Refer, "The classical theory of Fields" by Landau&Lifshitz (Chap 3). Consider a disk of radius R, then circumference is $2 \pi R$. Now, make this disk rotate at velocity of the order of c(speed of light). Since velocity is perpendicular to radius vector, Radius does not change according to the observer at rest. But the length vector at boundary of disk, parallel to velocity vector will experience length contraction . Thus, $\dfrac{\text{radius}}{\text{circumference}}>\dfrac{1}{2\pi}$ , when disc is rotating. But this violates rules of Euclidean geometry.

What is wrong here?

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    $\begingroup$ Closely related $\endgroup$ – Danu Aug 20 '15 at 11:33
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    $\begingroup$ Google Ehrenfest paradox $\endgroup$ – John Rennie Aug 20 '15 at 14:58
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    $\begingroup$ Do you need the disk? Is not the same problem that asking for the existence and properties of a circular orbit? $\endgroup$ – arivero Aug 20 '15 at 16:39
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    $\begingroup$ Alejandro, it's clearly not quite the same because the planet on the orbit may rather easily shrink in the direction of motion and no one cares whether it occupies a smaller fraction of the circumference than it did at rest. The problem with the disk is that the proper length of the circumference seems to be "prescribed" by the character of the object, and when this should get Lorentz-contracted at the same moment, a paradox follows. The resolution of the paradox is that no perfectly rigid objects are allowed by relativity. $\endgroup$ – Luboš Motl Aug 20 '15 at 17:19
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    $\begingroup$ That's not the resolution. Arivero, you, and I can zoom past a planet from orthogonal directions. Arivero claims it's length-contracted in the X direction, you claim it's length-contracted in the Y direction, and I claim it's length contracted in the Z direction. We can't all be right. $\endgroup$ – John Duffield Aug 22 '15 at 11:38

What is wrong is the idea that one can actually make the disk rotate; and it will remain perfectly rigid.

In reality, what this correct argument shows is that relativity doesn't admit the existence of any perfectly rigid bodies. This is a perfectly basic, settled, and indisputable textbook material that every mature physicist knows. The first sentence of this paragraph contains a link to the Gravity Probe B website. The thought experiment is known as the Ehrenfest paradox and Ehrenfest himself already offered the right basic answer – no rigid objects exist in relativity – when he outlined the thought experiment in 1909.

When one takes a solid disk and makes it rotate, it will do all kinds of things resulting from the "imperfection of the material". It will tear apart by the centrifugal force, and if it won't, it will either tear basically along radial lines, or it will bend (the disk won't be planar anymore) because the circumference really shrinks by the Lorentz factor. If there existed a material that is perfectly rigid and cannot stretch or bend or tear, then it would be impossible to make it spin. In any world governed by relativity, the proper distances between the individual points/atoms of the objects simply have to change when the object is brought to motion. (The definition of rigidity using the constant proper distances between points/atoms of the object was given by Max Born in 1909 and is known as the Born rigidity.)

However, the non-existence of such a material may be shown even microscopically. It is not possible to "order" any solid object to keep the proper distances at every moment because the distance between two atoms (or points on the solid object) may only be measured with a delay $\Delta t = \Delta x / c$ simply because no information may move faster than light. That's why it's always possible to squeeze any rod on one end and the opposite end of the rod won't move at least for this $\Delta t = \Delta x / c$. This relationship between the "limited speed of signals by $c$" and "non-existence of rigid objects in relativity" was already pointed out by Max von Laue in 1911.

