Invariant definition of acceleration in Newtonian physics vs proper acceleration in GR

[cianfa72]

TL;DR

About the invariant definition of acceleration in Newtonian
physics vs proper acceleration in GR

Does it exist an invariant way to define acceleration in Newton physics like the proper acceleration in GR ?

In Newton physics if an accelerometer attached to an object reads 0 it does not mean it is actually not accelerating (since gravity is a force).

To define inertial motion the concept of inertial frame of reference is actually needed, so going in circle (IMO).

 

[Dale]

You can use Newton Cartan gravity. That has a lot of the usual features of GR. There is a Newtonian curved spacetime, and the equivalence principle, and geodesics are free fall, and proper acceleration vs coordinate acceleration.

The differences are that the connection is more complicated and you have a pair of degenerate metrics instead of a single metric.

 

[cianfa72]

You can use Newton Cartan gravity. That has a lot of the usual features of GR.

Yes, but from a physical point of view is there the "equivalent of an accelerometer" to single out inertial motions ?

 

[Dale]

Uh, yes. An accelerometer

 

[cianfa72]

Uh, yes. An accelerometer

Ok, but using it you cannot detect for instance whether an object is coasting in an empty region of the Universe (i.e. is in inertial motion) or whether is free-falling in the gravitational field of a star (i.e. it is actually accelerating according Newtonian physics).

Edit: in the inertial frame in which the former body is at rest the latter is "really" accelerating according to Newton.

 

[Dale]

Ok, but using it you cannot detect for instance whether an object is coasting in an empty region of the Universe (i.e. is in inertial motion) or whether is free-falling in the gravitational field of a star (i.e. it is actually accelerating according Newtonian physics).

With Newton Cartan gravity that is the same as with GR. Gravity is locally a fictitious force. Free fall in the presence of gravity is inertial.

Newton Cartan is experimentally the same as Newtonian gravity, but with most of the math of GR. Including specifically curved spacetime and the equivalence principle.

 

[PeterDonis]

Does it exist an invariant way to define acceleration in Newton physics like the proper acceleration in GR ?

It depends on what formulation of Newtonian physics you use. As @Dale mentioned, there is at least one, Newton Cartan gravity, in which gravity is a fictitious force, just like in GR.

In Newton physics if an accelerometer attached to an object reads 0 it does not mean it is actually not accelerating (since gravity is a force).

This is the case for some formulations of Newtonian physics, but not for all (for example, as @Dale says, it is not true in Newton Cartan gravity).

To define inertial motion the concept of inertial frame of reference is actually needed, so going in circle (IMO).

If you mean this as a general claim, it's false; inertial motion in GR, and in Newton Cartan gravity, is defined by an accelerometer reading zero. This is a frame-independent definition.

 

[Bosko]

Does it exist an invariant way to define acceleration in Newton physics like the proper acceleration in GR ?

Also in Newtonian mechanics, acceleration is invariant in relation to the inertial reference system.
Did you mean something else by "invariant way"?

In Newton physics if an accelerometer attached to an object reads 0 it does not mean it is actually not accelerating (since gravity is a force).

(copy/paste):
In 1907, Einstein had the "happiest thought of my life" when he was sitting in a chair at the patent office in Bern:
If a person falls freely he will not feel his own weight.
This led him to the postulate the "equivalence principle", which states that one cannot tell the difference between an accelerating reference frame and a gravitational field.

To define inertial motion the concept of inertial frame of reference is actually needed, so going in circle (IMO).

In the simple case of a spherical source of gravity (for example the Earth) by moving in a circle, the gravitational acceleration will be canceled by the centrifugal acceleration.

On a straight line between the Earth and the Moon (closer to the Moon) there is a point where the gravitational attraction of the Earth and the Moon are equal.So there is no gravity at that point.

If you move in a circle around the Earth that passes through that point, gravity will disappear and centrifugal acceleration will throw you off course.

You can separate the causes of acceleration in Newtonian mechanics into those caused by gravity and all others caused by an engine, by turning the pedals faster on a bicycle...

If you put zero on all accelerations caused by gravity you will get "proper acceleration" as in general relativity.

Moderator's note: Off topic content deleted.

 

[PeterDonis]

In 1907, Einstein had the "happiest thought of my life"

Which resulted in a different theory of gravity from Newtonian physics.

On a straight line between the Earth and the Moon (closer to the Moon) there is a point where the gravitational attraction of the Earth and the Moon is equal. So there is no gravity at that point.

No net gravity, in Newtonian terms, from the Earth and the Moon. But that does not mean no gravity altogether; there are other sources of gravity besides the Earth and the Moon.

