B Is Gravity Really a Force? A Look at General Relativity's Perspective

[Jupiter60]

TL;DR Summary

Force or not?

According to general relativity, gravity is not a force, however it is referred as one of the four fundamental forces. This seems like a contradiction.

 

[Drakkith]

Classically (meaning non-quantum and non-GR) gravity is described by Newton's law of Universal Gravitation, in which gravity is certainly a force. However, General Relativity supersedes classical physics when it comes to gravity, and gravity is instead the manifestation of curved spacetime and is not a force.

Part of the reason we commonly call gravity a fundamental force is simply that we've called it a force in the past and many people continue to do so. The other reason is that very, very few people use GR compared to classical physics, so calling gravity a force is usually easier than trying to explain why it isn't. All these students in their undergrad physics classes treat gravity as a force, as do practically all engineers and scientists not working with GR, and nothing is incorrectly calculated for having done so. Indeed, the errors are so small when designing buildings, bridges, airplanes, and other non-GR-specific experiments/devices that it is immeasurably easier to treat gravity as a force and use classical physics.

So, to be accurate, gravity is not a force according to the theory that best predicts gravitational phenomena. But, the vast majority of the time, it is immensely useful to treat it as one.

 

[mpresic3]

Physics terminology looks at "forces" as interactions. Gravity is a force. Einstein showed that gravity can be thought of as a curvature in space-time. Hence, gravitational interaction may be more appropriate for gravitational phenomena between masses. If and when electromagnetism and weak and eventually strong forces are understood as well (and physicists believe they eventually will be), these forces will be considered interactions as well. Do not be surprised if the four forces in many authoritative sources are arguably more referred to as the gravitational interaction, the electromagnetic interaction, the weak interaction, and the strong interaction.

 

[Dale]

Summary:: Force or not?

Yes

 

[Drakkith]

If and when electromagnetism and weak and eventually strong forces are understood as well (and physicists believe they eventually will be), these forces will be considered interactions as well.

What do you mean? All of these are extremely well understood and are already considered interactions.

 

[Drakkith]

@Dale Correct me if I'm wrong, but isn't the concept of a 'force', meaning something that causes an acceleration of an object, just a subset of all the possible interactions in QED? In other words, when two particles interact in QED there are many more things that can happen other than an acceleration. The application of a force could be considered as a single type of interaction out of a larger set, correct?

 

[Dale]

I mean “yes” because you can consider it a force or spacetime curvature. It doesn’t really matter. It is just a label

 

[mpresic3]

Mainly a response to Drakkith

Electroweak and strong interactions are quite well understood but there is still much research into these areas. The last word has not been written. Of course, the same can be said for gravity. I do think electroweak and strong interactions are not as well established as a curvature in higher dimensions the way gravity is, and /or gravity is not so well established as an exchange of particles, the way the electromagnetic interaction with photons, the weak interaction with electrons, and taus, and their neutrinos, or the strong interaction with the mesons quarks etc.

 

[Jupiter60]

I guess the answer to "is gravity a force?" can be "yes and no".

In the context of general relativity it is not a force, however in most other situations like in designing buildings, bridges etc. it can be referred as a force.

 

[Gazi]

relative to another object in space which moves at same speed at the beginning, accelerating through Earth with 9.81m/s^2 is a force because both will not have same speed at the end. Space ships use gravity for sling shot so they increase speed with it thus it should be force

 

[Ibix]

relative to another object in space which moves at same speed at the beginning, accelerating through Earth with 9.81m/s^2 is a force because both will not have same speed at the end. Space ships use gravity for sling shot so they increase speed with it thus it should be force

This is rather confusingly worded, but seems to me to say that because gravity causes acceleration it must be a force. Of course, in the general relativistic picture, gravity does not cause acceleration in any meaningful sense. That is, objects in free-fall are moving inertially. It can be difficult to state precisely what you mean by changing speed in relativity, but you can certainly set up circumstances where two objects meet twice with different relative velocities when they meet. In relativity this is due to the curvature of spacetime, not due to a force. It's roughly analogous to the fact that if you start in front of me facing to my left, and we both walk straight ahead half way around the world and meet again, you will be facing to my right, although neither of us turned once.

