Second postulate of SR quiz question

[PeterDonis]

You seem to be using the word observer as if it was something physical instead of a mathematical abstraction

Exactly. An "observer" as I'm using the term is a physical thing that can carry an accelerometer and observe and record its readings. It's certainly not a mathematical abstraction. You and I can be observers.

apparently a distinction can be made between an inertial observer and an inertial frame

Yes, certainly. A physical thing that can carry an accelerometer has some physical reading on its accelerometer, independent of how we describe the thing and its motion mathematically. Without such a distinction, our physical theories would have no meaning, because we would have no way of linking them up with actual observations.

What if we make the distinction between an observer at rest in the rindler coordinates, that is noninertial because it has to accelerate to counter the "acceleration" of the curvilinear coordinates, and an observer following the hyperbolic motion

These two are the same. An observer at rest in Rindler coordinates is non-inertial, and is following hyperbolic motion in Minkowski coordinates.

a Minkowski observer in Rindler coordinates, is it inertial or noninertial?

If you mean an observer at rest in Minkowski coordinates, such an observer is inertial. This observer won't be at rest in Rindler coordinates, but that is irrelevant to the question of whether the observer (as opposed to the coordinates) is inertial or not.

I think there is a problem with observers and frames as objects with motion, so it is best to stick to coordinates. If a frame or observer is something physical rather than an abstract labeling the conclusions derived from them are different.

I'm not sure why you would think this. The fact that observers are physical things doesn't mean we can't describe them mathematically. One way is to describe observers by the worldlines they follow; then we can write equations for those worldlines in different coordinate charts. Another way is to describe observers by frame fields (mappings of sets of orthonormal basis vectors to points in spacetime). I don't see that mathematically modeling observers poses any particular problem; it just requires a clear understanding of what you're trying to model.

 

[loislane]

If you mean an observer at rest in Minkowski coordinates, such an observer is inertial. This observer won't be at rest in Rindler coordinates, but that is irrelevant to the question of whether the observer (as opposed to the coordinates) is inertial or not.

And yet it is possible to Lorentz transform from Minkowski coordinates to Rindler coordinates(in the wedge area of Minkowski space where both coordinates exist), what happens to the Minkowski observer when he tries to keep at rest in the new coordinates? Does he have to accelerate uniformly?

 

[stevendaryl]

And yet it is possible to Lorentz transform from Minkowski coordinates to Rindler coordinates(in the wedge area of Minkowski space where both coordinates exist), what happens to the Minkowski observer when he tries to keep at rest in the new coordinates? Does he have to accelerate uniformly?

I'm not sure what you're asking. Are you saying that an observer is initially at rest relative to Minkowsky coordinate system, and then changes his motion so that he is at rest relative to Rindler coordinates? If that's the case, that means that the observer at some point starts undergoing constant proper acceleration.

 

[PeterDonis]

what happens to the Minkowski observer when he tries to keep at rest in the new coordinates?

Um, what? A Minkowski observer, by definition, is inertial. He can't "try to keep at rest in the new coordinates". His state of motion is already specified. When you transform to Rindler coordinates, his worldline looks like whatever it looks like.

 

[Dale]

Well, but at least there is a "restframe of the measurement apparatus",

There is a center of momentum frame for all measurement devices, and many devices have a rest frame. But that frame is not uniquely privileged from a physics standpoint and any other frame can be used.

and you statement is not always true.

My point is that YOUR statement is not always true. In order to correct what you see as a mistake in physics (overemphasis of geometry) you go way too far by asserting that it is the measurement which introduces the reference frame, and further asserting that it applies to "any measurement".

Of course, physicists measure it in the rest frame of the detectors

No, this is simply false. The device functions according to the laws of physics, the laws of physics are the same in all frames, s the physicist can use any frame they like to analyze the measurement.

It often is convenient to use the rest frame of the detector, but sometimes it is easier to use a different frame (eg the GPS ECI). If a different frame is simpler then a good physicist will choose that one. The choice of frame is not forced on the physicist by the physics or the measurement device.

