Second postulate of SR quiz question

[loislane]

Isotropy in physics is always taken to be spatial. When I was formulating an assumption of isotropy (for SR) I explicitly said there exists a group (Poincare group of all global inertial coordinates) of global coordinates such that observed physical laws expressed ins such coordinates display manifest isotropy, with the latter meaning spatial because that is always assumed.

Then I don't understand what your discussion is about, rotational invariance in 3 dimensions is clearly a symmetry when fixing an origin(it clearly cannot be "not assumed"), but it is independent of anything related to synchronization or simultaneity conventions, or the one-way versus 2-way speed of light debate, that is related to the boosts part of the Lorentz group and the absence of a 4 dimensional spatial isotropy.

 

[PeterDonis]

Minkowski space is a real affine space with an indefinite bilinear form that determines certain symmetries in the vector space associated when fixing a point as the origin, namely the Lorentz group, a 6-parameter group of symmetry, 3 for the boosts and 3 for rotational invariance in three dimensions restricted to the tangent vector spaces at the points in affine space, not for affine space itself where there is no origin fixed.

Let me try to restate what I think you're saying in different terms. In order to give meaning to the term "isotropy", we need to pick a point to serve as the origin, so that we can pick out the particular group of transformations that count as "rotations" about that origin. But Minkowski spacetime does not have any particular point which is preferred as the origin; any point will do. So the term "isotropy" has no meaning if we just look at Minkowski spacetime itself; it only has meaning once we pick a particular point as the origin, and then it means "isotropy about this chosen point", not "isotropy of spacetime in general".

I agree with this as far as it goes; but I think what most people implicitly mean when they say that Minkowski spacetime is isotropic is that you can pick any point you like as the origin and you will have isotropy (in the sense of the 3-parameter group of spacelike Killing vector fields being present) about that point. Technically this is sloppy phrasing, but I think physicists are often sloppy in that way, at least from the viewpoint of mathematicians.

 

[harrylin]

other posters have manifested their preference for the meaning that makes the second postulate just a consequence of the first.

That misunderstanding was debunked in posts #22 and #67: There is no necessity to deviate from Galilean relativity on the basis of the relativity postulate alone, and there is no apparent contradiction between a consequence of something and that something.

 

[loislane]

Let me try to restate what I think you're saying in different terms. In order to give meaning to the term "isotropy", we need to pick a point to serve as the origin, so that we can pick out the particular group of transformations that count as "rotations" about that origin. But Minkowski spacetime does not have any particular point which is preferred as the origin; any point will do. So the term "isotropy" has no meaning if we just look at Minkowski spacetime itself; it only has meaning once we pick a particular point as the origin, and then it means "isotropy about this chosen point", not "isotropy of spacetime in general".

I agree with this as far as it goes; but I think what most people implicitly mean when they say that Minkowski spacetime is isotropic is that you can pick any point you like as the origin and you will have isotropy (in the sense of the 3-parameter group of spacelike Killing vector fields being present) about that point. Technically this is sloppy phrasing, but I think physicists are often sloppy in that way, at least from the viewpoint of mathematicians.

Yes, that's right.
Then my point about PAllen comments about "isotropy" and its relation with convention of synchronization for diferent frames, and how this determines the relations between one-way and two way speed of light is purely geometric.
The absence of a 4 dimensional-O(4)- rotational symmetry in Minkowski space that comes from not having a positive definite euclidean inner product leaves one degree of freedom in the affine space that is used in the arbitrary choice of simultaneity hyperplane-O(1.3) symmetry- for a certain time axis(i.e. tilting is allowed), that is what boosts are, while allowing also obviously the orthogonal coordinate choice. This specific choice is what an Einstein synchronization amounts to, and what I think PAllen refers to by "isotropy" convention that one is free to assume or not.

 

[loislane]

That misunderstanding was debunked in posts #22 and #67: There is no necessity to deviate from Galilean relativity on the basis of the relativity postulate alone, and there is no apparent contradiction between a consequence of something and that something.

I've stepped on a LET supporter' toe maybe?