 

[RobHerd]

Summary:: Force or not?

According to general relativity, gravity is not a force, however it is referred as one of the four fundamental forces. This seems like a contradiction.

The laws of general relativity break in the singularity like a black hole. That’s why physicists try to find a quantum relationship between gravity and general relativity.
https://www.quantamagazine.org/why-gravity-is-not-like-the-other-forces-20200615/

 

[vanhees71]

In relativistic physics we rather talk about the fundamental interactions than forces.

The usual view at the gravitational interaction is that it is in some sense special compared to the other three interactions (strong, weak, electromagnetic) that its corresponding dynamics can be reinterpreted as a dynamical spacetime manifold, i.e., the gravitational interaction is described as a pseudo-Riemannian spacetime manifold whose Lorentzian pseudometric is determined by the Einstein field equations. Wheeler dubbed this notion as "geometrodynamics", i.e., spacetime tells the matter and radiation how to move and matter (or rather any kind of energy, momentum, and stress of matter and radiation) tells spacetime "how to curve".

Another point of view is, however, that gravitation is described by a gauge theory as any of the other interactions too.

In the Standard Model the (non-Abelian) local gauge symmetries refer to intrinsic symmetries of the fields describing particles (leptons, quarks, and Higgs boson) and quantized gauge fields (corresponding to gluons, W+Z bosons, and photons), i.e., color, weak isospin, and weak hypercharge symmetries (where the electroweak part is Higgsed making the W+Z bosons massive and keeping the photon massless) based on the local gauge group $\text{SU}(3)_{\text{color}} \times \text{SU}(2)_{\text{wiso}} \times \text{U}(1)_{\text{Y}}$.

The gravitational interaction can be described in a similar way as a gauge theory, but the gauge group is special, because what's gauged in this approach to GR is the Lorentz group, i.e., the Lorentz group is made a local symmetry. This gauging then leads to the notion of both an affine connection and a Lorentzian pseudometric, which in general extends the pseudo-Riemannian (Lorentzian) spacetime manifold to a Einstein-Cartan space with torsion. The latter occurs necessarily as soon as matter fields with spin (particularly spin 1/2) are taken into account. Of course also here you can reinterpret the gravitational interaction as a dynamical spacetime. As long as you only take into account the macroscopically relevant matter (usually described by classical transport theory or hydrodynamics/dust) and the electromagnetic field you end up with the Lorentz manifold (i.e., a pseudo-Riemannian spacetime), i.e., standard GR.

 

[mpresic3]

Most of this by Vanhee's is what I was trying to say, but in rereading my earlier post, I did not express this well at all. Vanhee's however, expressed this with a better understanding than I currently have.

 

[mpresic3]

I anticipate, in the future all interactions will eventually be understood, possibly using differential geometry, perhaps using concepts that are to be determined (TBD).

 

[Frodo]

Summary:: Force or not?

According to general relativity, gravity is not a force, however it is referred as one of the four fundamental forces. This seems like a contradiction.

No and yes. It is also a matter of semantics and how people decide to describe things.

No because the force a body feels when attracted by a massive body is due to the body's interaction with the curved space time created by the massive body as is described by GR. However, describing it as gravitational force due to a law called Newton's Law of Gravity is a very, very, very good approximation to what you measure.

Yes because you can measure the strength of the force with which the body is attracted to the other body.

Also, gravity is a virtual, pseudo or fictitious force.

A virtual, pseudo or fictitious force is a force which can be made to disappear by choosing an appropriate reference frame.

If you are in free fall then you feel no force of gravity because, in your accelerating frame of reference, gravity has disappeared. An example: Imagine you are holding an apple in your hand and you jump off a high building. Assume there is no air resistance. Let go of the apple. The apple falls at exactly the same speed as you and hovers next to you exactly where you let go of it. The apple doesn't fall relative to you. This is exactly what you see in videos from the International Space Station.