Physicists do not perform an experiment in a reference frame, they perform an analysis in a reference frame.

So still, I think that the frame-free formulation of RT is a theoretical construct while to address observations you always have to introduce a reference frame

I am not disagreeing with you on this. I understand this point, but you are going way too far beyond this claim in making your case.

 

[atyy]

vanhees71 is probably thinking along the lines of MTW 13.6: "A physicist .. may use several different coordinates systems at once. But a coordinate system of special utility is one at rest relative to all the apparatus bolted into the floor and walls ... This proper reference frame ..."

 

[atyy]

I think this is the main difference between mathematicians, who are very much emphasizing the geometric aspects of spacetime.

A reference frame is somehow defined by real physical objects, be it a human being with his senses looking at a phenomenon or any fancy measurement device invented to discover accurate quantitative facts about nature that are not directly "detectable" by our senses.

I largely agree with DaleSpam's comments. But I would like to add that in a way, what you are proposing is more "geometric", not less. In your view, the laboratory walls serve as a physical manifestation of a proper reference frame and ideal clocks measure proper time, test particles and light rays follow geodesics. All those things are test objects or external rulers in the sense that they are assumed not to contribute to spacetime curvature. In conventional geometry, our rulers are rigid external objects, so I would say your view is more "geometric".

The "mathematical" view is not necessarily more "geometric". One can think of it as more correct than what you are proposing, because what they really mean is gauge-invariant (which is geometric in their sense, but not in the sense of having an external physical ruler or test particle). Of course it is not always physical in the sense of strict GR because there are neither observers nor measuring instruments in any vacuum spacetime. "It is the theory that says what can be observed"

So when martinbn says mercury follows a geodesic - it is not correct in the sense that it fails to account for mercury's perturbation of the background spacetime, but it is physical in the sense that he is talking about a gauge-invariant and hence physical object.

 

[vanhees71]

And yet it is possible to Lorentz transform from Minkowski coordinates to Rindler coordinates(in the wedge area of Minkowski space where both coordinates exist), what happens to the Minkowski observer when he tries to keep at rest in the new coordinates? Does he have to accelerate uniformly?

The transformation from Minkowski (i.e. pseudo-Caratesian coordinates in an inertial frame) to Rindler coordinates are clearly not Lorentz transformations. They are not even linear. I've given the formulae somewhere in this thread yesterday.

 

[vanhees71]

There is a center of momentum frame for all measurement devices, and many devices have a rest frame. But that frame is not uniquely privileged from a physics standpoint and any other frame can be used.

My point is that YOUR statement is not always true. In order to correct what you see as a mistake in physics (overemphasis of geometry) you go way too far by asserting that it is the measurement which introduces the reference frame, and further asserting that it applies to "any measurement".

I think it's semantics, but can you give me an example for a measurement that can be made without a clear specification of a reference frame? This is impossible, because measuring something means to have a reference you can compare the measured quantity to. Of course, measurement apparati are part of the physical system and thus follow the generally valid physical laws. Otherwise you couldn't define such a reference. Many quantities are defined in a certain reference frame by convention. E.g., cross sections in relativistic collisions are defined as if they were made in an fixed-target experiment and then expressed in a manifestly covariant way such that I don't need to think much anymore when doing a measurement and make my histograms in quantities everybody is used to.

Another convention is that phase-space distrubtion functions in relativistic kinetics are by definition Lorentz scalars. In (local) equilibrium intrinsic quantities like temperature and chemical potentials are, by convention, scalar (fields) defined in the (local) rest frame of the fluid. For details, see

http://fias.uni-frankfurt.de/~hees/publ/kolkata.pdf

All these definitions use "natural" reference frames, which are distinguished in the one or the other way by the physical situation discribed ("lab" frame of a two-body collision, (local) rest frame(s) of a fluid, etc.). All these quantities can written in manifestly covariant form and thus directly and conveniently measured in any reference frame.