 

[PAllen]

Then I don't understand what your discussion is about, rotational invariance in 3 dimensions is clearly a symmetry when fixing an origin(it clearly cannot be "not assumed"), but it is independent of anything related to synchronization or simultaneity conventions, or the one-way versus 2-way speed of light debate, that is related to the boosts part of the Lorentz group and the absence of a 4 dimensional spatial isotropy.

The argument about inability to prove isotropy of (one way) light speed is, in effect, there exists a different group of coordinates which lead to identical measurements of observables, where not only one way light speed is anisotropic, but (necessarily to match experiment) equations of EM and mechanics are also anisotropic in just the right way. This is taken to mean isotropy can't be proven. My argument is that the important physical point is existence of the large group that manifests isotropy for all laws. One could easily envision some universe where you have to establish coordinates different ways for different laws to show isotropy of that law. The ability to find such a large group that shows isotropy for all laws is, by my argument, the definition of what it means for the universe to show isotropy as fundamental symmetry. Then, IMO, the existence of coordinate schemes that don't manifest isotropy becomes irrelevant to the question of whether isotropy is an intrinsic symmetry of our universe.

 

[vanhees71]

Let me try to restate what I think you're saying in different terms. In order to give meaning to the term "isotropy", we need to pick a point to serve as the origin, so that we can pick out the particular group of transformations that count as "rotations" about that origin. But Minkowski spacetime does not have any particular point which is preferred as the origin; any point will do. So the term "isotropy" has no meaning if we just look at Minkowski spacetime itself; it only has meaning once we pick a particular point as the origin, and then it means "isotropy about this chosen point", not "isotropy of spacetime in general".

I agree with this as far as it goes; but I think what most people implicitly mean when they say that Minkowski spacetime is isotropic is that you can pick any point you like as the origin and you will have isotropy (in the sense of the 3-parameter group of spacelike Killing vector fields being present) about that point. Technically this is sloppy phrasing, but I think physicists are often sloppy in that way, at least from the viewpoint of mathematicians.

An affine space is in any case translation invariant. So if it is isotropic wrt. one point, it must be isotropic around any point. Minkowski space admits not an SO(4) group. So it is not isotropic in the strict sense.

 

[PeterDonis]

An affine space is in any case translation invariant. So if it is isotropic wrt. one point, it must be isotropic around any point.

Yes, agreed. This is consistent with what I said.

Minkowski space admits not an SO(4) group.

No, but once you've picked an origin, it does admit an SO(3) symmetry group about that origin (and, as you agree, any origin will do). That's what "isotropy" is usually taken to mean.

 

[vanhees71]

Sure, that's the isotropy of space for one (and thus all) inertial observers, which is tacitly assumed in Einstein's 1905 paper.

 

[loislane]

No, but once you've picked an origin, it does admit an SO(3) symmetry group about that origin (and, as you agree, any origin will do). That's what "isotropy" is usually taken to mean.

If by what physicists usually understand by isotropy is meant the concept in classical Newtonian physics, then I disagree, in classical mechanics the geometry the physics is based on is euclidean and in three dimensional euclidean geometry the SO(3) symmetry has a different mathematical sense from that in four dimensional Minkowskian geometry. In the euclidean case the rotational invariance is part of the isometry group, and isometries are distance-preserving maps between metric spaces. While the SO(3) symmetry of the vector spaces associated to Minkowski space has nothing to do with metric spaces, it is not a metric space property like the isotropy of classical mechanics.

 

[loislane]

The argument about inability to prove isotropy of (one way) light speed is, in effect, there exists a different group of coordinates which lead to identical measurements of observables, where not only one way light speed is anisotropic, but (necessarily to match experiment) equations of EM and mechanics are also anisotropic in just the right way. This is taken to mean isotropy can't be proven. My argument is that the important physical point is existence of the large group that manifests isotropy for all laws. One could easily envision some universe where you have to establish coordinates different ways for different laws to show isotropy of that law. The ability to find such a large group that shows isotropy for all laws is, by my argument, the definition of what it means for the universe to show isotropy as fundamental symmetry. Then, IMO, the existence of coordinate schemes that don't manifest isotropy becomes irrelevant to the question of whether isotropy is an intrinsic symmetry of our universe.

