Would the one accelerating please stand up?

[Dale]

Numbers 2 & 4 are the problem, and yet again (no surprise), this is due to what we think about the meaning of the equivalence principle.

All 5 points (including 2 and 4) are the experimental results that are predicted by GR. Any way you think about the equivalence principle MUST be consistent with them.

 

[MikeGomez]

All 5 points (including 2 and 4) are the experimental results that are predicted by GR. Any way you think about the equivalence principle MUST be consistent with them.

I don’t disagree that GR predict all those things.. It’s just that I thought that there might be a couple less restraints in the flat spacetime scenario such as the one regarding adherence to relative velocity. I guess my mistake, based on what everyone here is telling me is my lack of understand of what the equivalence principle means.

I have looked at Sean Carroll’s description of the equivalence principle and of course it seems reasonable, although there are a couple points which may come into consideration in our discussion.

When he says “…we can no longer speak with confidence about the relative velocity of far away objects…” this may come into play.

When he says “…It is the EEP which implies (or at least suggests) that we should attribute the action of gravity to the curvature of spacetime…” This might not be an issue, and it might not come up at alll, and even if it does come up it doesn’t mean that it will necessarily affect the argument one way or the other.

He says”… It is impossible to "prove" that gravity should be thought of as spacetime curvature, since scientific hypotheses can only be falsified, never verified…” That may or may not be an issue. I just feel more comfortable conceptually with the stress-energy tensor side of the EFE equations.

I’m afraid I won’t be able to keep up with a lot of that follows when I starts talking about the tensor calculus, but if it gets that far we probably will have settled the issue anyway.

 

[Dale]

When he says “…we can no longer speak with confidence about the relative velocity of far away objects…” this may come into play.

Which is why I specified the measurement process (Doppler radar). We can speak with confidence about the outcome of a Doppler measurement.

For a flat spacetime scenario to be equivalent it must reproduce all measurements, not just a chosen few. That includes Doppler shift. A flat spacetime scenario can be devised which reproduces anyone of the above measurements, but not all of them together. In fact, I think that even just 3 and 4 together are impossible.

 

[PeterDonis]

I think that even just 3 and 4 together are impossible.

Not by themselves, because they don't say anything about distance or (Doppler measured) relative velocity. But if you combine #3 and #4 with either #1 or #2, the three together are impossible in flat spacetime.

 

[MikeGomez]

That is not what uniform gravitational field usually means. Uniform gravitational field usually means that it produces no tidal effects. Free falling bodies (point masses) always have zero proper acceleration, regardless whether the gravitational field is uniform or not.

So uniform gravitational fields mean flat spacetime, and non-uniform gravitational fields are mean curved spacetime, but what would be the terminology of another type of non-uniform gravitational field which is non-uniform in the sense that the gravitational strength of the source various with time, not distance (i.e. like if the source is a pulsar)?

The proper accelerations of all involved objects must be reproduced, in order to have an equivalence of physical laws between the two cases.

Maybe the non-zero proper acceleration clock in the experiment would be able to transition from linear acceleration to rotational acceleration while maintaining its constant proper acceleration.

 

[A.T.]

what would be the terminology of another type of non-uniform gravitational field which is non-uniform in the sense that the gravitational strength of the source various with time,

Static vs. non-static?

Maybe the non-zero proper acceleration clock in the experiment would be able to transition from linear acceleration to rotational acceleration while maintaining its constant proper acceleration.

Every possible measurement has to be reproduced to have equivalence, including what a gyroscope would show.

 

[MikeGomez]

Not by themselves, because they don't say anything about distance or (Doppler measured) relative velocity. But if you combine #3 and #4 with either #1 or #2, the three together are impossible in flat spacetime.

I was also thinking that the direction of the proper acceleration of the second clock in #4 would not matter in order for the final clock reading to be the same. I was thinking there is a reduction in the accuracy of which one body may know about the other body when separated, but if each body from both scearios locally maintains the exact same results, and the end results from both scearios match up, then that would be an acceptable definition for reproducing the experiment.

 

[MikeGomez]

Static vs. non-static?

Every possible measurement has to be reproduced to have equivalence, including what a gyroscope would show.

Gyroscope would be difficult, but I am not sure if that would be impossible. Anyway I still think the experiment can be done for normal non-gyroscopic clocks that I thought we were talking about.

 

[A.T.]

for normal non-gyroscopic clocks that I thought we were talking about.

I thought we are talking about an equivalence in the sense of the equivalence principle: a general equivalence of all laws of physics. Some hand picked cases and measurements that happen to match are not of much interest.

 

[MikeGomez]

I thought we are talking about an equivalence in the sense of the equivalence principle: a general equivalence of all laws of physics. Some hand picked cases and measurements that happen to match are not of much interest.

I am not hand picking anything. If you take the limiting case to such extremes then the argument you are making would mean that the standard man in the elevator scenario is hand picked (and technically it is). There are approximations in the flatness of the spacetime in the elevator case, and there are approximations of flatness of spacetime on the Earth side of it (you need to neglect tidal forces).

 

[A.T.]

I am not hand picking anything.

You can't get beyond the most trivial case of a single point mass, and just a few measured quantities.

the argument you are making would mean that the standard man in the elevator scenario is hand picked

The EP states that all measurements in that elevator in the will be equivalent. That is a general local equivalence.

and there are approximations of flatness of spacetime on the Earth side of it (you need to neglect tidal forces)..

Without tidal (non-uniform) gravity the clock in the hole wouldn't even come back to the clock on the surface. How can you neglect that?

 

[PeterDonis]

the direction of the proper acceleration of the second clock in #4 would not matter in order for the final clock reading to be the same.

Even if the direction doesn't matter (I actually think it does, but I'll assume it doesn't here for the sake of argument), the magnitude certainly does. In your flat spacetime version where the "elevator" is the size of the Earth, and varies its acceleration in order to duplicate the relative motion of the Earth and the clock in the hole, the proper acceleration of every piece of the "elevator" must vary with time. That is obviously different from the actual Earth scenario, where the proper acceleration of a clock at one end of the hole is constant.

(Actually, it's even worse than that; in the Earth scenario, the proper acceleration of every single piece of the Earth is constant--the magnitude of the acceleration varies with distance from the center, but at any given distance it's constant. There's no way to match that in flat spacetime while also matching the relative motion of the clock and the "elevator". This has been pointed out before.)

if each body from both scearios locally maintains the exact same results, and the end results from both scearios match up

They won't. See above.