 

[vanhees71]

I largely agree with DaleSpam's comments. But I would like to add that in a way, what you are proposing is more "geometric", not less. In your view, the laboratory walls serve as a physical manifestation of a proper reference frame and ideal clocks measure proper time, test particles and light rays follow geodesics. All those things are test objects or external rulers in the sense that they are assumed not to contribute to spacetime curvature. In conventional geometry, our rulers are rigid external objects, so I would say your view is more "geometric".

The "mathematical" view is not necessarily more "geometric". One can think of it as more correct than what you are proposing, because what they really mean is gauge-invariant (which is geometric in their sense, but not in the sense of having an external physical ruler or test particle). Of course it is not always physical in the sense of strict GR because there are neither observers nor measuring instruments in any vacuum spacetime. "It is the theory that says what can be observed"

So when martinbn says mercury follows a geodesic - it is not correct in the sense that it fails to account for mercury's perturbation of the background spacetime, but it is physical in the sense that he is talking about a gauge-invariant and hence physical object.

This I don't understand. Of course, you can only measure gauge-invariant properties, but I think, we are really drifting too much apart from the original topic of this thread.

 

[atyy]

This I don't understand. Of course, you can only measure gauge-invariant properties, but I think, we are really drifting too much apart from the original topic of this thread.

Essentially there are two meanings of "geometric".

1) Gauge-invariant. Everyone agrees on this aspect of geometry.

2) Euclidean geometry is a model of measurements using straight edge, ruler, compass etc - external rigid instruments. When you say that a proper reference frame is physically realized, the physical realization is an external rigid apparatus, since it does not contribute to spacetime curvature. So in the sense of needing an external rigid apparatus, your view is more "geometric", not less.

 

[loislane]

I've obtained three different answers to the same question, maybe just superficially in disagreement with each other, could you perhaps give a consensus reply?

If that's the case, that means that the observer at some point starts undergoing constant proper acceleration.

Agreed, so once the Minkowski observer starts accelerating he is no longer inertial in Minkowski coordinates right, but would he then be inertial in Rindler coordinates?

Um, what? A Minkowski observer, by definition, is inertial. He can't "try to keep at rest in the new coordinates". His state of motion is already specified. When you transform to Rindler coordinates, his worldline looks like whatever it looks like.

With "looks like whatever it looks like" you mean it looks noninertial in Rindler coordinates?

The transformation from Minkowski (i.e. pseudo-Caratesian coordinates in an inertial frame) to Rindler coordinates are clearly not Lorentz transformations. They are not even linear. I've given the formulae somewhere in this thread yesterday.

The transformation from Rindler chart to Minkowski chart:

is not a Lorentz transformation? It certainly looks to me like a Lorentz transformation if gt acts as the hyperbolic angle, this is the usual geometric form of a Lorentz transformation written as hyperbolic rotation AFAIK.

 

[DrGreg]

Agreed, so once the Minkowski observer starts accelerating he is no longer inertial in Minkowski coordinates right, but would he then be inertial in Rindler coordinates?

You have a misunderstanding what "inertial" means. It does not mean "at constant coordinate velocity" (="at zero coordinate acceleration"). It means "free-falling", i.e. "under no external force" i.e. "with zero proper accleration". All of these descriptions do not depend on your current coordinate system. Something is either inertial or not; "inertial in Minkowski coordinates" or "inertial in Rindler coordinates" are meaningless expressions.

 

[DrGreg]

The transformation from Rindler chart to Minkowski chart:

is not a Lorentz transformation? It certainly looks to me like a Lorentz transformation if gt acts as the hyperbolic angle, this is the usual geometric form of a Lorentz transformation written as hyperbolic rotation AFAIK.

A Lorentz transformation looks like$$T = t \cosh \phi + x \sinh \phi; \, X = t \sinh \phi + x \cosh \phi; \, Y = y; \, Z = z.$$That's a lot different than the transformation you quoted.

