Lorentz Contraction Circular Motion

[yuiop]

I have found another worm to add to the can. To a non rotating observer the radius of the disc is r or 1 lightsecond in the numerical example. To an observer on the rotating disc, the radar measured radius (R) is r/gamma = 0.6 lightseconds. This makes the situation even less Euclidean to the disc observer, because by his measurements of circumference and radius using radar measurements, the circumference is 2*pi*R*gamma^2.

Fredrik, you have still failed to provide a practical method by which an observer on the disc will measure the proper circumference as being simply 2*pi*r and when you do find a method, I predict it will not have the desirable properties of transitive time synchronisation, isotropic speed of light, same distance measured in either direction, a local speed of light of c, etc, etc.

 

[atyy]

Edit: I took a look at Poisson's http://www.physics.uoguelph.ca/poisson/research/agr.pdf . It's not exactly relevant, since it deals with timelike geodesics (section 2.3), and my guess is the Langevin observers are not geodesic. However, he says that although there are congruences that are not hypersurface orthogonal (section 2.3.3), these congruences still have a "transverse metric" which is purely "spatial" in the limited sense that it is "locally spatial" (section 2.3.1, that last term is mine, see his notes for the real maths).

Yeah, this is pretty much right. The point about the moving lattice's simultaneity is subtle, since there's no way of doing an Einstein synchronization over any finite area. However, all you need for a radar-ruler measurement of distance is Einstein synchronization on a line segment, and since that doesn't enclose any area, you're OK. I personally think Dieks' treatment is a lot more transparent than Rindler's, and I would suggest starting there. Rindler's is more general and abstract. The thing that confused me at first was that I thought there was a surface of simultaneity, which would appear as an equation of constraint that would eliminate one variable from the metric. That's incorrect, for the reasons outlined in #91. If you read Dieks' treatment, I think it's more clear what's going on. There's no variable whose differential is dt in the rotating frame. You just substitute in for dt, which mathematically represents doing a local Einstein synchronization between the two ends of a radar-ruler of length dl. So this is the only thing that I think might not be quite right in your quote above. In the last sentence, it's not that you're making the time term vanish, it's that you're eliminating the variable dt.

I just read Gron's treatment, which is identical to Rindler's as far as I can tell, except he gives a picture of the surface, and also states in the preceding chapter "... there does not exist a single space of simultaneity encompassing the "rest spaces" of all observers in an arbitrary reference frame. In this sense the 3-dimensional space described by the spatial metrical tensor is local." So I suppose this is all very opaque from the SR point of view, but apparently is a useful precursor to thinking about things in GR, just like Fredrik's favourite index notation for tensors in SR

 

[Fredrik]

I think I did understand what you meant, and what you meant was incorrect. In the first quote, you're defining a space consisting of a subset of Minkowski space. That isn't what you want to do in this example, because there is no global time synchronization, and therefore no natural way to pick such a subset, and it doesn't define the metric you want by the usual process of restricting a space to a lower-dimensional subspace.

I was really confused by the first comments, but the last one shows that you have definitely misunderstood what I'm doing. I'll try to explain it in a different way. Minkowski spacetime can be defined in the following way:

* Choose the set \mathbb R^4.
* Choose the standard topology.
* Choose the standard coordinate systems (i.e. all C^\infty injective functions from \mathbb R^4 into \mathbb R^4)
* Choose the metric to have components \eta_{\mu\nu} in the coordinate system I, where I is the identity map on \mathbb R^4.

What I'm doing is to define a different spacetime manifold M':

* Choose the set \mathbb R\times [0,\infty)\times [0,2\pi)\times\mathbb R, i.e. the range of the specific coordinate system on Minkowski spacetime that we've been considering.
* Choose the topology inherited from \mathbb R^4.
* Choose the coordinate systems to be the same as the ones used in the construction of Minkowski spacetime, but restricted to this smaller subset of \mathbb R^4.
* Do not choose the metric that would be inherited from Minowski spacetime if we thought of this as a submanifold (it would have the same components as the Minkowski metric), but instead choose the metric that has components

g_{00}=1-r^2\omega^2,\ g_{11}=1,\ g_{22}=\frac{r^2}{1-r^2\omega^2},\ g_{33}=1

g_{\mu\nu}=0 \mbox{ when }\mu\neq\nu

in the coordinate system I, where I is now the restriction of the identity map of \mathbb R^4 to M'.

Edit: This may not be enough. Maybe we should use [0,2\pi] instead of [0,2\pi), and then identify the line \phi=0 with the line \phi=2\pi.

The reason I can tell that you're not understanding this correctly is that you keep on talking about spirals.

Not sure if that means that I've misunderstood something, or if it means that you have.

The only reason to talk about spirals is if you're imagining this as a process where you induce a new metric by restricting the parent space to a surface of simultaneity. All of your previous objections (that the subspace is flat) make perfect sense with this approach. That's why it's not a fruitful approach.

Then we agree about that. I thought this was the approach taken by Grøn & Hervik in their book. Are you saying it isn't?

I thought I made it clear that this is how I interpreted what they were doing, and no one has objected against that until now. A.T. even defended calling the spiraling hypersurface "space".

As I've been saying, Rindler appears to be doing something different. To be more precise, he appears to be doing what I described for the second time earlier in this post.

