Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

BRS: degenerate cases of Born-rigidity

  1. Sep 12, 2010 #1

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    There has been yet another thread on the Ehrenfest paradox, and one of the issues in that thread has been the question of whether a one-dimensional ruler can be subjected to Born-rigid angular acceleration. CH has suggested that I might want to explain my view on that here. We've PM'd back and forth a few times, which has been helpful in showing me what I need to clarify about my presentation.

    I believe that some of the crystallized knowledge that GR experts have about Born rigidity is inaccurate in certain degenerate cases. Baez's page on the rigid disk, http://math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html , quotes [Pauli 1958] as follows: "It was further proved, independently, by Herglotz and Noether that a rigid body in the Born sense has only three degrees of freedom... Apart from exceptional cases, the motion of the body is completely determined when the motion of a single of its points is prescribed." (Pauli's statement clearly assumes that we're talking about the body once it's already been constructed. That is, if it's constructed in a certain state of rotation, then specifying the motion of one of its points suffices to describe its future motion.) Note the qualifier, "Apart from exceptional cases," which is omitted, e.g., in Rindler's Essential Relativity.

    As an example that I think is one of these exceptional cases, consider a rod which, in the nonrotating laboratory frame, has a fixed length L, one end fixed at the origin, and an orientation [itex]\phi(t)[/itex], where L, [itex]\phi[/itex], and t are all measured in the lab-frame's Minkowski coordinates (a global coordinate chart). The rod is not pasted onto a disk or anything like that; the rod is all there is, and it's a one-dimensional rod. Trusting Pauli to be accurate, we have two possibilities if we want the rod to be Born-rigid: (1) Once the rod is constructed in a certain state of rotation, its motion is completely determined when the motion of its end-point at the origin is prescribed. In other words, [itex]\phi(t)=\omega t+\phi_o[/itex] is the only possibility. (2) The rod is one of the exceptional cases, and [itex]\phi(t)[/itex] need not be linear.

    I claim that #2 is true, and in fact any smooth function [itex]\phi(t)[/itex] is consistent with Born-rigidity. I take the definition of Born-rigidity, in a plane, to be as follows: Cover the object with a triangular network that is completely inside the object, such that no lines of the network intersect. Let a light signal make a round-trip from node A to adjacent node B and back. Then the object is Born-rigid if the time for this round trip, as measured by a clock comoving with A, is constant for all A and B (with an error that can be made as small as desired by making the network sufficiently fine-grained).

    In the case of the ruler rotating around its endpoint, the triangles of the triangular network are all degenerate triangles with zero interior area, and it suffices to check that there is constant proper round-trip time for a light signal between points A and B on the ruler at r and r+dr. This is true, because the proper round-trip time is simply 2dr, with errors of order dr2, independent of angular velocity or acceleration. To see this, consider the inertial frame instantaneously comoving with the point at r, at some initial time t, when the signal is emitted at r. The Lorentz boost from the lab frame to this frame is perpendicular to AB, so AB=dr exactly at time t; the only reason for the round-trip time, as measured by A's clock, to differ from 2dr is because A and B have radial and tangential accelerations. Since [itex]\phi(t)[/itex] is assumed smooth, the events at which the light signal is reflected and received depend on the angular acceleration [itex]d^2\phi/d t^2[/itex] only to order dr2. The curvature of the circles also leads to errors that are only of order dr2, while time dilation of the clock at A causes an error of order dr4. Therefore the difference in round-trip time between a case with angular acceleration and a case with no angular acceleration is of order dr2, which is negligible, but we already know that the case with no angular acceleration is consistent with Born-rigidity.

    By the way, if it were impossible to subject a one-dimensional ruler to an angular acceleration, then there would be foundational problems with Einstein's whole presentation of general relativity in [Einstein 1916], where he assumes that rulers can be picked up, rotated, and put down in another place.

    Another interesting case is a Born-rigid object shaped like a letter "C," initially at rest in the nonrotating lab frame and having the gap in the circle as small as you like. This object can then be given any angular acceleration about the center, while maintaining its Born-rigidity. However, the gap will be larger at higher angular velocities.

