Whether a non-inertial frame is absolute

In summary, according to the speaker, proper acceleration is invariant under a general Galilean boost. This means that inertial frames are always inertial, and that the magnitude of the acceleration is relative to some frame.
  • #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.
 
Last edited by a moderator:
Physics news on Phys.org
  • #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.
 
  • Like
Likes nasu, Dale and etotheipi
  • #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?
 
  • Like
Likes etotheipi
  • #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.
 
  • Like
Likes Dale, etotheipi and vanhees71
  • #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.
 
  • Like
Likes Dale and etotheipi
  • #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:
 
  • Like
Likes vanhees71
  • #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!
 
Last edited by a moderator:
  • #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.
 
  • Like
Likes vanhees71 and etotheipi
  • #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.
 
Last edited:
  • Like
Likes vanhees71
  • #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.
 
  • Like
Likes vanhees71, Ibix, Motore and 3 others
  • #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.
 
Last edited by a moderator:
  • #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?
 
  • Haha
Likes atyy
  • #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.
 
  • Like
Likes nasu, Dale and Vanadium 50
  • #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.
 
Last edited by a moderator:
  • Like
Likes hutchphd and vanhees71
  • #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.
 
  • Like
Likes PeterDonis, Vanadium 50 and vanhees71
  • #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.
 
  • Like
Likes vanhees71
  • #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.
 
  • Like
Likes vanhees71
  • #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.
 
  • Like
Likes stevendaryl
  • #61
etotheipi said:
This is personal preference :smile:

To me it is purely a model.
On this forum we use the professional scientific literature as the standard. It could be that there are professional sources that use the term “classical mechanics” in that way, but I have not seen them. Most sources describe classical mechanics as a scientific theory describing real-world macroscopic objects.

“The mathematical study of the motion of everyday objects and the forces that affect them is called classical mechanics.” https://www.livescience.com/47814-classical-mechanics.html

“Classical mechanics is used for describing the motion of macroscopic objects, from projectiles to parts of machinery, as well as astronomical objects, such as spacecraft , planets, stars, and galaxies. It produces very accurate results within these domains, and is one of the oldest and largest subjects in science, engineering and technology.” https://www.newworldencyclopedia.org/entry/Classical_mechanics

“Classical mechanics is the study of the motion of bodies (including the special case in which bodies remain at rest) in accordance with the general principles first enunciated by Sir Isaac Newton” http://farside.ph.utexas.edu/teaching/301/lectures/node3.html

“Classical mechanics is a physical theory describing the motion of macroscopic objects, from projectiles to parts of machinery, and astronomical objects, such as spacecraft , planets, stars, and galaxies” https://en.m.wikipedia.org/wiki/Classical_mechanics

And similarly for textbooks I have read. I have yet to see any standard source that claims that classical mechanics is to be identified with the mathematical model regardless of the presence or absence of any connection to experiment.

Have you seen such sources?
 
  • Like
Likes nasu
  • #62
stevendaryl said:
The third law only applies to proper forces (proportional to proper acceleration)
This is the General Relativity view. In standard Classical Mechanics, Newtonian Gravitation is a force that obeys the third Law, even tough it doesn't cause proper acceleration. Although Classical Mechanics could be formulated differently (as already mentioned here) this is the standard formulation.
 
  • #63
A.T. said:
This is the General Relativity view. In standard Classical Mechanics, Newtonian Gravitation is a force that obeys the third Law, even tough it doesn't cause proper acceleration. Although Classical Mechanics could be formulated differently (as already mentioned here) this is the standard formulation.
I wasn't actually talking about gravity. I was talking about other inertial forces such as the centrifugal force. There is no "equal and opposite" force to centrifugal force.

From the point of view of Newtonian physics, gravity is a real force, and acceleration due to gravity is proper acceleration.
 
  • #64
stevendaryl said:
From the point of view of Newtonian physics, gravity is a real force, and acceleration due to gravity is proper acceleration.
That is not how I see the term "proper acceleration" usually used. This would be "classical proper acceleration", which does not always correspond to the reading of an accelerometer.
 
  • #65
A.T. said:
That is not how I see the term "proper acceleration" usually used. This would be "classical proper acceleration", which does not always correspond to the reading of an accelerometer.

Good point. So we have two different notions of inertial coordinate system, and in Newtonian physics, they disagree in the presence of gravity:

  1. Proper acceleration means what an accelerometer shows. An inertial coordinate system is one in which objects at "rest" have zero proper acceleration.
  2. Proper acceleration is the acceleration as measured in an inertial coordinate system. An inertial coordinate system is one in which there are no "fictitious forces".

The second definition leaves a circularity that maybe can be resolved by applying Newton's third law. A "real" force is one for which there is a corresponding equal and opposite force?
 
  • #66
A.T. said:
This is the General Relativity view. In standard Classical Mechanics, Newtonian Gravitation is a force that obeys the third Law, even tough it doesn't cause proper acceleration. Although Classical Mechanics could be formulated differently (as already mentioned here) this is the standard formulation.
I wonder, whether the notion of "proper acceleration" or "proper xyz" makes much sense within Newtonian physics. There we have simply "absolute time" and "absolute space" which are postulated to exist and providing an inertial reference frame. Time is a directed 1D oriented manifold and space a Euclidean affine space. This is the modern, abstract formulation for Newton's Lex Prima.

