The rotating disk of Albert Einstein

[Nana Dutchou]

Hello

To demonstrate that we can deduce from special relativity the existence of an observer who finds that his three-dimensional space is not euclidean (a conclusion that is false in my opinion), in subsection "Behaviour of Clocks and Measuring Rods on a Rotating Body of Reference" of his book "Relativity: The Special and the General Theory", Albert Einstein wrote:

Hitherto I have purposely refrained from speaking about the physical interpretation of space- and time-data in the case of the general theory of relativity. As a consequence, I am guilty of a certain slovenliness of treatment, which, as we know from the special theory of relativity, is far from being unimportant and pardonable. It is now high time that we remedy this defect; but I would mention at the outset, that this matter lays no small claims on the patience and on the power of abstraction of the reader.
We start off again from quite special cases, which we have frequently used before. Let us consider a space-time domain in which no gravitational fields exists relative to a reference-body K whose state of motion has been suitably chosen. K is then a Galileian reference-body as regards the domain considered, and the results of the special theory of relativity hold relative to K. Let us suppose the same domain referred to a second body of reference K', which is rotating uniformly with respect to K. In order to fix our ideas, we shall imagine K' to be in the form of a plane circular disc, which rotates uniformly in its own plane about its centre. An observer who is sitting eccentrically on the disc K' is sensible of a force which acts outwards in a radial direction, and which would be interpreted as an effect of inertia (centrifugal force) by an observer who was at rest with respect to the original reference-body K. But the observer on the disc may regard his disc as a reference-body which is “at rest”; on the basis of the general principle of relativity he is justified in doing this. The force acting on himself, and in fact on all other bodies which are at rest relative to the disc, he regards as the effect of a gravitational field. Nevertheless, the space-distribution of this gravitational field is of a kind that would not be possible on Newton’s theory of gravitation. 1 But since the observer believes in the general theory of relativity, this does not disturb him; he is quite in the right when he believes that a general law of gravitation can be formulated—a law which not only explains the motion of the stars correctly, but also the field of force experienced by himself.
The observer performs experiments on his circular disc with clocks and measuring-rods. In doing so, it is his intention to arrive at exact definitions for the signification of time- and space-data with reference to the circular disc K', these definitions being based on his observations. What will be his experience in this enterprise?
To start with, he places one of two identically constructed clocks at the centre of the circular disc, and the other on the edge of the disc, so that they are at rest relative to it. We now ask ourselves whether both clocks go at the same rate from the standpoint of the non-rotating Galileian reference-body K. As judged from this body, the clock at the centre of the disc has no velocity, whereas the clock at the edge of the disc is in motion relative to K in consequence of the rotation. According to a result obtained in Section XII, it follows that the latter clock goes at a rate permanently slower than that of the clock at the centre of the circular disc, i.e. as observed from K. It is obvious that the same effect would be noted by an observer whom we will imagine sitting alongside his clock at the centre of the circular disc. Thus on our circular disc, or, to make the case more general, in every gravitational field, a clock will go more quickly or less quickly, according to the position in which the clock is situated (at rest). For this reason it is not possible to obtain a reasonable definition of time with the aid of clocks which are arranged at rest with respect to the body of reference. A similar difficulty presents itself when we attempt to apply our earlier definition of simultaneously in such a case, but I do not wish to go any farther into this question.
Moreover, at this stage the definition of the space co-ordinates also presents unsurmountable difficulties. If the observer applies his standard measuring-rod (a rod which is short as compared with the radius of the disc) tangentially to the edge of the disc, then, as judged from the Galileian system, the length of this rod will be less than 1, since, according to Section XII, moving bodies suffer a shortening in the direction of the motion. On the other hand, the measuring-rod will not experience a shortening in length, as judged from K, if it is applied to the disc in the direction of the radius. If, then, the observer first measures the circumference of the disc with his measuring-rod and then the diameter of the disc, on dividing the one by the other, he will not obtain as quotient the familiar number pi= 3.14 …, but a larger number, 2 whereas of course, for a disc which is at rest with respect to K, this operation would yield pi exactly. This proves that the propositions of Euclidean geometry cannot hold exactly on the rotating disc, nor in general in a gravitational field, at least if we attribute the length 1 to the rod in all positions and in every orientation. Hence the idea of a straight line also loses its meaning. We are therefore not in a position to define exactly the co-ordinates x,y,z relative to the disc by means of the method used in discussing the special theory, and as long as the co-ordinates and times of events have not been defined we cannot assign an exact meaning to the natural laws in which these occur.
Thus all our previous conclusions based on general relativity would appear to be called in question. In reality we must make a subtle detour in order to be able to apply the postulate of general relativity exactly. I shall prepare the reader for this in the following paragraphs.

