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]

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<br /> T = t \cosh \phi + x \sinh \phi; \, X = t \sinh \phi + x \cosh \phi; \, Y = y; \, Z = z.<br />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<br /> T = t \cosh \phi + x \sinh \phi; \, X = t \sinh \phi + x \cosh \phi; \, Y = y; \, Z = z.<br />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&#039; = x cosh(\theta) - ct sinh(\theta)
t&#039; = 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&#039; = x cos(\theta) + y sin(\theta)
y&#039; = 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.

 

[PeterDonis]

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).

We should probably avoid the words "global" and "local" since they are ambiguous. The transformation between the Minkowski chart and the Rindler chart covers more than an infinitesimal neighborhood (which is what "local" usually means); it covers a "wedge" of Minkowski spacetime to the right of the origin, bounded by the null lines $t = x$ and $t = -x$. Whereas a transformation between Minkowski charts covers all of Minkowski spacetime (which is what you are using "global" to mean).

However, a transformation that covers a particular open region of spacetime obviously covers any smaller open neighborhood within that region; so any such transformation is certainly "local".

Are you then saying that there are no local Lorentz transformations?

Certainly not. Any Lorentz transformation--i.e., any transformation between Minkowski charts--covers any local neighborhood of Minkowski spacetime. However, that doesn't make the transformation between a Minkowski chart and a Rindler chart a Lorentz transformation.

 

[Dale]

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.

The watch' rest frame is not preferred by the physical situation in any way. It is only preferred for computational convenience.

 

[vanhees71]

For me it's obvious, that there's no physics without reference frames. A (ideal) clock shows its proper time, and this is important to make sense of its readings. Without a clear definition with regard to which reference frame a quantity is measured and how the corresponding pointer readings of measurement device transform from one reference frame to another, these readings are just useless numbers!

 

[Demystifier]

For me it's obvious, that there's no physics without reference frames.

A non-trivial question is: Is there differential geometry (as a branch of mathematics) without reference frames? I could be wrong, but it seems to me that the so-called "coordinate free" formulations of differential geometry still use coordinates somewhere at the start where definitions and axioms are put down explicitly.

 

[vanhees71]

Sure, to define what a differentiable manifold is, you introduce charts and atlasses, i.e., coordinates. Of course, this the concept is diffeomorphism invariant by construction and thus you take the differentiable manifold as an abstract mathematical object, taking everything "modulo diffeomorphisms", and then it's "coordinate free".

The experimental physicist, however uses objects (measurement devices) with pointer readings mapping a physical quantity to (real) numbers, and these real numbers (in the ideal case) reflect the measured quantity in a specific frame of reference. In a sense the pointer reading is a coordinate rather than the abstract mathematical object (tensor) of the mathematical description.

That's even more clear in terms of quantum theory. There the mathematical objects used to describe the theory are even farther from real-world observables than in classical physics. In the basis-free and picture-free description you work with abstract rigged Hilbert-space objects (bras and kets in Dirac's formalism as well as operators and a $C^*$-operator algebra). In the real world, there are no such mathematical objects, but pointer readings from measurement apparati and Born's rule to map the abstract mathematical objects to real-world observables (in this case you have only probabilistic meanings of the abstract objects).

For a theoretical physicist it's important, not to loose contact with experiment(alists). You must know what is measured and sometimes even to a certain extent even how this is done. Of course, the abstract formalism is as important as well, but at the end you must "get the numbers out" to compare with real-world experimental/observational results. This is impossible without (a sometimes pretty subtle) clear definitions of appropriate reference frames.

 

[Dale]

A (ideal) clock shows its proper time, and this is important to make sense of its readings. Without a clear definition with regard to which reference frame a quantity is measured and how the corresponding pointer readings of measurement device transform from one reference frame to another, these readings are just useless numbers!

No. The proper time read by an ideal clock is an invariant quantity. It is not measured "with regard to" any reference frame. You can use any frame "to make sense of its readings".

For me it's obvious, that there's no physics without reference frames.

Note, I am not arguing that there is physics without reference frames. If you would stop there then your comments would be non objectionable. You start going wrong when you take that next step and claim that the device or the measurement defines the frame. This is false. A reference frame is a mathematical tool produced by and used in the analysis.

I feel like you still don't get the distinction between a measurement and an analysis.

 

[martinbn]

A non-trivial question is: Is there differential geometry (as a branch of mathematics) without reference frames? I could be wrong, but it seems to me that the so-called "coordinate free" formulations of differential geometry still use coordinates somewhere at the start where definitions and axioms are put down explicitly.

That may depend on what exactly one means by coordinates, but isn't it possible to avoid them? Where do you think they are unavoidable?

 

[vanhees71]

No. The proper time read by an ideal clock is an invariant quantity. It is not measured "with regard to" any reference frame. You can use any frame "to make sense of its readings".

Note, I am not arguing that there is physics without reference frames. If you would stop there then your comments would be non objectionable. You start going wrong when you take that next step and claim that the device or the measurement defines the frame. This is false. A reference frame is a mathematical tool produced by and used in the analysis.

I feel like you still don't get the distinction between a measurement and an analysis.

A proper time is defined with reference to the rest frame of the clock. How else. Two identical clocks show different proper times, depending on their time-like worldline. In a coordinate free way you can say proper time is a functional of the clock's worldline. That proper time can be formulated in a coordinate free way doesn't mean that the clock doesn't define a frame.

 

[harrylin]

A proper time is defined with reference to the rest frame of the clock. How else. [...] That proper time can be formulated in a coordinate free way doesn't mean that the clock doesn't define a frame.

How else? Proper time is a measure for the progress of a physical process - in the case of an ideal clock it can simply be the clock readout. That has as little to do with a frame as the temperature readout of a thermometer, or the color of your shirt. But of course any of those can be used for defining a frame.

And what does all that have to do with the second postulate?

 

[Demystifier]

That may depend on what exactly one means by coordinates, but isn't it possible to avoid them? Where do you think they are unavoidable?

See e.g. the post #180 by vanhees.

 

[vanhees71]

How else? Proper time is a measure for the progress of a physical process - in the case of an ideal clock it can simply be the clock readout. That has as little to do with a frame as the temperature readout of a thermometer, or the color of your shirt. But of course any of those can be used for defining a frame.

And what does all that have to do with the second postulate?

Temperature is another very good example for the importance of reference frames. That it is a scalar (field) is due to a careful convention, which is not that old in the history of relativity. I can't point the finger to one single paper, where a paradigm change occured, but you can google-scholar for it. You get tons of papers about the transformation properties of the thermodynamic quantities. If you read older papers and textbooks on relativity by Planck, von Laue et al (with no doubt people who completely understood relativity from the moment of Einstein's paper of 1905) the temperature is defined as a quantity that transforms with a Lorentz-$\gamma$ factor when changing from one inertial frame to another (note that I talk about SR only here). Within SR there is no trouble with this, and all the thermodynamics and statistical mechanics is consistent with such a definition, but it's very inconvenient, particularly when it comes to GR (and I wouldn't like to work with the old convention in my own research on heavy-ion collisions, which is complete within SR, either).

The modern definition of temperature (which, I think goes back to people like van Kampen, Israel, Stuart in the 1960ies) in the relativistic realm is very clearly using a reference frame preferred by the physical situation: Temperature makes sense in local thermal equilibrium of a substance (say a fluid like a liquid or gas). Local thermal equilibrium defines local rest frames of the fluid. This is a macroscopic quantity, i.e., (in any inertial frame you like) you put a spatial and temporal grid on spacetime which is coarse enough such that in each so defined "fluid cell" are many particles but also fine enough that each fluid cell is small against the typical time and lengths scales over which the macroscopic fluid properties change (you need such a separation of microscopic and macroscopic scales to make sense of a local thermal equilibrium (aka ideal hydro) description of the medium). Then the temperature is defined as the reading of an ideal thermometer at rest in the local rest frame of each fluid cell, defining the scalar temperature field.

In statistics and kinetics you also have to take care of the convenient definition of the phase-space distribution function in terms of a scalar field $f(t,\vec{x},\vec{p})$ for classical on-shell particles (I don't go into the even more complicated issue of the proper derivation from many-body quantum field theory). For local thermal equibrium the upshot is that for an ideal gas the classical phase-space distribution function is defined as the scalar quantity (Boltzmann-Jüttner distribution function)
$$f(t,\vec{x},\vec{p}) = \frac{1}{(2 \pi \hbar)^3} \exp\left [-\frac{u(x) \cdot p-\mu(x)}{T(x)} \right ], \quad p^0=E_{\vec{p}}=\sqrt{m^2+\vec{p}^2}.$$
Taking into account quantum statistics, instead of the exponential function you have Bose or Fermi distributions. Temperature and chemical-potential field (the latter referring to some conserved charge-like quantity like net-baryon number or electric charge) are scalar field, and $u$ is the four-velocity of the fluid flow.

A very illuminating paper about all this is

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

 

[martinbn]

See e.g. the post #180 by vanhees.

I did, but again it depends on what one understands by coordinates. He says you need to define charts and atlases i.e. coordinates. But a chart is a homeomorphism from and open set of a topological space to an open set in an Euclidean space. This by it self doesn't need coordinates. I am just wandering is not possible to avoid the use of coordinates.

 

[atyy]

I did, but again it depends on what one understands by coordinates. He says you need to define charts and atlases i.e. coordinates. But a chart is a homeomorphism from and open set of a topological space to an open set in an Euclidean space. This by it self doesn't need coordinates. I am just wandering is not possible to avoid the use of coordinates.

I think it depends on how one defines a manifold. In the physics view, roughly it is something that locally looks R^N, so the idea is that it is something on which one can put coordinates. Do the mathematicians have a definition of manifold that doesn't use coordinates?

 

[loislane]

I did, but again it depends on what one understands by coordinates. He says you need to define charts and atlases i.e. coordinates. But a chart is a homeomorphism from and open set of a topological space to an open set in an Euclidean space. This by it self doesn't need coordinates. I am just wandering is not possible to avoid the use of coordinates.

Charts(local homeomorphism from and open set of a topological space to an open set in an Euclidean space) are local coordinates, that is how local coordinates in a manifold are defined, what do you mean this doesn't need coordinates? It is how they are defined. You cannot say that you need to define them that way and then wonder if they can be avoided unless you are suggesting a new definition of manifolds that dispenses with local homeomorphisms to Rn.

 

[loislane]

I think it depends on how one defines a manifold. In the physics view, roughly it is something that locally looks R^N, so the idea is that it is something on which one can put coordinates. Do the mathematicians have a definition of manifold that doesn't use coordinates?

You would have to renounce to the local resemblance to Rn, but then that is what defines a topological space as a manifold.

 

[Demystifier]

I did, but again it depends on what one understands by coordinates. He says you need to define charts and atlases i.e. coordinates. But a chart is a homeomorphism from and open set of a topological space to an open set in an Euclidean space. This by it self doesn't need coordinates. I am just wandering is not possible to avoid the use of coordinates.

Let us be less abstract and try to do something more concrete in differential geometry. Consider a two-dimensional sphere with unit radius, immersed in the 3-dimensional space with Euclid metric. Can you prove that the area of the sphere is $4\pi$ without using any coordinates? (I can't.)

 

[loislane]

How else? Proper time is a measure for the progress of a physical process - in the case of an ideal clock it can simply be the clock readout. That has as little to do with a frame as the temperature readout of a thermometer, or the color of your shirt. But of course any of those can be used for defining a frame.

The clock readout(a local time) evidently defines the instantaneous rest frame of the clock, different at the different points and related between those points by Lorentz transformations. I don't know how this can generate disagreement.

And what does all that have to do with the second postulate?

It has to do with the discussion about considering how inertial frames are defined in the postulates, but it is true that the thread went a bit off topic already more than 100 posts ago with the discussion about Rindler coordinates introduced by PeterDonis.

 

[loislane]

No. The proper time read by an ideal clock is an invariant quantity. It is not measured "with regard to" any reference frame. You can use any frame "to make sense of its readings".

Note, I am not arguing that there is physics without reference frames. If you would stop there then your comments would be non objectionable. You start going wrong when you take that next step and claim that the device or the measurement defines the frame. This is false. A reference frame is a mathematical tool produced by and used in the analysis.

