A pulsar or spinning neutron star can reach relativistic angular velocities. Special relativity asserts that objects traveling near the speed of light contract in length. Therefore, it seems reasonable that a rapidly rotating object could exhibit a paradox, in which the surface has less area than the layer beneath it. If this is true, could this have an effect on other physical aspects of such an object (e.g. charge density distribution, etc.)?

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    $\begingroup$ Relativistically rotating objects were one of the things that initially led Einstein into thinking about space itself distorting. Special relativity deals with inertial frames of reference, which rotating objects do not have, and to describe a rotating object like that, you do need to move away from a normal euclidean geometry. A caveat: the ideal rigid body doesn't survive into special relativity. Great question, I look forward to reading the answers of the more knowledgeable. $\endgroup$ – CDCM May 18 '17 at 17:03
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    $\begingroup$ @CDCM: I think it's pretty clear that the OP intends to ask the question from the viewpoint of an inertial (and hence non-rotating) observer. $\endgroup$ – WillO May 18 '17 at 18:58
  • $\begingroup$ Possible duplicate: Euclidean geometry in non-inertial frame. See also An electromagnetic twist on Ehrenfest's paradox $\endgroup$ – AccidentalFourierTransform May 21 '17 at 18:05
  • $\begingroup$ You can find the treatment of this problem in Einstein's book "Relativity - The Special and The General Theory" in the cap. 23 "Behaviour of Clocks and Measuring Rods on a Rotating Body of Reference"; the short answer is that a clock on the surface will go slower than a clock near the center. There is another effect described in "The Meaning of Relativity" :"A material particle, moving perpendicularly to the axis of rotation inside a rotating hollow body, is deflected in the sense of the rotation (Coriolis field).". This also tells us that a rotating(spinning) pulsar/black hole loses energy... $\endgroup$ – Mihai B. May 21 '17 at 18:28
  • $\begingroup$ Since you've mentioned charges, take a look at this research regarding the "Relativistic electron spin motion in cycloatoms" treated in the framework of relativistic QM osapublishing.org/oe/abstract.cfm?uri=oe-8-2-51 $\endgroup$ – Mihai B. May 21 '17 at 18:37

Let me fully answer just the part which can be done with pen and paper: if we have a family of rigidly rotating observers in flat space-time, will the measured circumference ever be decreasing with respect to the distance from the axis of rotation? As we will see, the answer is yes, at least from the point of the rotating observers.

Because we are dealing with accelerated observers and thus, from the observers' point of view, curved space-time, I will use the tetrad formalism. The tetrads are nothing else than tiny local coordinate frames very much in the spirit of special relativity. The main idea is to have the time axis, which coincides with the observer four-velocity $u^\mu$ and then construct three "infinitesimal coordinate axes" which are 1) space-time orthogonal to the time axis, and 2) orthogonal to each other. The length of these vectors then tells us the physical lengths of lines such as the circumference.

Corotating circumference

Let us get started, flat space-time in cylindrical coordinates looks like this $$ds^2 = -dt^2 + \rho^2 d \phi^2 + d\rho^2 + dz^2$$ The rigidly rotating observers move with a coordinate angular velocity $\Omega = d\phi/dt$ which gives them a four-velocity $u^\mu = N(1,\Omega,0,0)$, where $N$ is a normalization factor equal to $N = 1/\sqrt{1-\Omega^2 \rho^2}$. The point where $\Omega \rho = 1$ is where the rigid rotation forces the observers to move at the speed of light and the rigid rotation must thus necessarily stop. (I use $c=1$ units.) We will be interested in regions below that.

Now let us construct the tetrad. Naturally, two axes are pointing in the $\rho,z$ direction and see no deformation. The one in the phi direction, however, will see a skewing and contraction. It is of the form $e_{(\phi)}^\mu = (e_{(\phi)}^t, e_{(\phi)}^\phi,0,0)$. From the condition of orthogonality to $u^\mu$ we get $$-e_{(\phi)}^t u^t + \rho^2 e_{(\phi)}^\phi u^\phi = 0$$ $$\to e_{(\phi)}^t = \rho^2 e_{(\phi)}^\phi \Omega $$ Now we have one undetermined number which is $e_{(\phi)}^\phi$. We get rid of it by requiring that the coordinate axis is normalized to one: $$e_{(\phi)}^\phi = \rho \sqrt{1 - \rho^2 \Omega^2}$$ Now we have constructed the local infinitesimal axes, but we need to specify the measurement process done by the observers to obtain the circumference.