 

[Dale]

Not by themselves, because they don't say anything about distance or (Doppler measured) relative velocity. But if you combine #3 and #4 with either #1 or #2, the three together are impossible in flat spacetime.

You are of course correct. I was thinking of the distance being 0 at two points, which is a condition beyond 3 and 4, but not quite as strong as 1.

 

[MikeGomez]

Ok, finally I see the source of confusion (predicable it’s my fault).
------------------------------------------------------------------------------------------
Case A: Standard Earth versus flat spacetime scenario.

All the physics in the Earth lab frame are the same as in the elevator lab frame.
-------------------------------------------------------------------------------------------
Case B: Hole in Earth, Mike’s versus

In the Earth scenario clock 1 drops down a hole in the Earth (freefall) and takes its lab frame with it on its journey. Clock 2 (proper acceleration) remains at the surface and it has a separate lab frame. When the two clocks meet up again they will show different time readings.

In the flat spacetime scenario clock 1 in its freefall lab frame must match the physics from clock 1 from the Earth scenario. Clock 2 in its proper acceleration lab frame must match the physics from the Earth clock 2 during its journey. When the clocks meet up again they show the same time difference as in the Earth scenario.

-----------------------------------------------------------------------------------------
Case C: Hole in Earth, Everybody else

In the Earth scenario, clock 1 in its freefall communicates with clock 2. In this way the size of the lab frame is the diameter of the entire earth.

The flatspace scenario here is impossible.
------------------------------------------------------------------------------------------

Based on the context of the conversation I should have realized the situation a long time ago, but I was kind of blinded because I was thinking in terms of smaller lab frames. Profuse apologies.

..In your flat spacetime version where the "elevator" is the size of the Earth,...

The giant Earth sized thing was just a random idea and I indicated after that, that I had other ideas, but I guess I was not clear enough.

You can't get beyond the most trivial case of a single point mass

I agree. I was confused about certain requirements that I thought people were making about the experiment. I hope that is cleared up now.

Without tidal (non-uniform) gravity the clock in the hole wouldn't even come back to the clock on the surface. How can you neglect that?

In my version of the flat spacetime scenario the clock in the hole stays in the same location, so it doesn't need to come back, but it does need a changing gravitational field in order to reproduce the proper time differences. The proper acceleration clock in the elevator does need to come back. That's why I was talking about transitioning from linear acceleration to rotational acceleration, so it could make the return journey. The rotational acceleration will produce small gyroscopic effects, and those need to be compensated for by manipulating its gravitational field (which is what I am also assuming we can do for the clock in freefall).

 

[PeterDonis]

finally I see the source of confusion

I still don't think you see all of it. See below.

clock 1 drops down a hole in the Earth (freefall) and takes its lab frame with it on its journey

Yes, but this "lab frame" is not a "local inertial frame" in the sense of the EP. It is localized in space but not in time. The technical term for this "lab frame" is "Riemann normal coordinates". If you look in the mainstream references we gave you, you will see that these are carefully distinguished from "local inertial coordinates". They are not the same thing.

Clock 2 (proper acceleration) remains at the surface and it has a separate lab frame.

Just as a note, this "lab frame" is non-inertial. More importantly, the caveats about it not being "local" apply here as well. In this case, since the "lab" is accelerated, the technical term for what you are calling the "lab frame" here is "Fermi normal coordinates". Again, in the mainstream references we gave you, these are carefully distinguished from "local inertial coordinates". They are even different from "accelerated coordinates in a local inertial frame". These are all important distinctions and you need to grasp them if you are going to understand what the mainstream references are saying.

In the flat spacetime scenario clock 1 in its freefall lab frame must match the physics from clock 1 from the Earth scenario. Clock 2 in its proper acceleration lab frame must match the physics from the Earth clock 2 during its journey. When the clocks meet up again they show the same time difference as in the Earth scenario.

No; this is impossible. If clock 2 in flat spacetime matches the proper acceleration of Earth clock 2, there is no way for it to meet up again with clock 1. If clock 2 in flat spacetime meets up again with clock 1, there is no way for its proper acceleration to match that of Earth clock 2. Not even if you throw in rotation. See below.

Please, instead of waving your hands about this scenario, take the time to actually do the math. It will not work out the way you apparently think it will.

In the Earth scenario, clock 1 in its freefall communicates with clock 2. In this way the size of the lab frame is the diameter of the entire earth.

The flatspace scenario here is impossible.

This is all true, but irrelevant to the actual issue with your version. See above.

I was thinking in terms of smaller lab frames

Yes, but that's not the problem. See above.

The proper acceleration clock in the elevator does need to come back. That's why I was talking about transitioning from linear acceleration to rotational acceleration, so it could make the return journey.

That won't work. See below.

The rotational acceleration will produce small gyroscopic effects, and those need to be compensated for by manipulating its gravitational field

In flat spacetime, there is no "gravitational field" except what is produced by the clock's proper acceleration. So if the clock's proper acceleration has a rotating component, the "gravitational field" will have a rotating component, i.e., it will not be uniform. There's no other "gravitational field" to adjust to compensate; spacetime is flat, so there are no other free parameters besides the magnitude and direction of the clock's proper acceleration.

which is what I am also assuming we can do for the clock in freefall

See, this is why I objected to your usage of the term "gravitational field" before. It has misled you into thinking that you can somehow "adjust the gravitational field" of a free-falling clock in flat spacetime. You can't; there's nothing to adjust. The clock is in free fall; the spacetime it is in is flat; there are no free parameters whatsoever that you can use to "adjust the gravitational field".

 

[MikeGomez]

“Yes, but this "lab frame" is not a "local inertial frame" in the sense of the EP. It is localized in space but not in time. The technical term for this "lab frame" is "Riemann normal coordinates". If you look in the mainstream references we gave you, you will see that these are carefully distinguished from "local inertial coordinates". They are not the same thing.”

That I did not know. Thank you.

“Just as a note, this "lab frame" is non-inertial. More importantly, the caveats about it not being "local" apply here as well. In this case, since the "lab" is accelerated, the technical term for what you are calling the "lab frame" here is "Fermi normal coordinates". Again, in the mainstream references we gave you, these are carefully distinguished from "local inertial coordinates". They are even different from "accelerated coordinates in a local inertial frame". These are all important distinctions and you need to grasp them if you are going to understand what the mainstream references are saying.”