You would need to make explicit what symmetry group you are referring to now when you say isotropy and also what large group you mean: SO(4), SO(1,3)? It is hard for me to understand your post without that information.

 

[PeterDonis]

If by what physicists usually understand by isotropy is meant the concept in classical Newtonian physics, then I disagree

But as you note, the classical concept is not applicable in Minkowski spacetime, so clearly in the context of SR "isotropy" must mean something different.

In the euclidean case the rotational invariance is part of the isometry group, and isometries are distance-preserving maps between metric spaces.

Agreed.

While the SO(3) symmetry of the vector spaces associated to Minkowski space has nothing to do with metric spaces

But there is a connection between the two. If we pick a point of Minkowski spacetime as the origin, and we pick a particular spacelike hypersurface of simultaneity through that origin, then that hypersurface is a metric space with the geometry of Euclidean 3-space, including an SO(3) isometry group about the chosen origin. And this SO(3) isometry group is the same one we get when we use the properties of Minkowski spacetime as an affine space to determine the symmetries of the vector space at the chosen origin. (More precisely, the restriction, if that's the right term, of the symmetries of that vector space to the chosen spacelike hypersurface.)

 

[loislane]

But there is a connection between the two. If we pick a point of Minkowski spacetime as the origin, and we pick a particular spacelike hypersurface of simultaneity through that origin, then that hypersurface is a metric space with the geometry of Euclidean 3-space, including an SO(3) isometry group about the chosen origin. And this SO(3) isometry group is the same one we get when we use the properties of Minkowski spacetime as an affine space to determine the symmetries of the vector space at the chosen origin. (More precisely, the restriction, if that's the right term, of the symmetries of that vector space to the chosen spacelike hypersurface.)

Of course there is a certain connection but it is ultimately conventional, the first conditional (picking an origin) simply reflects and undoes the generalization from a vector space to an affine space.
The second conditional you use to stablish the connection between SO(3) in both geometries is the requirement of an arbitrary choice of a particular plane of simultaneity in the second case, it reflects the generalization from the euclidean metric geometry to the affine non-metric geometry. This is a particular choice of local coordinates at a particular point and thus conventional.

 

[harrylin]

I've stepped on a LET supporter' toe maybe?

I think that you stepped on no toe; did I step on one?

 

[vanhees71]

SO(3) is of course a subgroup of SO(1,3), and this is due to an important additional assumption on the structure of relativistic space-time, namely that for each inertial observer space is a Euclidean 3-dimensional affine space as in Newtonian mechanics. The new thing is that this is indeed restricted to inertial observers, because accelerated observers find a non-Euclidean space in special relativity.

 

[PeterDonis]

accelerated observers find a non-Euclidean space in special relativity.

There are some caveats to this statement that I think are worth mentioning.

First, there is one particular family of accelerated observers for whom "space" is still Euclidean: Rindler observers. These have proper acceleration all in the same (fixed) direction, and with magnitude that varies in just the right way to keep the radar distance between them constant. In the accelerated coordinates in which these observers are at rest, the metric is

$$
ds^2 = - a^2 x^2 dt^2 + dx^2 + dy^2 + dz^2
$$

where the observers' proper acceleration is in the ##x## direction, and the observer at ##x = 1## has proper acceleration of magnitude ##a##. It is evident that spacelike slices of constant coordinate time ##t## are Euclidean for this metric.

Second, when we look at a family of accelerated observers for whom "space" is non-Euclidean, we have to be careful defining what "space" means. For example, consider the family of Langevin observers, who are all moving in circular trajectories about a common origin, with the same angular velocity ##\omega##. In the accelerated coordinates in which these observers are at rest (we use cylindrical coordinates here to make things look as simple as possible), the metric is

$$
ds^2 = - \left( 1 - \omega^2 r^2 \right) dt^2 + 2 \omega r^2 dt d\phi + dz^2 + dr^2 + r^2 d\phi^2
$$

Note that this metric is only valid for ##0 < r < 1 / \omega##; at larger values of ##r##, there are no Langevin observers (if there were, they would be moving around their circles faster than light).