    I'd previously thought that any object that enclosed zero area could be angularly accelerated, the reasoning being that essentially the only reason that angular acceleration violates Born-rigidity is that the impulses supplying the acceleration have to be applied simultaneously, but Einstein synchronization is non-transitive on a triangle that encloses a nonzero area. However, this does not quite work, because you can have cases that violate the condition that the lines of the triangular network should not intersect. For example, suppose the "C" is initially rotating at a high angular velocity, and the gap between its ends is very small. If we then try to slow the rotation, the gap will close up, and one end will intrude on the other.

    The converse is easy: objects that enclose nonzero area can't be angularly accelerated. We already know that a disk (whether or not the interior is included) can't have an angular acceleration about its center imposed on it while maintaining Born rigidity. But any object that encloses nonzero area encloses a disk. If the object were to undergo an angular acceleration, then we could transform into a frame in which the center of the enclosed disk was at rest, and then in that frame the disk would undergo angular acceleration about its center, which is inconsistent with Born-rigidity.

    Wolfgang Pauli, "Theory of Relativity", pages 130--134, Pergamon Press, 1958. Google books lets me peek through a keyhole at the relevant page, but I can't seem to get it to let me see the notes that contain the Herglotz and Noether references.

    A. Einstein, "The foundation of the general theory of relativity," Annalen der Physik, 49 (1916) 769; translation by Perret and Jeffery available in an appendix to the book at http://www.lightandmatter.com/genrel/ (PDF version)
     
    Last edited: Sep 13, 2010
  2. jcsd
  3. Sep 13, 2010 #2

    Chris Hillman

    User Avatar
    Science Advisor

    Is this correct? I think you are looking at the two dimensional world sheet formed by the world lines of certain Langevin observers, namely those which at t=0 intersect z=0 in a short segment with one endpoint at r=0, so that your ruler is rotating about one end with constant angular velocity. If so, then, since these world lines belong to a rigid congruence, I agree that your ruler can be called rigid.

    But it must be the far endpoint, not the one at r=0, whose motion you propose to specify as a helix. Then Pauli would say that you can "fit" that helix into a unique Langevin congruence. The corresponding Langevin observers can be said to be rotating rigidly with fixed angular velocity.

    (See what goes wrong when we try to do this for an entire disk?)

    BTW, the UseNet Physics FAQ entry
    Code (Text):

    math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html
     
    is by Michael Weiss. Baez did coauthor many entries, but not that one.

    In the old PF thread I cited, myself, pervect, and Greg Egan discussed "rigidly rotating rings", which is closely related to the thing you describe. From what you said about gaps, I think you already know what goes wrong if we try to define a naive "physical geometry of the rigidly rotating disk".

    Sounds like everyone here (not in the public threads!) understands that we can't spin up a nonrotating disk to a rigidly rotating disk without deforming the disk, which makes it difficult to speak sensibly of "two disks, identical except that one is static and one is rotating rigidly with angular velocity a" [sic].

    I think your definition of "rigid motion" is more or less equivalent to vanishing expansion tensor, which would be good!
     
    Last edited: Sep 13, 2010
  4. Sep 13, 2010 #3

    Dale

    Staff: Mentor

    Hi bcrowell,

    I just want to mention that I do not disagree with you, I just don't know enough to agree with you either. I had not intended to pursue it until you specifically asked, and I definitely prefer working it out here rather than in the main forum.

    Before we go on, do we have a reference for the mathematical equations that govern Born rigidity. A brief Google search revealed a lot of discussion, but not so much definition. I posted a definition in:

    https://www.physicsforums.com/showpost.php?p=2877530&postcount=30

    but now I don't know where I got that procedure and I am not certain that it is correct.
     
    Last edited: Sep 13, 2010
  5. Sep 13, 2010 #4

    Chris Hillman

    User Avatar
    Science Advisor

    @Dale:

    Born rigid should mean zero expansion tensor: nearby particles in the body are not moving toward or away from each other. And you are correct that (thankfully!) this doesn't require us to say anything about body forces or stresses. But in the context of linear elasticity, the strain tensor arises from considering the expansion tensor; see
    Code (Text):

    www.physicsforums.com/showthread.php?t=171079
     
     
    Last edited: Sep 13, 2010
  6. Sep 13, 2010 #5

    DrGreg

    User Avatar
    Science Advisor
    Gold Member

    For what it's worth, this paper gives a technical definition of rigidity (in the final section "Rigid motion"). The way it is phrased appears to indicate it is Born's definition. It involves Lie derivatives and to be honest I haven't worked my way through the definition to make full sense of it.

