left acting on it, as above. It moves with a constant velocity in either situation, in the frame of an ant clinging to the ball.

These two are NOT identical. There's no symmetry between the two.

I still don’t see it. How can the ant not be stationary with the ball in both situations? How could the situations possibly be different when, according to the equivalence principle, “it is theoretically impossible to distinguish between gravitational and accelerational forces by experiment” (Encarta). You haven't answered these questions.

This is the VERY reason why we can tell that it is US that is orbiting around the sun and NOT the sun orbiting around us even when it appears that way just because the sun rises and sets every day.

I thought both we and the Sun orbited around the Sun-Earth barycenter, in the absence of other bodies.

 

[ZapperZ]

You mean a spaceship in free fall that is far enough away that the ship essentially stays at rest with respect to the planet? I’ll assume that for now.
Agreed.
I did that already. I transform to the inertial frame of an ant sitting on the ball (well, clinging, since the ant is weightless). Then it’s possible and not NONE, right?

Excuse me? The ball is ACCELERATING. How is it possible to transform to an inertial frame of the ball? Does your dictionary have no definition of an "inertial frame"?

That’s the Newtonian viewpoint. It is not gravity we feel, Einstein says, but simply the ground pushing up on our feet. Likewise, in an accelerating rocket, it is not gravity the crew feels, but simply the rocket pushing up on their feet. Gravity is not a force in general relativity. No F acts on the ball in either situation.

You're contradicting yourself. You just agreed with me that from the point of view of an inertial frame, the ball is accelerating. Now you are telling me there's no "F". So the ball is accelerating all on its own due to... what? Remember, this is from the point of view of an observer in an inertial frame!

But then again, I don't think you know what an inertial frame is judging from your comments above. It makes the rest of your reply (and probably the whole thread) moot.

Zz.

 

[Zanket]

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.

...

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).

GR does not predict this. It predicts that an observable receding object recedes at an observable rate (proper distance over proper time) always less than c. That is the basis of the horizon problem in cosmology based on GR. What you describe, and what I describe in #3, is an observable rate of xc, where x is any factor >= 0. GR predicts that objects receding at a rate where x >= 1 lie beyond a cosmological horizon, where they are unobservable; i.e. there is no causal contact with them.

Now I say, and you seem to agree, that GR must predict the same for both situations (i.e. for both the crew and the observer on the planet). This is demanded by special relativity and the equivalence principle, components of GR. Then GR is inconsistent. When the same is predicted for both situations, the horizon problem can be explained away by gravity.

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.

I think #3 implies this already. The crew can travel from Earth to Andromeda in 20 proper years at a rate of acceleration just over 1 Earth gravity. At this relativity low rate of acceleration they can observe the buoy, once they pass it, to recede at an average observable (not actual) rate of one hundred thousand c. You are correct that the rate of acceleration is immaterial.

 

[Hurkyl]

GR does not predict this. It predicts that an observable receding object recedes at an observable rate (proper distance over proper time) always less than c.

As you've stated this, it's patently false. Remember that any coordinate chart is perfectly valid -- which includes coordinate charts where this distant object has a coordinate velocity millions of times of c.

(And don't forget that light is only guaranteed to travel at c near the origin of "good" coordinate charts. It could be anything at all near the origin of a "bad" coordinate chart, as well as anywhere that isn't at the origin of a "good" coordinate chart)

(I'll confess to not having read the entire thread in depth, so my apologies if you've agreed upon something that makes your statement true)

 

[pervect]

GR does not predict this. It predicts that an observable receding object recedes at an observable rate (proper distance over proper time) always less than c

This is where you are wrong. This is a somewhat confusing topic, but I've dug up a few references to demonstrate this. You might also want to read Hurkyl's post on the same point.

http://casa.colorado.edu/~ajsh/sr/postulate.html

In general relativity, arbitrarily weird coordinate systems are allowed, and light need move neither in straight lines nor at constant velocity with respect to bizarre coordinates (why should it, if the labelling of space and time is totally arbitrary?). However, general relativity asserts the existence of locally inertial frames, and the speed of light is a universal constant in those frames.

What needs to be added to the above is that the clocks and rulers that are used to measure the speed of light must be local clocks and rulers, i.e. clocks and rulers that are physically present in those "locally inertial frames".

Another point that should be obvious but probably should be mentioned is that the local clocks and rulers must be properly normalized / standardized. One should probably imagine lugging along an actual copy of the meter-bar in paris, plus a pair of atomic clocks, and distances being laid out with the bar and times measured with the clock(s). (Since we are talking about _measuring_ the speed of light, we are conceptually reverting to the days before the SI meter was re-defined in terms of the speed of light, so it's appropriate to imagine carrying around a platinum-alloy bar).