If we look at a spacelike slice of constant coordinate time ##t## in this metric, we find something unexpected: it is Euclidean! The metric of such a slice is simply ##dz^2 + dr^2 + r^2 d\phi^2##, which is the metric of Euclidean 3-space in cylindrical coordinates. Why, then, is it always said that "space" is not Euclidean for such observers?

The answer is that, although the observers are at rest (constant spatial coordinates ##z##, ##r##, ##\phi##) in this chart, the spacelike slices of constant coordinate time ##t## are not simultaneous spaces for those observers. That is, the set of events all sharing a given coordinate time ##t## are not all simultaneous (by the Einstein definition of simultaneity) for the observers. In fact, the set of events which are simultaneous, by the Einstein definition of simultaneity, to a given event on a given observer's worldline do not even form a well-defined spacelike hypersurface at all. So we can't even use that obvious definition of "space" for such observers.

In fact, the "space" that is non-Euclidean for these observers is a different kind of mathematical object: it is the 3-dimensional abstract space obtained by taking the quotient space of the 4-dimensional spacetime by the set of worldlines of the Langevin observers. In other words, we take each worldline and treat it as a "point", and look at the structure of the 3-dimensional space of such "points". We find that this "space" has a non-Euclidean metric, and that this metric gives a good description of the distances the observers would measure between themselves and neighboring observers. But this "space" does not correspond to any spacelike slice that can be taken out of the 4-dimensional spacetime.

Further discussion here:

https://en.wikipedia.org/wiki/Born_coordinates

 

[loislane]

SO(3) is of course a subgroup of SO(1,3), and this is due to an important additional assumption on the structure of relativistic space-time, namely that for each inertial observer space is a Euclidean 3-dimensional affine space as in Newtonian mechanics. The new thing is that this is indeed restricted to inertial observers, because accelerated observers find a non-Euclidean space in special relativity.

There are some caveats to this statement that I think are worth mentioning.

First, there is one particular family of accelerated observers for whom "space" is still Euclidean: Rindler observers. These have proper acceleration all in the same (fixed) direction, and with magnitude that varies in just the right way to keep the radar distance between them constant. In the accelerated coordinates in which these observers are at rest, the metric is

$$
ds^2 = - a^2 x^2 dt^2 + dx^2 + dy^2 + dz^2
$$

where the observers' proper acceleration is in the ##x## direction, and the observer at ##x = 1## has proper acceleration of magnitude ##a##. It is evident that spacelike slices of constant coordinate time ##t## are Euclidean for this metric.

Second, when we look at a family of accelerated observers for whom "space" is non-Euclidean, we have to be careful defining what "space" means. For example, consider the family of Langevin observers, who are all moving in circular trajectories about a common origin, with the same angular velocity ##\omega##. In the accelerated coordinates in which these observers are at rest (we use cylindrical coordinates here to make things look as simple as possible), the metric is

$$
ds^2 = - \left( 1 - \omega^2 r^2 \right) dt^2 + 2 \omega r^2 dt d\phi + dz^2 + dr^2 + r^2 d\phi^2
$$

Note that this metric is only valid for ##0 < r < 1 / \omega##; at larger values of ##r##, there are no Langevin observers (if there were, they would be moving around their circles faster than light).

If we look at a spacelike slice of constant coordinate time ##t## in this metric, we find something unexpected: it is Euclidean! The metric of such a slice is simply ##dz^2 + dr^2 + r^2 d\phi^2##, which is the metric of Euclidean 3-space in cylindrical coordinates. Why, then, is it always said that "space" is not Euclidean for such observers?

The answer is that, although the observers are at rest (constant spatial coordinates ##z##, ##r##, ##\phi##) in this chart, the spacelike slices of constant coordinate time ##t## are not simultaneous spaces for those observers. That is, the set of events all sharing a given coordinate time ##t## are not all simultaneous (by the Einstein definition of simultaneity) for the observers. In fact, the set of events which are simultaneous, by the Einstein definition of simultaneity, to a given event on a given observer's worldline do not even form a well-defined spacelike hypersurface at all. So we can't even use that obvious definition of "space" for such observers.