Of course, there's no way to really determine operationally Newton's absolute time and space, even within Newtonian mechanics, because of Galileo symmetry, i.e., any other reference frame moving with constant velocity against this fictitious absolute inertial frame has the very same properties as this absolute frame itself, but this doesn't matter for Newtonian mechanics at all.

Further the physical laws are all formulated wrt. the class of inertial reference frames (Lex II and III).

Acceleration is independent of the particular choice of an inertial reference frame, because it's a Galileo-invariant vector.

Gravitation in Newtonian physics is simply an interaction acting instantaneously at a distance, i.e., two bodies attract each other according to Newton's universal law of gravity,
$$\vec{F}_{12}=-\frac{G m_1 m_2}{|\vec{r}_1-\vec{r}_2|^3} (\vec{r}_1-\vec{r}_2)=-\vec{F}_{21}.$$
As any dynamical theory you can formulate also Newtonian mechanics in a generally covariant way, i.e., expressing the fundamental laws, as given in the class of inertial frames, in arbitrary coordinates, defined using a reference body that moves with arbitrary acceleration against the inertial frames. All you have to do is to introduce a "covariant time derivative" for vector components wrt. this non-inertial frame, if it is rotating against the inertial frames of reference (implying also how to take covariant time derivatives for any tensor components, but these you almost never need in Newtonian mechanics, except the tensor of inertia of a rigid body). Neither absolute time nor space as an affine Euclidean manifold for any non-inertial observer changes.

The funny thing is that accelerometers like the one in our smartphones indeed measure "proper accelaration" by construction. It shows ##\vec{a}=0## when in free fall in the gravitational field of the Earth which can indeed by taken as homogeneous in the usual cases of application. In a reference frame at rest wrt. the Earth it shows ##\vec{a}=\vec{g}## with ##\vec{g} \approx 9.81 \;\text{m}/\text{s}^2##.

Also this is very natural within Newtonian mechanics, because indeed a reference frame freely falling in a gravitational field of the Earth (as far as one can neglect tidal forces, i.e., can consider the gravitational field as homogeneous), is indeed an inertial frame of reference (due to the equivalence principle, which in Newtonian mechanics reduces to an empirical not further explainable fact that inertial an gravitational mass are equal).
 
Last edited:
  • Like
Likes atyy
  • #67
Dale said:
On this forum we use the professional scientific literature as the standard. It could be that there are professional sources that use the term “classical mechanics” in that way, but I have not seen them. Most sources describe classical mechanics as a scientific theory describing real-world macroscopic objects.

“The mathematical study of the motion of everyday objects and the forces that affect them is called classical mechanics.” https://www.livescience.com/47814-classical-mechanics.html

“Classical mechanics is used for describing the motion of macroscopic objects, from projectiles to parts of machinery, as well as astronomical objects, such as spacecraft , planets, stars, and galaxies. It produces very accurate results within these domains, and is one of the oldest and largest subjects in science, engineering and technology.” https://www.newworldencyclopedia.org/entry/Classical_mechanics

“Classical mechanics is the study of the motion of bodies (including the special case in which bodies remain at rest) in accordance with the general principles first enunciated by Sir Isaac Newton” http://farside.ph.utexas.edu/teaching/301/lectures/node3.html

“Classical mechanics is a physical theory describing the motion of macroscopic objects, from projectiles to parts of machinery, and astronomical objects, such as spacecraft , planets, stars, and galaxies” https://en.m.wikipedia.org/wiki/Classical_mechanics

And similarly for textbooks I have read. I have yet to see any standard source that claims that classical mechanics is to be identified with the mathematical model regardless of the presence or absence of any connection to experiment.

Have you seen such sources?
"Yet within its known slow-motion domain, Newton’s theory will undoubtedly continue to be used for reasons of conceptual and technical convenience. And as a logical construct it will remain as perfect and inviolate as Euclid’s geometry." Rindler, Relativity, 2e, 2006, p109
 
  • Like
Likes Dale and vanhees71
  • #68
A.T. said:
In standard Classical Mechanics, Newtonian Gravitation is a force that obeys the third Law, even tough it doesn't cause proper acceleration.
This is actually the main reason that I hesitate to say things like “gravity isn’t a force in GR”. At least not without adding “locally”.
 
  • Like
Likes vanhees71
  • #69
atyy said:
"Yet within its known slow-motion domain, Newton’s theory will undoubtedly continue to be used for reasons of conceptual and technical convenience. And as a logical construct it will remain as perfect and inviolate as Euclid’s geometry." Rindler, Relativity, 2e, 2006, p109
I don’t think that this is indeed claiming that classical mechanics refers to the logical construct alone, but I can see how someone with that opinion could read it thus.
 
Last edited:
  • Like
Likes vanhees71
  • #70
etotheipi said:
To me it is purely a model.
A model, by itself, is math, not physics. It's only physics if it makes predictions that can be compared with reality. If it's just a model with no possible connection to reality, nobody would call it "physics".
 
  • Like
Likes artis, Vanadium 50 and vanhees71

Similar threads

Replies
3
Views
2K
Replies
18
Views
1K
Replies
15
Views
3K
Replies
2
Views
1K
Replies
5
Views
4K
Replies
7
Views
2K
Replies
25
Views
2K
Replies
13
Views
2K
Back
Top