General relativity and accelerated frames

[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

 

[RiccardoVen]

Thanks A.T.,
it's clear now, I was ( as expected ) confusing real Vs fictituous forces. Probably this is due to some confusions I inherited from classical mechanics about inertial/non-intertial.
But I just noticed ( from other threads ) like this:

https://www.physicsforums.com/showthread.php?t=437784

inertial/non-inertia frames are different between classical mechanics and GR ( and I wasn't aware of that 'til now ).
I will look further to the thread above to align to that. Nevertheless now it's clear what force is real and what is fictitous and my guess using equivalence principle was correct.
thanks

EDIT: I was thinking about the "twin" case of the rocket about the obsever standing still on Earth: I see here the situation is abit reversed, i.e. the real force is the one pushing the observer downwards the Earth center ( a.k.a. as gravity ) so the fictititous one is the one exerted by the Earth upwards. Is this correct?

 

[A.T.]

I was thinking about the "twin" case of the rocket about the obsever standing still on Earth: I see here the situation is abit reversed, i.e. the real force is the one pushing the observer downwards the Earth center ( a.k.a. as gravity ) so the fictititous one is the one exerted by the Earth upwards. Is this correct?

No, the other way around. In General Relativity gravity is fictitious. The ground force on his feet is real:

rocket floor = ground
real upwards force from rocket floor = real upwards force from ground
fictitious downwards force in the rocket frame = fictitious downwards force in the ground frame

Here is a visual comparison of classical gravity and GR, in the frame of the ground. Fictitious forces are not shown, instead the non-inertiality of the ground frame is expressed geometrically via the distorted coordinates.

https://www.youtube.com/watch?v=DdC0QN6f3G4

 

[RiccardoVen]

No, the other way around. In General Relativity gravity is fictitious. The ground force on his feet is real:

rocket floor = ground
real upwards force from rocket floor = real upwards force from ground
fictitious downwards force in the rocket frame = fictitious downwards force in the ground frame

OK, it's clear I've to clean my mind a little bit. So:

1) in CM the Earth is an inertial frame, and so we can state 3rd Newton's law holds here, i.e. both gravity and upward Earth reaction are real forces. We don't need to introduce any fictitious force here, since the frame is inertial and the observer can easily explain it's being at rest using Newton's law.

2) in GR the Earth is not an inertial frame and gravity is a fictitious force ( since gravity is due to curved ST and it's not a real force ), while the upward force from the Earth is real. Since the frame is not inertial here, the observer needs to introduce a downwards fictitious force to balance the real upwards from the Earth, in order to justify it's being at rest.

Is this correct, please?

Here is a visual comparison of classical gravity and GR, in the frame of the ground. Fictitious forces are not shown, instead the non-inertiality of the ground frame is expressed geometrically via the distorted coordinates.

Thanks a lot for the video, I will look to it carefully.

 

[A.T.]

OK, it's clear I've to clean my mind a little bit. So:

1) in CM the Earth is an inertial frame, and so we can state 3rd Newton's law holds here, i.e. both gravity and upward Earth reaction are real forces. We don't need to introduce any fictitious force here, since the frame is inertial and the observer can easily explain it's being at rest using Newton's law.

2) in GR the Earth is not an inertial frame and gravity is a fictitious force ( since gravity is due to curved ST and it's not a real force ), while the upward force from the Earth is real. Since the frame is not inertial here, the observer needs to introduce a downwards fictitious force to balance the real upwards from the Earth, in order to justify it's being at rest.

Is this correct, please?

Yes. Some points regarding 2):

- The center of the Earth is still inertial in GR, but the surface of the Earth is not inertial (even if it wasn't rotating). Inertial frames exist only locally in curved space time, so despite center and surface being at relative rest, one can be inertial while the other one isn't (again, even if Earth wasn't rotating).