Blue highlight: That much I know. For the rest, thank you, I did not know that.

“In flat spacetime, there is no "gravitational field" except what is produced by the clock's proper acceleration. So if the clock's proper acceleration has a rotating component, the "gravitational field" will have a rotating component, i.e., it will not be uniform. There's no other "gravitational field" to adjust to compensate; spacetime is flat, so there are no other free parameters besides the magnitude and direction of the clock's proper acceleration.”

This I still have an issue with. That is why I was trying to establish the meaning of the relative relationship of a body in its field versus the rest of the universe, and why I was trying to establish the unambiguous terminology for what we mean when we say “gravitational field”.

I feel that the clock does not produce a field due to its proper acceleration, but rather a relative proper acceleration field exists for the clock due to its relative condition with respect to the rest of the universe.”

See, this is why I objected to your usage of the term "gravitational field" before. It has misled you into thinking that you can somehow "adjust the gravitational field" of a free-falling clock in flat spacetime. You can't; there's nothing to adjust. The clock is in free fall; the spacetime it is in is flat; there are no free parameters whatsoever that you can use to "adjust the gravitational field"..

Blue highlight: Yes, but at the same time that is why I was trying to establish the correct nomenclature for all the difference types of gravitational fields. For example when people say something along the lines of “gravity disappears for the body in freefall” or curvature is the “real” gravity, that kind of bothers me because we still have gravity in flat spacetime. In fact the flat spacetime gravity is what I would consider “real” because that type of gravity is the type of gravity directly due to the source of gravity as per the equivalence between gravity and inertia. As far as planets and stars go, then yes, curvature gravity is the “real” gravity in this arena.

As for the rest in non-highlighted blue, I was thinking that only the field in the immediate area of the individual lab frames is what is required. That is the only field “felt” by the bodies in their respective lab frames, so I thought as long as the local lab frames fields are correctly simulated (if possible) then that would satisfy the requirements of the experiment. But these issues are probably mute anymore considering we are all talking about the giant Earth frame now.

 

[Dale]

I was trying to establish the unambiguous terminology for what we mean when we say “gravitational field”.

Yes, in GR I believe that there is not a standard meaning for the term "gravitational field". There are several possible candidates. One would be the Riemann curvature tensor, another would be the metric, and a third would be the Christoffel symbols. As soon as you specify "flat spacetime" (as you do in using the equivalence principle locally) then you have fixed both the curvature tensor and the metric and there is no remaining freedom as Peter Donis mentioned.

There still is quite some freedom in the Christoffel symbols, but none of those degrees of freedom can change any experimentally measurable outcome. Since you cannot replicate the experimental results (1-5) in a standard inertial frame and standard coordinate system, you cannot replicate them by changing the Christoffel symbols.

My personal preference would be to equate the term "gravitational field" with the Christoffel symbols, but since there is a possibility of confusion I think that it is better to just use the correct GR terminology rather than to try to adapt Newtonian terminology. Any time you use the term "gravitational field" in a GR discussion you are inviting misunderstanding.

relative proper acceleration

Aack! If you are going to invent terms, at least don't make them oxymorons.

 

[PeterDonis]

I was trying to establish the meaning of the relative relationship of a body in its field versus the rest of the universe

What does "a body in its field" mean? Remember that if we're in flat spacetime, there are no sources of gravity at all (if there were, spacetime would be curved, not flat), and even in curved spacetime, the clocks are not sources of gravity; only the massive body (like the Earth) is. (More precisely, the clocks are such tiny sources of gravity by comparison with the Earth that we can ignore them as sources.) So the only possible "source of a field" in the scenarios we've been discussing is the Earth, and that is only present in the curved spacetime version.

I was trying to establish the unambiguous terminology for what we mean when we say “gravitational field”.

As DaleSpam pointed out (and as I pointed out a number of posts ago), there is no unambiguous meaning for that term. DaleSpam gave several possibilities; but let me focus on another point. It seems to me that you want the "gravitational field" to be something that is not frame-dependent; you want there to be a "gravitational field" for all observers, not just for some. If that is true, then there are only two possible things that "gravitational field" could mean: either the metric (the geometry of spacetime), or the curvature (the curvature of that geometry). They pretty much amount to the same thing for this discussion, so I'll assume that "gravitational field" means "the metric" so that there can be a "gravitational field" in flat spacetime as well as curved (flat spacetime has a metric--it's the flat one, with zero curvature).

But, if "gravitational field" means "the metric", then a key fact about it is that you can't change it by changing your state of motion. The metric is the same for all observers, whether they're in free fall or accelerating uniformly or accelerating non-uniformly according to some detailed prescription. (The same would be true for curvature.) So you can't think of a "gravitational field" as being something that depends on how a particular clock or observer is moving. See further comments below.

I feel that the clock does not produce a field due to its proper acceleration, but rather a relative proper acceleration field exists for the clock due to its relative condition with respect to the rest of the universe.

If "relative proper acceleration field" (which is indeed an oxymoron, as DaleSpam says) is supposed to mean "gravitational field", then see my comments above. If it means something else, what is it supposed to mean?

For example when people say something along the lines of “gravity disappears for the body in freefall” or curvature is the “real” gravity, that kind of bothers me because we still have gravity in flat spacetime.

We do in the sense that there is a metric, yes. But see my comments above about that.

When people say "gravity disappears for the body in free fall", they mean by "gravity" something like the Christoffel symbols, which correspond to what a Newtonian physicist would call the "acceleration due to gravity" in a frame fixed to the Earth. This kind of "gravity" is frame-dependent; you can make it vanish by choosing coordinates appropriately. So if you want "gravity" to be something that isn't frame-dependent, then you have to give up all your Newtonian intuitions about how "gravity" works. (You seem to want to do that anyway, since you want there to be "gravity" in flat spacetime; but see further comments below about that.)

In fact the flat spacetime gravity is what I would consider “real” because that type of gravity is the type of gravity directly due to the source of gravity as per the equivalence between gravity and inertia.

Um, what? In flat spacetime, there are no sources of gravity. The only sense in which "gravity" exists is that there is a metric, but the metric is flat, and a flat metric means no sources of gravity exist (the technical way of saying this is that the stress-energy tensor must be zero everywhere). So the "gravity" that exists in flat spacetime is not what I would call "real", because it doesn't have any sources. It's just another way of saying "flat spacetime has a flat metric".

