Naked Science Forum
Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: Pmb on 22/11/2013 08:03:51
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This thread is not about what an inertial force is, what Einstein thought of it, what you think GR has to say on it etc.
The definition of inertial force is well known and never debated. I know the definition quite well in fact so let’s skip any discussion on definition, okay? For those who don’t know what it is then please see
http://home.comcast.net/~peter.m.brown/gr/inertial_force.htm
In that page I've described various viewpoints from some well-known physics textbooks on mechanics general relativity and cosmology.
Before any of you claim to know how it's viewed in the physics community please see the quotes at the end which expresses one side of the viewpoint.
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?
Thanks
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You can translate it away, as you yourself pointed out, or alternatively consider yourself negating a gravity by constructing another, expressed through your 'free fall', although the last one is a hard nut to accept. But so is ordinary every day gravity too I think, the one I find here on Earth. The idea that a acceleration is equivalent to a constant uniform acceleration, when turned around, seems then to imply that we would have a 'accelerating' Earth constantly and uniformly accelerating at one G, ignoring spin.
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Definition of a fictitious force includes:
the force does not arise from any physical interaction between two objects, but rather from the acceleration of the non-inertial reference frame itself.
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. Since the force is observed by an observer in the same reference frame as the source and test objects, and is measurable when all objects and the observer are stationary with respect to each other, it must be a real force.
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" **Pre-reqs:** None.
I intend to talk about forces and force diagrams, but there is a more fundamental question to address first. What is a force? Most texts define it as a push or a pull. That really isn’t a bad definition. Maybe a better (or maybe worse) definition would be “forces are things that change the motion of an object” (change being the key word). If I had to choose one definition of force, it would be something like this:
**Force:** *A force is an interaction between two objects. There are 4 known forces:*
Gravitational force: An attractive long range force between objects with mass
Electromagnetic force: An attractive or repulsive long range force between two objects with charge
Strong Nuclear force: An attractive short range force between particles like protons and neutrons
Weak Nuclear force: A short range force responsible for beta decay. *Yes, I know that is a confusing force – for introductory physics, you won’t use this force*
All forces are some form of the above forces. " http://scienceblogs.com/dotphysics/2008/09/26/basics-what-is-a-force/
Don't know, it's contradictory to me. "“forces are things that change the motion of an object” (change being the key word)." Gravity is a geodesic.
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Take a look here: http://www.av8n.com/physics/fictitious-force.htm
It's really interesting this one. I could argue, I think? That a force becomes one thing as observed in a 'common for us all, container universe' but another if I only use local definitions. I'm not sure on that one though :) It just struck me, but I think it could be possible to argue.
Anyway: What the he* is a force :)
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At this point I would like to interject a few “hitch-hiker’s” thoughts and seek comments/correction. They are numbered for ease of response.
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.
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I think you was thinking something different at 3. A geodesic is not what happens when I lift that rock. When I throw it though, it should describe a geodesic. Lifting it I'm accelerating it. As for 6 there should be no general equivalence between a geodesic and the hills slope. But if you mean that gravity is the reason for the ball following the slope I would agree.
Also, and here's a tricky one, as I find it. When you're in a 'free fall' you're also in a geodesic, and no matter whether this will be a 'gravitationally accelerating' frame for some far observer (seeing me falling into a BH for example), or not (free falling in a 'flat space') you still will find a equivalent environment, defined from being inside in a 'black box scenario', ignoring tidal forces(spin). And I think we should ignore those, as we otherwise only have those to point at :)
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A black hole is not that good a example, too much time dilation, etc. I think Earth will do nicely for it though.
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What I mean is that 'potential energy' can only be described in a relation. You either need to define it relative some other 'frame of reference', as another object, or you need to have made a definition relative some anchor, once existing (predefined) as your point of reference.
Think of two uniformly moving 'black boxes' hurtling towards each other, inside each one there is nothing telling them of any motion, they are perfectly equivalent environments (ignoring tidal forces/spin). Whatever potential energy each of them may be assumed to contain, will only be apparent relative a 'third observer', seeing them being on a collision course, defining their speed and mass relative his local measurements. And in a scenario of two equivalent, uniformly moving objects, measuring each others speed and so inherent 'potential energy', each one is free to define the other as being still. So I would call potential energy a relation.