In fact, the "space" that is non-Euclidean for these observers is a different kind of mathematical object: it is the 3-dimensional abstract space obtained by taking the quotient space of the 4-dimensional spacetime by the set of worldlines of the Langevin observers. In other words, we take each worldline and treat it as a "point", and look at the structure of the 3-dimensional space of such "points". We find that this "space" has a non-Euclidean metric, and that this metric gives a good description of the distances the observers would measure between themselves and neighboring observers. But this "space" does not correspond to any spacelike slice that can be taken out of the 4-dimensional spacetime.

Further discussion here:

https://en.wikipedia.org/wiki/Born_coordinates

At the risk of making a pedantic point I must say that geometrically it makes little sense to talk about the spaces either in inertial or noninertial coordinates in terms of euclidean or non-euclidean unless is done in a purely metaphorical sense. The choice of observers or coordinates can never affect the intrinsic geometry of a space or a subspace of lower dimensions.
The Lorentz group of symmetry in the tangent space indeed has SO(3) as a subgroup, but it has other three dimensional subgroups that have nothing to do with euclidean geometry(like the group of isometries of the hyperbolic plane). What's important here is that these symmetries have in the context of points in an affine space nothing to do with euclidean or non-euclidean metric geometries.

 

[PeterDonis]

The choice of observers or coordinates can never affect the intrinsic geometry of a space or a subspace of lower dimensions.

Agreed. But it can affect which subspace you pick out as worthy of interest. When we talk about the spatial geometry of inertial coordinates being Euclidean, we mean that the 3-dimensional subspaces of Minkowski spacetime that are picked out as "spacelike hypersurfaces of constant coordinate time" in inertial coordinates have Euclidean geometry (if you pick a particular point as the spatial origin, so you have a metric space).

My point in the long post you quoted was that, for at least some non-inertial coordinates (Born coordinates in the case I described), saying that "space is non-Euclidean" for observers at rest in the coordinates isn't a statement about a 3-dimensional subspace of the spacetime at all. It's a statement about a quotient space.

 

[loislane]

Agreed. But it can affect which subspace you pick out as worthy of interest. When we talk about the spatial geometry of inertial coordinates being Euclidean, we mean that the 3-dimensional subspaces of Minkowski spacetime that are picked out as "spacelike hypersurfaces of constant coordinate time" in inertial coordinates have Euclidean geometry

When you pick inertial coordinates with standard synchronization you may describe the x, y, z coordinates as defining a cartesian space that one can identify with a Euclidean space.

(if you pick a particular point as the spatial origin, so you have a metric space).

No, you have a vector space, metric spaces are not related to fixing an origin but with determining distances, indefinite bilinear forms cannot determine distances.

My point in the long post you quoted was that, for at least some non-inertial coordinates (Born coordinates in the case I described), saying that "space is non-Euclidean" for observers at rest in the coordinates isn't a statement about a 3-dimensional subspace of the spacetime at all. It's a statement about a quotient space.

SR postulates don't hold for noninertial coordinates and again the possibility of assigning Euclidean or non-euclidean spatial relations from the choice of different curvilinear coordinates is trivial, but it doesn't say much about the underlying spaces, and it is my understanding that it cannot determine anything physical either. They are just labels.

 

[vanhees71]

True, it's a very bad habit of physicists to talk about "metrics" when dealing with the fundamental forms of pseudo-Riemannian (or pseudo-Euclidean) manifolds, but that's the jargon used. Usually people look a bit confused, when you start talking about pseudo-metrics or something like this, but strictly speaking it would be good to change this habit, particularly for beginners in the field!

 

[vanhees71]

There are some caveats to this statement that I think are worth mentioning.Second, when we look at a family of accelerated observers for whom "space" is non-Euclidean, we have to be careful defining what "space" means. For example, consider the family of Langevin observers, who are all moving in circular trajectories about a common origin, with the same angular velocity ##\omega##. In the accelerated coordinates in which these observers are at rest (we use cylindrical coordinates here to make things look as simple as possible), the metric is

$$
ds^2 = - \left( 1 - \omega^2 r^2 \right) dt^2 + 2 \omega r^2 dt d\phi + dz^2 + dr^2 + r^2 d\phi^2
$$

Note that this metric is only valid for ##0 < r < 1 / \omega##; at larger values of ##r##, there are no Langevin observers (if there were, they would be moving around their circles faster than light).