It's true that you can make "gravity" in the sense of Christoffel symbols (a sort of Newtonian "acceleration due to gravity") appear in flat spacetime by accelerated motion. But again, this kind of gravity doesn't have a "source". You haven't changed anything about the metric or the curvature of spacetime. You've just taken up a trajectory through flat spacetime such that free-falling objects have coordinate acceleration relative to you.

As far as the equivalence principle goes in this connection, the EP does not say that gravity "is" inertia. It says that inertial mass and (passive) gravitational mass have to be equal because they are both manifestations of the same thing: the fact that the spacetime metric determines what states of motion at a given event are freely falling and what states of motion are not. You can call the spacetime metric "gravity" if you like to reinforce this, but again, if you do that, you have to give up any connection between "gravity" and a source, because in flat spacetime there is no source.

I was thinking that only the field in the immediate area of the individual lab frames is what is required.

By "the field", do you mean the metric? (Please read the above before answering that question.) If you do, then yes, you only need the metric to be the same (to within a good enough approximation) within the lab frames. But that doesn't change what I said; if spacetime is flat, it's flat, in the lab frame as well as everywhere else, so there's still nothing to adjust. You can't change the metric by changing your state of motion.

these issues are probably mute anymore considering we are all talking about the giant Earth frame now.

I'm not. Everything I've said applies to your original version, where you only wanted to "simulate" things within a small lab frame.

 

[MikeGomez]

Many people see the man on the surface of the Earth as being pulled down and the man on the elevator is being pushed up, and that gravitation and inertia are different but only look the same. I see the two situations as identical, and that is a very satisfying feeling. The problem I am struggling with is that what people are telling me here doesn’t allow me to maintain my current view. I want to be corrected with my view, but I don't want to lose how I feel about inertia and gravity being the same.

For my understanding of the equivalence between gravity and inertial, as well as my argument for the hole in the Earth scenario, I require two things. A supposition of particles to calculate the field strength and direction, and a real force field. To understand the equivalence between gravity and inertial, the direction of the force between the Earth and the man must match the direction of the force between the man and the elevator.

I can start with the man in the accelerated elevator with a force of 1g. I can add a couple of particles of matter to the system, but I will need to ease up on the acceleration ever so slightly to maintain the constant force of 1g. I can add more particles and apply less force repeatedly in this manner until eventually the particles will add up to a planet and the man will be in the elevator on the surface of the Earth with the force pointing in the correct direction. The man doesn’t know whether he is in an accelerated elevator or at the surface of the Earth or anything in between. The original inertial energy of the elevator was gradually replaced with the mass energy of the planet, and the man didn’t notice, all due to Einstein’s mass energy equivalence (and his stress energy tensor?).

As soon as you specify "flat spacetime" (as you do in using the equivalence principle locally) then you have fixed both the curvature tensor and the metric and there is no remaining freedom as Peter Donis mentioned.

Is that what is meant by transposing away the field? If we transpose away the field, that sounds mathematical, not physical. It seems to me like we should be able to choose not to transform away the field if we wish, so that we can have a field, so that we can calculate the strength of the field using the principle of superposition.

As far the field being fictitious, in my opinion that doesn’t mean it’s not real.

Maybe the Einstein field equations are mathematical tool for calculating gravity, only after the equivalence of gravity and inertia are established, and only after the configuration of the source of gravity has been permanently fixed (ie we have a planet and it says a planet, or we have an elevator and it stays an elevator)?

“Einstein managed to blur forever the distinction between real and fictitious forces. General relativity is his theory of gravity, and gravity is certainly the paradigmatic example of a “real” force. The cornerstone of Einstein’s theory, however, is the proposition that gravity itself is itself a fictitious force (or, rather that it is indistinguishable from a fictitious force).”

-CIT theoretical physicist and 2004 Nobel laureate David Politzer.

 

[Dale]

I don't want to lose how I feel about inertia and gravity being the same.

The term "inertia" is, itself, a little vague. So you can probably use that wiggle room to make the statement "inertia and gravity are locally the same" justifiable. I don't know of any way to make it true beyond locally. The usual concept of inertia doesn't produce tidal effects.

The original inertial energy of the elevator was gradually replaced with the mass energy of the planet

Generally the mass energy of the planet will be far far far in excess of the original kinetic energy of the elevator. Particularly in the elevator's frame where the KE of the elevator is, by definition, 0.

Is that what is meant by transposing away the field? If we transpose away the field, that sounds mathematical, not physical. It seems to me like we should be able to choose not to transform away the field if we wish, so that we can have a field, so that we can calculate the strength of the field using the principle of superposition.

Again, if by "the field" you mean "the Christoffel symbols", then yes you can choose to transform them away in flat spacetime or you can choose not to transform them away. If you choose not to transform them away so that you have non-zero Christoffel symbols, then yes, you can calculate the "strength" of the Christoffel symbols (no superposition is involved AFAIK). The strength or even direction of the Christoffel symbols, however, cannot change any measurement.

“Einstein managed to blur forever the distinction between real and fictitious forces. General relativity is his theory of gravity, and gravity is certainly the paradigmatic example of a “real” force. The cornerstone of Einstein’s theory, however, is the proposition that gravity itself is itself a fictitious force (or, rather that it is indistinguishable from a fictitious force).”

-CIT theoretical physicist and 2004 Nobel laureate David Politzer.

I tend to feel exactly the opposite. I think that GR clarifies the distinction between real and fictitious forces. Real forces can be measured with an accelerometer, fictitious forces cannot. That demands a reclassification of gravitational forces, but the distinction becomes clear. I think that Dr. Politzer is exaggerating for effect here.

 

[PeterDonis]

I want to be corrected with my view, but I don't want to lose how I feel about inertia and gravity being the same.

That is certainly an achievable goal. Don't give up.

Many people see the man on the surface of the Earth as being pulled down and the man on the elevator is being pushed up

A simple way to see how the equivalence works in GR is to make a small change to the wording of the above. According to GR, the man on the surface of the Earth is being pushed up, and the man in the elevator is also being pushed up. That's why the situations are the same.

I can start with the man in the accelerated elevator with a force of 1g. I can add a couple of particles of matter to the system, but I will need to ease up on the acceleration ever so slightly to maintain the constant force of 1g. I can add more particles and apply less force repeatedly in this manner until eventually the particles will add up to a planet and the man will be in the elevator on the surface of the Earth with the force pointing in the correct direction.

