Whether a non-inertial frame is absolute

[feynman1]

If a frame is a non inertial frame, then it must have an acceleration. Then which reference frame is this acceleration with respect to? If this acceleration varies with the reference frame this acceleration is calculated with respect to, is this non inertial frame absolute?

 

[etotheipi]

This acceleration is most naturally quantified with respect to inertial frames. Exercise, for you: prove that the acceleration of a particle is invariant under a general Galilean boost.

 

[Ibix]

Then which reference frame is this acceleration with respect to?

Proper acceleration can be measured by a spring balance with no reference to any frame. A reference frame is inertial or not depending on whether objects defined to be at rest in the frame measure proper acceleration.

 

[Dale]

Then which reference frame is this acceleration with respect to?

The proper acceleration of an observer is an invariant. It is not "with respect to" anything.

The acceleration of a non-inertial frame is given as the proper acceleration of observers at rest throughout the non-inertial frame. Therefore, it is also not "with respect to" anything. It is invariant, and depending on the details of the specific non-inertial frame, it can vary from location to location throughout the frame.

Edit: thus also an inertial frame is inertial in an invariant sense and not "with respect to" anything, since inertial means roughly that the proper acceleration of a stationary object is 0 everywhere

 

[etotheipi]

Here is a question. How is the 'proper acceleration' of a point in classical mechanics to be defined, if not as the acceleration of that point with respect to an inertial co-ordinate chart (one of a class of co-ordinate charts defined by the restriction that a Galilean transformation of the co-ordinate space applied to all points in a mechanical system gives world-lines of the same system except with new initial conditions)?

 

[Dale]

Here is a question. How is the 'proper acceleration' of a point in classical mechanics to be defined, if not as the acceleration of that point with respect to an inertial co-ordinate chart (one of a class of co-ordinate charts defined by the restriction that a Galilean transformation of the co-ordinate space applied to all points in a mechanical system gives world-lines of the same system except with new initial conditions)?

I prefer to define it as the quantity that is measured by an accelerometer. But the two definitions are equivalent.

 

[feynman1]

This acceleration is most naturally quantified with respect to inertial frames. Exercise, for you: prove that the acceleration of a particle is invariant under a general Galilean boost.

If This acceleration is most naturally quantified with respect to inertial frames, then are inertial frames absolute and are they always inertial?

 

[feynman1]

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?

 

[Dale]

If This acceleration is most naturally quantified with respect to inertial frames, then are inertial frames absolute and are they always inertial?

Um, yes, inertial frames are always inertial.

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?

Not all kinematic quantities are relative. Proper time is not, nor is proper acceleration.

In fact, it is necessary that proper acceleration be invariant. Otherwise you could not unambiguously define an inertial frame.

 

[feynman1]

Um, yes, inertial frames are always inertial. Rainy afternoons are always rainy. And magical fairies are always magical.

Then this reference frame has an acceleration=0, then how is this acceleration measured, is it measured with respect to a reference frame?

 

[Dale]

Then this reference frame has an acceleration=0, then how is this acceleration measured, is it measured with respect to a reference frame?

It is measured with an accelerometer. As I said above.

 

[feynman1]

Um, yes, inertial frames are always inertial.

Not all kinematic quantities are relative. Proper time is not, nor is proper acceleration.

In fact, it is necessary that proper acceleration be invariant. Otherwise you could not unambiguously define an inertial frame.

Okay thanks but can we speak only Newtonian mechanics and forget about special relativity?

 

[Dale]

Okay thanks but can we speak only Newtonian mechanics and forget about special relativity?

No, I don’t think so. Newton didn’t have a very clear concept of reference frames at all, and some of his key ideas on the matter were wrong. The modern clarity on reference frames was largely due to relativity forcing us to confront and fix some Newtonian errors.

In any case, Newtonian physics has more invariants, not more relative quantities

 

[feynman1]

No, I don’t think so. Newton didn’t have a very clear concept of reference frames at all, and some of his key ideas on the matter were wrong. The modern clarity on reference frames was largely due to relativity forcing us to confront and fix some Newtonian errors.