 

[atyy]

Concerning the curved "spatial" hypersurface, are these statements true?
(i) it is everywhere locally orthogonal to the rotating observers
(ii) it is not a spacelike hypersurface, which is what disqualifies the rotating observers from being "hypersurface orthogonal" in the technical sense

 

[Fredrik]

If we're talking about the "spiral" hypersurface of Minkowski spacetime, then (i) is true and (ii) is not. (That's actually implied by (i)). I think it's also true for the alternative approach that I've been describing (as my interpretation of what Rindler is doing), but I'm not 100% sure about that one.

 

[atyy]

If we're talking about the "spiral" hypersurface of Minkowski spacetime, then (i) is true and (ii) is not. (That's actually implied by (i)). I think it's also true for the alternative approach that I've been describing (as my interpretation of what Rindler is doing), but I'm not 100% sure about that one.

Sorry, just to clarify, did you mean that (a) the spiral surface is not spacelike or (b) the spiral surface is spacelike?

BTW, I think Rindler and Gron are doing the same thing. Gron's 4.70, 4.71 and 4.76 are the same as Rindler's equations.

Edit: I see you meant that (b) the spiral surface is spacelike. So is Wiki wrong that the Langevin observers are not hypersurface orthogonal?

 

[Fredrik]

Yes, I meant that it's spacelike.

I don't know what a Langevin observer is. I haven't been reading all the articles and books that have been referenced. Just a couple of pages here and there.

Rindler and Grøn/Hervik are using the same definition of the rotating coordinate system. That's why they get the same result when they express the Minkowski metric in that coordinate system. But this doesn't mean that they're doing the same thing. They might be doing the same thing. Right now I'm not sure what either of them are doing. I think the choices are a) to take the last three terms to be the metric of a spacelike hypersurface of Minkowski spacetime, which consists of a bunch of spirals, and b) to define a new spacetime with a metric that has the same components in a cartesian coordinate system that the Minkowski metric has in rotating polar coordinates, and then consider the 3-dimensional submanifold (of the new spacetime) that's defined by a constant 0th coordinate.

 

[Fredrik]

If no one says anything in like two days, does that mean I won?

 

[yuiop]

If no one says anything in like two days, does that mean I won?

Nope, it just means we are waiting for you to answer this question:

Fredrik, you have still failed to provide a practical method by which an observer on the disc will measure the proper circumference as being simply 2*pi*r ...

As I understand it, the proper length of something, is a measurement that all observers will agree on and yet the learned members of this forum do not seem to be able to agree on what that measurement would be.

 

[DrGreg]

The "spiralling surface of simultaneity" (in spacetime) that some people have referred to doesn't exist.

Yes, if you consider simultaneity around a rotating circular ring, you get a spiral. And you can join lots of "concentric spirals" together to get a surface. But the surface you get coincides with the local definition of simultaneity (viz. of comoving inertial observer) only in the tangential direction and not in the radial direction. In the radial direction, a line of simultaneity would have to be horizontal (in the usual way of drawing things, with the centre of the circle's worldline vertical). The spirals don't join together "horizontally". It's not just a "global" problem of a discontinuity after a complete revolution, there are discontinuities around smaller loops, which vanish only when the loops shrink to zero.

So it's impossible to come up with a coordinate system in which

1. all points in the spinning disk have constant spatial coordinates
2. each surface of constant time coordinate coincides with every comoving inertial observer's local definition of simultaneity

In case you can't picture this geometrically, look at the maths of this. If (t,r,\theta) are inertial polar coordinates with the disk centre at r=0, and (t',r',\theta') are rotating polar coordinates which are supposed to meet the two criteria above (we can ignore the third spatial dimension), by considering a Lorentz transformation we require:

<br /> \begin{array}{rcl}<br /> dt&#039; &amp; = &amp; \gamma(r) (dt - r^2\omega\,d\theta / c^2)\\<br /> r\,d\theta&#039; &amp; = &amp; \gamma(r) (r\,d\theta - r\omega\,dt)\\<br /> dr&#039; &amp; = &amp; dr<br /> \end{array}<br />​

where

\gamma(r) = \frac{1}{\sqrt{1 - r^2 \omega^2 / c^2}}​

There is no solution to these equations.

For example, the second equation above is another way of saying

<br /> \begin{array}{rcl}<br /> \partial \theta&#039; / \partial \theta &amp; = &amp; \gamma(r)\\<br /> \partial \theta&#039; / \partial t &amp; = &amp; \gamma(r) \omega\\<br /> \partial \theta&#039; / \partial r &amp; = &amp; 0<br /> \end{array}<br />​

So you can't find the 3D space-metric of the lattice points as a restriction of the 4D Minkowski spacetime-metric, as no suitable coordinates exist.

Distance along any space-curve has to be determined by "local radar" between infinitesimally close points, which you then integrate to get the macroscopic length. Local radar implies two events that Einstein-simultaneous relative to the lattice points that are deemed to stationary in our frame. That means that in the equation quoted by Fredrik in post #94, the first term is zero, between any neighbouring events, and that's why the other terms represent the space-metric for the lattice points. Just don't bother trying to extend this 3D space-metric to a 4D spacetime-metric, because (as far as I can tell) that wouldn't mean anything useful.

 

[A.T.]

A.T. even defended calling the spiraling hypersurface "space".

I rather meant: Even if the 3D-spatial geometry as measured by rulers at rest in the rotating frame looks weird in 4D space-time, this doesn't stop me from calling it "spatial".

