BRS: degenerate cases of Born-rigidity

[bcrowell]

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 $\phi(t)$, where L, $\phi$, 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, $\phi(t)=\omega t+\phi_o$ is the only possibility. (2) The rod is one of the exceptional cases, and $\phi(t)$ need not be linear.

I claim that #2 is true, and in fact any smooth function $\phi(t)$ 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 $\phi(t)$ is assumed smooth, the events at which the light signal is reflected and received depend on the angular acceleration $d^2\phi/d t^2$ 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 let's 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)

 

[Chris Hillman]

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

math.ucr.edu/home/baez/physics/Relativity/SR/rigid_disk.html

is by Michael Weiss. Baez did coauthor many entries, but not that one.

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.

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!

 

[Dale]

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.

 

[Chris Hillman]

@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

www.physicsforums.com/showthread.php?t=171079

 

[DrGreg]

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.)

 

[bcrowell]

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 $\xi^a$ of tangents normalized such that $\xi^a \xi_a=1$ (in the +--- signature that I prefer). This can be done for a three-dimensional congruence embedded in 3+1 dimensions. Next we define $B_{ab}=\nabla_b\xi_a$. 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 $\theta$ 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 $\theta$ 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

 

[bcrowell]

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

 

[bcrowell]

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.

 

[Chris Hillman]

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.

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:

then the system rotates rigidly, i..e the fluid is n shearfree and expansionfree motion

i.e. the shear tensor and expansion scalar vanish, i.e. the expansion tensor vanishes.

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).

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

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.

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.

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.

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

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 $\xi^a$ of tangents normalized such that $\xi^a \xi_a=1$ (in the +--- signature that I prefer).

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.

This can be done for a three-dimensional congruence embedded in 3+1 dimensions.

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

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.

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.

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.

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

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

relativity.livingreviews.org/Articles/subject.html

also introduce these concepts, incidentally.

 

[Chris Hillman]

(A side note: Wald defines the expansion $\theta$ 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 $\theta$ Wald defines?)

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.

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

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 possesses 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.

 

[bcrowell]

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

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).

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

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.

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 possesses 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.

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.

 

[Fredrik]

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

$$B_{ab}=\nabla_b T_a$$

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

$$B(X,Y)=\nabla (g(T,\cdot))(Y,X)=\nabla_Y(g(T,X))$$

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

By this definition, we have

$$B(T,Y)=\nabla_Y(\underbrace{g(T,T)}_{=-1})=0$$

for all Y, so B(T,·)=0. This is what Wald writes as $B_{ab}T^a=0$. So far, so good. But Wald also claims that $B_{ab}T^b=0$, 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.

$$B(X,T)=\nabla_T(g(T,X))$$

Now what? This is

$$=g(\nabla_T T,X)+g(T,\nabla_T X)$$

right? And $\nabla_T T=0$ by definition of "geodesic"? So why do I have a term left?

 

[bcrowell]

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.

Can you relate your expansion tensor and vorticity tensor to Wald's B, $\theta$, $\sigma$, and $\omega$? MTW p. 566 seems to use the basically same notation as Wald.

 

[Fredrik]

Can you relate your expansion tensor and vorticity tensor to Wald's B, $\theta$, $\sigma$, and $\omega$? MTW p. 566 seems to use the basically same notation as Wald.

This is just my interpretation of Chris's verbal description: We have B_ab=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: $$B_{(ab)}=\sigma_{ab}+\frac{1}{3}\theta h_{ab}$$

 

[George Jones]

So why do I have a term left?

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 $B\left(X,Y\right)$ using index notation:

$$\begin{equation*} \begin{split} B \left( X , Y \right) &= X^a Y^b B_{ab} \\ &= X^a Y^b \nabla_b T_a \\ &= Y^b \nabla_b \left(X^a T_a \right) - Y^b T_a \nabla_b X^a \\ &= \nabla_Y g \left( T,X \right) - g \left( T, \nabla_Y X \right). \end{split} \end{equation*}$$

 

[Chris Hillman]

This is just my interpretation of Chris's verbal description: We have B_ab=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: $$B_{(ab)}=\sigma_{ab}+\frac{1}{3}\theta h_{ab}$$

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.

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.

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:

I can choose the world sheet of a one-dimensional filament such that nearby bits of matter in the filament "maintain constant proper distance measured along the filament". Then, I can fit the world lines in the world sheet carved out by the history of the filament into a timelike congruence which is rigid (vanishing expansion tensor).

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.

 

[Fredrik]

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

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, $\nabla\omega$ means $(X,Y)\mapsto\nabla_Y\omega(X)$.

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

$$B(X,Y)=\nabla T^\flat(Y,X)=\nabla_X T^\flat(Y)$$

where $T^\flat$ ("T-flat") is the 1-form corresponding to T: $T^\flat=g(T,\cdot)$. The "flat" symbol $\flat$ 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

$$=X(T^\flat(Y))-T^\flat(\nabla_X Y)=X(g(T,Y))-g(T,\nabla_X Y)$$

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

$$B(X,T)=X(g(T,T))-g(T,\nabla_X T)$$

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

$$0=\nabla_X(g(T,T))=2g(T,\nabla_X T)$$

$$B(T,Y)=T(g(T,Y))-g(T,\nabla_T Y)$$

This is =0 because the first term is

$$\nabla_T(g(T,Y))=g(\nabla_T T,Y)+g(T,\nabla_T Y)=g(T,\nabla_T Y)$$

 

[bcrowell]

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.

