
Original by Michael Weiss The Rigid Rotating Disk in RelativityThis old chestnut predates GR. The trail starts with Max Born's notion of an "SR rigid body", a relativistic replacement for the classical notion. It leads past Einstein's discussion in his great 1916 paper on GR, where he uses the rotating disk to introduce nonEuclidean geometry. It then twists and turns as Eddington, Lorentz, and lesser lights attempt to compute the fate of the rigid disk. In this entry, I summarize what I know of the literature, and invite others to fill in the gaps. Surely such a celebrated problem should have found a definitive resolution by now. But the tale I have to tell ends on an incomplete note. We look at SR first, then GR. Special RelativityBorn rigidity: in 1909, Born proposed a Lorentzinvariant definition of "rigid body". Pauli's monograph on relativity [1] gives a nice summary of Born's notion, and the responses it drew from Ehrenfest, Herglotz, Noether, and von Laue. (Pais's Einstein bio suggests that Born's 1909 paper may have helped set Einstein on the road to Riemannian geometry [2].) We know already that rigidity and SR don't mix just think of the Barn and the Pole (see below)! How could a physicist like Born, mathematically sophisticated, have made such an elementary error? A simple remark by Pauli clarifies things considerably:
Born thought he was defining a rigid body, but Pauli's rephrasing saves the mathematics while improving the physics. We have no rigid rods in SR, but if you accelerate every atom of an ordinary rod in just the right way, you can move the rod rigidly. And Born's definition is Lorentzinvariant. I won't plunge right into Born's definition (as Pauli does). Instead I'll approach it by thinking atomistically. Imagine our solid as made up of a large number of atoms A_1,...,A_n; between any two nearby atoms A_i and A_j there is a "natural distance" d_ij, natural in the sense that if A_i and A_j are pushed together or pulled apart, stresses result, trying to restore the distance d_ij. Of course there are propagation delays, but if we start with the solid at rest in some inertial frame, and accelerate it gently, the resulting elastic waves in the solid should die out pretty quickly. Or we can pretend that exactly the right force is applied to each atom at all times, so that natural distances are preserved. Let the number of atoms tend to infinity (continuum approximation), let the stress/strain ratio tend to infinity, and apply forces gently enough so that the elastic waves can be ignored this leads to Born's definition. Born used coordinates, but I'll try for a coordinatefree rephrasing. First, what is a solid? Is it just a swath in spacetime? This is not enough: the atomistic viewpoint suggests we should be able to "mark" a point inside the solid (call it a particle) and follow its worldline. An "event", as usual, is a point in spacetime. The events along a particle's worldline are parametrized by tau, the timelike interval.
If you don't like the notion of "infinitesimally close", there are ways to get around it, but I won't go into that. (The basic idea is that we are using particle worldlines to transport the metric from one tangent plane to another.) OK, now what? First, it should be pretty clear that Born's definition captures the idea that the body moves without internal stresses. Or if you prefer, you can say that we have nearly rigid motion when Young's modulus is large enough, and the accelerations are gentle enough, so that infinitesimal pieces of the body are barely deformed, when viewed in the comoving frame of reference. Bornrigidity is then the limiting case. If we accelerate a rod rigidly in the longitudinal direction, then the rod suffers the usual Lorentz contraction. More generally, rigid motion without any twisting corresponds to socalled FermiWalker transport (see MTW [3]). The acceleration of the front of the rod is less than the acceleration of the rear; this is a variation on Bell's Spaceship Paradox (see [4]; FAQ entry coming soon.) Ehrenfest noted that a disk cannot be brought from rest into a state of rotation without violating Born's condition. Integrating tau out of Born's condition, we see that infinitesimally close particles must keep the same proper distance. So in the original rest frame, they suffer Lorentz contraction in the transverse direction but none in the radial direction. The circumference contracts but the radius doesn't. But in the original rest frame, the circumference is a circle, sitting in a spatial slice (t=constant) of ordinary flat Minkowski spacetime. In other words, we would have a "non_Euclidean circle" sitting in ordinary Euclidean space. This is a contradiction. The issue of spatial slices deserves a few words. The particles in a rotating disk (not assumed rigid) cannot agree on a global notion of simultaneity. For if you make a circuit around the edge, joining up the infinitesimal planes of simultaneity, when you return to your starting point, the planes no longer match up. This makes it problematical to talk about geometry "as seen by the particles" (or by observers standing on the disk). I've talked about making complete circuits about the center, but both Ehrenfest's argument and the simultaneity problem have local versions. Say we have a ballbearing a lightyear from the Sun. We cannot put the ballbearing in orbit around the Sun, keeping one face to the Sun, without violating Born's condition. Nor can the particles of the ballbearing agree on a notion of simultaneity. The proofs are not hard. I'll sketch the local version of Ehrenfest's argument. Take a spatial slice in the rest frame and look at the metric. The simplest way to express it is to use polar coordinates inherited from when the ballbearing was at rest. Each particle was then labelled with coordinates (r,theta), and we can use these to label its whole worldline, and thus also the points in the spatial slice. The same "transverse vs. radial" argument that Ehrenfest used shows that: ds^2 = dr^2 + (1  r^2 w^2)r^2 (d theta)^2 where w is angular velocity. A routine computation shows that the curvature is nonzero. This contradicts the fact that the spatial slice is Euclidean space. Pauli states: "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." I haven't looked up the papers by Herglotz and Noether, but I dare say it's similar to the argument above. Max von Laue pointed out that a body made up of n pointparticles must have at least 3n degrees of freedom. Say we give an impulse to each particle at t=0. Because of the finite speed of light, there can be no constraints relating the velocities of different particles. Rigid motion can occur in SR only through a conspiracy of forces. So much for the rotating rigid disk in SR.