 

[Fredrik]

Nope, it just means we are waiting for you to answer this question:

Fredrik, you have still failed to provide a practical method by which an observer on the disc will measure the proper circumference as being simply 2*pi*r

He could set up a bunch of radar devices along the edge and have them all measure the distance (i.e. cT/2) to the next radar device at the precise moments when they receive the same spherical light signal from the point at the center, and then divide the results by \gamma before he adds them up (to compensate for the fact that each radar measurement gives him the proper length of a curve in spacetime that doesn't end where the next curve begins).

If you think this is somehow less valid than simply adding up the length measurements performed by a sequence of co-moving rulers (or radar devices) to 2\pi\gammar, then I request that you show me how this follows from a definition of "special relativity", or a reasonable definition of "length" or "circumference". See #88 for my definition of SR, and post your own if you don't like mine. The reason I consider my "measurement" a more appropriate representation of "the circumference in the rotating frame" is that it gives us the proper length of a continuous closed curve (a circle) in a hypersurface of constant time coordinate in the rotating frame. (Note that the rotating frame is defined with the same time coordinate as the inertial frame that's co-moving with the center).

As I understand it, the proper length of something, is a measurement that all observers will agree on and yet the learned members of this forum do not seem to be able to agree on what that measurement would be.

To me "proper length" is a mathematical concept that doesn't have anything to do with measurements until we have included an axiom in the theory that tells us how the two are related. If you prefer to use the term differently, that's not a real problem (except for you, who would have to think of another term for the mathematical concept), but you should at least agree that the issue here isn't what numbers the measuring devices are "displaying" to us, but how those numbers are related to things in the mathematical model.

 

[Fredrik]

The "spiralling surface of simultaneity" (in spacetime) that some people have referred to doesn't exist.

Yes, if you consider simultaneity around a rotating circular ring, you get a spiral. And you can join lots of "concentric spirals" together to get a surface. But the surface you get coincides with the local definition of simultaneity (viz. of comoving inertial observer) only in the tangential direction and not in the radial direction. In the radial direction, a line of simultaneity would have to be horizontal (in the usual way of drawing things, with the centre of the circle's worldline vertical). The spirals don't join together "horizontally". It's not just a "global" problem of a discontinuity after a complete revolution, there are discontinuities around smaller loops, which vanish only when the loops shrink to zero.

So it's impossible to come up with a coordinate system in which

1. all points in the spinning disk have constant spatial coordinates
2. each surface of constant time coordinate coincides with every comoving inertial observer's local definition of simultaneity

Yes, we have to drop number 2. I think what we should do is to define a new "time" coordinate

t&#039;=t-\frac{r^2\omega}{1-r^2\omega^2}\phi

Note that when \phi=0, we have t=t'. So the hypersurface of constant t' can be described as follows. Start with a straight line from the center to the radius, in a hypersurface of constant t. The hypersurface of constant t' is the union of all the spirals through that line, that have the property that at every event E on the spiral, the tangent of the spiral is Minkowski orthogonal to the world line of the point in the disc whose world line goes through E.

These spirals are "going up faster" near the edge than near the center. I can't really tell if this implies that the hypersurface of constant t' has positive or negative curvature.

 

[heldervelez]

I saw no reference to the 2002 paper of Rizzi & Ruggiero "Space geometry of rotating platforms: operational approach" at arxiv:gr-qc/0207104v2 13 Sep 2002
with their position and a review of methodologies, and at page 6:

"(s2) both the radius and circumference contract, so that their ratio remains 2*Pi (f.i. Lorentz, Eddington)"

...

 

[yuiop]

"(s2) both the radius and circumference contract, so that their ratio remains 2*Pi (f.i. Lorentz, Eddington)"

...

That should be qualified by "as measured in the non-rotating frame".

What the ratio would be measured as, by an observer in the rotating frame is still being debated here.

 

[yuiop]

... Suppose that you have two cars with odometers. I think the odometers match up with your notion of d\ell defined by laying down rulers. The cars also have clocks on their dashboards. You send one car out around the disk in the clockwise direction, and the other in the counterclockwise direction. When the cars meet up on the far side of the disk, their clocks will be out of sync due to the d\theta&#039; dt term in the metric, even though they've traveled an equal distance at an equal speed. You could just accept this, but it's uncomfortable, because it leaves you wondering where the funny asymmetry comes from. Someone who doesn't like your laying-down-rulers definition can say, "See? I told you that definition would lead to no good!"

I just did some quick calculations and concluded that a car going clockwise around the rim of the rotating disk will not return to the start point at the same time as a car that starts out at the same time and goes anti-clockwise, when both cars have the same velocity relative to the rim. Since they are both returning to the same point on the rim, there should be no issues about how clocks are synchoronised and they either return at the same time or not in any reference frame. The conclusion seems a bit shocking and counter intuitive and maybe I made a mistake. Can anyone confirm or refute this observation?

... Suppose that you have two cars with odometers. I think the odometers match up with your notion of d\ell defined by laying down rulers...

Actually, the odometers will read d\ell\ \gamma where \gamma = \sqrt{(1-u^2/c^2)} and u is the velocity of the car relative to the disc rim. In other words in linear example, two cars traveling from point A to point B at different velocities (where A and B are at rest wrt each other), will read different distances on their mechanical odometers and this will therefore not agree with laying down of rulers method. We analysed how a relativistic wheel rolls in an old thread and for each complete revolution of the wheel moving with linear velocity u relative to a road, the wheel moves forward a distance x in the frame comoving with the wheel axis and a distance x/ \gamma in the frame that the road is at rest in.