Thus, using the assumption that a body D is described by an inertial reference R as a rigid disc in uniform rotation about an axis perpendicular to the disc plane and passing through its center, it should be concluded :
(i) There exists an observer of D who can state that "D is actually a rigid disk".
(ii) The observer of D notices that the relationship which connects the circumference and diameter of D is not that of euclidean spaces in other words the ratio between these two quantities is not the number pi.


We will show that (i) is questionable. Indeed, consider two material points which are fixed on D and such that one is on the center and the other on his circumference. Then:
(a) By assumption, since D is a rigid body according to R, this inertial frame can assert that the spatial distance between these two points does not vary in time.
(b) The Lorentz transformation allows to state that there is at least one inertial frame R' who can say that the spatial distance between these two points varies in time in other words D is a deformable body according to R'.


Because R assert that D is an indeformable body and because R' asserts the opposite, knowing that all inertial frames are physically equivalent, it is impossible to express that there is an observer of D who notices that D is actually an indeformable body. Thus, (i) is questionable and the conclusion of Albert Einstein is wrong.


To demonstrate (b) it is sufficient to choose R' as an inertial reference frame whose velocity vector v (with respect to R) is in the plane of D and is therefore orthogonal to the axis of rotation of D. Under these conditions, the transformation of Lorentz teaches that the contraction of the lengthes enters R and R' is maximal when the radius vector between both material points is colinear to v and this contraction of the lengthes enters R and R' is worthless when the radius vector between both material points is orthogonal to v. Finally, we know that the radius vector between the two material points travel alternately these two configurations because D is rotating.

Cordially.

 

[Nana Dutchou]

If he had deeply noticed that even in classic kinematics we can easily build a coherent family of world lines that are not a set of fixed points according to a unique observer, he would have been able to deduct from his analysis that in a relativist framework :

- The family of trajectories [described with respect to an inertial coordinate system by equations that highlight the classical notion of rotational motion] does not constitute a set of fixed points according to a unique observer.

- It is therefore not surprising that we have difficulty in conceiving that regular digital clocks having these trajectories are synchronisables in the sense of the special relativity.

- It is necessary to reinvent the complexity of the equations which have to describe, with respect to an inertial coordinate system, a set of points continuously fixed with regard with respect to an accelerated experimenter. Do not PLAGIARISM equations of classical kinematics.

- The geometry of the three-dimensional space of an accelerated observer can remain euclidian if the fixed points which constitute this three-dimensional space are described (with respect to an inertial coordinate system) by the new complex equations.

What do you think ?
I can put a little time to respond to your answers.

Cordially.

 

[WannabeNewton]

(a) By assumption, since D is a rigid body according to R, this inertial frame can assert that the spatial distance between these two points does not vary in time.
(b) The Lorentz transformation allows to state that there is at least one inertial frame R' who can say that the spatial distance between these two points varies in time in other words D is a deformable body according to R'.

If you're going to attempt to prove Einstein wrong then please, please learn what "rigidity" of the disk means first in relativity. The disk is rigid if spatial distances between local points are constant in their instantaneous rest frames. This is called Born rigidity and is the sense in which we say the disk is rigid. It's trivial to show that the uniformly rotating disk is in fact Born rigid. This makes your entire "argument" moot.

It is also very easy to show that the local spatial geometry of the disk is non-euclidean relative to the local rest spaces of observers atop the disk. It requires 3 to maybe 4 steps of algebra at best. However it's important to note that this is only the local geometry. Nothing can be said of the global geometry from this specific local geometry because the observers atop the disk cannot Einstein synchronize their clocks.

