Rotating Earth as an inertial frame

[Dale]

Your example does not address the points I made in Posts # 30, 38 & 44.

I'm sorry; it appears that we are having a miscommunication because I thought I did exactly that. Could you please restate your question in the most clear and concise manner possible (e.g. 2 sentences or less) and without reference to any previous posts.

 

[truhaht]

$t'$ is a spacelike coordinate! Can anyone see why?

Um, is it because its variance results in accelerations, which are +1 higher order spatial flux then the linear time line progress?

Tell me why then.

I like your post too

 

[The Dagda]

As others have said (I think), there is nothing in either special or general that prohibits coordinate speeds from being greater than $c$ other than misunderstanding, and rhetorical skills do not change this elementary fact.

Consider rotating coordinates defined form standard inertial coordinates by

$$\begin{equation*}<br /> \begin{split}<br /> x' &= x \cos \left(\omega t \right) - y \sin \left( \omega t \right) \\<br /> y' &= x \sin \left(wt \right) + y \cos( \omega t) \\<br /> z' &= z \\<br /> t' &= t.<br /> \end{split}<br /> \end{equation*}$$

Then, where it looks like coordinate speeds exceed $c$, $t'$ is not even a timelike coordinate, i.e., $t'$ is a spacelike coordinate! Can anyone see why?

Because time an space are intrinsically linked, thus time/space, t', and it's dimensions are one and the same thing. Do I win a prize for stating the obvious?

Time/space in a non euclidian frame is a rotation about an axis, it's as simple as that.

 

[JesseM]

Because time an space are intrinsically linked, thus time/space, t', and it's dimensions are one and the same thing. Do I win a prize for stating the obvious?

Time/space in a non euclidian frame is a rotation about an axis, it's as simple as that.

In relativity there is a clear difference between paths that are timelike, spacelike, and lightlike, and which category a path falls into is a coordinate-independent fact (physically, a timelike path is the worldline of an object moving slower than light, a lightlike path is the worldline of a light beam, and a spacelike path cannot be treated as any actual object's worldline unless we allow FTL particles). The metric gives a coordinate-invariant notion of the "spacetime distance" along a path ds^2 (which for timelike paths is just -c^2 times the proper time along the path), in much the same way that we can talk about the geometric distance along a path on a curved 2D surface like a sphere even though there are different possible coordinate systems you could place on that surface. And ds^2 is negative for timelike paths, 0 for lightlike paths, and positive for spacelike paths. So to say a coordinate is "timelike" at a point means that if you consider the infinitesimal path created by varying that coordinate an infinitesimal amount from the point while keeping all the other coordinates constant, this path would be a timelike one.

 

[atyy]

Then, where it looks like coordinate speeds exceed $c$, $t'$ is not even a timelike coordinate, i.e., $t'$ is a spacelike coordinate! Can anyone see why?

I suppose you are asking for a non-brute force way to see this, since one should always guess before calculating: t' points along the worldline that is an upward spiral around the t axis, as x' gets bigger and bigger the spiral gets shallower so that t' eventually ends up less parallel to the t axis and more parallel to the x axis?

 

[The Dagda]

In relativity there is a clear difference between paths that are timelike, spacelike, and lightlike, and which category a path falls into is a coordinate-independent fact (physically, a timelike path is the worldline of an object moving slower than light, a lightlike path is the worldline of a light beam, and a spacelike path cannot be treated as any actual object's worldline unless we allow FTL particles). The metric gives a coordinate-invariant notion of the "spacetime distance" along a path ds^2 (which for timelike paths is just -c^2 times the proper time along the path), in much the same way that we can talk about the geometric distance along a path on a curved 2D surface like a sphere even though there are different possible coordinate systems you could place on that surface. And ds^2 is negative for timelike paths, 0 for lightlike paths, and positive for spacelike paths. So to say a coordinate is "timelike" at a point means that if you consider the infinitesimal path created by varying that coordinate an infinitesimal amount from the point while keeping all the other coordinates constant, this path would be a timelike one.

Can anything move only in time or only in space?

And how would you describe it?

 

[JesseM]

Can anything move only in time or only in space?

Not in any absolute, coordinate-independent way. Relative to a particular coordinate system, I suppose you could say that an object with a timelike worldline is "moving only in time" if its coordinate position remains constant, but obviously other coordinate systems would disagree. In contrast, the issue of whether a given worldline is timelike, lightlike or spacelike is a coordinate-independent fact that everyone can agree on.