    (Thanks to Frederik who brought this to my attention in an old thread.)
     
  7. Sep 13, 2010 #6

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Thanks, Dale and Chris, for your comments!

    A couple of notes:

    When you apply the definition of Born rigidity that I gave in #1 to a one-dimensional ruler, it does *not* guarantee that the ruler remains straight. It only guarantees that its proper length is preserved. This obviously does violence to the intuitive meaning of the word "rigid." When a Born-rigid object in the plane consists of n particles, the number of degrees of freedom can vary from 2 (in the case where all the particles are rigidly locked into the triangular network) to n+1 (when they're arranged in the form of a straight ruler).

    I agree that we shouldn't spin our wheels debating the correct definition, which has presumably been worked out by others. I think that in nondegenerate cases the triangular network definition I gave is equivalent to the zero-expansion definition that CH gave. What I'm not so sure about is whether either one is the optimal definition in the degenerate cases. A congruence is ordinarily assumed to have a union that is four-dimensional. Given the motion of the ruler I described in #1, there is no uniquely defined four-dimensional congruence; the world-lines of the points on the ruler only sweep out a three-dimensional space. I have no previous experience with congruences, but if I read the treatment on p. 217 of Wald, here's how it looks to me. The first step is to define a vector field [itex]\xi^a[/itex] of tangents normalized such that [itex]\xi^a \xi_a=1[/itex] (in the +--- signature that I prefer). This can be done for a three-dimensional congruence embedded in 3+1 dimensions. Next we define [itex]B_{ab}=\nabla_b\xi_a[/itex]. But this seems to break down in the degenerate case, because we can't evaluate covariant derivatives when b is a coordinate in the direction orthogonal to the ruler's world-sheet. (A side note: Wald defines the expansion [itex]\theta[/itex] as a scalar. CH, when you refer to it as the expansion tensor, do you just mean that in the sense that a scalar is a rank-0 tensor, or are you referring to something different than the scalar [itex]\theta[/itex] Wald defines?)

    It's a drag that the original references by Born, Herglotz, and Noether are a century old and presumably in German. Since [Pauli 1958] says that Herglotz and Noether explicitly considered exceptional cases, so it would make a heck of a lot of sense to find out what they said about them, rather than reinventing the wheel.

    It's possible that there is some interesting way to define a limiting process in which the thickness of a Born-rigid object approaches zero while its shape is in some sense kept rigid. In Newtonian mechanics, you could let the elastic modulus approach infinity during this limiting process. We wouldn't expect this to be compatible with relativity for an object kept rigid by its own internal stresses, but maybe there is some way of defining a similar notion when the shape is maintained, as in Born rigidity, by external forces.

    -Ben
     
    Last edited: Sep 13, 2010
  8. Sep 13, 2010 #7

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Aha! Thanks, Frederik and DrGreg, for the reference to the Giulini paper. Giulini gives the following references to the original papers. (The translations of the titles are my best guesses with help from Google Translate.) The Herglotz and Noether papers are freely available online. The Born one has been digitized, but is paywalled. The Ehrenfest paper is freely available in English translation (link below).

    Max Born. Die Theorie des starren Elektrons in der Kinematik des Relativitätsprinzips. (The theory of the rigid electron in relativistic kinematics) Annalen der Physik (Leipzig), 30:1–56, 1909.

    Gustav Herglotz. Über den vom Standpunkt des Relativitätsprinzips aus als "starr" zu bezeichnenden Körper. (On the status of so-called "rigid" bodies according to the principle of relativity) Annalen der Physik (Leipzig), 31:393–415, 1910. http://gallica.bnf.fr/ark:/12148/bpt6k15335v.image.f403

    Fritz Noether. Zur Kinematik des starren Körpers in der Relativitätstheorie (On the kinematics of rigid bodies in relativity theory) Annalen der Physik (Leipzig), 31:919–944, 1910. http://gallica.bnf.fr/ark:/12148/bpt6k15335v.image.f932

    Ehrenfest, Gleichförmige Rotation starrer Körper und Relativitätstheorie, Physikalische Zeitschrift, 10: 918, http://en.wikisource.org/wiki/Uniform_Rotation_of_Rigid_Bodies_and_the_Theory_of_Relativity

    It looks to me like the definition given in the Giulini paper is probably the same as the one CH gave, just dolled up in slightly different language, so it probably has the same problems when applied to degenerate cases.