If we look at a spacelike slice of constant coordinate time ##t## in this metric, we find something unexpected: it is Euclidean! The metric of such a slice is simply ##dz^2 + dr^2 + r^2 d\phi^2##, which is the metric of Euclidean 3-space in cylindrical coordinates. Why, then, is it always said that "space" is not Euclidean for such observers?

Thanks for the clarification, but this statement is a bit misleading. You cannot simply set ##\mathrm{d} t## to 0 to infer what the observer considers the geometry of his "space" (time slice). The local geometry is rather defined by a metric of a 3D submanifold, where ("infinitesimal") distances are defined via the two-way speed of light (this is most clearly explained in Landau-Lifshitz vol. 2).

So the rotating observer sends a light signal towards an infinitesimal distant point, where it is reflected and measures the time his signal needs to come back to him. This time determines the distance (modulo a factor ##c##, which I set to 1). The corresponding times for the light signal to travel forth and back is determined by the null-geodesics condition for the light ray:
$$g_{\mu \nu} \mathrm{d} q^{\mu} \mathrm{d} q^{\nu}=0.$$
Split in temporal and spatial components you get (setting ##q^0=t##)
$$g_{00} \mathrm{d} t^2+2g_{0i} \mathrm{d} t \mathrm{d} q^i + g_{ij} \mathrm{d} q^i \mathrm{d} q^j=0.$$
Latin indices run from ##1## to ##3## (i.e., sum over the spatial coordinates).

This has two solutions for ##\mathrm{d} t##, and then you define the spatial distance as
$$\mathrm{d} \ell=\frac{\mathrm{d} t_1-\mathrm{d} t_2}{2} = \sqrt{\left (g_{ij}-\frac{g_{i0} g_{j0}}{g_{00}} \right )\mathrm{d} q^i \mathrm{d} q^j}.$$
Then you get the spatial metric as
$$\mathrm{d} \ell^2=\mathrm{d} r^2 + \frac{r^2}{1-\omega^2 r^2} \mathrm{d} \varphi^2+\mathrm{d} r^2.$$
This is the metric of a non-Euclidean 3D Riemannian space (or more strictly speaking for a region of (in this case Minkowskian) spacetime, covered by these coordinates, which are restricted by ##\omega r<1##).

Of course, that's a convention, defining the geometry of the observer's 3D spacelike hypersurface, but it's the one which is (for infinitesimal distances) equivalent to the usual Einsteinian description in SR (where he, however, uses the one-way speed of light rather than the round-trip speed of light, but this is problematic for a general (non-static) metric).

 

[Demystifier]

My answer would be that I don't know, because the question refers to a particular formulation of an obsolete axiomatization, and I don't think anyone in the year 2015 should be memorizing that kind of historical trivia (what's postulate #1, what's postulate #2, etc.). I think it's unfortunate if people are still teaching their students SR using Einstein's postulates, because they reinforce various misconceptions, such as the belief that c has something to do with the speed of light, or that light plays some fundamental role in relativity. Since Einstein himself had a view of SR that, looking back from 2015, seems to have been in many ways hazy and incorrect, why would it be of interest to anyone other than historians of science to try to figure out exactly what he had in mind?

In that context I would also recommend reading the book "Einstein's mistakes":
https://www.amazon.com/dp/0393337685/?tag=pfamazon01-20
The mistakes include a mistake on a two-way clock synchronization procedure, as well as repeated mistakes in 7 (!) different derivations of ##E=mc^2##.