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:

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).

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 $L=\Sigma d_{n,n+1}$, where $d_{n,n+1}$ 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.)

 

[Chris Hillman]

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?):

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.

and then tried to restate that as

I can choose the world sheet of a one-dimensional filament such that nearby bits of matter in the filament "maintain constant proper distance measured along the filament". Then, I can fit the world lines in the world sheet carved out by the history of the filament into a timelike congruence which is rigid (vanishing expansion tensor).

If so, I take it you are saying I don't yet understand your filament. Do I have that right?

 

[bcrowell]

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

 

[Chris Hillman]

BRS: Seeking clarification

I'm getting confused trying to track down all these quotes, so please bear with me.

Re my Post #16, when I wrote

So, "space at a time, as experienced by a family of observers" makes sense ... I think the problem with your twisting filament may be that when you say "maintain constant proper spatial distance"

I think those were "scare quotes" intended to briefly convey that I am aware that the phrase are problematical. I apologize for any confusion.

Also in my Post #16, when I wrote

I can choose the world sheet of a one-dimensional filament such that nearby bits of matter in the filament "maintain constant proper distance measured along the filament". Then, I can fit the world lines in the world sheet carved out by the history of the filament into a timelike congruence which is rigid (vanishing expansion tensor).

that was my attempt to rephrase what I thought you were trying to say in a manner which seems closer to the notion of the expansion tensor of a congruence. So, does my attempted restatement seem OK to you?

By "denerate cases", you mean a one-dimensional object, correct? How does this differ (if it does) from the filament which can twist but not stretch?

Did you see what I was trying to get at when I suggested that I am not sure your twisting filament thought experiment will stand up?

 

[bcrowell]

Here is my incomplete attempt to translate the relevant portion of [Born 1909a]. Scans of the original paper are freely available at the URL given below. I don't speak German, so this is basically just my attempt to make the Google Translate translation intelligible. There may be lots of inaccuracies.

----

Therefore one must search for a way to generalize the definition of rigidity used in the old mechanics. One can use the fact that if the condition r=const. between distant world-lines is replaced with a differential condition between infinitely nearby world-lines, fulfilled throughout space, the result is that r=const.

For this purpose, we consider at time t the distance between two adjacent infinite world lines, i.e., the arc-element ds = sqrt(dx2+dy2+dz2). Putting this equal to a constant s, we have the equation ds2 = s2 of an infinitely small sphere. [The equations x=x(xi,eta,zeta,t), y=y(...), x=x(...), where xi, eta, and zeta are body-fixed coordinates] then represent the motion of an infinitesimal ellipsoid. This is obtained when the size ds2 is expressed by virtue of the equations dx = (partial x/partial xi)d xi +... as the quadratic form ds2=p11 dxi2+p22 deta2 + ... + 2p12 dxi deta +... This is the matrix of the "deformation parameters" P=(p_ab) from the matrix A = (partial x/partial xi ...) sewn together in the following manner: P = A*A. We call the motion of the very small parts rigid, if an infinitely small fabric is not changed during the motion, so so that all the p_ab are independent of time. We then have the infinitesimal rigid-body conditions (12) dp_ab/dt = 0. If xi, eta, zeta are the initial values of z, y, z, then at t = 0 the matrix A is equal to the unit matrix 1, (12) then becomes P = A*A = 1. It is now an elementary theorem of differential geometry that if this condition is fulfilled everywhere, the flow is represented by equations of the form x=a1+a11 xi + ..., which is to say that the motion is that of a rigid body.

These infinitesimal rigid-body conditions are now easily carried over to the relativistic kinematics.

In the following, the only lengths that should have physical meaning are those that are invariant under the Lorentz transformations.

We now consider a flow which is represented by equations, not of the form (1) but of the following form, which corresponds better to the symmetry of the coordinates x, y, z, t in the relativity principle: x=x(xi,eta,zeta,tau), ... where tau is the proper time, i.e., there is the identity (partial x/partial tau)^2+...=-c^2; tau would be measured at any given "cross-section" of the flow.

The eta,zeta,tau characterize the individual streamlines, but their significance is otherwise not determined. We now define for the moment x(0,0,0,tau)=frakx(tau), ...

----

He then goes on to generalize the nonrelativistic definition to relativity, using extremely cumbersome old-fashioned coordinate notation, and that's where my motivation to keep translating bogged down.

The general idea is that he makes body-fixed coordinates, and then defines a world-line for every body-fixed point. I think what it boils down to is that infinitesimally close world-lines are required to maintain constant proper distance from one another. There does not seem to be any prohibition against self-intersection, so I think that when the letter C undergoes angular deceleration, and the two end-points pass into one another, that is not ruled out by Born's definition. Although he clearly has in mind a space-filling body, I actually don't see anything in his treatment that requires the xi,eta,zeta coordinates to be nondegenerate, so it's possible that some or all of his math carries over without modification to one-dimensional or two-dimensional bodies. He doesn't use anything like a congruence, where one assumes that every point in spacetime maps onto a world-line; he only maps from the body-fixed coordinates into spacetime.

