Rotating Disk Physics: SR Reconciles Fast-moving Points?

[soothsayer]

Imagine we have a spinning disk. Points near the center of the disk rotate slowly while those increasingly far out are moving faster. Now suppose we have a sufficiently large disk such that the points on the edge of it are moving relativistically, and would classically be reaching or surpassing the speed of light. How does SR reconcile this?

 

[Fredrik]

You obviously can't build a disk that's already rotating that fast, so you'd have to build a stationary disk, or one that's rotating slowly, and then try to give it a spin. If you use SR to calculate the work it would take to give the edge speed v, you would find that the result goes to infinity as v goes to c.

 

[jtbell]

Assuming the disk is rigid enough so it can withstand the internal stresses, it has a limiting maximum angular speed:

$$\omega_{max} = \frac {c}{R}$$

where R is the radius of the disk. It can never actually reach this angular speed. As you make it spin faster and faster, the work you need to do increases without limit as Fredrik noted, and the internal stresses increase without limit. This is the ideal case of course. In practice the internal stresses tear the disk apart at some point.

 

[soothsayer]

Fredrik, I understand it's impossible to get the disk to rotate at speed c, what I'm asking though, is that after some non-infinite work, eventually, the disk would reach a relativistic speed, like 0.2 c? or even 0.9 c. The outer points move relativistically while the inner points are not. So how would a phenomena like length contraction work when the disk is supposed to be a rigid object? I realize this is all theoretical in the first place.

jtbell, thanks for that, that explains a lot!

 

[Ben Niehoff]

So how would a phenomena like length contraction work when the disk is supposed to be a rigid object?

The answer is that the idea of a "rigid object" is inconsistent with relativity. Or to put it another way, there are no such things as rigid objects in relativity, even in theory.

 

[jtbell]

So how would a phenomena like length contraction work when the disk is supposed to be a rigid object?

Consider a linear length-contraction scenario first, for simplicity. A rod with rest-length 10 m traveling at speed 0.6c, normally gets contracted to 8 m, right?

Suppose the rod is made of rubber, and is being carried by two runners, one holding each end. As they run, each runner pulls on his end of the rod so as to stretch it out to 12.5 m in their reference frame. The rod now measures 10 m in the "stationary" frame, the same as its original rest-length. According to a stationary observer the length of the rod has not changed, but if he can inspect it carefully as it flies past, he might be able to observe signs of the internal strains in the rod (little cracks in the surface, or whatever).

A ring-shaped section of a rotating disk similarly has its moving "length" (circumference) kept at its original "rest length" by internal strains inside the disk, until the rotational speed becomes large enough that this is no longer possible, for the particular material the disk is made of.

 

[soothsayer]

Ok, thanks Ben! I was actually just perusing wikipedia and came across the article on the Ehrenfest Paradox, which I've been reading up on and it's shed some light, though some of the mathematical/physical concepts used are a bit out of my league.

EDIT: ...Though it seems like actual physicists had enough problem with this paradox as it is, so I don't feel too bad ;)

Also, thank you to jtbell too.

 

[phinds]

The answer is that the idea of a "rigid object" is inconsistent with relativity. Or to put it another way, there are no such things as rigid objects in relativity, even in theory.

OK, but what is the implication of that? Does it mean the disk would come apart before reaching any significant portion of c ?

 

[PAllen]

OK, but what is the implication of that? Does it mean the disk would come apart before reaching any significant portion of c ?

Of course. Let's do a simple calculation. Suppose a disk of 10 cm. Suppose we hope to spin it at .1 c. The acceleration at the rim would be 1000 trillion times the surface gravity. Way before relativistic corrections come into play, you reach fundamental material limits.

 

[phinds]

Of course. Let's do a simple calculation. Suppose a disk of 10 cm. Suppose we hope to spin it at .1 c. The acceleration at the rim would be 1000 trillion times the surface gravity. Way before relativistic corrections come into play, you reach fundamental material limits.

