A The g_ij as potentials for the gravitational field

[PeterDonis]

I'm not sure why you guys feel that the deflection of light is terribly relevant for this discussion

Because you keep talking about it.

the confusion about which symbol is the gravitational field

There is no single quantity that corresponds to "the gravitational field" in GR. That's why, for clarity and precision, one should use math, not vague ordinary language, to describe what one is talking about.

the 'force' concept seems to be the most unambiguous thing to define

The way "force" is defined in GR, gravity is not a force; a force is something that is actually felt as a force. An object moving along a geodesic worldline, which is what you have been discussing, feels no force at all and is in free fall. So nothing you are discussing has anything to do with "force" in the GR sense.

You appear to be using "force" in a different sense, but that sense doesn't work for what you are trying to do. See below.

and the closest thing to the potentials are the forces, which are the gradients of the potentials

If the potentials in GR are the metric coefficients, their gradients can all be made to vanish at a particular event, or along a particular worldline, by an appropriate choice of coordinates. So the intuitive Newtonian idea of "the force is the gradient of the potential" doesn't work in GR.

Note that it doesn't work in general in electromagnetism either. The potential in EM is $A_\mu$, but the force is not the gradient of $A_\mu$, except in the special case of a static Coulomb potential.

 

[dx]

The way "force" is defined in GR, gravity is not a force; a force is something that is actually felt as a force. An object moving along a geodesic worldline, which is what you have been discussing, feels no force at all and is in free fall. So nothing you are discussing has anything to do with "force" in the GR sense.

But in any reference system which is not the 'inertial' free fall frame, there is an equivalent description in terms of fictitious forces. As this idea is embedded in the equivalence principle, there should be some way to retain this idea at least in some sense. See below.

You appear to be using "force" in a different sense, but that sense doesn't work for what you are trying to do. See below.If the potentials in GR are the metric coefficients, their gradients can all be made to vanish at a particular event, or along a particular worldline, by an appropriate choice of coordinates. So the intuitive Newtonian idea of "the force is the gradient of the potential" doesn't work in GR. Note that it doesn't work in general in electromagnetism either. The potential in EM is $A_\mu$, but the force is not the gradient of $A_\mu$, except in the special case of a static Coulomb potential.

I was using the word 'gradient' in a slightly wider sense than what you are talking about here. By 'gradient' I mean some combination of the space and time derivatives of the potentials. Both E = ∇φ and B = ∇×A are such things. (or in more pretty fancy language, F = dA)

In the corresponding equations for gravity

d/ds ( g_ij v^j) = 1/2 g_kl,i v^k v^l

d/ds vⁱ = -1/2[ g_il,k + g_kl,i - g_il,l ] v^j v^k

There are two candidates for the 'gradient of the potential', i.e.

g_il,k + g_kl,i - g_il,l

g_kl,i

to be the analog of the 'gradient' A_j,l - A_l,j which gives the electric and magnetic fields. The fact that this equation can be written in two ways, and there are two candidates for the 'gradient', is obviously connected to that fact that the idea of 'force' is more subtle in general relativity, but as I said before, provided we take the subtleties into account, we may be able to retain some force like interpretation for these things. In fact, these subtleties and what they mean is one thing I want to discuss.

 

[PeterDonis]

in any reference system which is not the 'inertial' free fall frame, there is an equivalent description in terms of fictitious forces

Locally, you could construct a local non-inertial frame centered on the worldline of an object with nonzero proper acceleration (such as a rocket whose engines are firing), and in that frame, the equation of motion for a geodesic would include a "fictitious force" term corresponding to the acceleration of the rocket. (Note that this means "gravity" itself, in GR, is a "fictitious force".)

However, in a curved spacetime (GR), this does not work globally. Even in flat spacetime, a non-inertial frame in which such a "fictitious force" interpretation works (such as Rindler coordinates centered on an object with constant proper acceleration) cannot cover the entire spacetime, only a region of it.

Note also that these non-inertial frames are carefully constructed ones; I don't think you could plausibly defend a "fictitious force" interpretation for any non-inertial frame, since many non-inertial frames will look very, very different from the ones you are intuitively imagining.

we may be able to retain some force like interpretation for these things

No, you can't, because none of "these things" are tensors. You have asked about general covariance, and a key message of general covariance is that anything that depends on your choice of coordinates has no physical meaning, since you can always make it disappear by some choice of coordinates, and general covariance means any choice of coordinates is just as valid, physically, as any other. The physical meaning of the theory is contained in invariants, not coordinate-dependent quantities. The only invariants that correspond to "forces" in GR are quantities associated with non gravitational forces and proper accelerations--things that are actually felt as forces.

 

[pervect]

Locally, you could construct a local non-inertial frame centered on the worldline of an object with nonzero proper acceleration (such as a rocket whose engines are firing), and in that frame, the equation of motion for a geodesic would include a "fictitious force" term corresponding to the acceleration of the rocket. (Note that this means "gravity" itself, in GR, is a "fictitious force".)

Yes, I agree. The other noteworthy effect is that such an accelerated frame, constructed in the standard manner, will have clocks ticking at different rates depending on their "heights". This is a new effect in special relativity that is not present in Newtonian physics. It's an effect that does not fall into the "force" paradigm.

The idea that the only difference between an accelerating frame and a non-accelerating one is the addition of fictitious forces works for Newtonian physics, but it doesn't work for special relativity.

 

[haushofer]

If the potentials in GR are the metric coefficients, their gradients can all be made to vanish at a particular event, or along a particular worldline, by an appropriate choice of coordinates. So the intuitive Newtonian idea of "the force is the gradient of the potential" doesn't work in GR.

Why is that an argument? The same argument applies to the Newtonian potential, i.e. the equivalence principle also holds there, albeit restricted to spatial accelerations.

 

[PeterDonis]

The same argument applies to the Newtonian potential

No, it doesn't. You can't make the gradient of the Newtonian potential vanish by changing coordinates in Newtonian physics.

the equivalence principle also holds there

The equivalence principle holds in Newtonian physics in the sense that gravitational mass is assumed to be equal to inertial mass. But in Newtonian physics, that is an added assumption that is not linked to anything else. In GR it is not an assumption at all; it is an unavoidable consequence of spacetime geometry.

 

[dx]

Einstein's viewpoint, which is not generally the one adopted in textbooks, is that even in a flat spacetime, if you choose curvilinear coordinates, there is no way to say that these coordinates are the truly non-inertial ones. The idea of an 'inertial frame' has absolutely no objective significance. Since the physical equations are generally covariant, any reference frame has equal right to claim that it is the 'nicest' one. The laws have the exact same form in any system of coordinates, so just by making measurements, you cannot say 'this system of coordinates is truly curvilinear.' Rather, we say, "This system is also inertial, except it is filled with a uniform gravitational field." The simple minded idea that 'gravity is simply the curvature of spacetime' I think, though convenient and good enough for most purposes, does not do justice to general relativity. Such an approach must be supplemented by caveats about gauge invariance, in order to make any sense. The reason I say this is the following: Take two systems of reference A and B. In both A and B, the equation of motion of particles is

maⁱ = -mΓⁱ_jk vⁱ v^j

If, in A, the connection coefficients all happen to be zero, we want to say that it is the 'more nice' system of coordinates. We want to say that the field Γⁱ_jk = 0 is somehow special, because it seems to be distinguished from the non-zero fields by being connected to the physical 'flatness' of this particular spacetime solution. This, according to Einstein, is foreign to the spirit of general relativity. The spirit of general relativity is, 'all coordinate systems are equally good.' Even though the system A seems more special, Einstein would consider both systems as having equal right to claiming that it is the 'most nice.' He came to this conclusion from something called the 'hole paradox' and the associated 'point coincidence argument.' The upshot of these arguments is that the coordinate system is simply a book keeping device in order to label these coincidences, and no way of doing this can claim to be 'more natural.'

No, it doesn't. You can't make the gradient of the Newtonian potential vanish by changing coordinates in Newtonian physics.

Both vanishing Γⁱ_jk and the non-zero Γⁱ_jk have equal right and equal standing as representing the physical situation, and neither can claim to be nicer than the other. I realize that this is not stressed in textbooks (which invariably never discuss gauge invariance at all), but let me share the following quote to support my claims about Einstein's views.