In fact, the delay will be much larger than that, dictated basically by the speed of sound, not by the speed of light. Whatever material you have, relativity guarantees that it can be squeezed as well as stretched as well as bent.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – David Z Aug 23 '15 at 1:01
  • $\begingroup$ Born Rigid objects won't exist.This is implied by relativity of simultaneity(which if would've be the only factor could've been overcome by appropriate material) and the finite speed of light(which is in principle, not overcomable), but the effect of failure of 'pi' is a different one than this. Your answer even though correct, I think is Pedagogically a bit unclear. The failure of $\pi\neq3.14$ is a real thing depicted in this answer here physics.stackexchange.com/questions/8659/…, and has to do with Lorentz Contraction $\endgroup$ – Viesr Dec 19 '15 at 1:03
  • $\begingroup$ Sorry, Viesr, but $\pi=3.14159...$ is a mathematical identity that is always valid, whether one talks about physics or not, and whatever model of physics one chooses. $\pi$ is the circumference/diameter ratio of a circle - we mean a real idealized circle in the Euclidean flat geometry. In physics of GR, we need other geometries but we don't generalize the geometry by redefining concepts like pi. We still need the constant pi in general relativity, with the same value. Instead, we appreciate that the space isn't Euclidean and most statements valid in flat space are no longer valid in GR. $\endgroup$ – Luboš Motl Dec 19 '15 at 8:44
  • $\begingroup$ The decision whether $\pi=3.14...$ is not a matter of pedagogy. It's a matter of basic mathematics and the answer is Yes, it's always the value of pi. There doesn't exist any remotely meaningful way to describe general relativity that would distort the value of pi. $\endgroup$ – Luboš Motl Dec 19 '15 at 8:45
  • $\begingroup$ I am not saying at all that your answer is wrong. It is actually 100% percent correct, these stresses inside the circle will appear, but it isn't true that pi=3.14 in all cases. One can easily see that it isn't true by considering a particle moving in a straight line, and at an instant write down its displacement as $rd\theta$ and divide it by r, to see that the two observers won't agree on their values of $d\theta$, and there is nothing special about $\pi$ over $d\theta$. It is true $\pi=3.14..$ if you have a circe which is at rest w.r.t. you $\endgroup$ – Viesr Dec 20 '15 at 20:19

No, it doesn't violate the rules of geometry, it violates the rules of Euclidean geometry. Simple conclusion: for an observer fixed to a disk rotating uniformly relative to an inertial frame, the spatial geometry is non-Euclidean; in particular, the ratio of a circle's circumference to its diameter depends both on the circle's diameter and center position. There is no longer a simple notion of $\pi$ for such an observer. The geometry is approximately that of the hyperbolic plane, as claimed by Kaluza (who made the unsubstantiated claim that it was exactly that of the hyperbolic plane). In short, there's nothing wrong with your reasoning: it's spot on.

This is actually the thought experiment of the famous Ehrenfest Paradox and I discuss it further in my answer here.

As in Lubos's Answer, it is impossible for the disk to stay rigid as its angular velocity rises from nought to its steady state. A rigid body (in the sense of something that moves by Euclidean isometries) is a concept that is utterly incompatible with either special or general relativity because we cannot accelerate a rigid body of nonzero extent: if we shove it at one end, the other end a distance $L$ away cannot begin to move until at least a time $L/c$ later without violating special relativity. The disk must end up in a strained state in its steady state to conform to the geometry demanded of it (as reasoned in my other answer), otherwise it will shatter or warp.

In such problems, at steady state (e.g. constant angular speed for the disk) a weaker, more generalized notion of Born Rigidity replaces the notion of a rigid body that can undergo Euclidean isometries.

For similar reasons, note that the radial co-ordinate $r$ in the Schwarzschild Metric marks the radial position where a circle centered on the origin has a circumference of $2\,\pi\,r$. This is by definition and is different from the notion of radius that one would get by taking a reference point at some "radius" $R_0$ (as measured by the Schwarzscild $r$ co-ordinate) and measuring out a radial distance $\Delta r$ (according to your local measuring instruments): the difference between the circumferences of concentric circles passing though the two points would not be $2\,\pi\,\Delta r$ - the "radiusses" don't add and subtract linearly and stay related to the corresponding circumference by the constant $2\,\pi$. The Schwarzschild $r$ co-ordinate of the point you reach by beginning at $R_0$ and radially walking a distance $\Delta R$ is different from $R_0+\Delta R$. Again, there is no simple notion of circumference to diameter ratio: the ratio depends on the circle's radius and center position.