Good. I assumed that to be the case but wasn't sure what the math might show. Thanks.

 

[HallsofIvy]

In fact, if a disk is rotating relativistically (strictly speaking " rotate at speed c" doesn't even makes sense. c is a linear speed, not angular rotation. It is impossible to have a disk rotating a sufficient angular speed that points on the circumference move at c.), then the circumference contracts while the radius (perpendicular to the direction of motion) does not. Result- the geometry of a rotating disk is NOT Euclidean.

 

[Fredrik]

Result- the geometry of a rotating disk is NOT Euclidean.

This depends on what exactly you mean by "the (spatial) geometry of a rotating disk". I think it's far from obvious how that should be defined. The mathematical object that has a non-Euclidean geometry in this scenario is a quotient manifold. It's a 3-dimensional manifold where each point corresponds to a world line of a particle component of the disk. If we define "the geometry of the disk" to be that manifold, then sure, it's not Euclidean. But in all these threads about this subject, I haven't seen a reason to think that this manifold is interesting other than as a mathematical curiosity. So I don't know why we would want to define that term that way. (There might be a good reason. I'm just saying that I haven't seen one).

I we instead define "space" as a hypersurface of simultaneity of a coordinate system, we don't get a non-Euclidean space in this scenario. The coordinate system that's rotating with the disk has the same hypersurfaces of simultaneity as the inertial coordinate system where the center of the disk is at rest.

 

[PAllen]

Maybe another way to put this would be that the spacetime here is assumed to effectively flat, but a flat spacetime cannot accommodate a rigid rapidly spinning disc: For each atom to maintain equal distance to its near comoving neighbors requires the circumference must be less than 2 pi r (from the point of view of an inertial observer). This is impossible in flat spacetime, therefore a rigid, relativistically spinning disc is impossible.

 

[harrylin]

Ok, thanks Ben! I was actually just perusing wikipedia and came across the article on the Ehrenfest Paradox, which I've been reading up on and it's shed some light, though some of the mathematical/physical concepts used are a bit out of my league.

EDIT: ...Though it seems like actual physicists had enough problem with this paradox as it is, so I don't feel too bad ;)

Also, thank you to jtbell too.

Perhaps it's less confusing to simply give the solution that Lorentz derived (Nature 1921) for a rotating disk:

with v= velocity at the rim, the contraction (of radius and circumference) is 1/4 of the standard Lorentz contraction factor.

And of course, for a thin rim the contraction would be nearly equal to the standard factor; and all without accounting for inertia.

Harald

 

[PAllen]

Maybe another way to put this would be that the spacetime here is assumed to effectively flat, but a flat spacetime cannot accommodate a rigid rapidly spinning disc: For each atom to maintain equal distance to its near comoving neighbors requires the circumference must be less than 2 pi r (from the point of view of an inertial observer). This is impossible in flat spacetime, therefore a rigid, relativistically spinning disc is impossible.

I would like to clarify and weaken this comment. It is perfectly possible to posit a steady state of 'atom world lines' representing a disc whose rim atoms are moving anything less than c. Forgetting physical plausibility, a la Born rigidity, we assume each atom is self propelled with whatever arbitrarily large proper acceleration is needed to follow a chosen path. Then the only issue is you can choose either of the following:

1) If the world lines have isotropic separation measured by an inertial observer stationary with respect to the disc center, the distribution looks increasing non-isotropic to atoms approaching the rim.

2) If the world lines are chosen so every atom sees its nearest neighbors having isotropic distances, then, to the inertial observer, the distribution is increasingly non-isotropic toward the rim.

but you can't have isotropy from both points of view. Other than the asymptotically infinite force required to hold each atom on the chosen path, either steady state is mathematically consistent with relativity.

 

[kapildverma]

I was also think of this since last few days during my vacation time. Let's assume that we have two discs close to each other rotating in opposite direction such that the edges of each of those are at 0.5c.