- Instead of fictitious forces you can model non-inertial frames with distorted coordinates. When you combine those local distorted patches shown in the video above, you get the global picture of spacetime curvature around the massive sphere, as shown here:
http://www.adamtoons.de/physics/gravitation.swf

 

[RiccardoVen]

Yes. Some points regarding 2):
- The center of the Earth is still inertial in GR, but the surface of the Earth is not inertial (even if it wasn't rotating). Inertial frames exist only locally in curved space time, so despite center and surface being at relative rest, one can be inertial while the other one isn't (again, even if Earth wasn't rotating).

OK I got the point.

- Instead of fictitious forces you can model non-inertial frames with distorted coordinates. When you combine those local distorted patches shown in the video above, you get the global picture of spacetime curvature around the massive sphere, as shown here:
http://www.adamtoons.de/physics/gravitation.swf

Both the video and the last link are really enlightling, Thanx for those.
Anyhow, I've tried, as told you above, to read the post
https://www.physicsforums.com/showthread.php?t=437784

It seems that post reports a different point of view ( the bcrowell one, which is still interesting ), which leads to conclude centrifugal force are real forces. Probably I'd better to avoid, since uses a definition of inertial/non-intertial frames quite different than the one "normally" used in GR.
It really confused me :-) Nevertheless I will keep my above points ( confirmed by you ) as a valid explanation.

 

[A.T.]

It seems that post reports a different point of view ( the bcrowell one, which is still interesting ), which leads to conclude centrifugal force are real forces.

The centrifugal force in rotating frames of reference is not "real" in the sense of Newton's 3rd Law : It is not an interaction force, so it's not part of a Newton's 3rd Law force pair.

 

[RiccardoVen]

I agree with you, but from that thread it doesn't seem so, if you read it.
You can see bcrowell is defining inertial frames in Newton's framework in a different way
than we are used to. In that way Peter Donis desumed that definition lead to centrifugal forces
as real. this is really confusing to me ( and the whole thread is quite messy or too technical to me ).
Anyhow I'm not going to reopen a closed thread ( and I don't knwo if this is the right way to do it )

 

[WannabeNewton]

Gravity is in a sense fictitious but it isn't fictitious in the same sense that the centrifugal and Coriolis forces are. You have to be careful about adopting such a mindset. Can you see why they aren't of the same character?

Next, I'm not seeing a proper definition of "real" here. What is meant by a "real" force? Until a proper definition is given, the confusion you're having cannot itself be properly identified. The centrifugal force is certainly not real if one defines "real" forces to be those that prevail in local inertial frames. This doesn't mean the observer under acceleration does not feel anything-certainly they feel something and in their frame attributes it to inertial forces but one can describe what it is the person is feeling using forces present in an inertial frame.

 

[A.T.]

Next, I'm not seeing a proper definition of "real" here.

The frame invariant forces.

... they feel something and in their frame attributes it to inertial forces ...

Non-inertial frames can attribute coordinate accelerations to inertial forces, but not frame invariant effects like strains/stresses that you "feel". These frame invariant effects have the same cause in all frames, inertial and non-inertial.

 

[RiccardoVen]

WBN and AT, I'm getting really confused. Not only by your answer, but it seems really Physics somehow is depending onto who is answering to me. Someone tells me about "real" forces, someone else is telling they are not really there.
Then someone else is telling about a way to define non-inertial frame in GR. Someone else ( like the post https://www.physicsforums.com/showthread.php?t=437784 ) is telling me yet another definition of inertial frame in classical mechanics ( WNB, please read that post eventually. I was just quoting a sentence there telling some assumption are making centrifugal forces as real ).

I know there's another way to tell things, in which fictitious forces disappear in GR, and curved ST can take care of them. May be I'm mixing too many approaches, but this is due mainly to different answer I got from here.
I seems to me ( not just in this case ) we are often talking about "words" and "interpretation of words" instead of "contents". Again , this is not just the only case, I can see it everywhere.
This post, for example, is the most "argued" with so many different point of views:

https://www.physicsforums.com/showthread.php?t=437784

Really confusing for someone trying to get the Physical basis of GR.

And you right always right WNB: the most demanding and difficult part of GR is BY FAR the basics "rules" or "axioms" from which it starts ( different from Newton's one, of course ). Tensor analysis is really less demanding ( at least to me ) than understanding the basic rules. I mean, most books start talking about tensor calculus dedicating only few paragraphs about Equivalence principle or so.