How sound, however, Mach's critique is in essence can be seen particularly dearly from the following analogy. Let us imagine people construct a mechanics, who know only a very small part of the Earth's surface and who also can not see any stars. They will be inclined to ascribe special physical attributes to the vertical dimension of space (direction of the acceleration of falling bodies) and, on the ground of such a conceptual basis, will offer reasons that the Earth is in most places horizontal. They might not permit themselves to be influenced by the argument that as concerns the geometrical properties space is isotropic and that it is therefore supposed to be unsatisfactory to postulate basic physical laws, according to which there is supposed to be a preferential direction; they will probably be inclined (analogously to Newton) to assert the absoluteness of the vertical, as preyed by experience as something with which one simply would have to come to terms. The preference given to the vertical over all other spatial directions is precisely analogous to the preference given to inertial systems over other rigid co-ordination

 

[PeterDonis]

The idea of an 'inertial frame' has absolutely no objective significance.

This is obviously false. An inertial frame is a frame in which a body with zero proper acceleration is either at rest or moving with constant coordinate velocity (zero coordinate acceleration). This is a perfectly objective criterion, and I have seen nothing to indicate that Einstein was unaware of it or did not appreciate its significance. The only difference between flat and curved spacetime in this respect is that inertial frames can be global only in flat spacetime; in curved spacetime they can only be local.

we say, "This system is also inertial, except it is filled with a uniform gravitational field."

I don't think anyone in the GR literature has ever said this, including Einstein. Einstein did describe frames "filled with a uniform gravitational field", but I am not aware that he ever called them "inertial".

Note that in Newtonian physics, the definition of an "inertial frame" is different from its definition in relativity, because in Newtonian physics, gravity counts as a force. So it makes sense in Newtonian physics to say that, for example, a frame in which you, sitting on the Earth's surface, are at rest is "inertial", because you can account for the coordinate acceleration of a dropped rock by saying that the force of gravity is pulling on it. But in GR, you can't do this; a frame in which you, sitting on the Earth's surface, are at rest is not inertial in GR, which includes not being "inertial with a uniform gravitational field".

let me share the following quote

From where? Please give a reference.

to support my claims about Einstein's views

It doesn't. Einstein is not saying that non-inertial frames are indistinguishable from inertial frames, or that there is any such thing as "an inertial frame with a uniform gravitational field", instead of simply calling such a frame non-inertial. He is only saying that the laws of physics should not be any different in inertial frames and non-inertial frames.

 

[dx]

This is obviously false. An inertial frame is a frame in which a body with zero proper acceleration is either at rest or moving with constant coordinate velocity (zero coordinate acceleration). This is a perfectly objective criterion, and I have seen nothing to indicate that Einstein was unaware of it or did not appreciate its significance.

The following quote is from "The Meaning of Relativity".

"The weakness of the principle of inertia lies in this, that it involves an argument in a circle: a mass moves without acceleration if there are no forces; we know that there are no forces only by that fact that it moves without acceleration" - Einstein

From where? Please give a reference.

"Autobigraphical Notes"

 

[PeterDonis]

The following quote is from "The Meaning of Relativity".

Einstein is talking about the Newtonian definition of "inertia" in terms of "a body that is not subject to any forces". Also, here Einstein is using "acceleration" to mean "coordinate acceleration". In short, Einstein is pointing out inherent problems with the Newtonian definition of "inertia".

In relativity, however, "inertial frame" is not defined this way. It is defined as I said: a body with zero proper acceleration is at rest or moving at a constant coordinate velocity. "Zero proper acceleration" is a direct observable: you can measure it with an accelerometer. It does not require any definition of "force" or any choice of coordinates, and so there is no circularity in defining an inertial frame this way: we can pick out which objects are "moving inertially" (have zero proper acceleration) without having to already know what an "inertial frame" or a "force" is and without having to measure any coordinate acceleration.

 

[pervect]

I'm not sure why you guys feel that the deflection of light is terribly relevant for this discussion ("cannot be ascribed to a force", connection of time dilation to forces etc.),

It's mainly a few specific example of where the force paradigm you seem to be attached to breaks down. Pointing out abstractly that it breaks down is one thing, but I thought a few more specific examples would help. So far, I can't say that I feel like you appreciate the point in quesiton.

The other example where the force paradigm breaks down is "gravitational time dilation". That phenomenon is one that occurs in accelerated frames in SR and some situations in GR, but does not naturally (or unnaturally, as far as I know) arise from any amount of musing about forces. It's an example of why there is more to Christoffel symbols than "forces".

but I have some remarks about this which may be relevant. The geodesic equation can also be written in the form

(d/ds) p_i = 1/2 ∂g^jk/∂xⁱ p_j p_k

Why $p_i$ and not $u^i$? $u^i$ being the 4-velocity. I believe it's also necessary that one impose $|u^i u_i| = 1$, but I'd have to look up the details.