I would suggest you read the profuse of literature first before attempting to disprove well-established properties of the rigidly rotating disk.

 

[A.T.]

Nothing can be said of the global geometry from this specific local geometry because the observers atop the disk cannot Einstein synchronize their clocks.

You can measure the spatial geometry of the disc with rulers, without clocks and their synchronization issues. Rulers laid out on the rotating disc will measure a non-Euclidean ratio between radius and circumference.

 

[WannabeNewton]

You can measure the spatial geometry of the disc with rulers, without clocks and their synchronization issues. Rulers laid out on the rotating disc will measure a non-Euclidean ratio between radius and circumference.

That's local geometry not global geometry so it doesn't change a word of what I said. It's equivalent to saying that the constant curvature Riemannian quotient manifold obtained by identifying points on the same orbits of the Killing field $\xi = \partial_t$ in the rest frame of the rotating disk is hyperbolic. When we speak of metric geometry it is always local. Clearly this local geometry cannot be foliated into a global geometry because $\xi$ fails to satisfy the Frobenius condition for integrability i.e. $\xi^{\flat}\wedge d\xi^{\flat} \neq 0$ which is equivalent to the inability to extend local Einstein synchronization to a global synchronization relative to $\xi$.

Read the attachment (taken from Gron's lecture notes on GR).

 

[A.T.]

That's local geometry not global geometry

So if I measure the total diameter and total circumference of the disc, using rulers placed on it, then it is still a "local" measure, even if it encompasses the entire disc?

 

[WannabeNewton]

So if I measure the total diameter and total circumference of the disc, using rulers placed on it, then it is still a "local" measure, even if it encompasses the entire disc?

Yes because it depends only on the Riemannian metric in the local rest spaces of observers on the rotating disk; placing infinitesimal tangential rulers along the circumference of the disk and adding up their lengths and comparing the result to the added up lengths of infinitesimal radial rulers placed from the center to the circumference is equivalent to integrating the Riemannian metric of the quotient manifold along appropriate curves in order to obtain the relevant metrical measurements. In order to have any kind of global spatial geometry relative to $\xi$ for the disk obtained from the local rest spaces you would need to be able to define global simultaneity slices everywhere orthogonal to $\xi$ which is impossible because in the local rest spaces we have $\vec{\nabla}\times \vec{\xi} \neq 0$. If you haven't already then read the attachment it will clear things up hopefully.

 

[pervect]

The approach used for measuring circumference is important. The standard approach for measuring it boils down to something equivalent to this procedure:

Pick a standard ruler of length L. Place as many of them around the circumference of the disk as you can, so that each end of the ruler is on the edge spinning disk, and the rulers do not overlap. Call the number of rulers you have placed N. Take the limit of L*N as L gets smaller and smaller, i.e. the limit of L*N as L goes to zero.

We need to define a "standard ruler". It is a pair of worldlines, that maintain a constant distance. In the limit as L goes to zero (which is all we need to define), we can define this constant distance, the length of the ruler, via the exchange of two-way radar signals between the ends, and the constancy of the speed of light.

There is a long history of people using different procedures to determine the circumference in the literature, one of the standard references is O. Gron, American Jouranl of Physics Vol. 43 No. 10 pg 869 (1975), which goes into some history of the different things that have been computed and called "the circumference" of the disk. IT may be hard to find this paper on the internet - other papers by Gron can be found more easily, but won't necessarily cover the history.

 

[WannabeNewton]

We need to define a "standard ruler". It is a pair of worldlines, that maintain a constant distance. In the limit as L goes to zero (which is all we need to define), we can define this constant distance, the length of the ruler, via the exchange of two-way radar signals between the ends, and the constancy of the speed of light.