 

[loislane]

You have a misunderstanding what "inertial" means. It does not mean "at constant coordinate velocity" (="at zero coordinate acceleration"). It means "free-falling", i.e. "under no external force" i.e. "with zero proper accleration". All of these descriptions do not depend on your current coordinate system. Something is either inertial or not; "inertial in Minkowski coordinates" or "inertial in Rindler coordinates" are meaningless expressions.

You might as well be right, I'm just asking but then I don't understand why the use of the terms "Rindler oberver" or "Minkowski observer" seen for instance in the Rindler coordinates page of wikipedia, where they seem to refer to observer at rest(wich is what I understood to mean inertial) with respect to Rindler coordinates and observer at rest with respect to Minkowski coordinates respectively.

 

[loislane]

A Lorentz transformation looks like$$T = t \cosh \phi + x \sinh \phi; \, X = t \sinh \phi + x \cosh \phi; \, Y = y; \, Z = z.$$That's a lot different than the transformation you quoted.

Yes, of course, that is in fact what I understand by an active hyperbolic rotation, but I was quoting above a passive change of the coordinate systems, the point remains and only the coordinates are changed so the terms corresponding to the active motion when the point changes of position($t \cosh \phi, t \sinh \phi$) are not present.

 

[PeterDonis]

so once the Minkowski observer starts accelerating

Then he's no longer a Minkowski observer.

he is no longer inertial in Minkowski coordinates right, but would he then be inertial in Rindler coordinates?

Once more: whether or not a given observer is inertial is independent of which coordinates you are using. This has been told to you repeatedly. A Minkowski observer, by definition, is inertial--that is his state of motion.

With "looks like whatever it looks like" you mean it looks noninertial in Rindler coordinates?

You are confusing changing coordinate charts with changing an observer's state of motion. If I have an observer in a given state of motion--for example, a Minkowski observer who always has a proper acceleration of zero--then the equation of this observer's worldline will look different if I change coordinate charts. But that's not a physical change in the observer's motion; it's just a change in the mathematics I'm using to describe the motion.

The transformation from Rindler chart to Minkowski chart:

is not a Lorentz transformation?

Certainly not. A Lorentz transformation is between two Minkowski coordinate charts sharing the same origin.

 

[Dale]

I think it's semantics, but can you give me an example for a measurement that can be made without a clear specification of a reference frame?

Sure. Start and stop a stopwatch. No frame was specified in making the measurement.

The clock has a rest frame, and for convenience you may arbitrarily choose to specify the clock's rest frame in your analysis. You may also choose to specify the ECI, or the sun's rest frame, or any other frame you like. The measurement can easily be made without any such specification, and after the measurement is made any frame may be specified for the analysis.

Do you understand the distinction I am drawing between "making" a measurement and "analyzing" a measurement?

This is impossible, because measuring something means to have a reference you can compare the measured quantity to.

A reference, yes. A reference frame, no. The kilogram is a reference, not a reference frame.

All these quantities can written in manifestly covariant form and thus directly and conveniently measured in any reference frame.

Yes. I believe that you are making my point here.

 

[stevendaryl]

Agreed, so once the Minkowski observer starts accelerating he is no longer inertial in Minkowski coordinates right, but would he then be inertial in Rindler coordinates?

The transformation from Rindler chart to Minkowski chart:

is not a Lorentz transformation?

No. Lorentz transformations transform from one inertial coordinate system to another. Rindler coordinates are a non-inertial coordinate system.

It certainly looks to me like a Lorentz transformation if gt acts as the hyperbolic angle, this is the usual geometric form of a Lorentz transformation written as hyperbolic rotation AFAIK.