[Born 1909a] 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), Annalen der Physik 30, 1; also referred to as 335 (11), 1-56 (vierte folge, band 30), 1909. -- http://gallica.bnf.fr/ark:/12148/bpt6k15334h.image.f7

 

[Chris Hillman]

BRS: An old Ph.D. thesis on rigid motion; plus, the chcentric universe

As further proof that God exists, and created the universe as a joke--- on me!:

Courtesy of

http://adsabs.harvard.edu/

Title: Relativistic Rigid Body Motion.
Authors: Bennett, Robert James
Affiliation: AA (STEVENS INSTITUTE OF TECHNOLOGY.)
Publication: Thesis (PH.D.)--STEVENS INSTITUTE OF TECHNOLOGY, 1971.
Source: Dissertation Abstracts International,
Volume: 32-06, Section: B, page: 3550.
Publication Date: 00/1971
Category: Physics: General
Origin: UMI

The author is currently being discussed over at slashdot, where he has another claim to fame. Guess what that is?

Aw, come on, guess!

OK, I'll tell you: he wrote one chapter in a lengthy book by Robert A. Sungenis of Catholic Apologetics International Publishing:

www.galileowaswrong.com/galileowaswrong/

(NOTE: I strongly suggest using NoScript and Tor when visiting any crank site!)

Galileo Was Wrong is a detailed and comprehensive treatment of the scientific evidence supporting Geocentrism, the academic belief that the Earth is immobile in the center of the universe. Garnering scientific information from physics, astrophysics, astronomy and other sciences, Galileo Was Wrong shows that the debate between Galileo and the Catholic Church was much more than a difference of opinion about the interpretation of Scripture.

Scientific evidence available to us within the last 100 years that was not available during Galileo's confrontation shows that the Church's position on the immobility of the Earth is not only scientifically supportable, but it is the most stable model of the universe and the one which best answers all the evidence we see in the cosmos.

...

Robert J. Bennett, Ph.D., holds a doctorate in General Relativity from Stevens Institute of Technology. He served as a physics instructor at Manhattan College and Bergen Community College from 1967-1983, and is presently doing private tutoring in physics and mathematics. Dr. Bennett has written Chapter 10, a detailed, technical and mathematical explanation of the various arguments for Geocentrism. He has served as a consultant for the entire Galileo Was Wrong: The Church Was Right project.

Before you ask: Sungenis has been at this a long time, and, no, there's every reason to think he's not just kidding. One of the authors of a blurb praising his book is Gerardus Bouw (Astronomy, Baldwin-Wallace College), who inherited custody of the Tychonian Society from Walter van der Kamp (and changed the name to Association for Biblical Astronomy). Van der Kamp has been called the father of the (pun alert!) geocentrism movement.

I have to say, in order of increasing nuttiness, and also increasing rarity, we have three recognizable styles of contemporary physics crackpottery:

• anti-Einstein
• anti-Newton
• anti-Galileo

 

[bcrowell]

Something funny happened in the slashdot discussion of the immobile-earth stuff. Note that there are actually two distinct logically different statements that someone who's not a kook could make: (1) The Earth moves, and we can prove it; therefore Bennett is wrong. (2) There is no way to prove that the Earth stands still; therefore Bennett is wrong. #1 makes sense in terms of Newtonian mechanics, and was enthusiastically asserted by various people. #2 is actually more like the general-relativistic picture, but attempts to assert it were unpopular.

 

[bcrowell]

Here is a summary of my current understanding of the rigid-body stuff.

Historically, Born-rigidity seems to have been defined by two different approaches. In approach (1), which historically came first, we have a body-fixed space B, and we map $\{(p,t)|p\in B\}\rightarrow M$, where M is a manifold such as Minkowski space. Infinitesimally close world-lines determined in this way have to maintain constant radar distance. It seems to be assumed that B is three-dimensional, but this doesn't seem to be necessary. Self-intersection (like the letter C undergoing angular deceleration) could be explicitly ruled out, but is usually not considered. In approach (2), we define a congruence in some open neighborhood, which effectively reverses the direction of the map, and we set conditions on the expansion tensor. This approach can be written down in simple and elegant form using tensors and covariant derivatives, but it isn't capable of being modified in order to discuss self-intersection or non-space-filling objects, because neither of these can be described appropriately using a congruence.

The Herglotz-Noether theorem says that the only Born-rigid motions in flat spacetime are: (i) accelerated motion without rotation and (ii) unaccelerated motion with uniform rotation about an axis that remains parallel to itself (which I assume means the axis is parallel-transported?). (In curved spacetime, ii allows any congruence defined by a Killing vector.)

The H-N theorem is proved using congruences, so it doesn't say anything about non-space-filling bodies. For example, the one-dimensional ruler with angular acceleration would be a counterexample to the H-N theorem if we took it as applying to non-space-filling bodies.