    BTW, Fritz Noether was Emmy's brother. He was executed by the Soviets.

    -Ben
     
    Last edited: Sep 13, 2010
  9. Sep 13, 2010 #8

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Looking through the historical definitions, here's what I find:

    Born 1909
    as described in Ehrenfest 1909: "Born has based this definition - in accordance to the basic idea of relativity theory - not on the system of measurement of a stationary observer, but on the (Minkowskian) measure-determinations of, say, a continuum of infinitesimal observers who travel along with the points of the non-uniformly moving body: for each of them in their measure the infinitesimal neighborhood should appear permanently undeformed."

    Ehrenfest 1909
    "A body is relatively rigid, that is: it is deformed continuously at any movement, so that for a stationary observer, each of its infinitesimal elements at any moment has just that Lorentz contraction (compared to the state of rest) which corresponds to the instantaneous velocity of the element's center." Ehrenfest says he thinks this is equivalent to Born 1909.

    Herglotz 1910
    Description by Giulini: "A continuum performs rigid motion if the world lines of all its points are equidistant curves." Described by Giulini as equivalent to Born 1909.

    Born 1910
    Described by Giulini as a different definition, not equivalent to Born 1909.

    Noether 1910
    Described by Giulini as a different definition, not equivalent to Born 1909.
     
  10. Sep 15, 2010 #9

    Chris Hillman

    User Avatar
    Science Advisor

    Agree. But let me stress that the standard definition of a rigid congruence is that the expansion tensor vanish. See section 21.2 of Stephani et al., Exact Solutions of the Einstein Field Equations, second edition:
    i.e. the shear tensor and expansion scalar vanish, i.e. the expansion tensor vanishes.

    Can you elaborate? Sorry if I seem a bit slow...

    Agreed, a congruence of curves should consist of nonintersecting curves, and precisely one curve in the congruence should pass through any point in the neighborhood where it is defined.

    You can fit a given world line into (many) congruences. By requiring this congruence to satisfy certain properties, in some cases you may obtain a unique congruence into which the given world line fits.

    You mean, the two-dimensional world sheet of a one dimensional object?

    I use -+++ so watch out for that, e.g. in defining the projection tensor I mentioned above.

    See also 6.1.1 in Stephani et al., Exact Solutions, which is probably the definitive source for such things. See also Poisson, A Relativist's Toolkit, and Hawking and Ellis, Large Scale Structure of Space-Time.

    Now I am not sure what you mean by "three-dimensional"!

    There might be some confusion about what I had in mind here. I didn't have in mind a discussion of the historical development of concepts of rigidity, but rather a discussion of how to use the notion which has been standard since the monograph of Hawking and Ellis, namely, the kinematic decomposition.

    The kinematic decomposition was introduced in gtr some time after the early work of Born, Herglotz, &c.--- I believe that Bondi and Ellis played a role in popularizing it among gtr researchers.

    In the long PF thread containing a good discussion between myself, pervect, and Greg Egan (plus noise from idgits) which I cited in the first post of the other BRS thread, I advocated making a careful study of the Newtonian limit; indeed, I advocated beginning with a fairly detailed Newtonian analysis, and I also suggesting developing and using spring models in the Newtonian case. We might be thinking along similar lines, because regarding issues of the "rigidly rotating disk", I also suggested focusing on the simpler case of a "rigidly rotating ring". This was in the context of discussing "memory effects" while spinning up (neccessarily with distortions) a disk or ring from a zero angular momentum state to a nonzero angular momentum but rigidly rotating state. I was interested in obtaining some information from a suitable perturbation analysis of how relativistic effects change the Newtonian predictions as the velocities of particles near the rim of the disk become relativistic.

    I don't think the paper by Guiliani will be very useful here, but the definition given there (Lie drag the projection tensor by the given vector field) is equivalent to the standard notion: vanishing expansion tensor.

