Second postulate of SR quiz question

PeterDonis

Mentor
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.

• vanhees71

Dale

Mentor
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

Gold Member
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!

• loislane

Demystifier

2018 Award
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

Gold Member
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.

• Demystifier

Dale

Mentor
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

Gold Member
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

2018 Award
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

Gold Member
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

• loislane

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

2018 Award
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..

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Dale

Mentor
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

Staff Emeritus
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

Mentor
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.

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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

Mentor
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

Gold Member
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.

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.

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