Second postulate of SR quiz question

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

Gold Member
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

Gold Member
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

Mentor
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

Mentor
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

Staff Emeritus
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

Gold Member
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

Gold Member
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 [Broken]
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 [Broken]
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.

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Demystifier

2018 Award
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

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.

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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.

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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

2018 Award
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

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

Gold Member
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

Staff Emeritus
I don't have mercy, because more confusion in the long run brings more understanding. Assuming that eventually the confusion clears up...

Demystifier

2018 Award
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

Staff Emeritus
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 travelling 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

Staff Emeritus
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

Gold Member
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.

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