The g_ij as potentials for the gravitational field

[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

$$x^{'i} = x^i + \xi^i (t)$$

the gradient transforms as

$$\partial^{'i} \phi'= \partial^{i} \phi - \ddot{\xi}^i (t)$$

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

$$\partial^{'i} \phi' = 0 \ \ \ \ (locally \ in \ space!)$$

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

$$\partial^{i} \phi = \ddot{\xi}^i (t)$$

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

$$x^{'i} = x^i + \xi^i (t)$$

the gradient transforms as

$$\partial^{'i} \phi'= \partial^{i} \phi - \ddot{\xi}^i (t)$$

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

$$\Gamma^i_{00} = \partial^i \phi$$

and under a coordinate transformation

$$x^{'i} = x^i + \xi^i (t)$$

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

$$\Gamma^{'i}_{00} = \Gamma^i_{00} + \ddot{\xi}^i$$

Hence (only!) locally, one can put

$$\Gamma^{'i}_{00} = 0$$

 

[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".

 

[haushofer]

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".

I don't see why not, considering the correspondence principle. The transformation of the Newtonian potential, (and its gradient) according to the Newtonian limit of GR, must be exactly the same as in "Newtonian gravity".

In Newtonian physics, according to you, how does the Newton potential or its gradient transform if you switch to an accelerating observer? I'd say you derive that from the equivalence principle: in free fall you're weightless, so locally you can set the gradient of the potential to zero by an acceleration. GR makes that clear via the transformation rules of the connection.

So I'd say gravity is also a fictious force in Newtonian physics.

 

[dx]

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.

That is right that you can write the index up or down, but let me explain the point in a different way. I wrote the 1-form as p_i using the letter p because this is indeed the momentum (m = 1 in my equations). Think of the Hamilton-Jacobi theory instead of WKB. The "WKB" approximation in this language corresponds to the high frequency limit or the low wavelength limit applied to expressions of the form exp(S), so the phase here is S, and the phase gradient or "phase velocity" is p_i = ∂_iS, which is just the i component of the 4-momentum. So when I write something like d/ds p_i, I do intend for people to think of it as the left hand side of the relativistic force law

d/ds p_i = F_i

and I want "v_i" "v^j" to be distinguished conceptually, one as velocity and one as momentum.

[ aside: Momentum is indeed in some sense a 1-form (because d/dt p = F and F = -dV), but it is also a vector in another sense (3-momentum is the flux of energy pc = Ev/c). ]

 

[PeterDonis]

The transformation of the Newtonian potential, (and its gradient) according to the Newtonian limit of GR, must be exactly the same as in "Newtonian gravity".

No, it mustn't. See below.

In Newtonian physics, according to you, how does the Newton potential or its gradient transform if you switch to an accelerating observer?

For me to even answer this question, you would have to show me such a transformation explicitly and explain how it is valid in Newtonian physics, i.e,. how the laws of Newtonian physics are invariant under such a transformation. I have not read the references you provided fully, but in my reading of them so far I have seen no such "transformation to an accelerating observer" given for Newtonian physics itself, only for Newtonian limits of relativistic theories, which, as I have said, is not the same thing.

In the Newtonian limit of GR, by contrast, we are not starting with the laws of Newtonian physics and their invariance properties; we are starting from the GR laws of physics and their invariance properties. And we already know that the latter are invariant under a "transformation to an accelerated observer", since they are invariant under any coordinate transformation. So to take the Newtonian limit of GR "for an accelerated observer", one merely needs to find an coordinate choice that can reasonably be called "the frame of an accelerated observer" and then take a limit which can reasonably be called "the Newtonian limit" in those coordinates. Which, as far as I can tell, is basically what is being done in the references you give.

This latter procedure is perfectly fine mathematically; it is just not Newtonian physics.

I'd say gravity is also a fictious force in Newtonian physics.

Sorry, but this is simply wrong. Gravity in Newtonian physics is the force that appears in the equation $F = G m_1 m_2 / r^2$. A fictitious force in Newtonian physics is a force that appears only in non-inertial frames. Gravity in Newtonian physics is not fictitious by that definition since the gravity force is present in inertial frames.

Note that "inertial frames" in Newtonian physics are not the same as inertial frames in GR. In GR, a frame in which you, standing on the Earth's surface, are at rest is not inertial. In Newtonian physics, it is.

 

[haushofer]

No, it mustn't. See below.
For me to even answer this question, you would have to show me such a transformation explicitly and explain how it is valid in Newtonian physics, i.e,. how the laws of Newtonian physics are invariant under such a transformation. I have not read the references you provided fully, but in my reading of them so far I have seen no such "transformation to an accelerating observer" given for Newtonian physics itself, only for Newtonian limits of relativistic theories, which, as I have said, is not the same thing.