I hadn't understood until today that Born-rigidity of space-filling objects was as restrictive as this. If a space-filling object is initially spinning, then its center of rotation can't undergo acceleration! Historically, the intense interest in Born-rigidity ca. 1910 was because they wanted to model the electron as a rigid sphere. They didn't know anything about electron spin then, but given the H-N theorem, I suppose they would have had to assume that rigid, spherical electrons were all created in a perfectly nonrotating state, because a rotating electron would have had infinite inertia.

I could be wrong about this, but I think I have a good intuitive example of why a rotating object's center can't accelerate. Suppose you have a cylinder rigidly rotating about its symmetry axis z. If you accelerated it along that axis, then a comoving observer at z would experience gravitational time dilation with respect to another observer at z'. But then there would be a mismatch between each one's perception of his own angular velocity and his perception of the other observer's angular velocity.

Pauli says: "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." If I'm understanding him correctly, he's saying that typically if you arbitrarily impose a world-line on one point in the body, the world-line is accelerating, so the body can't rotate; it has three d.f. because you can accelerate it at will. The exceptional cases would be those in which the single world-line is inertial, in which case the body can also have rotation; in this case, it has zero d.f.

I wonder if a zero-thickness rotating disk can have its center accelerated without violating Born-rigidity? This wouldn't violate the H-N theorem, because the H-N theorem only applies to space-filling bodies.

 

[Chris Hillman]

Hi, Ben,

Pictures would really help, but I am becoming more confident that I do understand what you are saying.

So by "degenerate cases", you mean bodies of dimension less than three, correct? Like thin wires (possibly nonlinear) and thin plates (possibly nonplanar) in E^3? Only in some possibly curved spacetime?

If so, I can try to give some explicit examples in Minkowski spacetime and maybe others, e.g. FRW dust.

In approach (2), we define a congruence in some open neighborhood, which effectively reverses the direction of the map, and we set conditions on the expansion tensor. This approach can be written down in simple and elegant form using tensors and covariant derivatives, but it isn't capable of being modified in order to discuss self-intersection or non-space-filling objects, because neither of these can be described appropriately using a congruence.

I don't think that's true, for at least two reasons:

• consider a thin wire or a thin plate as a limit of "thinnish but non-infinitesimally thin" wire or "thinnish but non-infinitesimally thin" plate,
• embed the world lines of the matter in a thin wire or thin plate in a congruence.

(See the figures below.) There is a possible issue involving

• show it doesn't matter too much how you take the limit
• show it doesn't matter too much how you make the exension to a congruence

but I expect that is not problematic.

The Herglotz-Noether theorem says that the only Born-rigid motions in flat spacetime are: (i) accelerated motion without rotation and (ii) unaccelerated motion with uniform rotation about an axis that remains parallel to itself (which I assume means the axis is parallel-transported?).

Yes. Equivalently, the cases are:

• Rindler congruence (zero expansion tensor and zero vorticity but nonzero acceleration vector, varying in a very specific way wrt one coordinate, if we use an appropriate chosen chart)
• Langevin congruence (zero acceleration vector and expansion tensor but nonzero vorticity tensor)

If you accelerated it along that axis, then a comoving observer at z would experience gravitational time dilation with respect to another observer at z'.

z' is another location on the symmetry axis? If not, picture please!

I hadn't understood until today that Born-rigidity of space-filling objects was as restrictive as this.

Right, there are fewer d.o.f. for rigid motion in relativistic physics than in non-relativistic physics. That was the surprise for Einstein and the other early workers.

If I'm understanding him correctly, he's saying that typically if you arbitrarily impose a world-line on one point in the body, the world-line is accelerating, so the body can't rotate; it has three d.f. because you can accelerate it at will.

Two for the unit vector giving the direction of acceleration, and one for the constant magnitude of acceleration. If you like.

The exceptional cases would be those in which the single world-line is inertial, in which case the body can also have rotation; in this case, it has zero d.f.

Rather, three d.o.f., up to translation (two for a unit vector giving the orientation of the symmetry axis and one for choice of the constant angular velocity, if you like).

Or IOW, compare the acceleration vector for any world line in the Rindler or Born congruences in the BRS "Those Damnable Paradoxes".

These "d.o.f.s" are simply the dimensions of certain complexions (certain manifolds formed from some coset space formed by the appropriate symmetry group, which is a Lie group, and a closed Lie subgroup). See the unfinished BRS thread "Exploring the Rubik Group with GAP. IV".

I wonder if a zero-thickness rotating disk can have its center accelerated without violating Born-rigidity? This wouldn't violate the H-N theorem, because the H-N theorem only applies to space-filling bodies.

I think not; consider a thinnish but not infinitesimally thin disk using the Langevin congruence and take a limit.

Figures:

• thin plate as limit of thinnish but not infinitesimally thin plate
• world lines of matter in a thin plate embedded in a congruence

 

[bcrowell]

So by "degenerate cases", you mean bodies of dimension less than three, correct? Like thin wires (possibly nonlinear) and thin plates (possibly nonplanar) in E^3?

Yes.

Only in some possibly curved spacetime?

I only found out today that anyone had ever worked seriously on rigid motion in curved spacetime, so I'm not confident that I understand anything at all about it. In cases where I come across a statement by others that is clearly not restricted to flatness, then I figure I might as well reproduce their statement in full generality, but personally I'd be happier just considering flatness for now.