The International Space Station orbits at a height of about 300 miles where gravity is about 84% as strong as on the ground. So why is everything weightless? Because everything is in free fall, and in that frame of reference, gravity disappears.

Other virtual, pseudo or fictitious forces include Coriolis force and centrifugal force, both of which can be made to appear or disappear at will by changing reference frame.

Newton's Law of Gravity is a very, very, very close approximation to General Relativity in everything other than extreme circumstances and it agrees extremely well with experiment. IIRC every space mission is calculated based on Newton's Law (apart from GPS, where orbits are calculated according to Newton, but where GR is needed to know what the correct time is as the orbiting clocks run at different rates at different heights).

 

[vanhees71]

I anticipate, in the future all interactions will eventually be understood, possibly using differential geometry, perhaps using concepts that are to be determined (TBD).

This was what Einstein and Schrödinger were hoping at the end of their lives too. Unfortunately it didn't work out.

 

[vanhees71]

A virtual, pseudo or fictitious force is a force which can be made to disappear by choosing an appropriate reference frame.

Then gravity is not a "fictitious force", because a true gravitational field cannot be made to disappear by choosing an appropriate reference frame since if there's a true gravitational field (due to some distribution of energy, momentum, and stress like a star) the curvature tensor doesn't vanish, and there's no way to get the tensor to vanish by choosing any reference frame, because the nonvanishing of the curvature tensor is a frame-independent property.

In terms of physics: Any true gravitational field leads to some tidal forces which are not vanishing in a free-falling reference frame.

 

[Frodo]

In terms of physics: Any true gravitational field leads to some tidal forces which are not vanishing in a free-falling reference frame.

Interesting quibble when one sees the generality of what the OP was asking in a sub-forum entitled Classical Physics and at Basic (High school) level.

Quibbling your quibble, does that apply at a point? Don't tidal forces require an extended region?

 

[Dale]

Just to hopefully nip a pointless semantic argument in the bud:

The Christoffel symbols can be made to vanish at any point by suitable choice of coordinates. This is the origin of the concept that gravity is an inertial force. Other inertial forces also show up as Christoffel symbols.

The Riemann curvature tensor is a coordinate geometric object that cannot be made to disappear by a change of coordinates.

There is no general agreement on what the term “gravity” refers to. It could be the Christoffel symbols (Einstein used that) it could be the curvature tensor or it could be the whole set of observations and theory together. It is perfectly legitimate to use either meaning.

 

[Frodo]

Just to hopefully nip a pointless semantic argument in the bud:

It does seem to me that a number of posters (I have one, not vanheese71, in mind!) are more keen to demonstrate their depth of knowledge than to put themselves in the shoes of the questioner and answer the question without bringing up minutiae which serve only to confuse and digress.

One gives different answers to primary school children, high school students, sixth formers, undergraduates, graduates and post-doctoral researchers.

 

[Vanadium 50]

One gives different answers to primary school children, high school students, sixth formers, undergraduates, graduates and post-doctoral researchers.

This is a B-level thread. I've seen answers with Christoffel symbols, pseudo-Riemannian spacetime, Lie groups, etc. Not exactly B-level stuff.

 

[PeroK]

I guess the answer to "is gravity a force?" can be "yes and no".

Or even: Yes. No. Maybe.

 

[vanhees71]

It's difficult to answer a question at high school level when it comes to a precise meaning and if there are different opinions about interpretational issues.

I never understood, why inertial forces are called fictitious. I never use the notion of "fictitious forces" for "inertial forces". Only with the appropriate math one can really explain, what's meant. As long as the non-inertial frame is not rotating against the inertial frames one can explain this math at the high-school level provided the use of calculus is allowed. For rotating frames you need rotation matrices and their derivatives, which is for sure beyond high-school level.

Concerning gravity it's clear that it is an interaction and not a "fictitious force". Of course you can make the Christoffel symbols vanish at anyone point (that's the mathematical precise formulation of the weak equivalence principle) and you can interpret the Christoffel symbols to represent a "gravitational force" on a test particle from the equations of motion of this test particle.

