difference between free fall and zero gravity?

[JanisChambers]

Forgive my ignorance of physics.. but if an object had zero gravity would it just fly apart?

 

[Schneibster]

I mentioned this in a previous debate we had: gravity is indeed not a 1/r² force. The force diverges at the event horizon, not at r=0 (in fact, force diverges at event horizons created by using an accelerated reference frame as well).

Maybe I'm not seeing it, but the event horizon for Earth should be the size of a pea (or would be, if the Earth were a black hole of the same mass- as it is, there's not enough mass inside the Schwarzchild radius for it to be one). So I'm having trouble seeing how that would affect this particular thing- I'm thinking that the falloff would be just a little bit steeper than 1/r², but not seeing why the event horizon of a nonexistent black hole would apply. Is this a calculation/visualization technique I'm not familiar with?

 

[Schneibster]

Forgive my ignorance of physics.. but if an object had zero gravity would it just fly apart?

If it were held together by gravity against a force trying to push it apart and the gravity went away, yes- but that's it's own gravity, not gravity from outside. Otherwise, most objects are held together by electric interactions between their atoms and molecules, which are called "Van der Waals forces," so for most objects, the answer is, "No."

 

[JanisChambers]

Hmm, this is really making me want to do more reading (science is so much more awe inspiring than any holy book)

 

[Thabiguy]

I disagree here. I intentionally did not say that they share a frame; what I said was, unless very minor (unless the gravity field is extreme, unmeasurable) differences are considered, what the inertial, far away observer sees appears the same as what the motionless observer accelerating in the gravity field sees. My point was not that they are the same; it was that they're not, but they look the same, and that confuses people.

The highlight in your quote shows precisely what I did not intend to say, and do not believe is the case.

I know you didn't say that; I was not trying to make it seem that you did. I was merely pointing out the circumstances under which the conclusion would be adequate and gravity could be fully modelled by a force.

They may; but unless it is a very strong gravity field, if they agree they are motionless relative to one another, the differences will be unmeasurable save by the most sensitive instruments we can conceive of. Gravity Probe B used such instruments to gather data for over a year (IIRC), and it's looking like it'll take most of a year, or even more than that, to analyze that data and extract the measurements out of it.

I think that's something slightly different. The Gravity Probe B experiment is trying to confirm frame-dragging, which is predicted to occur near rotating massive objects, and is indeed extremely small. I was talking about gravitational time dilation, the consequence of which is that observers at rest in each other's reference frame (as they can verify by measuring the distance over and over again and always finding it the same) can still experience time dilation. The difference between a distant inertial observer and an observer on the surface of the Earth should be about one part in a billion, which is quite measurable. (I hear that for GPS satellites this effect is actually greater than time dilation predicted by special relativity.)

I'm going to have to think about this a little while, and we may need an answer to the question of whether the inverse-square falloff is not precisely correct to get to this. I think that the definition of acceleration is a matter not of measurement, but of spacetime geometry- and the fact that inertial frames are possible in both says that your conjecture here may not be correct. Think of how two different inertial frames would see each other in a gravity field, and how that might vary from how they would see each other outside it, or from how one inside and one outside might see each other. I think that might be the key to this. I sense that it is, based on Einstein's use of this to get to the Equivalence Principle.

I think that's the essence of the problem: there is acceleration of an object, as measured in a given reference frame, and then there is local acceleration, or acceleration of a reference frame, and these two are different in general relativity. You can have an object with zero acceleration in an inertial reference frame but non-zero in its local reference frame (observer on the surface of a planet), or an object with zero local acceleration but non-zero acceleration in an inertial reference frame (object orbiting a planet). Note that in special relativity, this never happens; inertial reference frames are globally equivalent and will never accelerate with respect to each other, and an object that accelerates in an inertial reference frame is never an origin of another inertial reference frame.

The equivalence principle is only talking about the latter kind of acceleration, the local. Referring to "acceleration due to gravity" however invokes the former kind of acceleration, the measured. And the confusion arises when the equivalence principle is misapplied to the former through the inaccurate interpretation that "gravity is indistinguishable from acceleration", which for objects with mass is the same as "gravity is indistinguishable from effects of a force" or even "gravity is a force". But that's not true in general relativity (as opposed to special relativity, where it is indeed a force): gravity is distinguishable from effects of a force, and I have mentioned some experiments how this can be done.

The truth is that in general relativity (and as it happens, in our universe as well), when you try to model gravity with a force - any kind of force, even one that has complicated shape and falloff - you fail. You let a probe fly around a planet, and you measure its path. Then you send it through empty space and accelerate it so that its observed path is the same. You can pull on it with strings, you can push it with engines, you may charge it and tug on it with an electric field; even when you make it follow the same path, with the same acceleration at every moment, by definition the observed force is the same - and yet the result will still be different. This is because in general relativity, force does not correctly model gravity or explain its effects. In other words, the assumption that gravity is force is wrong.

Note that this doesn't mean that gravity can't manifest as an apparent force. But only in the same way that inertia can sometimes manifest as an apparent centrifugal force - a pseudo-force which is the sole product of choosing a non-inertial reference frame. Similarly, in general relativity, the observed force of gravity is a pseudo-force, which is the sole product of choosing either a non-inertial or a non-local reference frame. The old models (Newtonian and special relativity) where gravity is a real force, like EM force, and the model where gravity is not a force but spacetime curvature, the apparent "force" being merely an observational artifact, may seem equivalent, but they are not. There is a slight difference, and it is this difference that general relativity successfully predicts.