I don't think that's true, for at least two reasons:

• consider a thin wire or a thin plate as a limit of "thinnish but non-infinitesimally thin" wire or "thinnish but non-infinitesimally thin" plate,
• embed the world lines of the matter in a thin wire or thin plate in a congruence.

(See the figures below.) There is a possible issue involving

• show it doesn't matter too much how you take the limit
• show it doesn't matter too much how you make the exension to a congruence

but I expect that is not problematic.

Hmm...well, I'd claim that we have at least two inequivalent ways of taking the limit. One way gives, e.g., a one-dimensional ruler that can't flex or undergo angular acceleration, while the other gives a ruler that can flex and undergo angular acceleration. The issue is precisely how you make the extension from the world-sheet to a congruence. E.g., I could have a congruence in which the world-lines are of the form x=x_oe^t, y=const, z=const. This congruence isn't Born-rigid, and yet a line segment lying at rest in the y-z plane is Born-rigid. Given the world-sheet of this line segment, I don't see how you can apply the definition of the expansion tensor to determine whether the segment is Born rigid.

z' is another location on the symmetry axis? If not, picture please!

Yes, that's what I meant.

The exceptional cases would be those in which the single world-line is inertial, in which case the body can also have rotation; in this case, it has zero d.f.

Rather, three d.o.f., up to translation (two for a unit vector giving the orientation of the symmetry axis and one for choice of the constant angular velocity, if you like).

This may depend on what you mean by a degree of freedom. To make sense out of Pauli's statement, I defined a d.f. as something that can change due to external forces once the body's initial conditions have been set. Then the number of d.f. is 3 in case (i) and 0 in case (ii).

If you count both the initial d.f. and the d.f. that can be affected by external forces, then I guess it's 3 in case (i) and 6 in case (ii).

Regardless of which definition you choose, I guess the "exceptional cases" are the ones that fall under (ii).

 

[Chris Hillman]

One way gives, e.g., a one-dimensional ruler that can't flex or undergo angular acceleration, while the other gives a ruler that can flex and undergo angular acceleration.

Are these both rigid rulers?

Can you give an explicit example of (a)(b)? Using say world lines from the Langevin congruence?

E.g., I could have a congruence in which the world-lines are of the form $x=x_0 \, \exp(t), \; y=y_0, \; z=z_0$. This congruence isn't Born-rigid, and yet a line segment lying at rest in the y-z plane is Born-rigid.

But such a line segment doesn't have world lines which belong to your congruence.

[EDIT: actually, it is not a congruence. Some local interference prevented me from immediately recognizing that; sorry!]

This may depend on what you mean by a degree of freedom. To make sense out of Pauli's statement, I defined a d.f. as something that can change due to external forces once the body's initial conditions have been set. Then the number of d.f. is 3 in case (i) and 0 in case (ii).

It's true that the original researchers tend to use "d.o.f." loosely, but I don't think that's what Pauli had in mind.

Choose any world line from any Rindler congruence. In a Cartesian chart, the acceleration vector has form
$$\nabla_{\vec{e}_1} \vec{e}_1 = \hbox{scalar} \; \vec{e}_2$$
where the scalar depends on the coordinates and one parameter. So two d.o.f. for choice of unit vector and one d.o.f. for the parameter, and this acceleration vector will be constant for that world line and determine all the world lines in the congruence.

Choose any world line from any Langevin congruence. In a Cartesian chart, the acceleration vector has form
$$\nabla_{\vec{f}_1} \vec{f}_1 = \hbox{scalar} \; \vec{f}_2$$
where the scalar depends on the coordinates and one parameter. So two d.o.f. for choice of unit vector and one d.o.f. for the parameter, and this acceleration vector will be constant for that world line and will determine all the world lines in the congruence.

It might help to write that out in incredibly gory detail to see that, in the acceleration vector of an arbitrary event on an arbitrary world line in your rigid congruence, there is only one parameter in the scalar and then the other two determine the choice of unit vector.

So Pauli says three d.o.f. for both cases, correct?

 

[bcrowell]

One way gives, e.g., a one-dimensional ruler that can't flex or undergo angular acceleration, while the other gives a ruler that can flex and undergo angular acceleration.

Are these both rigid rulers?

It depends on which limiting process you use to define "rigid." According to the definition I gave in #1, both motions are rigid motions.

Can you give an explicit example of (a)(b)? Using say world lines from the Langevin congruence?

Sorry, what are (a) and (b)?

E.g., I could have a congruence in which the world-lines are of the form x=x_oe^t, y=const, z=const. This congruence isn't Born-rigid, and yet a line segment lying at rest in the y-z plane is Born-rigid.

But such a line segment doesn't have world lines which belong to your congruence.

Hmm...why do you say that? Maybe we're misunderstanding one another. Say the ruler's world lines are of the form x=0, y=k, z=0, where $0 \le k \le 1$. Then for all k we have a world line of the form given above, with x_o=0.

It's true that the original researchers tend to use "d.o.f." loosely, but I don't think that's what Pauli had in mind.

I don't know for sure what Pauli had in mind, but I don't think it's a matter of "loosely" versus "not loosely." A d.f. that is not free to vary over time isn't a d.f. For example, a roller-coaster on a track doesn't have degrees of freedom corresponding to transverse motion, because its transverse position isn't free to vary over time.