 

[stevendaryl]

My answer would be that I don't know, because the question refers to a particular formulation of an obsolete axiomatization, and I don't think anyone in the year 2015 should be memorizing that kind of historical trivia (what's postulate #1, what's postulate #2, etc.). I think it's unfortunate if people are still teaching their students SR using Einstein's postulates, because they reinforce various misconceptions, such as the belief that c has something to do with the speed of light, or that light plays some fundamental role in relativity. Since Einstein himself had a view of SR that, looking back from 2015, seems to have been in many ways hazy and incorrect, why would it be of interest to anyone other than historians of science to try to figure out exactly what he had in mind?

I think that in learning science, students are simultaneously learning two different things:

1. What is the current, best theory of gravity, relativity, electromagnetism, whatever.
2. How do people go about formulating and testing new theories.

The historical way that a theory developed is not relevant to the first, but it is certainly relevant to the second.

 

[vanhees71]

In that context I would also recommend reading the book "Einstein's mistakes":
https://www.amazon.com/dp/0393337685/?tag=pfamazon01-20
The mistakes include a mistake on a two-way clock synchronization procedure, as well as repeated mistakes in 7 (!) different derivations of ##E=mc^2##.

Here's a paper (perhaps a short version of the book)?
http://arxiv.org/abs/0805.1400

 

[PeterDonis]

No, you have a vector space, metric spaces are not related to fixing an origin but with determining distances, indefinite bilinear forms cannot determine distances.

The metric of a single spacelike hypersurface, with a given point chosen as the spatial origin, is positive definite. That's the metric I'm talking about, not the metric of Minkowski spacetime as a whole.

again the possibility of assigning Euclidean or non-euclidean spatial relations from the choice of different curvilinear coordinates is trivial, but it doesn't say much about the underlying spaces, and it is my understanding that it cannot determine anything physical either. They are just labels.

Strictly speaking, yes, but if there are particular physical measurements that happen to match up with the coordinates in a particular way, then sloppy physicists will tend to talk about the coordinates as though they were physical observables instead of labels.

 

[PeterDonis]

The local geometry is rather defined by a metric of a 3D submanifold

But for the case of Langevin observers, there is no single submanifold that is "shared" by all the observers. Each observer has a different one, and the same observer has different ones at different times (and they don't even define a consistent coordinate chart, since the same spacetime event can be contained in multiple such submanifolds, corresponding to different times, for the same observer). The 3D manifold you end up deriving a non-Euclidean metric for is the quotient space I spoke of; it is not a 3D submanifold of 4D Minkowski spacetime.

 

[vanhees71]

Why is this a quotient space? I thought, that's how you generally define the local geometry of any observer in both flat (Minkowski space) and General Relativity. It's analogous to what you do for inertial observers in Minkowski space, but restricted to local (infinitesimal) neighborhoods of any point in the domain of map defined by the coordinates. That's the distance measured by light signals bouncing back and force from a mirror located at the point of interest (sometimes called "radar distance"). I thought, that's a very useful concept when setting up a (1+3)D formalism in general spacetimes. I found the discussion of electrodynamics in curved spacetime in Landau-Lifshitz using this convention very illuminating. I don't see that there is something wrong with this mathematically, it's just a rewriting of the covariant Maxwell equations in a curved (background) spacetime in a not manifestly co-variant way in the sense of a (1+3)D formalism, as one does also in usual Minkowski space in inertial frames to solve more practical problems in E&M. Is there something wrong with it physically (or even mathematically)?

EDIT: Or do you refer to the problem to define distances of finitely separated point in the case of a non-stationary metric, i.e., where the ##g_{\mu \nu}## depend on time, and you cannot define a distance measure for them?

 

[PeterDonis]

Why is this a quotient space? I thought, that's how you generally define the local geometry of any observer in both flat (Minkowski space) and General Relativity.

If we are just using your method to construct local coordinates around the worldline of a single Langevin observer, then yes, each "slice" of constant coordinate time is a 3D submanifold (with the non-Euclidean metric you give) of the 4D manifold covered by the coordinates. But that 4D manifold is not all of Minkowski spacetime; it's just a narrow "world tube" around the chosen worldline.