General RelativityEinstein's 1916 paper on GR [5] makes no mention of elevators; instead, the Equivalence Principle is introduced via the rotating disk. Einstein reproduces Ehrenfest's argument, but with a different conclusion: since we are no longer assuming flat Minkowski space, Einstein asserts that geometry for the rigid rotating disk is nonEuclidean. The Equivalence Principle now implies that geometry in a gravitational field will also be nonEuclidean. (By "geometry", I mean spatial geometry, i.e., we're not concerned with the temporal components of the spacetime metric.) Can we make any sense of Einstein's argument? The simplest interpretation makes a couple of assumptions:
Assumption (2) seems a lot more dubious than (1), but it does allow us to talk about spatial slices (t=constant), and hence the geometry of the spinning disk. We appeal to axialsymmetry and steadystate conditions in making f,g, and h independent of theta and t. We have no such justification for leaving out z. Einstein doesn't give an explicit formula for the spacetime metric of the rigid spinning disk, but here's one obvious candidate (assuming the angular speed is w): (d tau)^2 = dt^2  dz^2  dr^2  r^2(1w^2 r^2)(d theta)^2 Turn to GR. Now all sorts of complications appear.
To settle the question definitively, it seems one has to perform a fullblown, hairy GR calculation. Perhaps someone has done this; perhaps someone has turned the vague notion of "infinitely rigid" into a formula for a stressenergy tensor, plugged that into the Einstein field equations, and solved. If the Gentle Reader knows of a reference, please let me know. [1] Wolfgang Pauli, "Theory of Relativity", pp. 130134, Pergamon Press, 1958. [2] Abraham Pais, "Subtle is the Lord: the Science and Life of Albert Einstein", p.???. [3] Misner, Thorne, and Wheeler, "Gravitation", pp.???. [4] J.S.Bell, "Speakable and Unspeakable in Quantum Mechanics", p.67, Cambridge University Press, 1987. [5] "The Principle of Relativity", Dover. [6] G. Cavalleri, Nuovo Cimento 53B pg. 415. [7] O. Gron, AJP Vol. 43 No. 10 pg 869 (1975) [8] C. Berenda Phys. Rev. 62 pg. 280 (1942)
updated 4AUG1992 by SIC; original by Robert Firth A Special Relativity Paradox: The Barn and the PoleThese are the props. You own a barn, 40m long, with automatic doors at either end, that can be opened and closed simultaneously by a switch. You also have a pole, 80m long, which of course won't fit in the barn. Now someone takes the pole and tries to run (at nearly the speed of light) through the barn with the pole horizontal. Special Relativity (SR) says that a moving object is contracted in the direction of motion: this is called the Lorentz Contraction. So, if the pole is set in motion lengthwise, then it will contract in the reference frame of a stationary observer. You are that observer, sitting on the barn roof. You see the pole coming towards you, and it has contracted to a bit less than 40m, in your reference frame. (Does it actually look shorter to you? See Can You See the LorentzFitzgerald Contraction?(below) for the surprising answer. But in any case, you would measure its length as a bit less than 40m.) So, as the pole passes through the barn, there is an instant when it is completely within the barn. At that instant, you close both doors simultaneously, with your switch. Of course, you open them again pretty quickly, but at least momentarily you had the contracted pole shut up in your barn. The runner emerges from the far door unscathed. But consider the problem from the point of view of the runner. She will regard the pole as stationary, and the barn as approaching at high speed. In this reference frame, the pole is still 80m long, and the barn is less than 20 meters long. Surely the runner is in trouble if the doors close while she is inside. The pole is sure to get caught. Well does the pole get caught in the door or doesn't it? You can't have it both ways. This is the "Barnpole paradox." The answer is buried in the misuse of the word "simultaneously" back in the first sentence of the story. In SR, that events separated in space that appear simultaneous in one frame of reference need not appear simultaneous in another frame of reference. The closing doors are two such separate events. SR explains that the two doors are never closed at the same time in the runner's frame of reference. So there is always room for the pole. In fact, the Lorentz transformation for time is t'=(tv*x/c^2)/sqrt(1v^2/c^2). It's the v*x term in the numerator that causes the mischief here. In the runner's frame the further event (larger x) happens earlier. The far door is closed first. It opens before she gets there, and the near door closes behind her. Safe again  either way you look at it, provided you remember that simultaneity is not a constant of physics. References: Taylor and Wheeler's Spacetime Physics is the classic. Feynman's Lectures are interesting as well. 12Oct1995 by Michael Weiss Can You See the LorentzFitzgerald Contraction? Or: PenroseTerrell RotationPeople sometimes argue over whether the LorentzFitzgerald contraction is "real" or not. That's a topic for another FAQ entry, but here's a short answer: the contraction can be