Let's say that one member among a ring of observers at a fixed $\rho, z$ has a rope, and starts to pass it out to the others in the ring. While the rope is passed around, she measures what has been used up. Once the rope goes around the loop and comes back to her, she postulates the length needed to enclose the loop as the circumference. This circumference will be $$C_{rope} = \int_0^{2\pi} e_{(\phi)}^\phi d\phi = 2 \pi \rho \sqrt{1-\rho^2 \Omega^2}$$ If you were to construct this circumference by a more naive length-contraction special-relativistic argument, you would get the same result. $2 \pi \rho$ is just the usual radius $\rho \Omega$ is the linear velocity in the lab frame, and $\sqrt{1 - \rho^2 \Omega^2}$ is thus simply the length-contraction factor $\sqrt{1 - v^2}$.

We see that for instance at $\Omega \rho = 1$ this circumference is zero, so there is a breaking point, where the larger $\rho$ definitely means smaller circumferences. We can find this point by $dC/d\rho = 0$ as $$\rho \Omega = \frac{\sqrt{2}}{2} \approx 0.71$$ This is an interesting result - the circumferences will stop growing and start shrinking instead when rigid rotation forces you above $71 \%$ of the speed of light!

Corotating surface area

If you want to compute the surface, things get a little bit more arbitrary because you first have to answer the question of how one even defines the surface in a coordinate-independent sense. If we just use a coordinate definition with $R = \sqrt{\rho^2 + z^2}$ constant and change to spherical coordinates, we get a surface area defined similarly to the circumference above as $$S(R) = 2\pi R^2 \int_0^\pi \sin \vartheta\sqrt{1 - \Omega^2 R^2 \sin^2\vartheta} d\vartheta$$ Which, with a little help of Mathematica yields a function which also has a tipping point somewhere at $1>\Omega R > 0.9$. I.e., these surfaces will also stop growing at a certain point. However, one should remember that the part of the surfaces at the axis is rotating with small linear velocity and thus has negligible contraction in this sense; it is mainly the shrinking of the circumference around the equator which causes the halt in the growth of the surface area.

So, strictly in the sense of measurements made by observers corotating with the matter of the pulsar, this kind of "surface-shrinking" geometry is very much possible. On the other hand, if a family of static observers at the same point makes analogous measurements with a rope, they will get simply a circumference and area corresponding to the usual $2 \pi \rho$ and $4 \pi R^2$ formulas.

Astrophysical pulsars

Now for real physics. Millisecond pulsars, the fastest rotating known neutron stars, rotate once per about millisecond and are estimated to have a radius of about $10 km$. This gives us an estimate of the linear velocity at the surface as $\sim 10 000 km/s$. But this gives us $v/c \sim 0.03$ which is an order smaller than the critical "circumference-shrinking" $v/c \sim 0.71$ which we have computed in flat space-time. So, in the pulsars we know this definitely does not happen.

The question is whether pulsars where this happens could be observed in the future. The answer is that probably not. First, the high rotation is quickly relaxed by radiation mediated by the magnetic field of the neutron star (this radiation is the "pulse" in the "pulsar"). Second, even if no magnetic field is present, $v/c \sim 0.7$ at the surface is most probably above the mass-shedding limit of the neutron star, where the gravitational force is unable to work against the centrifugal force and the matter flies away.

On the other hand, at the level of bare principles, there seems to be nothing which stops us from constructing a star which rotates at such a rate that this "corotating geometry" exhibits this exotic behavior.

Some speculation on associated physical effects

The properties of the corotating geometry do not seem to necessarily have any direct effect on the observational properties of the pulsar. This is because observational properties are defined with respect to observers at rest at infinity rather than the corotating ones. For instance, the invariant mass of the pulsar is defined by projecting the stress-energy tensor into the time-like Killing vector, which is simply the time-direction of observers at infinity, and integrating over space. The charge of the pulsar is measured by external observers in a similar way so there is also no connection of the behavior of the observed charge with the special behavior of the corotating geometry.

However, the corotating geometry is important for the local dynamics of the pulsar matter. For instance, the fact that a fluid element (ideally forced to stay in corotation by various physical effects) can slowly travel "up", away from the star, and ends up contracting into a smaller circumference rather than expanding means that our intuitions abut convective stability will be broken. This can ultimately lead to the conclusion that "corotating circumference contraction" is not consistent with a convectively stable star.

Another interesting effect might arise during some "breathing" pulsations of the neutron stars, which are pulsations which contract or expand the radii on which the fluid elements move. A typical effect of a contraction of any star is the growth of magnetic fields due to magnetic-flux conservation and the fact that the magnetic fields are effectively frozen-in into the fluid elements (and thus the corotating geometry is the relevant one for our discussion). However, if we cut out a surface in the equatorial plane between two radii $\rho_1$ and $\rho_2$ above the critical "circumference-shrinking" point, then pressing the fluid elements from $\rho_1,\rho_2$ to smaller $\rho$ will generically increase rather than decrease the surface area enclosed between them. To satisfy flux conservation, the magnetic fields then have to decrease between them! In other words, the circumference-shrinking property should generically lead to counter-intuitive behaviour of magnetic fields