I see where you are going with this thought experiment, and in principle what you are concluding from it about the equivalence principle is correct, but I don't think it's actually realizable. The problem is that, when you're at the stage where the mass behind the elevator is a lot less than the mass of the Earth, so the rocket in the elevator is still providing most of the acceleration, the rocket will be moving away from the mass, because the gravity of the mass is not enough to hold it in place, as the Earth's gravity is. In the limiting case of flat spacetime, you don't have to worry about this because there is no "mass behind the elevator"; but the problem starts as soon as you try to go continuously from that limiting case to the case of being at rest on the Earth's surface. The two extremes (flat spacetime and at rest on Earth) are "stable", so to speak, but none of the intermediate cases are.

The original inertial energy of the elevator was gradually replaced with the mass energy of the planet

I don't understand what you mean by "inertial energy", and the mass-energy of the planet is not "added" to the elevator. This looks to me like another case where you are using your own personal terminology instead of learning the standard terminology, and it is hindering your understanding.

The difference between the accelerating elevator in flat spacetime and the elevator at rest on the surface of the Earth (note that, as I said above, there is not a realizable continuous set of cases in between these two) is simply that spacetime is flat in the first case and curved in the second. Or, to put it in more physical terms, tidal gravity is absent in the first case and present in the second. Locally, there is no way to tell the difference; but measurements over a large enough region of spacetime (either a large enough distance in space, or a long enough interval of time, or both) can show the difference. It has nothing to do with the "kind of energy" of the elevator or anything else; the only role that the mass-energy of the Earth plays is as the source of the spacetime curvature.

It seems to me like we should be able to choose not to transform away the field if we wish, so that we can have a field, so that we can calculate the strength of the field using the principle of superposition.

The Einstein Field Equation is nonlinear, so in general you cannot superpose different solutions to get another solution. In the weak field approximation, you can get away with this because the nonlinear terms are small enough to be ignored. So, for example, you can get a good enough prediction of the tides on Earth by adding together the tidal gravity due to the Moon and the tidal gravity due to the Sun.

However, in the case you're talking about, "superposition" is irrelevant because there is, at most, one source of curvature (the Earth, which is only present in one of the two scenarios being compared). What you are doing when you choose a coordinate chart having nonzero Christoffel symbols (and therefore a nonzero "gravitational field" by one definition) is not "superposing" a field onto another field; you're just choosing coordinates such that there is a "field" in those coordinates. As DaleSpam said, doing this does not change any physical observables, so it's not at all the same as superposing the fields due to different sources.

Maybe the Einstein field equations are mathematical tool for calculating gravity, only after the equivalence of gravity and inertia are established, and only after the configuration of the source of gravity has been permanently fixed (ie we have a planet and it says a planet, or we have an elevator and it stays an elevator)?

In order to solve the EFE, you do have to choose what the stress-energy tensor is; flat spacetime and the curved spacetime around and within the Earth are two different solutions to the EFE, with different stress-energy tensors. The equivalence of gravity and inertia is part of the fundamental framework of GR, so it's there in all solutions of the EFE; in that sense it is "established" before you calculate any solutions, yes.

 

[MikeGomez]

I tend to feel exactly the opposite. I think that GR clarifies the distinction between real and fictitious forces. Real forces can be measured with an accelerometer, fictitious forces cannot. That demands a reclassification of gravitational forces, but the distinction becomes clear. I think that Dr. Politzer is exaggerating for effect here.

And I tend to feel exactly the opposite about that. When we analyze what is going on with the accelerometer at the atomic level, we see that each unit mass of the accelerometer is being attracted by gravity equally, and that is why there is no reading on the accelerometer. The same applies for an accelerometer of any construction, even a laser based one.

On the other hand, the accelerometer on Earth shows a reading simply due to contact forces at the surface. The atoms at the ground can’t go through the surface of the earth, but the atoms at the top of the accelerometer can move a little until the stress becomes too large. The same goes with the accelerometer in the elevator with respect to the elevator floor.

 

[MikeGomez]

The problem is that, when you're at the stage where the mass behind the elevator is a lot less than the mass of the Earth, so the rocket in the elevator is still providing most of the acceleration, the rocket will be moving away from the mass, because the gravity of the mass is not enough to hold it in place, as the Earth's gravity is. In the limiting case of flat spacetime, you don't have to worry about this because there is no "mass behind the elevator";.

It's just a thought experiment. It doesn't have to be a rocket accelerating the elevator the whole time. As it gets bigger and bigger we could just imagine more and more rockets.

but the problem starts as soon as you try to go continuously from that limiting case to the case of being at rest on the Earth's surface. The two extremes (flat spacetime and at rest on Earth) are "stable", so to speak, but none of the intermediate cases are.

Enough of the limiting case for each case should be sufficient. The elevator can become quite large and start acquiring curvature long before the technical difficulties are insurmountable.

In the limiting case of flat spacetime, you don't have to worry about this because there is no "mass behind the elevator";

Yes there is. In the elevator we are talking about the energy of the accelerating force being the equivalent to a very dense concentration of mass directly beneath the elevator floor. The amount of energy is equivalent, but the direction is in a straight line (no curvature), so the only curvature is the negligible curvature due to the physical nature of the elevator. As we add more mass to the system the curvature increases, and as we decrease the force of the rocket we decrease the amount of linearly directed energy (which afterall was why spacetime was flat here to begin with, so really are gradually decreaing the amount of "flatness".

I don't understand what you mean by "inertial energy",

the stress energy due to the force applied to the elevator.

and the mass-energy of the planet is not "added" to the elevator. This looks to me like another case where you are using your own personal terminology instead of learning the standard terminology, and it is hindering your understanding.

I’m not adding mass-energy of the planet to the elevator. I am adding mass to the elevator.

the only role that the mass-energy of the Earth plays is as the source of the spacetime curvature.

I would say that the curvature is due to the three dimensional configuration of the system. The density of the field flux for a given area is due to the inverse square law. If the nature of gravity per say had any curvature, then gravitational strength would fall off as the fourth power of distance.

What you are doing when you choose a coordinate chart having nonzero Christoffel symbols (and therefore a nonzero "gravitational field" by one definition) is not "superposing" a field onto another field;

I am not superposing one field onto another. I am superposing the composite energy of all the particles of Earth to calculate the overall field strength.