I really think we are really talking about Physics here more than many textbooks, but somehow it's difficult
to understand which point of view is the best among you.

You know, I've studied both "officially" at Politecnico SR and attended some Physics course
in Tourin, but all of them deal and spent a lot of time with tensors and stuff like this.
Which was very interesting, since I'm quite math addict. But now I feel the need to reload
myself exploring things starting from a stronger Physical POV.

Nevertheless, thanks for the help.

 

[RiccardoVen]

Gravity is in a sense fictitious but it isn't fictitious in the same sense that the centrifugal and Coriolis forces are. You have to be careful about adopting such a mindset. Can you see why they aren't of the same character?

From what I can have understood, gravity is fictitious because it's due to the curved ST, and so it's a sort of "consequence" from this, without being a "real" force like Newton's gravity. I mean, geodesics are accelerating towards each other or so, depending on curvature. This acceleration is actually on observers, without they can feel force on them. That's why they have somehow to explain those forces as fictitious.
I see when a body is free floating, it doesn't feel any force in GR but with Newton it would experience the gravitational force.

OTOH, centrifugal and Coriolis are fictitious in a different ways, i.e. they are not due to property of ST ( you can have centrifugal force with flat ST as well ).

But I'm not sure if I stated it properly.

 

[PeterDonis]

I'm not seeing a proper definition of "real" here. What is meant by a "real" force?

Since I introduced that term, I should respond to this. A "real" force is a force that is felt, i.e., a force that corresponds to a nonzero proper acceleration. (A.T.'s definition of real forces as frame invariant is basically saying the same thing.) By this definition, gravity, centrifugal, and Coriolis forces are all not real.

This doesn't mean the observer under acceleration does not feel anything-certainly they feel something and in their frame attributes it to inertial forces

Not if they're using the above definition. Take the example of the bus driver feeling a force when the bus accelerates. As A.T. pointed out, the force he feels is *not* an "inertial force"; it's the perfectly ordinary contact force of the seat pushing him. Or, if you're inside a rocket whose engines are firing, you feel the force of the engines pushing on you. Similarly, if you're standing on the surface of the Earth, the force you feel is not "gravity"; it's the perfectly ordinary contact force of the Earth's surface pushing on you.

It's true that there's a difference between these cases: in the case of standing on the Earth, the surface pushes in different directions on people at different locations, and yet they can all remain at rest relative to each other. That is because spacetime around the Earth is curved, whereas in the case of the bus or the rocket, spacetime is (or can be idealized to be, since spacetime curvature is not required for the observed phenomena) flat. However, that doesn't make gravity a "real" force; it just means that curved spacetime is different from flat spacetime. We can equate curved spacetime with *tidal* gravity, but that's not the same as the "force of gravity" that, in Newtonian terms, "pulls" objects towards the center of the Earth.

 

[A.T.]

Someone tells me about "real" forces, someone else is telling they are not really there.

The terms "real" & "fictitious" are just names. They have nothing to do with "really being there", which is rather philosophy than physics. Personally I prefer to talk about "interaction" & "inertial" forces to avoid this confusion.

 

[A.T.]

From what I can have understood, gravity is fictitious because it's due to the curved ST, and so it's a sort of "consequence" from this, without being a "real" force like Newton's gravity.

In your thinking you should replace "is" with "can be modeled as". Gravity can be modeled as an interaction force (as Newton did) or it can be modeled as an inertial force (as Einstein did).

I mean, geodesics are accelerating towards each other or so, depending on curvature.

That is tidal gravity in non-uniform fields.

 

[RiccardoVen]

Since I introduced that term, I should respond to this. A "real" force is a force that is felt, i.e., a force that corresponds to a nonzero proper acceleration. (A.T.'s definition of real forces as frame invariant is basically saying the same thing.) By this definition, gravity, centrifugal, and Coriolis forces are all not real.