That's what makes me say that in general relativity, gravity is not a force.

 

[Ziggurat]

Maybe I'm not seeing it, but the event horizon for Earth should be the size of a pea (or would be, if the Earth were a black hole of the same mass- as it is, there's not enough mass inside the Schwarzchild radius for it to be one). So I'm having trouble seeing how that would affect this particular thing- I'm thinking that the falloff would be just a little bit steeper than 1/r², but not seeing why the event horizon of a nonexistent black hole would apply.

The fact that it diverges at the event horizon just means that it cannot be 1/r², but must indeed fall off faster than that. 1/r² is a weak-field (large r) approximation. For anything within the solar system, we're in weak enough fields for the approximation to be quite good, but the point is it's only an approximation, no matter how good it is. Just as sin(x)=x is a good approximation for small x.

 

[69dodge]

I think that's the essence of the problem: there is acceleration of an object, as measured in a given reference frame, and then there is local acceleration, or acceleration of a reference frame, and these two are different in general relativity. You can have an object with zero acceleration in an inertial reference frame but non-zero in its local reference frame (observer on the surface of a planet), or an object with zero local acceleration but non-zero acceleration in an inertial reference frame (object orbiting a planet). Note that in special relativity, this never happens; inertial reference frames are globally equivalent and will never accelerate with respect to each other, and an object that accelerates in an inertial reference frame is never an origin of another inertial reference frame.

I've underlined a couple of phrases. They mean "a reference frame in which the planet isn't accelerating", correct?

Is "reference frame" a term usually used in general relativity, anyway? I don't really know, but somehow it implies to me something that can be extended globally in a unique way, like an inertial reference frame in special relativity. I've seen "coordinate system" used in GR.

I'm trying to reconcile in my mind the fact that in GR any coordinate system at all is "as good" as any other, with the fact that you can still say "inertial reference frame" when referring to a global sort of thing, and reasonably expect everyone to know what you mean. Is it just because the earth is not too massive, so that its gravity can be thought of approximately as a force even though not exactly as one?

I was talking about gravitational time dilation, the consequence of which is that observers at rest in each other's reference frame (as they can verify by measuring the distance over and over again and always finding it the same) can still experience time dilation.

Continuing with the all-coordinate-systems-are-equally-good idea: At first glance, it seems pretty clear what you mean when you say one observer can verify that another is at rest in his reference frame, but wouldn't it be more correct, strictly speaking (very strictly speaking), to say rather that the one assumes that the other is at rest, and that this assumption determines a coordinates system in which he can then verify the existence of, and the exact amount of, time dilation?

I think this is correct, but it does seem strange. Are things really as arbitrary as I'm claiming, or is there some way of picking a coordinate system that is objectively "more natural" than others?

 

[Ziggurat]

I think this is correct, but it does seem strange. Are things really as arbitrary as I'm claiming, or is there some way of picking a coordinate system that is objectively "more natural" than others?

What happens in special relativity, and even in Newtonian mechanics, if you pick a "bad" (non-inertial) reference frame is that you have to introduce fictitious forces (centrifugal, corriolis, etc.) to account for motion. In some sense, picking a fixed coordinate system (like the Schwarzchild metric) makes gravity a fictitious force in the same way (since a falling coordinate system is locally inertial). But it's unavoidable, since gravity isn't uniform, so you can't have your entire coordinate system fall uniformly. So there is no perfect coordinate system for GR, and which one you choose depends largely on what you want to do. There are certainly WORSE coordinate systems to use (meaning you have to introduce fictitious forces without any attendant simplifications), so it's not completely arbitrary in that sense. But there is no best coordinate system. Considering black holes, for example, the Schwarzchild metric is good for comparison with the flat Minkowski metric, but it's got a coordinate singularity at the event horizon. The Kruskal-Szekeres coordinates are good for covering the event horizon and the interior of a black hole, but are awkward for distances far away.

 

[Thabiguy]

I've underlined a couple of phrases. They mean "a reference frame in which the planet isn't accelerating", correct?

Well, I really meant a locally inertial frame. In the examples I mentioned, I had a theoretical frame of an infinitely distant observer in mind (this would be the only locally inertial frame in which the planet weren't accelerating). That's unrealistic of course, as is the assumption that the planet is not rotating or orbiting around anything etc.; it was a simplified example to illustrate the difference I wanted to point out.

Is "reference frame" a term usually used in general relativity, anyway? I don't really know, but somehow it implies to me something that can be extended globally in a unique way, like an inertial reference frame in special relativity. I've seen "coordinate system" used in GR.

I think you're right. Maybe someone more experienced can enlighten us both about the proper usage.

Continuing with the all-coordinate-systems-are-equally-good idea: At first glance, it seems pretty clear what you mean when you say one observer can verify that another is at rest in his reference frame, but wouldn't it be more correct, strictly speaking (very strictly speaking), to say rather that the one assumes that the other is at rest, and that this assumption determines a coordinates system in which he can then verify the existence of, and the exact amount of, time dilation?

