Whether a non-inertial frame is absolute

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A non-inertial frame is defined by its acceleration, which is invariant and can be measured using an accelerometer without reference to any other frame. Proper acceleration is not relative to any frame, and it is essential for defining inertial frames, which are characterized by having zero proper acceleration for stationary objects. The discussion emphasizes that while kinematic quantities are often frame-dependent, proper acceleration remains invariant across different inertial frames. The concept of proper acceleration is compatible with classical mechanics, although it may not be explicitly mentioned in introductory texts. Overall, the distinction between proper acceleration and coordinate acceleration is crucial for understanding motion in both inertial and non-inertial frames.
  • #31
etotheipi said:
Accelerometers are a nice heuristic, but you can't really define an accelerometer mathematically.
Why would you need to define it mathematically? We are doing science, not math.
 
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  • #32
Classical mechanics, especially, is very much a mathematical structure. There are many applied mathematics texts on the subject, e.g. Abraham/Marsden, Spivak, Arnold, et cetera. I'm not a mathematician myself, but I still don't think one should be using mythical objects such as accelerometers in definitions of physical quantities :wink:
 
  • #33
etotheipi said:
accelerometers are a nice heuristic, but you can't really define an accelerometer mathematically.
Then you have a problem. Ultimately, all the maths we do is supposed to support predictions like "if the dial on this machine reads ##x\pm \delta x## units then the dial on that machine will read ##y\pm \delta y## units". If it's not capable of doing that, what's it got to do with the real world?

Don't get me wrong - all of the mathematical abstraction is incredibly powerful and can lead to remarkable insights. But if it can't relate to direct observables then it's either unfinished or useless for physics.
 
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  • #34
etotheipi said:
I still don't think one should be using mythical objects such as accelerometers in definitions of physical quantities
I 100% disagree with this. In my opinion not only are operational definitions necessary for science, they are the most important ones.
 
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  • #35
Dale said:
Why would you need to define it mathematically?
Agreed. Why do you need to define it mathematically if you can just buy one?
 
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  • #36
Mhm. This is turning into something philosophical and uninteresting, but I might as well say that in my opinion you are looking at things in the wrong way. Classical mechanics is fundamentally a mathematical structure, a study of certain differential equations, differential and symplectic geometry, et cetera.

In the context of this discussion, the mathematical concept is one of a Galilean co-ordinate chart ##\varphi: A^4 \longrightarrow \mathbf{R} \times \mathbf{R}^3##, where ##A^4 \cong \mathbf{R}^4## is an four-dimensional affine space. In this chart one can describe motions of a system of ##n = N/3## particles by a function ##\boldsymbol{x} : I \longrightarrow \mathbf{R}^N## where ##I \subseteq \mathbf{R}##. The acceleration of the system is nothing but the second derivative ##\ddot{\boldsymbol{x}}(t_0) = \mathrm{d}^2 \boldsymbol{x} / \mathrm{d} t^2 \big{|}_{t_0}##. And finally Newton's principle of determinacy ensures the existence of a unique function ##\mathbf{F} : \mathbf{R}^{N} \times \mathbf{R}^N \times \mathbf{R} \longrightarrow \mathbf{R}^N## such that ##\ddot{\boldsymbol{x}} = \mathbf{F}(\boldsymbol{x}, \dot{\boldsymbol{x}}, t)##.

The above describes a purely mathematical and self-contained framework. In order to make physical predictions we must somehow map from the mathematical model to the real world; this is the point at which one implements operational procedures. One uses rulers and clocks to realize position and time respectively. One realizes an inertial frame by ensuring that an isolated particle advances at a constant rate in a fixed direction. And through experiment one determines the form of the function ##\mathbf{F}## which reproduces the motions observed in the real word.

For example, for a suitable range of extensions the length of a vertical spring changes is observed to change in proportion to the mass attached to the end; this constitutes an operational definition of force. One then lays the spring horizontally on an ice-rink, and pulls the spring in such a way that its extension is always constant. By calculating the rate at which the velocity of the mass increases the acceleration may be deduced, and by measuring the extension of the spring the force exerted on the mass may also be deduced. It is then confirmed through repeated experiments that, to good accuracy, their ratio is constant and Newton's equation holds good for ##\mathbf{F} = \text{constant vector}##.