However, this does not mean that there are no forces acting on the test particle at all also in this local inertial reference frame, because in general the particle will move and then the tidal forces in the neighborhood of the one point where the Christoffel symbold vanish are necessarily there. That's clear from my argument concerning the curvature tensor components above. Of course in many cases the tidal forces are negligible and in this sense in a sufficiently small neighborhood you can say that the particle is approximately force free.

I think, it's important also at the high school level to be precise. Of course you cannot argue with tensor calculus at the high school level, but you can make as precise as possible qualitative statement about the gravitational interaction, and it should be made clear that inertial reference frames only exist locally at any point in spacetime. I don't say, that's a simple task for a high school teacher having not the sharp mathematical language of tensor calculus.

 

[PeroK]

I think, it's important also at the high school level to be precise. Of course you cannot argue with tensor calculus at the high school level, but you can make as precise as possible qualitative statement about the gravitational interaction, and it should be made clear that inertial reference frames only exist locally at any point in spacetime. I don't say, that's a simple task for a high school teacher having not the sharp mathematical language of tensor calculus.

I think it's important to understand the context and limitations of your knowldege. Someone at high school must accept that they have a limited knowledge and it that there is a deeper understanding. What's important is that they think flexibly when confronted with additional knowledge, refinements or subtleties.

I would use IT as an analogy. I worked in database support for many years (Adabas and Oracle). The software developers were taught one thing about the way the database worked, but we knew the next level of detail. It was often a shock when, in the process of resolving a problem, it was revealed to a developer that the database software didn't quite work in the simplistic way they had been taught. Then, occasionally, I would learn something from the technical support at Adabas or Oracle that revealed yet a deeper layer of understanding.

You can follow this in terms of computer hardware as well, back through the layers of understanding of how a disk drive works, all the way back to the QM that underpins the whole thing. There's nothing wrong with a developer seeing a disk as just something to hold data. And, there's nothing wrong with a technical support guru - who knows lots about disk technology - knowing nothing about QM.

 

[Frodo]

I never understood, why inertial forces are called fictitious. I never use the notion of "fictitious forces" for "inertial forces".

May I suggest you start by reading Fictitious force as I think it will greatly assist you in understanding fictitious forces. I gave a few examples in my post above - they may also help your understanding.

If you always do your analyses using an inertial frame of reference you never see fictitious forces.

But, if you do your analysis using a non-inertial frame of reference then you must invoke forces which are not present in inertial frames to calculate the motions. They are called fictitious forces.

A fictitious force (also called a pseudo force, d'Alembert force, or inertial force) is a force that appears to act on a mass whose motion is described using a non-inertial frame of reference, such as an accelerating or rotating reference frame. An example is seen in a passenger vehicle that is accelerating in the forward direction – passengers perceive that they are acted upon by a force in the rearward direction pushing them back into their seats. An example in a rotating reference frame is the force that appears to push objects outwards towards the rim of a centrifuge. These apparent forces are examples of fictitious forces.

Do this simple experiment which I did many, many years ago. Find a childrens' playground with a roundabout on which you can stand. Stand on the roundabout facing the centre pole. Now swing your leg and try kick the central pole. It's easy.

Now have the roundabout set in motion and repeat. You will now personally feel a fictitious force. It comes in from the side and it deflects your leg making it impossible for you to kick the central pole. I suggest you hold on tightly as you are likely to fall over.

Or imagine a reference frame set up on a roundabout. Tie an apple to a piece of string and hold the string.

Stand on the stationary roundabout: the apple hangs vertically in the frame of reference of the roundabout.

Now set the roundabout in motion: the apple no longer hangs vertically. The apple moves away from the centre and, if you get fast enough, the string will be almost horizontal. Now, still using the roundabout as your reference frame, explain why the apple moves away from the centre. You need to invoke a force which only appears when the roundabout is rotating and whose strength is dependent on the angular velocity. Physicists call it a centrifugal force (literally centre-fleeing).

Slow the roundabout and the force disappears. Why don't you now feel the force? Where has it gone?

