General relativity and accelerated frames

[RiccardoVen]

Hi,
I'm going deeper in basics points on general relativity but, instead of swimming directly between the differential geometry, I'm trying to base my knowledge on strong physics bases first. I'm studying both on Wheeler's stuff ( I'm collecting almost all his books ), directly on Einstein's papers and on Thorne's books as well.

The most 2 important things I think to have grasped are:

1) "in a gravitational fields ( of small spatial extension ) things behave as they do in space free of gravity". After having thought a while, it seems soooo simple but soooo strong as principle. If the frame is sufficient small ( i.e. you can neglect tidal forced ) particles free falling within this frame cannot notice any gravity effect on them, and so gravity can be removed from the picture.
This also mean these particle can be seen as inertial and so we can apply special relativity principle to this situation ( turning free falling into free floating).

2) when we are standing still on Earth surface, we receive a repulsive force which is opposite to the gravitational pull in that point. This force is not intertial at all, since it's due to columbian and nuclear forces. This makes our "true" frame of reference as not inertial indeed. This is why we usually see bullets following a parabolic trajectory. It's not because the bullet is not inertial, is because WE are not in a inertial frame. If we'd remove the Earth below us, we would float together the bullet, seeing its trajectory being a geodesic in flat space ( i.e. a straight line ).

It seems something is missing in my picture to me and I will try to explain it. so far we are able to get rid of gravity in a sufficient small "local inertial free float frame" and apply special relativity rules. I know, and it makes a lot of sense to me, a rocket with engines ignited producing a 1g upward acceleration in a gravity free far space is not distinguishable from a man standing still on Earth, experiencing costant downward acceleration. This is another aspect of the equivalence principle ( it should be the strong version of it ).

But now my doubt and missing brick is asking me: "does this mean we can consider the physics on the rocket as there would be gravity? If yes, it means I would standing still experiencing a sort of non inertial force on my feet, exactly if I were on Earth."
So this would mean I could not get rid of gravity, since my frame is not intertial?
And finally "how I could apply special relativity to accelerated frames if I'm not able to get rid of gravity?".
This sounds also like asking "are these "special" local inertial free float frames still appliable to "pure" accelerated rockets and frames? ( or at least "how are they related?" ).

I mean: it's clear to me when the I'm free float I can get rid of gravity and I can apply special relativity, but I cannot see how general relativity applies to an accelerated frame.
It sounds like "we can replace the engine acceleration with a place holder gravitational field" ( and it sound good to me ) but I cannot see how we can apply special relativity to those frames which are not inertial...

I remember also from an old post of mine the accelerated frames are already encompassed somehow in special relativity, but no one is usually talking about that ( eventually please depict them a bit, if you are going to ). Since it's true, it seems the general relativity is really a theory of mainly gravity, since accelerated frames take already a good place in special relativity framework.

I know I've been a bit not enough precise in my sentences above, so I apologize in advance for that. These idea are so "simple" but also so difficult to put together coherently.
this is probably their strength.

Thanks

 

[mfb]

"does this mean we can consider the physics on the rocket as there would be gravity? If yes, it means I would standing still experiencing a sort of non inertial force on my feet, exactly if I were on Earth."

Right.

So this would mean I could not get rid of gravity, since my frame is not intertial?

You do not have gravity in the rocket. You feel the same non-inertial force, but due to a different reason (rocket engines instead of a solid, non-falling floor). You can use special relativity without any problems.

 

[RiccardoVen]

Right.
You do not have gravity in the rocket. You feel the same non-inertial force, but due to a different reason (rocket engines instead of a solid, non-falling floor). You can use special relativity without any problems.

Thanks for you quick answer. The main problem is that I cannot see how to use special relativity in that situation. The main point ( or at least the one I'm confusing ) is the rocket frame, which is accelerated, is not intertial at all ( exactly as the one with gravity , so how can I apply special relativity?

Of course, you can see easily from here, my main problem is how accelerated frames are encompassed in special relativity. People always claim about GR and how it extends special relativity encompassing accelerated frames, but it turns out to be not true to me, since it encompass "only" gravity and not all accelerated frames.


So the point is: I've 2 frames with a relative acceleration. How SR can model them?

 

[DrGreg]

Rindler coordinates cover a rocket rigidly accelerating with constant proper acceleration in a straight line.

Born coordinates cover an object rigidly rotating with constant angular velocity.

 

[WannabeNewton]

In SR, the rest frame of an accelerating observer can be idealized as a continuous sequence of inertial frames instantaneously at rest with respect to the observer (the spatial orientations of the instantaneous inertial frames must be fixed separately).

In GR the same holds but with local inertial frames; that's all there is to it.

 

[jtbell]

So the point is: I've 2 frames with a relative acceleration. How SR can model them?

By using calculus. In an analogous situation, in elementary classical mechanics, students learn that work is force times distance: W = FΔx. However, this works only if F is constant. If the force varies at different points, you have to use integral calculus:
$$W = \int_{x_1}^{x_2} {F(x)dx}$$

 

[RiccardoVen]

In SR, the rest frame of an accelerating observer can be idealized as a continuous sequence of inertial frames instantaneously at rest with respect to the observer (the spatial orientations of the instantaneous inertial frames must be fixed separately).