 

[haushofer]

No, it doesn't. You can't make the gradient of the Newtonian potential vanish by changing coordinates in Newtonian physics.

You mean globally? Under an acceleration

<br /> x^{&#039;i} = x^i + \xi^i (t)<br />

the gradient transforms as

<br /> \partial^{&#039;i} \phi&#039;= \partial^{i} \phi - \ddot{\xi}^i (t)<br />

such that locally (where we can approximate $\partial^{i} \phi$ to be constant) we can put

<br /> \partial^{&#039;i} \phi&#039; = 0 \ \ \ \ (locally \ in \ space!)<br />

by solving, for a given $\partial^i \phi (x)$,

<br /> \partial^{i} \phi = \ddot{\xi}^i (t)<br />

for the acceleration $\xi^i (t)$. But maybe I'm misunderstanding your statement.

 

[PeterDonis]

You mean globally?

No, I mean locally. Consider just the radial direction for simplicity. If $x$ is the spatial coordinate in a frame in which the Earth is at rest, then $x' = x + v t$ is a Galilean transformation, and $v$ is the relative velocity, not an acceleration. An object that is free-falling downward due to the Earth's gravity will only be at rest in the transformed frame for a single instant. The gradient transforms like this:

$$
\frac{\partial}{\partial x'} = \frac{d x}{dx'} \frac{\partial}{\partial x} + \frac{d t}{dx'} \frac{\partial}{\partial t}
$$

Since $dx / dx' = 1$ and $dt / dx' = 0$, the gradient is unchanged by this transformation.

 

[PeterDonis]

Under an acceleration

<br /> x^{&#039;i} = x^i + \xi^i (t)<br />

the gradient transforms as

<br /> \partial^{&#039;i} \phi&#039;= \partial^{i} \phi - \ddot{\xi}^i (t)<br />

What kind of transformation is this? Certainly not a Galilean transformation. But the Newtonian laws of physics are only invariant under Galilean transformations.

 

[dx]

Why $p_i$ and not $u^i$? $u^i$ being the 4-velocity. I believe it's also necessary that one impose $|u^i u_i| = 1$, but I'd have to look up the details.

It is essentially to distinguish between group velocity and phase velocity. The group velocity v = g(_,p) where p is the phase velocity. There are several ways to work with these problems, but for the point I am trying to make, it is convenient to treat the motion of particles in the WKB approximation, or the 'geometrical optics limit' of waves described by the 'dispersion relation'

g(p, p) = 1

So the group velocity

d/ds = -g^ijp_j (∂/∂xⁱ)

represents the 4-velocity of the particle, i.e. the tangent vector to the curve s → xⁱ(s), i.e. vⁱ = -g^ijp_j. Differentiating the dispersion relation, we get

∂_k (g^ij p_i p_j) = g^ij_,k p_i p_j + 2g^ijp_i ∂_kp^j

Because p is a phase velocity, we have ∂_kp_j = ∂_jp_k, so we get the equation

∂_jp_k (g^ij p_i) = -1/2 g^ij_k p_ip_j

The left hand side is the derivative of p_k in the direction of the tangent vector d/ds = -g^ijp_j (∂/∂xⁱ), so we are left with

d/ds ( g_ij p^j) = 1/2 g_kl,i p^k p^l

The only reason I was talking about phase velocity and group velocity is to distinguish between 1-forms p_j and vectors vⁱ = -g^ijp_j. This is what allows us to write ∂_kp_j = ∂_jp_k. If you keep this distinction in mind, you can forget about the whole 'geometrical optics limit' interpretation.

 

[PeterDonis]

It is essentially to distinguish between group velocity and phase velocity.

We are not talking about waves, we are talking about particles. There is no such thing as "group velocity" vs. "phase velocity" for particles.

The only reason I was talking about phase velocity and group velocity is to distinguish between 1-forms p_j and vectors vⁱ = -g^ijp_j.

That distinction has nothing to do with "phase velocity" vs. "group velocity" for particles. See above.

Also, there is no minus sign in the transformation from 1-form to vector.

 

[dx]

The WKB approximation treats particles as wave-packets, so there is indeed a distinction. phase velocity is a 1-form. But as I said, this is not the essential thing in my post. An object with a superscript is a tangent vector, and an object with a subscript is a 1-form. This is all that is needed to understand my post.