    A very good resource for cosmological models is

    Code (Text):

    arxiv.org/abs/gr-qc/9812046
     
    Cosmological models (Cargèse lectures 1998)
    George F R Ellis, Henk van Elst

    Been a while since I looked at it, but it must introduce the expansion and vorticity tensors. No doubt several of the review papers at
    Code (Text):

    relativity.livingreviews.org/Articles/subject.html
     
    also introduce these concepts, incidentally.
     
    Last edited: Sep 15, 2010
  11. Sep 15, 2010 #10

    Chris Hillman

    User Avatar
    Science Advisor

    Sorry, this got lost above. The expansion scalar is the trace of the expansion tensor, and the shear tensor is the traceless part of the expansion tensor. So the expansion scalar and shear tensor contain precisely the same information as the expansion tensor. The vorticity tensor is the antisymmetric part of the covariant derivative, projected as above. The vorticity vector is the vector construced from the (three-dimensional, antisymmetric!) vorticity tensor in the usual way; it contains the same information as the vorticity tensor.

    Notice that we should think of the expansion and vorticity tensors as three dimensional tensors. At each event, they live in the three-dimensional spatial hyperplane orthogonal to the tangent vector to the curve in the congruence which passes through that event--- we can't say "three dimensional hyperslice everywhere orthogonal to the world lines in the congruence" because such a thing exists precisely in case the vorticity tensor vanishes!

    This is standard terminology, but Wald doesn't bother to go into much detail, which I think is a mistake.

    I didn't know that. Max Noether, the father, was a leading mathematician in his own right.

    Well, the historical development is probably interesting, so I'd be happy to learn what you find out if you try to read all these papers and follow the detailed development. My guess is that Born was trying to give the now standard definition, but didn't possess the language. I guess I am not sure who first wrote down the now standard definition in the context of gtr, but I would guess it might have been Bondi or a contemporary c. 1960.
     
    Last edited: Sep 15, 2010
  12. Sep 15, 2010 #11

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Hi, CH -- As always, you are very knowledgeable, helpful, and generous with your time.

    The definition of rigidity I'm using is that we have a network, and the proper lengths of all the lines in the network stay constant. When one applies that to a ruler, modeled as a one-dimensional set of segments, then essentially it becomes a model of a flexible chain, not a rigid rod. The chain maintains its length, but is not constrained to stay straight. If there are n nodes, then it has 2n momenta (in the plane), but there are n-1 constraints on the proper lengths, so you have n+1 degrees of freedom. On the other hand, for an object like a disk you recover the standard result that a Born-rigid object's motion, once it has been constructed in a certain state of rotation, is completely specified by giving the motion of one of its points; in the plane, this means 2 degrees of freedom. There are various intermediate cases. E.g., I could attach a chain to the edge of a disk.

    I won't have time to dig into the old papers until this weekend. I don't speak German, so it may be a tedious business to try to decode them with computer assistance. The quick impression I have so far, based on paraphrases in later papers written in English, is that although all of the definitions are probably equivalent for an object that encloses a nonzero area, they are probably inequivalent in the degenerate cases, and in fact the standard modern definition in terms of the expansion tensor is probably not a useful one in the degenerate cases.
     
  13. Sep 15, 2010 #12

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Not sure where to ask this. I'll put it here since we're talking about definitions of rigidity. I'm trying to study Wald's presentation of the relevant tensors, and I almost immediately got stuck on a detail that looks trivial. Wald considers a smooth congruence of timelike geodesics, parametrized by proper time. This is on page 217. I will call the tangent vector T instead of ξ. Wald defines

    [tex]B_{ab}=\nabla_b T_a[/tex]

    I'm much more comfortable with coordinate independent notation than abstract index notation, so I usually try to translate stuff like this into coordinate independent notation. The coordinate independent definition of B is

    [tex]B(X,Y)=\nabla (g(T,\cdot))(Y,X)=\nabla_Y(g(T,X))[/tex]

    Perhaps I should explain the notation. [itex]g(T,\cdot)[/itex] is the map [itex]X\mapsto g(T,X)[/itex]. [itex]\nabla[/itex] is the connection, and I'm writing the map [itex](X,Y)\mapsto\nabla_X\omega(Y)[/itex] as [itex]\nabla\omega[/itex], so that I can write [itex]\nabla\omega(X,Y)=\nabla_X\omega(Y)[/itex].