In short; an accelerometer is something physical which is described by the theory, but must not be part of the theory itself.
 
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  • #37
etotheipi said:
an accelerometer is something physical which is described by the theory
Which means the theory has to be capable of predicting what the accelerometer will read; which means there has to be some theoretical entity that corresponds to that prediction. So call whatever that theoretical entity is in classical (non-relativistic) mechanics "proper acceleration" and you're done.
 
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  • #38
PeterDonis said:
Which means the theory has to be capable of predicting what the accelerometer will read; which means there has to be some theoretical entity that corresponds to that prediction. So call whatever that theoretical entity is in classical (non-relativistic) mechanics "proper acceleration" and you're done.
Ah, but that theoretical entity corresponding to it via the inverse physical map, which I'll light-heartedly call ##\mathcal{P}^{-1}## for fun, is precisely ##\mathbf{a} \big{|}_K##! It is nothing but the acceleration with respect to an inertial reference system.
 
  • #39
etotheipi said:
that theoretical entity corresponding to it via the inverse physical map, which I'll light-heartedly call ##\mathcal{P}^{-1}## for fun, is precisely ##\mathbf{a} \big{|}_K##! It is nothing but the acceleration with respect to an inertial reference system.
Ok, then call that "proper acceleration". What's the problem?
 
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  • #40
PeterDonis said:
Ok, then call that "proper acceleration". What's the problem?
That's exactly it - what others here are referring to as 'proper acceleration' is nothing but a special case of co-ordinate acceleration in classical mechanics (as opposed, however, to relativistic theories). There's no need for this extra terminology:
etotheipi said:
The point I'm making is that I haven't really noticed any classical mechanics textbooks use 'proper acceleration', and there isn't particularly a need to introduce it.
 
  • #41
etotheipi said:
There's no need for this extra terminology
I can see at least two good reasons to have it:

(1) "Proper acceleration" is shorter and easier to use than "coordinate acceleration in an inertial frame".

(2) It makes use of the fact that classical mechanics is an approximation to relativistic mechanics, and helps to illustrate an important connection between them.
 
  • #42
etotheipi said:
What others here are referring to as 'proper acceleration' is nothing but a special case of co-ordinate acceleration in classical mechanics
Actually, there is one important case where this is not true: in classical mechanics (as opposed to relativity), gravity is a force. So in classical mechanics, the acceleration of a rock dropped from a height above the surface of the Earth is coordinate acceleration in an inertial frame (because the frame in which the surface of the Earth is at rest is an inertial frame), so it would qualify as "proper acceleration" under the correspondence suggested in posts #38 and #39. But in relativity, it isn't.

And note that in this case, the mathematical entity in the theory that corresponds to the actual accelerometer reading is not ##\mathbf{a} \big{|}_K##, because that is nonzero, but the accelerometer reading for an accelerometer attached to the rock is zero.
 
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  • #43
etotheipi said:
Classical mechanics, especially, is very much a mathematical structure. There are many applied mathematics texts on the subject, e.g. Abraham/Marsden, Spivak, Arnold, et cetera. I'm not a mathematician myself, but I still don't think one should be using mythical objects such as accelerometers in definitions of physical quantities :wink:
Measurement devices are not mythical, but the real-world equipment we use to observe the world quantitatively and only this operational definition of observables enables us to relate the mathematical descriptions of our beloved theories with the observed phenomena in Nature.

Nowadays most of us have pretty precise accelerometers with us since any Smart or iPhone contains accelerometers, and my colleagues from the physics-didactics department (and many more worldwide) use it to develop nice experiments directly illustrating these sometimes confusing subjects about inertial (global in Newtonian and special-relativistic physics, local in general relativistic physics) frames of reference and how the physical laws look in non-inertial reference frames.
 
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  • #44
PeterDonis said:
Actually, there is one important case where this is not true: in classical mechanics (as opposed to relativity), gravity is a force. So in classical mechanics, the acceleration of a rock dropped from a height above the surface of the Earth is proper acceleration (because the frame in which the surface of the Earth is at rest is an inertial frame). But in relativity, it isn't.

