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Lorentz Contraction of a Rotating Object

  1. Aug 10, 2005 #1
    Let's make it simple and suppose you have a very, very, perfectly sturdy record or something, and a record player that can spin it at enough RPMs to give the edge of the disk a speed that is a significant fraction of the speed of light.

    Since length shortens along the direction of motion, what would happen? Every spot on the circumference of the disk is moving perpendicular to the radius, which means that the length of the radius stays the same?

    Do you now have a circle whose circumference isn't it's radius * 2 * pi?

    What about points alongside the inside of the record? Their length contracts as well, but if it's a really big record, let's say it's radius is the size of the sun, the points along the inside are going to have drastically lower speeds, meaning less contraction.

    It hurts my head thinking about it, but I was looking at my fan while I was lying down yesterday and the question came to me.

    Can anyone set my mind at ease here?

    EDIT: I should mention I've been discussing it with my friends (my teacher didn't have an answer), and the theories we came up with were that space bends in such a way that the circle is resting on a dome (like you pushed an elastic circle with a rigid circumference up against the side of a ball). That would effectively make it so that the circumference is less than *pi times the diamater, but it seems kind of hokey. :/. Someone suggested that the rotating object collapses, but that seems even less viable.

    Do either of these trains of thought have any merit?
    Last edited: Aug 10, 2005
  2. jcsd
  3. Aug 10, 2005 #2
    I find it a very good question.

    There is a paper submited to the Am. J of Phys. of july 2005 (vol 73, NO 7) named:
    "he Appearance, apparent speed and removal of optical effects for relativistically moving objects"
    Robert Deissler from Cleveland State University

    I guess it may help.
    Although I haven't perfomed the calculations, I guess after removing the optical effects the disk should be sensed as kind of deformed, possibly an ellipsoid seen from the closest approach perspective.

    Best Regards

  4. Aug 10, 2005 #3
    I don't know a lot of relativity but this is an example that my professor gave us and I asked the exact question you did. He said it would be a dome shape, as the radius does infact stay the same but the circumference would be measured as smaller. I believe he also said it would follow the Kerr geometry.
  5. Aug 10, 2005 #4
    The disk would break apart or stretch circumferentially.

    Yes, the radius stays the same.

    If the disk could stretch circumferentially, then yes. Otherwise it would break circumferentially, and no longer be a circle.


    Supposedly Einstein pondered the same things. His thoughts led him to the idea of an excess radius, a derivative prediction of gravitational length contraction. (Have you heard of gravitational time dilation? Time dilation and length contraction always go hand-in-hand in relativity. Excess radius is the way that gravitational length contraction is typically referenced.) Excess radius is analogous to excess circumference; in the former case it is gravity that causes it, and in the latter case rotation. For more info, check out the pinwheel paradox here on PF. Reading that, you will see that both your example and the pinwheel paradox boil down to Bell’s spaceship paradox along any infinitesimal segment of the circumference. I think your train of thought is a great way to learn about excess radius in general relativity. Some of my laymen’s books use the spinning disk idea to teach that.

    The "dome shape" mentioned above is shown when a 2D plane that bisects the center of gravity of an object is modeled as 3D to show the excess radius as a dome. It's just a way of perceiving the excess radius. There's not really a dome in that 2D plane, but the excess radius can be physically measured (although no experiment has done that directly). I’m not sure how an excess circumference could be modeled as 3D.
  6. Aug 10, 2005 #5
    From a reference frame which views the disc as spinning, the circumference will have to contract, which means, as Zanket stated, that it will either have to stretch or break. Assume it stretches. If you measure the diameter and the circumference, you will still find the ratio between them to be [tex]\pi[/tex]. The only thing that will happen in your reference frame is that you will know that the circumference has to stretch or break because the atoms making it up will get smaller along the direction of motion.

    However, from the disc's reference frame things will be different. If you are standing on the disc, from the reference frame in which the disc is observed to be spinning you will be pulled toward the outside of the disc by a centrifugal force. But from your perspective, the disc isn't spinning; the force is gravitational. As Zanket points out, this (and the equivalence principle) are what led Einstein to the conclusion that lengths contract and time dilates in gravitational fields. If you measure the diameter and circumference of the disc while riding along on top of it, you will indeed find the ratio between them to be larger than [tex]\pi=3.14159[/tex]. This is easy to understand from the reference frame in which the disc is observed to be spinning; your meter stick will shorten as you turn it to measure the circumference, so although someone who observes the disc to be spinning will still measure a ratio of [tex]\pi=3.14159[/tex] they can easily understand why you will measure a larger ratio, since your meter stick shrinks when you turn it to measure the circumference. From your perspective riding along the disc, your meter stick doesn't shrink, but space expands in a larger gravitational field. You no longer live in Euclidean geometry.