 

[Dale]

I suppose you are asking for a non-brute force way to see this, since one should always guess before calculating: t' points along the worldline that is an upward spiral around the t axis, as x' gets bigger and bigger the spiral gets shallower so that t' eventually ends up less parallel to the t axis and more parallel to the x axis?

To follow up with the brute force approach. The metric in George's rotating frame is:
$$ds^2=dt'^2 \left(c^2 - \omega ^2 (x'^2 + y'^2) \right)+2 \omega dt' (dy' x' -dx' y') -dx'^2-dy'^2-dz'^2$$

For an object "at rest" in this frame (dx'=dy'=dz'=0) this simplifies to:
$$ds^2=dt'^2 \left(c^2 - \omega ^2 (x'^2 + y'^2) \right)$$

Which is clearly spacelike for any $$\omega ^2 (x'^2 + y'^2) > c^2$$

 

[atyy]

To follow up with the brute force approach. The metric in George's rotating frame is:
$$ds^2=dt'^2 \left(c^2 - \omega ^2 (x'^2 + y'^2) \right)+2 \omega dt' (dy' x' -dx' y') -dx'^2-dy'^2-dz'^2$$

For an object "at rest" in this frame (dx'=dy'=dz'=0) this simplifies to:
$$ds^2=dt'^2 \left(c^2 - \omega ^2 (x'^2 + y'^2) \right)$$

Which is clearly spacelike for any $$\omega ^2 (x'^2 + y'^2) > c^2$$

With this in hand, would you like to comment on my guess? I suspect my guess wasn't right, because to answer Tam Hunt's question, shouldn't the worldline at large radii be timelike?

 

[RandallB]

I'm sorry; it appears that we are having a miscommunication because I thought I did exactly that. Could you please restate your question in the most clear and concise manner possible (e.g. 2 sentences or less) and without reference to any previous posts.

Option one:
As best as I can Tell GR requires that an Observer that observes rotations; can use any rotating frame of reference with themselves at the center of rotation they Like to simplify how something is observed. But cannot be required to hold such a selection as “the preferred” rotating frame of reference. Thus if an alternate item of is considered for observation and the same “non-preferred” frame of rotation is continued to be used – then FTL “coordinate” violations should be expected as possible.

Option two:
I cannot tell if you are saying GR is allowing the defining of a “preferred frame of rotation”.
If so it would seem an observer could be defined not only at the center of rotation but also at any radius from the center of such a preferred frame of rotation. As a “preferred frame” it should not see FTL events due to rotation.

Under the Option one understanding of GR the OP question has no standing.

Under the Option two understanding of GR I do not see how anyone has yet to resolve the OP question.

I don't know how to make the question any simpler.

 

[JesseM]

As best as I can Tell GR requires that an Observer that observes rotations; can use any rotating frame of reference with themselves at the center of rotation they Like to simplify how something is observed.

What do you mean by "can use"--are you implying the observer can't use other coordinate systems where they aren't the center of rotation? As I've asked before, do you understand that as far as making physical predictions goes, anyone can use absolutely any coordinate system whatsoever? For example, an inertial observer in SR has no obligation to use their own rest frame when making predictions, they could just as easily use a frame where they are moving at 0.99c.

 

[Dale]

Option one:
As best as I can Tell GR requires that an Observer that observes rotations; can use any rotating frame of reference

Yes, in GR you can use absolutely any coordinate system you choose.

with themselves at the center of rotation they Like to simplify how something is observed.

GR does not require an observer to be at the center of rotation.

But cannot be required to hold such a selection as “the preferred” rotating frame of reference.

Correct, no coordinate system (frame of reference) is preferred over any other. That is why you can use anyone you choose.

Thus if an alternate item of is considered for observation and the same “non-preferred” frame of rotation is continued to be used – then FTL “coordinate” violations should be expected as possible.

Yes, timelike worldlines may have coordinate speeds > c in some coordinate systems.

Option two:
I cannot tell if you are. saying GR is allowing the defining of a “preferred frame of rotation”.

No, there is no preferred coordinate system.

If so it would seem an observer could be defined not only at the center of rotation but also at any radius from the center of such a preferred frame of rotation.

I don't follow. Why would the freedom to choose an observer's distance from the center of rotation imply that a frame is prefered?

If anything, I would think the opposite would be true: any required restriction to a coordinate system would define a preferred set of frames - those that fulfilled the requirements. I think you have this backwards.

As a “preferred frame” it should not see FTL events due to rotation. .

There is no preferred frame in GR.

Under the Option one understanding of GR the OP question has no standing.

