# Rotating sphere

1. Jul 6, 2010

### natski

Hi all,

I am thinking about neutron stars and wondering about rotations at relativistic speeds. Consider a perfect sphere of radius R rotating about the Y-axis with the observer located along the +Z-axis.

The outer edge of the sphere rotates with tangential velocity V and so from the observer's perspective the fastest transverse velocity occurs for the portion of the sphere located at (0,0,R) which has velocity (+V,0,0).

So if you took a small finite strip of material along the X-direction at this location, with length L, it should get contracted to length L' = L/gamma. Now the next portion along will not be contracted by as much because the projected velocity in the X direction has now decreased. So does that mean that there exists a 'black spot' between these two shortened portions? How can the sphere be continuous, which of course it should be?

Also, what would the general appearance of the sphere be? Does it becomes prolate and if so with what axes lengths?

Finally, what about the Penrose-Terrel rotation? If each "strip" gets rotated by the Terrel angle, then it's projected length would be additionally shortened on-top of the length contraction effect.

Natski

2. Jul 6, 2010

### bcrowell

Staff Emeritus
This part of your reasoning doesn't make sense to me. As long as the observer is stationary relative to the axis of rotation, there is nothing to break the symmetry. Every part of the sphere should be length-contracted an equal amount, in the direction of its own motion.

I don't think there is an unambiguous answer to this question. It depends on the frame of reference of the observer and the method used by the observer for measuring.

Terrell rotation is an optical effect, so I assume you're now talking about the shape perceived optically by an observer at rest with respect to the axis of the sphere. I know that if an observer goes in circles around a stationary sphere and observes it optically, the sphere remains a sphere, but the areas of various parts of the sphere get increased or decreased. I would think that it would work the same way for a rotating sphere and an observer stationary with respect to the sphere's axis, but I could be wrong about that.

3. Jul 6, 2010

### bcrowell

Staff Emeritus
4. Jul 6, 2010

### natski

Yes exactly, but the direction of motion is not in the Z-direction except at the coordinate (0,0,R). So you can either say the projected velocity is reduced by \cos\theta and so the length contraction is reduced by \cos\theta... or you can say the length contraction is along the vector of motion, but in the next step we have to account for the projected view of the strip which requires multiplying by \cos\theta.... either way you get the same answer.

Point is... the strip moving fastest gets squashed a lot in the X-direction but the strip next door appears less squashed in X, and thus it seems that gaps should appear between strips - which of course cannot be the case...

But the observer's position and method are well defined here. We are just looking at a sphere from a very large distance away along the Z-axis and the sphere rotates with an axis of rotation along the Y-axis.

Interesting. This problem is really quite simple, is it not? A sphere rotating. I mean it does get much simpler than that, and yet a simple answer seems elusive. I have tried Googling this and have come up with surprisingly recent articles on similar types of problems. Surely a solution must exist because neutron stars are known to rotate at relativistic speeds and they are essentially spherical in nature.

Natski

5. Jul 6, 2010

### natski

In regard to the "gaps" - I ahve just realized the solution, I think. It is not special to rotation, even a moving piece of metal, say, could be considered as a sequence of pieces of metal joined together and each piece is length contracted. So why don't gaps appears?

The answer is of course that the distance between each strip also contracts, as well as the strip lengths themselves.

So now the outstanding question I pose is what is the appearance of the rotating sphere? My feeling is still a prolate spheroid... but I would like to see a proof.

Natski

6. Jul 6, 2010

### yuiop

It is known that the Earth is a prolate sphere and that the diameter at the Equator is significantly larger than the Pole to Pole Diameter. This is pretty much what you would expect from newtonian cetripetal considerations. What is interesting from the GR point of view is that even though the tangetial velocity is greatest at the Equator (and therefore the time dilation due to kinematic effects is greatest there), clocks at sea level anywhere in the world all run at the same rate. This is no coincidence. Sea water effectively moves from where time runs fastest (high effective gravitational potential) to where the time dilation is the greatest (low effective gravitational potential) until the effective potential at sea level is the same everywhere and then the process stops.

7. Jul 6, 2010

### bcrowell

Staff Emeritus
http://www.anu.edu.au/Physics/Searle/Commentary.html: "In 1959 R. Penrose observed that a sphere would present a spherical outline to all observers regardless of their relative motion, [...]"

The reference is to R. Penrose, "The apparent shape of a relativistically moving sphere," Proc. Camb. Phil. Soc. 55, 137 (1959).

8. Jul 6, 2010

### starthaus

No, it is a sphere, see here

9. Jul 7, 2010

### Austin0

10. Jul 7, 2010

### Austin0

If I am understanding correctly the soccer ball example deals with a ball in translation but without rotation. Is that correct?
If I remember Penrose right; that also dealt with translation and the effect was based on propagation time offsetting contraction ???