In any case, Newtonian physics has more invariants, not more relative quantities

Got it thanks. So is my question unanswerable at all within the Newtonian framework?

 

[etotheipi]

The question is quite answerable completely within the classical framework. Did you try to prove the invariance of acceleration under a general Galilean transformation, as in post #2?

 

[Dale]

Got it thanks. So is my question unanswerable at all within the Newtonian framework?

Which of your many questions are you specifically referring to?

 

[feynman1]

Which of your many questions are you specifically referring to?

the original post

 

[feynman1]

The question is quite answerable completely within the classical framework. Did you try to prove the invariance of acceleration under a general Galilean transformation, as in post #2?

As per the galilean transformation, two systems need to move relative to each other with a constant speed. Then what if there’s a relative acceleration between them and how to apply the galilean transformation?

 

[Dale]

the original post

I already answered it in post 4, which is valid in Newtonian physics too

 

[feynman1]

I already answered it in post 4, which is valid in Newtonian physics too

Is proper acceleration even defined in classical physics/introductory physics before relativity?

 

[etotheipi]

As per the galilean transformation, two systems need to move relative to each other with a constant speed. Then what if there’s a relative acceleration between them and how to apply the galilean transformation?

The point is that in classical mechanics there are certain distinguished frames of reference, so-called inertial frames, where Newton's law of inertia holds (free particles move at constant velocities). A Galilean transformation maps from one inertial frame to a different inertial frame. (And of course a transformation between two frames undergoing non-zero relative acceleration is not Galilean...!)

Take, for instance, a particle with a trajectory $\mathbf{x} = \mathbf{X}(t)$ in one inertial frame. It's acceleration in this frame is nothing but $\mathbf{a}(t) = \ddot{\mathbf{X}}(t)$. Under a Galilean transformation, for example to a second inertial frame moving at a constant velocity $\mathbf{v}$ with respect to the first and with initial displacement $\mathbf{s}$, the trajectory in this new inertial frame is $\mathbf{x}' = \mathbf{X}'(t) = \mathbf{X}(t) - \mathbf{v}t - \mathbf{s}$, which again implies $\mathbf{a}'(t) = \ddot{\mathbf{X}}(t)$. The acceleration of the particle as measured by any inertial frame is then the same, i.e. $\mathbf{a}(t) = \mathbf{a}'(t)$.

That's partly why inertial frames are distinguished, because the accelerations of particles as measured by any frame in that class are invariant. Whilst, by a suitable choice of non-inertial frame, you can make the acceleration whatever you want

 

[Vanadium 50]

Okay thanks but can we speak only Newtonian mechanics and forget about special relativity?

An odd complaint, since the person who dragged in SR was...you.

I count five answers to your original question. Clearly these answers don't satisfy you. It might help if you explained how and why they don't satisfy you. Otherwise we will keep going around in circles.

 

[Dale]

Is proper acceleration even defined in classical physics/introductory physics before relativity?

Proper acceleration is just the acceleration measured by an accelerometer. It is perfectly compatible with Newtonian physics.

 

[PeterDonis]

Aren't all kinematic quantities measured w.r.t a reference frame?

No.

 

[A.T.]

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?

Proper acceleration can be thought of as relative to a local free falling frame. But this boils down to what an accelerometer measures, so you don't need to introduce that frame.

 

[Nugatory]

Is proper acceleration even defined in classical physics/introductory physics before relativity?

Yes. It’s essential to understanding what’s going on with centripetal and centrifugal forces when you’re swinging a weight on a string in a circle, for example.

Introductory physics texts often simplify things by choosing coordinates in which the frame-dependent coordinate acceleration (which is “relative”) happens to be equal to the proper acceleration. This simplification is one of the things that you’ll unlearn when you come to a more complete treatment of the subject.

 

[etotheipi]

Yes. It’s essential to understanding what’s going on with centripetal and centrifugal forces when you’re swinging a weight on a string in a circle.