I'm not so sure about this. It seems to me that the observer shouldn't need to assume stuff about other objects in order to define his own coordinate system. He should be able to define it locally, without looking at distant objects. I think it's more a matter of assuming that the local coordinate system can be extended non-locally. And I think it always can be (otherwise we couldn't make any useful measurements of non-local objects, ever), it's just that with such assumption the result of the observation will appear to violate special relativity and effects of general relativity must be taken into account to explain it. - But I may be wrong about this, so if someone knows better, let us know.

And about the rest, I think Ziggurat offered a better answer than I ever could.

 

[Bodhi Dharma Zen]

That's what makes me say that in general relativity, gravity is not a force.

Exactly my feelings, even when Im far from being articulated in the subject Am I right assuming that the only way to show gravity "as a force" is having another force in the place, which will manifest as acceleration (considering that there are no other forces involved)?

For me, this would explain that the gravity "itself" its not "experienciable", this is, a subject will have the same experience being in between two planets (without being pulled in by any of them) and "falling" rapidly towards one of them.

No force involved in both circumnstances, unless a same frame based acceleration occurs. This is the only way to differenciate if the subject is in a strong "gravity field" or not (other than watching outside the spaceship for example).

 

[Thabiguy]

Am I right assuming that the only way to show gravity "as a force" is having another force in the place, which will manifest as acceleration (considering that there are no other forces involved)?

It's one way, but not the only way. If you and your local buddy fall together in a gravitational field, you may start your jetpack and observe that your buddy accelerates away. You may decide to claim yourself stationary and say that your buddy is accelerated by the force of gravity.

Another way is observing non-locally. If you and your buddy fall in tandem, separated by some distance, you'll observe that you'll accelerate away from each other. You may say that your buddy is accelerated by a tidal force. If you are far away from the planet and your buddy close to it, you may not call it a tidal force but instead the force of gravity.

For me, this would explain that the gravity "itself" its not "experienciable", this is, a subject will have the same experience being in between two planets (without being pulled in by any of them) and "falling" rapidly towards one of them.

Well, yes. I think everybody here agrees that except for non-local phenomena (such as tides or observing your surroundings), the experience would be the same.

 

[Schneibster]

I know you didn't say that; I was not trying to make it seem that you did.

I know; I was pointing out that I deliberately avoided saying it.

I was merely pointing out the circumstances under which the conclusion would be adequate and gravity could be fully modelled by a force.

Well, the thing is, you're defining a force in a spacetime-geometry-distortion-free manner, and gravity is a distortion of spacetime geometry. I disagree that it's not a force, because if you factor in the differences that spacetime distortion makes in what different observers see, it acts like a force. I'll get into that in a little bit.

I think that's something slightly different. The Gravity Probe B experiment is trying to confirm frame-dragging, which is predicted to occur near rotating massive objects, and is indeed extremely small. I was talking about gravitational time dilation, the consequence of which is that observers at rest in each other's reference frame (as they can verify by measuring the distance over and over again and always finding it the same) can still experience time dilation. The difference between a distant inertial observer and an observer on the surface of the Earth should be about one part in a billion, which is quite measurable. (I hear that for GPS satellites this effect is actually greater than time dilation predicted by special relativity.)

Billionths of a second is a high-precision measurement. I wouldn't make a distinction between a maser (atomic clock) and the instruments on Gravity Probe B in terms of sensitivity that you do, when compared with the instrumentation available in the 17th century. Both are many orders of magnitude more sensitive than anything available then; yet, gravity was clear then. These types of instruments are far closer to one another in sensitivity than they are to anything available then, and that was my point. The effects of gravity are obvious; measured to three or four significant figures, they are indistinguishable from a truly inverse-square law force. Even the effects on Mercury's orbit, which are relatively gross compared to what you're talking about, weren't realized until the late 18th century. You need a very strong gravitational field, or very sensitive instrumentation, to see these effects.

I think that's the essence of the problem: there is acceleration of an object, as measured in a given reference frame, and then there is local acceleration, or acceleration of a reference frame, and these two are different in general relativity. You can have an object with zero acceleration in an inertial reference frame but non-zero in its local reference frame (observer on the surface of a planet), or an object with zero local acceleration but non-zero acceleration in an inertial reference frame (object orbiting a planet). Note that in special relativity, this never happens; inertial reference frames are globally equivalent and will never accelerate with respect to each other, and an object that accelerates in an inertial reference frame is never an origin of another inertial reference frame.

You're missing the point of GR. GR is a description of accelerated reference frames. That it also describes gravity is a bonus, but the point is, an inertial reference frame in a gravity field is only measurably different from one not in a gravity field because of the spherical nature of all real gravity fields, and because of the inverse square falloff. If the gravity field were planar, rather than spherical, and didn't change over distance, then there would be no experiment you could perform in such a frame that would allow you to differentiate between such an inertial frame and one not in a gravity field. Yet, that frame's coordinate system would neither remain constant relative to the other's, nor move at a constant velocity. It would accelerate. But the real point is, it would still be inertial, easily measured as such, and thus easily proven to be inertial. THAT is the point of GR. People often say that SR doesn't deal with acceleration. That's not true. What it doesn't deal with is accelerated frames of reference.

That said, I think you made some errors:

I think that's the essence of the problem: there is acceleration of an object, as measured in a given reference frame, and then there is local acceleration, or acceleration of a reference frame, and these two are different in general relativity.