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A geodesic is not what happens when I lift that rock.
OK, but isn't it following a geodesic when it returns to the surface under gravity?
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Yep, that was what I suspect you was thinking.
And I've done the same, several times.
Thank God for the edit function :)
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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.
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. :)
You’re thinking in terms of Newtonian mechanics. What you say here is not true in Einstein’s general theory or relativity. In that theory the gravitational force requires only the presence of a gravitational field and an object on which the field is acting, i.e. exerting a force on. If you're moving with constant velocity relative to and far removed from all matter, e.g. in interstellar space, then you're in an inertial frame of reference. Now change to a non-inertial frame, the particle, which was originally at rest, will be accelerating in this new frame. According to Einstein’s general theory or relativity there is a gravitational force now acting on the particle since gravitational forces are identical in nature to inertial forces.
See http://hem.bredband.net/b153434/Works/Einstein.htm
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.
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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.
Correct.
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.
Correct.
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.
While it’s true that the rock moves on a worldline such a worldline is not a geodesic. See - http://home.comcast.net/~peter.m.brown/math_phy/geodesics.htm
4. Does this mean that gravity is a force that requires/involves expenditure of energy?
Yes.
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.
The hill exerts a force on you and you exert a force on the ball. It can only roll down the hill if the hill exerts a force on the ball.
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.
There is a relationship between the height of the hill and the gravitational potential of an object sitting on the hill.
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.
Gravity can distort spacetime. It doesn’t always do so. A uniform gravitational field doesn’t and neither does a straight cosmic string or a vacuum domain wall.
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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.
I changed the title of this thread because I'm really more interested in inertial forces in general. The gravitational force just happens to be an example of an inertial force.
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The inertial forces such as Gravitational, Centrifugal and Coriolis forces? they are measurable, and definable.. I think its hard to say its not real if you can measure it.. you can even predict them, so how can they not be real.
Maybe the question is of the definition of 'real'.
"Notice that all inertial forces have the mass as a constant of proportionality in them. The status of inertial forces is again a controversial one. One school of thought describes them as apparent or fictitious which arise in non-inertial frames of reference (and which can be eliminated mathematically by putting the terms back on the right hand side). We shall adopt the attitude that if you judge them by their effects then they are very real forces."
How else would you judge them?
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Thanks for the response, Pete. Good to know I have a few accurate thoughts in my head. Then I followed your link and began wondering if I had anything at all in there. :(
While it’s true that the rock moves on a worldline such a worldline is not a geodesic.
As I commented in response to yor_on's point: if it is following a geodesic when it returns to the surface under gravity, why is the same path not a geodesic when traversed in the opposite direction? I bet the answer is on that linked page, but it would need it to be a lot simpler than that before I could even pretend to understand it.
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You mean if you keep to the same path as you would expect it to have when falling, but lifting it up instead? In SR a geodesic is defined as a path of shortest SpaceTime interval, in flat spacetime (no gravity) a straight line with a constant velocity. In GR, as I understand it, it is defined as all particles 'free falling', are following a geodesic in a 'curved SpaceTime' meaning gravity's influence. A 'proper acceleration' in relativity is one where you inside a 'black box scenario' can measure a acceleration by a accelerometer. (https://en.wikipedia.org/wiki/Accelerometer).
In a gravitational acceleration, being in a 'free fall' towards Earth, there is no acceleration measurable by a accelerometer. And I think that connects to what Pete discussed earlier, translating away a gravitational field. Someone on Earth will find you accelerating, but for you inside that black box there will only be a 'weightlessness', as you're in a free fall.
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This is ignoring Earths, or the box, rotation (Coriolis force, tidal forces) btw, maybe Pete has a better definition.
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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.
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.
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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.
Pete.
Do you mean the G wave leaves the particle in the same state in which it was found? That is...the particle has not moved out of place after the wave as gone?
