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

[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".

 

[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.