Or place a golf ball on the roundabout and strike it towards the centre. When the roundabout is stationary, as expected the ball moves in a straight line in the frame of the roundabout - Newton's Laws of Motion. Now set the roundabout in motion and strike the ball. It no longer travels in a straight line in the frame of the roundabout. How do you explain its motion - you imagine up a fictitious force called the Coriolis Force which magically appearers when the frame rotates and deflects the ball.

A fictitious force can be made to appear or disappear at will by changing the frame of reference. Change to a non-rotating frame and the fictitious forces above all disappear.

Back to gravity. Do a thought experiment with a semi-infinite slab of matter an infinite plate of zero thickness and uniform mass density and investigate motion above the plane where the strength of gravity is constant at every location. Go into free-fall. Where does the gravity go? There are now no tidal forces to help you ... [Correction - see below]

If you are wondering why anyone would ever want to do an analysis using a rotating frame of reference consider the problems faced by a helicopter designer. What is the stress on the rotors when the helicopter goes through various aerobatic manoeuvrers?

 

[vanhees71]

I know that fictitious forces is a synonym for inertial forces, and I know what they are. You just shuffle parts of the left-hand side "mass times acceleration" of the Newtonian EoM to the right-hand side "force" and call them inertial forces. They are not fictitious in any sense because they are part of the EoM for coordinates defining a non-inertial frame.

A homogeneous gravitational field is the one example, where the naive interpretation of a gravitational force as an inertial force is correct, but that's an approximation for a sufficiently small region of space. Again, sufficiently small means a region small enough such that tidal forces can be neglected.

 

[Frodo]

Again, sufficiently small means a region small enough such that tidal forces can be neglected.

Did you see my suggestion of considering a semi-infinite slab of mass? You can have your region as big as you want without tidal forces.

 

[vanhees71]

I don't think that a semi-infinite slab would not be subject to tidal forces. Do you have a concrete model in mind?

 

[mpresic3]

It is almost hard to believe this question was asked in May of last year. Despite the importance of strict definitions, physicists need to use language to express their ideas to others. If you are sending a rocket to the moon and addressing physicists and engineering colleagues, it is unhelpful to "explain away" gravity as a pseudoforce. Maybe in an astrophysical or cosmological context, it may be unhelpful to consider gravity as a force. In geodesy, "gravity" is not even the same as "gravitation". The most important point is that all (or at least the majority) of the audience can agree (mostly) with the ideas and arguments presented by the speaker.
It is interesting to speculate whether other languages (I know mostly English, small amounts of Spanish, and Italian), have different words for forces. I understand Eskimos, Aleuts have 19 different words for snow?, and 12 different words for caribou.(?) Posing an well defined physics problem to Eskimos in the context of reindeer and snow might be challenging.

 

[A.T.]

... it is unhelpful to "explain away" gravity as a pseudoforce. ...

The term "inertial force" sounds less like "explaining away".

 

[mpresic3]

To parphrase Vanadium, regarding the B-level thread. I do not know this term with regard to the physics forums. Can one find the level. I suspect many answers including mine are at a level that leaps across levels. Moreover, I think most of the advanced thread writers agree. Clearly, I think most readers should describe a force as a push or pull to a middle to high school student. With this understanding gravity is a force. A few students will go on to take general relativity and learn the refinements.
I remember one of my high school teachers told me not to read about the Bohr atom in a chapter of my biology textbook because I would have to learn the quantum mechanical model in Chemistry the next year. I have read some similar thoughts on these threads that claim that learning a simplified and (slightly) misleading models will cause great harm and limit the student's learning in the future.
I have to say I have never had much of a problem "unlearning" material, and I have not met many people, who felt injured by learning a simple model to begin with. Most of us learn plane geometry a la Euclid in high school, and I do not think it hinders us when (a few of us) learn non-euclidean geometry, but this is material for a different thread

 

[Halc]

Did you see my suggestion of considering a semi-infinite slab of mass? You can have your region as big as you want without tidal forces.