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In general I personally prefer the pulling paradigm over pushing, but that's just personal preference. A body on Earth is attracted to the center of the earth. In fact all the components of the earth/body system all tend towards the center of greatest energy. For the elevator situation the source of energy is beneath the elevator floor, and that is where is the body is trying to get to. The floor of the elevator prevents that. In both cases you just have to get the direction of the force vector correct, but there is an overall relative relationship in both cases which is identical

 

[PeterDonis]

It's just a thought experiment.

Thought experiments are idealizations, but they still have to be consistent with the laws of physics. You can't just ignore consequences of those laws that you don't like.

As it gets bigger and bigger we could just imagine more and more rockets.

How would that help? The point is that the observer inside the elevator has to experience a constant proper acceleration; that is what defines the scenario. As soon as you specify that, you have implicitly required that, as you add a gravitating object of gradually increasing mass to the scenario, the observer must move away from that mass. How the proper acceleration is caused is immaterial.

The elevator can become quite large and start acquiring curvature long before the technical difficulties are insurmountable.

I don't understand what you mean by this.

In the elevator we are talking about the energy of the accelerating force being the equivalent to a very dense concentration of mass directly beneath the elevator floor.

Not if we are going to be consistent with the laws of physics. In flat spacetime, once again, there are no sources of gravity; the stress-energy tensor is zero everywhere. A "very dense concentration of mass" is a source of gravity, with a nonzero stress-energy tensor, and no such thing can exist in flat spacetime. You are not allowed to just handwave one into existence. If you are thinking about the flat spacetime scenario this way, then you are thinking about it wrong; what you are imagining is not consistent with the laws of physics.

the stress energy due to the force applied to the elevator.

There can't be any stress-energy in flat spacetime; see above.

I would say that the curvature is due to the three dimensional configuration of the system.

Spacetime has four dimensions, not three, and spacetime curvature is four-dimensional curvature, not three-dimensional curvature.

The density of the field flux for a given area is due to the inverse square law. If the nature of gravity per say had any curvature, then gravitational strength would fall off as the fourth power of distance.

This looks like a personal theory again. Do you have a reference for this statement? Or can you show a mathematical derivation of it?

I am superposing the composite energy of all the particles of Earth to calculate the overall field strength.

This works in Newtonian physics, but it doesn't really work in GR. You can't calculate the spacetime curvature due to the Earth as a whole by imagining it as a collection of particles that each produces some curvature, and adding up their contributions. The Einstein Field Equation is nonlinear, so solutions to it don't work this way.

In fact all the components of the earth/body system all tend towards the center of greatest energy.

What does this even mean? Once again, you should not be using your own personal terminology; you need to learn the standard terminology for what you are talking about, so we can understand what you are saying. Can you give a mainstream reference for what you mean by "the center of greatest energy"? Or restate it in standard terminology?

For the elevator situation the source of energy is beneath the elevator floor, and that is where is the body is trying to get to.

If, as I suspect, by "source of energy" you mean the same as the "stress-energy due to the force applied to the elevator", see my comments on that above.

there is an overall relative relationship in both cases which is identical

No, there isn't. You are misunderstanding how the flat spacetime scenario works. See above.

 

[MikeGomez]

Sorry I misspoke. The inverse square law is due the gravitation extending radially from a point source, slightly different from being due to being a sphere. The linear nature of the accelerated rocket is a directed force, not radial.

Will respond to other comments but must go now.

 

[PeterDonis]

For my understanding of the equivalence between gravity and inertial, as well as my argument for the hole in the Earth scenario, I require two things. A supposition of particles to calculate the field strength and direction, and a real force field.

I went back and re-read this in your earlier post, and I want to comment on it because it may be part of what is getting in the way. Unfortunately, according to the standard view of the equivalence principle, neither of the two things you say you want are present in both scenarios (flat spacetime accelerating elevator, and elevator at rest on Earth).

(1) In flat spacetime, there are no "particles" generating a field; spacetime is flat, there is zero stress-energy everywhere, and there is thus no "source" of any field.

(2) In flat spacetime, corresponding to (1), there is no "real force field", because there is no source to generate one. You can make an "apparent force field" by accelerating (firing your rocket, for example), but it is only "apparent", not "real".

The way that GR deals with this apparent problem is this: rather than insist on a "real force field" in flat spacetime, admit that what in Newtonian physics we call "the force of gravity" is not a real force field. It's only an "apparent" one, just like the "apparent" force field that appears when you accelerate. In other words, the reason a rock accelerates downwards relative to us, standing on Earth, is the same reason that a rock released in an elevator accelerating in flat spacetime accelerates relative to someone at rest in the elevator. In both cases, it's simply that we, the observer, have nonzero proper acceleration, and that nonzero proper acceleration creates an "apparent force field" in our vicinity. That's really all there is to it.

On this view, the equivalence principle is just obvious: the two situations are equivalent because they are the same thing: proper acceleration causing an apparent force field. The only difference is what is causing the proper acceleration, but as far as measurements made inside the elevator are concerned, the cause of the elevator's acceleration is immaterial. It only comes into play when we want to go beyond the EP and look at phenomena not restricted to a single local inertial frame.

Of course, once we go beyond a single local inertial frame, it is obvious that the two situations are different. But what is different? It's not that there is a "real force field" in one and not the other. It's that spacetime is curved in one and not the other. Spacetime curvature means that the local inertial frames at different events do not "line up" with each other the way they do in flat spacetime. But you can only see this by comparing local inertial frames at different events--for example, at opposite ends of the hole in the Earth, where the proper accelerations of observers at rest relative to the Earth point in opposite directions, making it obvious that the local inertial frames in these two regions of spacetime do not "line up" with each other. The Einstein Field Equation links together this "failure to line up" of local inertial frames with the presence of stress-energy, in this case that of the Earth.

When we analyze what is going on with the accelerometer at the atomic level, we see that each unit mass of the accelerometer is being attracted by gravity equally, and that is why there is no reading on the accelerometer.

This is the Newtonian view, yes, but in GR, it's not correct. And there is a good reason why the GR view works better: how do we measure how each individual atom of the accelerometer is being "attracted by gravity"? Do we measure each electron, each quark inside the nucleus? At what point do we expect to see an actual force "pulling" on something, i.e., some actual measurable difference between being in free fall in a "gravity force field" and just floating at rest in empty space with no gravitating masses anywhere?

