A The g_ij as potentials for the gravitational field

[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

<br /> m\ddot{x}^i = F^i<br />

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

<br /> x^{&#039;i} = R^i{}_j x^j + v^i t + \zeta^i<br />

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

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

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

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

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

<br /> m\ddot{x}^i = m\partial^i \phi \ \ \rightarrow \ \ \ \ddot{x}^i = \partial^i \phi<br />

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

<br /> \ddot{x}^{&#039;i} = \partial^{&#039;i} \phi&#039;<br />

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:

<br /> \partial_i \partial^i \phi = 4 \pi G \rho<br />

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

<br /> \partial^i \phi = (0,0,g)<br />

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

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

with

<br /> \ddot{x}^{&#039;i} = (0,0,-g)<br />

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

<br /> \ddot{z}&#039;(t) = -g \ \ \rightarrow \ \ \ z&#039; = z - \frac{1}{2}gt^2<br />

(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

<br /> m\ddot{x}^i = m\partial^i \phi<br />

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

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

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.

 

[dx]

Thanks to everyone who participated in this thread. It was interesting to hear the different views on gauge invariance in GR. I have been reading up on this during the course of this discussion, and I realized that I need to spend some more time studying certain aspects of differential geometry before proceeding further, especially Cartan's approach. The question I posed in the OP seems to be related to the following formula of Cartan. Just like F is obtained from the electromagnetic potential A by taking its exterior derivative dA, there is a way to take the exterior derivative of the gravitational potentials using Cartan's methods:

dg_ij = ω_ij + ω_ji

where the ω_ij are the 'connection 1-forms' ωⁱ_j = Γⁱ_jk θ^k.

The θ^k are Cartan's "moving frames"

 

[haushofer]

Thanks to everyone who participated in this thread. It was interesting to hear the different views on gauge invariance in GR. I have been reading up on this during the course of this discussion, and I realized that I need to spend some more time studying certain aspects of differential geometry before proceeding further, especially Cartan's approach. The question I posed in the OP seems to be related to the following formula of Cartan. Just like F is obtained from the electromagnetic potential A by taking its exterior derivative dA, there is a way to take the exterior derivative of the gravitational potentials using Cartan's methods:

dg_ij = ω_ij + ω_ji

where the ω_ij are the 'connection 1-forms' ωⁱ_j = Γⁱ_jk θ^k.

The θ^k are Cartan's "moving frames"

If you want an (hopefully clear :P ) exposition of how GR can be treated as a gauge theory, see

https://arxiv.org/abs/1011.1145

This paper applies a similar method to obtain so-called Newton-Cartan theory, but the relativistic case is also repeated for comparison. I based this Insight on that part of the paper,

https://www.physicsforums.com/insights/general-relativity-gauge-theory/

This method can also be applied to obtain, e.g., N=1 supergravity.

There are different approaches to regard GR as a gauge theory, but coming from a supergravity-background I find this approach the most clear one.

 

[haushofer]

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.

What I meant, was: how would you say that the equation

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

transforms under the transformation

<br /> x^{&#039;i} = x^i + \frac{1}{2}at^2<br />

for a constant acceleration a? If in Newtonian gravity a freely falling observer feels weightless, this should translate to $\partial^{'i} \phi'(x')=0$ for this accelerated observer, right? I don't get the reason why you doubt that this transformation is "valid" in the Newtonian context. The Newtonian equations of motion are tensorial under a certain group of transformations, and you can just calculate how these equations transform under whatever transformation you'd like to consider. I mean, the concept of "tensors", "tensor equations" and "coordinate transformations" is not inherently GR. If I can perform a boost on Newton's 2nd law and the Poisson equation, why not an acceleration?

 

[PeterDonis]

how would you say that the equation

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

transforms under the transformation

<br /> x^{&#039;i} = x^i + \frac{1}{2}at^2<br />

for a constant acceleration a?

