SR flat Lorentzian manifold and Anderson simultaneity convention

[cianfa72]

TL;DR Summary

About the relation between the SR flat Lorentzian manifold and simultaneity conventions other than Einstein's convention.

Hi,
I was thinking about the following.

From a mathematical point of view, SR assumes the following postulate: spacetime is a flat Lorentzian smooth manifold.

From the above and a minimal interpretation (i.e. a minimal set of "rules" to define the correspondence between mathematical objects and physical things) it follows there exist non-accelerating clocks (zero proper acceleration as measured by accelerometers attached to them) filling the entire space mutually at rest (constant round-trip time as measured by any single clock using bouncing light pulses) and Einstein's synchronizated (the latter is equivalent to say that the one-way speed of light in the frame being defined is isotropic with invariant speed $c$).

Now let's change the simultaneity convention for the above clocks using for instance Anderson convention such that the one-way speed of light is no longer isotropic ($c+ \neq c-$).

The frame/coordinate chart defined that way is no longer inertial, yet the two-way speed of light over any closed path is always isotropic with the same constant invariant speed $c$ (as measured by any single clock).

So, the fact that in a frame/chart the two-way speed of light is isotropic over any closed path with the same speed $c$ doesn’t rule out it as a non-inertial frame ?

 

[Ibix]

I would have said Anderson coordinates were inertial coordinates (all things with zero proper acceleration also have zero coordinate acceleration) but not frames (the coordinate basis is not orthonormal).

 

[cianfa72]

I would have said Anderson coordinates were inertial coordinates (all things with zero proper acceleration also have zero coordinate acceleration) but not frames (the coordinate basis is not orthonormal).

With frames you mean frame field(tetrad) defined at each spacetime point.

 

[Ibix]

Yes.

 

[cianfa72]

I would have said Anderson coordinates were inertial coordinates (all things with zero proper acceleration also have zero coordinate acceleration).

Here the coordinate acceleration of an object is the derivative of object's spacelike coordinates w.r.t. the timelike coordinate in the given coordinate chart.

 

[cianfa72]

I would have said Anderson coordinates were inertial coordinates (all things with zero proper acceleration also have zero coordinate acceleration)

Sorry, does that mean that in an SR global inertial frame/coordinate chart the one-way speed of light might be anisotropic ?

 

[Ibix]

I think you are worrying way too much about details of language.

There have been arguments on here before about the definition of "frame", let alone "inertial frame", and there are multiple possible viewpoints. You and I can talk for a while and pin down an agreement on whether "inertial chart" only includes orthonormal coordinates - but you may then pick up a book that uses a different standard. What you and I think is "right" is irrelevant to understanding that.

I personally think that it is reasonable to call any chart where inertial motion has a constant coordinate speed "inertial". I wouldn't call it an inertial frame unless it also had an orthonormal coordinate basis. But I am not an authority. Nobody is. You need to have a certain flexibility of definitions around this.

 

[cianfa72]

I personally think that it is reasonable to call any chart where inertial motion has a constant coordinate speed "inertial". I wouldn't call it an inertial frame unless it also had an orthonormal coordinate basis. But I am not an authority. Nobody is. You need to have a certain flexibility of definitions around this.

Yes, from my point of view "frame" is synonym of coordinate chart. What you call frame to me is actually "frame field (tetrad)".

So you draw a distinction: the above coordinate chart is inertial however the frame (associated to it) is not.

 

[PeterDonis]

So you draw a distinction: the above coordinate chart is inertial however the frame (associated to it) is not.

No, he's saying that the chart and the frame field associated with it are not orthonormal. The basis vectors are not all orthogonal to each other.

"Inertial" is usually taken to mean that observers at rest are in free fall. That is still true for the chart and frame field you defined.

In any case, as @Ibix has said, it is a mistake to worry about details of language. Physics is not done in words. It's done in math. If you have a consistent mathematical definition of your coordinate chart and frame field, you have all that is needed to do physics. How it is labeled in ordinary language is irrelevant.