In GR the same holds but with local inertial frames; that's all there is to it.

Hi WannabeNewton, it's always a pleasure to meet your explanations. I was totally lacking about the part of sequence of inertial frames in SR ( I has a SR course in Politecnico of Tourin, but no mention was done about it ). Now you are depicting it it sounds really really like the sequence of local frames in GR.
So, again, it seems to me GR is about gravity and not about "generic" accelerated frames, since they are already encompassed in SR. Don't know why some textbooks report that "general" refers to accelerated Vs uniform frames. this now sounds really odd to me.

More than this, I cannot see why the rocket Vs Earth non-inertial frame example is so widely used in GR ( or at least again on some texts ). In that accelerated rocket frame the SR is still enough to describe things, so no GR is needed at all.
So why it's so used as main example? Probably because it's the Einstein version of equivalence principle ( Galileo weak one ) which states the equivalence between gravitational and inertial mass?

From a mere historical point of view, was Einstein going to encompass only gravity in order to move from SR to GR? I mean, there were no need to model accelerated frames since they were already there in SR, isn'it?

 

[RiccardoVen]

By using calculus. In an analogous situation, in elementary classical mechanics, students learn that work is force times distance: W = FΔx. However, this works only if F is constant. If the force varies at different points, you have to use integral calculus:
$$W = \int_{x_1}^{x_2} {F(x)dx}$$

OK this makes a lot sense as well, thanks

 

[RiccardoVen]

Rindler coordinates cover a rocket rigidly accelerating with constant proper acceleration in a straight line.

Born coordinates cover an object rigidly rotating with constant angular velocity.

Yes I've seen Rindler coordinates within a Susskind GR lecture ( when talking about observers falling or pasting near a black hole horizon ) but I dind't immediately realized they were good within SR as well. Thinking of it now, it was straighforward, and Susskind described accelerated wordlines as hyperbolas in Minkowsky space.

 

[WannabeNewton]

So, again, it seems to me GR is about gravity and not about "generic" accelerated frames, since they are already encompassed in SR.

Yes that is correct.

Don't know why some textbooks report that "general" refers to accelerated Vs uniform frames. this now sounds really odd to me.

They might be outdated books. Regardless, it's just flat out wrong so definitely don't waste your time with books that make such claims.

From a mere historical point of view, was Einstein going to encompass only gravity in order to move from SR to GR? I mean, there were no need to model accelerated frames since they were already there in SR, isn'it?

People were already working with rotating systems in SR well before Einstein formulated GR (c.f. Ehrenfest's paradox).

 

[adrian_m]

Yes that is correct.
...
People were already working with rotating systems in SR well before Einstein formulated GR (c.f. Ehrenfest's paradox).

A bit confused. How does this fit in with the Equivalence Principle? Does it mean that a non-gravitational accelerated frame (even if tidal/central) is not equivalent to a true gravitational field? (If not, in what way?)

 

[WannabeNewton]

Does it mean that a non-gravitational accelerated frame (even if tidal/central) is not equivalent to a true gravitational field? (If not, in what way?)

What is a "gravitational accelerated frame" or a "non-gravitational accelerated frame"? I suppose you mean instead accelerated frames that are in curved space-time as opposed to those in flat space-time? A rigidly rotating frame in flat space-time defined up to the light barrier can, through the equivalence principle, be considered instead as a stationary (but not static!) gravitational field i.e. one which possesses angular momentum. The equivalence principle simply says we can do physics using either paradigm: that of a rigidly rotating frame in free space or that of a rotating gravitational field. Of course the gravitational field so obtained will not correspond to a curved space-time geometry i.e. there will be no tidal forces.

 

[king vitamin]

A bit confused. How does this fit in with the Equivalence Principle? Does it mean that a non-gravitational accelerated frame (even if tidal/central) is not equivalent to a true gravitational field? (If not, in what way?)

An accelerated frame in Minkowski space still has a zero Reimann tensor everywhere, and thus must correspond to a trivial stress-energy tensor everywhere by Einstein's equations. Also, there can be no tidal forces (since only the Reimann tensor enters the geodesic deviation equation). Therefore, no physical (produced by sources) gravitational field corresponds to an accelerated Minkowski frame.

In a small enough region that there are no tidal forces (we are measuring zero Reimann tensor), we are equivalent to Minkowski space, possibly with acceleration. We can go to geodesic normal coordinates, where the metric is Minkowski; this is the equivalence principle.

In this way, "true gravity" is distinguished by the existence of tidal forces.

 

[adrian_m]

What is a "gravitational accelerated frame" or a "non-gravitational accelerated frame"? I suppose you mean instead accelerated frames that are in curved space-time as opposed to those in flat space-time? A rigidly rotating frame in flat space-time defined up to the light barrier can, through the equivalence principle, be considered instead as a stationary (but not static!) gravitational field i.e. one which possesses angular momentum. The equivalence principle simply says we can do physics using either paradigm: that of a rigidly rotating frame in free space or that of a rotating gravitational field. Of course the gravitational field so obtained will not correspond to a curved space-time geometry i.e. there will be no tidal forces.

