# Whether a non-inertial frame is absolute

Apologies for a late reply; the pubs were open again today 😄

Given any frame ##F = P \zeta \eta \xi## one can define the time derivative with respect to ##F## of an arbitrary vector ##\mathbf{u}##$$\mathcal{D}_F \mathbf{u} := \frac{\mathrm{d}\mathbf{u}}{\mathrm{d}t} \big{|}_{F} = \frac{\mathrm{d}{u}^{\zeta}}{\mathrm{d}t} \mathbf{e}_{\zeta} + \frac{\mathrm{d}{u}^{\eta}}{\mathrm{d}t} \mathbf{e}_{\eta} + \frac{\mathrm{d}{u}^{\xi}}{\mathrm{d}t} \mathbf{e}_{\xi}$$It's easy to show, given another frame ##G = Q \rho \sigma \tau##, that ##\mathcal{D}_F \mathbf{u} = \mathcal{D}_G \mathbf{u} + \boldsymbol{\omega} \times \mathbf{u}## where ##\omega_i = \frac{1}{2} \epsilon_{ijk} \dot{\mathcal{R}}_{jl} \mathcal{R}_{kl}## and ##(\mathcal{R}_{ij})## is the orthogonal, time-dependent matrix such that ##\mathbf{e}^G_i = \mathcal{R}_{ij} \mathbf{e}^F_j##.

Now let us show how to obtain the equations of motion for a particle in an arbitrary reference system. Begin with an inertial reference system ##K = Oxyz##, in which the Lagrangian is simply$$\mathscr{L} = \frac{1}{2} m v^2 - \varphi$$where ##\mathbf{v} := \mathcal{D}_K \boldsymbol{x}##. Let's now consider a second reference system ##L = O'x'y'z'## and define ##\mathbf{R}(t) := \overrightarrow{O(t)O'(t)}##, such that ##\boldsymbol{x} = \mathbf{R} + \boldsymbol{x}'##. Operating with ##\mathcal{D}_K## yields, upon defining ## \boldsymbol{\mathcal{V}} := \mathcal{D}_K \mathbf{R}## and ##\mathbf{v}' := \mathcal{D}_L \boldsymbol{x}'##\begin{align*}

\mathcal{D}_K \boldsymbol{x} &= \mathcal{D}_K \mathbf{R} + \mathcal{D}_K \boldsymbol{x}' \\

\mathbf{v} &= \boldsymbol{\mathcal{V}} + \left[ \mathcal{D}_L \boldsymbol{x}' + \boldsymbol{\omega} \times \boldsymbol{x}' \right] \\

\mathbf{v} &= \boldsymbol{\mathcal{V}} + \mathbf{v}' + \boldsymbol{\omega} \times \boldsymbol{x}'

\end{align*}It follows that\begin{align*}v^2 = \mathcal{V}^2 + v'^2 + (\boldsymbol{\omega} \times \boldsymbol{x}')^2 + 2 \mathbf{v}' \cdot \boldsymbol{\omega} \times \boldsymbol{x}' + 2 \boldsymbol{\mathcal{V}} \cdot \mathbf{v}' + 2 \boldsymbol{\mathcal{V}} \cdot \boldsymbol{\omega} \times \boldsymbol{x}'

\end{align*}The new Lagrangian is then written\begin{align*}
\mathscr{L}' = \frac{1}{2} m \left( \mathcal{V}^2 + v'^2 + \boldsymbol{\omega} \cdot (\boldsymbol{x}' \times (\boldsymbol{\omega} \times \boldsymbol{x}')) + 2 \mathbf{v}' \cdot \boldsymbol{\omega} \times \boldsymbol{x}' + 2 \boldsymbol{\mathcal{V}} \cdot \mathbf{v}' + 2 \boldsymbol{\mathcal{V}} \cdot \boldsymbol{\omega} \times \boldsymbol{x}' \right) - \varphi
\end{align*}where we have re-written ##(\boldsymbol{\omega} \times \boldsymbol{x}')^2 = \boldsymbol{\omega} \cdot (\boldsymbol{x}' \times (\boldsymbol{\omega} \times \boldsymbol{x}'))##. Let us now compute the derivatives of the Lagrangian\begin{align*}

