The g_ij as potentials for the gravitational field

[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

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

transforms under the transformation

$$x^{'i} = x^i + \frac{1}{2}at^2$$

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

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

transforms under the transformation

$$x^{'i} = x^i + \frac{1}{2}at^2$$

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

$$\partial_{\mu} \Phi(t,x)$$

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

$$x^{'i} = x^i + v^i t$$

we demand that

$$i \partial'_t \psi'(t',x') = - \frac{1}{2m} \nabla^{'2} \psi'(t',x')$$

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

$$\psi'(t',x') = e^{if(t,x)}\psi(t,x)$$

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.

 

[dx]

I should add that PeterDonis's remarks about phi(x) being a "spacetime scalar" is problematic for the following reason. In generally covariant or gravitational theories, local operators are not gauge invariant. The idea of "value of phi at the point x" would be an ambiguous notion because of the active diffeomorphism gauge transformations.

 

[haushofer]

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

Your confusion (and mine...) arises because of the differences between symmetries and pseudosymmetries, as I explain in my thesis. I guess you're right that the coordinate transformation (acceleration) does not induce the transformation of phi I propose in the usual way of tensor fields. Of course, I see that and I mention this explicitly in my thesis. In that sense the transformation belonging to this pseudosymmetry can be regarded as a "redefenition". In my thesis I motivate this by regarding the point particle action as a (quite unusual) sigma-model, in which these pseudosymmetries arise more often, and I give an analogy with the relativistic case. But of course, that already takes us way out of what we normally consider as "Newtonian gravity theory". But again, this interpretation is motivated by the Newtonian limit of GR and the correspondence principle.

But let me stress again that I see your point, and that it was confusing of me to state that this is just ordinary Newtonian mechanics with the usual tensor calculus.

 

[haushofer]

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.

Yes, but nobody was claiming that phi isn't a scalar under Galilei-transformations ;)

 

[vanhees71]

So what are you disputing then with @PeterDonis ?

 

[PeterDonis]

I should add that PeterDonis's remarks about phi(x) being a "spacetime scalar" is problematic for the following reason. In generally covariant or gravitational theories, local operators are not gauge invariant. The idea of "value of phi at the point x" would be an ambiguous notion because of the active diffeomorphism gauge transformations.

Adjusting the gauge of $\phi$ in Newtonian physics corresponds to choosing a "zero point" for the potential. The standard choice is to set $\phi = 0$ at infinity, but of course that is not the only possible choice (another fairly common one is to set $\phi = 0$ at the Earth's surface). None of that affects what I am saying about transformation properties; the only additional condition you need is that you must fix a gauge for $\phi$ before applying any coordinate transformations.

 

[PeterDonis]

So what are you disputing then with @PeterDonis ?

If he's agreeing that, in standard Newtonian mechanics, Newton's laws, including the law of gravity, are only invariant under Galilean transformations, then I think there is no dispute. I was not expressing any opinion about what was in the thesis he referenced except that it isn't standard Newtonian mechanics.

 

[pervect]

Newton's law is F = ma. This law is supposed to work only in inertial frames.

Especially in the context of a discussion of gauge symmetries, why F=ma (force = mass * acceleratio) rather than L = T-V (expressing the Lagrangian L as the difference between "kinetic energy" and "potential energy"), as vanhees71 suggested?

The gauge symmetries are the symmetries of the Lagrangian. The Lagrangian formulation of Newtonian mechanics is equivalent to Newton's original formulation, and would seem to be more relevant to questions about the gauge symmetries of the theory.

Moving on to special and/or general relativity, we still have a Lagrangian L, though it's no longer in the form T-V.

 

[vanhees71]

Adjusting the gauge of $\phi$ in Newtonian physics corresponds to choosing a "zero point" for the potential. The standard choice is to set $\phi = 0$ at infinity, but of course that is not the only possible choice (another fairly common one is to set $\phi = 0$ at the Earth's surface). None of that affects what I am saying about transformation properties; the only additional condition you need is that you must fix a gauge for $\phi$ before applying any coordinate transformations.