 

[PeterDonis]

The WKB approximation treats particles as wave-packets

This is the relativity forum, not the quantum physics forum. Discussion of this kind of thing belongs in the quantum physics forum.

this is not the essential thing in my post

Yes, and it is also off topic for this forum, as above.

An object with a superscript is a tangent vector, and an object with a subscript is a 1-form.

Yes, anyone participating in an "A" level thread in the relativity forum already knows this.

 

[dx]

Yes, and it is also off topic for this forum, as above.

The algebra is easier to do in the WKB approximation picture, talking about 'dispersion relation' etc. (It is useful conceptually also, but I won't use this if you feel it is off topic) The 'dispersion relation' is of course nothing but the mass-shell condition, so you can do it that way also. The equation I wrote can be derived by a different procedure, but I wanted to avoid that. It has nothing to do with the WKB thing, it is equivalent to the usual form of the geodesic equation, and general relativity books derive it in a slightly different way.

 

[PeterDonis]

The algebra is easier to do in the WKB approximation picture.

It's not the "WKB approximation picture". It's the "1-form picture". This is the relativity forum.

The equation I wrote can be derived by a more complicated procedure, but I wanted to avoid that.

This is an "A" level thread in the relativity forum. That means all participants are assumed to know what the geodesic equation is and how to raise and lower indexes and what vectors and 1-forms are. There is no need to belabor any of these points.

What @pervect was asking was why you prefer the 1-form picture to the vector picture for the geodesic equation. The only substantive answer you have provided so far that is even applicable to a discussion in this forum is "the algebra is easier to do", which is just your subjective opinion and doesn't make much sense to me--raising and lowering indexes is a simple operation and doesn't change the algebraic complexity of any equations.

 

[PeterDonis]

@love_42 this thread seems to be losing focus. You have said you want to discuss general covariance, but it's not clear what specific issue regarding general covariance you want to discuss, and we have gone off on several side issues now. If you do not have a clear, specific question that can be answered, this thread will be closed.

 

[dx]

The reason I "prefer the 1-form picture" is that it is necessary to understand the difference between the two expressions d/ds vⁱ and d/ds ( g_ij v^j) which occur on the left hand sides of the two forms of the geodesic equation:

d/ds ( g_ij v^j) = 1/2 g_kl,i v^k v^l

d/ds vⁱ = -1/2[ g_il,k + g_kl,i - g_il,l ] v^j v^k

The fact that one of them has a g_ij within the d/ds derivative is of course the essential thing to understand their difference.

 

[PeterDonis]

The reason I "prefer the 1-form picture" is that it is necessary to understand the difference between the two expressions d/ds vⁱ and d/ds (v_ig^ij) which occur on the left hand sides of the two forms of the geodesic equation:

d/ds ( g_ij v^j) = 1/2 g_kl,i v^k v^l

d/ds vⁱ = -1/2[ g_il,k + g_kl,i - g_il,l ] v^j v^k

There is no difference physically; you're just raising and lowering indexes.

The actual physics is that you have a geodesic worldline, and that worldline has a tangent vector. It does not have a tangent 1-form. Playing tricks with index gymnastics does not change that.

If you want to analyze a classical wave, instead of a classical particle, then you are using the wrong equation to begin with. A classical wave does not have a worldline at all, and the geodesic equation does not describe a classical wave. A classical wave has wave fronts, which can be described by a 1-form, but it makes no sense to say a 1-form is "traveling on a geodesic".

 

[dx]

The difference between these two equations is not just a matter of raising and lowering indices. They are the euler-lagrange equations obtained from the actions

S₁ = -m ∫ ds

S₂ =1/2 ∫ ds²

One involves a square root, and one doesn't. These correspond to the 'nambu-goto' and 'polyakov' actions which are analogous to S₁ and S₂ respectively. One is reparametrization invariant and one is not, so obviously this is relevant for our discussions of general covariance. The actions are indeed equivalent, but conceptually they are different.

 

[PeterDonis]

These correspond to the 'nambu-goto' and 'polyakov' actions which are analogous to S1 and S2 respectively.

Ah, I see now what you intended. Yes, mathematically these two actions are different, although physically they are the same. However, the difference has nothing to do with 1-forms vs. vectors; you can just as easily write the equation of motion arising from either action with the free index either raised or lowered.