That's true enough; but remember that the equivalence principle says that all inertial frames of reference are equivalent, whether their coordinate systems are accelerating in a gravity field or not; and they are different from accelerated frames of reference, which are also all equivalent, whether they are moving (outside a gravity field) or still (inside one). That's the entire point of the equivalence principle.

You can have an object with zero acceleration in an inertial reference frame but non-zero in its local reference frame (observer on the surface of a planet),

Acceleration is always measured by experiment within the local frame. The fact that it appears to be motionless relative to a frame outside the field is immaterial; the ONLY material fact is what you measure in that frame. That's what the equivalence principle says. That's the only way you can measure it. Are the paths of inertial objects in your frame straight, or curved? If they're straight, you're in an inertial frame. If they're curved, you're in an accelerated frame. That is the one and only way of determining whether the frame is inertial or accelerated.

or an object with zero local acceleration but non-zero acceleration in an inertial reference frame (object orbiting a planet).

Again, inertial is inertial. You can measure it, in the local frame. It doesn't matter what a distant frame sees in the definition of whether you are in an inertial frame or not. All that matters is what you measure in that frame. The entire point of all of relativity is that that's all that matters. Einstein once said that the Theory of Relativity was misnamed; it should have been the Theory of Absoluteness. An inertial frame is an inertial frame, wherever it happens to be. That's the meaning of relativity; inertial is an absolute, and is always locally measurable.

Note that in special relativity, this never happens; inertial reference frames are globally equivalent and will never accelerate with respect to each other, and an object that accelerates in an inertial reference frame is never an origin of another inertial reference frame.

But what you're trying to say is that an inertial frame in a gravity field is somehow different from an inertial frame outside one; and that's contrary to the equivalence principle. I have to point out that the equivalence principle has the same standing as the speed-of-light limit, or the existence of inertial frames, or the postulate of a spacetime continuum. This is taken as proven, by the correct calculations that you can do with relativity math. This is the nature of the universe. If it's not, we'll know in December; Gravity Probe B's results won't show frame dragging.

The equivalence principle is only talking about the latter kind of acceleration, the local. Referring to "acceleration due to gravity" however invokes the former kind of acceleration, the measured. And the confusion arises when the equivalence principle is misapplied to the former through the inaccurate interpretation that "gravity is indistinguishable from acceleration", which for objects with mass is the same as "gravity is indistinguishable from effects of a force" or even "gravity is a force". But that's not true in general relativity (as opposed to special relativity, where it is indeed a force): gravity is distinguishable from effects of a force, and I have mentioned some experiments how this can be done.

You've turned it on its head. An inertial frame is an inertial frame, no matter where it is, in a gravity field or not. An accelerated frame is an accelerated frame, in a gravity field or not. Whether the frame is inertial or accelerated is not determined by how it moves. It's determined by local experiment. If inertial objects move in straight lines, it's inertial. If they do not, it's accelerated. That is the one and only guide. You can't measure a frame; you can only define a frame, and then measure the motions of objects. A frame doesn't have real existence; it's a metric, an imposed conceptual structure. Whether it is straight or curved can only be determined by the behavior of inertial objects with reference to it; it is defined by that behavior.

The difference between being in a gravity field and being outside one cannot be determined by the behavior of inertial objects nearby compared to that far away. For example, suppose inertial objects nearby are moving in straight paths, but objects in a certain direction over thataway are moving in curved paths. Now, are you in a gravity field, shared by objects close to you, whose paths are therefore straight because they are inertial; or are the objects over there in the gravity field? You can't tell. All you can tell is whether you are inertial or not. Gravity is relative.

The ONLY way you can determine it is by the fact that the gravity field in the real world is always spherical, and always falls off approximately by the inverse square law. Thus, if you observe that inertial objects close to you move in straight lines, but ones far away in all directions move in curved ones, then you are correct in assuming you are in an inertial frame in a gravity field. And if objects over there move in curved lines, but they move in straight lines in every other direction, then you are equally correct in assuming you are not, and the objects over there are. But only by making such observations, which rely on the spherical nature and inverse-square falloff of the field can you make these conclusions; in a uniform field, you could not. The entire universe could be in such a uniform field, and we would never notice, because the only ways we can measure such things is by these peculiarities of the real world manifestation of gravity.

THAT's what the equivalence principle says.

The truth is that in general relativity (and as it happens, in our universe as well), when you try to model gravity with a force - any kind of force, even one that has complicated shape and falloff - you fail. You let a probe fly around a planet, and you measure its path. Then you send it through empty space and accelerate it so that its observed path is the same. You can pull on it with strings, you can push it with engines, you may charge it and tug on it with an electric field; even when you make it follow the same path, with the same acceleration at every moment, by definition the observed force is the same - and yet the result will still be different. This is because in general relativity, force does not correctly model gravity or explain its effects. In other words, the assumption that gravity is force is wrong.

I also don't agree with this. From the point of view of GR, you're correct- gravity isn't a force. But we also have QM, and in QM, gravity is the ultimate force- the only one that we can't model as a gauge boson quantum exchange. Furthermore, by the simplest most basic definition of a force, that which imposes acceleration, gravity is unquestionably a force. In other words, it follows the action principle, whatever the reason.