Well, it needs to be actually infinite since the finite case is only locally a more or less uniform field. It defies the 'as big as you want' part.

I don't think that a semi-infinite slab would not be subject to tidal forces. Do you have a concrete model in mind?

Been pondering this question myself, but it seems the correct answer needs to be some solution to Einstein's field equations. Feel free to point out why I'm spouting nonsense, because I'm bound to do that somewhere in this post.

Newtonian physics says the field will be absolutely uniform with no tidal effects. Both (Isaac and Al) say that escape velocity from the slab is infinite, but it isn't like a black hole with some defined event horizon at any point since there's no altitude where escape becomes possible.

Two clocks, one 'above' the other, but resting on (at rest relative to) the slab will run at different rates and thus objects placed at different altitudes will accelerate (relative to a slab observer) at different rates. It isn't an inertial frame, but sort of the equivalent of an accelerated one. Trying to draw on the equivalence principle here. The uniform gravitational field is not similar to a Rindler coordinate system since there is definite spacetime distortion going on here and the Rindler coordinates map what is flat Minkowski spacetime, but relative to a continuously accelerating point object. So the comparison is invalid in my opinion.

Let's see: If there are tidal effects, then a rod with a mass at each end, oriented at 45 degrees to the gravity, and supported (or not) in the center will have a torque on it tending to align it vertical. Such a rod will rotate if held on a tower on Earth, or if in free fall in orbit. Such a rod will tend to rotate if accelerated in flat spacetime, which is Rindler coordinates. But does the rod (supported/accelerated or not) in the gravitational field formed by the infinite slab have any torque on it? I think if it was supported, then yes, since that's a form of acceleration and a gravitational tidal field isn't required for such torque. But in free-fall? The mathematics is beyond me.

 

[Halc]

If you are sending a rocket to the moon and addressing physicists and engineering colleagues, it is unhelpful to "explain away" gravity as a pseudoforce.

That's right. The guys putting a ship on the moon need to get it there with simple computations. They're not looking for the correct answer to this question, they're looking for an easy one that gets their ship there. The Newtonian answer works far better for them.

Clearly, I think most readers should describe a force as a push or pull to a middle to high school student. With this understanding gravity is a force. A few students will go on to take general relativity and learn the refinements.

The OP (who is undoubtedly no longer paying attention to this thread or even this site) did not ask if using gravity-as-a-force mathematics will suffice to get a ship to the moon. He asked which is the more correct answer, which is definitely a relativity question even if he's not actually in a graduate level general relativity course. That correct answer should not detract the student from using the simplified models when appropriate, which for practical purposes, is almost always. An exception is the precise timing needed for function of the GPS system, which must use GR mathematics for both satellite and cell phone, or it just doesn't work.

 

[vanhees71]

The term "inertial force" sounds less like "explaining away".

Once more: Gravity is a fundamental interaction and not a purely inertial force. The latter is an approximation for sufficiently small spacetime regions. The weak equivalence principle tells you that you can in any spacetimepoint use a locally inertial reference frame but not that there exists a global inertial reference frame. If this were true, then all of physics would just be special relativity with a fixed Lorentzian affine space, but GR (in its geometrical interpretation) tells you that at presence of any kind of energy, momentum, and stress distribution (the energy-momentum tensor of all matter and radiation) spacetime is described by a pseudo-Riemannian manifold with non-zero curvature and thus the always existing local inertial frames are only local, which is why they are called local!

 

[Ibix]

I would say that the layman thinks of gravity as "what makes this ball fall if I let go of it". That's precisely the bit of the interaction (equivalent to the force $mg$ in Newtonian physics) that can be transformed away by an appropriate choice of coordinates, so is reasonably called an "inertial force".

In a broader sense, of course, not all effects that we attribute to gravity can be transformed away. With sufficient precision you can detect tidal effects in a small room, and you cannot make those vanish by picking a free falling frame. Non-uniform gravitational fields (such as the Earth's on a relatively large scale) are the explanation for orbits, tides, the shape of celestial bodies, galaxies, and more. It's still not a force (at least not in the usual sense of the word), but it isn't explicable as a coordinate transform (i.e. it's not an inertial force) either.