Will the two observers standing on the edges of the two discs see each other moving at c?

 

[A.T.]

This depends on what exactly you mean by "the (spatial) geometry of a rotating disk".

The spatial geometry of a rotating disk is measured by rulers at rest relative to the disc.

So I don't know why we would want to define that term that way. (There might be a good reason. I'm just saying that I haven't seen one).

See my comments in this post:

If you were building a giant rotating space station from prefabricated elements, you would have to produce more elements for the circumference, than for a non rotating space-station with the same radius. That is the physically relevant spatial geometry.

I we instead define "space" as a hypersurface of simultaneity of a coordinate system,

I don't know why we would want to define spatial geometry that way in this case. Simultaneity is irrelevant here because you don't need clocks to measure constant distances, just rulers.

 

[jtbell]

Lets assume that we have two discs close to each other rotating in opposite direction such that the edges of each of those are at 0.5c.

Will the two observers standing on the edges of the two discs see each other moving at c?

As they pass each other, each one will see the other going past at a speed < c which can be calculated from the usual relativistic "velocity addition" equation. It's no different from having two spaceships traveling in opposite directions in a straight line.

 

[phinds]

I was also think of this since last few days during my vacation time. Let's assume that we have two discs close to each other rotating in opposite direction such that the edges of each of those are at 0.5c.

Will the two observers standing on the edges of the two discs see each other moving at c?

No. velocities do not add that way at relativistic speeds.

 

[yuiop]

I was also think of this since last few days during my vacation time. Let's assume that we have two discs close to each other rotating in opposite direction such that the edges of each of those are at 0.5c.

Will the two observers standing on the edges of the two discs see each other moving at c?

No they will each see the other observer moving at (0.5+0.5)/(1-0.5*0.5) = 0.8c.

See relativistic velocity addition http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html

 

[yuiop]

... then the circumference contracts while the radius (perpendicular to the direction of motion) does not. Result- the geometry of a rotating disk is NOT Euclidean.

Let's say we have metal hoop that is somehow spun up to a rim speed proportional to a gamma factor of 10. We have two observers, One is in a non rotating inertial reference frame in which the centre of the hoop is at rest. Call this frame S. The other is on the hoop and is in a rotating non inertial reference frame. Call this frame S'. Irrespective of whether the hoop expands or contracts as measured in either frame, an observer in frame S will always measure the circumference to be equal to 2*pi*r and the observer in frame S' will always measure the circumference to be equal to 20*pi*r. No observer sees the circumference contract relative to the radius.

In practice, for almost all known materials, a disc, hoop, cylinder or sphere will self destruct under rapid rotation long before relativistic effects become significant. Large massive bodies bound together by gravity may fare a little better.

 

[PeterDonis]

The coordinate system that's rotating with the disk has the same hypersurfaces of simultaneity as the inertial coordinate system where the center of the disk is at rest.

Which coordinate system are you referring to as "the coordinate system that's rotating with the disk", and how are you determining its hypersurfaces of simultaneity? I take it you are *not* using Einstein clock synchronization; if you do that between two observers rotating with the disk, the hypersurfaces of simultaneity you obtain will not be the same ones as you would get by doing Einstein clock synchronization with two observers at the center of the disk and not rotating with it.

 

[Fredrik]

The spatial geometry of a rotating disk is measured by rulers at rest relative to the disc.See my comments in this post:
I don't know why we would want to define spatial geometry that way in this case. Simultaneity is irrelevant here because you don't need clocks to measure constant distances, just rulers.

I just find it very strange to use the term "spatial geometry" for the geometry of a quotient manifold, instead of for the geometry of "space". To me, the term "space" can only refer to a subset of spacetime. The quotient manifold that has this "spatial geometry" isn't a subset of spacetime.