Also, "at that point" is misleading; gravity doesn't suddenly change at just that point. See below.

If you move in a circle around the Earth that passes through that point, gravity will disappear and centrifugal acceleration will throw you off course.

This is a very garbled description. Gravity doesn't just "disappear" at that one point while being fully Earth gravity at other points. The net gravitational force, in Newtonian terms, on an object is a continuous function of position. The correct analysis of the situation you are describing is that there is no circular free-fall orbit around the Earth alone at that altitude at all; to compute free-fall orbits in this region you must take both the Earth's and the Moon's gravity into account, and since there is no known exact solution for this case even in Newtonian gravity, you must solve for the motion numerically. This is, for example, what is done to compute trajectories for space probes in this region.

If you put zero on all accelerations caused by gravity you will get "proper acceleration" as in general relativity.

I'm not sure what you mean by this.

 

[PeterDonis]

I hope I'm not getting too off topic

You are. I have deleted that portion of your post.

 

[cianfa72]

It depends on what formulation of Newtonian physics you use.

At the undergraduate level (at least here in Italy), Newtonian physics is introduced starting from inertial motion defined by Newton's first law. As in wiki this definition is actually going in circle as noted from Einstein himself.

 

[Dale]

At the undergraduate level (at least here in Italy), Newtonian physics is introduced starting from inertial motion defined by Newton's first law. As in wiki this definition is actually going in circle as noted from Einstein himself.

Yes, that is how it is introduced at the undergraduate level here in the USA too. But your question was

Does it exist an invariant way to define acceleration in Newton physics like the proper acceleration in GR ?

And as I have said multiple times now, the answer is yes. That is what is done in Newton Cartan gravity.

We don’t teach Newtonian gravity as curved Newtonian spacetime at the undergraduate level, but Newtonian gravity can nevertheless be formulated in that manner.

 

[vanhees71]

I'm a bit puzzled by this complicated discussion since indeed in standard Newtonian mechanics acceleration is invariant under Galilei transformations. A Galilei boost of the spacetime coordinates of Newtonian mechanics reads
$$t'=t, \quad \vec{x}'=\vec{x}-\vec{v} t,$$
where $\vec{v}=\text{const}$ is the velocity of the IF $\Sigma'$ wrt. the IF $\Sigma$.

Let $\vec{x}(t)$ be the trajectory of a particle. Then
$$\vec{a}(t)=\ddot{\vec{x}}(t)=\ddot{\vec{x}}'(t)=\vec{a}'(t).$$

 

[A.T.]

I'm a bit puzzled by this complicated discussion since indeed in standard Newtonian mechanics acceleration is invariant under Galilei transformations.

I think there is some confusion about the term "invariant":
- You mean the value doesn't change under Galilean transformations.
- The OP means the value is the same in all frames (including non-inertial ones).

 

[cianfa72]

Btw I meant a coordinate-free definition of acceleration in Newtonian physics like the proper acceleration in relativity (measured by an accelerometer).

 

[FactChecker]

Btw I meant a coordinate-free definition of acceleration in Newtonian physics like the proper acceleration in relativity (measured by an accelerometer).

Do you mean "coordinate free" in the sense of a definition that makes no reference to a coordinate system, or a coordinate agnostic definition that gives the same value regardless of which inertial coordinate system is used? @vanhees71's definition in post #13 is uses coordinates but is coordinate agnostic.

 

[Dale]

It is good to highlight that the words can be used multiple ways. I think I answered the question as intended by @cianfa72, but indeed I could have missed the mark

 

[cianfa72]

I found this lecture Newton-Cartan on the topic. In section 2 it defines the Newtonian spacetime and the axioms of Newton-Cartan gravity. Free particles moves along timelike geodesics of a torsion-free Newtonian connection $\nabla$. According to @Dale these are the paths taken from particles having zero acceleration as measured by an accelerometer attached to them.

 

[cianfa72]

Btw I'm not sure that in the Newton-Cartan setting with absolute time (i.e. closed one-form $\tau$, $d\tau=0$), the time elapsed along timelike paths between two given spacelike hypersurfaces of the foliation is always the same regardless of the specific path taken (locally it is since $\tau$ is locally exact but possibly not globally).

 

[PeterDonis]

I'm not sure that in the Newton-Cartan setting with absolute time (i.e. closed one-form $\tau$, $d\tau=0$), the time elapsed along timelike paths between two given spacelike hypersurfaces of the foliation is always the same regardless of the specific path taken (locally it is since $\tau$ is locally exact but possibly not globally).