 

[Chris Hillman]

Your counterexample is invalid

Hi, Ben,

You wrote:

I'd claim that we have at least two inequivalent ways of taking the limit. One way gives, e.g., a one-dimensional ruler that can't flex or undergo angular acceleration, while the other gives a ruler that can flex and undergo angular acceleration. The issue is precisely how you make the extension from the world-sheet to a congruence. E.g., I could have a congruence in which the world-lines are of the form $x=x_0 \, \exp(t), y=y_0, z=z_0$. This congruence isn't Born-rigid, and yet a line segment lying at rest in the y-z plane is Born-rigid.

Here you are, I think, claiming to give an example of a non-rigid congruence in Minkowski vacuum, which contains a "degenerate" configuration of world lines corresponding to a Born rigid object.

But what you wrote does not actually define a congruence of parameterized curves in Minkowski vacuum!

To fix up the counterexample, you need to first of all define a congruence with nonzero expansion tensor, and then to find a one or two dimensional configuration of world lines belonging to the congruence which you can show are "rigid" according to some definition you'll need to explain.

I started to try to fix your example, along these lines: in the usual Cartesian chart on Minkowski vacuum, consider the frame field
$$\begin{array}{rcl} \vec{e}_1 & = & \sqrt{1+k^2 \, \exp(2t)} \; \partial_t + k \, \exp(t) \; \partial_x \\ \vec{e}_2 & = & k \, \exp(t) \; \partial_t + \sqrt{1+k^2 \, \exp(2t)} \; \partial_x \\ \vec{e}_3 & = & \partial_y \\ \vec{e}_4 & = & \partial_z \end{array}$$
The integral curves of $\vec{e}_1$ are then given by solving
$$\dot{t} = \sqrt{1+k^2 \, exp(2t)}, \; \dot{x} = k \, \exp(t)$$
The acceleration vector of $\vec{e}_1$ is
$$\nabla_{\vec{e}_1} \vec{e}_1 = k \, \exp(t) \; \vec{e}_2$$
the vorticity tensor vanishes, and and the only nonzero component of the expansion tensor is
$${H\left[ \vec{e}_1 \right]}_{22} = \frac{k^2 \, \exp(2t)}{\sqrt{1+ k^2 \, \exp(2t)}}$$
Now this is nonzero (so that the congruence is nonrigid) iff k is nonzero.

Can you find a "degenerate" selection of world lines belonging to this congruence, which are rigid according to some definition (which you will need to explain)?

And still seeking clarification: "degenerate" means a selection of world lines from a timelike congruence which can be interpreted as the "world sheet" of a one or two dimensional object, right?

 

[bcrowell]

I've been having too much fun on PF this weekend, and now I need to grade lab notebooks (ugh), but I thought I'd post the general outlines of some thoughts I had while out running today. This has to do with how to use congruences to talk about the degenerate cases.

Let a k-congruence be a congruence in a manifold with signature (k,1). We can embed a k-congruence in a manifold with spatial dimension m>k. We can also define a notion of extending such an embedding by adding more world-lines to the collection, such that every point on a world-line in the embedded congruence is surrounded by an open neighborhood where the m-congruence is defined.

Let's describe a k-congruence embedded in a higher-dimensional spacetime as intrinsically rigid (i-rigid) if the k-congruence is Born rigid, and say that it's extrinsically rigid (e-rigid) if there exists an extension of the embedding such that the extension is Born rigid. When m=3, I think e-rigidity holds exactly under the conditions given by the Herglotz-Noether theorem.

I decided I wanted more vivid names for the examples, so I'm now referring to the angularly accelerating ruler as the "baseball bat," and the wriggling filament as the "snake."

The bat and the snake are both i-rigid, but neither is e-rigid.

One thing that i-rigidity is sort of good enough for is giving a rigorous definition of the measuring rods that Einstein refers to all over the place in his early writings. I-rigidity gives you a measuring chain, not a measuring rod, but that's probably good enough. The usual way of handling this is to assume that the rod deforms when you accelerate it, but returns to its equilibrium shape when you stop accelerating it. This doesn't quite work in general, because it means that you can't use such a rod to measure some dimension of an object while that object is undergoing acceleration. Born-rigidity of a space-filling rod is too strict, because such a rod can't be put freely into any desired orientation -- in general, you can't even verify that one space-filling rod is the same length as another space-filling rod, because you may not be able to match their orientations without violating the H-N theorem.

The letter "C" with angular deceleration about its center isn't e-rigid for all time, but it's e-rigid everywhere in the past of some spacelike hypersurface that has in it the event where the tips intrude on one another. This is a reasonably useful way of dealing with the self-intersection issue within the framework of congruences.

There are some seemingly silly physical consequences. For example, say we have a zero-thickness disk that is rotating at constant angular velocity. If the disk's center accelerates in its own plane, we don't even need an embedding, and it's clearly not rigid, by the H-N theorem. If it accelerates along its own axis, it's i-rigid but not e-rigid. What this means is that an e-rigid object can have infinite inertia along some axes, but finite inertia along others -- i.e., its mass isn't even a scalar!