I feel like you still don't get the distinction between a measurement and an analysis.

Why do you think that a measurement is incompatible with an invariant quantity? The measurement understood as the proper time readout of the clock is obviously changing from point to point, and at ech point the reading(the measure) defines an instantaneous rest frame or local coordinates. This is not incompatible with an interval between different events being an invariant, on the contrary since the local measurements are related using Lorentz transformations between events..

 

[Dale]

Why do you think that a measurement is incompatible with an invariant quantity?

I don't.

A measurement being frame invariant is incompatible with the idea that the measurement itself is what defines the reference frame.

 

[stevendaryl]

That may depend on what exactly one means by coordinates, but isn't it possible to avoid them? Where do you think they are unavoidable?

Well the very definition of a smooth manifold involves continuous and differentiable mappings between open subsets of the space into R^n. But such a mapping basically is a coordinate system.

 

[Dale]

A proper time is defined with reference to the rest frame of the clock. How else.

This is simply factually false. Proper time is defined as $\int ds$

Again, I am not claiming that physics can be done without using a reference frame. I am just saying that it is not the measurement or the physical apparatus which determines the frame. The reference frame is not a physical object, it is a mathematical tool. If it were a physical object then we would not have the freedom to change frames as we see fit.

Frankly it seems to me that you don't understand the first postulate at all.

 

[loislane]

I don't.

A measurement being frame invariant is incompatible with the idea that the measurement itself is what defines the reference frame.

You should make more precise what you mean by reference frame. What I understand in this particular debate is an instantaneous rest frame, and it is evident that clocks readouts(if you consider them as time measurements) define those particular frames. I don't think you can disagree with this, so you must be referring to some other definition of frame?

 

[loislane]

This is simply factually false. Proper time is defined as $\int ds$

No, proper time as what a clock measures is $d\tau$ and the interval is computed $Δ\tau=\int \frac{ds}{c}=\int d\tau$

 

[Dale]

No, proper time as what a clock measures is $d\tau$ and the interval is computed $Δ\tau=\int \frac{ds}{c}=\int d\tau$

That is fine. I didn't clarify that i was using units where c=1.

 

[vanhees71]

This is simply factually false. Proper time is defined as $\int ds$

Again, I am not claiming that physics can be done without using a reference frame. I am just saying that it is not the measurement or the physical apparatus which determines the frame. The reference frame is not a physical object, it is a mathematical tool. If it were a physical object then we would not have the freedom to change frames as we see fit.

Frankly it seems to me that you don't understand the first postulate at all.

I guess, I'm just unable to make this obvious point. I try one last time, and then keep quiet.

A clock doesn't do an abstract integral; it is a real-world device like an atomic clock at the national bureaus of standard like NIST in the US or the PTB in Germany. In an idealized simplified way it is providing a "pointer reading" that allows you to read off "time marks", defining the duration of time between two events. This reading is, again for an idealized clock, its proper time.

Another example is an unstable particle like a muon, which provides also a time, but only in a statistical sense, by its mean lifetime. The measured mean lifetime (e.g., by muons somehow stored in a storage ring or some trap) of the muon clearly depends on the reference frame. If you have an ideal clock described above, you can (again idealized) its rest frame as an inertial frame of reference and the proper time of the clock is then identical with your "coordinate time". Comparing the lifetime of the muon with the readings of this clock (as already Einstein emphasized measuring time are always coincidence observations, i.e., you have a coincidence of a pointer reading of the clock and a measurement telling you that you have a muon and another one showing its decay. The duration gives, repeated over many muons at rest, the proper lifetime of the muon. According to the laws of physics this proper life time of the muon is a natural constant, but it's defined in the muon's rest frame. If you have muons moving relative to your clock, you'll find a longer average lifetime, because the proper lifetime is time dilated relative to your frame, and this proper lifetime is a functional of the world line of the particle, which is evaluated by your integral.

Besides these physical trifles, you even need a reference frame (or even coordinates) to evaluate your integral for a given world line (or better quantum state) of your muons, but that's not besides my point.