By the same chain rule reasoning used before, since $dx / dx' = 1$ and $dt / dx' = 0$, this transformation should leave the spatial gradient invariant. (I'm limiting to one spatial dimension for simplicity.) So we should have $\partial / \partial x' = \partial / \partial x$.

Since $x'$ is a function of both $x$ and $t$, $\phi'$ must be a function of both $x'$ and $t'$ (whereas in the original coordinates $\phi$ was only a function of $x$ since the field is static in those coordinates). But in any surface of constant $t'$, which will be the same as a surface of constant $t$, because the coordinate transformation is $t' = t$, the spatial gradient of $\phi'$ will be nonzero because, as above, the spatial gradient is left invariant by the transformation, and the spatial gradient of $\phi$ in the same surface is nonzero everywhere (for every value of $x$).

Where your transformation gets complicated is the time derivative; by the chain rule, we have

$$
\frac{\partial}{\partial t'} = - a t' \frac{\partial}{\partial x} + \frac{\partial}{\partial t}
$$

So transforming the equation of motion is messy, and I haven't worked through that. But the reasoning above is sufficient to show that the spatial gradient of $\phi'$ cannot vanish.

 

[haushofer]

By the same chain rule reasoning used before, since $dx / dx' = 1$ and $dt / dx' = 0$, this transformation should leave the spatial gradient invariant. (I'm limiting to one spatial dimension for simplicity.) So we should have $\partial / \partial x' = \partial / \partial x$.

Since $x'$ is a function of both $x$ and $t$, $\phi'$ must be a function of both $x'$ and $t'$ (whereas in the original coordinates $\phi$ was only a function of $x$ since the field is static in those coordinates). But in any surface of constant $t'$, which will be the same as a surface of constant $t$, because the coordinate transformation is $t' = t$, the spatial gradient of $\phi'$ will be nonzero because, as above, the spatial gradient is left invariant by the transformation, and the spatial gradient of $\phi$ in the same surface is nonzero everywhere (for every value of $x$).

Where your transformation gets complicated is the time derivative; by the chain rule, we have

$$
\frac{\partial}{\partial t'} = - a t' \frac{\partial}{\partial x} + \frac{\partial}{\partial t}
$$

So transforming the equation of motion is messy, and I haven't worked through that. But the reasoning above is sufficient to show that the spatial gradient of $\phi'$ cannot vanish...

... if you assume that the potential transforms homogeneously.

But I agree it's not straight forward , and the details have been a while for me, so I have to check my thesis again for that ;)

But let's assume your conclusion is right. That would mean a freely falling observer would still measure some sort of gravity, right? For me that's a clear hint your conclusion should be wrong (If we choose the acceleration to be g, the local value of the gradient of phi in the z-direction)

 

[haushofer]

Let me rephrase it even differenty, under the adagium "first think, then calculate". We know that the gradient of phi gives us the acceleration for a mass m in a gravitational field. This is independent of m, because inertial mass equals gravitational mass. So going to an accelerated observer with exactly that gravitational acceleration, i.e. freely falling, should allow us to transform the gradient of phi away (like fictitious forces can be transformed away by going to an inertial frame).

We also know how the left hand side of Newton's second law transforms under an acceleration. From this we can deduce the inhomogeneous transformation of the gradient of phi (and phi itself).

Is there something wrong with this reasoning? Or inherently "non-Newtonian"?

 

[PeterDonis]

That would mean a freely falling observer would still measure some sort of gravity, right?

No, it means Newton's laws are not invariant under your proposed transformation.

 

[PeterDonis]

We know that the gradient of phi gives us the acceleration for a mass m in a gravitational field if we are working in an inertial frame as Newton defines that term.

See the bolded addition above. You are simply assuming that that bolded addition can be removed. But you can't just help yourself to that assumption. The math I posted previously suggests that that assumption is wrong.