This is, I think, a restatement of @Dale's #20 in less mathematical terms.

 

[Dale]

Once more:

I am pretty sure that nobody here either disagrees or is unaware of all of this (in particular @A.T. understands). Do you have a specific reason why you are repeating it?

 

[Frodo]

I don't think that a semi-infinite slab would not be subject to tidal forces. Do you have a concrete model in mind?

I apologise profusely as my fingers did not type what my mind was thinking.

I meant an infinite plate of zero thickness and of uniform mass density.

The gravitational force is then uniform throughout the region and there are no tidal forces either parallel or perpendicular to the plate. The proof is given directly or by the equivalent electrical analogy with charge - eg Feynman 5–6 A sheet of charge; two sheets

IIRC your objection was, essentially, that a gravitational force cannot be made to disappear by a change of reference frame because tidal forces are always present. This example is one which disproves your assertion as the gravitational force here can be made to disappear merely by using a frame in free fall and there are no tidal forces.

Again, sufficiently small means a region small enough such that tidal forces can be neglected.

I also questioned "point" and extended "region". Here you can have a region as large as you wish without tidal forces.

Anticipating a possible objection, an infinite plane of zero thickness and uniform mass density is just as valid for analysis as is a light inextensible string passing over a frictionless pulley.

 

[vanhees71]

Sure, that's what (not only) I said above several times.

So the short answer simply is: Gravitation is a fundamental interaction and thus a "true force" and not (only) an "inertial force".

It's even clear in Newtonian mechanics: You can transform away a homogeneous gravitational field by just using the free-falling reference frame. The proof is very simple. If you have a gravitational field $\vec{g}$ that can be considered homogeneous the equation of motion is
$$m \ddot{\vec{x}}=m \vec{g}=\text{const}.$$
Now transform to the freely-falling reference frame, for which
$$\vec{x}=\vec{x}'+\frac{1}{2} \vec{g} t^2$$
you get
$$m \ddot{\vec{x}}=m \ddot{\vec{x}}' + m \vec{g}=m \vec{g} \; \Rightarrow \; m \ddot{\vec{x}}=0.$$
In this sense a homogeneous gravitational field is inertial, because it can be transformed away by going to an accelerated reference frame.

Of course, the true gravitational field of the Earth is not homogeneous and this field cannot be transformed completely away by changing to an accelerated reference frame.

 

[Frodo]

We are in violent agreement.

 

[Halc]

I meant an infinite plate of zero thickness and of uniform mass density.

Why the zero thickness? If the field it produces is uniform then it is not a function of distance from the mass, so the field is a function only of the mass per unit area, not mass per unit volume. That means an zero thickness plate of a million kg/m² (infinite density) exerts the exact same force (at any positive altitude) as a slab of 10,000 km thick foam with mass of a million kg/m² which is a density of 0.1kg/m³

 

[vanhees71]

Well, yes. You can always invent some highly idealized academic example, which cannot be realized in nature ;-)).

 

[Frodo]

Well, yes. You can always invent some highly idealized academic example, which cannot be realized in nature ;-)).

And I thought I had anticipated that objection

First, I'm just following Feynman, someone for whom I have great respect.

Secondly, in what sense is "a light inextensible string passing over a frictionless pulley" not a "highly idealized academic example"? It, and a "point mass" and "a smooth frictionless surface" appear in countless excercises.

I was attempting to validate my assertion that gravity was a virtual force against all those who quibbled with me.

Remember this is a B, or High school level thread.

 

[vanhees71]

I've also the highest respect for Feynman, but what has this to do with the argument that gravity is an interaction and not just an inertial force? Does Feynman make the statement gravity was just an inertial force? How is this then consistent with his marvelous "Feynman lectures on gravitation"?

 

[Dale]

The OP was long since answered and, since this has now degenerated into the semantic argument I had tried to avoid, it is time to close the thread.

It is simply not necessary that everyone use the same terminology, as long as we are all using terminology that is consistent with the literature.