  • $\begingroup$ Newtonian rigidity isn't impossible because of how bodies interact. Newtonian rigidity is not a relativistic invariant. So it doesn't make sense as a property of any system. $\endgroup$ – Timaeus Aug 20 '15 at 14:37
  • $\begingroup$ @Timaeus I think I'm saying the same thing from a different angle: rather than saying that rigidity (property of a body) isn't a meaningful relativistic invariant, I'm saying that motion by Euclidean isometry is incompatible with STR, unless everything is at steady state. $\endgroup$ – Selene Routley Aug 20 '15 at 14:49
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    $\begingroup$ But things can move in a way that seems Newtonian rigid to one observer, it just won't seem rigid to other observers. And that is 100% the issue going on in this example. If you are moving as a Newtonian rigid object to a central observer you won't be Newtonian rigid according to other observers. You make it sound like it isn't possible to exert forces on a disk to make it rigidly move according to some observer. As long as the edge goes at sunlight speed it is possible. Making it sound like you can't do that makes it seem magical. $\endgroup$ – Timaeus Aug 20 '15 at 14:54
  • $\begingroup$ @Timaeus Perhaps I'm not being clear enough: the problem is when the disk's angular speed is increasing: not when it is steadily rotating: yes the motions of the points on the disk from a central observer will seem to rigidly rotate at steady state. Another way to say this is that the speed of sound in a material cannot be infinite - this is what conveys the motion to everything as it spins up to speed. $\endgroup$ – Selene Routley Aug 20 '15 at 15:08
  • $\begingroup$ I read the question as very very general. The phrase "make this disk rotate at velocity of the order of c" leaves open many ways to get it up to speed. I could shoot lasers at it from above and below to that each part gets an outwards acceleration of exactly what I want when I want it, or through balls, same thing. I can get it up to speed in any way I want. If the disk has bonds instead of being a disk of dust then there is a question about what happens if or when I stop making it do whatever I want. But I could keep making it go at a steady speed if I want. $\endgroup$ – Timaeus Aug 20 '15 at 15:37

This is just relativity of simultaneity again. A similar thing happens if you have a bunch of spaceships in a line that fire their thrusters at a fixed time. Different observers will disagree about whether they fired at the same time and will disagree about the spacing. Always in a consistent way.

So I'd like to address the concept of geometry by not having a disk. Imagine a large section of empty space. Then position one ship in the center of a ring of other ships. All at rest. All synchronize their clocks. All ships are identical.

They could make a ring of radius one lightyear. And they could be spaced out equally each about 100m apart. Right now everything is at rest and nobody is disagreeing about anything.

Since the clocks are synchronized they could sit and wait for a few years to confirm that everyone was in position and then as per prearranged plans the ships in the ring could fire rockets to move in a circle until they get to a predetermined speed (revolutions per second) and if you think they can't determine their revolutions per second we could have labeled buoys around the circle so they know which buoys they are passing as they pass. Or they can just follow the agreed thrust pattern and it can be a plan that just leads to the observer at the center saying they are going at a fixed speed.

When they get to the final speed they change to firing to make them go in a circle at a steady rate (buoys per proper time, revolutions per second, or just fire thruster as programmed so that central observer says it is at constant speed).

So they originally thought they were equally spaced around the circle about 100 meters apart and that they fired their thrusters at the same time at first. But as they start to speed up the comoving frames of the ships in the ring no longer agree that the ships are making their moves at the same time. They think the the ship that originally was farthest away for instance is executing their maneuvers in slow motion for instance.

The central observer is inertial and has every ring ship executing equal motions with each firing their rockets outwards at the same rate and orbiting at the same angular velocity.

The ring ships are not inertial. They could use radar time and radar distance to not have discontinuous changes in how they measure distances to and times of separated events, and those reduce to inertial coordinates when they are move inertially. Or at each instant we can consider the instantaneously comoving inertial frame. But the point is those frames for the ring ships don't observe the other ships as using their rockets the same way. They agree that the proper time on the ships as recorded by the onboard ships has the same readings when the controls on the engines are at the same settings. They agree that the subjective experience of each ship is the same as theirs (except for the labels of the buoys if you use those). But what they observe according to the instantaneously comoving frame doesn't record the other ships doing the same things at the same time (time according to the instantaneously comoving frame).

This is not different than the ships in a line that accelerate at a fixed rate schedule according to onboard clocks (say firing one rocket for the first minute then two then three and then two then just one rocket then none). In the line version they observe the ones behind accelerating later and the ones in front as accelerating early. This isn't a change in geometry. This is each ship going on a curved path and each one having a comoving inertial frame that has a plane a simultaneity that intersects the world lines of the other ships at different points of their journey.