If this is right then neither would seem to apply in this situation where the observer is at rest wrt the rotation axis , so if there was optical or physical deformation the difference in propagation time wouldn't have any effect.

11. Jul 7, 2010

### starthaus

No, the ball is translating and rotating at the same time.

It is based on the compounded effect of the differences in light time propagation and the application of Lorentz transforms. I am quite sure I have answered this same exact question before, in another thread.

Not true, see above.

Last edited: Jul 7, 2010
12. Jul 7, 2010

### bcrowell

Staff Emeritus
It also applies when the observer is circling around a stationary sphere. That's what the commentary at the link in #7 is referring to. At the end of the video, they show what the earth would look like if you circled around it at relativistic speeds.

If the OP wants to know whether it applies when the sphere is rotating and the observer is at rest with respect to the axis, he should definitely look up the Penrose paper at the nearest university library.

This pdf physnet2.pa.msu.edu/home/modules/pdf_modules/m44.pdf reproduces the abstracts of several old papers on this topic.

13. Jul 7, 2010

### natski

Thanks for the Penrose reference.

It the Penrose sphere is indeed translating and rotating, then the solution is that the sphere remains a sphere. Is this generally true of any ellipsoid?

A further question regarding pure spheres... imagine a grid of longitude and latitude on the sphere. Even if the sphere's outline does not become distorted, does this grid become distorted tom degree?

Natski

14. Jul 7, 2010

### starthaus

Don't know, must do the calculations.

Yes, the grid gets distorted to a certain degree.

15. Jul 7, 2010

### natski

Ah-ha... well if the grid gets distorted, the only logical way it can be distorted is a length contraction, as opposed to an expansion.

However, if the centre of the sphere gets contracted, there must be some counter-balancing expansion at a different part of the gird or the sphere would no longer remain a sphere. This therefore means that a) the sphere does become distorted or b) the grid does not become distorted.

The two cannot be commensurable.

16. Jul 7, 2010

### starthaus

the center of a sphere is a point, so it can't get "contracted"

No

No. Do the calculations and you'll find out. Writing literary essays doesn't solve the problem.

17. Jul 7, 2010

### bcrowell

Staff Emeritus
That's not what I said, and I don't think that's what Penrose's theorem says. If you're really interested in this topic, you should look up the paper.

I don't see how it could be.

It's frustrating that I keep on pointing you to sources of information, but you keep asking questions like this one that show you haven't looked at them.

Having thought about this a little more, I think I understand this better. Let's distinguish four cases:

A. The sphere is not rotating. The sphere's center is at rest. The observer is moving in a straight line.

B. The sphere is not rotating, but its center is moving in a straight line. The observer is at rest.

C. The sphere is at rest and not rotating. The observer moves around it in a circle with its center coinciding with the center of the sphere.

D. The sphere is rotating, with its center at rest. The observer is at rest.

A and B are equivalent under a Lorentz transformation. The Penrose result clearly includes these cases. The outline of the sphere is still spherical.

C is also equivalent to A and B, because there are only two effects (Lorentz contraction and optical aberration), and both of them depend only on the observer's instantaneous velocity, not on his history of motion. This proves the result asserted in the commentary to the ANU video.

D is not a well-defined question. When asking this question, we're implicitly assuming that the sphere has some well-defined "real" shape, which appears different because the sphere has been set into motion. But you can't impart an angular acceleration to a perfectly rigid body in relativity.

A more well defined case is:

E. A rotating, self-gravitating body is in hydrostatic equilibrium. What is its shape as perceived by an observer at rest with respect to the body's center of mass, and how does this compare with the shape that would be inferred by observers surveying the body's surface with co-moving meter-sticks?

In case E, the observer clearly sees an oblate ellipsoid, because the oblateness is a nonrelativistic effect, which is much, much stronger than any of the relativistic effects. The question is then whether relativistic effects modify this slightly, and in what way. I don't know whether it's valid here to make the approximation of ignoring gravitational aberration of light.

18. Jul 8, 2010

### Austin0

Semantics strikes again. You are quite right . I was applyiing the definition of the original query I.e. physical rotation of the sphere. I had forgotten that Penrose used the same word for apparent rotation due to visual distortion.

Austin0
If I remember Penrose right; that also dealt with translation and the effect was based on propagation time offsetting contraction ??? [I.e. Lorentz transformation]

Could you tell me what difference you see in what you said and in what I had already said??

19. Jul 8, 2010

### starthaus

Lorentz transforms and Lorentz contraction are not one and the same thing.

20. Jul 8, 2010

### Austin0

Austin0
If I remember Penrose right; that also dealt with translation and the effect was based on propagation time offsetting contraction ??? [I.e. Lorentz transformation]

starthaus
It is based on the compounded effect of the differences in light time propagation and the application of Lorentz transforms. I am quite sure I have answered this same exact question before, in another thread.

Self evidently but in this case contraction is the only transform that is relevant so contraction and applied Lorentz transform are equivalent .

No?