And note that in this case, the mathematical entity in the theory that corresponds to the actual accelerometer reading is not ##\mathbf{a} \big{|}_K##, because that is nonzero, but the accelerometer reading for an accelerometer attached to the rock is zero.
Right, sure, but everything I've been saying in this thread is justified by the assumption of sticking well within the realm of classical physics, so here gravity well and truly is a force ##- \nabla \varphi## :smile:
 
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  • #45
Having just said that I was restricting myself to classical physics, here's another example from special relativity: what is time?

In some instances it is a scalar field ##t : U \longrightarrow \mathbf{R}## defined on the manifold ##M## where ##U \subseteq M## is some open subset of ##M##. In other instances it is a parameter of a curve ##\gamma: I \longrightarrow M## taking values in some interval ##I \subseteq \mathbf{R}##.

It's often also said that "time is what a clock measures". For similar reasons to above, to me this is not really a satisfactory definition, at least. It's much better in my opinion to go the other way and define an ideal clock as a set of events ##\{ E_i \}## along a timelike worldline such that the interval of the worldline parameter between two tick events is ##\tau(E_{i+N}) - \tau(E_{i}) = \kappa N## where ##\kappa## is some constant. Realisation of that physically amounts to having a physical object, of negligible spatial size, whose tick events along its worldline more or less approximate the behaviour of this ideal clock defined above.

But the clock, as a physical device, really ought not to factor into the definition of time!
 
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  • #46
etotheipi said:
here's another example from special relativity: what is time?
There is no single definition for the ordinary language word "time" in relativity. The two most common meanings are "coordinate time" (which requires choosing a coordinate chart, and which at least strongly implies that the coordinate you are labeling as "time" is timelike) and "proper time" (which requires choosing a timelike curve). Your "scalar field" and "curve parameter" definitions correspond to these two cases.

etotheipi said:
It's often also said that "time is what a clock measures".
Which is true for the "proper time" (curve parameter) definition: the clock measures proper time along its worldline.

etotheipi said:
the clock, as a physical device, really ought not to factor into the definition of time!
It doesn't; "time is what a clock measures" is what defines a clock, just as you suggest, not what defines time.
 
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  • #47
feynman1 said:
I don't understand why the acceleration can be invariant. Aren't all kinematic quantities measured w.r.t a reference frame? Then the magnitude of the acceleration should also be relative to some frame?
@etotheipi gave the answer in post #2. Acceleration is measured with respect to an inertial frame, but you can show that the answer is the same in all inertial frames.

There are important distinctions between accelaration in Newtonian mechanics and proper acceleration of relativistic physics that @PeterDonis gives in post #42.
 
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  • #48
etotheipi said:
'proper acceleration' is nothing but a special case of co-ordinate acceleration in classical mechanics
Emphasis added.

It is very often useful to have special terms for special cases.

etotheipi said:
Classical mechanics is fundamentally a mathematical structure, a study of certain differential equations, differential and symplectic geometry, et cetera.
No. Classical mechanics is fundamentally a scientific theory that uses those mathematical structures to make accurate predictions of classical physics experiments. You cannot remove the connection to experiment and still claim to be doing classical mechanics. Hence operational definitions are essential to the theory. Those operational definitions are precisely what make it classical mechanics instead of just symplectic geometry etc.
 
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  • #49
PeterDonis said:
in classical mechanics (as opposed to relativity), gravity is a force. So in classical mechanics, the acceleration of a rock dropped from a height above the surface of the Earth is coordinate acceleration in an inertial frame (because the frame in which the surface of the Earth is at rest is an inertial frame), so it would qualify as "proper acceleration" under the correspondence suggested in posts #38 and #39. But in relativity, it isn't.
Of course, there is more than one way to formulate classical mechanics. With the Newton Cartan formulation of Newtonian gravity you get the relativistic definitions of inertial frames, the equivalence principle, and you can consider gravity to be a fictitious force in a local inertial frame. It is a little cumbersome to actually use, but it is nice to know that these specific good features of GR are “backwards compatible”.
 