Maybe that manifold is useful somehow, and in that case I guess we should be talking about it, but why not call its metric something like "the ruler metric" instead of "the spatial metric"? It's not the metric of anything I would be comfortable calling "space".

Which coordinate system are you referring to as "the coordinate system that's rotating with the disk",

I'm talking about a rotating coordinate system, e.g. the one defined by the following transformation from an inertial coordinate system in which the entire disk is at rest before it's given a spin.
$$\begin{pmatrix}t'\\ x'\\ y'\\ z' \end{pmatrix}=\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & \cos\omega t & -\sin\omega t & 0\\ 0 & \sin\omega t & \cos\omega t & 0\\ 0 & 0 & 0 & 1\end{pmatrix}\begin{pmatrix}t\\ x\\ y\\ z \end{pmatrix}$$

and how are you determining its hypersurfaces of simultaneity?

By noting that by definition, t'=t.

I take it you are *not* using Einstein clock synchronization; if you do that between two observers rotating with the disk, the hypersurfaces of simultaneity you obtain will not be the same ones as you would get by doing Einstein clock synchronization with two observers at the center of the disk and not rotating with it.

A "hypersurface of simultaneity" is defined by a coordinate system. The mathematical object that describes the local experiences of each point of the disk is a frame field, not a coordinate system. I don't think this particular frame field determines anything that can be called a hypersurface of simultaneity.

Instead, the congruence of world lines of component parts of the disk determines a frame field, and the world lines are the points of a quotient manifold with a funny geometry. Since this manifold isn't a subset of spacetime, I find it very odd that people want to call it "space".

 

[PeterDonis]

I'm talking about a rotating coordinate system, e.g. the one defined by the following transformation from an inertial coordinate system in which the entire disk is at rest before it's given a spin.
$$\begin{pmatrix}t'\\ x'\\ y'\\ z' \end{pmatrix}=\begin{pmatrix}1 & 0 & 0 & 0\\ 0 & \cos\omega t & -\sin\omega t & 0\\ 0 & \sin\omega t & \cos\omega t & 0\\ 0 & 0 & 0 & 1\end{pmatrix}\begin{pmatrix}t\\ x\\ y\\ z \end{pmatrix}$$

Are these basically the Cartesian version of Born coordinates? See the Wiki page here:

http://en.wikipedia.org/wiki/Born_coordinates

A "hypersurface of simultaneity" is defined by a coordinate system. The mathematical object that describes the local experiences of each point of the disk is frame field, not a coordinate system.

Ah, got it. My terminology was confused.

I don't think this particular frame field determines anything that can be called a hypersurface of simultaneity.

Not a single one that is common to different points in the frame field, no. At each point of the frame field, the spatial vectors define a spacelike "hypersurface of simultaneity" (by means of the Einstein simultaneity convention) for that point, but it's a different hypersurface for each point.

 

[A.T.]

but why not call its metric something like "the ruler metric" instead of "the spatial metric"?

That's the same. Rulers laid out at rest are measuring the spatial metric.

A "hypersurface of simultaneity" is defined by a coordinate system. The mathematical object that describes the local experiences of each point of the disk is frame field, not a coordinate system. I don't think this particular frame field determines anything that can be called a hypersurface of simultaneity.

Well, then the concept of hypersurfaces of simultaneity is not usefull here, and appraently not generally applicable. Maybe it is better to use a more general one. From the other thread:

To put it technically, the 3D space we are constructing here is not the orthogonal complement of the worldlines; it is a quotient space. For static worldlines, the two concepts coincide, but for stationary non-static worldlines they do not.

 

[Fredrik]

That's the same. Rulers laid out at rest are measuring the spatial metric.

You're saying this as if you're stating a fact, but in your previous post you said the same thing in response to my "that depends on what you mean by the spatial geometry". So you have already indicated that this statement is just your definition of the term "spatial geometry". If that's the case, then you're not stating a fact at all. You're just repeating your definition. So your statement isn't a useful response to my concern that it's really weird to use the term the term "spatial" when we're not even talking about a subset of spacetime.