Yes, it is. $d\tau$ in Newton-Cartan spacetime is globally exact, because it is the exterior derivative of a scalar, $\tau$. (Btw, it makes no sense to say that $\tau$ is "exact"; $\tau$ is a scalar, not a 1-form.)

 

[Dale]

(locally it is since $\tau$ is locally exact but possibly not globally).

Hmm, that’s a good question. I don’t know that. I hadn’t considered the possibility of strange global geometries or topologies.

I suppose if you had a loop in time that was larger or smaller in different parts of the spatial manifold that you could get path dependencies.

 

[cianfa72]

Yes, it is. $d\tau$ in Newton-Cartan spacetime is globally exact, because it is the exterior derivative of a scalar, $\tau$. (Btw, it makes no sense to say that $\tau$ is "exact"; $\tau$ is a scalar, not a 1-form.)

In Definition 1.1 $\tau$ is a nowhere vanishing 1-form and the absolute time condition requires it to be closed (hence locally exact).

 

[PeterDonis]

I hadn’t considered the possibility of strange global geometries or topologies.

There can't be any in Newton-Cartan spacetime, since the time function $\tau$ is a globally defined scalar function. In fact, each spacelike surface of constant $\tau$ is the Newton-Cartan analogue of a Cauchy surface, so Newton-Cartan spacetime is effectively globally hyperbolic, which rules out any strangeness that would make $d\tau$ not globally exact.

 

[PeterDonis]

In the Definition 1.1

Meaning, in the lecture you referenced. But you should be explicit about these things.

$\tau$ is a nowhere vanishing 1-form and the absolute time condition requires it to be closed.

More precisely, from the lecture you referenced:

The "absolute time condition" means $d\tau = 0$, from the text right after 1.1. And from the text right after 1.3, that means that locally $\tau = dt$, where $t$ is a time function.

However, those things are not specific to Newton-Cartan gravity; they are just general properties of Galilei connections. In chapter 2, where Newton-Cartan gravity specifically is formulated, and particularly in section 2.2 about recovering Newtonian gravity, it seems to me that the constructions given imply that $t$ must be a global time function (or rather that such a function $t$ can always be found), which requires that $\tau$ be globally exact, not just locally exact.

There might be an issue here of terminology. The term "Newton-Cartan gravity" is ambiguous: does it mean just the specific formulation from which standard Newtonian gravity is recovered, or does it mean the general class of formulations described in the notes, which include theories that are not exactly the same as standard Newtonian gravity? The lecture you referenced seems to me to use the term both ways in different places. In any event, the statements I am making are only about the specific formulation that gives standard Newtonian gravity, not the more general framework.

 

[PAllen]

Getting back to Newtonian gravity as defined before GR (Newton-Cartan only came later), the typical dodge for defining inertial frames was to reference "the distant stars". This language is quite common in physics papers of the era, as a way out of circularity. Then, it was certainly true that there were circumstances where an accelerometer could read zero while Galilean invariant acceleration was not zero.

 

[Dale]

The term "Newton-Cartan gravity" is ambiguous: does it mean just the specific formulation from which standard Newtonian gravity is recovered, or does it mean the general class of formulations described in the notes, which include theories that are not exactly the same as standard Newtonian gravity?

Personally I would use it only in the first way. Otherwise it seems like just Cartan rather than Newton Cartan.

 

[cianfa72]

it seems to me that the constructions given imply that $t$ must be a global time function (or rather that such a function $t$ can always be found), which requires that $\tau$ be globally exact, not just locally exact.

Ok, so to get the same elapsed time along a timelike path between two given spacelike hypersurfaces regardless of the specific path, $\tau$ must be globally exact not just locally (i.e. there must be a global smooth function $t$ such that $\tau=dt$).

 

[cianfa72]

Then, it was certainly true that there were circumstances where an accelerometer could read zero while Galilean invariant acceleration was not zero.

Such for instance a particle in free-falling in a gravitational field produced by a planet/star.

 

[PeterDonis]

Ok, so to get the same elapsed time along a timelike path between two given spacelike hypersurfaces regardless of the specific path, $\tau$ must be globally exact not just locally (i.e. there must be a global smooth function $t$ such that $\tau=dt$).

If the "same elapsed time" condition holds globally, instead of just in a local neighborhood, yes. In the general framework developed in the paper (i.e., not restricted to standard Newtonian gravity), the "same elapsed time" condition is discussed as just being within a local neighborhood, so one could conceivably have different local neighborhoods with different elapsed times. But in standard Newtonian gravity, it is a global condition.

 

[cianfa72]

But in standard Newtonian gravity, it is a global condition.

So, as you pointed out in post #24, in section 2 in that paper (in particular in 2.1) a global condition on $\tau$ is actually implicitly assumed.