No. If you want to write the Lorentz transform in terms of hyperbolic angles, you can write it this way:

$x' = x cosh(\theta) - ct sinh(\theta)$
$t' = t cosh(\theta) - \frac{x}{c} sinh(\theta)$

You can think of this as analogous to rotations in two spatial dimensions. If you have a coordinate system $(x,y)$, then you can transform to a rotated coordinate system by:

$x' = x cos(\theta) + y sin(\theta)$
$y' = y cos(\theta) - x sin(\theta)$

But that's very different from a transformation from rectangular coordinates to polar coordinates:

$x = R cos(\theta)$
$y = R sin(\theta)$

A rotation is a linear transformation. Converting from rectangular to polar coordinates is nonlinear.

 

[vanhees71]

I've obtained three different answers to the same question, maybe just superficially in disagreement with each other, could you perhaps give a consensus reply?

Agreed, so once the Minkowski observer starts accelerating he is no longer inertial in Minkowski coordinates right, but would he then be inertial in Rindler coordinates?

No! This is easy to see, because a light ray (in the sense of ray optics, i.e., the eikonal approximation of the Maxwell equations) is not a straight line from his point of view anymore. An observer can objectively figure out that he is moving accelerated with respect to the family of inertial coordinate systems.

This holds true even in GR, but only in a local sense. A free falling body defines a local inertial reference frame. All local laws are precisely the same as in an inertial frame of reference. Roughly speaking "local" should mean something like "space-time distances small compared to any curvature measure around the free-falling observer".

With "looks like whatever it looks like" you mean it looks noninertial in Rindler coordinates?

The transformation from Rindler chart to Minkowski chart:

is not a Lorentz transformation? It certainly looks to me like a Lorentz transformation if gt acts as the hyperbolic angle, this is the usual geometric form of a Lorentz transformation written as hyperbolic rotation AFAIK.

No! It's not a linear transformation. The boost velocity depends on the coordinate, i.e., the rapidity is $\eta=g t$. See my long posting on this point.

 

[vanhees71]

Sure. Start and stop a stopwatch. No frame was specified in making the measurement.

My stopwatch is a massive body and thus defines its rest frame which is in a sense a reference frame preferred by the physical situation. This is very important for the entire "relativity business", because if you have an ideal stopwatch, it precisely defines a measure of time, namely its proper time. Of course, you can observe the watch from any other reference frame and convert between your own proper time and the proper time of the stopwatch. Nevertheless the stopwatch defines a frame (if it's accelerated in Minkowski space or in GR a local frame) of reference.

The clock has a rest frame, and for convenience you may arbitrarily choose to specify the clock's rest frame in your analysis. You may also choose to specify the ECI, or the sun's rest frame, or any other frame you like. The measurement can easily be made without any such specification, and after the measurement is made any frame may be specified for the analysis.

Yes, but as I said above, all these frames are somehow realized by the phsyical situation (what's "ECI"?).

Do you understand the distinction I am drawing between "making" a measurement and "analyzing" a measurement?

A reference, yes. A reference frame, no. The kilogram is a reference, not a reference frame.

Yes. I believe that you are making my point here.

I don't understand this distinction. We are doing physics not mathematics. Physics is about measurements in the real world, which I want to analyze as a theorist. I must make a connection between the mathematical concepts (here the spacetime geometry, which of course I can describe in a frame-independent way) and the real-world measurements. The measurement apparti define a frame of reference, and the various quantities they measure are related to this frame of reference. You must now, how to map the pointer readings of your apparti to the quantities you define (conveniently as some tensor quantities, whose components have simple transformation between different reference frames) in your "calculational frame".

This is of utmost importance in the relativistic realm. Dealing with relativistic many-body systems (in my case little fireballs of quark-gluon-plasma evolving into a hot hadron gas and finally freezing out as hadron or lepton/photon spectra in the detectors at RHIC, LHC, GSI, and hopefully in the future at FAIR), I know that this is often a source of confusion. Already the definition of a scalar phase-space-distribution function in relativistic kinetic theory and (as the limit of local thermal equibrium) hydrodynamics, is not trivial. If you want a taste of the difficulties these issues were still in the not too far past, see one of the ground-breaking papers related to it:

Fred Cooper and Graham Frye. Single-particle distribution in the hydrodynamic and statistical thermodynamics models of multiparticle production. Phys. Rev. D, 10:186, 1974.
http://dx.doi.org/10.1103/PhysRevD.10.186

For the details of the point of view from kinetic theory, see my Indian lecture notes:

http://fias.uni-frankfurt.de/~hees/publ/kolkata.pdf

For the quantum-field theoretical approach, see

O. Buss, T. Gaitanos, K. Gallmeister, H. van Hees, M. Kaskulov, et al. Transport-theoretical Description of Nuclear Reactions. Phys. Rept., 512:1–124, 2012.
http://dx.doi.org/10.1016/j.physrep.2011.12.001
http://arxiv.org/abs/1106.1344

or

W. Cassing. From Kadanoff-Baym dynamics to off-shell parton transport. Eur. Phys. J. ST, 168:3–87, 2009.
http://dx.doi.org/10.1140/epjst
http://arxiv.org/abs/arXiv:0808.0715

and the very good textbooks

C. Cercignani and G. M. Kremer. The relativistic Boltzmann Equation: Theory and Applications. Springer, Basel, 2002.
http://dx.doi.org/10.1007/978-3-0348-8165-4

S. R. de Groot, W. A. van Leeuwen, and Ch. G. van Weert. Relativistic kinetic theory: principles and applications. North-Holland, 1980.

 

[Demystifier]

The transformation from Rindler chart to Minkowski chart:

is not a Lorentz transformation? It certainly looks to me like a Lorentz transformation if gt acts as the hyperbolic angle, this is the usual geometric form of a Lorentz transformation written as hyperbolic rotation AFAIK.

As others said, Rindler transformation is not a Lorentz transformation. However, it can be written as a continuous sum of infinitely many infinitesimal Lorentz transformations:
http://lanl.arxiv.org/abs/gr-qc/9904078 [Phys.Rev.A61:032109,2000] (see Sec. 2)

 

[loislane]

Certainly not. A Lorentz transformation is between two Minkowski coordinate charts sharing the same origin.

A global Lorentz transformation can only be performed between Minkoski charts, that is for sure. But there is no global transformation between a Minkowski chart and a Rindler chart for the simple reason that the Rindler chart is local(and doesn't include the origin). Are you then saying that there are no local Lorentz transformations?

 

[loislane]

No! It's not a linear transformation. The boost velocity depends on the coordinate, i.e., the rapidity is $\eta=g t$. See my long posting on this point.

As explained above I'm referring to a local linearization(constant Jacobian determinant change of coordinates) preserving time and space orientation, the global change is certainly not linear but again there is no possible global change of coordinates here as the Rindler chart doesn't cover all of Minkowski spacetime.
On the other hand the Poincare transformations in the affine Minkowski space including translations and proper orthochronous Lorentz transformations are not strictly linear either but affine and projective.

 

[loislane]

As others said, Rindler transformation is not a Lorentz transformation. However, it can be written as a continuous sum of infinitely many infinitesimal Lorentz transformations:
http://lanl.arxiv.org/abs/gr-qc/9904078 [Phys.Rev.A61:032109,2000] (see Sec. 2)

Exactly, it is not a general Lorentz transformation, but as you say a continuous composition of infinitesimal boosts, the continuous identity component of the group of Lorentz transformations:proper orthochronous Lorentz transformations.

 

[martinbn]

As others said, Rindler transformation is not a Lorentz transformation. However, it can be written as a continuous sum of infinitely many infinitesimal Lorentz transformations:
http://lanl.arxiv.org/abs/gr-qc/9904078 [Phys.Rev.A61:032109,2000] (see Sec. 2)

Aaaah, that's only going to bring more confusion and a new branch of this thread with more explations. Don't you have mercy on Peter, Dale, Steven, VanHees and a few more?

 

[Demystifier]

Aaaah, that's only going to bring more confusion and a new branch of this thread with more explations. Don't you have mercy on Peter, Dale, Steven, VanHees and a few more?

