Second postulate of SR quiz question

[loislane]

As stressed by stevendaryl in the posting before, a rindler observer is noninertial, because he is accelerated relative to the class of inertial frames.

In the wikipedia page on rindler coordinates it can be read that the Rindler observers are in inertial motion with respect to Minkowskian observers at rest. On the other hand they are stationary in relation to the rest of Rindler observers.

It's the most simple example of a nonrotating accelerated (with constant proper acceleration) observer. He is non-rotating, because the infinitesimal boosts are in a fixed direction.

How do you fix the direction? A composition of boosts always includes pure rotations. Thomas rotation is an example of this.

 

[vanhees71]

A composition of boosts in different directions is equivalent to another pure boost followed by a rotation (or a rotation followed by a pure boost), the socalled Wigner rotation. See my treatment of the Thomas precession in my (still unfinished) SRT FAQ article:

http://fias.uni-frankfurt.de/~hees/pf-faq/srt.pdf

I don't understand, why an observer that is accelerated relative to an inertial reference frame (and thus relative to any inertial reference frame) should be inertial. I can't find such a statement in the Wikipedia article

https://en.wikipedia.org/wiki/Rindler_coordinates#The_Rindler_observers

In section

https://en.wikipedia.org/wiki/Rindler_coordinates#Minkowski_observers

you find a somewhat overcomplicated derivation that "free falling observers" are inertial observers, but that's clear, because we are in Minkowski space, where no gravitation is present. In other words the time-like geodesic congruences define not only local but global inertial frames, namely the Minkowski frames. An observer at rest relative to a Minkowskian reference frame is of course inertial. In the more general case with a real gravitational field present, the free-falling observers are locally inertial but not globally. In Minkowski space the latter are of course globally inertial.

 

[stevendaryl]

In the wikipedia page on rindler coordinates it can be read that the Rindler observers are in inertial motion with respect to Minkowskian observers at rest. On the other hand they are stationary in relation to the rest of Rindler observers.

I think you got the exactly wrong impression from that sentence. In section https://en.wikipedia.org/wiki/Rindler_coordinates#Minkowski_observers, the article is talking about Minkowski observers (that is, inertial observers) as viewed by Rindler coordinates. It's not talking about Rindler observers.

Definitely, a Rindler observer is not inertial.

 

[Dale]

A Rindler observer is inertial(it is in inertial motion with respect to an observer at rest in Minkowski coordinates), but for him space is not euclidean.

As was mentioned above, a Rindler observer is not inertial. An accelerometer attached to a Rindler observer reads non zero.

 

[loislane]

That is going to depend on the specific definition of inertial and noninertial frames, I get the impression that the concept of frames is at the least problematic in physics(I just spotted a thread discussing the validity of the concept of frame), so I'm not going to get involved in a semantics debate.

I'll simply add that proper acceleration is supposed to be coordinate invariant so its presence or absence cannot depend on labels. And no you cannot measure acceleration in a label.

 

[PeterDonis]

A Rindler observer is inertial(it is in inertial motion with respect to an observer at rest in Minkowski coordinates), but for him space is not euclidean.

Both halves of this statement are incorrect. As others have pointed out, a Rindler observer is accelerated, not inertial. But also, spacelike slices of constant Rindler coordinate time are Euclidean. This is the simplest counterexample to the common (incorrect) belief that space must always be non-Euclidean for accelerated observers.

 

[PeterDonis]

I'll simply add that proper acceleration is supposed to be coordinate invariant so its presence or absence cannot depend on labels.

It doesn't. The nonzero proper acceleration of a Rindler observer is indeed an invariant. Nobody is disputing that.

 

[loislane]

A composition of boosts in different directions is equivalent to another pure boost followed by a rotation (or a rotation followed by a pure boost), the socalled Wigner rotation. See my treatment of the Thomas precession in my (still unfinished) SRT FAQ article:

Since I'm considering only the identity component of Lorentz transformations(the proper orthochronous Lorentz transformations that are continuous and the ones usually considered physical) I don't consider boosts in the same direction as different boosts, it's just one boost.