The reason for this is, of course, that the "gauge freedom" for the Poisson equation boils down to just that pretty trivial invariance of Newtonian "field theory of gravitation" under changing the gravitational potential by an additive constant. This is in accordance with the fact that absolute values of energies (or energy densities) don't play any role in Newtonian and special relativistic physics.

 

[vanhees71]

Especially in the context of a discussion of gauge symmetries, why F=ma (force = mass * acceleratio) rather than L = T-V (expressing the Lagrangian L as the difference between "kinetic energy" and "potential energy"), as vanhees71 suggested?

The gauge symmetries are the symmetries of the Lagrangian. The Lagrangian formulation of Newtonian mechanics is equivalent to Newton's original formulation, and would seem to be more relevant to questions about the gauge symmetries of the theory.

Moving on to special and/or general relativity, we still have a Lagrangian L, though it's no longer in the form T-V.

One should, however, distinguish between gauge symmetries and usual symmetries. Gauge symmetries imply that you use more degrees of freedom than physically observable independent quantities, and the "unphysical" degrees of freedom are not determined uniquely by the equations of motion. E.g., in electrodynamics you use the vector potential which is defined only modulo gauge transformations. You need an additional constraint (like the Lorenz gauge, which fixes the gauge partially or the Coulomb gauge which fixes the gauge completely together with the necessary boundary and initial conditions) to make the solutions for the potentials unique, but this gauge freedom is not an incompleteness of the description physically observable phenomena, because the indetermined unphysical degrees of freedom do not occur in physical observables (or, in other words, for a quantity defined within a gauge theory to make sense as an observable, it must be gauge invariant).

Usual symmetries describe just the fact that the 1st variation of the action is invariant under a group of transformations of physically observable quantities, implying conservation laws, if the group is a Lie group (each independent one-parameter subgroup defines a generator of the symmetry, which is a conserved quantity).

The difference between these two kinds of symmetry is particularly important in quantum field theory where the breaking of a usual (global) Lie symmetry implies the existence of massless scalar or pseudoscalar particles, the Nambu-Goldstone bosons of the symmetry and the degeneracy of the ground state of the theory, while local gauge symmetries cannot be spontaneously broken, and the ground state is (usually) non-degenerate and there are no Nambu-Goldstone bosons (but rather massive gauge bosons). That's known as Elitzur's theorem.

 

[haushofer]

So what are you disputing then with @PeterDonis ?

Well, disputing...let's call it discussing :P : How the Newtonian potential transforms if you go beyond the Galilei group of transformations.

Let me formulate how I see things after the (enlightening!) discussion with Peter.

In the textbook Newtonian limit of GR, the Einstein field equations become the Poisson equation, and the geodesic equation becomes Newton's 2nd law. The gradient of the Newton potential shows up in the Christoffel connection,

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

So one can easily deduce that under a transformation, which we can call an acceleration,

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

we have in the Newtonian limit of GR that

$$\Gamma^{'i}_{00} = \Gamma^{i}_{00} - \ddot{\xi}^i \ \ \rightarrow \ \ \partial^{'i} \phi' = \partial^i \phi - \ddot{\xi}^i$$

Such a transformation leaves both the geodesic equation and Einstein's field equations in the Newtonian limit covariant. This transformation can be physically understood as the fact that locally in spacetime, where the Newtonian force is approximately constant, one can choose a coordinate frame such that

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

by choosing the acceleration appropriately ($\ddot{x}^i = (0,0,g)$ if the gravitational field is point in the z-direction). This is the interpretation of the statement that "a freely falling observer feels itself being weightless".

So, the correspondence principle suggests that this is how $\partial^i \phi$ should transform under accelerations in the Newtonian theory. However, in the Newtonian theory, $\phi$ transforms as a scalar. The induced transformation on $\partial^i \phi$ by accelerating does not reproduce the transformation that the GR-limit gives. In GR, $\partial^i \phi$ shows up in the connection, and we know how connection coefficients transform. In Newtonian theory, $\partial^i \phi$ is just the gradient of a scalar.