\frac{\partial \mathscr{L}'}{\partial x'^i} &= -m \epsilon_{ijk} \epsilon_{klm} \omega^j \omega^l x'^m -m \epsilon_{ijk}\omega^j (\mathcal{V}^k + \dot{x}'^k) - \partial_i \varphi \\ \\

\frac{\mathrm{d}}{\mathrm{d}t} \frac{\partial \mathscr{L}'}{\partial \dot{x}'^i} &= m \dot{\mathcal{V}}^i + m \ddot{x}'^i + m\epsilon_{ijk} ( \dot{\omega}^j x'^k + \omega^j \dot{x}'^k)

\end{align*}The Euler-Lagrange equation ##\partial \mathscr{L}' / \partial \boldsymbol{x}' = d_t \left(\partial \mathscr{L}' / \partial \dot{\boldsymbol{x}}' \right)## thus implies the following equation of motion in the ##L## reference system, after converting from suffix notation back into vector notation,$$m \mathbf{a}' = - \nabla \varphi - m \boldsymbol{\mathcal{A}} - 2m \boldsymbol{\omega} \times \mathbf{v}' - m \boldsymbol{\alpha} \times \boldsymbol{x}' -m \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{x}')$$where ##\boldsymbol{\mathcal{A}} := \mathcal{D}_K \boldsymbol{\mathcal{V}} = \mathcal{D}_L \boldsymbol{\mathcal{V}} + \boldsymbol{\omega} \times \boldsymbol{\mathcal{V}} ## is the translational acceleration of the ##L##-system with respect to the ##K##-system and ##\boldsymbol{\alpha} := \mathcal{D}_{K} \boldsymbol{\omega}## is the rotational acceleration of the ##L##-system with respect to the ##K##-system. And of course ##\mathbf{a}' = \mathcal{D}_L^2 \boldsymbol{x}'## is the acceleration of the particle with respect to the ##L##-system.

This is the equation of motion of a particle in an arbitrary reference system ##L##; the additional terms arising from the co-ordinate transformation are not forces, although they are sometimes affectionally referred to as 'inertial forces'.

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Haborix, atyy and vanhees71
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Apologies for a late reply; the pubs were open again today 😄
Good to see you have your priorities in order.

vanhees71, berkeman, Haborix and 2 others
artis
@etotheipi Of all the people I know that are fans of pubs you seem to have the best mathematical skills I must say.
I would like to have a drink with you in a nice inertial frame but only if you promise me you wouldn't introduce any centrifugal forces. Otherwise it might end badly for my stomach.

vanhees71
2022 Award
@etotheipi Of all the people I know that are fans of pubs you seem to have the best mathematical skills I must say.
I would like to have a drink with you in a nice inertial frame but only if you promise me you wouldn't introduce any centrifugal forces. Otherwise it might end badly for my stomach.
"The Inertial Frame" would be an excellent name for a pub. People would naturally remain at rest inside.

artis
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).
But in GR, one doesn't even need to neglect tidal forces. For geodesic motion, the proper acceleration is zero even if tidal forces are not zero. So maybe in that sense, the more natural language might be Newton-Cartan, as @Dale was perhaps suggesting, though I don't know that theory well enough to know if the proper acceleration carries over from GR.

Can such a thing as an infinitesimal accelerometer exist (so that it measures proper acceleration even when tidal forces are not zero)? If it cannot, then what an accelerometer measures is not the proper acceleration.

In the standard Newtonian formulation. I agree the proper acceleration is not so natural a concept, so I would prefer to use the term specific force for what an accelerometer measures. Wikipedia does use "proper acceleration" as a synonym for "specific force". But is the specific force (defined in standard Newtonian language) the same as the proper acceleration in Newton-Cartan (defined by inheritance from GR)?