The GR answer is "never", because there is no difference. Gravity is not a force. It does not "pull" on anything. We say this because we measure no "pull" at all when an object is moving solely under gravity. That is the experimental fact. It is no answer to this to say that "well, the pull is really there, but we can't measure it because it affects everything equally". The only reason to hypothesize a "pull" in the first place would be to account for some measurement that is different when gravity is "pulling" than when there is no gravity present. But there is no such measurement, so the simplest thing to do is to just admit that gravity is not a force at all.

 

[Dale]

we see that each unit mass of the accelerometer is being attracted by gravity equally, and that is why there is no reading on the accelerometer

Which is exactly the feature that gravity has in common with fictitious forces. You can say exactly this same comment about any fictitious force.

On the other hand, the accelerometer on Earth shows a reading simply due to contact forces at the surface.

This is also true of any scenario where there is a fictitious force and a real contact force.

 

[MikeGomez]

Which is exactly the feature that gravity has in common with fictitious forces. You can say exactly this same comment about any fictitious force.

It has the same features as all common fictitious forces because it is a fictitious force. I agree that you can say the same comment about any fictitious force. All fictitious forces work exactly the same way.

 

[MikeGomez]

(1) In flat spacetime, there are no "particles" generating a field; spacetime is flat, there is zero stress-energy everywhere, and there is thus no "source" of any field.

Agree there are no particles generating a field. The spacetime is flat. A small concentrated field is linearly transmitted at the floor of the elevator. That is stress energy due to the force applied.

(2) In flat spacetime, corresponding to (1), there is no "real force field", because there is no source to generate one.

The source is the rocket.

You can make an "apparent force field" by accelerating (firing your rocket, for example), but it is only "apparent", not "real".

It is real.

The way that GR deals with this apparent problem is this: rather than insist on a "real force field" in flat spacetime, admit that what in Newtonian physics we call "the force of gravity" is not a real force field. It's only an "apparent" one, just like the "apparent" force field that appears when you accelerate.

It’s not an apparent force. It is a fictitious force, and just like any other force there is a reaction pair. One component is the particle, and the other component is in the strongest direction of the energy of the field. The particle seeks the direction of greatest energy, In the case of the Earth that is the center of the earth. For the accelerated elevator that is from the direction of the force applied at the floor of the elevator.

In other words, the reason a rock accelerates downwards relative to us, standing on Earth, is the same reason that a rock released in an elevator accelerating in flat spacetime accelerates relative to someone at rest in the elevator. In both cases, it's simply that we, the observer, have nonzero proper acceleration, and that nonzero proper acceleration creates an "apparent force field" in our vicinity. That's really all there is to it.

It is relative, yes, but I do not believe all there is to it.

On this view, the equivalence principle is just obvious: the two situations are equivalent because they are the same thing: proper acceleration causing an apparent force field. The only difference is what is causing the proper acceleration, but …

Agree except again for the “apparent” part.

as far as measurements made inside the elevator are concerned, the cause of the elevator's acceleration is immaterial.

I disagree.

It only comes into play when we want to go beyond the EP and look at phenomena not restricted to a single local inertial frame.

I think it comes into play everywhere you look.

Of course, once we go beyond a single local inertial frame, it is obvious that the two situations are different. But what is different? It's not that there is a "real force field" in one and not the other. It's that spacetime is curved in one and not the other.

As per Sean Carroll, curvature can not be proved.

Spacetime curvature means that the local inertial frames at different events do not "line up" with each other the way they do in flat spacetime. But you can only see this by comparing local inertial frames at different events--for example, at opposite ends of the hole in the Earth, where the proper accelerations of observers at rest relative to the Earth point in opposite directions, making it obvious that the local inertial frames in these two regions of spacetime do not "line up" with each other. The Einstein Field Equation links together this "failure to line up" of local inertial frames with the presence of stress-energy, in this case that of the Earth.

The Einstein field equations make everything line up and that is fantastic. It is a tool for us to use. A gift from Einstein.

This is the Newtonian view, yes, but in GR, it's not correct. And there is a good reason why the GR view works better: how do we measure how each individual atom of the accelerometer is being "attracted by gravity"? Do we measure each electron, each quark inside the nucleus?

I already talked about superposition in a hand wavy kind of way. I did not mean exactly the same as for electromagnetism. It would be much more difficult. I am just about inept at math as I am with physics, but I will look into it a little.

At what point do we expect to see an actual force "pulling" on something, i.e., some actual measurable difference between being in free fall in a "gravity force field" and just floating at rest in empty space with no gravitating masses anywhere?

With no stress, no momentum, no relativistic effects, the attraction between two neutral particles is the same as the gravitational mass attraction, based on the mass energy equivalence of each, wouldn't it? All of these components must be considered in the superposition calculation. [/QUOTE]

As far as measuring individual particles are concerned, that is difficult, as it always it. Is this a valid reason for denying the existence of the field. We can’t calculate it, we can’t measure it, so it doesn’t exist?

The GR answer is "never", because there is no difference. Gravity is not a force. It does not "pull" on anything. We say this because we measure no "pull" at all when an object is moving solely under gravity. That is the experimental fact. It is no answer to this to say that "well, the pull is really there, but we can't measure it because it affects everything equally". The only reason to hypothesize a "pull" in the first place would be to account for some measurement that is different when gravity is "pulling" than when there is no gravity present. But there is no such measurement, so the simplest thing to do is to just admit that gravity is not a force at all.

This has pretty much been gone over before, and I’ll go over some of it again briefly in the following.

 

[MikeGomez]

@DaleSpam
Regarding accelerometers:

In the elevator the accelerometer measures inertia. At Earth the situation looks to be reversed, but the magnitude of the reading is the same. Is that not “relative inertia”?

On Earth the accelerometer shows a reading of proper acceleration reading. The situation on the elevator appears to be reversed, but the magnitude of the reading of proper acceleration is the same. Is that not “relative proper acceleration”?

In freefall you transform the gravity away.

I say you only set the gravitational potential to zero relative to the accelerometer, and that it still accelerates and comes crashing to the ground.

You say that is coordinate acceleration, due to fictitious forces.

I say “fictitious” does not mean fictional. The forces are real. They originate from the interaction between the particles and the field created by the particles.

You say the field does not exist.