I would prefer to pick a new term instead of changing the meaning of old one. And if we're going to change the meaning of terms like "space" and "spatial", then someone should at least say that that's what we're doing.

Well, then the concept of hypersurfaces of simultaneity is not usefull here, and appraently not generally applicable. Maybe it is better to use a more general one.

It's not generally applicable to frame fields, but it is generally applicable to coordinate systems. So it's certainly correct to say that spatial geometry associated with the rotating coordinate system is flat. It wouldn't make sense to define it as anything else than the geometry of a hypersurface of constant time coordinate. I'm certainly not disputing that there's also a Riemannian manifold associated with the frame field associated with the motion of a solid rotating disk, or that its geometry is curved. I just think it's weird to call that manifold "space" and its geometry "the spatial geometry".

 

[Fredrik]

Are these basically the Cartesian version of Born coordinates? See the Wiki page here:

http://en.wikipedia.org/wiki/Born_coordinates

I haven't made the effort to fully understand Born coordinates, but the rotating coordinate system I defined is just what you get if you take an inertial coordinate system and rotate its spatial part by a time-dependent angle. I could have written the defining equation as
$$\begin{pmatrix}t'\\ x'\\ y'\\ z' \end{pmatrix}=\begin{pmatrix}1 & 0 & 0 & 0\\ 0 \\ 0 & & R_z(\theta(t))\\ 0 \end{pmatrix}\begin{pmatrix}t\\ x\\ y\\ z \end{pmatrix},$$ where $R_z(\theta(t))$ is an SO(3) matrix that performs a rotation around the z axis by a t-dependent angle $\theta(t)$. This is the simplest way to define a rotating coordinate system. It completely ignores all the physical stretching and other weird stuff that goes on in a solid rotating disk.

 

[PeterDonis]

I haven't made the effort to fully understand Born coordinates, but the rotating coordinate system I defined is just what you get if you take an inertial coordinate system and rotate its spatial part by a time-dependent angle.

As I understand it, that's basically what the Born chart does. Another way of putting is that, in your rotating coordinate system, the worldlines of observers who are rotating with angular velocity omega (where omega is a constant) appear as vertical lines--those worldlines are the orbits of "time translations" in your rotating chart (i.e., they are worldlines of constant x', y', z').

 

[A.T.]

And if we're going to change the meaning of terms like "space" and "spatial",

Change? I for my part have always understood "spatial geometry" as "what rulers at rest measure". So just to clarify:

- Does an ideal ruler at rest (in some frame) measure the "spatial distance" (in that frame)?

- Does a lattice of such rulers at rest (in some frame) measure the "spatial metric" (in that frame)?

- Would such a lattice at rest around a massive sphere measure the "spatial metric" as given by the spatial part of the Schwarzschild solution?

It wouldn't make sense to define it as anything else than the geometry of a hypersurface of constant time coordinate

To me it doesn't make sense to define it via a constant time coordinate in the rotating frame, because It has nothing to do with time. The rulers are at constant positions and have constant lengths. They are measuring a time independent metric. How much more "spatial" can it get?

 

[pervect]

If you draw the set of points that are Einstein-synchronized on a rotating cylinder, you get a non-closed spiral.

See for instance http://en.wikipedia.org/wiki/File:Langevin_Frame_Cyl_Desynchronization.png

Talking about the spatial geometry of such a non-closed surface (which I gather can be thought of also as a quotient manifold) is definitely odd and tends to cause confusion. Specifically, it's generally assumed the circumference of a disk is a closed curve, and we can clearly see from the diagram that this is not the case if one uses Einstein clock synchronization.

It also seems to me that actual rulers measure distances between worldlines, rather than points in space-time. If you have an actual physical ruler, each end traces out a different worldline in space-time and when you try to use rulers to measure the distances between events, you ensure that the beginning and ending events are on the respective front and back worldlines of the ruler.