In the Newtonian limit of GR, by contrast, we are not starting with the laws of Newtonian physics and their invariance properties; we are starting from the GR laws of physics and their invariance properties. And we already know that the latter are invariant under a "transformation to an accelerated observer", since they are invariant under any coordinate transformation. So to take the Newtonian limit of GR "for an accelerated observer", one merely needs to find an coordinate choice that can reasonably be called "the frame of an accelerated observer" and then take a limit which can reasonably be called "the Newtonian limit" in those coordinates. Which, as far as I can tell, is basically what is being done in the references you give.

This latter procedure is perfectly fine mathematically; it is just not Newtonian physics.
Sorry, but this is simply wrong. Gravity in Newtonian physics is the force that appears in the equation $F = G m_1 m_2 / r^2$. A fictitious force in Newtonian physics is a force that appears only in non-inertial frames. Gravity in Newtonian physics is not fictitious by that definition since the gravity force is present in inertial frames.

Note that "inertial frames" in Newtonian physics are not the same as inertial frames in GR. In GR, a frame in which you, standing on the Earth's surface, are at rest is not inertial. In Newtonian physics, it is.

I agree that the wording "fictitious force" is a bit misleading and confusing in the Newtonian context; I merely mean that Newtonian gravity can always be transformed away by going to an appropriate (accelerating!) frame.

But I'm puzzled by your other remarks. I'd say the correspondence principle dictates that the resulting transformation of the Newton potential in the Newtonian limit of GR is exactly the same as in (what we call) "Newtonian physics". So let me ask you to understand it better: if a student asks you during a class of classical mechanics how the term of Newtonian gravity (gradient of the potential), or Newton's second law for a particle in a gravitational field, changes if you go to an accelerating frame, how would you answer that?

In those references, the action of a particle coupled to the Newton potential is written down, including the behaviour under accelerations. I'd say that pretty much defines the dynamics of a particle in a Newtonian gravity field.

I guess most of our discussion is mainly categorical ;)

 

[PeterDonis]

Newtonian gravity can always be transformed away by going to an appropriate (accelerating!) frame

You still have not shown me a transformation to such an "accelerating frame" that preserves the Newtonian laws of physics. The Newtonian ones, not the "Newtonian limit" ones. Unless and until you do, I will continue to disagree with this claim.

if a student asks you during a class of classical mechanics how the term of Newtonian gravity (gradient of the potential), or Newton's second law for a particle in a gravitational field, changes if you go to an accelerating frame, how would you answer that?

The burden is not on me to answer this question. The burden is on you to show me what an "accelerating frame" is and how transforming to it preserves the Newtonian laws of physics.

I have asked for this in previous posts. You have not provided it.

the action of a particle coupled to the Newton potential is written down. I'd say that pretty much defines the dynamics of a particle in a Newtonian gravity field

Not in an "accelerating frame" unless you can show me what one is. See above.

 

[haushofer]

You still have not shown me a transformation to such an "accelerating frame" that preserves the Newtonian laws of physics. The Newtonian ones, not the "Newtonian limit" ones. Unless and until you do, I will continue to disagree with this claim.
The burden is not on me to answer this question. The burden is on you to show me what an "accelerating frame" is and how transforming to it preserves the Newtonian laws of physics.

I have asked for this in previous posts. You have not provided it.
Not in an "accelerating frame" unless you can show me what one is. See above.

These transformations are given in the references I gave you :)

But let me repeat them here, as I see it (I hope I'm not hijacking this topic; if it's offtopic, feel free to move this discussion to a separate thread). Newton's second law for a particle with fixed mass following a trajectory $x^i(t)$ which undergoes a force $F^i$reads

$$m\ddot{x}^i = F^i$$

This equation is a tensor equation under the Galilei group of transformations on the spatial coordinates, as you can easily check;

$$x^{'i} = R^i{}_j x^j + v^i t + \zeta^i$$

Here, R denotes a constant rotation, v denotes the parameter of the Galilei boost and $\zeta$ denotes a constant translation. Of course, the full Galilei group also contains temporal translations, but let's ignore those. But in the presence of Newtonian gravity, this group is extended. Namely, under an acceleration

$$x^{'i} = x^i + \xi^i (t)$$

the Newtonian expression for the gravitational force $\partial^i \phi$ transforms as

$$\partial^{'i}\phi'(x') = \partial^{i}\phi(x) + \ddot{\xi}^i$$

Namely, if you consider Newton's second law for a point particle (with fixed mass) in a Newtonian gravitational field, then

$$m\ddot{x}^i = m\partial^i \phi \ \ \rightarrow \ \ \ \ddot{x}^i = \partial^i \phi$$