How so? You can do all the analysis simply with the definition of acceleration with respect to a frame $F$ $\mathbf{a} \big{|}_{F} = \frac{\mathrm{d^2} \boldsymbol{x}}{\mathrm{d}t^2} \big{|}_F$ and if you're using a non-inertial frame then the additional relationship $\frac{\mathrm{d} \mathbf{u}}{\mathrm{d}t} \big{|}_{Oxyz} = \frac{\mathrm{d} \mathbf{u}}{\mathrm{d}t} \big{|}_{O'x'y'z'} + \boldsymbol{\omega} \times \mathbf{u}$ for any arbitrary vector $\mathbf{u}$, which is required in order to obtain the equation of motion in the non-inertial frame (either by finding the Lagrangian in the rotating frame, or by deriving the relationship between $\mathbf{a} \big{|}_{Oxyz}$ and $\mathbf{a} \big{|}_{O'x'y'z'}$ and then substituting for $\mathbf{a} \big{|}_{Oxyz}$ in Newton's equation, where $Oxyz$ constitute inertial co-ordinates).

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.

 

[Dale]

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.

The term may not be used, but accelerometers are classical devices and their readings are perfectly compatible with classical mechanics. There is nothing inherently non-classical about proper acceleration, and borrowing the term aids greatly in clarity.

 

[Ibix]

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.

The $\left.\mathbf{a}\right|_F$ you defined is coordinate acceleration, and physical accelerometers will not measure this in general. So you clearly have two concepts here - coordinate acceleration and whatever it is the accelerometers are measuring. The latter is proper acceleration, whether it's called that or not, and I would say that it's actually the truly important concept because it's where the maths corresponds to something directly measurable. (Edit: there's a gag about using the proper terms for things here somewhere, but it'd add more confusion than it'd be worth.)

 

[etotheipi]

The $\left.\mathbf{a}\right|_F$ you defined is coordinate acceleration, and physical accelerometers will not measure this in general. So you clearly have two concepts here - coordinate acceleration and whatever it is the accelerometers are measuring. The latter is proper acceleration, whether it's called that or not, and is actually the truly important concept because it's where the maths corresponds to something directly measurable.

But that can just be stated: let $K$ be any inertial reference system, then what you call the proper acceleration is what I would denote $\mathbf{a} \big{|}_K$ - accelerometers are a nice heuristic, but you can't really define an accelerometer mathematically. And I don't see the point of introducing extraneous terminology.

It's nicer to start from the principle of Galilean relativity, which implies a Galilean structure defined by the property that Newton's equation is invariant with respect to the group of Galilean transformations, and from that derive that a mechanical system of one free particle has zero acceleration in any inertial reference system in this class^*. Then it's already clear that non-zero accelerations with respect to inertial frames have physical significance, and not least because the values of these accelerations are invariant under Galilean transformations.

Relativity is different, because there the four-acceleration $a^{\mu} = \mathrm{d}^2 x^{\mu} / \mathrm{d}s^2$ can be defined without even acknowledging any reference system, and is an entirely different quantity to co-ordinate acceleration. Whilst in classical mechanics, 'proper acceleration' is just a special case of co-ordinate acceleration!

^* [i.e. use that in an inertial reference system, the acceleration $\ddot{\boldsymbol{x}} = \mathbf{\mathcal{F}}(\boldsymbol{x}, \dot{\boldsymbol{x}}, t)$ of a mechanical system can only depend on relative positions and velocities $\ddot{\boldsymbol{x}} = \varphi( \{ \boldsymbol{x}_i - \boldsymbol{x}_j, \dot{\boldsymbol{x}}_i - \dot{\boldsymbol{x}}_j \})$ for some $\varphi$, and is also rotationally invariant $\mathcal{F}(G\boldsymbol{x}, G\dot{\boldsymbol{x}}) = G\mathcal{F}(\boldsymbol{x}, \dot{\boldsymbol{x}})$ any for orthogonal matrix $G$]