Nothing weird is going on. If your comoving inertial frame observes your neighbor ships as firing their rockets differently than you fire yours then of course that frame computes the distance between events in that hyperplane of simultaneity to be different.

But there is no change in geometry. There is a flat Minkowski geometry. The ring ships are not moving inertially and the comoving inertial frames do not observe the other ships going in a circle even though the inertial central observer does.

You can do the acceleration up to speed in one giant impulse and then there are two comoving frames for that point. The pre impulse frame and the post impulse frame.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. (However I've deleted a few comments that were inappropriate for various reasons, as well as some replies to them.) $\endgroup$ – David Z Aug 23 '15 at 1:00

To someone moving on the edge of the disk, the disk is length contracted in the direction of travel. So to them, it is not a circle, no problems with geometry.

Euclidean geometry only holds for an inertial plane of simultaneity. By the end of this answer you should know when and how to use Euclidean geometry and what it means (or doesn't). And a collection of objects that together collectively looks like a uniformly rotating disk in one frame won't look like it in a different inertial frame. So it isn't a disk in the another frame.

Imagine the worldlines of the parts of a rotating disk. Draw time as the vertical axis. In the frame of the center the center worldline is vertical. In the frame of the center the other parts go in helixes. In the frame of the center is possible to have the parts all go in a circle (this requires forces to be applied to the parts, but that is how real disks rotate too).

Now in an inertially moving frame the hyperplane of simultaneity is tilted so that the spatial origin of the frame makes an equal angle with the 45 degree light cone as the observers wordline does. And it is vectors in this plane of simultaneity where euclidean geometry holds. (Minkowksi geometry holds everywhere, but it reduces to Euclidean geometry in a spacelike plane of simultaneity.)

In such a frame that is instantaneously comoving with some part of the disk the disk is an object that looks in the drawing like a stretched circle (we will get to the geometry later right now I want you to see which events we are talking a out in spacetime). The intersection of the worldlines of the object with the hyperplane of simultaneity of the frame is like a cylinder cut with a plane through the center of the cylinder. Only one plane gives a circular cross section, that is the frame of the center of rotation.

Now you know where Euclidean geometry applies and what it applies to. You can see from the spacetime diagram that these little vectors are all in some plane of simultaneity. And they trace out different events for different planes.

If instead of going straight to using euclidean geometry which only holds for events in some hyperplane of simultaneity for some inertial frame you could talk about spacetime geometry. This is always for events. For instance the world line of a particle is made up events so you can find the proper time of such a curved between two events. Or you can talk about the curve of spacelike separated points in some plane of simultaneity and find their proper length. And then you get that euclidean circumference we talked about. But not for a circle unless you pick the one frame where it is a circle.

But if you compare these spacelike curves in a plane of simultaneity they are clearly different events for different frames hence different curves so there is no reason to expect them to be the same in circumference. We are developing intuition for the correct Minkowski geometry to know when and if we can use Euclidean geometry.

For different points on the edge of the disk you get different planes of simultaneity each one making an equal angle with the the light cone there as the worldline makes with the light cone at that event. So each point on the edge and each time generates a different comoving frame.

I promised to talk geometry and you asked about length contraction.

If you have a rod at rest then its worldlines go straight up but to a moving observe the ends of the rod are the intersection of the worldlines of the ends with a hyperplane of simultaneity. The line segment might look longer in the drawing. But it is shorter metrically. That's because when we say the plane of simultaneity is Eeuclidean it means there are three independent vectors in the plane that are all positive unit length and they span the plane. For their point of view everything is euclidean. But we have to know which three vectors are the unit orthogonal vectors.

If the unit vectors in on plane (say, the xy plane) point in two directions then we can draw time as the z axis. Then to a frame moving in the x direction the unit y vector is still a unit spatial vector but now a vector that to the first frame looks like it points to the future and in the x direction is now a spatial vector (lives in the hyperplane of simultaneity). And it is a vector that looks longer in the original drawing that has unit length.

So the point is that spacelike vectors that look quite long in one frame can be unit length. The spacelike vectors that are unit length can be written as having tails at the origin and heads at $(\sqrt{x^2+y^2+z^2-1},x,y,z),$ which looks like a hyperboloid. The point is those vectors that in the drawing are close to the light cone are still unit length even though they look long.