I haven't seen any published papers take this particular approach, but it makes sense to me. Furthermore, when you take this approach that rulers measure distances between worldlines and not points, it becomes clear that the circumference is measured by adding together the distances between successive wordlines until one returns to the beginning worldline.

You still won't have a closed curve for the circumference, but you'll have a well-defined notion of its length.

 

[PeterDonis]

The rulers are at constant positions and have constant lengths.

But the spacelike curves realized by the rulers are *not* curves in the quotient manifold; they are curves in spacetime. More precisely, if you look at the spacelike curve that represents all the points on a given ruler at a given instant of that ruler's proper time, that curve is a curve in spacetime, *not* the quotient manifold. So I think Fredrik's uneasiness with calling the quotient manifold "space" is well founded.

 

[yuiop]

No one has yet explicitly mentioned that it is impossible to synchronise clocks all the way around the perimeter of a spinning disc using the Einstein convention, so I thought I would . (If it has already been mentioned, then I missed it and my apologies in advance). If we assume the speed of light is isotropic and use a suitable set of mirrors all the way around the perimeter together with a single clock on the perimeter to measure the distance around the disc then the distance depends on which direction we send the timing signal. If we take the average distance of the clockwise and anticlockwise measurements (effectively a radar distance measurement) we obtain the ruler distance obtained by laying rulers end to end all the way around the perimeter. It soon becomes clear, using a single clock on the perimeter, that the one way speed of light in a rotating reference frame is not c and is direction dependent.

 

[DrGreg]

But the spacelike curves realized by the rulers are *not* curves in the quotient manifold; they are curves in spacetime. More precisely, if you look at the spacelike curve that represents all the points on a given ruler at a given instant of that ruler's proper time, that curve is a curve in spacetime, *not* the quotient manifold. So I think Fredrik's uneasiness with calling the quotient manifold "space" is well founded.

But to measure the length of a stationary object, you chop it up into "infinitesimal" bits, measure each bit and add all the bits together (to put it in crude terms -- mathematically you perform an integral). As the object is stationary, each bit has a constant length over time, so there's no need to measure all the bits simultaneously (whatever your definition of simultaneity). So there's no need to stick the bits together in 4D spacetime. You just take each bit's proper length and add the proper lengths together.

 

[DrGreg]

I haven't seen any published papers take this particular approach, but it makes sense to me. Furthermore, when you take this approach that rulers measure distances between worldlines and not points, it becomes clear that the circumference is measured by adding together the distances between successive wordlines until one returns to the beginning worldline.

You still won't have a closed curve for the circumference, but you'll have a well-defined notion of its length.

See what I said last year...

This topic is covered in more detail in Rindler W (2006, 2nd ed), Relativity: Special, General, and Cosmological, Oxford University Press, ISBN 978-0-19-856732-5 Section 9.7 p.198. Earlier in that chapter, Rindler proves a more general result to show an arbitrary stationary spacetime's metric can be expressed in the form$$ds^2 = e^{2\Phi/c^2}\left( c\,dt - \frac{w_i}{c^2}dx^i\right)^2 - k_{ij}\,dx^i\,dx^j$$where $\Phi,\,w_i,\,k_{ij}$ are all independent of t. Here, $k_{ij}\,dx^i\,dx^j$is the time-independent metric of 3D space, but $c\,dt - w_i\,dx^i/c^2$ isn't the differential of any coordinate, except in the static case when $\textbf{w} = \textbf{0}$.

 

[PeterDonis]

As the object is stationary, each bit has a constant length over time, so there's no need to measure all the bits simultaneously (whatever your definition of simultaneity). So there's no need to stick the bits together in 4D spacetime.

So basically, the quotient manifold isn't a "spacelike slice" taken directly out of the spacetime, but it does represent a "spatial geometry" indirectly derived from ruler measurements via this procedure.