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

[Zanket]

Numbered for reference:

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

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

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

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

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

 

[Doc Al]

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

 

[Zanket]

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

 

[Doc Al]

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

 

[Zanket]

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

 

[Doc Al]

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

 

[selfAdjoint]

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

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

 

[Zanket]

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

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

 

[Zanket]

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

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

 

[ZapperZ]

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

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

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

Zz.

 

[Doc Al]

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

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

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

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

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

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

 

[Zanket]

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

Zz.

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

Let's try this:

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

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

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

Which of these statements do you object to?

 

[Zanket]

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

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

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

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

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

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

 

[ZapperZ]

Regardless of what #3 depends on,

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

Acceleration is not velocity.

Zz.

 

[George Jones]

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

This is roughly true.

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

I agree completely.

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

Here, I'm a little lost.

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

Regards,
George

 

[Zanket]

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

Acceleration is not velocity.

Zz.

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

Then what is your objection?

 

[ZapperZ]

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

Then what is your objection?

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

Zz.

 

[Doc Al]

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

Then what is your objection?

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

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

 

[Zanket]

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

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

 

[Zanket]

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

An apple thrown upwards.

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

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

 

[George Jones]

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

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

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

But I'm not sure what the problem is.

Regards,
George

 

[George Jones]

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

No, this isn't true.

Regards,
George

 

[ZapperZ]

An apple thrown upwards.



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

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

Zz.

 

[Zanket]

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

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

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

... GR is inconsistent.

 

[Zanket]

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

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

But I'm not sure what the problem is.

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

 

[Zanket]

No, this isn't true.

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

 

[Zanket]

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

Zz.

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

 

[ZapperZ]

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

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

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

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

Zz.

 

[derz]

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

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

 

[George Jones]

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

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

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

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

Regards,
George

 

[Zanket]

Where does it say that?

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

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

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

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

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

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

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

 

[Zanket]

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

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

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

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

 

[Zanket]

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

What do you disagree to:

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

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

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

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

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

The final distance is one million light years.

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

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

 

[pervect]

Here is my $.02 on the issue.

If we adopt the metric

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

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

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

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

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

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

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

But now we address the original question.

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

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

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

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

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

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

 

[ZapperZ]

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

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

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

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

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

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

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

Zz.