GR is perfectly capable of dealing with coordinate systems that are not normalized or standardized in the above or any other manner - it is actually a convention of physicists that such normalization be be done (and be done at the origin of a coordinate system). This is what Hurkyl is talking about when he talks about "good" vs "bad" coordinate systems.

http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html

also addresses this question.

Einstein went on to discover a more general theory of relativity which explained gravity in terms of curved spacetime, and he talked about the speed of light changing in this new theory. In the 1920 book "Relativity: the special and general theory" he wrote: . . . according to the general theory of relativity, the law of the constancy of the velocity of light in vacuo, which constitutes one of the two fundamental assumptions in the special theory of relativity [. . .] cannot claim any unlimited validity. A curvature of rays of light can only take place when the velocity of propagation of light varies with position. Since Einstein talks of velocity (a vector quantity: speed with direction) rather than speed alone, it is not clear that he meant the speed will change, but the reference to special relativity suggests that he did mean so. This interpretation is perfectly valid and makes good physical sense, but a more modern interpretation is that the speed of light is constant in general relativity.

The problem here comes from the fact that speed is a coordinate-dependent quantity, and is therefore somewhat ambiguous. To determine speed (distance moved/time taken) you must first choose some standards of distance and time, and different choices can give different answers. This is already true in special relativity: if you measure the speed of light in an accelerating reference frame, the answer will, in general, differ from c.

In special relativity, the speed of light is constant when measured in any inertial frame. In general relativity, the appropriate generalisation is that the speed of light is constant in any freely falling reference frame (in a region small enough that tidal effects can be neglected).

Again, the clocks and rulers that are used to measure the speed of light in a freely falling frame must be local clocks and rulers if one is to measure the speed of light as 'c'. And they must also be appropriately standardized (no fair using a clock that ticks at a non-standard rate, or a meter-bar that is too short or too long).

Because the tidal forces are neglgible, the clocks will all be running at the same rate, and the rulers will all measure the same distance, in the "locally inertial" region.

It is unfortunate that this important point has not been stressed more in the references I could find on-line.

Our hypothetical black hole observer is both accelerating, and in a region of curved space-time, so he cannot be expected (and will not) measure the speed of light to be globally equal to 'c'.

The black-hole observer will, like the accelerating observer, measure the speed (coordiante speed) of light to be 'c' only at the origin of the coordinate system, and find that it will vary with position.

One can also make the following statement about both observers. The arbitrarily high rate of change of position coordinates with respect to time coordinate of fast-moving free-falling objects will still always be lower than the rate of change of the position coordinate with respect to the time coordinate of a light beam at the same location, as long as the regions are causally connected.

Add:
(This can be deduced from the time independence of the problem, and the fact that light does get from point A to point B. Time independence is an important part of the argument here, without it the above statement would be incorrect).

In this case, the regions are causally connected, as you have already noted. (And the metric can be written in a time-independent manner). Thus the arbitrarily fast coordinate-speed of physical objects will be mirrored by an even faster arbitrarily fast coordinate-speed of light.

Also note that the specific numerical value of this "coordinate velocity" is also totally dependent on the coordinate system chosen, i.e. it is not a coordinate independent quantity.

 

[Zanket]

We aren’t talking a dictionary world here, this is physics!

Agreed. I said that the dictionary definition of “rising” is not changed in physics to “falling.”

You need to decide what you mean by “FREE RISING” that's not in the dictionary so you tell us.
When you hit a pop-up in baseball how is it not “Free-Rising”. Just like your space buoy?
Yes or no is your object slowing down or not, does stop rising when you reach Andromeda or not.

I said that by “free-rising” I mean the object is in free fall, rising upward. So yes to both questions.

You agree that if they don’t turn their engines off at Andromeda then the buoy falls toward them, but some how that is irrelevant?
Why?

Why not? You haven’t shown how it is relevant as far as I can tell. Yes, the buoy’s motion in the crew’s frame is just like that of a pop-up baseball. It rises and then, if they keep their engines running at Andromeda, it falls. I just don’t see how the fall affects #4. Only the rise is needed to make #4 a valid question.

From that point, you must agree then that it “falls” just like the baseball does for the pop-up once it reaches it high point.

Agreed.

