General relativity and accelerated frames

[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?

 

[Nugatory]

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

"flat at a point" is a statement about the metric tensor at that point, not the Riemann tensor.
I can find coordinates in which the components of the metric tensor at a point are diag(-1,1,1,1) but Riemann still won't vanish.

 

[RiccardoVen]

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.

OK this clarifies a bit my doubt. It's really rally interesting the distinction you are making betweem gravity = Christoffel symbols and Riemann = tidal forces.
I've not read something like this in any book, and I think it's really clarifying things, at least a bit.
I've to work it out properly, since I think this view can really help me.

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

OK, I will look at it as well.

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

Thanks for these one as well

 

[RiccardoVen]

"flat at a point" is a statement about the metric tensor at that point, not the Riemann tensor.
I can find coordinates in which the components of the metric tensor at a point are diag(-1,1,1,1) but Riemann still won't vanish.

Yes that's exactly what WBN pointed out in the above post. This was leading confusion to me.
Not it's ok, thanks for your help.