The study of highly rotating neutron stars is impossible analytically, and is also non-trivial numerically. Since we generally observe pulsars at rather small rotations (as compared to relativistic scales), only a handful of research groups currently invests time and effort into the realistic modelling of highly rotating neutron stars. This means that, at least to my knowledge, there are no concrete examples where a connection between these properties of the corotating geometry and associated physical effects would be investigated and documented. (But I think it would be a great topic for a master's thesis or even a part of a dissertation if anyone is interested :))

  • $\begingroup$ I've made a few edits to your response to make a some statements clearer, but overall, I think it is a good answer. When calculating the simplified, ring scenario, you mention that "[i]f you were to construct this circumference by a more naive length-contraction special-relativistic argument, you would get the same result." Why would you find this method to be particularly 'naive', as it seems to be a logical approach. I will wait until the bounty period is over to award this answer, should no better one come along. Thank you! $\endgroup$ – Master Drifter May 23 '17 at 18:45
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    $\begingroup$ @MasterDrifter Thanks for the edits. Special relativity, the theory of transformations between uniformly moving frames, does not really tell you about deformations of objects which are both accelerated and move with a different velocity from point to point. I could easily say "the velocity is $\rho \Omega$ and thus the circumference will be contracted by $\sqrt{1 - \Omega^2 \rho^2}$ as we know from special relativity" - and it would work as a simple argument. But I wanted to make sure that it is clear what is the circumference measurement in an operational sense. $\endgroup$ – Void May 23 '17 at 19:44
  • $\begingroup$ Your answer did a great job of explaining how to approach the problem and demonstrating that pulsars most likely cannot exhibit such a paradoxical physical geometry. However, I am still interested in whether or not such a geometry could exist, and if so, what effect it could have on the other properties mentioned in the OP. $\endgroup$ – Master Drifter May 26 '17 at 8:22
  • $\begingroup$ @MasterDrifter Well, in principle there is really nothing which restricts the existence of such a corotating geometry, but the physical consequences are essentially impossible to compute by hand and nobody has worked on this so far. I have made an edit which makes a few proposals on what could happen, but this is essentially an open research question and I don't think anyone can give you a more precise answer right now. $\endgroup$ – Void May 27 '17 at 11:24
  • $\begingroup$ In spite of stack exchange etiquette, I"d like to say "Thanks so much!" $\endgroup$ – Master Drifter May 27 '17 at 14:39

The apparent paradox that you mention is known as Ehrenfest's paradox, although unfortunately I don't think that the Wikipedia article explains it very well. Ehrenfest's paradox was historically important for the development of general relativity, although resolving it doesn't actually require any GR.

The short answer is that an inertial observer does not see the surface area of a relativistically rotating solid as being less than when it's not rotating. To understand this, note that the notion of an extended "solid" is not well-defined in special relativity, because signals can't travel faster than the speed of light, so any force applied at one point of the solid will necessarily deform the solid, at least until the information about the force has had time to cross the entire solid so that it can restore its shape. If you were to start with the sphere at rest and then apply a uniform torque everywhere simultaneously that sets it rotating, then an observer riding along the surface would perceive the torque as being applied at different times, in such a way that causes the proper surface area to stretch by a factor of $\gamma$. Back in the inertial frame, the Lorentz contraction exactly cancels this stretching effect and restores the surface area back to its non-rotating value. See the answer to my previous question for more info.


I wonder if the light produced by a pulsar is actually escaping from the poles of it's fastest rotation. That would mean that the millisecond pulsar is the the slower of two rotations. The pressure generated by the lorentz contraction about it's equator of faster rotation might be what is converting its mass into the light we see emitted as it spins slower along another access.

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    $\begingroup$ Does this answer the question though? $\endgroup$ – MannyC May 7 '19 at 4:49

As I understand it, relativity applies between two moving frames of reference. Therefore, if look at a disk from the z axis, I will see a circle whose diameter will not change (not moving towards or away from me). If I look at the disk from the X (or Y) axis, I will see a line, whose length will not not change ( for the same previous reason).
The results are the same, even if the disk is rotating at $wr = c$

  • $\begingroup$ I think there is a flaw in your argument. Viewed from the z-axis, a perfectly circular disk may appear not to move, but in fact each infinitesimal component of the disk is moving with respect to your position, in its tangential direction. Consider the differential element of the circumference $dl=r d \phi$, moving at relativistic angular speeds in a circular orbit of radius $r$, as viewed from your z-axis perspective. You are forced to consider effects of length contraction on this tiny length. The tangential velocity $v=r \frac{d\phi}{dt}$ ensures a contracted length $\frac{dl}{\gamma} $. $\endgroup$ – Master Drifter May 26 '17 at 7:44
  • $\begingroup$ A similar argument can be made for the x and y-axis reference frame. $\endgroup$ – Master Drifter May 26 '17 at 7:47

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