One is reparametrization invariant and one is not, so obviously this is relevant for our discussions of general covariance.

Reparameterization invariance is not the same as invariance under coordinate transformations. They are two different things.

 

[haushofer]

What kind of transformation is this? Certainly not a Galilean transformation. But the Newtonian laws of physics are only invariant under Galilean transformations.

These are accelerations. In the presence of gravity the Galilei group of symmetries is extended with these accelerations. If you write down the action of a point particle coupled to the Newton potential, this action is invariant under the group of Galilei-transformations plus accelerations, considering the transformation of the Newton potential.

But of course, I agree that a boost won't do to cancel gravity. The Newton potential is a scalar field under the Galilei group, but not under accelerations.

Just a question: acording to you, if you take the textbook Newtonian limit of GR and gauge fixing coordinates, to what group of coordinate transformations are the gct's broken down?

 

[PeterDonis]

In the presence of gravity the Galilei group of symmetries is extended with these accelerations. If you write down the action of a point particle coupled to the Newton potential, this action is invariant under the group of Galilei-transformations plus accelerations, considering the transformation of the Newton potential.

Do you have a reference for this? It's not something I am familiar with.

if you take the textbook Newtonian limit of GR and gauge fixing coordinates, to what group of coordinate transformations are the gct's broken down?

I don't understand. If you've gauge fixed the coordinates, there is no remaining freedom in the choice of coordinates.

 

[haushofer]

Do you have a reference for this? It's not something I am familiar with.

I don't understand. If you've gauge fixed the coordinates, there is no remaining freedom in the choice of coordinates.

For the point particle coupled to the Newtonian potential, see e.g. eqns 2.10 and 2.11 of

https://arxiv.org/abs/1206.5176

If you take the Newtonian limit, you gauge fix the coordinates, but not completely (!). So the gauge is only partially fixed. So in the Newtonian limit the the group of gct's breaks down to, if I remember correctly, the Galilei group plus arbitrary accelerations. These accelerations are the reason why the equivalence principle stil holds. It would be quite odd if, in this Newtonian limit, suddenly one cannot use the equivalence principle anymore, right? One can still be in free fall, eliminating every appearance of the Newton potential. Since the Newton potential is a scalar under Galilei transformations, you need something more ;)

See also section 3.4 and 3.5 of my thesis,

https://www.rug.nl/research/portal/...ed(fb063f36-42dc-4529-a070-9c801238689a).html

From this perspective, it's a bit strange that most textbook-treatments of the Newtonian limit in GR don't explicitly mention the remaining group of coordinate transformations. In the end, you end up with the only non-zero component of the connection being

<br /> \Gamma^i_{00} = \partial^i \phi<br />

and under a coordinate transformation

<br /> x^{&#039;i} = x^i + \xi^i (t)<br />

one gets (don't pin me down on the sign)

<br /> \Gamma^{&#039;i}_{00} = \Gamma^i_{00} + \ddot{\xi}^i<br />

Hence (only!) locally, one can put

<br /> \Gamma^{&#039;i}_{00} = 0<br />

 

[PeterDonis]

@haushofer thanks for the references, I'll take a look.

 

[PeterDonis]

If you take the Newtonian limit

Taking the "Newtonian limit" of some relativistic theory (I note that in the references you gave, you talk about strings and branes and don't seem to be solely discussing Newtonian limits of standard GR) is not the same thing as Newtonian physics. In Newtonian physics, gravity is a force, and is the gradient of the Newtonian potential; those things are not coordinate-dependent quantities in Newtonian physics. The property of "fictitious forces" that can be transformed away by changing coordinates does not apply to gravity in Newtonian physics. The fact that it possibly does apply in "Newtonian limits" of relativistic theories does not change that fact.

The original comment that I made that started this subthread was about "the Newtonian idea of force as the gradient of a potential" as applied to gravity; I said that idea doesn't apply in GR. You are basically trying to argue that it doesn't apply in Newtonian physics either; but your arguments are based on conflating actual Newtonian physics with the "Newtonian limit" of some relativistic theory. In actual Newtonian physics, as I said above, gravity is a force, and is the gradient of the Newtonian potential, and those quantities cannot be transformed away by changing coordinates. If you want to construct a "Newtonian limit" in which those statements are not true, that's fine, but you cannot claim that your construction is "Newtonian physics".