Can a Rotating Sphere at Relativistic Speeds Create a 'Black Spot'?

[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

 

[bcrowell]

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?

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.

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

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.

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.

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.

 

[bcrowell]

You might want to check out the final segment of this video: http://www.youtube.com/watch?v=JQnHTKZBTI4&feature=related

 

[natski]

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.

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...

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.

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.

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.

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

 

[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

 

[yuiop]

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

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.

 

[bcrowell]

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.

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).

 

[starthaus]

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

No, it is a sphere, see here

 

[Austin0]

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 1) the process stops.[/QUOTE]

Agreed it is interesting that clocks at sea level all have an equal dilation factor but : )
how is it possible that 1 ) the process stops? At the equator there is always the sea level difference in altitude (r) isn't there ? It seems like it just moves continually relative to the terrain and at rest wrt the moon.
Or am I misunderstanding you ??

 

[Austin0]

No, it is a sphere, see here

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.

 

[starthaus]

If I am understanding correctly the soccer ball example deals with a ball in translation but without rotation. Is that correct?

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

If I remember Penrose right; that also dealt with translation and the effect was based on propagation time offsetting contraction ?

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.

If this is right then neither would seem to apply in this situation where the observer is at rest wrt the rotation axis ,

Not true, see above.

 

[bcrowell]

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 ?

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.

 

[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

 

[starthaus]

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?

Don't know, must do the calculations.

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

Yes, the grid gets distorted to a certain degree.

 

[natski]

Yes, the grid gets distorted to a certain degree.

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.

 

[starthaus]

However, if the centre of the sphere gets contracted,

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

No

or b) the grid does not become distorted.
.

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

 

[bcrowell]

It the Penrose sphere is indeed translating and rotating, then the solution is that the sphere remains a sphere.

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.

Is this generally true of any ellipsoid?

I don't see how it could be.

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?

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.

 

[Austin0]

=starthaus;2790681]No, the ball is translating and rotating at the same time.

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]

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.

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

What is the question you think I was asking if your answer was just a repetition of what I had just said?

 

[starthaus]

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??

What is the question you think I was asking if your answer was just a repetition of what I had just said?

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

 

[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.

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

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

No?

 

[starthaus]

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

No?

No. Try reading the papers bcrowell recommended.

 

[Austin0]

No. Try reading the papers bcrowell recommended.

I have tried web search and the links in this thread but haven't had any luck.

What do you think are the other relevant Lorentz effects??

As I said I have only distant memories of reading Penrose . I also remember understanding the concept but questioning the conclusion so I would very much like to refrsh my memory.
Just based on contraction and light propagation I think it can be demonstrated that the constant sphericity would not apply if contraction is assumed .

Perhaps??

 

[starthaus]

I have tried web search and the links in this thread but haven't had any luck.

What do you think are the other relevant Lorentz effects??

As I said I have only distant memories of reading Penrose . I also remember understanding the concept but questioning the conclusion so I would very much like to refrsh my memory.
Just based on contraction and light propagation I think it can be demonstrated that the constant sphericity would not apply if contraction is assumed .

Perhaps??

You need to find all the rays of equal transit time in order to determine the image of the object. In order to do that you need both full-fledged Lorentz transforms, applying length contraction only doesn't solve the problem. your local library will get you the paper for about 1$.

 

[bcrowell]

I have tried web search and the links in this thread but haven't had any luck.

How about going to a university library and reading the papers there?

 

[yuiop]

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.

We can address the physical problems inherent in case D by doing the following. Design an oblate spheroid (like a Rugby ball stood on end) made of a suitable tough but elastic material such that when the spheroid is spun around its long axis it expands into a perfect sphere (due to centripetal considerations) as measured in the non-rotating frame.Now the spinning we can consider the visual effects due to light transmission times only.

After carrying out the analysis (effectively ray tracing) this is what I obtain. Consider the long axis of the original spheroid to be North and South and the sphere is rotating anti-clockwise as seen from above the North pole. To an observer high above the equator, the objects on the equator are moving left to right or West to East. To this observer, if the sphere is rotating at relativistic speeds, parts of the map or grid on the left of the sphere will visually appear to greatly stretched and parts of the map or grid to the right of sphere will appear greatly compressed visually.

[EDIT]Seeing as no one has responded to this post yet, I have edited this final part to make it more accurate.

The visible portions on the right (the receding edge) are greatly compressed and not very clear. The portions on the left and at the centre are greatly magnified and dominate the view and this may give an overall appearance of "zooming in" or "bulging out" similar to (and possibly identical to) the view featured in the the last part of the video Ben linked to (See this video: http://www.youtube.com/watch?v=JQnHTKZBTI4&feature=related) for the case of an observer orbiting rapidly around a non-rotating globe.

 

[bcrowell]

@kev: Re your #25, I don't see anything logically wrong with your way of approaching my case D, but the thing is that the results will be completely dependent on what elastic properties you assume for the rotating sphere. IMO this means that case D lacks the universal, geometrical interest of cases A-C.

 

[yuiop]

@kev: Re your #25, I don't see anything logically wrong with your way of approaching my case D, but the thing is that the results will be completely dependent on what elastic properties you assume for the rotating sphere. IMO this means that case D lacks the universal, geometrical interest of cases A-C.

I agree that case D is physically awkward but it is the case that the OP has specifically expressed an interest in. I was trying to make the case that we could in principle (by suitable design and choice of materials) end up with a rapidly rotating object that is physically a sphere in the non rotating frame. (It is true that there is only one magnitude of angular velocity that will satisfy the requirements for a given material but that is not a deal breaker.) After that it becomes a purely geometric consideration of light transmission times and ray paths. We do not have to consider length contraction or time dilation or anything like that. Technically, if we are being really picky, we might have to consider the visual effects of frame dragging and gravitational lensing for a really massive spinning object, but that is probably going beyond the spirit of this thread.