I don't have mercy, because more confusion in the long run brings more understanding.

 

[loislane]

A rotation is a linear transformation. Converting from rectangular to polar coordinates is nonlinear.

Absolutely. There is no global way to linearly go from polar coordinates to cartesian ones. But in the vector space associated to Euclidean space(or to Minkowski space in the case discussed above), the linear Jacobian map moves tangent vectors at a point between the coordinate systems. That is all what's needed for a passive change of variables where all it changes is the coordinate basis vectors.

 

[vanhees71]

As explained above I'm referring to a local linearization(constant Jacobian determinant change of coordinates) preserving time and space orientation, the global change is certainly not linear but again there is no possible global change of coordinates here as the Rindler chart doesn't cover all of Minkowski spacetime.
On the other hand the Poincare transformations in the affine Minkowski space including translations and proper orthochronous Lorentz transformations are not strictly linear either but affine and projective.

Sure, but the boost rapidity is time dependent and thus it's not a linear transformation between Minkowski coordinates. That's very obvious. I don't understand the problem!

 

[stevendaryl]

I don't have mercy, because more confusion in the long run brings more understanding.

Assuming that eventually the confusion clears up...

 

[Demystifier]

Assuming that eventually the confusion clears up...

Of course. And in the paper I mentioned it was the case. To resolve a relatively narrow problem (Ehrenfest paradox associated with a uniformly rotating disk), a more confusion was introduced by considering a much more general problem (arbitrary motion of a non-rigid set of particles), which eventually resolved the original narrow problem, in a way which would be much more difficult to understand without considering the general problem.

 

[loislane]

Sure, but the boost rapidity is time dependent and thus it's not a linear transformation between Minkowski coordinates. That's very obvious. I don't understand the problem!

I don't understand what the problem is either since I'm not talking about any linear transformation between Minkowski coordinates, the transformation I'm referring to doesn't include reflections in space or time. And AFAIK Minkowski spacetime is time-independent, meaning it as a timelike killing vector field.

 

[stevendaryl]

I don't understand what the problem is either since I'm not talking about any linear transformation between Minkowski coordinates, the transformation I'm referring to doesn't include reflections in space or time. And AFAIK Minkowski spacetime is time-independent, meaning it as a timelike killing vector field.

I think you're veering quite a bit off-topic. What you asked was this:

Agreed, so once the Minkowski observer starts accelerating he is no longer inertial in Minkowski coordinates right, but would he then be inertial in Rindler coordinates?

The answer is emphatically "No". If an observer has nonzero proper acceleration, then he is not inertial. It doesn't make sense to say he is inertial relative to Rindler coordinates, but not relative to Minkowski coordinates. Being inertial has nothing to do with coordinates.

Now, what's nice about Minkowski coordinates (also called "inertial coordinates") is that you can tell whether an object is traveling inertially by computing the components of its coordinate acceleration: $\frac{d^2 x^\mu}{d\tau^2}$. If this quantity is zero, the object is traveling inertially. If it's nonzero, the object is not.

That equivalence between

the object is traveling inertially $\Leftrightarrow$ the object's coordinate acceleration is zero​

only works for inertial coordinate systems. It does not work for the Rindler coordinate system.

 

[stevendaryl]

Being inertial has nothing to do with coordinates.

What I mean is that being inertial is a coordinate-independent property: If it's true in one coordinate system, then it's true in every coordinate system.

 

[vanhees71]

I don't understand what the problem is either since I'm not talking about any linear transformation between Minkowski coordinates, the transformation I'm referring to doesn't include reflections in space or time. And AFAIK Minkowski spacetime is time-independent, meaning it as a timelike killing vector field.

Then you contradict yourself: A Lorentz transformation is a linear transformation between Minkowski coordinates. Rindler coordinates are non-Minkowskian, because they depend non-linearly on the Minkowski coordinates you used to describe it, and this clearly shows that an observer at rest in the sense of the Rindler coordinates is a non-inertial (i.e., accelerated) observer.