Exactly. And this is precisely why what we're measuring is the non-euclidean (local) metric geometry of the rotating disk relative to the instantaneous comoving inertial frames of points of the disk. In other words we're defining lengths of infinitesimal rulers using local Einstein (radar) simultaneity and using this to show that the curvature of the resulting metric is a constant negative value (hyperbolic). This is not the same thing as the global spatial geometry of the rotating disk obtained from a hypersurface orthogonal space-like foliation relative to the tangent field (in space-time) of the disk. Such a foliation is impossible to define for reasons already stated.

IT may be hard to find this paper - other papers by Gron can be found, but won't necessarily cover the history.

http://areeweb.polito.it/ricerca/relgrav/solciclos/gron_d.pdf A very instructive read indeed.

 

[A.T.]

In order to have any kind of global spatial geometry relative to $\xi$ for the disk obtained from the local rest spaces you would need to be able to define global simultaneity slices everywhere orthogonal to $\xi$

Why is simultaneity relevant to measuring a static spatial geometry? In the rotating frame the disc and rulers are static. Nothing changes over time about the geometry that they measure. So why care about time and simultaneity at all?

 

[WannabeNewton]

Did you read the attachment at all? I don't know how many more times to repeat it. There's a huge difference between local spatial geometry and global spatial geometry in relativity whenever we talk about extended bodies. Let the extended body be described by a time-like vector field $\eta$ in space-time and let $\xi$ be the tangent field to a family of observers. Then relative to $\xi$ we can define a spatial metric $h_{\mu\nu} = g_{\mu\nu} + \xi_{\mu}\xi_{\nu}$ where $g_{\mu\nu}$ is the space-time metric. This $h_{\mu\nu}$ describes the local spatial geometry of the disk relative to the observers following orbits of $\xi$. On the other hand if we want to describe the global spatial geometry of the disk relative to $\xi$ then we need to be able to define a one-parameter family of global simultaneity slices $\Sigma_t$ orthogonal to $\xi$ everywhere. Each $\Sigma_t$ intersects $\eta$, that is to say the entire world-tube of the disk, in some fashion and allows us to talk about the global spatial geometry of the disk i.e. the entire spatial shape of the disk relative to $\xi$.

If $\xi$ represents the inertial observers at rest with respect to the axis observer then we can define such simultaneity slices in the usual way. On each such slice the disk just takes the form of...well...a flat disk. But if $\xi$ represents the observers at rest on the disk, meaning $\xi = \eta$, then you clearly cannot define such simultaneity slices and the best you can do is probe the local spatial geometry of the disk using local rulers as defined by $h_{\mu\nu}$ i.e. you cannot define the entire spatial shape of the disk relative to these observers.

 

[DrGreg]

WannabeNewton, I understand the difference between what you are calling "local spatial geometry" and "global spatial geometry", but I've not seen them described with those phrases before. Is that standard terminology? It seems to me that what you calling the "local spatial geometry" is actually global in extent: you can use it to define a (non-Euclidean) global metric on a 3D quotient manifold, the "points" of which correspond to worldlines in 4D spacetime. This is a perfectly valid 3D manifold that we could call the "space" for the rotating disk, but it's not a surface of simultaneity within the 4D spacetime manifold.

I think this is the point A.T. was getting at.

The quotient manifold construction is discussed in Rindler's Relativity: Special, General and Cosmological 2nd Ed sections 9.6 & 9.7. It applies to any stationary but non-static congruence. Also in the Wikipedia article Stationary spacetime.

 

[WannabeNewton]

WannabeNewton, I understand the difference between what you are calling "local spatial geometry" and "global spatial geometry", but I've not seen them described with those phrases before. Is that standard terminology?

I don't know whether it is standard but I was simply paraphrasing Gron's terminology in the attachment (see post #5).

It seems to me that what you calling the "local spatial geometry" is actually global in extent: you can use it to define a (non-Euclidean) global metric on a 3D quotient manifold, the "points" of which correspond to worldlines in 4D spacetime. This is a perfectly valid 3D manifold that we could call the "space" for the rotating disk, but it's not a surface of simultaneity within the 4D spacetime manifold.