So, on the ship how do you make a ball do the same thing? How fast and how high would you have to “pop it up” for it to fall just like the buoy will fall if you let the “gravity” continue after reaching Andromeda?

To make a ball track with the buoy, the crew need only throw up the ball at the moment they pass the buoy, and with the buoy’s relative (to them) velocity. The ball will then remain at rest with respect to the buoy.

Just how close will it be to the buoy?

If you mean how close to buoy’s motion will the ball’s motion be, it will be the same motion. The buoy and the ball will essentially be the same system, a single buoy-ball object.

If you can not solve this problem you certainly cann’t work the one you’ve put forward.

Now that I have solved this, do you have any objections to #1, 2, or 3 in the original post, and if not, can you answer #4?

Can I assume that the reason you brought up the buoy’s fall is because you needed further convincing that the buoy is rising in the crew’s frame just like a pop-up ball could?

 

[Zanket]

Excuse me? The ball is ACCELERATING. How is it possible to transform to an inertial frame of the ball? Does your dictionary have no definition of an "inertial frame"?

Before I answer this and the rest of your post, let’s be clear on something: Are you suggesting that there is no inertial frame possible in a uniform gravitational field of a planet?

 

[RandallB]

If you mean how close to buoy’s motion will the ball’s motion be, it will be the same motion. The buoy and the ball will essentially be the same system, a single buoy-ball object.

Now that I have solved this, do you have any objections to #1, 2, or 3 in the original post, and if not, can you answer #4?

Can I assume that the reason you brought up the buoy’s fall is because you needed further convincing that the buoy is rising in the crew’s frame just like a pop-up ball could?

Why would I have a problem now, as per your solution and explanation in this post “The buoy and the ball will essentially be the same system, a single buoy-ball object.”
Now that the ball on earth, and the buoy in space are moving identically, you’ve shown the two to behave equivalently.

YOU have finally hit on your confirmation that the GR equivalence principal applies just as predicted. And therefore there is not “A Flaw of General Relativity” on this issue.

Congratulations good job – sometimes it just takes working though the details to get it eh – but it feels good when you do, doesn’t it.

 

[Zanket]

As you've stated this, it's patently false. Remember that any coordinate chart is perfectly valid -- which includes coordinate charts where this distant object has a coordinate velocity millions of times of c.

Sure, if the coordinate chart is stretching, stretching the distance. That’s not the case here.

 

[Zanket]

YOU have finally hit on your confirmation that the GR equivalence principal applies just as predicted.

That was the point of #1, 2, and 3. That it should apply.

And therefore there is not “A Flaw of General Relativity” on this issue.

Unfortunately GR does not let the equivalence principle apply to the observer on the planet just as predicted. GR does not predict that the same observation is possible for the observer on the planet. For that observer, a pop-up ball cannot recede to a distance of one million proper light years in ten proper years while causal contact is maintained. For that observer GR predicts that it can recede only less than ten proper light years in ten proper years while causal contact is maintained.

 

[Zanket]

This is where you are wrong. This is a somewhat confusing topic, but I've dug up a few references to demonstrate this. You might also want to read Hurkyl's post on the same point.

I read your whole post. Good info. Unless I missed something, it mostly boils down to Hurkyl’s comment that GR can predict the observation in #3 (in the original post) for the observer on the planet if the coordinate system is expanding, stretching the proper distance to an object thrown upward. But #3 can occur in a static coordinate system. Then #4 remains an open question. Making the two situations as equal as possible includes using the same coordinate system in all respects (like static for both).

Our hypothetical black hole observer is both accelerating, and in a region of curved space-time, so he cannot be expected (and will not) measure the speed of light to be globally equal to 'c'.

The workaround to this issue of measurements being skewed within an accelerated frame is to have the observer fall through a trap door when the object thrown upward reaches its highest point. The spacetime of the experiment is flat (negligibly curved) because we let the tidal force throughout the region enclosing the experiment be negligible (including below the observer, who can be on top of a tower if necessary), and the now free-falling observer can do good measurements and measures the speed of light to be c throughout the region. The observer and the object will remain at rest with respect to each other during the measurement, just like the crew remains at rest with respect to the buoy when they reach Andromeda and shut off their engine. Falling through the trap door is analogous to shutting off the engine.

To show that GR is inconsistent, the observer need measure a proper distance over proper time (between throwing the object up and falling through the trap door) rate >= c. The observer should be able to do that given that the crew can do that (according to the equivalence principle).

An observer with the same acceleration felt as the crew in #3 feels, who throws an object upward at the same velocity as the crew measures for the buoy when they pass it, and waits ten proper years before falling through the trap door, should find that the object is one million proper light years away.