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Dale and vanhees71
Gold Member
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If a real gravitational field is present also within GR you do not have a global inertial frame of reference. A free-falling reference frame is inertial strictly speaking only in one point, where all the Christoffel symbols are made to vanish. Because at presence of a gravitational field the curvature tensor is not 0, and this is true independently of the choice of the reference frame or coordinates, i.e., you always have tidal forces in a free-falling reference frame. It depends on the accuracy you aim to measure them how large the region in spacetime is you choose to neglect them.

Dale
Mentor
But is the specific force (defined in standard Newtonian language) the same as the proper acceleration in Newton-Cartan (defined by inheritance from GR)?
As far as I can tell, yes.

So maybe in that sense, the more natural language might be Newton-Cartan, as @Dale was perhaps suggesting, though I don't know that theory well enough to know if the proper acceleration carries over from GR.
Proper acceleration, being the reading on an accelerometer, is not a theory-based concept. It is an operational definition. You can use it in any theory.

Can such a thing as an infinitesimal accelerometer exist (so that it measures proper acceleration even when tidal forces are not zero)? If it cannot, then what an accelerometer measures is not the proper acceleration.
This is a practical issue, but not an in-principle problem. The usual alternative approach for identifying an inertial frame relies on isolated particles. That also is rife with practical problems, but is not in-principle problematic.

atyy
Proper acceleration, being the reading on an accelerometer, is not a theory-based concept. It is an operational definition. You can use it in any theory.
But is this true even when we consider that the accelerometer must be calibrated? Wouldn't one need Newtonian mechanics and inertial frames (or GR and proper acceleration) to calibrate an accelerometer?

vanhees71 and Dale
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But is this true even when we consider that the accelerometer must be calibrated?
It is not true that accelerometers must be calibrated. You can build accelerometers that don’t require calibration, but their manufacturing is more demanding and expensive. Such expensive non-calibrated accelerometers are mostly used for gravimetry.

Building a device that requires calibration is mostly an economic decision. By requiring calibration you can shift some of the device’s cost from the manufacturer to the user. So for a similar precision a calibrated device can be less expensive to purchase than an un calibrated device. That applies for accelerometers as well as other devices.

The thing with accelerometers is that a stable and accurate reference standard is free and readily available. So most accelerometers are designed requiring calibration because customers are willing to accept the burden of calibration for an accelerometer whereas they may not for other devices with less available standards.

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Mentor
For geodesic motion, the proper acceleration is zero even if tidal forces are not zero.
But you can't treat a frame in which all of the geodesics in question have constant spatial coordinates as an inertial frame if tidal forces are not zero, because in an inertial frame, geodesics that start out at rest relative to each other must remain at rest relative to each other. That will not be the case if tidal forces are not zero.

Mentor
Can such a thing as an infinitesimal accelerometer exist (so that it measures proper acceleration even when tidal forces are not zero)? If it cannot, then what an accelerometer measures is not the proper acceleration.
I'm not sure what you're trying to say here. In terms of the math of GR, an accelerometer measures the path curvature of a single worldline. It does not measure any relationship between multiple worldlines. So in terms of the math of GR, all accelerometers are infinitesimal. (Note that a geodesic worldline has zero path curvature, so if tidal gravity is present, accelerometers following multiple geodesic worldlines will all read zero even as those worldlines converge or diverge due to tidal gravity.)

In practice, of course, any actual accelerometer must have a finite size ##L##; so any actual accelerometer has two practical limitations: (1) it can only measure accelerations that are much smaller than ##1 / L##; and (2) it will not work at all unless whatever spacetime curvature (tidal gravity) is present has magnitude much less than ##1 / L^2##. (I am here using units in which ##G = c = 1##, which is common in GR.)