    By this definition, we have

    [tex]B(T,Y)=\nabla_Y(\underbrace{g(T,T)}_{=-1})=0[/tex]

    for all Y, so B(T,·)=0. This is what Wald writes as [itex]B_{ab}T^a=0[/itex]. So far, so good. But Wald also claims that [itex]B_{ab}T^b=0[/itex], and that's not the result I'm getting. To show that B(·,T)=0, we need to show that B(X,T)=0 for all X.

    [tex]B(X,T)=\nabla_T(g(T,X))[/tex]

    Now what? This is

    [tex]=g(\nabla_T T,X)+g(T,\nabla_T X)[/tex]

    right? And [itex]\nabla_T T=0[/itex] by definition of "geodesic"? So why do I have a term left?
     
  14. Sep 15, 2010 #13

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Can you relate your expansion tensor and vorticity tensor to Wald's B, [itex]\theta[/itex], [itex]\sigma[/itex], and [itex]\omega[/itex]? MTW p. 566 seems to use the basically same notation as Wald.
     
  15. Sep 16, 2010 #14

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    This is just my interpretation of Chris's verbal description: We have Bab=B(ab)+B[ab]. The first term is the expansion tensor and the second term is the vorticity/twist/rotation tensor. The expansion tensor can be further decomposed into the shear tensor and the expansion scalar: [tex]B_{(ab)}=\sigma_{ab}+\frac{1}{3}\theta h_{ab}[/tex]
     
  16. Sep 16, 2010 #15

    George Jones

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You have a term left because your coordinate-independent definition of B is not equivalent to Wald's abstract index definition of B. To see what went wrong, calculate [itex]B\left(X,Y\right)[/itex] using index notation:

    [tex]
    \begin{equation*}
    \begin{split}
    B \left( X , Y \right) &= X^a Y^b B_{ab} \\
    &= X^a Y^b \nabla_b T_a \\
    &= Y^b \nabla_b \left(X^a T_a \right) - Y^b T_a \nabla_b X^a \\
    &= \nabla_Y g \left( T,X \right) - g \left( T, \nabla_Y X \right).
    \end{split}
    \end{equation*}
    [/tex]
     
  17. Sep 16, 2010 #16

    Chris Hillman

    User Avatar
    Science Advisor

    Exactly. Where we use 1/3 not 1/4 because we are detracing a three-dimensional tensor to form the shear tensor. And where we have to be careful to distinguish between indices which refer to a coordinate basis or a frame field, and also to be careful with Wald's abstract index notation when comparing with other books, as George said.

    OK, I think I see what you are getting at now. When I get a chance, maybe tonight, I'll try to come back and draw a picture. I think that what you are saying is:

    A key point I probably haven't yet emphasized is that the following are equivalent conditions on a timelike congruence:
    • vanishing vorticity
    • hypersurface orthogonal
    So, "space at a time, as experienced by a family of observers" makes sense exactly in case the corresponding congruence has vanishing vorticity. For example, the Doran family of inertial infalling observers in the Kerr vacuum (which corresponds to the Lemaitre family of radially infalling observers in the Schwarzschild vacuum) is in the nonzero angular momentum case not vorticity-free, so that no family of orthogonal hyperslices exists in that case.

    Recall too that Born seems to have been thinking of fitting the world lines inside a material object into a congruence defined on an open neighborhood, and in effect computing its expansion tensor, and saying, if this vanishes, that the object is rigid.

    I think the problem with your twisting filament may be that when you say "maintain constant proper spatial distance"--- think of taking an integral of ds over the filament "at a time"--- you need to specify (presumably) ds "in the rest frame of a bit of matter in the filament". But if the filament is "twisting", I think that any congruence you fit its world lines into will have nonzero vorticity, so the orthogonal hyperslices won't exist, so "filament at a time" won't make sense.

    I'll have to think about this more when I have time and get back to you.
     
    Last edited: Sep 16, 2010
  18. Sep 16, 2010 #17

    Fredrik

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Thank you George. That explains it. My mistake was to think that [itex]\nabla\omega[/itex] in coordinate-independent notation means [itex](X,Y)\mapsto\nabla_X(\omega(Y))[/itex] when in fact it means [itex](X,Y)\mapsto\nabla_X\omega(Y)[/itex].