But, if there is no interaction, why does the G wave lose energy?
And to echo JP, the G wave had to have a source.
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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?
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It's been bugging me for a while, the definition of a 'frictionless gravity'. If I would define gravity as a 'force', it seems to me that I also have to assume it acting on bodies following geodesics through a curved space.
And if that would be true then it seems to me, assuming a straightforward propagation of light, that this also should be relevant for light paths, as we have the equivalence between energy and mass? As I think of it it shouldn't matter for the definition of a geodesic? If space is 'curved' (gravity), or not (flat space). But thinking of binary stars interacting, gravity seem to bleed away 'energy' from them, described as a system?
"Einstein's general theory of relativity explains gravity as a consequence of the curvature of spacetime created by the presence of mass and energy. As two stars orbit each other, gravitational waves are emitted - wrinkles moving out in spacetime. As a result, the binary slowly loses energy, the stars move closer, and the orbital period shortens." http://www.spacedaily.com/reports/Bizarre_binary_star_system_pushes_study_of_relativity_to_new_limits_999.html
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Ok, the mass quadrupole moment, is that what explains why binary stars is expected to lose energy, making waves, and why a sole body in a curved space, won't? (But if the universe is a 'container of it all', why can't I define that, or a gravitational field, relative a sole body's geodesic in a curved space too?)
"An object's gravitational monopole is just the total amount of its mass.
An object's gravitational dipole is a measure of how much that mass is distributed away from some center in some direction. It's a vector, since it had to convey not only how much the mass is off-center but also which way. Considering some object in the abstract, the natural 'center' to pick is the center of mass, which is the point around which the dipole is zero.
The quadrupole represents how stretched-out along some axis the mass is. A sphere has zero quadrupole. A rod has a quadrupole. A flat disk also has a quadrupole, with the opposite sign of the quadrupole of a rod pointing out from its flat sides. The rod is a sphere stretched along that axis and the disk is a sphere squashed along that axis. In general, objects can have quadrupole moments along three different axes at right angles to each other. (The quadrupole moment is something called a tensor.) ...
... for gravitational radiation you need an oscillating quadrupole moment. The difference is that electric charge comes in two varieties of charge, plus and minus. When you interchange the two charges, as in an oscillating electric dipole, you get a change in the electric field distribution. Gravitational mass, on the other hand, comes in only one sign: plus. There are no minus values. So if you interchange two masses you don't get a change in the gravitational field. Hence, no dipole radiation. "
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On the other tentacle, the universe seem nearly 'flat', as I've seen it defined.
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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.
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. :)
Long ago and far away does not mean nonexistent. Einstein abolished (or at least redefined) simultaneity early in the last century. If I die before you read this message, does it mean that messages do not require a source?
Now change to a non-inertial frame, the particle, which was originally at rest, will be accelerating in this new frame.
There is an infinity of possible noninertial frames for every particle. Therefore there is an infinite amount of energy in the universe since the particle is accelerating in each one. Or is someone talking out of an unconventional orifice?
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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?
This isn't that odd. A charged particle moving in a straight line doesn't slow down, but a charged particle moving in a curve will radiate away energy and slow down.
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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.
I disagree. Please recall what I was responding to
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.
There's no source object whose mass the gravitational field is proportional to.
The point is not to worry about the source of the field. It's the presense of the field itself which determines the gravitational force and we don't need to know the source. E.g. the gravitational field in a region of space does not tell you what created it. There are more than one ways to create the same field in some cases.
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Nobody is "worried" about the source, but everyone knows it has to exist. We have no evidence of a gravitational field with no mass at its centre.
the gravitational field in a region of space does not tell you what created it.
The presence of a red bus in the middle of London does not tell you where it was built, but we know that someone, somewhere, made it, and drove it to the city. We have no evidence of the spontaneous appearance of buses without factories.
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Pete, when you write "There's no source object whose mass the gravitational field is proportional to." Do you think of it as gravity's reach is limitless? There being no way, practically, to define who it 'belongs too'. Or are you thinking of it as a result of what frame I am in, finding a field existing for me? It's interesting. And it always seem to come down to comparisons between frames of reference. But even in a black box scenario, you can find if you're in gravitational field, assuming tidal forces acting on you. But that's no real proof, is it? You could imagine a gravitational field without tidal forces, me in a 'free fall' inside it. It's like there is a 'tension' of some sort, related to mass, motion, and energy, creating 'gravity' when locally accelerating, possibly? Because that is the equivalence principle, isn't it? That earth is constantly uniformly 'accelerating' at about one Gravity.
And if it is correct? Then I might imagine it as properties of a 'field' preexisting, although not presenting itself, for my measurements, except in certain circumstances. Or maybe I'm bicycling in the great younder again :)
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Nobody is "worried" about the source, ..
That’s not true. It was you who were worried about it. That’s why you said
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.
and I was saying that this is simply not true. You’re stuck in a Newtonian frame of mind where one object exerts a force on another object whereas in modern physics 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.
And even when there is a source the force isn't proportioal to the mass since there's a velocity dependance factor there, i.e. in GR the gravitational force is velocity dependant. See Eq. (14a) in http://home.comcast.net/~peter.m.brown/gr/grav_force.htm
And these are two very different conversations. You’re stuck with thinking that all fields require a source and I was talking about a totally different subject, i.e. the force on a particle due to the gravitational field.
..but everyone knows it has to exist. We have no evidence of a gravitational field with no mass at its centre.
And that assumption is just plain wrong in general since there need not be a source. For example; if you're in an inertial frame of reference in flat spacetime. A particle is in that frame and is at rest. Now you change to a uniformly accelerating frame of reference. Then in that frame of reference there is a uniform gravitational field and there is a gravitational force on that particle. Yet there is no source of gravity to be concerned with other than the distant stars.
Then again there are ways to generate a gravitational field with a finite amount of matter – See http://home.comcast.net/~peter.m.brown/gr/grav_cavity.htm
So because you know the field in a region of space it doesn’t mean that you know what the source of that field is.
Go out and look at the CMBR. That radiation came from the birth of the universe and was not created by other charges. Therefore the force that it exerts on charges today is not a result of the presence of a second object, i.e. the electromagnetic force on a "test" charge is not proportional to the charge of any source so if you tried such an argument with the electromagnetic force then that argument would also fail. There is nothing in the laws of electrodynamics which requires a non-zero charge density everywhere to create an electromagnetic wave so its not charge itself which generates EM waves. You can have electrically neutral systems, i.e. systems with the same amount of positive and negative charges, to create EM waves.
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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.)
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Heh, a 'flickering' universe, locally defined.
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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.)
A charged particle moving in a circle in flat spacetime is not moving on a geodesic. However if that charge is orbiting the earth and in that sense its moving on a cirlce then yes, it's moving on a geodesic since in this case the circle is a geodesic.
In flat spacetime a charged particle moving in a circle emits what is known as synchrotron radiation.
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true Pete, but what it made me think about was actually electrons 'orbitals'.
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true Pete, but what it made me think about was actually electrons 'orbitals'.
Where did you mentionor indicate that you were thinking about orbitals? Or did you? You wrote
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'.
Electrons in orbitals are not moving in circles and are not accelerating. Those are classical ideas which don't belong in the realm of quantum mechanics.
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The laws of physics don’t require the existence of gravitational objects for gravitational waves to exist.
Trying not to take your opening question too of course here, but...
Can you explain something of what you mean there.
I stuck with thinking of either mass or energy or momentum as having something to do with the origin of gravitational waves. Or are you saying not to worry about the origin of the waves?
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Or 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.
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Or 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.
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. :)
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I know that Pete, on the other hand you have 'photographic evidence', although created by taking several images/snapshots, of something called a particle (electron) in that Swedish experiment (think it was Lund's university). So we have both in some weird way, or not so weird if propagation can be questioned, instead using the idea of a field, keeping a local 'constant' arrow for any observer, allowing this 'particle' to be depicted through repeated 'snapshots'. It's the way my mind works Pete :) Also called galloping senility.
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(As well as thinking of it as if a 'acceleration' can be translated to something at a particle level, we do it with a planet, do we not? Earth 'accelerates', so the question I started to ask, and still ask, myself is if one could apply the same sort of view on a atom and its constituents. Accelerations versus uniform motion. )
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And to come a little clearer :) Not talking about a acceleration as something 'spending a energy' for this. As far as I know Earth do not spend any 'energy' by 'constantly uniformly accelerating' at about one gravity
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Maybe you can express it this way. If Earth is 'accelerating', then it is in no measurable direction for us. Instead we find a uniform motion that we can measure, through the universe. And that motion is a geodesic as I understands it, spending no 'energy', never mind how much matter, neutron stars, etc, I involve in Earths path.
That one gives me one headache, no 'friction', the other is what a 'acceleration' can be, if I define Earth as accelerating. And then we have particles, they should have a gravitational field too, shouldn't they? And they, just as Earth, should then 'accelerate'.
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Maybe the question could be. Do a acceleration spend 'energy'? I sort of automatically have assumed it must do so, it seems the reasonable assumption to me. But what about the equivalence principle then, and Earth 'accelerating'?
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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.
Spot the selfcontradiction.
If you can generate a gravitational field without a source object, you can have all the Nobel prizes at once. But until then, I think anyone observing the effect of a gravitational field will (rightly) infer the existence of a massive object. That is, after all, how astronomers conduct their business, and to date, they have been right every time.
Not sure how the laws of physics determine the existence of gravitational waves. Scientific laws are discovered by observation, not imposed by Brussels.
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.
But what did Einstein know about relativity? My friend pmb says you don't need mass. So there.
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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. :)
Well I can't be good at everything. :) But could have worded what exactly, better?
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It's a really tricky one Alan. But think of 'energy' for it. If there is a equivalence between energy and mass, then 'energy' should be enough for making a defined 'container of a universe, even in the absence of matter, as long as we assume gravity to have a relation to 'mass'. Light can be referred to as having mass too. So you're right in that we need a relation, but what that relation need to be? Then there is the way you can translate away a 'gravity' by changing your reference frame. If I imagine a gravitational field as a defined color, then this statement is the same as saying that the color disappear just by me changing some parameter 'inside' it, as stopping my acceleration, free falling. And if we involve different frames of reference, 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. Or if we define it as some sort of grid lines, then me accelerating either becomes a very local experience of 'gravity or, assuming gravity's infinite reach, those locally created 'field lines' reach forever, as long as I keep accelerating.
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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
Oh yes there is! 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.
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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
Oh no there isn’t
Oh yes there is! yor_on was attempting to explain Einstein’s equivalence principle to you. That principle states that at any event in spacetime it’s possible to transform the gravitational field away. And likewise it’s possible to create a gravitational field at any point in spacetime. If you really want to learn Einstein’s general theory of relativity then you should pick up a GR textbook and read it rather than making all of these false assumptions about it attempting to prove it wrong. Start here The Foundation of the General Theory of Relativity by Albert Einstein (1916), Annalen der Physik, 49. See
http://hem.bredband.net/b153434/Works/Einstein.htm
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.
That’s a misinterpretation of what yor_on said. He didn’t say that it’s possible to transform the gravitational field away at all events. He actually meant that at even event you an transform it away. However in a curved spacetime like that around earth it can’t be transformed away everywhere. In this case the observer on the satellite will not be in a gravitational field if he stays close to the (small) satellite. However in this case it’s the earth that is in free fall in his frame of reference. I.e. the earth is in the gravitational field now, not the satellite.
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If you are in a box or an elevator, maybe the walls are hiding the truth...
Nice article Pete!
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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. So where does the mass of the bodies come from, in our carefully measured "Gm1m2/r^2" forces?
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So you are saying that according to GR, gravitational fields exist in the absence of mass.
It's certainly allowed by the field equation, that's for sure.
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.
This is what I've been talking about. If you actually sat down and chose to learn general relativity then you'd no longer make mistakes like this.
In this case you're confusing the notion of incommensurate theories with the correspondence principle and then you thought you found a problem but the problem doesn't exist. The problem is with your poor knowledge of GR.
What you're trying to refer to is the correspondence principle in general relativity that is described here -
http://en.wikipedia.org/wiki/Physical_theories_modified_by_general_relativity
That principle demands that the weak field approximation of GR be what Newtonian gravity predicts. This is one of the things you'd learn correctly if you actually took our word for it and studied GR formally by picking up a good GR text and hitting it hard. Perhaps you don't have the math skills? Is that why you've been refusing to learn it? The math is something you learn as part of learning physics. It's not a roadblock to it but part of the education.
Or what is the reason that you refuse to learn GR before you continue to claim that it’s wrong?
However the correspondence principle doesn't apply to this. For example; in Newtonian gravity there are no gravitational waves so in this respect the correspondence principle doesn't apply. Then there's the fact that frames of reference in a freely falling frame are non-inertial frames in Newtonian mechanics but are inertial frames in GR. Also a place where the correspondence principle doesn't apply.
So where does the mass of the bodies come from, in our carefully measured
"Gm1m2/r^2" forces?
That's the problem. You're wondering where the mass of the bodies comes from but
don't even ask yourself what body is it that you're referring to. In Newtonian gravity all
gravitational fields had massive bodies as sources. However that's Newtonian gravity,
not general relativity. Again you're confusing the two theories.
You also don't seem to be able to understand anything except the gravitational fields outside of planets and stars. This is how you're stuck in Newtonian gravity. If you learned GR then you'd learn the error in making such assumptions.
Are you familiar with Mach's principle? I don't believe so since if you did then you wouldn't have made such a statement. Please learn about it at http://en.wikipedia.org/wiki/Mach's_principle
It explains things like this. E.g. suppose you were in a frame of reference that's rotating relative to an inertial frame of reference. In the rotating frame there'd be gravitational fields. However this is only true in GR and not in Newtonian mechanics where those forces are referred to as "inertial forces" and go by names such as the Coriolis force and the centrifugal force. In GR those are gravitational forces. The source of those gravitational forces is the matter from the distant stars. Consider how Einstein viewed such forces/fields. The following comes from an article that Einstein wrote which appeared in the February 17, 1921 issue of Nature
Can gravitation and inertia be identical? This question leads directly to the General Theory of Relativity. Is it not possible for me to regard the earth as free from rotation, if I conceive of the centrifugal force, which acts on all bodies at rest relatively to the earth, as being a "real" gravitational field of gravitation, or part of such a field? If this idea can be carried out, then we shall have proved in very truth the identity of gravitation and inertia. For the same property which is regarded as inertia from the point of view of a system not taking part of the rotation can be interpreted as gravitation when considered with respect to a system that shares this rotation. According to Newton, this interpretation is impossible, because in Newton's theory there is no "real" field of the "Coriolis-field" type. But perhaps Newton's law of field could be replaced by another that fits in with the field which holds with respect to a "rotating" system of co-ordinates? My conviction of the identity of inertial and gravitational mass aroused within me the feeling of absolute confidence in the correctness of this interpretation.
You could learn more about this if you really wanted to learn about GR by looking at http://home.comcast.net/~peter.m.brown/gr/inertial_force.htm
Search for the term inertial induction.
I've had you in my kill file for a very long time. Do you want to know why? If so then it's because of the fact that you refusing to learn GR means that you keep making mistakes and when I saw a mistake I felt compelled to correct it. Since there are so many mistakes and it takes a lot of work to correct you and you have no desire to go out and learn this for yourself it's a waste of my time.
Since this is getting to be way to much work again to keep correcting your mistakes I'm putting you back in mh kill file. Sorry but this is supposed to be a two way street, not just one way. You have to be willing to learn and you're not willing to read a good book on GR
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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.
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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?
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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.
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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.
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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.
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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/ (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.
:)
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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.
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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/ (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. :)
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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/ (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. :)
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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? :)
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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.
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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 :)
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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
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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!!
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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. :)
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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.
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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.
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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.
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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.
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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.
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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.