[Pmb (OP)]

Relax, man.

I am relaxed. This isn't a question of me not being relaxed either. Also please try to understand that I'm not saying these things to be a jerk or to come off as being arrogant. I'm truly and honestly trying to help you get to learn everything it is that you're seeking to learn and I know as fact and from experience that all of our answers are found by learning general relativity. So you can imagine my confusion at your absolute refusal to learn the theory which was created to answer just the kinds of questions that you're seeking answers to. 

I haven't questioned the validity of GR, nor would I bother to do so. I'm concentrating on the question of whether gravitation is an inertial force, that is, one that exists between two bodies.

Inertial forces are not defined as a force between two bodies. An inertial force is defined here http://home.comcast.net/~peter.m.brown/gr/inertial_force.htm 

An inertial force is a force which is proportional to the mass of the object the force is acting on just like the Coriolis and censtrifugal force. Those have nothing to do with a source body. And my point here is that all of this is explained in any decent GR text. After all that's exactly what the theory is explaining to you. So your choosing not to go directly to the source of where your answers are designed to be found in. Why is that?

[Mike_Fontenot]

My question for you is  - Do you believe that the gravitational force cannot be thought of as a "real" force and must therefore be called, at best, a pseudo force? Or to phrase it another way - How many of you believe that if a particle is accelerating under the action of a field for which the 4-acceleration on the particle is zero that any attempt to define a "force" on the particle must imply that it should be thought of/defined as a pseudo-force?

As to your first phrasing of the question, I think real gravitational fields are those that are caused by specific, identifiable masses (and not by "the masses at infinity", as in Mach's explanation of inertia).  I think the spatially-uniform gravitational field that one gets, when the equivalence principle is used to explain the perspective of a traveler who is being accelerated by a rocket (and who is far from any large masses), is a FICTITIOUS gravitational field.

As to your second phrasing of the question, it's been too long since I've studied GR, and 4-acceleration in particular, for me to comment on that specific question. But I will just say that I think the two very different concepts of acceleration in GR vs SR cause a HUGE amount of misunderstanding in many discussions I've witnessed. For example, the notion, that someone at rest on the surface of a (non-rotating and non-orbiting) earth, "is accelerating", and that someone who is free-falling into a deep hole in the earth "is not accelerating", is quite different from the notion that masses accelerate when there is a net force on them, and otherwise they don't accelerate. I don't remember if that difference is purely due to the difference between the concepts of spatial acceleration (3-acceleration?) vs 4-acceleration, or not.

Also, I certainly don't take seriously the notion that, if I were floating in empty space (far from any significant masses), and if I decided to turn on my rocket at some thrust level and direction, that a spatially-uniform gravitational field would suddenly come into existence, which exactly counteracts my rocket thrust, preventing me from accelerating, and causing all the other masses (anywhere in the universe) to accelerate. Such a notion has, in my opinion, little or no value in explaining the twin "paradox" ... its value lay in pointing the way to GR, by constraining the results that GR must give, and also as a verification of the resulting GR theory.

[Pmb (OP)]

As to your first phrasing of the question, I think real gravitational fields are those that are caused by specific, identifiable masses (and not by "the masses at infinity", as in Mach's explanation of inertia). etc

Mike - You already answered this question in my forum so you don't need to repeat it here unless you're doing so for the benefit of sharing your point of views with the other members here? If so then it’d make sense to repeat yourself here.  Otherwise it's redundant.

[Pmb (OP)]

It's been too long since I've studied GR, and 4-acceleration in particular, for me to comment on your specific question.  But I will just say that I think the two very different concepts of acceleration in GR vs SR cause a HUGE amount of misunderstanding in many discussions I've witnessed.

What are these misunderstandings that you’re referring to?

   For example, the notion, that someone at rest on the surface of a (non-rotating and non-orbiting) earth, "is accelerating", and that someone who is free-falling into a deep hole in the earth "is not accelerating", is quite different from the notion that masses accelerate when there is a net force on them, and otherwise they don't accelerate.