 

[PeterDonis]

the two-way speed of light over any closed path is always isotropic

This is an invariant (it must be because it is directly observable), so it cannot depend on any choice of coordinates.

 

[Dale]

I think either convention is fine as long as you clarify whenever it matters. I would call both non-inertial.

 

[cianfa72]

I think either convention is fine as long as you clarify whenever it matters. I would call both non-inertial.

From the point of view of SR Einstein's 2nd postulate I would say that the above chart (with its implied synchronization convention) is not inertial since the one-way speed of light is not isotropic (even though observers at rest in it have zero proper acceleration).

Edit: BTW even though one changes the coordinates to Anderson, the congruence of such clocks is still timelike geodesic, hypersurface orthogonal with zero expansion and shear (that is an invariant/coordinate-free geometric property).

 

[Sagittarius A-Star]

From the point of view of SR Einstein's 2nd postulate I would say that the above chart (with its implied synchronization convention) is not inertial since the one-way speed of light is not isotropic (even though observers at rest in it have zero proper acceleration).

Edwards resolved this issue as follows:

Edwards replaced Einstein's postulate that the one-way speed of light is constant when measured in an inertial frame with the postulate:

The two-way speed of light in a vacuum as measured in two (inertial) coordinate systems moving with constant relative velocity is the same regardless of any assumptions regarding the one-way speed.​

Source:
https://en.wikipedia.org/wiki/One-w...ansformations_with_anisotropic_one-way_speeds

 

[PeterDonis]

Edwards resolved this issue as follows

That still leaves open the issue of what "inertial coordinate system" means. Would a non-orthonormal chart like Anderson coordinates still count as "inertial"? The two-way speed of light would still be the same in any such chart. But I don't think Edwards had that kind of chart in mind as an "inertial coordinate system" any more than Einstein did.

 

[cianfa72]

Would a non-orthonormal chart like Anderson coordinates still count as "inertial"? The two-way speed of light would still be the same in any such chart.

The line element $ds^2$ in Anderson coordinate chart is (one gets the Einstein's convention for $\kappa = 0$) $$ds^2 = - dt^2 - 2 \kappa \ dt dx + (1-\kappa^2) dx^2 + dy^2 + dz^2$$ It is evident that the associated coordinate basis vectors are non-orthonormal (there is also a mixed term in $dtdx$, yet the timelike congruence described by fixed spacelike coordinates $x,y,z$ in this chart is hypersurface orthogonal).

 

[PeterDonis]

the timelike congruence described by fixed spacelike coordinates in this chart is hypersurface orthogonal

That's because whether or not a timelike congruence is hypersurface orthogonal is an invariant, independent of the choice of coordinates.

 

[cianfa72]

The line element $ds^2$ in Anderson coordinate chart is (one gets the Einstein's convention for $\kappa = 0$) $$ds^2 = - dt^2 - 2 \kappa \ dt dx + (1-\kappa^2) dx^2 + dy^2 + dz^2$$

The transformation from global inertial (Einstein's synchronization convention) coordinates $(\bar t,x,y,z)$ to Anderson coordinates $(t,x,y,z)$ should be $$\bar t= t + \kappa x$$
This way Anderson coordinates only change the coordinate time assigned to events, the spacelike coordinates remain the same (hence the spacetime paths of constant $x,y,z$ represent the same members of the timelike geodesic congruence in both the coordinate charts).

 

[cianfa72]

Sorry I've a doubt. Suppose to transform from SR standard inertial coordinates $(t,x,y,z)$ to $(\bar t,x,y,z)$ coordinates with $$\bar t = f(t,x)$$
the inertial timelike geodesic congruence (hypersurface orthogonal with zero expansion and shear) described by fixed $(x,y,z)$ is at rest in both coordinates. In the latter coordinates, however, timelike geodesic paths (zero curvature/proper acceleration) not at rest in it do not have in general zero coordinate acceleration (using the chain rule for derivative w.r.t. the coordinate time $\bar t$).

Therefore one gets a coordinate chart $(\bar t, x,y,z)$ from a geodesic timelike congruence hypersurface orthogonal with zero expansion and shear at rest in it, that does not count as inertial, however.