 

[haushofer]

No, it means Newton's laws are not invariant under your proposed transformation.

Well, I have to think about that, but If we assume you're right then we have to conclude that the Newtonian limit of GR predicts another transformation law for the gradient of phi then the Newtonian theory itself. That would make this Newtonian limit empirically different from Newton's theory. Also, standard Newton-Cartan theory would also be empiricaly different from ordinary Newtonian gravity; there the exact same transformation law for the gradient of phi (and phi itself) can be derived.

 

[haushofer]

See the bolded addition above. You are simply assuming that that bolded addition can be removed. But you can't just help yourself to that assumption. The math I posted previously suggests that that assumption is wrong.

No, it doesn't, because you assumed that the gradient of phi transforms homogeneously under accelerations. But that's exactly the point; according to the equivalence principle, it can't.

If I remember correctly, the subtlety we are dealing with can be classified under "symmetries" v.s. "pseudosymmetries" of the action. The latter are symmetries which are not induced by gct's, see eqn.3.53 of my thesis.

I think I can state the situation as follows: one can introduce the transformation of the gradient of phi by virtue of the equivalence principle. This symmetry is, in terms of field theory, a s-called "pseudosymmetry" of the action in which the particle is coupled to the potential. The transformation can also be derived in the context of the Newtonian limit of GR (by virtue of the non-homogeneous transformation of the connection), Newton-Cartan theory and the gauging of the Bargmann algebra which is described in my thesis. Because al of these theories should be empirically indistinguishable, the proposed pseudosymmetry of phi (with its corresponding transformation, see eqns. 2.16, 3.52 and 3.53, 5.8-5.10, and the discusson after eqn.5.22 of my thesis) is justified.

I hope I made myself clear, and will think more about your comments. Anyway, if you're fed up with it I can understand, but thanks anyway for the nice discussion :)
It's fun to see that already Newtonian gravity can be quite subtle and that you don't need any GR for that ;)

 

[PeterDonis]

you assumed that the gradient of phi transforms homogeneously under accelerations

No, I didn't assume anything about how the gradient transforms. I demonstrated how it transforms, using the properties of the coordinate transformation you specified.

according to the equivalence principle, it can't.

Wrong. The equivalence principle makes no guarantee that your coordinate transformation leaves Newton's laws invariant. If it turns out that $a = \nabla \phi$, which is one of Newton's laws, does not hold in your transformed coordinates, the proper conclusion to draw is that your transformation does not leave Newton's laws invariant, not that you have somehow magically dictated that $\nabla \phi$ transforms differently from how the math of your coordinate transformation says it does.

 

[PeterDonis]

If we assume you're right then we have to conclude that the Newtonian limit of GR predicts another transformation law for the gradient of phi then the Newtonian theory itself.

Not at all. GR is not required to leave Newton's laws invariant under arbitrary coordinate transformations. It is only required to leave the Einstein Field Equation invariant under arbitrary coordinate transformations. If your proposed coordinate transformation is valid, GR will accept it just fine, and if that transformation does not leave Newton's laws invariant, GR will tell you so.

 

[haushofer]

No, I didn't assume anything about how the gradient transforms. I demonstrated how it transforms, using the properties of the coordinate transformation you specified.

But how can you do that? You don't know a priori how $\phi$ transforms! That's the whole point. I would call it magic if somehow you can tell me, without physical principles, how $\phi$ transforms under accelerations. So I don't see how you can "demonstrate how it transforms". That's why we have to deduce from the equivalence principle how we expect it to transform. In Newtonian physics, this cannot be deduced from the coordinate transformation alone (what I explain in my thesis), and in GR it follows naturally from the transformation of the connection components.

Another example from relativistic field theory: imagine I give you the expression

<br /> \partial_{\mu} \Phi(t,x)<br />

and I tell you how the field $\Phi(t,x)$ transforms under Lorentz transformations (i.e. in which representation of the Lorentz group it sits) but not how it transforms under accelerations. How would you then conclude that an acceleration cannot be used to put $\partial_{\mu} \Phi(t,x)= 0$? You can't.