  • #50
  • #51
Dale said:
No. Classical mechanics is fundamentally a scientific theory that uses those mathematical structures to make accurate predictions of classical physics experiments. You cannot remove the connection to experiment and still claim to be doing classical mechanics. Hence operational definitions are essential to the theory. Those operational definitions are precisely what make it classical mechanics instead of just symplectic geometry etc.
This is personal preference :smile:

To me it is purely a model. The solution to any mechanical problem is the complete history of the mechanical system, the phase space: a symplectic manifold together with a Hamiltonian vector field whose integral curves ##\xi(t) = (\boldsymbol{q}(t), \boldsymbol{p}(t))## are solutions to Hamilton's equations of motion.

If, overnight, the laws of Physics were to suddenly change and become completely unrecognisable, then assuming we're still alive we could still do classical physics problems for fun. That's because it's an abstract structure, and must still make sense in the absence of a real world to compare it to.
 
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  • #52
etotheipi said:
If, overnight, the laws of Physics were to suddenly change and become completely unrecognisable, then assuming we're still alive we could still do classical physics problems for fun. That's because it's an abstract structure, and must still make sense ...
How do you know 'making sense' would still be the same?
 
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  • #53
etotheipi said:
To me it is purely a model.
A model of what, though? Why do people spend so much time studying it if it's just a system of equations? Why that one and not some other extremisation problem?

You seem to me to be being overly purist, to the extent that you are kind of missing the point. Certainly the equations of classical mechanics (like any other system of equations) must be internally consistent and can be played with without any intention of connecting them to our own experience. But the reason we study them (and why they have the name mechanics) is that some concepts in them behave the same way that things in the real world do. We could apply the Euler Lagrange equations to any arbitrary function and see what the results are (there are probably people who do), but the reason we apply it to the particular action(s) that we do is that those systems relate to quantitative measurements of the real world.

You can't have experiments without theory, but without connecting specific concepts in a theory to quantitative real world measurements there's nothing to pick one system of equations out from the infinitely many other logically consistent systems of equations.
 
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  • #54
Ibix said:
You can't have experiments without theory, but without connecting specific concepts in a theory to quantitative real world measurements there's nothing to pick one system of equations out from the infinitely many other logically consistent systems of equations.
The point is that you have to mentally separate the the model, classical mechanics, from real world realisations. You can merely put things from the model into correspondence with things from the real world, and hope that they are in good enough agreement. And on that note, classical mechanics doesn't agree with experiment, if you look close enough - but that doesn't at all imply that it isn't a self-consistent and perfectly nice mathematical theory!

In the context of this thread, it is completely unsatisfactory to appeal to a real, physical device - an accelerometer - in the theory. But you can of course define theoretical objects (cf. discussion of the 'ideal clock' a few posts ago).

But as I said before, it's really just personal preference and this whole digression is to a strong degree completely useless. It's really no different to asking that infamous question of whether a reference frame is something purely mathematical (a co-ordinate chart) or instead an actual physical realisation (some clocks and rulers). I'd say it's the former, but it's really just a matter of taste.
 
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  • #55
etotheipi said:
The point is that you have to mentally separate the the model, classical mechanics, from real world realisations.
Classical mechanics (or physics in general) is not just the mathematical model, but also the description how it relates to real world observation. Otherwise it's not physics, just math.

etotheipi said:
In the context of this thread, it is completely unsatisfactory to appeal to a real, physical device - an accelerometer - in the theory.
You seem to confuse "theory" with "mathematical model". A physical theory also includes the description of the relationship between the mathematical model and the observation.
 
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  • #56
feynman1 said:
If This acceleration is most naturally quantified with respect to inertial frames, then are inertial frames absolute and are they always inertial?
In General Relativity, there are no global inertial frames. What there is, instead, are "local inertial frames". At any point in spacetime, you can create a coordinate system that is approximately inertial in a small enough region around that point.

The criterion for a system ##(x,y,z,t)## to be inertial is, roughly speaking, that a point mass that is not affected by any non-gravitational forces will travel along straight lines: ##\frac{dx}{dt} = \text{constant}##, ##\frac{dy}{dt} = \text{constant}##, ##\frac{dz}{dt} = \text{contant}##. You can't make this absolutely true in the real universe, because of spacetime curvature. But what you can do is make it approximately true, which means that for any desired level of accuracy in the measurement of velocities, you can choose an appropriately small region of spacetime and an appropriate coordinate system such that ##\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}## for test particles don't change within that region, to that level of accuracy. Or something like that.
 