That’s not a quite precise way to describe what’s going on. It certainly isn’t how Einstein perceived the situation. Here’s how he phrased it. From The Foundation of the General Theory of Relativity by Albert Einstein, Annalen der Physik, 49, (1916) which can be found at http://hem.bredband.net/b153434/Works/Einstein.htm

Let K be a Galilean system of reference, i.e. a system relatively to which (at least in the four-dimensional region under consideration) a mass, sufficiently distant from other masses, is moving with uniform motion in a straight line. Let K' be a second system of reference which is moving relatively to K in uniformly accelerated translation. Then, relatively to K', a mass sufficiently distant from other masses would have an accelerated motion such that its acceleration and direction of acceleration are independent of the material composition and physical state of the mass.
Does this permit an observer at rest relatively to K' to infer that he is on a "really" accelerated system of reference? The answer is in the negative; for the above-mentioned relation of freely movable masses to K' may be interpreted equally well in the following way. The system of reference K' is unaccelerated, but the space-time territory in question is under the sway of a gravitational field, which generates the accelerated motion of the bodies relatively to K'.

So in Einstein’s viewpoint it’s valid to refer to a body sitting on the surface of the earth to be considered to be at rest. It’s only accelerating if you state with what it’s accelerating with respect to. If you’re in a free-fall frame then the body at rest on the surface of the earth is accelerating with respect to your free-fall frame.

Also, I certainly don't take seriously the notion that, if I were floating in empty space (far from any significant masses), and if I decided to turn on my rocket at some thrust level and direction, that a spatially-uniform gravitational field would suddenly come into existence, which exactly counteracts my rocket thrust, preventing me from accelerating, and causing all the other masses (anywhere in the universe) to accelerate.

I believe that the reason you reject what general relativity says on this point is because you’re trying to use the Newtonian theory of gravity to interpret what’s being described with general relativity. You’re stuck in the Newtonian frame of mind where all gravitational fields have a definite source to that is close by and used to calculate what’s happening to a body in the field. You haven’t gone over to general relativity as a result of being stuck with the Newtonian thought process. If you don’t study the reasons for this radically different viewpoint how can you justify rejecting it? You’re stuck with old definitions and interpretations and haven’t made the transition to GR.

Do you actually know why Einstein said what he said? What was on his mind? What was he thinking? What was his justification for it?

  Such a notion has, in my opinion, little or no value in explaining the twin "paradox" ... its value lay in pointing the way to GR, by constraining the results that GR must give, and also as a verification of the resulting GR theory.

You already stated your opinion on this in my forum so there’s no reason to put it here too. Especially out of context like this since nobody here knows what you’re talking about.

[lean bean]

Pete, You may remember this pdf found at the Mit site
Don't know if this has relevance on this thread, but...

From pdf here  http://ocw.mit.edu/courses/physics/8-224-exploring-black-holes-general-relativity-astrophysics-spring-2003/assignments/
Pick pdf “How Gravitational Forces Arise from Curvature” Page 2.

It say’s 

1. Introduction: Extremal Aging and the Equivalence Principle
These notes supplement Chapter 3 of EBH (Exploring Black Holes by Taylor and Wheeler).
They elaborate on the discussion of the Principle of Extremal Aging and the motion of massive bodies in curved spacetime.

Do you agree with the idea that the acceleration of a rock near earth, is just the outward appearance of the rock following a path of maximum aging on its wristwatch (rock’s proper time) in an altered spacetime near mass. Maximum aging defining the geodesic path of rock.

However, according to general relativity it still works out that the correct path is the one that maximizes proper time!
It seems astonishing that a result from special relativity carries over directly to general relativity without modification. The key is that, in the paradigm of general relativity, free-fall motion arises not from acceleration but from the effects of spacetime curvature. As we will see, the appearance of acceleration arises naturally from extremal paths in a curved spacetime.
We say “appearance of acceleration" because ordinary acceleration depends on the motion of one's reference frame. In an inertial reference frame in Newtonian gravity, a body moves at a constant velocity if no forces act on it. In Newtonian theory, an inertial reference frame can be extended over all of spacetime. But we have already argued in the first set of notes that there are no global inertial reference frames in curved spacetime. Consequently the notion of acceleration is ambiguous! Acceleration depends on frame, and if there are no preferred frames, there is no preferred concept of acceleration.