 

[DrGreg]

I just did some quick calculations and concluded that a car going clockwise around the rim of the rotating disk will not return to the start point at the same time as a car that starts out at the same time and goes anti-clockwise, when both cars have the same velocity relative to the rim. Since they are both returning to the same point on the rim, there should be no issues about how clocks are synchronised and they either return at the same time or not in any reference frame. The conclusion seems a bit shocking and counter intuitive and maybe I made a mistake. Can anyone confirm or refute this observation?

I agree with your conclusion and can explain it without any calculation.

To ease the explanation, I'd like to consider a slight modification. Instead of a circular disk, consider a tightly-wound helical rod. By "tightly-wound" I mean the distance between two adjacent turns is tiny, we can consider it negligible compared with the radius (and ultimately we can consider the limit as the distance drops to zero and the helix "collapses" to a circle). The helix is rotating about its axis. The reason I want to introduce this is so that I can unambiguously refer to "simultaneity along the rod", determined by local Einstein-simultaneity between pairs of nearby points and then "daisy chaining" along the rod. By this definition, the simultaneity is the same as if it were unwound into a straight line and moving along its length. Clearly the rod-simultaneity is not the same as the simultaneity of an inertial observer fixed on the axis, because of the relative motion.

Consider two points on the helix which are exactly one circumference apart on the rod: the points are right next to each other in 3D space. It should be clear that the local 3D definition of simultaneity disagrees significantly with daisy-chained-round-the-rod-simultaneity for these two points. (When the two points are a negligible distance apart we can talk of absolute local simultaneity instead of relative simultaneity.)

So now we consider kev's experiment above adapted to my helix. Two cars start at the same point and each travel one revolution in opposite directions at the same speed relative to the rod. They end up two turns apart on the helix, but right next to each other. It should be clear that they reach their destinations rod-simultaneously. (Consider the equivalent journey on an unwound rod; they travel the same distance, one circumference, at the same speed.) But as we've seen, rod-simultaneous is not locally-simultaneous for points that are as good as coincident in 3D space, so the cars don't actually get to the destination at the same event, as kev asserted.

 

[Fredrik]

I have another explanation. Suppose the two cars both start their laps at the same event A. We either imagine that they pass right through each other on the opposite side, or that the disc is so large that we can neglect the little detour that one of them has to make to avoid a collision. Let B be the event where they meet up for the second time after they separated. (The first is the near collision). Now look at the events from the inertial frame that's co-moving with the center. Let t be the time between event A and event B. Since both cars have been traveling for t seconds, they will have traveled the same distance if and only if they've been traveling at the same speed. But the relativistic velocity addition law implies that they haven't. (Because of that, they can't be at the starting point on the disc at B, so this isn't exactly the scenario that kev described).

 

[DrGreg]

(Because of that, they can't be at the starting point on the disc at B, so this isn't exactly the scenario that kev described).

It's pretty much the same scenario, because if they get to the starting point at two different times then (assuming the one who gets there first doesn't stop but keeps on going) they will meet somewhere other than the starting point i.e. after unequal distances measured on the disk.

 

[yuiop]

I have another explanation. Suppose the two cars both start their laps at the same event A. We either imagine that they pass right through each other on the opposite side, or that the disc is so large that we can neglect the little detour that one of them has to make to avoid a collision. Let B be the event where they meet up for the second time after they separated. (The first is the near collision). Now look at the events from the inertial frame that's co-moving with the center. Let t be the time between event A and event B. Since both cars have been traveling for t seconds, they will have traveled the same distance if and only if they've been traveling at the same speed. But the relativistic velocity addition law implies that they haven't. (Because of that, they can't be at the starting point on the disc at B, so this isn't exactly the scenario that kev described).

One small problem with this cut down analysis, is that it might lead a casual reader to the wrong conclusion. For a clockwise rotating disc, a clockwise going car returns first in your scenario, while in my scenario, the anti-clockwise going car returns first (still a clockwise rotating disc).

This leads on to an interesting alternative view of the circumference of a rotating disc. Since light and cars going clockwise take longer to go around the disc than light and and cars going anticlockwise (for a clockwise rotating disc) there is a sense that the clockwise circumference is longer than the anticlockwise circumference of the same clockwise rotating disk. Inhabitants of the disc might place mileage signs on roads saying things like "New York to Old Town 50 miles" and "Old Town to New York 10 miles" pointing in opposite directions. This might be pleasing to those who desire the speed of light to be the same in all directions, but the fact that it requires different size rulers to prove that, might not be convincing to everyone. This directional dual definition of length might be in agreement with distances defined by measuring angles using theodlites, but that is a bit complicated to work out.

Of course we could just stick to the usual cT/2 radar measurement of length and just accept that the speed of light is anisotropic in a non inertial frame over non infitesimal distances. Inhabitants of a rotating frame would be aware of the strange behavior of light in there system by observing that a photon going from A to B does not pass a photon going from B to A, as they follow two distinct spatial paths. To clarify what I mean, I have shown the light path AB in red and the light path BA in blue on a clockwise rotating disc in the attached sketch. (This is the viewpoint of observers on the disc.)

He could set up a bunch of radar devices along the edge and have them all measure the distance (i.e. cT/2) to the next radar device at the precise moments when they receive the same spherical light signal from the point at the center, and then divide the results by \gamma before he adds them up (to compensate for the fact that each radar measurement gives him the proper length of a curve in spacetime that doesn't end where the next curve begins).