These practical limitations are easily met here on Earth, or even in all the other places we have so far sent accelerometers. A typical accelerometer we use might have, say, ##L = 0.1 \ \text{m}##. This means ##1 / L## is about ##10^{17} \ \text{g}## (it's ##c^2 / L## in conventional units), and ##1 / L^2## is the spacetime curvature corresponding to an energy density of about ##10^{44} \ \text{kg m}^{-3}##, or ##10^{41}## times the density of water (the energy density ##\rho## in conventional units is given by ##c^4 / 8 \pi G L^2##). So the accelerometers we use in practice present no problem at all as far as treating them as infinitesimal for the accelerations and tidal gravity conditions in which we use them.

etotheipi
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In practice, of course, any actual accelerometer must have a finite size ##L##; so any actual accelerometer has two practical limitations: (1) it can only measure accelerations that are much smaller than ##1 / L##; and (2) it will not work at all unless whatever spacetime curvature (tidal gravity) is present has magnitude much less than ##1 / L^2##. (I am here using units in which ##G = c = 1##, which is common in GR.)
I left out a third limitation, which is actually the most restrictive one for us here on Earth (and anywhere else we have so far sent accelerometers): the size ##L## of the accelerometer must be much smaller than the distance over which the local "acceleration due to gravity" varies by an amount large enough that the accelerometer is supposed to be able to detect the difference. This is not quite the same as the second limitation, because here we're not concerned with spacetime curvature being large enough to disrupt the internal structure of the accelerometer (which is what the second limitation is about); here we're concerned with the accelerometer being large enough that there won't be a single value of spacetime curvature everywhere inside it, to whatever degree of accuracy it can measure.

So, for example, spacetime curvature in Earth's vicinity is given roughly by ##G M / R^3## (I'll switch to conventional units here--there is variation with direction but we'll ignore it for this calculation since we're just looking at rough orders of magnitude), where ##M## is the Earth's mass and ##R## is the distance from the Earth's center. The equivalent "acceleration" this produces across an accelerometer of size ##L## is ##G M L / R^3##; the smallest acceleration the accelerometer can distinguish will have to be significantly larger than this. If we plug in ##L = 0.1 \ \text{m}## and assume we're on the Earth's surface, we get about ##10^{-7} \ \text{m s}^{-2}##, or about ##10^{-8} \ \text{g}##; so an accelerometer of that size here on Earth will only be able to distinguish accelerations that differ by significantly more than that.

etotheipi
Homework Helper
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Proper acceleration, being the reading on an accelerometer, is not a theory-based concept. It is an operational definition. You can use it in any theory.
I need to understand better: this seems totally disingenuous to me. Operationally how do you construct this instrument without theoretical reference? Starting from a world without accelerometers, what does this process look like? Unless they are identical how do we know they measure the same thing?

Mentor
Operationally how do you construct this instrument without theoretical reference?
With blueprints or schematics. I am not being disingenuous, but it really is that simple. We can construct reference accelerometers from known designs, or we can just purchase them commercially with known accuracy and precision. They are standard pieces of lab equipment, just like clocks.

Starting from a world without accelerometers, what does this process look like?
We are not starting from a world without accelerometers. They do exist and regardless of your theory they produce some set of numbers. The numbers that they produce are in fact independent of the theory used to analyze the device. If your theory predicts different numbers then the theory is falsified (or the device is broken).

This is not unique to accelerometers, but applies to any measurement device. Scales, clocks, rulers, voltmeters, ammeters, thermometers, etc. all produce measurements that are theory independent in the same sense as I intended above.

[accelerometers] do exist and regardless of your theory they produce some set of numbers. The numbers that they produce are in fact independent of the theory used to analyze the device
Thus the real-world accelerometers cannot be part of any particular theory itself! It seems really unnatural to reference a real-world, manufactured device when formulating the foundations of classical mechanics.

It'd be okay, I think, to define an ideal accelerometer as a theoretical notion (similar in spirit to how one defines an ideal clock) which simply measures the curvature at a point of a world line on a manifold equipped with a connection. And a real, physical accelerometer would more or less approximate the ideal accelerometer.

But trying to incorporate actual real-world stuff into the theory itself will never work, IMO! One must have a fully internally consistent theory which an alien in a parallel universe with different physical laws could, hypothetically, make perfect sense of.