Thanks PD for having pointed out your definition. Since you are here, there's no better person could confirm my idea on it:

https://www.physicsforums.com/showthread.php?t=437784

I've mentioned several times this thread above, and you can notice I was talking about the bcrowell POV "against" other people. You seemed the only one who was able to understand and explain properly his definition.
In that thread, he was taking a slighlty different definition of intertial frame of reference, instead of talking about "real" force, i.e.:

In Newtonian mechanics, I would say that an inertial frame is one in which every force is caused by one material object acting on another material object

So, please correct me if I'm wrong, the kind of force encompassed in this definition really reminds to me your definition of "real" force. You were explain bcrowell POV like this:

I can't speak for bcrowell, but one way I could put the point I think he was trying to make is that, according to GR, gravity itself is kind of a "fictitious force"! That is, if an object is moving solely under the influence of what in Newtonian terms we call "gravity", that object will *feel* no force at all--it will be in free fall. GR uses this to define the notion of "local inertial frames"; they are the frames, at a given event, in which objects at rest are in free fall. A "real" force, according to GR, is a force that is actually *felt*, physically, by the object being subjected to it--in other words, it's a force that corresponds to a real, "proper" acceleration, one that can be measured with accelerometers, and which has an invariant geometric definition (the covariant derivative of the object's 4-velocity with respect to proper time is nonzero)

But the most important thing you have pointed out in that thread, and still a bit foggy to me, is:

Part of the reason bcrowell may have used the term "mind-blowing" for this is that, under the above definition, a force like centrifugal force might actually be considered "real", not fictitious!

This was as highligthing of

Exactly. The really mind-blowing thing that Einstein realized was that this distinction between fictitious and non-fictitious forces ultimately doesn't work

This sounds really really interesting. I'm not going to re-open that thread, but I think my doubt here fits exactly in the "feeling" of this current thread, in which we are talking about "real" or "fictitious".

May you explain better you exchange between you and brcrowell, please? Why that distinction is not working with Einstein's GR?

Thanks

 

[RiccardoVen]

The terms "real" & "fictitious" are just names. They have nothing to do with "really being there", which is rather philosophy than physics. Personally I prefer to talk about "interaction" & "inertial" forces to avoid this confusion.

It was clear to me PD was not talking about Philosophy, but Physics. I always try to take care to avoid "centered human being definition", like "the obverser sees the other frame". I know "see" and "real" are dangerous words ( at least in this forum ) until you define them.
Nevertheless words are words, and PD was correct in pointing out their meaning from his POV. As you noticed I'm Italian, and we are used to use "forze inerziali", i.e. "inertial forces" instead of "forze fittizie", i.e. "fictitious" one.
This is quite effective to me.

 

[RiccardoVen]

In your thinking you should replace "is" with "can be modeled as". Gravity can be modeled as an interaction force (as Newton did) or it can be modeled as an inertial force (as Einstein did).

I'm not among people thinking Physics is the reality. Of course GR and other theories are "just" models of reality. So "to be" is already in mind translated as "to be modeled as".

 

[RiccardoVen]

From what I can have understood, gravity is fictitious because it's due to the curved ST, and so it's a sort of "consequence" from this, without being a "real" force like Newton's gravity. I mean, geodesics are accelerating towards each other or so, depending on curvature. This acceleration is actually on observers, without they can feel force on them. That's why they have somehow to explain those forces as fictitious.
I see when a body is free floating, it doesn't feel any force in GR but with Newton it would experience the gravitational force.

OTOH, centrifugal and Coriolis are fictitious in a different ways, i.e. they are not due to property of ST ( you can have centrifugal force with flat ST as well ).

But I'm not sure if I stated it properly.

WBN, since you were posing that question, I'd like to hear from you about my answer. Is that properly, I mean, is your distinction similar as mine?
Thanks

 

[WannabeNewton]

I don't think that's quite right. Gravity is fictitious only in the sense that the Christoffel symbols can always be made to vanish at a given point or everywhere along a geodesic by transforming to a local inertial frame. We identify the Christoffel symbols with the gravitational field because the two are in fact equal in the Newtonian limit. Similarly the centrifugal and Coriolis forces in a rotating frame can also be transformed away in a local inertial frame. In fact a rigidly rotating frame is equivalent to a rotating gravitational field by the equivalence principle and indeed the centrifugal and Coriolis forces will show up as components of the Christoffel symbols in the rotating coordinates and this rotating gravitational field can be transformed away in a local inertial frame in exactly the same manner. This applies to both flat space-time and curved space-times.