It seems that all the observer does by "compensating" is work out what the disc circumference would be from the point of view of a non rotating observer. This is a bit like an observer in an accelerating rocket in in flat space time claiming that since he knows he is accelerating he should adjust the proper length measurement, of his rocket because he must be length contracting. It also requires the disc observer to have knowledge of his rotational velocity to work out out proper lengths in in his reference frame. In SR, the observer in a reference frame does not need any knowledge of his velocity to work out distances in his own reference frame.

One disadvantage of your compensated length measurement is that is the rotational velocity of the disk is changed to a new constant rotational velocity, all the road mileage markings have to be changed. The same applies to my dual directional method of defining circumferential distances. The simple uncompensated cT/2 method means that any road marking can remain permanent and are accurate even when the rotational velocity of the disc is changed. It only requires that the rulers remain unstressed along their length before and after any change. In SR, the proper length of a rocket remains unchanged before and after an acceleration period using either the cT/2 radar method or the layed down rulers method and surely the same should apply in the case of the disc?

If you think this is somehow less valid than simply adding up the length measurements performed by a sequence of co-moving rulers (or radar devices) to 2\pi\gammar, then I request that you show me how this follows from a definition of "special relativity", or a reasonable definition of "length" or "circumference". See #88 for my definition of SR, and post your own if you don't like mine.

Well your definition in #88 seems fine and I quote it again here:

3. A radar device measures infinitesimal lengths in the following way: If the roundtrip time is T, then cT/2 is the approximate proper length of the spacelike geodesic from the midpoint of the timelike geodesic through the emission event and the detection event to the reflection event. The approximation becomes exact in the limit T→0. (I haven't found a way to say this that isn't really awkward).
...
3'. A radar device moving as represented by a timelike geodesic measures lengths in the following way: If the roundtrip time is T, then cT/2 is the proper length of the spacelike geodesic from the midpoint of the worldline between the emission event and the detection event to the reflection event.

With 3', we have a theory that's at least as worthy of the name "special relativity" as anything Einstein could have written down in 1905, but it doesn't make any prediction at all about the circumference of the disc in the rotating frame. This theory simply doesn't tell us how to make measurements with non-inertial measuring devices. This is of course exactly why we should prefer 3 over 3'.

Neither method 3 nor 3' as defined by you, mention that the observer needs to work out his velocity and compensate by gamma to work out proper lengths in his own reference frame.

My first thought was that it doesn't make sense to call the result obtained this way "the measured circumference in the rotating frame". I thought that it made no sense to describe the sum of many measurements made by measuring devices in different states of motion as the result of a single measurement in a frame where all devices have constant spatial coordinates. But then I realized that this is exactly what we do when we claim to have used axiom 3 to measure something (non-infinitesimal) in an inertial frame. All the measuring devices have the same velocity, but not the same world lines, so we're definitely adding up results from measuring devices in different states of motion. If we allow ourselves to say that we have measured a non-infinitesimal length in an inertial frame (using axiom 3 rather than 3'), then we have no reason not to allow ourselves to say that we have measured non-infinitesimal lengths in the rotating frame...

Consider the following practical demostration. Draw the worldlines of points A and B moving at constant and equal velocity in a straight line in flat space on a time space diagram. Indicate the proper length distance between A and B taking simultaneity into account on the same diagram. Now roll the paper the diagram is drawn on into a cylinder. Essentially nothing has changed and the methods for calculating proper length in Minkowski spacetime apply equally in the rotating disc case. I know you do not actually need to draw the diagram, so count it as a thought experiment.

 

[Fredrik]

Well your definition in #88 seems fine and I quote it again here:

Neither method 3 nor 3' as defined by you, mention that the observer needs to work out his velocity and compensate by gamma to work out proper lengths in his own reference frame.

3 implies that we have to do that, if we're talking about a circle in a hypersurface of simultaneity of the rotating coordinate system. I still don't see a reason why anyone would use the term "the circumference of the disc in the rotating coordinate system" about something completely different, like the proper length of a spiral in Minkowski spacetime, or the arc length of a circle in a 3-dimensional Riemannian manifold that isn't a hypersurface of Minkowski spacetime. (I have been told that this manifold is actually the quotient space of Minkowski spacetime and the set of world lines of points on the disc. I'm expecting to receive more information about that soon).

 

[Demystifier]

I said a lot on this stuff more than 10 years ago in
http://xxx.lanl.gov/abs/gr-qc/9904078 [Phys.Rev. A61 (2000) 032109].
I hope someone may still find it usefull.

 

[yuiop]

I said a lot on this stuff more than 10 years ago in
http://xxx.lanl.gov/abs/gr-qc/9904078 [Phys.Rev. A61 (2000) 032109].

All quotes below are from your paper:

If the disc is constrained to have the same radius r as the same disc when it does not rotate, then L is not changed by the rotation, but the proper circumference L0 is larger than the proper circumference of the nonrotating disc. This implies that there are tensile stresses in the rotating disc.

Correct, there would be tensile stresses in the rotating disc, if the radius is constrained to be the same as when it is not rotating.

However, there is something wrong with this standard resolution of the Ehrenfest paradox.

I don't think there is.

Consider a slightly simpler situation; a rotating ring in a rigid nonrotating circular gutter with the radius r = r0. ...
This means that an observer on the ring sees that the circumference is L' = \gammaL. The circumference of the gutter seen by him cannot be different from the circumference of the ring seen by him, ...

