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The rotating disk of Albert Einstein

  1. Feb 24, 2014 #1
    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:


    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.
     
  2. jcsd
  3. Feb 24, 2014 #2
    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.
     
  4. Feb 24, 2014 #3

    WannabeNewton

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    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.
     
  5. Feb 24, 2014 #4

    A.T.

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    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.
     
  6. Feb 24, 2014 #5

    WannabeNewton

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    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).
     

    Attached Files:

  7. Feb 24, 2014 #6

    A.T.

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    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?
     
  8. Feb 24, 2014 #7

    WannabeNewton

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    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.
     
  9. Feb 24, 2014 #8

    pervect

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    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.
     
    Last edited: Feb 24, 2014
  10. Feb 24, 2014 #9

    WannabeNewton

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    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.

    http://areeweb.polito.it/ricerca/relgrav/solciclos/gron_d.pdf A very instructive read indeed.
     
  11. Feb 24, 2014 #10

    A.T.

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    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?
     
  12. Feb 24, 2014 #11

    WannabeNewton

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    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.
     
  13. Feb 24, 2014 #12

    DrGreg

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    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.
     
  14. Feb 24, 2014 #13

    WannabeNewton

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

    But the geometry that we describe using a metric tensor is local in the sense that it lets 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.
     
  15. Feb 24, 2014 #14

    WannabeNewton

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    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 :smile:
     
  16. Feb 25, 2014 #15
    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.
     
  17. Feb 25, 2014 #16
    # A mathematical formulation?

    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 spacial 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 ?

    # 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.
     
  18. Feb 25, 2014 #17

    A.T.

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    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?
     
  19. Feb 25, 2014 #18

    WannabeNewton

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    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##.

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

    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.
     
    Last edited: Feb 25, 2014
  20. Feb 25, 2014 #19
    Thank you WannabeNewton. Please read the text of Albert Einstein in post #1.

    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:
    Einstein said:
    Sorry, but I do not play with words.

    You said:
    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.
     
  21. Feb 25, 2014 #20

    Dale

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    What is a "proper geometry"?
     
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