We know that in free fall, an observer becomes weightless due to the equivalence principle (inertial mass equals gravitational mass; see above). If we plug in the transformation, we see that

$$\ddot{x}^{'i} = \partial^{'i} \phi'$$

i.e. Newton's second law for a point particle in a gravitational field is not only tensorial under the Galilei group, but also under accelerations. The same goes for the Poisson equation:

$$\partial_i \partial^i \phi = 4 \pi G \rho$$

To take a concrete example: around the surface of the earth, we know that

$$\partial^i \phi = (0,0,g)$$

for an inertial observer (in the Newtonian sense: having constant velocity) on the ground. So if I go to an accelerating reference frame then via

$$x^{'i} = x^i + \xi^i (t)$$

with

$$\ddot{x}^{'i} = (0,0,-g)$$

i.e. by naming $x^i = (x,y,z)$,

$$\ddot{z}'(t) = -g \ \ \rightarrow \ \ \ z' = z - \frac{1}{2}gt^2$$

(and x'=x, y'=y; this transformation could also contain a Galilei boost and spatial translation, but let's put them to zero) then I'm transforming to an observer in free fall, for which $\partial^{'i}\phi'(x') = 0$. Again, you can also check that the Poisson equation is invariant under this transformation.

As I see it, but maybe this is confusing: compare this with the Schrödinger equation. A Galilei-boost shows that the Schrödinger equation, with the wave function assumed as being scalar, is not invariant under boosts. Then you realize the interpretation of the wave function: it doesn't need to be a scalar under boosts; because of Born's rule, the wave function is allowed to transform under boosts up to a phase factor. With that interpretation ("the wave function is not a scalar under the Galilei group, but forms a projective represenation of the Galilei group" (or actually: Bargmann group)), the Schrödinger equation transforms tensorially under the Galilei group.

In a similar sense, from the equivalence principe (inertial mass equals gravitational mass) one can deduce the transformation of $\partial^{i} \phi$ under accelerations, showing that it is not a scalar under accelerations.

But as I said, maybe we're thinking categorically different about what "Newtonian physics" exactly exhibits. That's why I asked you the question about the student. It's a perfectly legitimit question to ask how the Newtonian equation

$$m\ddot{x}^i = m\partial^i \phi$$

transforms under the coordinate transformation ("an arbitrary acceleration")

$$x^{'i} = x^i + \xi^i (t)$$

right? So I'm just curious how you would answer that question, if you don't agree with my reasoning :)

(added some stuff about the Poisson equation and an explicit example; I did it quickly, so there could be some sign errors)

 

[dx]

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

Actually, the reparametrization invariant action (S₂ = 1/2 ∫ ds²)

$$S_2 = \frac{1}{2}\int (e^{-1} \dot{x}^2 - em^2) d\tau $$

implies that the generally covariant form of the mass shell condition is

$$\dot{x}^2 + e^2 m^2 = 0$$

The introduction of 'e' (which ensures reparametrization invariance) is necessary to understand how to make the mass-shell condition generally covariant.

 

[PeterDonis]

These transformations are given in the references I gave you :)

I already addressed this point. I am not looking for transformations that are valid in a relativistic theory that you take the Newtonian limit of. I am looking for transformations that are valid in Newtonian physics. The two are not the same.

maybe we're thinking categorically different about what "Newtonian physics" exactly exhibits.

I think we're thinking differently about what "Newtonian physics", as opposed to "the Newtonian limit of some relativistic theory", is.

I'm just curious how you would answer that question

I'm not convinced that such a transformation is valid in Newtonian physics (in the sense of leaving all of the laws of Newtonian physics invariant), as opposed to the Newtonian limit of whatever relativistic theory you are considering.

 

[haushofer]

I already addressed this point. I am not looking for transformations that are valid in a relativistic theory that you take the Newtonian limit of. I am looking for transformations that are valid in Newtonian physics. The two are not the same.

But I don't take a limit. These are simply coordinate transformations of the Newtonian equations of motion under which they don't transform tensorially. This is how fictious forces pop up in Newton's second law. Something similar can be done by changing coordinates by performing a time-dependent rotation; in that case, the centifugal and coriolis force pop up in Newton's second law. These rotations are also not part of the Galilei group, but who cares?

These are perfectly legitimate things to do, and not a single reference is made to GR or limits thereof (the transformation only becomes more natural in the context of the connection coefficients) so I don't see your problem. I also don't get how you would explain how a freely falling person feels weightless by considering Newton's 2nd law. But anyway, this is how I view it ;)

 

[PeterDonis]

I also don't get how you would explain how a freely falling person feels weightless by considering Newton's 2nd law.

In Newtonian physics, gravity is simply declared by fiat to be a force that is not felt as weight. In that respect it is like "fictitious forces", even though it is not "fictitious" in Newtonian physics.

If you want to say this is a questionable aspect of Newtonian physics, I would not disagree. GR can be viewed as correcting this questionable aspect by admitting that gravity is a "fictitious force", and taking the view that only forces that are felt as weight are "real" forces.