But the geometry that we describe using a metric tensor is local in the sense that it let's us compute geometric properties of the manifold such as curvature point-wise and, in the sense elucidated by Gron, in the local rest space of each inertial frame instantaneously comoving with a point on the disk. Sure the metric tensor is defined everywhere on the quotient manifold but that doesn't change the fact that it describes local geometry-this is true in general of any metric tensor on any space-time or space. The metric doesn't tell us what the overall geometric spatial shape of the disk actually is-which is something handled rather by the simultaneity surfaces.

 

[WannabeNewton]

The quotient manifold construction is discussed in Rindler's Relativity: Special, General and Cosmological 2nd Ed sections 9.6 & 9.7.

As an aside, thank you for the wonderful reference. I never even knew Rindler discussed the matters of rigidly rotating coordinates in stationary space-times and vorticity of stationary congruences as local rotation relative to a compass of inertia.

I unfortunately never went beyond the special relativity parts of Rindler's text. This certainly would have been a great reference to have had for some of the previous threads on rotation

 

[TrickyDicky]

Why is simultaneity relevant to measuring a static spatial geometry? In the rotating frame the disc and rulers are static. Nothing changes over time about the geometry that they measure. So why care about time and simultaneity at all?

But that's the key here, as WN has been saying it is not static, but stationary, in the differential geometry meaning. That's what makes impossible to talk about the global spatial geometry of the disk as hyperbolic, because we can't cleanly separate a spatial hypersurface.

 

[Nana Dutchou]

If you're going to attempt to prove Einstein wrong then please, please learn what "rigidity" of the disk means first in relativity. The disk is rigid if spatial distances between local points are constant in their instantaneous rest frames.

# A mathematical formulation?

It is also very easy to show that the local spatial geometry of the disk is non-euclidean relative to the local rest spaces of observers atop the disk.

For me, if E is a set of points (a mathematical set), a curve segment on E can be represented by a function that is defined on an open subset of the real line and takes values ​​in E.

By definition, to define a geometry on E we have to allocate a numerical value (a length) to each of its segments of curves.

In spatial relativity each inertial coordinate system possesses a unique three-dimensional space which is a very particular family of world lines (each point of this space is a world line) and there is a favored euclidian geometry on this space.

# What is a local geometry on E ?

you cannot define the entire spatial shape of the disk relative to these observers.

To demonstrate that we can deduce from special relativity the existence of an observer who finds that his three-dimensional space is not euclidean (a conclusion that is false in my opinion), in subsection "Behaviour of Clocks and Measuring Rods on a Rotating Body of Reference" of his book "Relativity: The Special and the General Theory", Albert Einstein wrote:

# If we cannot define the global shape of the disk with respect to the accelerated observer who is placed in its center, why try to verify the accuracy or the inaccuracy of the relation circumference = diameter * pi? How can we say that this relation is false?

Cordially.

 

[A.T.]

The metric doesn't tell us what the overall geometric spatial shape of the disk actually is

Doesn't this depend on how you define "actually is"? What is wrong with assuming that rulers laid out at rest on the disc tell us what the overall geometric spatial shape of the disk actually is?

 

[WannabeNewton]

# A mathematical formulation?

Let the tangent field (in space-time) of the disk be $\eta$ as above and let $h = g + \eta^{\flat}\otimes \eta^{\flat}$ where $g$ is the space-time metric and $\eta^{\flat}$ is the musical isomorphism. Intuitively the spatial metric $h$ represents the spatial distances between neighboring points on the disk in their instantaneous rest frames. We say the disk is Born-rigid if $\mathcal{L}_{\eta}h = 0$ where $\mathcal{L}_{\eta}$ is the Lie derivative along the flow generated by $\eta$.

Now it's easy to see why the disk is Born rigid. Go to the coordinates corotating with the disk so that $\eta = \partial_t$ and $g = -(1 - \omega^2 r^2)dt\otimes dt + \omega r^2 dt\otimes d\phi + \omega r^2 d\phi \otimes dt + dr\otimes dr + r^2 d\phi \otimes d\phi$. Then clearly $\mathcal{L}_{\eta}h = 0$ because $\eta$ only generates time translation flows whereas $h$ is independent of time translation flows.