I beg to differ. Let us consider the numerical example I mentioned in an earlier post. A ring of radius 1 lightsecond rotating clockwise with an instantaneous rim velocity of 0.8c. Here I will assume the rotating and non-rotating observers measure the radius to be the same. Using your statement "This means that an observer on the ring sees that the circumference is L' = \gammaL" the circumference of the ring measured by the observer on the ring is 2*pi*r*gamma = 10.4719. Now to measure the circumference of the gutter, the observer on the ring would note a mark on the gutter and time how long it takes the mark to complete a revolution and multiply the time by the relative velocity of the gutter to him. This method uses one clock so we can avoid argument about how spatially spearated clocks on a ring are synchronised. The result is 2*pi*r/gamma = 3.7699. I am dividing by gamma because the time measured by the observer on the ring is reduced by time dilation of his clock. The circumference of the gutter seen by him is not the same as the circumference of the ring seen by him, as you claim.

... so the observer on the ring sees that the circumference of the relatively moving gutter is larger than the proper circumference of the gutter, ...

The observer sees the circumference of the relatively moving gutter as 3.7699 and the proper circumference of the gutter is 2*pi*r = 6.2831 so your above statement is simply wrong.

... whereas we expect that he should see that it is smaller. This leads to another paradox.

It does not lead to another paradox, because your assumptions and calculations are wrong in the first place.

Note that the observer on the ring sees the gutter as length contracted and the observer on the gutter sees the ring as length contracted, exactly as relativity predicts. No paradox.

 

[Demystifier]

The observer sees the circumference of the relatively moving ring as 3.7699 and the proper circumference of the gutter is 2*pi*r = 6.2831 so your above statement is simply wrong.

The ring completely fills the gutter, so one observer should see that the two circumferences are equal. That's why I disagree with you.

 

[yuiop]

The observer sees the circumference of the relatively moving ring as 3.7699 and the proper circumference of the gutter is 2*pi*r = 6.2831 so your above statement is simply wrong.

The ring completely fills the gutter, so one observer should see that the two circumferences are equal. That's why I disagree with you.

I made a typo in my last post. I should have said:

The observer (on the ring) sees the circumference of the relatively moving gutter as 3.7699 and the proper circumference of the gutter is 2*pi*r = 6.2831

(The edits are in bold.)

To summarise what the various observers measure:

Observer on the rotating ring measures:
Circumference of the ring = 10.4719
Circumference of the gutter = 3.7699

Observer on the non rotating gutter measures:
Circumference of the ring = 6.2831
Circumference of the gutter = 6.2831

From the above measurements, it can be seen that only the observer on the gutter measures the circumference of the rotating ring and the non rotating gutter to be the same. The observer on the rotating ring does not see it that way. I admit that does seem slightly unintuitive, but that is relativity for you.

 

[Fredrik]

Looks like you're both confusing "an observer on the ring" with "a rotating observer". The coordinate system we would associate with the motion of a little guy riding the ring is a co-moving inertial frame. The ring isn't even circular in such a frame. A "rotating observer" on the other hand, isn't an observer at all. In operational terms, it's a bunch of measuring devices spread out all over the ring, and mathematically, it isn't represented by a coordinate system, but by a frame field.

 

[A.T.]

A "rotating observer" on the other hand, isn't an observer at all.

I guess the rotating observer would disagree with that.

In operational terms, it's a bunch of measuring devices spread out all over the ring,

Isn't every observer who observes more than his local surrounding a "bunch of spread out measuring devices"?

and mathematically, it isn't represented by a coordinate system, but by a frame field.

What prevents the rotating observer from using a rotating coordinate system.

 

[Fredrik]

I guess the rotating observer would disagree with that.

If you mean a guy riding on the disc/ring/whatever, then no, he wouldn't, as he only performs local measurements which agree with the coordinate assignments of a co-moving inertial frame. The only coordinate systems that can be naturally associated with his motion are co-moving inertial frames.

Isn't every observer who observes more than his local surrounding a "bunch of spread out measuring devices"?

You could say that, but there's definitely a coordinate system (a co-moving inertial frame) that we can associate with an inertial observer's motion and orientation (his notion of "right", "forward" and "up"). In addition to that, all of those other devices would agree with that coordinate system about all lengths and times.

What prevents the rotating observer from using a rotating coordinate system.

He could use any coordinate system he wants of course, but the measurements of the "bunch of spread out measuring devices" won't agree with the lengths assigned by the rotating coordinate system, as I have pointed out many times in this thread. The rotating coordinate system assigns the length 2*pi*r to the circumference of the disc at all times, while the measurements add up to 2*pi*gamma*r.

The motion and orientation of each of the measuring devices determine a co-moving inertial frame at each point on its world line. The inertial frame defines an orthonormal basis for the tangent space at that point. Orthonormal bases are bijective with frames at that point. (The technical definition of a "frame" at a point p in a manifold M is a linear function f:\mathbb R^n\rightarrow T_pM, where n=dim M). So the collection of world lines determine a frame at each point on any of the world lines. Hence, we have a "frame field" defined on the region of spacetime that's filled up with world lines of points in the disc.

 

[yuiop]

Looks like you're both confusing "an observer on the ring" with "a rotating observer". The coordinate system we would associate with the motion of a little guy riding the ring is a co-moving inertial frame. The ring isn't even circular in such a frame. A "rotating observer" on the other hand, isn't an observer at all. In operational terms, it's a bunch of measuring devices spread out all over the ring, and mathematically, it isn't represented by a coordinate system, but by a frame field.