See here: https://www.physicsforums.com/showthread.php?t=738936

The thing that sets apart gravity, that is, the thing that makes it not so fictitious in comparison to the Newtonian inertial forces, is that even in local inertial frames one can measure the gravitational tidal forces exerted on a swarm of surrounding particles and these forces will manifest exactly as the components of the Riemann tensor in that local inertial frame. So, to put it concisely, gravity still has a frame-independent existence through the non-vanishing of the Riemann tensor even in local inertial frames even if the gravitational field itself can be made to vanish in local inertial frames (including the rotating gravitational fields due to the centrifugal and Coriolis forces).

 

[RiccardoVen]

I don't think that's quite right. Gravity is fictitious only in the sense that the Christoffel symbols can always be made to vanish at a given point or everywhere along a geodesic by transforming to a local inertial frame.

This can also be read as "it's always possible to find a chart ( normal Gaussian coordinates, as far as I know ) in which the metric becomes "simple", i.e. flat. Is this correct?

We identify the Christoffel symbols with the gravitational field because the two are in fact equal in the Newtonian limit. Similarly the centrifugal and Coriolis forces in a rotating frame can also be transformed away in a local inertial frame. In fact a rigidly rotating frame is equivalent to a rotating gravitational field by the equivalence principle and indeed the centrifugal and Coriolis forces will show up as components of the Christoffel symbols in the rotating coordinates and this rotating gravitational field can be transformed away in a local inertial frame in exactly the same manner. This applies to both flat space-time and curved space-times.

See here: https://www.physicsforums.com/showthread.php?t=738936

OK this sounds quite interesting, since it's really math oriented. I will read it carefully, thanks.

EDIT: I've tried to read this thread but, to be honest, I think it's a bit too advanced to me for now. Is there a base explanation of this topic in some book, may be Hartle or Gravitation, please?

The thing that sets apart gravity, that is, the thing that makes it not so fictitious in comparison to the Newtonian inertial forces, is that even in local inertial frames one can measure the gravitational tidal forces exerted on a swarm of surrounding particles and these forces will manifest exactly as the components of the Riemann tensor in that local inertial frame.

OK, that's quite clear

So, to put it concisely, gravity still has a frame-independent existence through the non-vanishing of the Riemann tensor even in local inertial frames even if the gravitational field itself can be made to vanish in local inertial frames (including the rotating gravitational fields due to the centrifugal and Coriolis forces).

So, if the spacetime was flat, the Riemann tensor would vanish in every inertial frame, while if curved only in the normal Gaussian chart? Sorry if I give things the wrong names, eventually. You are (correctly ) the most rigorous in math and sometimes it's diffucult to express myself in the proper way.
Is correct, from a strict math point of view, to tell the local frame of reference can be called normal Gaussian charts as well?

 

[PAllen]

So, if the spacetime was flat, the Riemann tensor would vanish in every inertial frame, while if curved only in the normal Gaussian chart? Sorry if I give things the wrong names, eventually. You are (correctly ) the most rigorous in math and sometimes it's diffucult to express myself in the proper way.
Is correct, from a strict math point of view, to tell the local frame of reference can be called normal Gaussian charts as well?

If a tensor vanishes in one chart it vanishes in all. If gravity is present, no coordinates, even at one point, will make the Riemann tensor vanish.

 

[RiccardoVen]

If a tensor vanishes in one chart it vanishes in all. If gravity is present, no coordinates, even at one point, will make the Riemann tensor vanish.

Yes you are right, sorry, that's the key point of using tensors in SR and GR, i.e. guarateeing Physics equation are the same in all frames.