Of course, if there is no physical reason to expect that one could, it would be ad hoc to impose such a weird inhomogeneous transformation rule. But in the case of Newtonian gravity it isn't; see below.

Wrong. The equivalence principle makes no guarantee that your coordinate transformation leaves Newton's laws invariant. If it turns out that $a = \nabla \phi$, which is one of Newton's laws, does not hold in your transformed coordinates, the proper conclusion to draw is that your transformation does not leave Newton's laws invariant, not that you have somehow magically dictated that $\nabla \phi$ transforms differently from how the math of your coordinate transformation says it does.

No, not "magically", but by using the physical principe of the equivalence principle. I take that as the defining principle of how the field $\phi$ should transform, both in Newton's 2nd law and in the Poisson equation.

I don't see any problem with that. For instance, we also dictate on physical grounds (!) that the Newton potential $\phi$ transforms as a scalar under the Galilei group. Is that also by magic? I would call that physical principles. By stating that $\phi$ transforms as a scalar under the Galilei group, we then can deduce the transformation law for $\partial_i \phi(t,x)$ under the Galilei transformations (rotations, spatial and temporal translations, and boosts) and check that Newton's second law for a particle in a gravitational field is invariant under the Galilei group.

If you want to go beyond the Galilei group, by extending the transformations to include accelerations, you have to tell how the field $\phi$ transforms under that acceleration-extended group. Well, then we invoke the equivalence principle and state that it cannot transform tensorially under accelerations; it has to transform inhomogeneously under accelerations (without messing up the scalar transformations under the Galilei-transformations). And indeed, both the equations of motion, Newton's 2nd law and the Poisson equation (or their corresponding actions as you wish) are invariant under this transformation. But because we imposed this transformation on $\phi$, we call it a "pseudosymmetry", and as a result it doesn't have any accompanying Noether charges.

(edit: my relativistic example was confusing, changed it a bit)

 

[haushofer]

@PeterDonis

I took a long walk with my daughter to think about it, and I see your "magic"-point now.

In GR we propose, on basis of the equivalence principle, that the equations of motion (Einstein Field Equations and geodesic equation) are general-covariant. But the objections we work with (metric, connection) already have definit transformation laws; they follow mathematically.

What I propose , on basis of the equivalence principle, is that the Newtonian equations of motion (Newton's 2nd law and Poisson equation) are covariant under arbitrary accelerations (as I defined earlier). From that I deduce the transformation law for $\partial_i \phi$, which is different from GR. We totally agree on that.

Anyway, I don't want to hijack this topic, so my apologies to the topic starter if he/she feels like that.

 

[PeterDonis]

You don't know a priori how $\phi$ transforms!

Sure I do; it's a scalar. Transforming scalars is the easiest possible transformation: they just stay the same.

The question is how the spatial gradient transforms.

we also dictate on physical grounds (!) that the Newton potential transforms as a scalar under the Galilei group

No. We dictate on physical grounds that the Newtonian potential is a scalar function of space and time. That means it must transform as a scalar under any allowed transformation, not just the Galilei group: its value at a given point in spacetime must be the same no matter what coordinates we use.

The key thing here is really what the structure of "spacetime" is in Newtonian physics, as opposed to GR. In Newtonian physics, surfaces of constant time are absolute. That means only transformations that have $t' = t$ are allowed at all, independent of the question of whether they leave Newton's laws invariant. Your transformation meets that requirement, so it is an allowed transformation. But whether it leaves Newton's laws invariant is a separate question.

In GR we propose, on basis of the equivalence principle, that the equations of motion (Einstein Field Equations and geodesic equation) are general-covariant. But the objects we work with (metric, connection) already have definite transformation laws; they follow mathematically.