But still, the action of a Newtonian point particle coupled to the Newton potential (and it's equation of motion, Newton's 2nd law) is covariant under the transformation

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

This transformation of $\partial^i \phi$ is not "induced by the coordinate transformation". But there is a formalism in which we can interpret this kind of transformations: we call it the formalism of pseudosymmetries, which has its origin in sigma-models. The transformation of $x^i(t)$ does not induce the transformation on $\partial^i \phi$; instead the transformation of $\partial^i \phi$ can be regarded as a redefinition of $\partial^i \phi$. This "redefinition" (which for PeterDonis was the "magic"-part) is still a symmetry of the action, but because this redefinition is not induced by the transformation of $x^i(t)$, the corresponding symmetry does not have an accompanying Noether charge. That's why we call it a "pseudosymmetry". It's the price to pay if you want to connect the Newtonian limit of GR to the usual Newton theory on the level of symmetries.

 

[haushofer]

If he's agreeing that, in standard Newtonian mechanics, Newton's laws, including the law of gravity, are only invariant under Galilean transformations, then I think there is no dispute. I was not expressing any opinion about what was in the thesis he referenced except that it isn't standard Newtonian mechanics.

Indeed. I fully agree with your earlier statements and I was expressing myself not clearly. I'm also aware that the view I represent here on "Newtonian physics" is not standard.

Having said that, I'm still curious how you (or Vanhees or someone else) see the transformation of the wave function under Galilei boosts of the Schrodinger equation as transforming with an extra phase factor. It probably has to do something with projective representations of the Bargmann algebra. But somehow, that example always reminded me a bit of this discussion of the Newton potential.

 

[vanhees71]

Sure, you can write all equations in a general covariant way. E.g., for Newtonian mechanics and many other theories you can just use the action principle, leading to a formalism that is forminvariant under arbitrary diffeomorphisms in configuration space (in all kinds of generalized variables the EoM are given by the Euler-Lagrange equation) or general canonical transformations in the Hamiltonian formulation.

Still, the symmetry group of standard Newtonian mechanics is the Galilei group, because for a symmetry all the transformations making up (a representation of) the symmetry group the 1st variation of the action must be invariant. You know better than me that you can extend Galilean mechanics to a kind of Newton Cartan theory:

https://en.wikipedia.org/wiki/Newton–Cartan_theory

The realization of the Galilei group in non-relativistic QT is subtle. The proper unitary transformations of the classical (10-dim) Galilei group does not lead to a useful quantum theory, at least nothing that would apply to the real world (Inönü and Wigner). Rather you need the ray representation of a non-trivial central extension with the mass as an additional non-trivial "central charge" of the Lie algebra. In any case there's no reason not to also extend the ray representations of the symmetry group to its covering group. So in non-relativistic QT the classical Galilei group is realized as a unitary representation of the central extension of the covering group (the latter just implies that instead of the rotation subgroup SO(3) you use its covering group SU(2), allowing for half-integer spin, which obviously is needed to describe everything we call "matter", i.e., the leptons and baryons which are all fermions with half-integer spin). The central extension with the mass as an additional independent observable leads to a superselection rule forbidding to superimpose state vectors from representations of different mass eigenvalues, and this establishes an 11th independent conservation law from the Galilei symmetry, the conservation of mass.

E. In¨on¨u and E. P. Wigner, Representations of the Galilei group, Il Nuovo Cimento 9, 705 (1952),
https://doi.org/10.1007/BF02782239.

 

[PeterDonis]

one can easily deduce that under a transformation

No, you can't deduce this. And you admit that you're not:

This transformation of $\partial^i \phi$ is not "induced by the coordinate transformation". But there is a formalism in which we can interpret this kind of transformations: we call it the formalism of pseudosymmetries, which has its origin in sigma-models. The transformation of $x^i(t)$ does not induce the transformation on $\partial^i \phi$; instead the transformation of $\partial^i \phi$ can be regarded as a redefinition of $\partial^i \phi$. This "redefinition" (which for PeterDonis was the "magic"-part) is still a symmetry of the action, but because this redefinition is not induced by the transformation of $x^i(t)$, the corresponding symmetry does not have an accompanying Noether charge. That's why we call it a "pseudosymmetry". It's the price to pay if you want to connect the Newtonian limit of GR to the usual Newton theory on the level of symmetries.