I made it clear that I was specifying what an observer on the ring would measure and I do not think I was getting that mixed up with anything else.

You have probably noticed we have narrowed the discussion down to an infinitessimally thin ring rather than a disc, because I am sure we would all agree that there is no natural way for observers at different radii to synchronise clocks with each other and hopefully that will simplify things a bit.

Now we come to issue of specifying a common frame for observers spread out on a ring that are all equidistant from the axis of the ring. In linear SR, observers that are not moving parallel to each other, obviously do not share a common reference frame even if the magnitude of their velocities are equal. However in this situation it is also obvious that these observers do not measurethemselves to be at rest with each other as there spatial separation is continually changing over time. The case of the rotating ring is different because multiple observers on the ring can consider themselves to be at rest with respect to each other, as their mutual spatial separation does not change over time, so it should not be too difficult to construct a common coordinate system for those observers "on the ring".

Using the EP as a guiding principle it might be well to consider the issue from a gravitational point of view. Do you agree that observers equidistant from a central gravitational mass can consider themselves to be sharing a common reference frame? Do you agree that the Schwarzschild coordinate system is a "coordinate system" that suitably describes the measurements of a bunch of observers/measurement devices spread out all over the place at different radii with clocks running at different rates or would you say the Schwarzschild coordinate system or Schwarzschild metric would be better described as the "Schwarzschild frame field"? Please do not assume I am saying you are wrong, because I readily concede your knowledge of the terminology is much greater than mine and I am just trying to understand better what you are saying and maybe learn something.

 

[yuiop]

Here is a observation about rotational motion that may be of interest.

In linear SR, if a series of synchronised clocks spread out along a rod, are accelerated using Born rigid acceleration, the clocks naturally go "out of sync" because the clocks do not follow "parallel" paths through spacetime.

If a ring with a similar series of synchronised clocks spread out on it, is rotationally accelerated, the clocks remain in sync with each other naturally, because all the clocks follow a similar path through spacetime.

In other words, "transitive sychronisation" (peripheral clocks synchronised by a central clock) rather than "Einstein synchronistation" would seem to be the natural condition of rotating clocks. This seems to be an important difference between the linear and circular cases, but I am not quite sure what the significance of that observation is at the moment.

 

[Fredrik]

I made it clear that I was specifying what an observer on the ring would measure and I do not think I was getting that mixed up with anything else.

The observer on the ring only measures an infinitesimal segment at his own location, and you talked about the measured value of the circumference, so you're clearly not just talking about what he is measuring. Also, the ring isn't even shaped like a circle in the coordinate system that we would associate with his motion and orientation.

so it should not be too difficult to construct a common coordinate system for those observers "on the ring".

The common coordinate system is the rotating coordinate system, and it doesn't agree with their measurements.

Do you agree that observers equidistant from a central gravitational mass can consider themselves to be sharing a common reference frame?

Are they in orbit? In free fall? Held at fixed spatial coordinates in the Schwarzschild coordinate system? And what's a "reference frame"? In SR it's usually used synonymously with "inertial frame", which really means "inertial coordinate system". But then we usually only consider observers that are, always have been, and always will be, moving with the same constant velocity. When we're talking about an observer (still in SR) with a world line that isn't a geodesic, the obvious generalization is the concept of local inertial frame, which works in GR too. A local inertial frame in SR is actually just a co-moving global inertial frame.

"Reference frame" doesn't seem to be a well-defined concept to me. Maybe there is a definition, but if there is one, I'm still unaware of it. If we really want to consider a bunch of measuring devices spread out all over the place, then I think we need to be talking about frame fields instead of coordinate systems, for the reasons I've mentioned.

In other words, "transitive sychronisation" (peripheral clocks synchronised by a central clock) rather than "Einstein synchronistation" would seem to be the natural condition of rotating clocks.

Why? It doesn't work all the way round. I also wouldn't call them "rotating clocks". They're just clocks on circular paths (in "space", as defined by the rotating coordinate system, or equivalently, by the inertial frame in which the point at the center is at rest).

 

[Demystifier]

The coordinate system we would associate with the motion of a little guy riding the ring is a co-moving inertial frame. The ring isn't even circular in such a frame. A "rotating observer" on the other hand, isn't an observer at all. In operational terms, it's a bunch of measuring devices spread out all over the ring, and mathematically, it isn't represented by a coordinate system, but by a frame field.

I agree with you.

 

[Demystifier]

I
Observer on the rotating ring measures:
Circumference of the ring = 10.4719
Circumference of the gutter = 3.7699

If, by observer, one means a local observer staying at a single point of the ring, then I disagree. Such an observer observes that the circumference of the ring is equal to the circumference of the gutter.

 

[Demystifier]

The observer on the ring only measures an infinitesimal segment at his own location

That is exactly one of my points in the paper I mentioned above.

Yet, it is not in contradiction with my post #133, because this local observer may also watch the whole ring/gutter, which may also be counted as a sort of measurement. Of course, this is a measurement of a different kind.

 

[Demystifier]

What prevents the rotating observer from using a rotating coordinate system.

He can use any coordinates he wants. However, the particular coordinate system you mention does NOT correspond to his PROPER frame of coordinates, unless he is sitting in the center of rotation.

 

[Demystifier]

The common coordinate system is the rotating coordinate system, and it doesn't agree with their measurements.

