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

[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.