My bold.

[Pmb (OP)]

Pete, You may remember this pdf found at the Mit site
Don't know if this has relevance on this thread, but...

Yes. I remember it very well since the premise is wrong, i.e. curvature is not a necessary condition for the presence of a gravitational force on a body in a gravitational field.

Do you agree with the idea that the acceleration of a rock near earth, is just the outward appearance of the rock following a path of maximum aging on its wristwatch (rock’s proper time) in an altered spacetime near mass. Maximum aging defining the geodesic path of rock.

Only if the rock is in free fall is that true. But yes, of course I agree.

[lean bean]

Yes. I remember it very well since the premise is wrong, i.e. curvature is not a necessary condition for the presence of a gravitational force on a body in a gravitational field.

Are you saying it is not possible to transform away the gravitational field even locally ?
Do you need to transform away the field completely for it to not effect the test point/particle?


Is it this part you don’t agree with…
From link given in my last post.

What then is the path of a freely-falling body in the presence of gravity? According to Newtonian physics, the answer is given by solving ~F = m~a = md2~x=dt2 as a differential equation for ~x(t).
Gravity causes acceleration, and one might expect therefore that the maximal-aging result breaks down. A body obviously accelerates in a gravitational field, and so an unaccelerated path cannot be the correct one. However, according to general relativity it still works out that the correct path is the one that maximizes proper time!

equation symbols don’t copy correctly.

Besides spacetime curvature / space curvature and time dilation/gravitational redshift, what other evidence shows the presence of a field? Also how does this other evidence of field effect the Equivalence Principle if at all?

Have you let the MIT people know that their premise is wrong?
Given MIT are ‘teaching’ this, you would be doing them a service informing them.  Does E. Taylor still ‘reside’ at MIT?

Only if the rock is in free fall is that true. But yes, of course I agree.

Yes, the rock is un-powered and following a geodesic, defined by maximum aging on the rock’s watch and the spacetime metric, and so giving the appearance of acceleration when moving through an altered spacetime near mass. Why is the spacetime altered near mass? Don’t know.
What is the mechanism of an attracting ‘force’? Don’t know. Both being convenient models of observations.

From “Exploring Black Holes” Pick chapter titled “Diving” PDF page 2
Here…http://exploringblackholes.com/

Newton says a “force of gravity" leads to the parabolic trajectory. But  Einstein declares that Newton's “force of gravity" does not exist. Instead spacetime shouts, “Go straight!" The free stone obeys. What does “straight" mean?  Straight with respect to what? We know the answer: The path of the stone is straight with respect to every local free-fall (inertial) frame through which it passes.

I don’t know why I’m telling you this pete? I know you were one of those who proof read EBH or perhaps still are the 2nd edition.

[Pmb (OP)]

Are you saying it is not possible to transform away the gravitational field even locally ?

No. How did you get that from what I said?

Do you need to transform away the field completely for it to not effect the test point/particle?

No. So long as the particle isn’t charged. For a charged particle the field is part of the charge and that acts to “probe spacetime.” A charged particle does not follow a geodesic in a curved spacetime.

Is it this part you don’t agree with…

No. As I said above, the part that is wrong is that the gravitational force is not a manifestation of spacetime curvature since you can have a gravitational force with no spacetime curvature.

Besides spacetime curvature / space curvature and time dilation/gravitational redshift, what other evidence shows the presence of a field?

If a particle subjected to no other than the gravitational force accelerates then there is a gravitational field present. If the particle does not accelerate then there is no gravitational field. The presence of spacetime curvature can be detected by looking for geodesic deviation, i.e. when two particles in free-fall accelerate relative to each other.

Have you let the MIT people know that their premise is wrong?

Edwin knows that it’s wrong. I don’t think he’ll tell the man who wrote it though, i.e. Bertschinger. I personally don’t like Bertschinger. To me he’s kind of a snob. He gives me the impression that I’m not important enough to want to know or talk to. That’s why I really don’t like that guy at all. I hear he’s like that with a lot of people. That’s why I have no respect for him whatsoever.

Given MIT are ‘teaching’ this, you would be doing them a service informing them.

That’s their problem, not mine. I do my service by proof reading the text that they’re learning out of.

Does E. Taylor still ‘reside’ at MIT?

He still has an office there, yes. He’s retired. He teaches when he has the time and energy.

Newton says a “force of gravity" leads to the parabolic trajectory. But  Einstein declares that Newton's “force of gravity" does not exist.

Einstein never said that and the author knows that. He wrote it in there anyway. I wasn’t too happy about it.

I don’t know why I’m telling you this pete?

Perhaps you feel the need to cover all your bases?

I know you were one of those who proof read EBH or perhaps still are the 2nd edition.

Yep. I’m starting over from chapter 1 which I’m finishing up this weekend.

Yes. I remember it very well since the premise is wrong, i.e. curvature is not a necessary condition for the presence of a gravitational force on a body in a gravitational field.

Are you saying it is not possible to transform away the gravitational field even locally ?
Do you need to transform away the field completely for it to not effect the test point/particle?


Is it this part you don’t agree with…
From link given in my last post.

What then is the path of a freely-falling body in the presence of gravity? According to Newtonian physics, the answer is given by solving ~F = m~a = md2~x=dt2 as a differential equation for ~x(t).
Gravity causes acceleration, and one might expect therefore that the maximal-aging result breaks down. A body obviously accelerates in a gravitational field, and so an unaccelerated path cannot be the correct one. However, according to general relativity it still works out that the correct path is the one that maximizes proper time!

equation symbols don’t copy correctly.

Besides spacetime curvature / space curvature and time dilation/gravitational redshift, what other evidence shows the presence of a field? Also how does this other evidence of field effect the Equivalence Principle if at all?

Have you let the MIT people know that their premise is wrong?
Given MIT are ‘teaching’ this, you would be doing them a service informing them.  Does E. Taylor still ‘reside’ at MIT?

Only if the rock is in free fall is that true. But yes, of course I agree.

Yes, the rock is un-powered and following a geodesic, defined by maximum aging on the rock’s watch and the spacetime metric, and so giving the appearance of acceleration when moving through an altered spacetime near mass. Why is the spacetime altered near mass? Don’t know.
What is the mechanism of an attracting ‘force’? Don’t know. Both being convenient models of observations.

From “Exploring Black Holes” Pick chapter titled “Diving” PDF page 2
Here…http://exploringblackholes.com/

Newton says a “force of gravity" leads to the parabolic trajectory. But  Einstein declares that Newton's “force of gravity" does not exist. Instead spacetime shouts, “Go straight!" The free stone obeys. What does “straight" mean?  Straight with respect to what? We know the answer: The path of the stone is straight with respect to every local free-fall (inertial) frame through which it passes.

I don’t know why I’m telling you this pete? I know you were one of those who proof read EBH or perhaps still are the 2nd edition.

[lean bean]

Are you saying it is not possible to transform away the gravitational field even locally ?

No. How did you get that from what I said?

I don't know, that just entered my head at the time.

the part that is wrong is that the gravitational force is not a manifestation of spacetime curvature since you can have a gravitational force with no spacetime curvature.

If a particle subjected to no other than the gravitational force accelerates then there is a gravitational field present. If the particle does not accelerate then there is no gravitational field. The presence of spacetime curvature can be detected by looking for geodesic deviation, i.e. when two particles in free-fall accelerate relative to each other.

LATE MODIFICATION...Is your ‘cavity’ inside a large mass an example of gravitational field without curvature?  If so, In the cavity, from what direction does the force act on a test particle to accelerate it.
Or, are you saying the forces in the cavity cancel each other? If so, where’s the  evidence of the presence of a gravitational field if there is no acceleration of a test particle?

Newton says a “force of gravity" leads to the parabolic trajectory. But  Einstein declares that Newton's “force of gravity" does not exist.

Einstein never said that and the author knows that. He wrote it in there anyway. I wasn’t too happy about it.

Pete, I’m not sure it is Bertschinger’s total doing…

 There is a figure and caption on page 2-5 of “Exploring Black Holes” edition one only, The references at the end of the chapter state the figure plus caption are taken from the book “Gravitation” back in 1979, By Misner, Wheeler and thorne.

Part of the caption reads…

In Newtonian theory this effect is ascribed to gravitational force acting at a distance from a massive body.
According to Einstein, a particle gets its moving orders locally, from the geometry of spacetime right where it is. Its instructions are simple: “Go straight! Follow the straightest possible worldline (geodesic).” Physics is as simple as it could be locally. Only because spacetime is curved in the large do the tracks diverge or converge.

And in “Exploring Back Holes” edition one, page 3-4 there is…

So what is new about relativity? On the theory side, Einstein says that you can do away entirely with Newton’s gravitational force.

That edition is just J.Wheeler and E.Taylor, not Bertschinger.
So, these othes are wrong too?

[Pmb (OP)]

LATE MODIFICATION...Is your ‘cavity’ inside a large mass an example of gravitational field without curvature?

Yes. In fact that’s the entire purpose of the example.

If so, In the cavity, from what direction does the force act on a test particle to accelerate it.

Construct vector whose tip is at the center of the body and whose tail is the center of the cavity. The force is parallel to and in the direction of this vector.

Pete, I’m not sure it is Bertschinger’s total doing…

It doesn’t matter. He should know better. Thinking in terms of spacetime curvature is what leads people to make mistakes like this.

There is a figure and caption on page 2-5 of “Exploring Black Holes” edition one only, …. That edition is just J.Wheeler and E.Taylor, not Bertschinger.
So, these othes are wrong too?

Yes. It doesn’t matter though. Neither of these two men should be making an error as careless as this. They should know better. Thinking in terms of spacetime curvature is what leads people to make mistakes like this. They might be justifying it to themselves by saying that tidal forces are a result of gravitational force and that’s how they can justify claiming that force is a manifestation of spacetime curvature but not only is it wrong its very misleading. As you know from the cavity example you can have a gravitational force without spacetime curvature.

[lean bean]

tip is at the center of the body and whose tail is the center of the cavity. The force is parallel to and in the direction of this vector.

Be patient here, so, if I have a ring of test particles around the cavity walls would they converge to centre of cavity? What will they do at the centre? I may be getting my tip and tail confused...

Or, other way round. If I have a sphere of test particles centred on the cavity centre, would each particle travel radially outwards to the walls?

Thank goodness for the modify button

[Pmb (OP)]

tip is at the center of the body and whose tail is the center of the cavity. The force is parallel to and in the direction of this vector.

Be patient here, so, if I have a ring of test particles around the cavity walls would they converge to centre of cavity? What will they do at the centre? I may be getting my tip and tail confused...

Or, other way round. If I have a sphere of test particles centred on the cavity centre, would each particle travel radially outwards to the walls?

Neither. That would imply the presense of tidal force. Remember that the field is uniform which means that no matter where you place a particle in the field the magnitude and direction of the force on the particle will be indepenant of where the particle is. Therefore the field lines are parallel and all point in the same direction. I updated the page with a new diagram to illustrate this fact. See Figure 2 below

http://home.comcast.net/~peter.m.brown/gr/grav_cavity.htm

[lean bean]

Neither. That would imply the presense of tidal force. Remember that the field is uniform which means that no matter where you place a particle in the field the magnitude and direction of the force on the particle will be indepenant of where the particle is. Therefore the field lines are parallel and all point in the same direction. I updated the page with a new diagram to illustrate this fact. See Figure 2 below
http://home.comcast.net/~peter.m.brown/gr/grav_cavity.htm

Yes, I did 'see' the element of tidal accelerations there in my own descriptions and  was confused by that.
You must have updated just as I come away from that page, I will have a chew on the new page now. And, yes, I see what you mean now about the placing of tip and tail giving the vector direction, I had the vector joining particle and centre!!

[lean bean]

Pete, I have looked at your updated page and have to say it's way above my head.
So, all I can say is, I’m obviously  not qualified to say whether your selection and use of the equations is correct or not, but it would be the cherry on the cake if you could have explained the mechanism by which the force attracts, and not just the behaviour of an assumed force of attraction.