 

[Dale]

That is going to depend on the specific definition of inertial and noninertial frames,

No, it doesn't. An inertial observer is one that has 0 proper acceleration. As you mention, that is an invariant fact which is unaffected by the choice of reference frame. An inertial observer is inertial regardless of whether or not you are using an inertial frame to describe him.

 

[stevendaryl]

I'll simply add that proper acceleration is supposed to be coordinate invariant so its presence or absence cannot depend on labels. And no you cannot measure acceleration in a label.

I'm not sure what the last sentence means, but there is a coordinate-independent notion of a particle traveling inertially, and that is that it shows nonzero proper acceleration, according to an accelerometer. A simple mechanical accelerometer might be a box in the shape of a cube, with 6 identical springs attached to the center of each face. Where the 6 springs meet, there is a mass. If all 6 springs are the same length, then the box has zero proper acceleration. If some of the springs are stretched more than others, then the box has nonzero proper acceleration.

 

[loislane]

I have no problem withdrawing my claim that Rindler is an inertial observer, the basic reason is that the Rindler coordinates don't cover the whole spacetime, which is a condition for an inertial observer. I have serious doubts that an observational reference frame is a physical object one can attach an accelerometer to, I think it is something more abstract than that.

 

[Dale]

I have serious doubts that an observational reference frame is a physical object one can attach an accelerometer to, I think it is something more abstract than that.

I agree with that. This is a distinction between a frame (abstract mathematical object) and an observer (concrete physical object). An inertial reference frame is not synonymous with an inertial observer.

There is, of course, a standard convention for associating a particular inertial frame to a given inertial observer, so sometimes the distinction becomes blurred in discussions.

 

[PeterDonis]

I have no problem withdrawing my claim that Rindler is an inertial observer, the basic reason is that the Rindler coordinates don't cover the whole spacetime, which is a condition for an inertial observer.

You're mixing up observers and coordinates. An observer is inertial if he has zero proper acceleration; that is a direct physical observable that can be measured with an accelerometer. A Rindler observer is not inertial because he has nonzero proper acceleration; his accelerometer does not read zero. That's true regardless of what coordinates you are using to describe the Rindler observer's motion.

 

[vanhees71]

I think this is the main difference between mathematicians, who are very much emphasizing the geometric aspects of spacetime. This has some justification since modern physics hinges on geometric concepts in a very general sense, discovered by Riemann and Klein in the 19th century, namely the importance of symmetries (of spacetime and even abstract "flavor spaces" in QFT). This point of view in physics was one of the most important and usually overlooked breakthroughs in Einstein's paper on SR of 1905, where the first sentence can be read as a research program going on until today, namely to figure out the basic symmetries of the physical laws.

However, one can also overemphasize the geometrical aspects and forget that besides the elegant formulations in terms of geometric objects (differentiable manifolds, affine Euclidean and pseudo-Euclidean, Riemann- and pseudo-Riemann, varous fiber bundles,...) we still do physics, and physics is about what you can really observe (in a very broad sense, from naive looking at things with our senses to high-precision quantitative measurements with very tricky technology). Usually we don't realize it anymore, but any measurement (more or less tacitly) uses and introduces a reference frame. This can be simply the edges of our laboratory or the geometric setup of a particle detector or some fancy optical device (like a Michelson-Morley interferometer) etc. etc. 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.Now, in a gedanken experiment a Rindler observer can be seen as sitting in a rocket that is accelerated with constant proper acceleration. Now to cover some finite part of Minkowski space you need a whole family of such (pointlike) observers. This family trajectories is defined by a real spatial parameter $\xi$ and temporal parameter $\tau$ according to
$$t=\xi \sinh \tau, \quad x=\xi \cosh \tau.$$
The four-velocity of each observer (labeled by the parameter $\xi$, with $\tau$ the parameter of the trajectory) is given by
$$(u^0,u^1)=\frac{1}{\sqrt{\partial_{\tau} t^2-\partial_{\tau} x^2}}(\partial_{\tau} t,\partial_{\tau} x)=(\cosh \tau,\sinh \tau).$$
As you see, each observer in the family is clearly accelerated, because $u^1$ is not constant. The proper acceleration for each observer is
$$\alpha(\xi)=\frac{\mathrm{d} u^1}{\mathrm{d} t}=\frac{\mathrm{d} u^1}{\mathrm{d} \tau} \left (\frac{\mathrm{d} t}{\mathrm{d} \tau} \right)^{-1}=\frac{1}{\xi}.$$