More precisely, the rotating coordinate system agrees only with the measurements of the observer sitting in the center of rotation.

What many people here (Fredrik is excluded) do not seem to understand is that rotation is not the same as circular motion. The observer sitting in the center of the rotating disc rotates but does not move circularly. The observer sitting at the rim of the rotating disc both rotates and moves circularly. A gyroscope orbiting around a planet moves circularly but does not rotate.

 

[A.T.]

Held at fixed spatial coordinates in the Schwarzschild coordinate system?

Assume that for example. Can they consider themselves to be sharing a common "reference frame"?

And what's a "reference frame"? In SR it's usually used synonymously with "inertial frame",

It is used inaccurately then. The general term "reference frame" includes "non-inertial reference frames" as well. Rotating frames for example.

"Reference frame" doesn't seem to be a well-defined concept to me.

That is interesting, given the wide use of the term in physics.

 

[Demystifier]

Assume that for example. Can they consider themselves to be sharing a common "reference frame"?

There is a general way to associate a proper reference frame to a local observer moving in a specified way. (See Misner, Thorne, Wheeler, Sec. 13.6)
In particular, observers (static with respect to a Schwarzschild black hole) who sit at DIFFERENT points do not share a common proper reference frame.

 

[A.T.]

Assume that for example. Can they consider themselves to be sharing a common "reference frame"?

There is a general way to associate a proper reference frame to a local observer moving in a specified way. (See Misner, Thorne, Wheeler, Sec. 13.6)
In particular, observers (static with respect to a Schwarzschild black hole) who sit at DIFFERENT points do not share a common proper reference frame.

By "proper reference frame" do you mean "local inertial frame for freely falling and non-rotating observer"? I'm asking about a "reference frame" in general, not specifically an inertial one. The one defined by the Schwarzschild coordinates seems to be widely applied and useful. Is it a "reference frame"?

The concept of "rotating reference frames" seems to be accepted and used in physics as well. Inertial forces were introduced to make them handleable, and Relativity added some extra issues to them. But a rotating reference frame is still a "reference frames" isn't it?

 

[Demystifier]

By "proper reference frame" do you mean "local inertial frame for freely falling and non-rotating observer"?

No, I mean local noninertial frame for arbitrarily moving and rotating observer.

The one defined by the Schwarzschild coordinates seems to be widely applied and useful. Is it a "reference frame"?

It is a reference frame, but not a proper reference frame associated with a local observer.

But a rotating reference frame is still a "reference frames" isn't it?

Yes it is. Moreover, it is the proper reference frame associated with a local observer who, e.g., sits in the center of a rotating disc.

 

[Fredrik]

So a "reference frame" is a synonym for "coordinate system", i.e. a function from an open subset of the spacetime manifold into \mathbb R^4, and a "proper reference frame" is the non-inertial coordinate system that's associated with the motion and orientation of an object, in the most natural way. (The orientation of the object defines an orthonormal basis of the tangent space at each point on the world line. One of the basis vectors is equal to the tangent vector of the curve. The time axis of the coordinate system is the world line, labeled by proper time. A hypersurface of constant time consists of geodesics through a point on the curve, that are orthogonal to the curve at the point where they intersect it. The x,y,z axes are defined as the geodesics in the t=0 hypersurface that have the three remaining basis vectors as tangent vectors. They are labeled by the usual synchronization procedure).

I'm OK with that terminology. No real surprises there.

Note that this coordinate system will not extend very far from the world line. It's only well-defined in a region where the spacelike geodesics I just talked about don't intersect. Also note that "a bunch of measuring devices spread out over the ring/disc" don't define a proper reference frame according to this definition.

 

[Demystifier]

Fredrik, that's all correct.
(Provided that an inertial coordinate system is viewed as a special case of non-inertial coordinate systems.)

 

[Fredrik]

Yes, that's how I meant it. I wrote "non-inertial" mainly because I had previously said that a local inertial frame is the natural coordinate system to associate with a physical observer's motion and orientation, and I wanted to emphasize that I wasn't still talking about that. I guess I did it in a confusing way.

(Yes, I realize that this "proper reference frame" is at least as natural as the local inertial frame).

 

[Demystifier]

I'm glad that I allways eventually get to an agreement with you. :-)

 

[yuiop]

The ring completely fills the gutter, so one observer should see that the two circumferences are equal. That's why I disagree with you.

This is far too simplistic. This is like saying that because the observer at rest with the barn in the "pole and barn paradox" sees the pole completely filling the barn, that the observer on the pole should also conclude that the pole and barn are the same length. In fact the pole rider measures the barn to be shorter than the pole and he reasons that he passes safely through the barn because the front and back doors do not shut simultaneously from his point of view.

In the case of the ring and the gutter, the observer at rest with the gutter does indeed measure the ring and gutter to have the same circumference (analogous to the barn observer) and the observer on the ring measures the gutter to have a smaller circumference than the ring (analogous to the observer on the pole).

One other (not directly related) observation that may be of interest. If the observer in the gutter measures the velocity of the ring to be v relative to the gutter, the observer on the ring will not measure the velocity of the gutter relative to the ring to be exactly v, if he uses transitive (synchronised by a central clock) rather than Einstein clock synchronisation.

 

[heldervelez]

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I will take back the mention to the paper because when I checked again the publication, I realized that I've messed up with other paper with similar title in that Journal (the title also started by NONINVARIANT...and is from Bari Univ." .
My apologies to everyone.
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