In fact we can show that the above condition of Born-rigidity is equivalent to $\eta$ having a vanishing expansion (deformation-rate) tensor $\theta_{\mu\nu} = h^{\alpha}{}{}_{\mu}h^{\beta}{}{}_{\nu}\nabla_{(\alpha}\eta_{\beta)} = 0$. In our case this is trivial because $\eta$ is a Killing field i.e. $\nabla_{(\mu}\eta_{\nu)} = 0$.

# What is a local geometry on E ?

See the attachment in post #5 from Gron's GR lecture notes.

# If we cannot define the global shape of the disk with respect to the accelerated observer who is placed in its center, why try to verify the accuracy or the inaccuracy of the relation circumference = diameter * pi? How can we say that this relation is false?

Well for starters the observer at the center is not accelerated. This observer has vanishing proper acceleration; the observer can be described using either a rotating or non-rotating frame relative to coincident gyroscopes but that doesn't change the fact that the observer is non-accelerating. Secondly, relative to this observer the global shape of the disk is well-defined and is simply the flat circular shape you imagine when you think of a disk. Lastly, there is nothing inaccurate or accurate involved here-we simply have different spatial metric geometries relative to different families of observers.

The global shape of the disk cannot be defined using orthogonal simultaneity slices relative the family of observers at rest on the rigidly rotating disk, simple as that. As TrickyDicky noted this is a straightforward consequence of having stationary but non-static geometry of the gravitational field in which the disk is at rest. This in itself is a consequence of the easily verified fact that $\eta_{[\gamma}\nabla_{\mu}\eta_{\nu]} \neq 0$ i.e. the rigid coordinate lattice corotating with the disk has, at each point of the lattice, an angular velocity relative to gyroscopes fastened to that point. We can do it through other means but amounts to synchronization conventions other than Einstein synchronization.

But there's no problem defining the global shape of the disk using orthogonal simultaneity slices relative to the family of inertial observers at rest with respect to the axis (central) observer. As already stated, this gives us back the usual circular disk shape.

 

[Nana Dutchou]

Thank you WannabeNewton. Please read the text of Albert Einstein in post #1.

the observer can be described using either a rotating or non-rotating frame relative to coincident gyroscopes

The observer performs experiments on his circular disc with clocks and measuring-rods. In doing so, it is his intention to arrive at exact definitions for the signification of time- and space-data with reference to the circular disc K', these definitions being based on his observations. What will be his experience in this enterprise?

Consider the observer at the center of the disc. Is that his observations are the same when rotated around itself and when it does not revolve around himself?
Is that his observations are the same when the disc is rotating and when the drive is not rotaton?

I wrote: why try to verify the accuracy or the inaccuracy of the relation circumference = diameter * pi ?.
You said:

Lastly, there is nothing inaccurate or accurate involved here-we simply have different spatial metric geometries relative to different families of observers.

Einstein said:

If, then, the observer first measures the circumference of the disc with his measuring-rod and then the diameter of the disc, on dividing the one by the other, he will not obtain as quotient the familiar number pi= 3.14 …, but a larger number, 2 whereas of course, for a disc which is at rest with respect to K, this operation would yield pi exactly.

Sorry, but I do not play with words.

You said:

we simply have different spatial metric geometries relative to different families of observers.

We know that the world lines of a family of observers constantly at rest in an inertial coordinate system represent a three-dimensional mathematical space E on which we can define an euclidean geometry which is the proper geometry of this space. To define a geometry on E we have to allocate a numerical value (a length) to each of its segments of curves. Mathematically we can define several euclidean geometries on E. We can also define non-euclidean geometries on E! It's easy! But E has only one proper geometry.

How does one do to demonstrate that the proper geometry of the rotating disk is not euclidean? In other words, how does one do to demonstrate that trimensionnel space consisting of world lines of observers constantly at rest on the rotating disk is not an euclidean space?

We must understand that we are only interested in determining the observations of an observer who is accelerated relative to an inertial coordinate system. If you know how to determine his observations, then we can mathematically formulate the equivalence principle.

I do not play with words.

Cordially.

 

[Dale]

ThBut E has only one proper geometry.