In the frame comoving with a point on the edge the direction that looks elongated when you drew everything in the frame of the center is actually closer together. This is because those events if you trace them along the hyperboloid of same size spacelike vectors from the center of the disk end up smaller than the spacelike vectors going to the other points and the ones of largest size are just normal size the ones that didn't change as you switched frames.

So if you are moving in the x direction then your disk is the same length from end to end from $R\hat y$ of the disk to the $-R\hat y$ end of the disk but in the other direction it is shorter. This is for the frame comoving with that one point of the edge for that one instant.

But in that frame the object that is a disk on one frame isn't a disk. So you don't expect a formula for a circle to hold in another frame when you aren't applying it to a circle. And you are measuring different events anyway.

  • $\begingroup$ I find your other answer far clearer, and maybe you should focus on tweaking it, rather than this one which is too difficult for a beginner to take in on first reading. $\endgroup$ – John McAndrew Aug 22 '15 at 19:09
  • $\begingroup$ I tried tweaking the other one and this is what I got and it didn't seem better so I didn't want to hide the original one. This one describes in more detail how to find planes of simultaneity. And the first paragraph is really really clear. $\endgroup$ – Timaeus Aug 22 '15 at 19:11

I am tempted to give an experimentalist's point of view, though of course I agree with Lubos and Savanah's answers.

Consider a disk of radius R, then circumference is 2πR. Now, make this disk rotate at velocity of the order of c(speed of light).

Here it is evident that you are defining a center of mass system for a disk and a person/machinery that will give it an angular acceleration to reach a speed close to the speed of light. Thus you are implicitly defining a classical rigid body.

But the underlying level of nature is quantum mechanical. At this level each atom is attached to another atom/molecule through electromagnetic forces that fulfill quantum mechanics equations, not classical mechanics ones.

An analogue is as if all those atoms/molecules are attached to each other with springs which at low accelerations (acceleration transfers forces electromagnetically at the atomic level, hence the speed of light limit) and at classical distances and velocities are rigid and thus we have the classical rigid bodies , but when accelerations grow high no rigid body modeling can work. A classical model of a disk with springs would show deformations of its shape too.

Thus one cannot apply Euclidean geometry to a system where it is even hard to define a center of mass against which special relativity transformations can be applied. At each radius different forces obtain and no common center of mass system can be defined where the disk shape will be circular, as at each radius different forces are applied and the rotational deformation is inevitable.

  • $\begingroup$ If I understand your answer, the above answer by Selene Routley, and physics.stackexchange.com/questions/147505/… all suggest that the geometry is non Euclidean. I have a question here. Just changing coordinates can’t change the geometry of the manifold. If we do SR in an accelerating frame, we do get a pseudo gravity but the whole manifold is still flat, that is it’s Reimann tensor vanishes. $\endgroup$ – Shashaank Mar 4 at 7:47
  • $\begingroup$ Similarly here,wouldn’t the whole metric, including 𝑔𝑡𝜇 be still of flat space time, Minkowski space time? Wouldn’t the Reimann tensor vanish just like it does for the metric written in an accelerating coordinates ( hyperbolic motion) in flat space time ? The whole space time would still be flat, Minkowskian,right? Just changing coordinates can’t change the flat Minkowskian space time to a curved one? $\endgroup$ – Shashaank Mar 4 at 7:48
  • $\begingroup$ @Shashaank I am arguing that there is no one inertial frame to describe the disk rotating at velocity of light, as it is not a rigid body. Every atom with respect to the other atoms defines a different system. $\endgroup$ – anna v Mar 4 at 8:54
  • $\begingroup$ ok, but in the other answer, and specifically the one I linked in my last comment, are they saying that by changing to the rotating coordinates the whole space time manifold becomes curved or only the spatial submanifold becomes ( only the Reimann tensor of the 3 dimensional sub manifold doesn’t vanish) curved ( while the whole space time is still flat)? $\endgroup$ – Shashaank Mar 4 at 9:06
  • $\begingroup$ @Shashaank They are talking in four vectors, one has to in special and general relativity. $\endgroup$ – anna v Mar 4 at 9:12

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