I was just trying to understand better WBN words:

So, to put it concisely, gravity still has a frame-independent existence through the non-vanishing of the Riemann tensor even in local inertial frames even if the gravitational field itself can be made to vanish in local inertial frames (including the rotating gravitational fields due to the centrifugal and Coriolis forces)

If I'm not wrong, if you take a local frame of reference enough limited in spacetime, you can neglect tidal effects there, and so you can get rid of gravity ( this is the key of equivalence principle ). So, there exist some special frames in which the metric tensor can be turned to flat. I was just saying this char is usually called "special normal coordinates".
I know Riemann tensor is vanishing when their components are constant, so its vanishing it's a necessary condition for turning it in a flat metric.

So actually I'm confused, since it's seems in this frame Riemann is vanishing even if the ST is curved. Also WBN words are confused to me, expecially the part:

So, to put it concisely, gravity still has a frame-independent existence through the non-vanishing of the Riemann tensor even in local inertial frames even if the gravitational field itself can be made to vanish in local inertial frames (including the rotating gravitational fields due to the centrifugal and Coriolis forces)

 

[Nugatory]

So, there exist some special frames in which the metric tensor can be turned to flat.

At a given point in the spacetime... But only that one point. In the neighborhood of that point, "flat" is an approximation that can be made arbitrarily good by choosing an arbitrarily small neighborhood.

 

[WannabeNewton]

If I'm not wrong, if you take a local frame of reference enough limited in spacetime, you can neglect tidal effects there, and so you can get rid of gravity ( this is the key of equivalence principle ). So, there exist some special frames in which the metric tensor can be turned to flat.

No this is incorrect. You cannot turn a curved metric into a flat one, ever, not even at a single event. Putting the metric into Minkowski form at a given event, and making the Christoffel symbols vanish there does not mean the metric is flat. The Riemann tensor will still be non-zero there. A handful of GR textbooks will use the terminology "locally flat coordinates" but this is a sacrilegious and downright incorrect term. The equivalence principle only says the gravitational field (= Christoffel symbols) will vanish at the chosen event and that the metric can be put into Minkowski form there. It does not say that the Riemann tensor (= tidal forces) will vanish because they certainly won't-they are a physical characterization of the curvature of space-time. The stipulation of neglecting tidal forces so as to apply the equivalence principle (i.e. the stipulation of a local inertial frame) simply means the Christoffel symbols vanish only at a single event as opposed to on an extended neighborhood of that event and the metric is only Minkowski at this single event. If we start moving away from this event and into an extended neighborhood of it then we will find non-zero terms in the Christoffel symbols and non-trivial terms in the metric that scale with the curvature (i.e. the tidal forces).

As for Gaussian normal coordinates, there is a slight distinction between coordinates and frames but for our purposes here yes a Gaussian normal system is the same thing as a freely falling frame (the local inertial frame has the added property that it is non-rotating which requires the notion of Fermi-Walker transport). One can find explicitly what the Riemann tensor is in this freely falling frame both at its origin (where the metric is Minkowski and the Christoffel symbols = gravitational field vanish) and near the origin up to second order in the characteristic curvature set by the Riemann tensor. See exercise 13.13 in MTW.

EDIT: I've tried to read this thread but, to be honest, I think it's a bit too advanced to me for now. Is there a base explanation of this topic in some book, may be Hartle or Gravitation, please?

I can try to see if MTW has anything on that topic. Chapter 13 would be a good place to start.

In the meantime you can read the following papers:

http://articles.adsabs.harvard.edu//full/1990MNRAS.245..733A/0000735.000.html
http://adsabs.harvard.edu/full/1990MNRAS.245..720A

 

[RiccardoVen]

At a given point in the spacetime... But only that one point. In the neighborhood of that point, "flat" is an approximation that can be made arbitrarily good by choosing an arbitrarily small neighborhood.

Yes that's exactly what I wrote above. But reading from PAllen words:

If a tensor vanishes in one chart it vanishes in all. If gravity is present, no coordinates, even at one point, will make the Riemann tensor vanish.

it seems from here "even at a point" it's not possible to find such a chart. where am I wrong, please?