Einstein does more that simple say that the man in the chest will come to the conclusion that he is in a gravitational field. He explicitly says in (Relativity, A. Einstein, pg. 69)

…A gravitational field exists for the man in the chest, despite the fact that there was no such field for the co-ordinate system first chosen. Now we might easily suppose that the existence of a gravitational field is always only an apparent one…

I say the field exists as per Einstein, QM, and simple logic. I have seen no evidence here to the contrary.

 

[MikeGomez]

@PeterDonis

I believe your objections to my example of transitioning from the accelerated elevator to the massive body is unfounded. The idea works in principle. I can construct the experiment to my liking. My planet can be much less massive than the earth, and of whatever construction I think suitable. It doesn’t have to be dirt and water andmolten lava. In be something nice and manageable like Hard plastic or aluminum or something.

Let’s say the force of gravity on my small planet is 1/1000000000 times as strong as at the surface of the earth. Let’s call that g2. Now the rocket only needs to apply a thrust force of g2, and I need an accelerometer accurate enough to measure that small an amount of acceleration. With these parameters I can smoothly transition from the accelerated elevator to the massive body, maintaining a constant reading on the accelerometer all the while.

At no point does the reading on the accelerometer reduce to zero and then go back up, as it would have to do if the polarity of the force vector were to be inverted. The rockets maintain their direction and the thrust gradually decreases to zero, so they can not be the cause in change of direction. The planet maintains its direction the entire journey, so it also is not capable for changing the direction of the system. The polarity of the gravitational charge is exactly the same in both cases. The only way to make sense of this outcome is with the aid of the fictitious force vector. The fictitious force vector is determined in each case by the relative relationship between the accelerated body and the source of energy, that being the center of the planet in one case, and at the surface of the floor of the elevator in the other case. That is what is in alignment.

BTW, when I say that the Earth is below the elevator, I should clarify. This is not the entire mass of the Earth but much less. The ratio of the gravitational flux for an area (approximately) occupied by the accelerometer, as to the total gravitational flux through the total surface area of the Earth is approximately 9.618E-13 by my calculations, so that is a much smaller "earth under the floor of the elevator".

 

[Dale]

It has the same features as all common fictitious forces because it is a fictitious force. I agree that you can say the same comment about any fictitious force. All fictitious forces work exactly the same way.

Yes. It seems that we agree with each other on the classification and maybe only disagree on whether or not it is "clear".

Accelerometers do not detect fictitious forces, accelerometers do not detect gravity, therefore Gravity is a fictitious force. Seems clear to me, but that is a personal opinion in the end.

 

[PeterDonis]

MikeGomez, I think you need to make up your mind: do you want to learn what GR actually says, about the EP or anything else, or do you want to argue for your own personal interpretation of gravity? You have to pick one; you can't have both, because your personal interpretation just doesn't fit with what GR says. You could still reach your stated goal of being able to feel that inertia and gravity are the same: but it looks to me like you are going to have to discard a lot of your personal interpretation if you want to understand how that is true in the context of GR.

A small concentrated field is linearly transmitted at the floor of the elevator.

That is stress energy due to the force applied.

The source is the rocket.

All wrong; as I said before, in flat spacetime there is no source of gravity. Stress-energy is zero. Period. If you are thinking of a "field", or "stress energy due to force applied" from the rocket, you are thinking of it wrong. You are not going to understand GR unless you stop doing that.

It’s not an apparent force. It is a fictitious force, and just like any other force there is a reaction pair.

First, I am using "apparent force" as a synonym for "fictitious force". If you like the latter term better, that's fine, I'll start using that instead.

What you say about a reaction pair is wrong. Fictitious forces do not obey Newton's Third Law; that's one thing that distinguishes them from "real" forces (the other being, as DaleSpam says, that you can't measure them with accelerometers). In Newtonian gravity, the force of gravity is declared by fiat to be a special case: even though you can't measure it with an accelerometer, it still obeys the Third Law, so it is not considered a fictitious force. But in GR, this ad hoc special case is not needed; what we call the "force of gravity" is a fictitious force like any other (centrifugal, Coriolis, etc.) and does not obey the Third Law. This eliminates all the conceptual issues with Newtonian gravity being an "action at a distance" force, which could be finessed before relativity was developed, but which cannot be accommodated in a relativistic theory.

As per Sean Carroll, curvature can not be proved.

Please give a specific reference for this statement. Spacetime curvature is tidal gravity, which is a direct observable. I find it very difficult to believe that Sean Carroll says tidal gravity is not a direct observable.

With no stress, no momentum, no relativistic effects, the attraction between two neutral particles is the same as the gravitational mass attraction, based on the mass energy equivalence of each, wouldn't it?

No. With no stress-energy, there is no attraction. Spacetime is flat.

Is this a valid reason for denying the existence of the field. We can’t calculate it, we can’t measure it, so it doesn’t exist?

If I said (taking an example from Carl Sagan) that there was a dragon in my garage, but it was invisible and unmeasurable by any means, in what sense does the dragon "exist"?

 

[PeterDonis]

Let’s say the force of gravity on my small planet is 1/1000000000 times as strong as at the surface of the earth. Let’s call that g2. Now the rocket only needs to apply a thrust force of g2, and I need an accelerometer accurate enough to measure that small an amount of acceleration. With these parameters I can smoothly transition from the accelerated elevator to the massive body, maintaining a constant reading on the accelerometer all the while.

Not without the rocket moving relative to the massive body as you smoothly transition from zero mass (flat spacetime) to the mass necessary to hold the rocket in place if its thrust is g2. If you have an accelerometer accurate enough to measure such a small acceleration, then you have measuring tools accurate enough to detect the rocket's motion.

As I think I've said before: instead of waving your hands about this or any other scenario, do the math. It won't work out the way you are imagining it will.

 

[PeterDonis]

Einstein does more that simple say that the man in the chest will come to the conclusion that he is in a gravitational field. He explicitly says in (Relativity, A. Einstein, pg. 69)

…A gravitational field exists for the man in the chest, despite the fact that there was no such field for the co-ordinate system first chosen. Now we might easily suppose that the existence of a gravitational field is always only an apparent one…

I say the field exists as per Einstein, QM, and simple logic.

Do the math. Find out what Einstein actually meant by "gravitational field" in this passage. Find out what role it plays in the theory as a whole. Then you will know whether what he is saying in this passage means what you think it means.