What I did not make clear in my last post is that a rapidly rotating object, that is physically a sphere in the non rotating frame, will still appear to be a sphere in the non rotating frame. I have also deleted the bit about being able to see "around the back", because after further consideration that does not appear to be true. The main effect is the visual distortion of any surface patterns and casual analysis suggests the distortions will be basically the same as those shown in the video you linked to.

 

[bcrowell]

What I did not make clear in my last post is that a rapidly rotating object, that is physically a sphere in the non rotating frame, will still appear to be a sphere in the non rotating frame.

Did you make a mistake typing this? You have "in the non rotating frame" twice.

 

[yuiop]

Did you make a mistake typing this? You have "in the non rotating frame" twice.

No it is not a typo, but what I meant might not be clear. I meant that a rapidly rotating object that is "physically" a sphere (as measured using rulers in the non rotating frame) will "visually" still appear to be a sphere (as photographed in the non rotating frame).

What is physically measured and visually "seen" in a given frame is not always the same.

In the classic Penrose-Tyrell case, the sphere is moving inertially in a straight line relative to inertial frame S (for example). In this case the moving sphere is "physically" measured to be length contracted (and non-spherical) by the observers at rest in S using rulers and clocks, but "visually" seen by the same observers at rest in S as being perfectly spherical.

It is perhaps worth adding that for a non-spherical object such as a long rod moving inertially in a straight line in frame S, the length contraction of the rod, can not only be measured using rulers and clocks by observers at rest in S, but can also be visually seen to be length contracted when photographed by the same observers in frame S.

P.S. I had to say "is physically a sphere in the non-rotating frame" because in other frames the object is not physically a sphere. For example observers at rest on the surface of the sphere will measure the same object to be an oblate spheroid physically.

 

[bcrowell]

No it is not a typo, but what I meant might not be clear. I meant that a rapidly rotating object that is "physically" a sphere (as measured using rulers in the non rotating frame) will "visually" still appear to be a sphere (as photographed in the non rotating frame).

What is physically measured and visually "seen" in a given frame is not always the same.

I see. Sure, that makes sense. You could generalize this to any motion generated by a symmetry of the object, e.g., a line moving along its own length, or a cylinder rotating about its axis. And note that there is nothing relativistic about your statement. It would be equally true, for example, if you were imaging something underwater using sonar.

 

[yuiop]

Looking at this the last section of this video again http://www.youtube.com/watch?v=JQnHTKZBTI4&feature=related it seems to me that the assymetrical visual compression and stretching of the star background in the video is the wrong way around. I think the visual stretching of the stars background should be happening on the left and and visual compression should be happening on the right. I might be wrong as this coclusion is only based on informal ray tracing analysis. Any second opinions on what the inside of a rotating shell would look like from the inside by a stationary observer near (but not exactly at the centre) looking towards the centre?

 

[Austin0]

You need to find all the rays of equal transit time in order to determine the image of the object. In order to do that you need both full-fledged Lorentz transforms, applying length contraction only doesn't solve the problem. your local library will get you the paper for about 1$.

Hi
Unfortunately I am in S E Asia and I doubt that a local library [not that there are any]
can help.
I have found several articles on the web but none containg the the basic parameters of the derivation.

Could you possibly tell me how time dilation and/or simultaneity would apply.

Nutshell answer would be fine.

Thanks

 

[starthaus]

Hi
Unfortunately I am in S E Asia and I doubt that a local library [not that there are any]
can help.
I have found several articles on the web but none containg the the basic parameters of the derivation.

Could you possibly tell me how time dilation and/or simultaneity would apply.

Nutshell answer would be fine.

Thanks

Here is an excellent website that gives you all the mathematical details.

 

[yuiop]

Hi
Unfortunately I am in S E Asia and I doubt that a local library [not that there are any]
can help.
I have found several articles on the web but none containg the the basic parameters of the derivation.

Could you possibly tell me how time dilation and/or simultaneity would apply.

Nutshell answer would be fine.

Thanks

Hi Austin,

I assume you came across this link in your search:

http://math.ucr.edu/home/baez/physics/Relativity/SR/penrose.html

These two links, might be helpful but unfortunately they imply the common misconception that any object (Not just a sphere) can not have its length contraction photographed:

http://arxiv.org/PS_cache/arxiv/pdf/0901/0901.0309v1.pdf
http://www.math.ubc.ca/~cass/courses/m309-01a/cook/terrell1.html

This final link goes into more detail about the visual appearance of moving objects in general and contains some of the maths pertaining to the Lorentz transforms:

http://cdsweb.cern.ch/record/913692/files/0512054.pdf

Unfortunately, none of the freely available documents present the information with much clarity. I am tempted to produce something myself.

 

[Austin0]

Hi Austin,

I assume you came across this link in your search:

http://math.ucr.edu/home/baez/physics/Relativity/SR/penrose.html

These two links, might be helpful but unfortunately they imply the common misconception that any object (Not just a sphere) can not have its length contraction photographed:

not http://arxiv.org/PS_cache/arxiv/pdf/0901/0901.0309v1.pdf
not http://www.math.ubc.ca/~cass/courses/m309-01a/cook/terrell1.html

This final link goes into more detail about the visual appearance of moving objects in general and contains some of the maths pertaining to the Lorentz transforms:

http://cdsweb.cern.ch/record/913692/files/0512054.pdf

Unfortunately, none of the freely available documents present the information with much clarity. I am tempted to produce something myself.

Thanks kev the last one was informative. It looks like contraction is the only relevant Lorentz effect.