What is a "proper geometry"?

 

[A.T.]

What is a "proper geometry"?

To me, the "proper geometry" of the disc is the physically relevant geometry. If you wanted to build the rotating disc, and wanted to know how much material you will need, you would have to consider its proper geometry. The amount of material needed is a physical fact. It doesn't depend on clock synchronization, surfaces of simultaneity or other conventions that have no physical consequences.

 

[Dale]

I have never heard the term "proper geometry" before. In keeping with the usual meanings of proper time and proper length, I would guess that it is the geometry in the object's rest frame. So I think that may be related to what you are saying A.T.

However, one thing that I see as problematic in your "amount of material" bit is that the amount of material depends on the strain that the structure is under.

 

[WannabeNewton]

Consider the observer at the center of the disc. Is that his observations are the same when rotated around itself and when it does not revolve around himself?

They will differ as far as centrifugal and coriolis forces go but both cases correspond to non-accelerating observers. Only the non-rotating case will correspond to an inertial observer of course.

Is that his observations are the same when the disc is rotating and when the drive is not rotaton?

By "drive" I assume you mean disc? And no not all the observations will be the same for the observer at the center of the disk if the disk is rotating as opposed to non-rotating-some will and some won't. The observer can measure in an absolute sense whether or not the disk is rotating-this is simply given by $\eta_{[\gamma}\nabla_{\mu}\eta_{\nu]}$ as remarked earlier. What will not differentiate itself for this observer is the spatial metric geometry of the disk-the rigidly rotating case and the non-rotating case both yield Euclidean geometries for the central observer.

I wrote: why try to verify the accuracy or the inaccuracy of the relation circumference = diameter * pi ?.

It is not inaccurate. The spatial metric geometry of the disk relative to the inertial observers obeys $C = \pi D$ whereas the spatial metric geometry of the disk relative to observers at rest on the disk obeys $C > \pi D$ when the disk is rotating. Neither is any more correct than the other because spatial metric geometry is relative.

Sorry, but I do not play with words.

Oh but you are.

How does one do to demonstrate that the proper geometry of the rotating disk is not euclidean?

There is no such thing as "proper" geometry. See above.

As a bit of friendly advice I would suggest you learn more relativity and math before attempting a more detailed analysis of the rotating disk. Then you can read the plethora of literature on it and then start threads on why you think there are issues (there aren't of course but that's a different story).

 

[PAllen]

To me, the "proper geometry" of the disc is the physically relevant geometry. If you wanted to build the rotating disc, and wanted to know how much material you will need, you would have to consider its proper geometry. The amount of material needed is a physical fact. It doesn't depend on clock synchronization, surfaces of simultaneity or other conventions that have no physical consequences.

But this does show the various ambiguities:

- do you construct it in the inertial frame and spin it up?
- do you build a little platform circling around a center, and build the disk out from their?
- do you start with a very small spinning disk and enlarge it radially (pulling a spaghetti strand down the axis and spinning and smushing it out from the center).

These are nonequivalent procedures, leading to different answers. Each of these procedures corresponds to a different simultaneity approach.

 

[WannabeNewton]

What is wrong with assuming that rulers laid out at rest on the disc tell us what the overall geometric spatial shape of the disk actually is?

Because it's a false assumption unless you want to play semantics and come up with a self-serving definition of "geometric spatial shape". This is basic differential geometry so I really don't want to keep repeating myself. The metric tensor is a characterization of the local geometry of a manifold. The overall (global) geometric spatial shape of an extended object relative to a family of observers doesn't even have meaning if you can't define "space at an instant of time" relative to this family. "Space at an instant of time" is purely conventional based upon simultaneity or synchronization. As already stated one cannot define "space at an instant of time" relative to the family of observers at rest on the disk if we want these "spaces" to be everywhere orthogonal to the tangent field of the family (propagation of local Einstein simultaneity throughout the family).