Neither i-rigidity nor e-rigidity is really the right physical notion. Basically i-rigidity is too permissive and e-rigidity too lax. We want to be able to distinguish between the snake and the bat. I think there is probably some other useful notion that can be defined, which I'll call n-rigidity, according to which the bat is n-rigid and the snake isn't. If we want to define such a thing using congruence methods, then the usual expansion, shear, and vorticity tensors exhaust all the possibilities that can be written down in terms of the velocity field and its derivative. However, in the case of an embedding we have a new vector available, which is the normal to the world-sheet (or more than one normal, if m is greater than k by 2 or more). I think it's probably possible to cook up something tensorial that describes n-rigidity by taking the contraction of the shear with the normal vector, or doing something with the covariant derivative of the normal vector, or something along those lines.

-Ben

 

[Chris Hillman]

Ben, as often happens with new ideas, your definitions are not yet precise, I think, and it is not yet clear (at least not to me) that your Post #31 makes sense either physically or mathematically. But if you are on to something valid, I'd like to know!

I appreciate that time is an issue, and clarifying/fixing these problems might take time, so I guess we'll all need to be patient... maybe pick this up next weekend?

To clarify what you really have in mind (both to me and to yourself), I think it would be really helpful for you to write down some valid explicit examples. Do you see why your alleged counterexample didn't even define a congruence?

(A timelike congruence consists of parameterized timelike curves which do not intersect in U such that precisely one curve passes through each event in U, i.e. they are the integral curves of a suitable timelike vector field on U. Sometimes it is convenient to assume the curves are proper time parameterized, and some authors make that assumption without spelling it out, so be careful.)

Re

www.physicsforums.com/showthread.php?t=430381

I tend to doubt you will get better answers from non-SA/Ms, but in any case, FWIW:

Another good reference for the definitions is section 6.1.1 in Stephani et al., Exact Solutions of Einstein's Field Equations, second ed., Cambridge University Press, 2003.

Am I right in thinking that Wald's reason for restricting to geodesic congruences is that under these circumstances he gets the simpler expressions shown above, rather than the more complex ones that Hawking gives?

Yes. But for charged particles in an EM field, or fluid elements in a fluid with nonzero pressure or stress, &c., the accelleration vector will usually vanish so you need the extra term.

The definition of the spatial metric would clearly have to have the + sign flipped if you were using the +--- signature (since the purpose of the term is to punch the time-time component out of the metric).

Right,
$$h_{ab} = g_{ab} + u_a \, u_b$$
is just the projection tensor, which might look more familiar as the projection operator
$${h^a}_b = {g^a}_b + u^a \, u_b = {\delta^a}_b + u^a \, u_b$$
which projects to the spatial hyperplane element orthogonal to $\vec{u}$. This is just linear algebra; Halmos, Finite Dimensional Vector Spaces, has a superb discussion of projections with and without a notion of orthogonality. Algebraically, projection operators are idempotent and in a euclidean inner product space they are represented by symmetric matrices, while in a Lorentzian inner product space, they are self-adjoint using the Lorentzian adjoint, which is slightly different from the euclidean adjoint.

Would any other signs have to be changed for +---, like the sign in the definition of the shear?

Unfortunately, various authors choose various signatures (but -+++ is best for comparing E^3 and E^{1,3}), various normalizations for the quadratic invariants of the shear and vorticity tensors, and some authors, such as Lauritzen, even flip the sign of the expansion tensor. So I hesitate to say "no" for fear of creating some further misunderstanding.

So let me say this: a good way of making a "reality check" that your sign conventions are not awful is to compute your quantities for a judiciously chosen suite of examples for which it is clear what the correct answers should be. If I ever get to the proposed BRS, I plan to provide such examples.

There are currently too many threads related to expansion/vorticity to follow easily, unfortunately, but regarding the issue implicitly raised in "differing definitions of expansion, shear, and vorticity"

www.physicsforums.com/showthread.php?t=430381

I plan to explain why Wald (and some other authors) can get away with dropping the projection tensor from their definitions, provided one applies them only to timelike geodesic congruences. But we want to consider non-geodesic congruences too (for example, the Langevin and Rindler congruences).

Gawk, I can see that trying to organize a comparision of notation in some standard books will be one of the things I'd have to do in the proposed BRS on timelike congruences.

 

[bcrowell]

Hi, Chris,

Re your #30, it looks like we're doing a great job of confusing one another!

Here you are, I think, claiming to give an example of a non-rigid congruence in Minkowski vacuum, which contains a "degenerate" configuration of world lines corresponding to a Born rigid object.

Right.

But what you wrote does not actually define a congruence of parameterized curves in Minkowski vacuum!

Here you lost me right away. I wrote x=x_oe^t, y=const, z=const. Are you saying this doesn't define a congruence at all?? Why not? My understanding of a congruence is that it's simply a set of nonintersecting, timelike world-lines whose union covers some open set of a manifold. I think mine satisfies all of those criteria on the open set defined by t<0. (For t>0, the world-lines become spacelike.) Are you saying that it doesn't satisfy all those criteria, or are you using some other definition of a congruence...?

To fix up the counterexample, you need to first of all define a congruence with nonzero expansion tensor, and then to find a one or two dimensional configuration of world lines belonging to the congruence which you can show are "rigid" according to some definition you'll need to explain.