[Pmb (OP)]

Pete, I have looked at your updated page and have to say it's way above my head.
So, all I can say is, I’m obviously  not qualified to say whether your selection and use of the equations is correct or not, but it would be the cherry on the cake if you could have explained the mechanism by which the force attracts, and not just the behaviour of an assumed force of attraction.

I don't understand. All I did was include a new figure at the bottom of the page to illustrate the field inside the cavity.

Here's how it works; the electric field is defined as force per unit charge, i.e. E = F/q. In gravity the same thing holds true. The "gravitational charge" of the gravitational force is the passive gravitational mass, m. So F/m is the gravitational field. It so happens that F/m is gravitational acceleration.

Nobody knows what the force of attraction is. Even general relativity can't tell us that.

[lean bean]

I don't understand. All I did was include a new figure at the bottom of the page to illustrate the field inside the cavity.

Sorry, What I meant was... I didn't understand the whole page even before the new figure. Like many people, I may be able to do equations as such (do the equation and get the right answer), but that does not mean I understand the logic in the process. Or is that just me I'm not saying your wrong or right.

[Pmb (OP)]

I don't understand. All I did was include a new figure at the bottom of the page to illustrate the field inside the cavity.

Sorry, What I meant was... I didn't understand the whole page even before the new figure. Like many people, I may be able to do equations as such (do the equation and get the right answer), but that does not mean I understand the logic in the process. Or is that just me I'm not saying your wrong or right.

I see. I've been thinking for a while now that I should do the full fledged derivation anyway. I'll work on that tomorrow.

[jeffreyH]

Pete, I have looked at your updated page and have to say it's way above my head.
So, all I can say is, I’m obviously  not qualified to say whether your selection and use of the equations is correct or not, but it would be the cherry on the cake if you could have explained the mechanism by which the force attracts, and not just the behaviour of an assumed force of attraction.

I don't understand. All I did was include a new figure at the bottom of the page to illustrate the field inside the cavity.

Here's how it works; the electric field is defined as force per unit charge, i.e. E = F/q. In gravity the same thing holds true. The "gravitational charge" of the gravitational force is the passive gravitational mass, m. So F/m is the gravitational field. It so happens that F/m is gravitational acceleration.

Nobody knows what the force of attraction is. Even general relativity can't tell us that.

This may be the mechanism. As the wave ripples through the mass the contraction exerts a pull n the opposite direction. How does this relate to the stress-energy tensor? This would explain the mechanism of length contraction. The stronger the wave front the more compression is induced. In a balanced system such as the hollow planet example the effects would be of special interest and may involve unique tidal forces.

If the gravitational field increases in strength then the rate of compression increases. At near light speed a mass will encounter more gravitational waves simply by travelling trough them faster. This situation is equivalent to being in the stronger field around a mass of significant size. I agree with Pete that the gravitational field does not need a source. Gravity had to already exist during or immediately after the big bang as did everything else. It was just there along with all the other energy.

[Pmb (OP)]

This may be the mechanism. As the wave ripples through the mass the contraction exerts a pull n the opposite direction.

Jeff - If you're going to make an argument like this you should at least explain the terms that you're using. E.g. why would anybody reading your response know what this wave that you're talking about is?

In a balanced system such as the hollow planet example the effects would be of special interest and may involve unique tidal forces.

Again please define your terms. What tidal forces are you talking about? Inside the cavity, where the field that this conversation is about resides, there are no tidal forces present.

[jeffreyH]

This may be the mechanism. As the wave ripples through the mass the contraction exerts a pull n the opposite direction.

Jeff - If you're going to make an argument like this you should at least explain the terms that you're using. E.g. why would anybody reading your response know what this wave that you're talking about is?

In a balanced system such as the hollow planet example the effects would be of special interest and may involve unique tidal forces.

Again please define your terms. What tidal forces are you talking about? Inside the cavity, where the field that this conversation is about resides, there are no tidal forces present.

The wave is a gravitational wave. I don't see why there wouldn't be tidal forces as the mass surrounding the cavity would be exerting an influence on the matter in the cavity from all directions. This would be pulling the mass outward in all directions. It can be claimed that these forces cancel out but what physical evidence do we have to support that proposition.