Ultimately, it is overwhelmingly likely that gravity will be successfully quantized. When that happens, we will see it as a force in the quantum sense: an exchange of gauge bosons.

If string physics is correct, all the forces are matters of warping of spacetime, not just gravity. The fact that we cannot directly observe the extra dimensions that the EM force, for example, warps, does not make the warping any less real than it is for gravity. And never ever forget that Theodor Kaluza managed to derive Maxwell's equations directly from the assumption of a fifth spatial dimension, in a manner directly analogous to the way that Einstein derived the ten field equations of gravity in our four dimensional spacetime. This extremely compelling demonstration has haunted theoretical physicists for decades.

Note that this doesn't mean that gravity can't manifest as an apparent force. But only in the same way that inertia can sometimes manifest as an apparent centrifugal force - a pseudo-force which is the sole product of choosing a non-inertial reference frame.

And again I disagree. One can observe the force of gravity from an inertial frame. Obviously.

Similarly, in general relativity, the observed force of gravity is a pseudo-force, which is the sole product of choosing either a non-inertial or a non-local reference frame.

If I choose to observe objects in a gravity field from an inertial frame that is not in the field, I will observe it as a force. Again, it matters what the location of the frame is relative to the location of the field. But in the case of a uniform field, of course there is no way to tell.

The old models (Newtonian and special relativity) where gravity is a real force, like EM force, and the model where gravity is not a force but spacetime curvature, the apparent "force" being merely an observational artifact, may seem equivalent, but they are not. There is a slight difference, and it is this difference that general relativity successfully predicts.

Again, the differences GR predicts are based solely on the fact that gravity is inverse-square and spherical. You have not yet shown a difference that would apply in a uniform gravity field.

That's what makes me say that in general relativity, gravity is not a force.

In general relativity, you are correct. But GR is not all of physics, nor even half of it. And ultimately, if string physics is correct, none of the forces will be forces, by that definition. So I still say, if they're all not forces, then we'll have to find some other name to call them; but whatever that name is, they'll all be that together, gravity right along with the other three.

 

[Schneibster]

The fact that it diverges at the event horizon just means that it cannot be 1/r², but must indeed fall off faster than that. 1/r² is a weak-field (large r) approximation. For anything within the solar system, we're in weak enough fields for the approximation to be quite good, but the point is it's only an approximation, no matter how good it is. Just as sin(x)=x is a good approximation for small x.

Ah. A visualization technique. OK, I'm with you. Technically, of course, you're correct.

 

[Thabiguy]

You're missing the point of GR.

I'll freely admit that , but I don't see how it follows from the rest.

GR is a description of accelerated reference frames. That it also describes gravity is a bonus, ...

I disagree. Special relativity fully describes accelerated reference frames. It is curvature of spacetime (=gravity) that makes the difference between special relativity and general relativity. If you take away curvature of spacetime (=gravity) from general relativity, you get special relativity. I can't agree with calling the very essence of a difference, "a bonus".

... but the point is, an inertial reference frame in a gravity field (1) is only measurably different from one not in a gravity field (2) because of the spherical nature of all real gravity fields, and because of the inverse square falloff. If the gravity field were planar, rather than spherical, and didn't change over distance, then there would be no experiment you could perform in such a frame that would allow you to differentiate between such an inertial frame and one not in a gravity field. Yet, that frame's coordinate system would neither remain constant relative to the other's, nor move at a constant velocity. It would accelerate. But the real point is, it would still be inertial, easily measured as such, and thus easily proven to be inertial. THAT is the point of GR. People often say that SR doesn't deal with acceleration. That's not true. What it doesn't deal with is accelerated frames of reference.

GR defines inertial reference frames in that way. It differs in this definition from special relativity and classical mechanics. In GR, frame (1) is inertial, in SR, it is accelerated. The predictions of the two theories differ about how (1) and (2) would see each other. If this is what you're saying, then we agree. - Also note that in GR, frame (1) is inertial only if it accelerates with respect to (2) because of being in a gravitational field. If it were accelerating because of any other kind of field, it wouldn't be inertial. This again demonstrates that it's the explanation of gravity that makes GR distinct from SR, not some special handling of acceleration per se. Thus my disagreement above.

I fail to see in what sense SR doesn't "deal" with accelerated frames of reference. I would say that an accelerating object is an origin of an accelerated reference frame which SR deals with just fine: see here. Can you clarify?

That's true enough; but remember that the equivalence principle says that all inertial frames of reference are equivalent, whether their coordinate systems are accelerating in a gravity field or not; and they are different from accelerated frames of reference, which are also all equivalent, whether they are moving (outside a gravity field) or still (inside one). That's the entire point of the equivalence principle.

I agree, but can't understand how this truth applies to what I said. Can you clarify in what way this is related to my words?

[Physical] acceleration[, as opposed to coordinate acceleration,] is always measured by experiment within the local frame. The fact that it appears to be motionless relative to a frame outside the field is immaterial; the ONLY material fact is what you measure in that frame. That's what the equivalence principle says. That's the only way you can measure it. Are the paths of [local] inertial objects in your frame straight, or curved? If they're straight, you're in an inertial frame. If they're curved, you're in an accelerated frame. That is the one and only way of determining whether the frame is inertial or accelerated.

With the reservations that I inserted in bolded text in square brackets, I agree. But again I fail to see the relevance of this to my words. I don't dispute the equivalence principle; I was highlighting the difference between physical and coordinate acceleration, in order to subsequently show the error in misapplying the equivalence principle to the latter, which commonly happens and which I have seen on this forum. Could it be that you misunderstood this point of mine as saying that there is something wrong with the equivalence principle?

Again, inertial is inertial. You can measure it, in the local frame. It doesn't matter what a distant frame sees in the definition of whether you are in an inertial frame or not. All that matters [for the purpose of defining an inertial frame] is what you measure in that frame. The entire point of all of relativity is that that's all that matters. Einstein once said that the Theory of Relativity was misnamed; it should have been the Theory of Absoluteness. An inertial frame is an inertial frame, wherever it happens to be. That's the meaning of relativity; inertial is an absolute, and is always locally measurable.

With the reservation that I inserted in bolded text in square brackets, I agree. Again though, I feel this has little to do with what I said.

But what you're trying to say is that an inertial frame in a gravity field is somehow different from an inertial frame outside one; and that's contrary to the equivalence principle.

No, it is not. Equivalence principle says that these two frames are locally equivalent. Non-locally, they may differ all they like. That's why I pointed out the difference from SR, where inertial reference frames are equivalent globally.

I have to point out that the equivalence principle has the same standing as the speed-of-light limit, or the existence of inertial frames, or the postulate of a spacetime continuum. This is taken as proven, by the correct calculations that you can do with relativity math. This is the nature of the universe. If it's not, we'll know in December; Gravity Probe B's results won't show frame dragging.

I fail to understand why you are defending the equivalence principle. What made you think that I am questioning it?

You've turned it on its head. (3) An inertial frame is an inertial frame, no matter where it is, in a gravity field or not. An accelerated frame is an accelerated frame, in a gravity field or not. Whether the frame is inertial or accelerated is not determined by how it moves. It's determined by local experiment. If [local] inertial objects move in straight lines, it's inertial. If they do not, it's accelerated. That is the one and only guide. You can't measure a frame; you can only define a frame, and then measure the motions of objects. A frame doesn't have real existence; it's a metric, an imposed conceptual structure. Whether it is straight or curved can only be determined by the behavior of [local] inertial objects with reference to it; it is defined by that behavior.

With the reservations that I inserted in bolded text in square brackets, I agree with all sentences but the first, (3). Unless you demonstrate in what way this conflicts with what I said, you have no basis for concluding (3). What you are talking about does not address the point that I was trying to make. Maybe your perception that I've "turned it on its head" comes from the misperception of this point.

The difference between being in a gravity field and being outside one cannot be determined by the behavior of inertial objects nearby compared to that far away. For example, suppose inertial objects nearby are moving in straight paths, but objects in a certain direction over thataway are moving in curved paths. Now, are you in a gravity field, shared by objects close to you, whose paths are therefore straight because they are inertial; or are the objects over there in the gravity field? You can't tell. All you can tell is whether you are inertial or not. Gravity is relative.

You can tell that the faraway objects are accelerating in your reference frame. The very act of observing faraway objects makes this a non-local experiment. As you are engaging in that, you can as well communicate with them and ask them about their local acceleration (which they can always determine, as you have emphasized). From this, you can work out their spacetime curvature relative to yours, or your spacetime curvature relative to theirs. You can't work out the absolute value, but you can tell the difference. That allows you (in principle) to map the spacetime curvature in all universe up to a global constant (uniform acceleration of the entire universe).

This would make it wrong to say that all you can tell is whether you are inertial or not. Apparently you can tell a lot more than that. Here we approach again the misapplication of the equivalence principle, which talks about local experiments, to non-local experiments such as those that involve observing faraway objects. The equivalence principle does not restrict those to have the same result in all inertial frames.

The ONLY way you can determine it is by the fact that the gravity field in the real world is always spherical, and always falls off approximately by the inverse square law.

No. You can determine differences in gravitational field (=curvature of spacetime). Period. It doesn't matter if the fields are spherical or cubical, it doesn't matter whether they fall off in steep or shallow or bumpy fashion. If there are differences, you can determine them. And if there are none, you can determine that there are none. And in that case, you are in a flat spacetime and special relativity applies globally. What more is there to find out?

Thus, if you observe that inertial objects close to you move in straight lines, but ones far away in all directions move in curved ones, then you are correct in assuming you are in an inertial frame in a gravity field. And if objects over there move in curved lines, but they move in straight lines in every other direction, then you are equally correct in assuming you are not, and the objects over there are. But only by making such observations, which rely on the spherical nature and inverse-square falloff of the field can you make these conclusions; ...

The symmetry and falloff of spacetime curvature around masses allow to "normalize" the differences so that spacetime is flat far away from the mass, not near it. But they are not needed to observe the differences.

... in a uniform field, you could not. The entire universe could be in such a uniform field, and we would never notice, because the only ways we can measure such things is by these peculiarities of the real world manifestation of gravity.

Yes, in a globally uniform field, no gravity would be detected. Again, I fail to see the relevance of this to what I've said.

THAT's what the equivalence principle says.

Oh, come on. You know what the equivalence principle says. It has many consequences, and this is one of them. It's not THAT what the equivalence principle says. - More importantly though, as you seem to be using this as an argument against my words, please clarify where it conflicts with what I've said.

Furthermore, by the simplest most basic definition of a force, that which imposes acceleration, gravity is unquestionably a force. In other words, it follows the action principle, whatever the reason.

Only in the same sense that inertia is a force. But I think physicists would generally disagree with that. - Nevertheless, you are being inconsistent here. After all that talk about physical acceleration and how it's always measured locally and is the only material fact and really all that matters in all of relativity, you smoothly jump to coordinate acceleration being the basis for the "simplest most basic definition". But the importance of the distinction between these two concepts of acceleration was the entire point of my post.

In general relativity, gravity does not cause physical acceleration and is therefore, by the simplest most basic definition of a force, not a force.

And again I disagree. One can observe the force of gravity from an inertial frame. Obviously.

If I choose to observe objects in a gravity field from an inertial frame that is not in the field, I will observe it as a force. Again, it matters what the location of the frame is relative to the location of the field. But in the case of a uniform field, of course there is no way to tell.

I have said that the pseudo-force of gravity is the sole product of choosing a non-inertial or a non-local reference frame. The only apparent basis for disagreeing here would be ignoring my words.

In order for the frame not to be in the field, the observation must be non-local, by definition. In the case of a uniform field (flat spacetime), the pseudo-force of course won't show. That's not an argument against gravity being a pseudo-force. Coriolis force, which is a pseudo-force, also will manifest in some accelerated frames but not in others.

The point is that even when the spacetime is not flat, the force won't be observed except for experiments in a non-inertial frame or non-local experiments in an inertial frame. That's what makes it a pseudo-force and differentiates it from real forces that can be observed locally in an inertial frame.

Again, the differences GR predicts are based solely on the fact that gravity is inverse-square and spherical. You have not yet shown a difference that would apply in a uniform gravity field.

No. The differences GR predicts in no way rely on those particular properties of gravity. GR predicts those differences to occur anytime that curvature of spacetime is not uniform. But if it is uniform, then the spacetime is flat, there is no gravity to explain and GR becomes equivalent to its limit case, SR. Why should I show a difference that I'm not saying there is? Do you think I'm making some other kind of argument than I am?

In general relativity, you are correct. But GR is not all of physics, nor even half of it. And ultimately, if string physics is correct, none of the forces will be forces, by that definition. So I still say, if they're all not forces, then we'll have to find some other name to call them; but whatever that name is, they'll all be that together, gravity right along with the other three.

Okay then, I'm glad we agree here. I'm just surprised at the amount of disagreement in this agreement.

 

[NobbyNobbs]

Yes, and also if you feel it in a changing direction. Acceleration need not be constant to remain acceleration. Consider rotational motion.

Um...I thought rotational motion (circular motion, anyway) was constant acceleration. The direction of acceleration if towards the center of the circle, and constant.

Hmm, this is really making me want to do more reading (science is so much more awe inspiring than any holy book)

Amen.

 

[Ziggurat]

Um...I thought rotational motion (circular motion, anyway) was constant acceleration. The direction of acceleration if towards the center of the circle, and constant.

Acceleration is a vector quantity, which means that if the direction is changing (and it is for circular motion: "towards the center" isn't the same direction at different points on the circle), then the acceleration isn't constant. It's exactly the same situation as velocity (a vector quantity) changing during circular motion even if the speed (a scalar quantity) isn't.

 

[Schneibster]

Um...I thought rotational motion (circular motion, anyway) was constant acceleration. The direction of acceleration if towards the center of the circle, and constant.

Well, aside from Zig's caveat that the direction of the center isn't constant, more or less yes. Where did you get the idea I said it wasn't, considering I was using it as an example of precisely that?

 

[Schneibster]

I'll freely admit that , but I don't see how it follows from the rest.

Because SR is about motion, and GR is about curvature of spacetime.

I disagree. Special relativity fully describes accelerated reference frames. It is curvature of spacetime (=gravity) that makes the difference between special relativity and general relativity. If you take away curvature of spacetime (=gravity) from general relativity, you get special relativity. I can't agree with calling the very essence of a difference, "a bonus".

The difference between special and general relativity, if you want to get really technical, is that accelerated frames are treated on a different basis in SR than inertial frames, whereas GR treats all frames on an equal basis. In other words, in GR, the equations of motion take the same form in all frames of reference, but in SR, a special term must be used for accelerated frames; this is because SR cannot describe curved spacetime. In other words, it is possible to compensate for the curvature of spacetime introduced by acceleration of a frame, but it requires an extra tensor to do so; but in GR, the curvature of spacetime is present in the equation, so the same equation is used even if the curvature is zero.

This is the reason that GR is different; it accounts for accelerated frames not by the addition of a term that is not present in the base equations of motion, that accounts for the curvature of spacetime (but does not represent it), but by the use of a term that directly describes the curvature of spacetime- and it doesn't matter whether that curvature is introduced by acceleration, or by gravity. GR treats it the same way. Of course, because gravity is an approximately inverse-square force, radiating from a mass, the form of the "field" of gravity is different from the "form" of the "field" of acceleration, but that is not discernible at the local level. Locally, they are indistinguishable in the equations of GR.

Einstein tried and failed to use the SR term that describes an accelerated frame to describe the curvature of spacetime. It was only when he realized that gravity and acceleration were equivalent, and that the term he was adding to SR was the wrong way 'round, and that this directly implied that he needed a term not to describe how the frame varied, which he had but which did not work, but how the spacetime it was embedded in varied, that he developed GR. And GR worked. And the equivalence principle is the basis of GR.

GR defines inertial reference frames in that way. It differs in this definition from special relativity and classical mechanics. In GR, frame (1) is inertial, in SR, it is accelerated. The predictions of the two theories differ about how (1) and (2) would see each other. If this is what you're saying, then we agree. -

Not really. Actually, both GR and SR define inertial frames in the same way: in an inertial frame, local inertial objects move in straight lines. SR defines this only locally; GR defines a globally inertial frame, from which all inertial frames appear inertial. SR and GR both define frame 1 as inertial, because, in SR, local inertial objects move in straight lines in that frame, because they are within the same gravity field (otherwise they would not be local); in GR, the curvature of spacetime is an attribute of spacetime, not of the frame, and the movement of objects in spacetime of the same curvature will be in straight lines as observed from that frame, whereas the correction that will make inertial objects in spacetime of different curvature move in lines that do not appear straight, but really are, will be applied not to the frame, but to spacetime itself. GR, to repeat myself, describes curved spacetime, and how to correct for it; SR describes frames whose curvature changes in different locations, and how to correct for that. You also state that the predictions of SR and GR would be different for these situations; that is also incorrect. Correctly applied, the metric tensor would vary over the expanse of the frame, and this would give the correct prediction; however, it would be no more a description of gravity than Newton's universal gravitation, merely a statement that there was a twist in the frame. GR, on the other hand, because it describes the curvature of spacetime, directly describes the nature of gravity, and precisely describes how it varies from a true inverse-square-law force, and how it varies as a spherical field from the planar "field" of acceleration.

Also note that in GR, frame (1) is inertial only if it accelerates with respect to (2) because of being in a gravitational field. If it were accelerating because of any other kind of field, it wouldn't be inertial. This again demonstrates that it's the explanation of gravity that makes GR distinct from SR, not some special handling of acceleration per se. Thus my disagreement above.

And yet again you completely miss the point. The definition of an inertial frame is made in SR, not GR- GR inherits it from SR. The difference is, in SR, it's a locally inertial frame, whereas in GR, it's a globally inertial frame.

As far as whether gravity is a force, by your definition, the only force that "really is" a force is EM. Neither the weak nor the color forces fit your definition. GR provides field equations for gravity. Now I don't know about you, but where I learned this stuff, field equations describe the action of a force. You know, like Maxwell's equations, which are the field equations of EM. Or like Einstein's equations, which are the ten equations which describe the action of the force of gravity. I figure, if it waddles like a duck, quacks like a duck, and flies like a duck, it's probably a duck. You might call it a merganser. Fine, whatever. It's a duck.

I would say that an accelerating object is an origin of an accelerated reference frame which SR deals with just fine: see here. Can you clarify?

I don't accept your reference, and I've already said why. It's a very non-traditional way of teaching and understanding the subject, and I don't see that it makes it any easier. It certainly doesn't make it easier from the point of view of understanding how Einstein came up with it, and IMO it doesn't make it clear what the difference is between classical mechanics and spacetime mechanics; this is an important distinction today because quantum mechanics uses SR, not GR. GR is not incorporated in quantum mechanics. Go look it up.

With the reservations that I inserted in bolded text in square brackets, I agree. But again I fail to see the relevance of this to my words. I don't dispute the equivalence principle; I was highlighting the difference between physical and coordinate acceleration, in order to subsequently show the error in misapplying the equivalence principle to the latter, which commonly happens and which I have seen on this forum. Could it be that you misunderstood this point of mine as saying that there is something wrong with the equivalence principle?

Well, you've been talking about the equivalence principle like it had to do with accelerated objects pretty much since you started posting in this thread. And that's wrong; it has nothing to do with accelerated objects. It has to do with accelerated frames, and it is the primary differentiating postulate between SR and GR. The equivalence principle, plus the postulates of SR and SR itself, are the foundation of GR. The equivalence principle is not part of SR.

Look, this is boring. I keep saying the same thing, and you keep saying the same thing. Try again, but this time try to put it in terms everyone else will understand. In my opinion, while we've basically been patting ourselves on the backs about how smart we are, everyone else is getting very little from it. That's not what I'm interested in here.

 

[NobbyNobbs]

Well, aside from Zig's caveat that the direction of the center isn't constant, more or less yes. Where did you get the idea I said it wasn't, considering I was using it as an example of precisely that?

Well, you said acceleration need not be constant to be acceleration, and then cited rotational motion as an example. And I was going on the (wrong) idea that rotational motion is constant acceleration. I forgot about the fact that acceleration is a vector. Mea culpa.

 

[69dodge]

As far as whether gravity is a force, by your definition, the only force that "really is" a force is EM. Neither the weak nor the color forces fit your definition. GR provides field equations for gravity. Now I don't know about you, but where I learned this stuff, field equations describe the action of a force. You know, like Maxwell's equations, which are the field equations of EM. Or like Einstein's equations, which are the ten equations which describe the action of the force of gravity. I figure, if it waddles like a duck, quacks like a duck, and flies like a duck, it's probably a duck. You might call it a merganser. Fine, whatever. It's a duck.

Where did you learn this stuff?