    Edit: Lee puts the variable that goes into the subscript position last instead of first, so I'll switch to that convention. From now on, [itex]\nabla\omega[/itex] means [itex](X,Y)\mapsto\nabla_Y\omega(X)[/itex].

    Wald's [itex]B_{ab}=\nabla_bT_a[/tex] then translates to

    [tex]B(X,Y)=\nabla T^\flat(Y,X)=\nabla_X T^\flat(Y)[/tex]

    where [itex]T^\flat[/itex] ("T-flat") is the 1-form corresponding to T: [itex]T^\flat=g(T,\cdot)[/itex]. The "flat" symbol [itex]\flat[/itex] looks a bit too much like a "b" for my taste, but that shouldn't cause any confusion here. By the standard definition of the covariant derivative of a 1-form, the above is

    [tex]=X(T^\flat(Y))-T^\flat(\nabla_X Y)=X(g(T,Y))-g(T,\nabla_X Y)[/tex]

    We want to prove that B(T,X)=B(X,T)=0 for all X.

    [tex]B(X,T)=X(g(T,T))-g(T,\nabla_X T)[/tex]

    The first term is =0 because g(T,T) is constant. The second is =0 because

    [tex]0=\nabla_X(g(T,T))=2g(T,\nabla_X T)[/tex]

    [tex]B(T,Y)=T(g(T,Y))-g(T,\nabla_T Y)[/tex]

    This is =0 because the first term is

    [tex]\nabla_T(g(T,Y))=g(\nabla_T T,Y)+g(T,\nabla_T Y)=g(T,\nabla_T Y)[/tex]
     
    Last edited: Sep 16, 2010
  19. Sep 16, 2010 #18

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Hey, wait, I don't think I said what you put in quotes and attributed to me! At least, I didn't say it in this thread, or in the thread titled "Ehrenfest / rod thought experiment." The notion I'm discussing is purely local, and I think I made that pretty explicit in #1, where the definition I gave was:

    The final part in parentheses was to make it clear that I was talking about constant local distance, not constant integrated distance. I think the definition also makes it clear that we are talking about distance "in the rest frame of a bit of matter in the filament," since the time is specified to be measured "by a clock comoving with A." So it seems to me that the two criticisms you raised are both criticisms of your paraphrase of my definition, not of my definition.

    Constant integrated distance would be impossible to define in general, both for the reason you gave and because the integration would in general be path-dependent. However, in the case of the chain, I think these problems both vanish. Path-dependence isn't an issue, because it's one-dimensional. If you decided that you wanted to evaluate [itex]L=\Sigma d_{n,n+1}[/itex], where [itex]d_{n,n+1}[/itex] was the distance defined in my definition, then it wouldn't matter whether you evaluated the sum according to some notion of simultaneity, because all of the terms in the sum are completely time-independent. But in any case, my definition never refers to this integrated length L, and never requires any notion of simultaneity. (If the chain were not rigid -- actually "unstretchable" would be a better word in this degenerate case -- according to my definition, then L certainly could be impossible to define unambiguously, for the reasons you gave.)
     
  20. Sep 17, 2010 #19

    Chris Hillman

    User Avatar
    Science Advisor

    Did I misattribute something to you?! If so, I apologize, but I can't find it. Maybe you were referring to the bit where I quoted (misquoted?):
    and then tried to restate that as
    If so, I take it you are saying I don't yet understand your filament. Do I have that right?
     
  21. Sep 17, 2010 #20

    bcrowell

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Hi, Chris,

    I think the confusion here is this:

    In #1, I attempted to give a rigorous definition of Born-ridigidy that would be applicable to degenerate cases as well as nondegenerate ones.

    In #11, I gave a loosely worded interpretation of what that definition implies in the case of a one-dimensional network -- basically that it does not imply "rigidity" at all, but only unstretchability.

    In #16, you said, 'when you say "maintain constant proper spatial distance,"' which made it sound as if you were quoting me, but the words in quotes were not mine. I think you were commenting on my #11 as if it was intended to be my definition of rigidity, and finding it lacking as a definition. That would have been a well justified criticism, if I had intended #11 as a definition, but #11 was merely an interpretation of the implications of the definition in #1, as applied to a particular degenerate case.

    -Ben
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook