[Pmb (OP)]

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

[yor_on]

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.

[alancalverd]

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.

[yor_on]

" **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

[Bill S]

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.

[yor_on]

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.

[Bill S]

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?

[yor_on]

Yep, that was what I suspect you was thinking.
And I've done the same, several times.

Thank God for the edit function

[Pmb (OP)]

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.

[Pmb (OP)]

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.

[Pmb (OP)]

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.

[SimpleEngineer]

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?

[Bill S]

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.   

[yor_on]

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..

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.
==

This is ignoring Earths, or the box, rotation (Coriolis force, tidal forces) btw, maybe Pete has a better definition.

[JP]

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.

[lean bean]

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.

[yor_on]

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.

[alancalverd]

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?

[JP]

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. 

[Pmb (OP)]

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.