Personally, I'm not familiar with this notion of "pseudosymmetries" or with sigma-models, so I can't comment on them. But you admit that whatever these notions involve, it is not "deducing" anything; it's just declaring by fiat that $\partial^i \phi$ works the way you want it to. That's not deducing. It's assuming.

 

[haushofer]

No, you can't deduce this. And you admit that you're not:

Of course you can. We're doing the limit of GR, and you know how the Christoffel connection transforms. That's the whole point: in good old Newtonian gravity, $\partial^i \phi$ is "just a gradient of a scalar"; in GR it turns out to be $\Gamma^i_{00}$. And we know how $\Gamma^i_{00}$ transforms under a gct, so we also know how it transforms under the proposed accelerations (copy from Landau&Lifshitz):

Naming i=i, k=l=0 gives you the transformation of $\Gamma^i_{00}$; the $\ddot{\xi}^i$ which pops up is just the inhomogenous term of the transformation. On top of that, $\Gamma^i_{00}$ transforms as a vector under constant rotations, it gives you the Coriolis and centrifugal force under rotations which depend on time, it transforms as a scalar under Galilei boosts, etc.

But if you don't believe me, you can check the textbook Newtonian limit of GR yourself and check which coordinate transformations you're left with after taking the limit. The covariance under gct's is broken down to covariance under the Galilei-group plus accelerations.

I admit that in the good-old Newtonian theory, we can't deduce this transformation of $\Gamma^i_{00}$ from its tensorial properties.

Personally, I'm not familiar with this notion of "pseudosymmetries" or with sigma-models, so I can't comment on them. But you admit that whatever these notions involve, it is not "deducing" anything; it's just declaring by fiat that $\partial^i \phi$ works the way you want it to. That's not deducing. It's assuming.

Well, I'm OK with that naming. My fiat is the correspondence principle. It depends on how comfortable you are with bending the rules.

 

[haushofer]

Sure, you can write all equations in a general covariant way. E.g., for Newtonian mechanics and many other theories you can just use the action principle, leading to a formalism that is forminvariant under arbitrary diffeomorphisms in configuration space (in all kinds of generalized variables the EoM are given by the Euler-Lagrange equation) or general canonical transformations in the Hamiltonian formulation.

Still, the symmetry group of standard Newtonian mechanics is the Galilei group, because for a symmetry all the transformations making up (a representation of) the symmetry group the 1st variation of the action must be invariant. You know better than me that you can extend Galilean mechanics to a kind of Newton Cartan theory:

https://en.wikipedia.org/wiki/Newton–Cartan_theory

The realization of the Galilei group in non-relativistic QT is subtle. The proper unitary transformations of the classical (10-dim) Galilei group does not lead to a useful quantum theory, at least nothing that would apply to the real world (Inönü and Wigner). Rather you need the ray representation of a non-trivial central extension with the mass as an additional non-trivial "central charge" of the Lie algebra. In any case there's no reason not to also extend the ray representations of the symmetry group to its covering group. So in non-relativistic QT the classical Galilei group is realized as a unitary representation of the central extension of the covering group (the latter just implies that instead of the rotation subgroup SO(3) you use its covering group SU(2), allowing for half-integer spin, which obviously is needed to describe everything we call "matter", i.e., the leptons and baryons which are all fermions with half-integer spin). The central extension with the mass as an additional independent observable leads to a superselection rule forbidding to superimpose state vectors from representations of different mass eigenvalues, and this establishes an 11th independent conservation law from the Galilei symmetry, the conservation of mass.

E. In¨on¨u and E. P. Wigner, Representations of the Galilei group, Il Nuovo Cimento 9, 705 (1952),
https://doi.org/10.1007/BF02782239.

Ok. But how does this relate exactly to the fact that the Schrodinger equation is only covariant with respect to boosts if the wave function transforms under it with an extra phase factor? Is that somehow a hint at the level of equations of motion of the group-theoretical result from the Inönü and Wigner paper?