We can define "space at an instant of time" relative to this family as being the same as "space at an instant of time" relative to the inertial observers and we get back the flat circular shape for the disk as its overall spatial shape but this is clearly conventional and corresponds to synchronizing the standard clocks in the rest frame of the disk with the central clock i.e. we replace the standard clocks with the coordinate clocks in this rest frame following orbits of $\nabla^{\mu}t$. This is distinct from the fact that the spatial metric relative to the family of observers atop the disk on the quotient manifold obtained by identifying points on single orbits of the tangent field to this family (the stationary Killing field) yields a hyperbolic geometry-this is derived entirely from local radar simultaneity and does not care about global simultaneity conventions.

 

[A.T.]

However, one thing that I see as problematic in your "amount of material" bit is that the amount of material depends on the strain that the structure is under.

Once you have build your rotating disc, you want to paint it. The amount of paint is determined by the proper geometry of the disc.

 

[A.T.]

unless you want to play semantics and come up with a self-serving definition of "geometric spatial shape".

I want a useful definition of "geometric spatial shape".

The overall (global) geometric spatial shape of an extended object relative to a family of observers doesn't even have meaning if you can't define "space at an instant of time" relative to this family.

So defining "geometric spatial shape" based on "space at an instant of time" is not useful for a rotating disc.

 

[PAllen]

So defining "geometric spatial shape" based on "space at an instant of time" is not useful for a rotating disc.

Correct. There is space at an instant of time for a particular rim observer (if they don't push it too far). There is space at an instant of time for a central observer. But there is no collective instant of time that has any relevance to all disk observers.

 

[WannabeNewton]

I want a useful definition of "geometric spatial shape".

The usual definition in relativity is the set of intersection points obtained by slicing the world-tube of the object by a space-like hypersurface.

So defining "geometric spatial shape" based on "space at an instant of time" is not useful for a rotating disc.

The definition is completely general just like the definition of Einstein synchronization is completely general so it's not being defined on a case by case basis. However the main point is if we want the space-like hypersurfaces to be orthogonal to the world-tube of the disk then it's impossible to achieve this because the striations of the cylindrical world-tube are twisting around one another. In other words this is a special case wherein the general definition cannot be applied, just like the general definition of Einstein synchronization cannot be applied to certain special cases-that's all there is to it. So yes in that sense it is not useful simply because it cannot be applied. I hope that helps clear it up.

 

[Dale]

Once you have build your rotating disc, you want to paint it. The amount of paint is determined by the proper geometry of the disc.

But again, that depends on the strain of the disk. If the disk is in tensile strain then it will require more paint than if not.

 

[A.T.]

But again, that depends on the strain of the disk. If the disk is in tensile strain then it will require more paint than if not.

You know the radius of the disc and you want to know its proper area, to buy enough paint. To get the proper area you have to use the hyperbolic geometry. I don't see why the strains would matter here.

 

[pervect]

I want a useful definition of "geometric spatial shape".


So defining "geometric spatial shape" based on "space at an instant of time" is not useful for a rotating disc.

So are you defining your spatial geometry as a quotient manifold, or what? My understanding of the quotient manifold idea is that it maps a worldline on the disk to a point in an abstract space, the "quotient space".

 

[Dale]

You know the radius of the disc and you want to know its proper area, to buy enough paint. To get the proper area you have to use the hyperbolic geometry. I don't see why the strains would matter here.

If you have two identically constructed disks and one is unstrained but the other is in tensile strain then the one that is in tensile strain will require more paint than the other. Also, the paint itself can be under strain or not.

You are defining the proper geometry in terms of the amount of material (paint or disk material) needed to cover a given area or volume. That amount depends not only on the geometry but also the strain of the material. I don't see how you can think strains don't matter given how you are trying to define geometry.

 

[A.T.]

If you have two identically constructed disks and one is unstrained but the other is in tensile strain

If they are identically constructed spinning at the same rate, the strains will be the same.

 

[Dale]

If they are identically constructed spinning at the same rate, the strains will be the same.

Yes, but what if they are not spinning at the same rate? The same amount of material will then cover different geometries under different strains. According to your definition, two identically constructed disks will have the same proper geometry, regardless of their state of strain.

This is clearly not what is intended. It defines a fresh and a chewed piece of bubble gum as having the same proper geometry.