measured, but the measurement is framedependent. Whether that makes it "real" or not has more to do with your choice of words than the physics. Here we ask a subtly different question. If you take a snapshot of a rapidly moving object, will it look flattened when you develop the film? What is the difference between measuring and photographing? Isn't seeing believing? Not always! When you take a snapshot, you capture the lightrays that hit the film at one instant (in the reference frame of the film). These rays may have left the object at different instants; if the object is moving with respect to the film, then the photograph may give a distorted picture. (Strictly speaking snapshots aren't instantaneous, but we're idealizing.) Oddly enough, though Einstein published his famous relativity paper in 1905, and Fitzgerald proposed his contraction several years earlier, no one seems to have asked this question until the late '50s. Then Roger Penrose and James Terrell independently discovered that the object will not appear flattened [1,2]. People sometimes say that the object appears rotated, so this effect is called the PenroseTerrell rotation. Calling it a rotation can be a bit confusing though. Rotating an object brings its backside into view, but it's hard to see how a contraction could do that. Among other things, this entry will try to explain in just what sense the PenroseTerrell effect is a "rotation". It will clarify matters to imagine two snapshots of the same object, taken by two cameras moving uniformly with respect to each other. We'll call them his camera and her camera. The cameras pass through each other at the origin at t=0, when they take their two snapshots. Say that the object is at rest with respect to his camera, and moving with respect to hers. By analysing the process of taking a snapshot, the meaning of "rotation" will become clearer. How should we think of a snapshot? Here's one way: consider a pinhole camera. (Just one camera, for the moment.) The pinhole is located at the origin, and the film occupies a patch on a sphere surrounding the origin. We'll ignore all technical difficulties(!), and pretend that the camera takes full spherical pictures: the film occupies the entire sphere. We need more than just a pinhole and film, though: we also need a shutter. At t=0, the shutter snaps open for an instant to let the lightrays through the pinhole; these spread out in all directions, and at t=1 (in the restframe of the camera) paint a picture on the spherical film. Let's call points in the snapshot pixels. Each pixel gets its color due to an event, namely a lightray hitting the sphere at t=1. Now let's consider his & her cameras, as we said before. We'll use t for his time, and t' for hers. At t=t'=0, the two pinholes coincide at the origin, the two shutters snap simultaneously, and the light rays spread out. At t=1 for his camera, they paint his pixels; at t'=1 for her camera, they paint hers. So the definition of a snapshot is framedependent. But you already knew that. (Pop quiz: what shape does he think her film has? Not spherical!) (More technical difficulties: the rays have to pass right through one film to hit the other.) So there's a oneone correspondence between pixels in the two snapshots. Two pixels correspond if they are painted by the same lightray. You can see now that her snapshot is just a distortion of his (and vice versa). You could take his snapshot, scan it into a computer, run an algorithm to move the pixels around, and print out hers. So what does the pixel mapping look like? Simple: if we put the usual latitude/longitude grid on the spheres, chosen so that the relative motion is along the northsouth axis, then each pixel slides up towards the north pole along a line of longitude. (Or down towards the south pole, depending on various choices I haven't specified.) This should ring a bell if you know about the aberration of light: if our snapshots portray the nightsky, then the stars are white pixels, and aberration changes their apparent positions. Now let's consider the object let's say a galaxy. In passing from his snapshot to hers, the image of the galaxy slides up the sphere, keeping the same face to us. In this sense, it has rotated. Its apparent size will also change, but not its shape (to a first approximation). The mathematical details are beautiful, but best left to the textbooks [3,4]. Just to entice you if you have the background: if we regard the two spheres as Riemann spheres, then the pixel mapping is given by a fractional linear transformation. Wellknown facts from complex analysis now tell us two things. First, circles go to circles under the pixel mapping, so a sphere will always photograph as a sphere. Second, shapes of objects are preserved in the infinitesimally small limit. (If you know about the doublecovering of SL(2), that also comes into play. [3] is a good reference.)
References[1] and [2] are the original articles. [3] and [4] are textbook treatments. [5] has beautiful computergenerated pictures of the PenroseTerrell rotation. The authors of [5] later made a video [6] of this and other effects of "SR photography".
Other categories will be implemented upon request.