An accelerated frame in Minkowski space still has a zero Reimann tensor everywhere, and thus must correspond to a trivial stress-energy tensor everywhere by Einstein's equations. Also, there can be no tidal forces (since only the Reimann tensor enters the geodesic deviation equation). Therefore, no physical (produced by sources) gravitational field corresponds to an accelerated Minkowski frame.

In a small enough region that there are no tidal forces (we are measuring zero Reimann tensor), we are equivalent to Minkowski space, possibly with acceleration. We can go to geodesic normal coordinates, where the metric is Minkowski; this is the equivalence principle.

In this way, "true gravity" is distinguished by the existence of tidal forces.

OK, I am trying to get a simple answer, and an example may help.

Situation (a): A pebble is moving in a circular orbit around Earth at a distance R from center of Earth, under Earth's gravitational influence.

Situation (b): An alien sitting in a rocket in deep space is whirling about a pebble attached to a longish string of length R in a circular trajectory.

The orbital velocity of both pebbles are the same (i.e. the accelerations on both are the same).

Would there be a difference between the two pebbles in any relativistic way - e.g. would their time dilations turn out to be different?

 

[WannabeNewton]

(A) and (B) are not equivalent scenarios. Yes their time dilations will turn out to be different. Situation (B) involves only kinematical time dilation due to orbital rotation in free space whereas (A) involves both kinematical and gravitational time dilation due to both orbital rotation and the intrinsic gravitational field of the Earth. If you apply the equivalence principle to (B) then you are simply reinterpreting the kinematical time dilation as a pseudo-gravitational time dilation but you are not introducing anything new.

 

[PeterDonis]

Situation (a): A pebble is moving in a circular orbit around Earth at a distance R from center of Earth, under Earth's gravitational influence.

Situation (b): An alien sitting in a rocket in deep space is whirling about a pebble attached to a longish string of length R in a circular trajectory.

The orbital velocity of both pebbles are the same

Assuming the alien adjusts his rate of whirling appropriately, yes. There's no requirement for him to do that, however; he could whirl the pebble at any angular velocity he likes, so long as its tangential velocity is less than the speed of light.

(i.e. the accelerations on both are the same).

No, they're not, even if the orbital velocities match. The pebble in (b) feels a force; the pebble in (a) does not, it's in free fall. This means the *proper* accelerations of the two pebbles are different, and in relativity, the most useful definition of acceleration is proper acceleration.

There is something called "coordinate acceleration", which is what you are thinking of here, and which is the same for both pebbles (assuming we choose coordinates appropriately). But in relativity, coordinate acceleration isn't very useful because it depends on how you choose coordinates; you could choose coordinates centered on the pebble and its coordinate acceleration would be zero. Proper acceleration, OTOH, is a direct observable; just attach an accelerometer to the pebble. It will read zero in (a) and nonzero in (b).

Would there be a difference between the two pebbles in any relativistic way

Definitely. See above.

would their time dilations turn out to be different?

Yes, because gravity is present in (a) and not in (b).

 

[adrian_m]

(A) and (B) are not equivalent scenarios. Yes their time dilations will turn out to be different...

...No, they're not, even if the orbital velocities match. The pebble in (b) feels a force; the pebble in (a) does not, it's in free fall...Yes, because gravity is present in (a) and not in (b).

Then let us consider the 'pebbles' to be 'satellites' with human beings inside. Would they be able to tell the difference in the two scenarios (without any reference to the outside)? I feel in the second case also the centrifugal force will match the pull on the string, so inside observers would float freely and not be able to tell the difference... may be I am wrong.

In other words... are we saying that the following WEP statement (Wikipedia) is wrong?
"The local effects of motion in a curved space (gravitation) are indistinguishable from those of an accelerated observer in flat space, without exception."

 

[WannabeNewton]

Then let us consider the 'pebbles' to be 'satellites' with human beings inside. Would they be able to tell the difference in the two scenarios (without any reference to the outside)?

Yes because in one case you have freely falling observers in curved space-time, thus observers who are locally inertial, and in the other you have accelerated observers in flat space-time. Accelerometers on board the satellites will allow observers in the satellites to tell apart the two scenarios.

In other words... are we saying that the following WEP statement (Wikipedia) is wrong?

Nothing stated thus far contradicts Wikipedia's statement.

 

[PeterDonis]

Then let us consider the 'pebbles' to be 'satellites' with human beings inside. Would they be able to tell the difference in the two scenarios (without any reference to the outside)?

Certainly. In case (a) the humans will be free-floating inside the satellites, just as astronauts do inside the Space Shuttle or the International Space Station.

In case (b), the humans will feel a force pressing them against the outer wall of the satellite; they won't be free-floating.

I feel in the second case also the centrifugal force will match the pull on the string

What "centrifugal force"? In the global frame you were implicitly using (the one where "acceleration" in the sense of coordinate acceleration is present), there is no such thing as "centrifugal force". In case (a), there is the "force" of gravity, and that is the only "force" present; but "force" is in scare-quotes because objects moving solely under the influence of gravity feel no force; they are in free fall, weightless.