The definition is the one I gave in #1. The one-dimensional configuration of world-lines is the one I gave in #27, "a line segment lying at rest in the y-z plane," as clarified in #29: "Say the ruler's world lines are of the form x=0, y=k, z=0, where $0 \le k \le 1$. Then for all k we have a world line of the form given above, with x_o=0."

And still seeking clarification: "degenerate" means a selection of world lines from a timelike congruence which can be interpreted as the "world sheet" of a one or two dimensional object, right?

Right.

-Ben

 

[Chris Hillman]

Almost there!; more later...

Hi, Ben, glad you haven't given up on me yet, since I think we are making progress.

Congruence: a family of nonintersecting smooth curves filling up some domain (open neighborhood) U in some smooth manifold M, i.e. through every point in U passes precisely one curve in the family.

(But I think you already know that.)

Very briefly: I now see that I was expecting you to write down proper time parameterized curves forming a congruence. Reason first: easiest way to define a congruence of (proper time parameterized) timelike curves is generally to write down a timelike unit vector field $\vec{X}$ and say "take the integral curves of $\vec{X}$". Particularly if you are looking ahead to extending said unit vector field to a frame field. Reason second: the formalism of the decomposition of a congruence of timelike curves works with timelike unit vector fields. Reason third: there is no reason third, I just got carried away.

However, it is true that one should always be able to convert a family of curves given by equations relating the coordinates into a family of proper time parameterized curves. (But I think you know that also.)

I plead exhaustion but I'll try to check your alleged counterexample more carefully when I have more energy.

<gentlehint>

If you can possibly install xfig, I find that pictures really help in preventing "preventable misunderstandings", which is particularly useful when participants are tired or rushed! You might have noticed that I have been trying to provide some sketches to illustrate my own BRS posts.

</gentlehint>

More later...

 

[George Jones]

I am going to abuse notation and suppress two dimensions for much of this post. I hope Chris doesn't faint.

Ben has defined a bunch of curves parametrized by $t$,

$$t\mapsto \left( t\left( t\right) ,x\left( t\right) \right) =\left( t,k\exp \left( t\right) \right).$$

The tangent vectors $V$ to events on these curves are given by

$$V=\frac{d}{dt}\left( t,k\exp \left( t\right) \right) =\left( 1,k\exp \left( t\right) \right) =\left( 1,x\right) ,$$

or

$$V=\frac{d}{dt}=\partial _{t}+x\partial _{x}.$$

where $x_{0}$ has been changed to $k$ to avoid confusion with index notation.

At each event $\left( t,k\exp \left( t\right) \right)$, draw a vector with components $\left( 1,k\exp \left( t\right) \right)$. Ben's curves are the integral curves of the vector field $V$.

Now find curves parametrized by proper time $\tau$ that have the same images as Ben's curves. First, normalize the vector field $V$, to find the tangent vectors (4-velocities) of the new curves, which will then be worldlines.

$$V^{a}V_{a}=-1+x^{2}=-1+k^{2}\exp \left( 2t\right) ,$$

where $0<x<1$ restricts to a region where $V$ is timelike. Define

$$X=\frac{V}{\sqrt{-V^{a}V_{a}}}=\left( \frac{1}{\sqrt{1-x^{2}}},\frac{k\exp \left( t\right) }{\sqrt{1-x^{2}}}\right) =\left( \frac{1}{\sqrt{1-x^{2}}},\frac{x}{\sqrt{1-x^{2}}}\right) ,$$

or

$$X=\frac{d}{d\tau }=\frac{dt}{d\tau }\frac{d}{dt}=\frac{dt}{d\tau }V=\frac{1}{\sqrt{1-x^{2}}}\left( \partial _{t}+x\partial _{x}\right).$$

Clearly, $X^{a}X_{a}=-1$.

Now find the integral curves of $X$; find curves

$$\tau \mapsto \left( t\left( \tau \right) ,x\left( \tau \right) \right)$$

that have $X$ as tangent vector,

$$\begin{equation*} \begin{split} X&=\left( \frac{dt}{d\tau },\frac{dx}{d\tau }\right) \\ \left( \frac{1}{\sqrt{1-x^{2}}},\frac{x}{\sqrt{1-x^{2}}}\right) &=\left( \frac{dt}{d\tau },\frac{dx}{d\tau }\right) , \end{split} \end{eqaution*}$$

so,

$$\frac{dt}{d\tau }=\frac{1}{\sqrt{1-x^{2}}},\qquad \frac{dx}{d\tau }=\frac{x}{\sqrt{1-x^{2}}}.$$

This gives

$$\frac{dx}{dt}=\frac{\frac{dx}{d\tau }}{\frac{dt}{d\tau }}=x,$$

and integrating

$$\int \frac{dx}{x}=\int dt$$

over appropriate intervals produces (the images of) Ben's original curves. Integrating

$$\frac{dx}{d\tau }=\frac{x}{\sqrt{1-x^{2}}}$$

will give $\tau$ as a function of $x$ (and thus as a function of [/itex]t[itex].

I should go and calculate an orthonormal frame, 4-acceleration, vorticity, expansion, but I'm a little bit winded, and I have to do some work that puts bread on the table.