The Minkowski pseudometric reads in the new local coordinates $(\tau,\xi)$
$$\mathrm{d} s^2=\mathrm{d}t^2-\mathrm{d} x^2=(\mathrm{d} \tau \partial_{\tau} t+\mathrm{d}\xi \partial_{\xi} t)^2-(\mathrm{d} \tau \partial_{\tau} x+\mathrm{d} \xi \partial_{\xi} x)^2=\xi^2 \mathrm{d} \tau^2-\mathrm{d} \xi^2.$$
Obviously at $\xi=0$ we have a coordinate singularity. Since for any $\xi>0$ we have
$$\frac{t}{x}=\tanh \tau$$
for $\xi \rightarrow 0$ we have $\tau \rightarrow \pm \infty$ and thus the limit $\xi \rightarrow 0^+$ defines the Rindler wedge $x=\pm t>0$, which is an event horizon for the Rindler map of the so defined part of the Minkowski space. For a picture, see the Wikipedia article

https://en.wikipedia.org/wiki/Rindler_coordinates#Relation_to_Cartesian_chart

There you clearly see, that each Rindler observer is in hyperbolic motion with his proper acceleration $1/\xi$ and thus accelerated.Note that in the Wikipedia article they have a (to may taste a bit confusing) convention our $(\tau,\xi)$ are their $(t,x)$ and our $(t,x)$ are their $(T,X)$. We have also set conveniently their $g=1$ and used the west-coast instead of the east-coast convention.

The Christoffel symbols are not vanishing, which is another hint that the Rindler coordinates are not locally inertial:
$$\Gamma^0_{01}=\Gamma^{0}_{10}=1/\xi, \quad \Gamma^{1}_{00}=\xi,$$
and all other Christoffel symbols are vanishing.

The geodesics of the spacetime are no straight lines wrt. the Rindler frame, and thus again we see that the Rindler frame is not inertial. Of course, the geodesics are straight lines wrt. any inertial frame.

 

[martinbn]

I think the last paragraph can be misleading. The geodesics of a space-time depend only on the space-time and they are what they happen to be, for example the geodesics of Minkowski space-time are stright lines (in the usual sense), and that's that. They are not geodesics with respect to one frame and non-geodesics wrt another.

 

[stevendaryl]

I think the last paragraph can be misleading. The geodesics of a space-time depend only on the space-time and they are what they happen to be, for example the geodesics of Minkowski space-time are stright lines (in the usual sense), and that's that. They are not geodesics with respect to one frame and non-geodesics wrt another.

vanhees71 didn't say anything to suggest that something was geodesic with respect to one frame and not another. He was saying that the geodesics are "straight" relative to one frame versus the other. I'm not sure if there is a standard definition of what "straight" means, but relative to a coordinate system, one could say that a path is straight or not depending on whether there is a parametrization $x^\mu(s)$ such that $\dfrac{d^2 x^\mu}{ds^2} = 0$

 

[martinbn]

Well, that's why it can be misleading. For me geodesic and straight are the same thing, and it is a reference independent property of lines.

 

[Dale]

However, one can also overemphasize the geometrical aspects and forget that besides the elegant formulations in terms of geometric objects (differentiable manifolds, affine Euclidean and pseudo-Euclidean, Riemann- and pseudo-Riemann, varous fiber bundles,...) we still do physics, and physics is about what you can really observe (in a very broad sense, from naive looking at things with our senses to high-precision quantitative measurements with very tricky technology).

I don't think that this is a real problem. I think that physicists are well aware that the purpose of the math is to analyze and predict the outcome of measurements.

I think that you are making the opposite mistake, which is overly identifying the math with the measurements.

Usually we don't realize it anymore, but any measurement (more or less tacitly) uses and introduces a reference frame.