In case (b), the only force present is the pull on the string; and since there are no scare-quotes this time, this force is actually felt; it is transmitted by the outer wall of the satellite, which pushes radially inward on the humans, who therefore feel as though they are being pressed against the wall.

If you want to analyze these scenarios using "centrifugal force", you have to pick a frame where that force is present, i.e., a frame rotating with the "satellite". In this frame, the satellite is at rest. In case (b), it's at rest because there are two forces on it which balance: the pull on the string, and centrifugal force. But the latter is a "force" with scare-quotes, because it's like gravity: objects moving solely under the influence of centrifugal force don't feel any force. So in this case the humans inside the satellite feel the force exerted by the string and transmitted to them by the wall of the satellite.

In case (a), in the rotating frame, the satellite is at rest because there are two "forces" on it which balance: gravity and centrifugal force. But here *both* "forces" are in scare-quotes; no force is actually felt. So both the satellite and the humans inside it are just freely floating.

If you're wondering how the satellite and the humans inside it can be following a curved path while feeling no force in case (a), see below.

In other words... are we saying that the following WEP statement (Wikipedia) is wrong?
"The local effects of motion in a curved space (gravitation) are indistinguishable from those of an accelerated observer in flat space, without exception."

No, you're just mistaken about what "the local effects of motion in a curved space (gravitation)" are. You, standing on the surface of the Earth, feel the same local effects as you would if you were in a rocket in deep space accelerating at 1 g. The local effect of feeling 1 g acceleration is the "local effect of motion in a curved space (gravitation)" that the Wikipedia statement is referring to. (Granted, the Wikipedia statement doesn't make that very clear; that's why Wikipedia is not a good authoritative source. )

In other words, the reason case (a) is different from case (b) is that spacetime in case (a) is curved, so the path followed by the satellite, even though it looks "circular" (in space, not spacetime), just like the circular path in case (b), is actually straight (but straight in spacetime, not space), so an object following it feels no force. In case (a), spacetime is flat, so the path that looks curved *is* curved (circular in space, but a helix in spacetime because the time dimension has to be included), and an object following it must be feeling a force that is pushing it out of its natural straight line path.

 

[adrian_m]

Yes because in one case you have freely falling observers in curved space-time, thus observers who are locally inertial, and in the other you have accelerated observers in flat space-time. Accelerometers on board the satellites will allow observers in the satellites to tell apart the two scenarios.



Nothing stated thus far contradicts Wikipedia's statement.

I think what you are saying is right, and I am getting a sense of it, but there's something I am still not quite able to understand...

First, how are the two scenarios different from Einstein's elevator example... and maybe I am missing the understanding of the word 'tidal' ... will help if you can explain how that might make a difference...

Secondly, you are saying 'freely falling observers in curved space-time' will have a different experience than 'accelerated observers in flat space-time'. That seems to contradict the statement 'local effects of motion in a curved space (gravitation) are indistinguishable from those of an accelerated observer in flat space, without exception', which you are saying is still correct. I don't get the difference.

Perhaps I am understanding something wrongly or missing a nuance... is it 'space' vs. 'space-time'...?

 

[PeterDonis]

you are saying 'freely falling observers in curved space-time' will have a different experience than 'accelerated observers in flat space-time'.

Yes, because being in free fall is not the same as feeling acceleration. That's true regardless of whether spacetime is flat or curved.

That seems to contradict the statement 'local effects of motion in a curved space (gravitation) are indistinguishable from those of an accelerated observer in flat space, without exception', which you are saying is still correct. I don't get the difference.

Re-read the last part of my previous post. The "local effects of motion" are the feeling of being accelerated at 1 g, for example when you're standing on the surface of the Earth. That's what's indistinguishable from being accelerated at 1 g in, say, a rocket. But the equivalence principle certainly doesn't say that either of those states of motion are indistinguishable from being in free fall. That would be silly: it would be saying that feeling weight is indistinguishable from being weightless.

 

[adrian_m]

Yes, because being in free fall is not the same as feeling acceleration. That's true regardless of whether spacetime is flat or curved.



Re-read the last part of my previous post. The "local effects of motion" are the feeling of being accelerated at 1 g, for example when you're standing on the surface of the Earth. That's what's indistinguishable from being accelerated at 1 g in, say, a rocket. But the equivalence principle certainly doesn't say that either of those states of motion are indistinguishable from being in free fall. That would be silly: it would be saying that feeling weight is indistinguishable from being weightless.

That helps. I think I mostly understand. Will have to think further about it...

 

[adrian_m]

Thanks PD, WBN. I think you have clarified the situation.

Now let me ask you a more "thorny" question. I am not sure the question is appropriate, and if it is whether you have the answer...

Replace the second scenario of 'pebble on string' to a scenario where there is a central negative charge (replacing the string) and a 'pebble' with a positive charge, such that the acceleration at distance R is the same as the Earth-pebble scenario. The masses (and gravitational effects) of the negative and positive charged bodies are assumed negligible.

What will happen in this case? Will the scenario be more like the 'Earth-pebble' situation or the 'string-pebble' situation from a relativistic (say time dilation) perspective?

 

[WannabeNewton]

If the only force present is electromagnetic then that setup isn't even possible classically. The positive charge emits dipole radiation so it cannot self-sustain a circular orbit.

 

[RiccardoVen]

I had realized the rocket Vs Earth non intertial frame is mainly useful for 2 things:

1) it can be viewed as a sort of Einstein equivalence principle, which states ( as Galileo's weak one ) the equivalence between inertial and gravitational mass

2) it's useful to have a quick and intuitive grasp about light bending within a gravitational field.

Thanks wannabeNewton for your support about "pure" gravity theory. I know old books are probably too old fashioned. Nevertheless I've just bought the Hartle's one you strongly suggested in a a previous post of yours.

 

[bahamagreen]

I know, and it makes a lot of sense to me, a rocket with engines ignited producing a 1g upward acceleration in a gravity free far space is not distinguishable from a man standing still on Earth, experiencing costant downward acceleration.

Standing on the Earth you experience an upward acceleration.

I'm wondering if "local" includes a limiting stipulation on time duration.

Comparing the situations of the accelerating rocket and the gravitating Earth, one can distinguish them with enough time to detect the difference, even though the "local" stipulation seems to be spatially satisfied by observing a small region...

Assuming both the rocket and the Earth have an evacuated column through which something may fall, a spherical arrangement of marbles falling through the rocket will continue to remain spherical (but may come slightly closer together from mutual gravitation among the marbles), but the same sphere of marbles dropped above the Earth's surface will additionally distort. Two effects will combine to distort the sphere.

One effect is that each marble will be falling toward the center of the Earth, so their paths are not parallel, which will compress the sphere of marbles toward its center line path as it falls.

The other effect is that the gradient acts to make the bottom marbles fall faster than the top ones, so as to stretch the sphere along the longitudinal axis in the direction of its fall.

The combined effect is that the sphere takes on the distorted shape somewhat of an egg with its pointy end pointing in the direction of fall.

In both cases, the rocket and the Earth's gravity, the sphere of marbles seems spatially "local" in that the range of space defining the sphere is quite small, but it takes some time for the distortion in the later case to appear measurable. Wondering how much time is allowed for this kind of distinction to emerge before the time period is considered "non-local"...?

 

[RiccardoVen]

I'm wondering if "local" includes a limiting stipulation on time duration.

As far as I know, "local" means limited in a spacetime region, so not only in space but in time as well, of course. This means you should both guarantee your frame is limited in space extensions, i.e. its "average" dimension is less than gravitational spatial gradient, and to be sure your experiment duration is limited as well, in order to avoid seeing tidal forces.

[EDIT]: there are a couple of exercices in Taylor & Wheeler SR about it, if you are interested in:
http://www.eftaylor.com/pub/spacetime/STP1stEdExercP60to80.pdf, ex 32, Size of a inertial frame.


But I'm not sure if I've fit your question completely, sorry in case.

 

[WannabeNewton]

Wondering how much time is allowed for this kind of distinction to emerge before the time period is considered "non-local"...?

Just to add on to Riccardo, this depends entirely on the characteristic curvature scales of the space-time set by the metric. It should be noted however that while one can use freely falling frames instantaneously at rest at an event with respect to any observer whatsoever, one can also construct a freely falling rest frame specifically for a freely falling observer that remains centered on said observer's world-line indefinitely in the proper time of the observer's clock. In this frame the gravitational field will vanish all along the world-line of the observer but will be non-vanishing off of the world-line to orders comparable with the characteristic curvature scales in the neighborhood of the world-line (c.f. exercise 13.13 in MTW). With regards to your comment about a falling sphere eventually being deformed into an ellipsoid, the characteristic time scales set by the tidal forces for geodesic congruences don't matter for the above because we are only considering a single observer idealized as a point, described by a single time-like curve in space-time. The story for extended bodies (time-like congruences) is different and more complicated, not the least of which is due to issues such as Born rigidity and the fact that initially freely falling spinning bodies deviate from geodesic motion due to gravitational tidal torques.

 

[PeterDonis]

Will the scenario be more like the 'Earth-pebble' situation or the 'string-pebble' situation from a relativistic (say time dilation) perspective?

WBN is correct that, classically, the orbiting charge will radiate EM waves, so the orbit won't be stable. But if we restrict attention to a time scale much shorter than the time scale of energy loss by radiation, it will be like the string-pebble situation; the orbiting charge will feel acceleration, and will not be in free fall.

 

[RiccardoVen]

PD and WBN, I've read many many times all this topic, expecially the last part on pebbles and so on, and I have to say I've learned more Physics from those few rows that from hundred of textbookpages.
Thanks for that.

Nevertheless, this reading has raised few doubts which highlight a bit of confusion is still there ( which is good, since it means there's always plenty of room to improve my knowledge ).

The scenario I'm depicting is really "trivial" and simple, i.e.: let's consider a little ball which is dropping towards the Earth undergoing simply gravitiy, i.e. in free float/fall. Just below it, there's an observer standing still on Earth. the falling speed is non relativistic.
I'd like to analyse things from within each frame of reference, so:

1) the little ball is in free fall into a curved spacetime, due to Earth gravity. It's then in free float, experiencing no force on it, so its frame of reference is inertial. Its trajectory is then straight in spacetime being a geodesic in curve ST.
My first point is how the little ball would interpret "seeing" the observer accelerating towards it.
My idea is the ball frame of reference must be kept local in order to avoid tidal forces, so it cannot encompass the observer lying thousand of meter below it: does this clash with my question about how the little ball would interpret the observer accelerating towards it?

2) the obverser is standing still on Earth, which is a not inertial frame of reference, since Earth is preventing the observer to keep its free falling pushing against its feet. Its wordline it's hence a straight vertical wordline with position fixed and time passing. Is this correct? I mean the ST is curved here, so it's correct to say this wordline is vertical?
"Looking" at the little ball, it would see it accelerating towards him. Since he can look around him, he can conclude he's on a gravity field on Earth ( and not in free-gravity space ) so is it correct to say he can say the little ball is experiencing just the gravity force as cause for its acceleration towards him?

I've learned from Susskind a ghwellsjr ( who gently helped me in the past with the art of ST diagrams ) that the best approach to relativity problems is always to draw a ST diagram of the problem. I've then realized this is quite a different situation than "simpler" cases I've faced of two inertial frame of reference pasting to each other.
The main problem I have here is:

1) how to properly depict the curved spacetime due to Earth gravity. And better: how the little ball geodesic will look like into this "curved" ST? Have I to use Swartzchild metric ofr that, i.e. using Penrose diagrams or so?

2) the problem really resemble me the known case of an observer ( little ball ) free falling into a black hole ( ok, no event horizon here, but we have a similar situation I think ).

3) how to draw both frame of reference, one inertial while the other not? Probably I have to draw one of the 2 geodesic as an hyperbola in ST, since accelerated...

As you can see, from a relative simple case, I've still a lot of problems on that. May you eventually help me with them, please?

Thanks as always for your patience and devotion to Physics.

 

[PeterDonis]

PD and WBN, I've read many many times all this topic, expecially the last part on pebbles and so on, and I have to say I've learned more Physics from those few rows that from hundred of textbookpages.
Thanks for that.

You're welcome!

1) the little ball is in free fall into a curved spacetime, due to Earth gravity. It's then in free float, experiencing no force on it, so its frame of reference is inertial. Its trajectory is then straight in spacetime being a geodesic in curve ST.

Ok so far except for one key point: the little ball's "frame of reference" is only inertial if we define it locally--that is, we have to pick some particular event on the ball's worldline and set up a local inertial frame centered on that event (i.e., that event is the origin of the frame, with t = x = y = z = 0), which can only cover a small piece of spacetime near that event--"near" in both space and time. In a curved spacetime, that's the best you can do; there is no way to set up a global inertial frame that covers all of spacetime.

My first point is how the little ball would interpret "seeing" the observer accelerating towards it.

As an observer accelerating towards it. Within the local inertial frame, physics works the same as it does in flat spacetime; and in flat spacetime, the little ball would just be floating freely and the object would accelerate past it.

My idea is the ball frame of reference must be kept local in order to avoid tidal forces, so it cannot encompass the observer lying thousand of meter below it

Well, yes, but the observer won't always be thousands of meters below. There will come a time when the little ball and the observer will be close enough together that the observer is within the local inertial frame of the little ball. Then things will work as I said above. If the observer is *not* close enough to be within the local inertial frame of the little ball, then of course you can't use that frame to describe the motion of the observer (for that portion of spacetime).

the ST is curved here, so it's correct to say this wordline is vertical?

In a non-inertial frame in which the observer is at rest, yes. But of course this is a non-inertial frame, so it doesn't work the same as an inertial frame.

"Looking" at the little ball, it would see it accelerating towards him. Since he can look around him, he can conclude he's on a gravity field on Earth ( and not in free-gravity space ) so is it correct to say he can say the little ball is experiencing just the gravity force as cause for its acceleration towards him?

Sure, but since he's using a non-inertial frame, this "gravity force" won't work like a "real" force; it is a "fictitious" force, like centrifugal force or coriolis force, that doesn't cause an object being influenced by it to feel any acceleration. A "real" force is always felt.

how to properly depict the curved spacetime due to Earth gravity.

You can't depict all of the properties of curved spacetime in a single diagram. Which diagram you use depends on which properties you want to focus on.

And better: how the little ball geodesic will look like into this "curved" ST?

It depends on which diagram you're using.

Have I to use Swartzchild metric ofr that, i.e. using Penrose diagrams or so?

For some purposes these work fine. (Note, btw, that the Schwarzschild metric is not a spacetime diagram; it's part of the underlying math that can be depicted in various different ways by different diagrams. A Penrose diagram is one such diagram, but not the only possible one.)

the problem really resemble me the known case of an observer ( little ball ) free falling into a black hole ( ok, no event horizon here, but we have a similar situation I think ).

Yes, these cases are similar for the portion of the black hole spacetime that is outside the horizon.

how to draw both frame of reference, one inertial while the other not?

You can't draw them both on a single diagram.

Probably I have to draw one of the 2 geodesic

There's only one geodesic, the worldline of the little ball. The worldline of the observer at rest on the Earth's surface is not a geodesic. The fact that it's vertical in a particular diagram using a non-inertial frame does not make it a geodesic; non-inertial frames don't work like inertial frames, and non-geodesic worldlines can appear straight in them.

as an hyperbola in ST, since accelerated...

It won't necessarily be a hyperbola; that will depend on which diagram you use.

 

[RiccardoVen]

Ok so far except for one key point: the little ball's "frame of reference" is only inertial if we define it locally--that is, we have to pick some particular event on the ball's worldline and set up a local inertial frame centered on that event (i.e., that event is the origin of the frame, with t = x = y = z = 0), which can only cover a small piece of spacetime near that event--"near" in both space and time. In a curved spacetime, that's the best you can do; there is no way to set up a global inertial frame that covers all of spacetime.

Good, this was more or less exactly my feeling when I wrote "My idea is the ball frame of reference must be kept local in order to avoid tidal forces, so it cannot encompass the observer lying thousand of meter below it". I was trying to give a sort of meaning to "local".

As an observer accelerating towards it. Within the local inertial frame, physics works the same as it does in flat spacetime; and in flat spacetime, the little ball would just be floating freely and the object would accelerate past it.

Well, yes, but the observer won't always be thousands of meters below. There will come a time when the little ball and the observer will be close enough together that the observer is within the local inertial frame of the little ball. Then things will work as I said above. If the observer is *not* close enough to be within the local inertial frame of the little ball, then of course you can't use that frame to describe the motion of the observer (for that portion of spacetime).

So, basically, I have to wait the observer entering the ball local frame to tell the ball is seeing the observer pasting towards it? In that case the ball, since doesn't feel any force, is interpreting the observer acceleration as due to a real force in its local frame ( which is in turn fictitous in observer non-intertial frame )?

In a non-inertial frame in which the observer is at rest, yes. But of course this is a on-inertial frame, so it doesn't work the same as an inertial frame.

Were there TYPO there? Did you really mean:

"In a inertial frame in which the observer is at rest, yes." ?

Sure, but since he's using a non-inertial frame, this "gravity force" won't work like a "real" force; it is a "fictitious" force, like centrifugal force or coriolis force, that doesn't cause an object being influenced by it to feel any acceleration. A "real" force is always felt.

OK, that's clear

You can't depict all of the properties of curved spacetime in a single diagram. Which diagram you use depends on which properties you want to focus on.

It depends on which diagram you're using.

How can I find some in deep analysis of different diagrams related to the properties yuo would depict? Or better, which is the one you'd use to represent, for example, the free falling ball?

For some purposes these work fine. (Note, btw, that the Schwarzschild metric is not a spacetime diagram; it's part of the underlying math that can be depicted in various different ways by different diagrams. A Penrose diagram is one such diagram, but not the only possible one.)

Yes, sorry I didn't mean Schwarzschild metric was a ST diagram. I meant it was a possible metric depicting a central gravity fields like the one created by Earth. A part Penrose diagrams, I've roughly seen Kruskal–Szekeres ones, within some Susskind's lectures, which have the advantage to work better on extended Schwarzschild solution.
Nevertheless, probably, this is a too advanced topic to me now, I mean I need stronger basis on basic GR physics, before going on with tensors and diagrams. Like WBN told in one post, I agree GR is more challanging in its physics dynamics, than in the math involved in solving it.

Yes, these cases are similar for the portion of the black hole spacetime that is outside the horizon.

OK, I guessed that. The most important thing to me, like Wheeler always said, is to guess a qualitative solution tring to use the instinct. And the when guess is correct, is always a good hint.

You can't draw them both on a single diagram.

So this is because I was trying to draw both an inertial and a non-inertial frame altogether? So, if both were inertial there would be no problem, but the non-intertial one is actually forbid it.

There's only one geodesic, the worldline of the little ball. The worldline of the observer at rest on the Earth's surface is not a geodesic. The fact that it's vertical in a particular diagram using a non-inertial frame does not make it a geodesic; non-inertial frames don't work like inertial frames, and non-geodesic worldlines can appear straight in them.

Rethinking about it, you are right for sure, of course. Only free float particles can actually following geodesics in ST, no matter if curved or not. So the ball is the only candidate here.
The observer lies in a non-inertial frame, so it actually feels some forces acting on it which make its wordline not a geodesic. Hope this is the correct picture.

It won't necessarily be a hyperbola; that will depend on which diagram you use.

Being the space curved, in which diagrams it would appear an hyperbola?

That said, I was thinking about this picture would change considering a free float ball within an accelerated rocket in free gravity space ( the usual Einstein's rocket used for equivalence principle analysis ). In that case the spacetime is flat, since there would be no gravity.
So, if I understood correcty your above reasoning, this time we could theoretically define a "global" frame of reference encompassing both the ball and an observer standing still on rocket floor. This is because ST is flat and there's no gravity, at all. And, refining it better, we could even use SR in this case, since ST is flat.
Am I right in stating this, please.

Thanks for your precious time you spend with my being a GR noob.

 

[PeterDonis]

So, basically, I have to wait the observer entering the ball local frame to tell the ball is seeing the observer pasting towards it?

To be able to use the rules of special relativity to tell that, yes. There are other ways to construct a frame centered on the ball's worldline that cover a larger region of spacetime than a local inertial frame, which can cover the observer even when the observer is not in the ball's local inertial frame; but you can't use the rules of special relativity for those other types of frames. ("Coordinate chart" is actually a better term than "frame" here.)

Also, don't forget that someone who is free-falling with the ball has ways of detecting the presence and relative motion of the observer on Earth that reach beyond a local inertial frame: measuring the round-trip travel time and Doppler shift of light signals, for instance.

In that case the ball, since doesn't feel any force, is interpreting the observer acceleration as due to a real force in its local frame

This is not a "interpretation". The presence of a "real" force, one which is felt, is not frame-dependent. The force on the observer is a real force in any frame, or even if you don't define a frame at all. The fact that the observer feels a force is a direct observable.

( which is in turn fictitous in observer non-intertial frame )?

No, it's not. See above.

Were there TYPO there? Did you really mean:

"In a inertial frame in which the observer is at rest, yes." ?

No, no typo. The observer is not at rest in any inertial frame. The observer feels a force, and no observer who feels a force can be at rest in any inertial frame. So if you have a frame in which the observer is at rest, it must be a non-inertial frame.

How can I find some in deep analysis of different diagrams related to the properties yuo would depict? Or better, which is the one you'd use to represent, for example, the free falling ball?

Look up Painleve coordinates, for example here:

http://en.wikipedia.org/wiki/Gullstrand–Painlevé_coordinates

So this is because I was trying to draw both an inertial and a non-inertial frame altogether? So, if both were inertial there would be no problem, but the non-intertial one is actually forbid it.

No, it's because whenever you draw a spacetime diagram, the diagram is representing things as they are in one particular frame (again, "coordinate chart" is a better term here than "frame"). There's no such thing as a diagram that represents two coordinate charts at once; any time you draw a diagram, you have to pick which single coordinate chart determines the axes of the diagram.

The observer lies in a non-inertial frame, so it actually feels some forces acting on it which make its wordline not a geodesic. Hope this is the correct picture.

Yes.

Being the space curved, in which diagrams it would appear an hyperbola?

I'm not sure; I don't think the ball's worldline would be a hyperbola in any of the diagrams we've mentioned.

I was thinking about this picture would change considering a free float ball within an accelerated rocket in free gravity space ( the usual Einstein's rocket used for equivalence principle analysis ). In that case the spacetime is flat, since there would be no gravity.
So, if I understood correcty your above reasoning, this time we could theoretically define a "global" frame of reference encompassing both the ball and an observer standing still on rocket floor.

Yes, since the ball is in free fall, in flat spacetime there is a global inertial frame in which the ball is at rest, and since this frame is global, it can be used to describe the entire worldline of the observer as well.

This is because ST is flat and there's no gravity, at all. And, refining it better, we could even use SR in this case, since ST is flat.
Am I right in stating this, please.

Yes.

 

[RiccardoVen]

This is not a "interpretation". The presence of a "real" force, one which is felt, is not frame-dependent. The force on the observer is a real force in any frame, or even if you don't define a frame at all. The fact that the observer feels a force is a direct observable.

No, it's not. See above.

I got your point. So the force the observer is feeling, i.e. the one under its feet preventing him to fall, is real and not fictitituos, and this is true in all frames or even without a frame.
But now I'm asking to myself: we know from equivalence principle we cannot distinquish the rocket Vs Earth situations, and the rocket observer in this case will feel a force under its feet like the Earthian observer.
My point is: since this situations are indistiguishable, I guess the rocket observer felt force is real as well, so it's not fictitious. But now it came in mind this rocket case sounds me no different from a bus accelerating driver. In this case we know the driver is at rest in his non-inertial frame, and must take into account a force which is slamming him back towards its seat.
This force, from what I can see, is a fictitious one, since the driver has to introduce it to explan why it's at rest while the bus is accelerating.
So, I cannot distinguish in mind the force felt by the rocket observer and the bus driver and so I'm asking ( wrongly ) if my conclusion using the equivalence principle above is right or not. Using it I would state that force is real, but using the driver analogy I would this would be fictitiuous.

I'm aware my reasoning is wrong in between, but I cannot see completely the pitfall.
Probably this is due to some fog still present about inertial vs. non-inertial frame of reference ( which is a "simple" classical mechanics point of view, of course ).

No, no typo. The observer is not at rest in any inertial frame. The observer feels a force, and no observer who feels a force can be at rest in any inertial frame. So if you have a frame in which the observer is at rest, it must be a non-inertial frame.

Thanks, this sentence is almost enlighting

 

[A.T.]

I guess the rocket observer felt force is real as well, so it's not fictitious.

The force from the rocket floor on the astronaut is real.

But now it came in mind this rocket case sounds me no different from a bus accelerating driver. In this case we know the driver is at rest in his non-inertial frame, and must take into account a force which is slamming him back towards its seat.

That is not the force that the driver feels. The driver feels the real force from the seat pushing him forward.

So, I cannot distinguish in mind the force felt by the rocket observer and the bus driver

rocket floor = bus seat
real upwards force from rocket floor = real forward force from bus seat
fictitious downwards force in the rocket frame = fictitious backwards force in the bus frame