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the force does not arise from any physical interaction between two objects, but rather from the acceleration of the non-inertial reference frame itself.

A geodesic is not what happens when I lift that rock.

Gravitation requires the presence of a second object, and the gravitational force on a "test" object is proportional to the mass of the "source" object.

This view is made possible for us by the teaching of experience as to the existence of a field of force, namely, the gravitational field, which possesses the remarkable property of imparting the same acceleration to all bodies. The mechanical behaviour of bodies relatively to K' is the same as presents itself to experience in the case of systems which we are wont to regard as "stationary" or as "privileged." Therefore, from the physical standpoint, the assumption readily suggests itself that the systems K and K' may both with equal right be looked upon as "stationary" that is to say, they have an equal title as systems of reference for the physical description of phenomena. It will be seen from these reflexions that in pursuing the general theory of relativity we shall be led to a theory of gravitation, since we are able to "produce" a gravitational field merely by changing the system of co-ordinates. It will also be obvious that the principle of the constancy of the velocity of light in vacuo must be modified, since we easily recognize that the path of a ray of light with respect to K' must in general be curvilinear, if with respect to K light is propagated in a straight line with a definite constant velocity.

1. When on the surface of the Earth a rock has sufficient gravitational potential energy to take it to the centre of the Earth if nothing impedes its passage.

2. If I pick up the rock from the surface of the Earth I impart to it more gravitational potential energy. The higher I take the rock the more gravitational potential energy it has.

3. In terms of GR, in picking up the rock, I am simply moving it along a geodesic in spacetime. This action requires an input of energy, which is transferred to the rock.

4. Does this mean that gravity is a force that requires/involves expenditure of energy?

5. On the Earth’s surface, if I carry a ball up a hill and put it down, it will roll back to the bottom of the hill. Obviously, the hill is in some way responsible for the fact that the ball rolls down; but the hill is not a force.

6. Is there any validity in equating the hill and gravity? I.e. neither is a force, but both respond to the addition of gravitational potential energy to an object by causing that object to move towards the local centre of gravity.

7. Both gravity and the hill are distortions. Gravity distorts spacetime. The hill distorts a sphere that would represent a specific energy level around the local centre of gravity.

While it’s true that the rock moves on a worldline such a worldline is not a geodesic.

First off it's wrong to think of a force as an interaction between two objects. For example; if a charged object was at rest in the inertial frame S and an electromagnetic wave were to pass through that frame then there will be a non-zero Lorentz Force on that particle. The same holds true for a gravitational wave.

Quote from: alancalverdGravitation requires the presence of a second object, and the gravitational force on a "test" object is proportional to the mass of the "source" object.That's incorrect. For example; if a particle is at rest in the frame of reference S and a gravitational wave were to pass through S then there'd be a gravitational force on that particle. In such case there is no second source object in the immediate area. It could have been created a long long time ago in a galaxy far far away.

Now change to a non-inertial frame, the particle, which was originally at rest, will be accelerating in this new frame.

Is a geodesic only 'frictionless' in a flat space? gravity 'steals energy' of objects, right? Two heavenly bodies circling each other loses energy to their interacting gravitationally, so what about one body, moving through a curved space? Will that one also lose energy?

Yes, but the source of fields and waves are other objects, so in those cases it is interaction-at-a-distance that is mediated by fields.

the gravitational field in a region of space does not tell you what created it.

Nobody is "worried" about the source, ..

I don't know JP. A charged particle moving in a circle, is that a geodesic? If it isn't, then it is a acceleration. And if it accelerates it must lose 'energy'. (Thinking of it as 'fields', also assuming light to not 'propagate', you should get shapes describing it instead of a motion.)

true Pete, but what it made me think about was actually electrons 'orbitals'.

A charged particle moving in a circle, is that a geodesic? If it isn't, then it is a acceleration. And if it accelerates it must lose 'energy'.

The laws of physics don’t require the existence of gravitational objects for gravitational waves to exist.

Or are you saying not to worry about the origin of the waves?

Quote from: beanyOr are you saying not to worry about the origin of the waves?Yes. Anytime I've ever had to calculate the force exerted on a particle I never had to know the source of the field. All I had to know was the field.

one body generates a field and it’s the field that interacts with objects to exert forces on them. So all we need to know, and all that needs to exist, is the field, not the source. The laws of physics don’t require the existence of gravitational objects for gravitational waves to exist.

In Einstein's theory of general relativity, gravity is treated as a phenomenon resulting from the curvature of spacetime. This curvature is caused by the presence of mass.

Ok, now I know what you mean, but, I do think you could have worded that better... confused for a moment thinking what other laws, if not those of physics, do you require to make G waves.

then me on earth may define a satellite as 'free falling' inside a gravitational field, but for the satellite itself there is no gravitational field to be measured

Quotethen me on earth may define a satellite as 'free falling' inside a gravitational field, but for the satellite itself there is no gravitational field to be measuredOh no there isn’t

It will be seen from these reflexions that in pursuing the general theory of relativity we shall be led to a theory of gravitation, since we are able to "produce" a gravitational field merely by changing the system of co-ordinates.

The observer on the satellite will note that he is accelerating towards a massive object, and deduce that he is therefore in a convergent gravitational field.

So you are saying that according to GR, gravitational fields exist in the absence of mass.

Now here's a fine mess, because one principle of all non-newtonian physics is that it must approximate to the newtonian at the mesoscopic level, because that is what we observe and we don't like arbitrary discontinuities in our theories.

Relax, man.

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.

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). etc

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.

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.

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'.

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.

1. Introduction: Extremal Aging and the Equivalence PrincipleThese 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.

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.

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

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.

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.

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!

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

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.

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…

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

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?

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

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.

Quote from: Pmb on 30/11/2013 12:36:58Yes. 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.QuoteWhat 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?Quote from: Pmb on 30/11/2013 12:36:58Only 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 2Here…...sorry, you cannot view external links. To see them, please REGISTER or LOGINQuoteNewton 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.

Quote from: beanyAre you saying it is not possible to transform away the gravitational field even locally ?No. How did you get that from what I said?

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.

Quote from: beanyNewton 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.

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.

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

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.

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