This is simply false. No specific reference frame is required to analyze any given measurement. Many measurements are more conveniently or simply analyzed in the rest frame of the measurement apparatus, but it is not a requirement. The physicist retains complete freedom in which mathematical reference frame they prefer to use for the analysis of any measurement. The measurement itself does not introduce the frame, the physicist does.

 

[vanhees71]

Well, but at least there is a "restframe of the measurement apparatus", and you statement is not always true. E.g., the temperature of a flowing fluid is measured in its local restframe, i.e., with a co-moving thermometer. That's why in the modern definition of temperature for relativistic fluids temperature is a scalar (no Lorentz-$\gamma$ factors as in the older definition)!

Also cross sections are defined in the lab frame and then written covariantly. Of course, physicists measure it in the rest frame of the detectors and then recalculate it to the invariant cross section they like to. But to be able to do so you need clearly defined reference frames (here the lab-frame and the center-of-momentum frame in a collider experiment).

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, of course you can just calculate from one frame to the other, but you need to define a frame (or the different frames you want to analyze your data in), to make sense of your measurement results.

 

[vanhees71]

I think the last paragraph can be misleading. The geodesics of a space-time depend only on the space-time and they are what they happen to be, for example the geodesics of Minkowski space-time are stright lines (in the usual sense), and that's that. They are not geodesics with respect to one frame and non-geodesics wrt another.

Yes, maybe this was an unfortunate formulation. Still, for Rindler observers the geodesics appear not as straight lines, but for any inertial observer they appear as such. As an invariant object the geodesics are of course always straight lines.

 

[Demystifier]

For me geodesic and straight are the same thing

If you accept that language, then you must say, e.g., that trajectory of the planet Mercury around the Sun is straight. I think it is quite obvious that such a language would not satisfy an astronomer, for example. Thus for physicists (who are supposed to be able to speak with both astronomers and mathematicians) it makes a lot of sense not to treat geodesic and straight as the same thing.

Or are we talking only about special relativity? In that case my remark may be irrelevant.

 

[martinbn]

If you accept that language, then you must say, e.g., that trajectory of the planet Mercury around the Sun is straight. I think it is quite obvious that such a language would not satisfy an astronomer, for example. Thus for physicists (who are supposed to be able to speak with both astronomers and mathematicians) it makes a lot of sense not to treat geodesic and straight as the same thing.

Or are we talking only about special relativity? In that case my remark may be irrelevant.

Well, the world line of Mercury is a straight line in space-time, yes. The trajectory, which is a projection in space, of some choice of space and time split of the space-time need not be straight.

 

[Demystifier]

Well, the world line of Mercury is a straight line in space-time, yes. The trajectory, which is a projection in space, of some choice of space and time split of the space-time need not be straight.

OK, but how about the following problem? Consider non-relativistic physics, where everything happens in the 3-dimensional Euclidean space. (The concept of time, according to such a non-relativistic theory, is not related to the geometry of space.) In such a space, consider observers A and B. A is an inertial observer, so his trajectory is straight. B moves along a circle, so his trajectory is not straight. However, using a standard folk english, one can say that from the point of view of B, it looks as if the trajectory of A is not straight. How would you express this fact in a more precise language?

 

[stevendaryl]

Well, the world line of Mercury is a straight line in space-time, yes. The trajectory, which is a projection in space, of some choice of space and time split of the space-time need not be straight.

I would say that "straight" is a concept from Euclidean geometry, and that "geodesic" is a generalization to non-Euclidean geometry. They're not synonyms--one is a generalization of the other.

 

[vanhees71]

With straight line I mean a straight line in an affine space. It's the geodesic of the affine space, but there cannot be a straight line in the Schwarzschild metric, describing spacetime around the sun. Mercury is on a timelike geodesic in this spactime, but it's not in a straight line. In curved space there are no straight lines but geodesics. I think, one should not confuse the issue by making "straight line" a synonym with geodesic.