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  • #57
feynman1 said:
If a frame is a non inertial frame, then it must have an acceleration.
Didn't read the entire thread, but there are non-inertial frames that do not involve acceleration.
A rotating frame is just one of them. Rotation, like acceleration, is absolute although technically needs a relation to an inertial frame to specify the center of rotation.
The cosmological frame (the one used when saying that visible galaxy X is currently 25 billion light years away) is also non-inertial and yet does not involve proper acceleration of anything.

There are a host of other more exotic frames (coordinate systems) that may or may not involve proper acceleration.
 
  • #58
etotheipi said:
The point I'm making is that I haven't really noticed any classical mechanics textbooks use 'proper acceleration', and there isn't particularly a need to introduce it.
I disagree. Let's look at Newton's laws of motion. The first two just say that the acceleration of an object is proportional to the force acting on that object. That is true whether or not you consider "inertial forces" to be real forces, or not.

But then look at the third law, which can be stated informally as this: "If an object has a force ##\vec{F}## acting on it, then it must also exert a force on the rest of the universe that is equal to ##-\vec{F}##".

The third law is the basis for the conservation of momentum. Strictly speaking, Newton's formulation only applied to forces between objects and didn't consider the possibility of forces due to fields, but in later formulations of classical mechanics, this was extended to fields. The fields themselves carry momentum, and the third law applies to the interaction between fields and particles.

The third law only applies to proper forces (proportional to proper acceleration). In a rotating coordinate system, the "forces" such as the centrifugal force and Coriolis force don't obey the third law. Centrifugal force seems to pull an object away from the center of rotation, but there is no corresponding "equal and opposite" force exerted by that object.
 
  • #59
stevendaryl said:
I disagree. Let's look at Newton's laws of motion. The first two just say that the acceleration of an object is proportional to the force acting on that object. That is true whether or not you consider "inertial forces" to be real forces, or not.

But then look at the third law, which can be stated informally as this: "If an object has a force ##\vec{F}## acting on it, then it must also exert a force on the rest of the universe that is equal to ##-\vec{F}##".

The third law is the basis for the conservation of momentum. Strictly speaking, Newton's formulation only applied to forces between objects and didn't consider the possibility of forces due to fields, but in later formulations of classical mechanics, this was extended to fields. The fields themselves carry momentum, and the third law applies to the interaction between fields and particles.

The third law only applies to proper forces (proportional to proper acceleration). In a rotating coordinate system, the "forces" such as the centrifugal force and Coriolis force don't obey the third law. Centrifugal force seems to pull an object away from the center of rotation, but there is no corresponding "equal and opposite" force exerted by that object.
It might be true that classical physics doesn't make a big deal about proper acceleration. But that's because making the distinction between proper acceleration and coordinate acceleration is equivalent to first formulating the laws of motion in an inertial coordinate system, and then transforming to see what they are in a noninertial coordinate system.
 
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  • #60
stevendaryl said:
In General Relativity, there are no global inertial frames. What there is, instead, are "local inertial frames". At any point in spacetime, you can create a coordinate system that is approximately inertial in a small enough region around that point.

The criterion for a system ##(x,y,z,t)## to be inertial is, roughly speaking, that a point mass that is not affected by any non-gravitational forces will travel along straight lines: ##\frac{dx}{dt} = \text{constant}##, ##\frac{dy}{dt} = \text{constant}##, ##\frac{dz}{dt} = \text{contant}##. You can't make this absolutely true in the real universe, because of spacetime curvature. But what you can do is make it approximately true, which means that for any desired level of accuracy in the measurement of velocities, you can choose an appropriately small region of spacetime and an appropriate coordinate system such that ##\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt}## for test particles don't change within that region, to that level of accuracy. Or something like that.
...and "appropriately small" means that the gravitational field in the region under consideration is sufficiently homogeneous, i.e., tidal forces are negligible.
 
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