They "follow mathematically" in the sense that, if we know an object is a vector, or a (0, 2) tensor, or whatever, we know its transformation properties, yes.

What I propose, on basis of the equivalence principle, is that the Newtonian equations of motion (Newton's 2nd law and Poisson equation) are covariant under arbitrary accelerations (as I defined earlier).

I don't think you are free to propose this, because whether or not the Newtonian laws of physics are covariant under a given coordinate transformation is not freely specifiable; once you have the transformation, covariance or lack of it is a mathematical fact.

From that I deduce the transformation law for $\partial_i \phi$, which is different from GR. We totally agree on that.

No. You can't "deduce" a transformation law for a quantity from anything other than the coordinate transformation you are using. The coordinate transformation law you specified already tells you how $\partial_i \phi$ transforms; you can't change it to something else because you would like Newton's laws to be invariant.

 

[haushofer]

Sure I do; it's a scalar. Transforming scalars is the easiest possible transformation: they just stay the same.

No. We dictate on physical grounds that the Newtonian potential is a scalar function of space and time. That means it must transform as a scalar under any allowed transformation, not just the Galilei group: its value at a given point in spacetime must be the same no matter what coordinates we use.

Why do you propose that the Newton potential is a scalar of "space and time" (by that I assume you mean general coordinate transformations gct's)? I don't see any reason for that. A scalar (or for that matter, tensor in general) under one group of transformations is not automatically a tensor under another group of transformations. For me, if you call something a tensor, you must always specify the group of transformations under which it transforms as a tensor. Of course, for GR this is done implicitly, because there we work with gct's, but non-relativistically it becomes a more subtle matter.

E.g., we also know that the 3-acceleration of a particle is a vector under constant rotations, but not under time-dependent rotations or accelerations. For the same matter, we know that the Christoffel connection is a tensor under Lorentz transformations, but not under transformations for which the inhomogeneous term becomes non-zero (do these transformations have a name? I don't know). Of course, for those expressions we can derive these transformation-behaviours, so I guess that's where your objection lies.

The transformation law of the Newton potential cannot not be derived, as it can be in (the Newtonian limit of) GR; there $\Gamma^i_{00} = \partial^i\phi$. So in that sense you could call it "magic" to propose that $\partial^i\phi$ also transforms as such in the Newtonian theory. I call that "magic" the correspondence principle ;) But I don't see any inconsistency with it; for me, it's the natural thing to do. But I guess we differ on that matter. ;)

No. You can't "deduce" a transformation law for a quantity from anything other than the coordinate transformation you are using. The coordinate transformation law you specified already tells you how $\partial_i \phi$ transforms; you can't change it to something else because you would like Newton's laws to be invariant.

Well, let me come back to my example of the Schrödinger equation for a free particle. I'm curious how you regard that. This equation is proposed to be covariant under the Galilei group (actually, Bargmann group). In particular, under a boost

<br /> x^{&#039;i} = x^i + v^i t<br />

we demand that

<br /> i \partial&#039;_t \psi&#039;(t&#039;,x&#039;) = - \frac{1}{2m} \nabla^{&#039;2} \psi&#039;(t&#039;,x&#039;)<br />

This can only be achieved (because of the pesky time derivative, as you pointed out) if

<br /> \psi&#039;(t&#039;,x&#039;) = e^{if(t,x)}\psi(t,x)<br />

for some function $f(t,x)$ (which we can solve for, but which is not important now). I.e., the wavefunction is not a simple scalar under Galilei boost. How do you interpret that? Where does that phase factor $e^{if(t,x)}$ come from, according to you?

I must confess I'm not really into the whole "projective interpretation"-stuff (and, being out of academia for a while can make my mind a bit rusty on this matter), so maybe there is a formal group-theoretical way of deriving this transformation behaviour of the wavefunction under boosts, but this is how I understand it; regarding Borns rule there is no problem with taking the freedom to make the wavefunction transform with a phase factor under boosts. But we do that on physical grounds.