 

[martinbn]

OK, but how about the following problem? Consider non-relativistic physics, where everything happens in the 3-dimensional Euclidean space. (The concept of time, according to such a non-relativistic theory, is not related to the geometry of space.) In such a space, consider observers A and B. A is an inertial observer, so his trajectory is straight. B moves along a circle, so his trajectory is not straight. However, using a standard folk english, one can say that from the point of view of B, it looks as if the trajectory of A is not straight. How would you express this fact in a more precise language?

Well, that seems fine to me, but trajectories are frame dependent, while geodesics are not.

I would say that "straight" is a concept from Euclidean geometry, and that "geodesic" is a generalization to non-Euclidean geometry. They're not synonyms--one is a generalization of the other.

That is a possible and acceptable convention, but why the distinction. If you consider a seven dimensional Euclidean space, then are the geodesics straight lines or not? After all higher dimensional spaces are generalizations of Euclidean geometry too.

 

[loislane]

You're mixing up observers and coordinates. An observer is inertial if he has zero proper acceleration; that is a direct physical observable that can be measured with an accelerometer. A Rindler observer is not inertial because he has nonzero proper acceleration; his accelerometer does not read zero. That's true regardless of what coordinates you are using to describe the Rindler observer's motion.

You seem to be using the word observer as if it was something physical instead of a mathematical abstraction, according to what Dalespam wrote above , apparently a distinction can be made between an inertial observer and an inertial frame, I was not aware of that distinction, I'm only used to talk in terms of coordinates that is something well defined mathematically.

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 ("slidng" along the coordinate hyperbolas), would you say the latter is an inertial or a noninertial observer?

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

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.

 

[stevendaryl]

That is a possible and acceptable convention, but why the distinction

Because "straight" has connotations that don't apply to geodesics in general.

If you consider a seven dimensional Euclidean space, then are the geodesics straight lines or not?

Yes.

After all higher dimensional spaces are generalizations of Euclidean geometry too.

Yes, and the concept of "straight" applies equally well to higher dimensional spaces, but does not apply equally well to curved spaces.

 

[martinbn]

Because "straight" has connotations that don't apply to geodesics in general.

Why not?

Yes, and the concept of "straight" applies equally well to higher dimensional spaces, but does not apply equally well to curved spaces.

Why not?

 

[stevendaryl]

Why not?

Left as an exercise.

 

[martinbn]

A line is straight if its tangent vector is parallel transported along the line. It seems to apply equally well to Euclidean and curved spaces, no?

 

[stevendaryl]

A line is straight if its tangent vector is parallel transported along the line. It seems to apply equally well to Euclidean and curved spaces, no?

No. I would say that the concept of a "straight line" is described by Euclid's geometry, which includes several properties that don't hold of geodesics (or autoparallels) including:

1. Between any two points, there is a unique line connecting them.
2. If two distinct lines can intersect at at most one point.

Geodesics (or autoparallels) in curved space don't satisfy the notion of "straight line". The clue is the word "curved". A space with a connection is curved if its autoparallels are not straight lines.

Look, nobody uses the word straight in any context other than Euclidean space, as far as I know. Unless it's in quotes, where it's meant to remind you that geodesics are analogous to the straight lines of Euclidean geometry. But I don't really care what terminology you use, I'm just telling you that other people don't use that terminology.

 

[vanhees71]

I'd say for a straight line an arbitrary affine space is sufficient. You don't need a metric. It also makes sense in Minkowski space, which is only pseudo-Euclidean. Also in curved manifolds it's sufficient to have an affine connection to define geodesics. It doesn't need to be Riemannian.

 

[martinbn]

@stevendaryl, ok, no need for me to argue and derail the thread over terminology. All I wanted to say is that it is a matter of convention. You seem to insist on your preferred choice, but it is still a choice. It is not true that it is as universally accepted that straight line is used only in the context of Euclidean geometry as you suggest. For example non-Euclidean geometry (Lobachevsky).

 

[stevendaryl]

@stevendaryl, ok, no need for me to argue and derail the thread over terminology. All I wanted to say is that it is a matter of convention. You seem to insist on your preferred choice, but it is still a choice.

Yes, since the words "autoparallel' and "geodesic" are already used for the concept, I don't see any reason to use a different word, such as "straight"