Under the Option two understanding of GR I do not see how anyone has yet to resolve the OP question.

I would say that your option 1 is closer to correct, with the only modification being that there is no restriction on coordinate systems in GR, including no restriction on the location of observers.

I don't know how to make the question any simpler.

Thanks for the attempt. I hope my responses answered your question.

 

[RandallB]

There is no preferred frame in GR.
I would say that your option 1 is closer to correct,
with the only modification being that there is no restriction on coordinate systems in GR, including no restriction on the location of observers.

Thanks for the attempt. I hope my responses answered your question.

Well I guess that is close enough for me. As I could not see how option 2 could be the case. And certainly under the option 1 (mine or as refined by you) it is clear the OP problem is not a issue for GR.

The only part that I do not quite get is how a GR observer can assume a frame rotation centered somewhere else than at their own location.
1) It seems like requiring information from a second observer – where SR; for example, requires using just one frame of referance at a time.
2) Plus what little math I've seen on rotating GR applications do not seem to use rotations dislocated from the observer in used (seems like defining two points of rotation with one at the observer point holding an alignment to the a distant point of rotation).

But as I said before I don't know GR applications that well especially related to rotating systems – not important that my understanding get that advanced, it may be buried inside background independence anyway.
Thanks for your input; I think we have over killed the OP issue here.

 

[JesseM]

The only part that I do not quite get is how a GR observer can assume a frame rotation centered somewhere else than at their own location.
1) It seems like requiring information from a second observer – where SR E.G. requires using just one.

Once again, do you understand that even in SR it is merely a matter of linguistic convention that we refer to an observer's rest frame as "their" frame, that there is absolutely nothing stopping an inertial observer from using an inertial frame where they are moving at 0.99c in order to make physical predictions?

 

[Dale]

With this in hand, would you like to comment on my guess? I suspect my guess wasn't right, because to answer Tam Hunt's question, shouldn't the worldline at large radii be timelike?

Your guess was exactly correct. The time coordinate is spacelike at large radii, and a massive object's worldline must always be timelike. Therefore, no object can be "at rest" in these coordinates at large radii.

 

[JustinLevy]

[quote="JustinLevy]Please be careful what you say here.
Locally they are the same. If a local lorentz frame has a metric with diagonal -1,1,1,1 at the origin, then so too does the origin in a rotating system defined with the origin following a geodesic.

A caveat: I am a GR potzer. That said, what you just said appears to conflict with my understanding of the Born metric for a rotating frame. For example, see http://arxiv.org/abs/gr-qc/0305084 equation (43) (pdf page 22).[/QUOTE]
Well, yes, due to using non-rectilinear coordinates the metric isn't diagonal -1,1,1,1. I guess I should have just said the metric is the same locally at r=0 for that rotating frame and an inertial frame.

For instance if you look at the equation you referenced, setting $\Omega=0$ would be an inertial frame. At r=0, the metric is the same regardless of the value of omega. So the inertial frame and rotating frame are the same locally.

If I'm somehow misunderstanding something, please do let me know.

 

[D H]

Please be careful what you say here.
Locally they are the same. If a local lorentz frame has a metric with diagonal -1,1,1 at the origin, then so too does the origin in a rotating system defined with the origin following a geodesic.

A caveat: I am a GR potzer. That said, what you just said appears to conflict with my understanding of the Born metric for a rotating frame. For example, see http://arxiv.org/abs/gr-qc/0305084 equation (43) (pdf page 22).

Well, yes, due to using non-rectilinear coordinates the metric isn't diagonal -1,1,1,1. I guess I should have just said the metric is the same locally at r=0 for that rotating frame and an inertial frame.

Thanks. I wasn't talking about r=0. I was implicitly talking about non-zero distances, such as the distance to a quasar. Or at least the radius of a ring laser gyroscope. Speaking of which,

DH, I'm going a little beyond Wikipedia here. How does one distinguish between an inertial frame and a rotating frame of reference? And please don't appeal to the fixed stars, which suggests instantaneous action at a distance.

A Foucault pendulum or a even better, a set of three orthogonal ring laser gyros will do exactly what you asked. In fact, spacecraft do exactly this. One of the myriad pre-launch checks performed by spacecraft avionics is to answer the question "do my inertial measurement units report that the vehicle is in the expected non-inertial frame?" The accelerometers should report that the vehicle is accelerating upward at about 9.8 meters/second² and the gyros should report that the vehicle is rotating at 2*pi/sidereal day about the Earth's rotation axis. Failure to detect these known conditions scrubs the mission.