 

[RandallB]

That was the point of #1, 2, and 3. That it should apply.

Unfortunately GR does not let the equivalence principle apply to the observer on the planet just as predicted. GR does not predict that the same observation is possible for the observer on the planet. For that observer, a pop-up ball cannot recede to a distance of one million proper light years in ten proper years while causal contact is maintained. For that observer GR predicts that it can recede only less than ten proper light years in ten proper years while causal contact is maintained.

What are you talking about here ? What do you mean “GR does not predict that the same observation is possible for the observer on the planet.” ?

In your post 61 YOU already have shown that that is exactly predicted to be the same. Your not agreeing with yourself now.

 

[Zanket]

What are you talking about here ? What do you mean “GR does not predict that the same observation is possible for the observer on the planet.” ?

In your post 61 YOU already have shown that that is exactly predicted to be the same. Your not agreeing with yourself now.

It is GR that disagrees with itself. It is inconsistent. GR predicts that the same observation is possible for the observer on the planet, according to its own equivalence principle. But it also predicts that such is impossible. For example, look up "cosmological horizon" to see that GR predicts that causal contact with objects receding at a proper distance over proper time rate >= c is impossible. (In the absence of any stretching of the coordinate system, which stretches proper distances.)

Here is a theory:

Birds can fly.
Birds cannot fly.

If I show you that birds can fly, am I disagreeing with myself or is the theory inconsistent?

 

[RandallB]

It is GR that disagrees with itself. It is inconsistent. GR predicts that the same observation is possible for the observer on the planet, according to its own equivalence principle. But it also predicts that such is impossible. For example, look up "cosmological horizon" to see that GR predicts that causal contact with objects receding at a proper distance over proper time rate >= c is impossible. (In the absence of any stretching of the coordinate system, which stretches proper distances.)

Here is a theory:

Birds can fly.
Birds cannot fly.

If I show you that birds can fly, am I disagreeing with myself or is the theory inconsistent?

Holly Cow what are you doing with all this. Birds??
The only advice I can give you is from Fynnman, and that’s
“SHUT UP AND CALCULATE”

He wasn’t being insulting, just saying stop forcing things to be the way you think. And do the calculations in detail to show what’s really happening.

So find a way to put REAL numbers on your problem.
Defining what you’re looking for has got to be most of your trouble. So here:

Earth “E” to Mid point “M” to Andromada “A” change to any distances easy for you to work with.

On the ship start acceleration “a” at M to slow to a stop at A.
KEEP “a” going to return to M.
On ship at M hit ball at speed v to “stay” at M what is v.
V of ball will be Zero when batter reaches A.
Define height of ball in reference frame of batter while he is at A.

THEN move to the planet, hit a ball up at same starting speed v as batter at M did.
Solve for “g” gravity, required to have ball go up to same height as spaceship batter saw when he reached A.

Then compare “a” to “g”.

If don’t have time to do the math – quit arguing.

 

[JesseM]

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.

Yes, locally--in an arbitrarily small region of space and in an arbitrarily small time-interval.

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.

Sure.

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.

If you want to talk about what they "observe" as opposed to what they actually see using light signals, then presumably you must be talking about the velocity of the object in some coordinate system where the observer is at rest. The problem is that if you try to create a coordinate system where an observer is at rest throughout an extended period of acceleration, this will not be an inertial coordinate system, and thus it won't be relevant to the equivalence principle, nor is there any restriction on objects traveling faster than c in non-inertial coordinate systems. On the other hand, if you look at the inertial frame where the observer's instantaneous velocity is zero at a particular moment during the acceleration, in this instananeous inertial rest frame the velocity of outside objects will never be greater 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.

Only if by "perspective" you mean a non-inertial coordinate system where the crew is at rest throughout the journey. Again, not relevant to the equivalence principle. If you look at the crew's instantaneous inertial frame as they pass various buoys, and restrict it to purely local observations so that this will be relevant to the equivalence principle (ie measuring the velocity of a buoy in the crew's instantaneous inertial rest frame at the moment they pass next to that buoy), then no buoy will be observed to travel faster than c.

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

In formulating this paradox, you seem to have forgotten the part of the definition of the equivalence principle in point 1) that says the equivalence only applies to local observations, and you forgot to specify that it is only an equivalence between what is observed in a gravitational field and what is observed in an inertial frame by the observer moving in flat spacetime, not a non-inertial one. If you keep both of these in mind, you will see there is no paradox.