Maybe someone else who is more into this representation-stuff can comment on this issue; maybe it helps me (us) to understand the case for the Newton potential also a bit better.

 

[PeterDonis]

Why do you propose that the Newton potential is a scalar of "space and time"

Because it is. At a given distance from the Earth, for example, at a given time, the potential has a particular value which is a simple number, i.e., a scalar. Your coordinate transformation labels that point at a given distance from the Earth with a different $x$ value at different times, but that doesn't change the value of the potential at that point. It does make the potential a function of time as well as $x$ in your coordinates, whereas in coordinates fixed to the Earth (at least to an idealized Earth that is perfectly stationary), the potential is a function of $x$ only, not time. But it's a scalar--a number--in both cases.

(by that I assume you mean general coordinate transformations gct's)?

Not at all. Please read my posts more carefully; I explicitly said this was not the case for Newtonian physics.

The transformation law of the Newton potential cannot not be derived

Sure it can. We know what $\phi$ and its gradient are in the original coordinate chart, the Newtonian inertial frame in which the Earth's center is at rest. You have specified a coordinate transformation that takes that chart to a different chart. Doing that also specifies how $\phi$ and its gradient transform: they both have to transform such that, at the same physical point in space (for example, at the same distance from the Earth's center in the same direction, such as towards the star Polaris), $\phi$ has the same numerical value. There is no freedom left to pick anything about how $\phi$ or its gradient transform.

I don't see any inconsistency with it

You seriously don't see that, since we know $\phi$ and its gradient in one chart (the Newtonian inertial frame in which the Earth's center is at rest), once you specify a coordinate transformation from that chart to a different chart, which you have, you have already specified what $\phi$ and its gradient are in the new chart? You seriously don't see that you do not have the freedom to say what $\phi$ or its gradient are in addition to specifying the coordinate transformation?

I am very confused as to how you could possibly not see that.

 

[PeterDonis]

let me come back to my example of the Schrödinger equation for a free particle. I'm curious how you regard that.

I'm not even ready to talk about it, since what you are saying about the simpler case of Newtonian physics doesn't make sense to me.

 

[vanhees71]

I think it's pretty easy to see that within Newtonian mechanics the gravitational interaction is described by a scalar potential when taking the Lagrangian of the Kepler problem
$$L=\frac{m_1}{2} \dot{\vec{x}}_1^2 + \frac{m_2}{2} \dot{\vec{x}}_2 + \frac{\gamma m_1 m_2}{|\vec{x}_1-\vec{x}_2|}.$$
It is of the general form for a Galilei-invariant Lagrangian and obviously the interaction potential
$$V(\vec{x}_1,\vec{x}_2)=-\frac{\gamma m_1 m_2}{|\vec{x}_1-\vec{x}_2|}$$
is a scalar field under Galilei transformations.

Of course, Newtonian mechanics is not generally covariant as GR is.

 

[dx]

Newton's law is F = ma. This law is supposed to work only in inertial frames. Let us say we go to an accelerated frame. Provided we take the fictitious forces (uniform gravitational field in the accelerated frame) as real, and we include those forces on the F side, the law F = ma is covariant under accelerations. The key point is of course that we must regard the 'fictitous force' as a force which we include on the F side of F = ma. If we do this, the group of transformations under which F = ma is invariant is extended to include accelerations. There is nothing to prevent us from regarding the accelerated frame as "an inertial frame filled with a uniform gravitational field" and we can use F = ma in the accelerated frame also.

So there is definitely some sense in which haushofer's remarks are correct. This type of argument is routinely used in elementary physics also. Consider a box moving on an inclined plane, and the inclined plane is accelerating. Solving problems like this, it is very useful to consider the inclined plane to be at rest, and there is a uniform gravitational field. It is much more difficult to think about or solve such problems in the original frame.