Mentor
One must have a fully internally consistent theory which an alien in a parallel universe with different physical laws could, hypothetically, make perfect sense of.
It depends on what you want the word "theory" to mean. Do you want it to mean just a mathematical structure, which might or might not describe anything in the actual world, but which is consistent within itself? Or do you want it to mean such a structure plus the predictions it makes and the operational rules that tell you how to compare its predictions with actual data?

You are using the word in the first sense. @Dale is using it in the second. I personally would prefer the second usage myself, since to me, calling something a "theory" when we're talking about physics, or any observational science, only makes sense if we include the predictions and the comparison with data. I could concoct all sorts of wacky but internally consistent mathematical structures and call them "theories", but I suspect you would object to such usage on the grounds that those wacky mathematical structures obviously have nothing to do with physics. The reason you're concentrating on what you call "classical mechanics" is that it does have something to do with physics: it does make predictions and there are operational rules for comparing those predictions with data, and because of that we now know that this theory is not exactly correct; it's only an approximation that works in weak gravitational fields for relative motion at speeds much less than the speed of light. But if that's why you care about "classical mechanics" as a mathematical structure, then you can't have it both ways: you can't both care about it because it does make physical predictions and does give operational rules for comparing its predictions with data, but then insist that those predictions and operational rules aren't part of the "theory". Maybe they aren't part of the mathematical structure you're looking at, but they are part of why you care about it and why the term "theory" is justified when talking about it.

Motore
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Thus the real-world accelerometers cannot be part of any particular theory itself!
Why not? Why shouldn’t a theory reference things outside of the theory?

It seems really unnatural to reference a real-world, manufactured device when formulating the foundations of classical mechanics.
Why? Don’t you want the theory to predict their behavior?

But trying to incorporate actual real-world stuff into the theory itself will never work, IMO! One must have a fully internally consistent theory which an alien in a parallel universe with different physical laws could, hypothetically, make perfect sense of.
I don’t share your opinion here. I think that operational definitions are an intrinsic part of the theory. That is the part that forms the so-called minimal interpretation. Why shouldn’t such essential concepts be part of the foundations of the theory?

Motore
It depends on what you want the word "theory" to mean. Do you want it to mean just a mathematical structure, which might or might not describe anything in the actual world, but which is consistent within itself? Or do you want it to mean such a structure plus the predictions it makes and the operational rules that tell you how to compare its predictions with actual data?
I really do mean that it should solely refer to an internally-consistent mathematical structure. The theory-to-reality mapping is something you need to build on top of that. A translator. 😄

It's really important not to conflate the real world with abstract structures that live solely in one's mind 😉

Mentor
I really do mean that it should solely refer to an internally-consistent mathematical structure.
Then, while I understand your preference, I do not think it is a very common preference among physicists. Nor do I think you can validly claim that the other preference, the one @Dale and I favor, doesn't make sense.

It's really important not to conflate the real world with abstract structures that live solely in one's mind
It's also really important not to ignore the fact that the reason we care about particular abstract structures is the predictions they make about the actual world. If the abstract structure you call "classical mechanics" didn't make such predictions, and didn't have many people believing for about two centuries that those predictions were exact, it would not be called "classical mechanics".

Dale
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It's really important not to conflate the real world with abstract structures that live solely in one's mind 😉
So let a scientific theory ##(F,I)## consist of an abstract mathematical framework ##F## and a minimal interpretation ##I:P\rightarrow R## mapping predictions ##P\subset F## to real world experimental results ##R##. Now we are in no danger of conflating ##F## with ##R##, and yet we can see that ##F## by itself is insufficient to form a theory.