But if we try to view the non-Euclidean metric you give as the metric of a global "space" that includes all of the Langevin observers--i.e., not limited to a narrow world tube around one observer's worldline--then that "space" is a quotient space; it does not correspond to any 3D submanifold of the 4D manifold that includes the worldlines of all the Langevin observers. (Even that 4D manifold is not, strictly speaking, all of Minkowski spacetime, since it only covers the region of finite radius around the origin in which there are such observers. But that's a much larger region than the narrow world tube of a single observer's worldline.)

 

[loislane]

an important additional assumption on the structure of relativistic space-time, namely that for each inertial observer space is a Euclidean 3-dimensional affine space as in Newtonian mechanics.

I was discussing this assumption at the beginning of the thread. So you mean that the footnote that Einstein added to his first sentence in the kinematical part of his seminal 1905 paper was wrong? IOW that in SR inertial coordinates Newtonian mechanics holds good fully, not just to the first approximation?

 

[stevendaryl]

I was discussing this assumption at the beginning of the thread. So you mean that the footnote that Einstein added to his first sentence in the kinematical part of his seminal 1905 paper was wrong? IOW that in SR inertial coordinates Newtonian mechanics holds good fully, not just to the first approximation?

I don't see the connection between Einstein's comment and whether spacetime can be described as time + Euclidean space.

Obviously, Newton's laws don't hold exactly, because SR gives relativistic corrections. I interpret Einstein's words to mean that he's talking about a coordinate system such that Newton's laws hold, in the limit where all velocities are slow compared to the speed of light.

 

[loislane]

I don't see the connection between Einstein's comment and whether spacetime can be described as time + Euclidean space.

The Newtonian case is time+Euclidean space, SR according to Einstein correction is that only to first order.

Obviously, Newton's laws don't hold exactly, because SR gives relativistic corrections. I interpret Einstein's words to mean that he's talking about a coordinate system such that Newton's laws hold, in the limit where all velocities are slow compared to the speed of light.

Yes, that is what I understand the footnote to mean. But that limit is hard to see in the second postulate, where not all speeds are slow compared to c since it deals with c in inertial coordinates so it would seem here Newtonian mechanics must hold good exactly.

 

[vanhees71]

We are talking about Minkowski space. Here, for any inertial observer, space is Euclidean. I still don't understand this confusion.

 

[loislane]

We are talking about Minkowski space. Here, for any inertial observer, space is Euclidean. I still don't understand this confusion.

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.

 

[stevendaryl]

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.

A Rindler observer is the best-studied example of a NONinertial observer. So I think that what you're saying is completely wrong.

Roughly speaking, a frame, or coordinate system, is inertial if an object at rest relative to it has zero proper acceleration. Proper acceleration is measurable using an accelerometer.

 

[vanhees71]

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 stressed by stevendaryl in the posting before, a rindler observer is noninertial, because he is accelerated relative to the class of inertial frames. 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. You can define it as the sequence of rest frames of a particle in a globally homogeneous electric field. The solution for the equation of motion for the four-velocity
$$\frac{\mathrm{d} u^{\mu}}{\mathrm{d} \tau}=\frac{q}{m} F^{\mu \nu} u_{\nu}$$
can be written as
$$u(\tau)=\exp(\frac{q \tau}{m} \hat{F}) u(\tau=0).$$
For a homogeneous electric field you have ##F^{jk}=0## for ##j,k \in \{1,2,3\}## and ##F^{j0}=-F^{0f}=E^j=\text{const}##, which discribes a boost in a fixed direction ##\vec{n}=\vec{E}/|\vec{E}|##. The corresponding rapidity is growing with time, according to ##y(\tau)=q E \tau/m##.

This already shows that this is a pretty unphysical case, and it's only apparently simple. If you use it in other than kinematical context, you can have a lot of trouble with it. E.g., if you study the radiation of a so accelerated point charge, you run into trouble with the usual Lienart-Wiechert potentials. For that example and how the trouble is resolved, see

J. Franklin, D. J. Griffiths, The fields of a charged particle in hyperbolic motion, Am. J. Phys. 82, 755 (2014); erratum Am. J. Phys. 83, 278 (2015)
http://dx.doi.org/10.1119/1.4875195
http://dx.doi.org/10.1119/1.4906577
http://arxiv.org/abs/1405.7729