It's also really important not to ignore the fact that the reason we care about particular abstract structures is the predictions they make about the actual world. If the abstract structure you call "classical mechanics" didn't make such predictions, and didn't have many people believing for about two centuries that those predictions were exact, it would not be called "classical mechanics".
But all this is precisely by construction. The universe isn't actually a four-dimensional affine space, et cetera; these are just artefacts of attempts to build a mathematical theory that resembles certain aspects of reality. That doesn't make it any less abstract! Once formulated it exists independently of any puny reality

So let a scientific theory ##(F,I)## consist of an abstract mathematical framework ##F## and a minimal interpretation ##I:P\rightarrow R## mapping predictions ##P\subset F## to real world experimental results ##R##. Now we are in no danger of conflating ##F## with ##R##, and yet we can see that ##F## by itself is insufficient to form a theory.
I enjoyed reading that, but it is certainly not a proof of anything. It's because all these terms are ill-defined anyway, and this whole disagreement is really one of opinion more than anything else 😄

Mentor
I enjoyed reading that, but it is certainly not a proof of anything. It's because all these terms are ill-defined anyway, and this whole disagreement is really one of opinion more than anything else 😄
It does show that insisting that a scientific theory contain more than merely an abstract mathematical framework in no way implies that we are conflating the abstract framework with real world things.

If you want to make an argument that your use of the word “theory” is better than mine then you need to avoid strawman arguments like the “conflate the real world” argument.

The point is really what @PeterDonis said. I don’t care about most abstract mathematical frameworks. I only care about the abstract mathematical frameworks that can be mapped to the outcomes of experiments. Furthermore, in order to use the scientific method I need both the framework and the mapping. The framework by itself neither helps me design nor analyze an experiment. So since I always use them together in science then as a scientist I want a word to refer to them together. That word is “theory”.

If you wish to recommend the word theory apply only to the framework then to convince many scientists you will also need a word to refer to the framework and the minimal interpretation together. Then you will have to explain why the term “mathematical framework” is insufficient for describing the abstract math part. Finally you will have to convince a large enough group of scientists to drop the use of “mathematical framework” adopt “theory” instead and replace the current usage of “theory” with whatever term you invented above.

To do that will require some good practical reasons why your new terminology is beneficial. So far I don’t see any benefit to it.

Mentor
But all this is precisely by construction. The universe isn't actually a four-dimensional affine space, et cetera; these are just artefacts of attempts to build a mathematical theory that resembles certain aspects of reality. That doesn't make it any less abstract! Once formulated it exists independently of any puny reality
Nobody is disputing any of this with regard to the mathematical structures alone. You don't need to explain to us that abstract mathematical structures are not the same as the real world phenomena that we use them to model. We all know that. That is not the point at issue.

The point at issue is the usage of the word "theory". Nothing that you have said, as far as I can see, gives any justification for your usage as opposed to the usage @Dale and I have described (and which I think is the usage most physicists would agree with). Saying that abstract mathematical structures are different from real world phenomena, by itself, is not an argument for your preferred usage. To make it into one, you would have to show evidence that using the word "theory" to describe the mathematical structures plus the predictions they make plus the operational rules for comparing predictions with data, causes people to confuse abstract mathematical structures with real world phenomena. You have produced no such evidence. Do you have any?

italicus
@feynman1
let’s imagine you are driving your car on a rectilinear motorway , at constant speed. Ignore, for a short time, that the Earth isn’t an inertial reference frame. At a certain moment, you accelerate the car. LEt’s suppose that the acceleration wrt the motorway is constant, as measured by an observer standing aside.
You’ll feel a force in your back, applied by the seat. That force is exerting on you what can be called “proper acceleration”. In classical mechanics , there is no difference between proper acceleration and coordinate acceleration, the one measured by the observer. So you experience a proper constant acceleration.
But when you come to Special Relativity, which can manage objects which are accelerated wrt to a coordinate frame, there is a substantial difference between proper acceleration ##\alpha## and coordinate acceleration ## {du}/{dt} ## . It can be shown that the relation between them is :

$$\alpha = \gamma(u)^3 {du}{/{dt} }$$

here ##\gamma(u)## is the Lorentz factor , which increases with the speed “u” .

The LHS is the proper acceleration, which can be assumed constant, for example. So the coordinate acceleration ## {du}/{dt} ## cannot increase indefinitely, and the speed of the object ( for example a space ship) cannot reach the speed of light.

This is called “hyperbolic motion” , because the line of universe on a Minkowski diagram is an hyperbole. This was studied by Wolfgang Rindler deeply. Look for “Rindler coordinates” .