    For a better understanding of this, refer to Chapter 23 of Einstein's Relativity: The Special and General Theory. Chapter 24 talks some about non-Euclidean geometry, so you may find that helpful as well.

    As Zanket points out, the dome idea is just to help you visualize a non-Euclidean 2D geometry in Euclidean 3D space. This is known as an "embedding diagram". No one (neither someone riding along the disc nor someone watching the disc spin) will actually observe a dome. Both observers will still see a circle (ignoring possible optical illusions).

    Also as Zanket points out, the embedding diagram for a spinning disc's geometry is not a dome because a dome has a smaller ratio between the diameter and the circumference than [tex]\pi=3.14159[/tex]. The correct embedding diagram would be one of a saddle, which has negative curvature.

    edit: However, now that I think about it, rather than stretching or breaking, the easiest thing for a spinning disc to do may be to bend into a dome shape in order to keep its same dimensions. If the disc bent into a dome shape, it would make up for the fact that it needs to fit in more circumferential space by contracting its circumference. However, a disc with the properties that would allow it to bend into a dome may be pretty extreme. Since the centrifugal forces will be trying to keep the disc from forming a dome, it would probably have to be pretty rigid in the radial direction, but malleable in a perpendicular direction, off the plain of the disc (up and down). My guess is that most realistic discs would probably break, and when you talk about a perfectly rigid disc you assume it keeps its form. And of course any disc that forms a dome will no longer be a disc, will it?

    It is important to note that what I am talking about here is not space and time forming a dome, but the actual atoms of the disc possibly aligning to form a dome shape, rather than a disc shape. If you want to talk about spacetime curvature, the curvature of the reference frame of the disc would have negative curvature; a dome has positive curvature. A saddle is an example of something with negative curvature. But I think caution needs to be exercised in thinking about embedding diagrams in this particular case. Space and time will not warp the disc into a saddle shape as observed by anybody. But if you tried to draw a diagram of the disc in visualized 3 dimensional Euclidean hyperspace, you would end up with needing to draw a saddle in order to give the correct ratio between the diameter and the circumference. But the disc will exist in non-Euclidean space, and will never take on a saddle-like form as observed by anyone.
    Last edited: Aug 10, 2005
  7. Aug 10, 2005 #6
    I'm a bit confused by these two parts. In the first part, would you not measure a greater [tex]\pi[/tex] than someone mearsuring from ontop of the disk? You would both agree on the radius, but not the circumference, as you would measure the circumference of the roatating disk as less than the guy riding around on it would you not?

    For the second part, a disk in euclidean geometry has c=[tex]\pi[/tex]d, when the disk is in spherical geometry isn't [tex]\pi[/tex] larger? So if we have a disk that we can measure [tex]\pi[/tex] as larger than than euclidean results it seems pretty practicle to deduce it could be explained as a 2d disk in a spherical geometry to me. :confused:
    Last edited: Aug 10, 2005
  8. Aug 10, 2005 #7


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    It's worth looking at the sci.physics.faq on the topic:


    The notion of the circumference of a rotating disk is somewhat ill-defined, because there isn't any global notion of simultaneity on the surface of a rotating disk - at least not an isotropic notion of simultaneity.

    I don't know if Einstein realized this or not, but later authors certainly did.

    Here's a quote from the sci.physics.faq

    There's a paper by Tartaglia that makes much the same point, though it's probably harder to read than the FAQ.


    When you measure the legnth or circumference of a moving object, you need to ensure that you measure the length or circumference of a set of points that are all "at the same time". This becomes problematical on the rotating disk with standard notions of simultaneity.

    People usually use a non-Einsteinian notion of simultaneity to deal with rotating disks (it's handier than drawing the equivalent of an international date-line on the disk).

    This is the ultimate origin of the Sagnac effect, and also the source of a great deal of confusion.
    Last edited: Aug 10, 2005
  9. Aug 10, 2005 #8
    I was fortunate enough to talk to Professor Lewin at MIT about this, here's what he said:

    Unfortunately, he didn't give me any more information than that, but I can't blame him. He's a very busy man and I'm happy he responded at all.

    He brings up the same point that's been brought up in this thread: any experimentation would rely on simultaneous measurements, which each person in the non-moving reference frame noting a different thing. What would it look like to each individual? Would it be an elliptical shape, as someone mentioned?
  10. Aug 10, 2005 #9
    I don't see how Lewin's getting "a circle whose circumference is less than its radius * 2 * pi". Shouldn't the circumference be greater? An excess circumference?
    Last edited: Aug 11, 2005
  11. Aug 10, 2005 #10
    Interesting. Thanks for the links.
  12. Aug 10, 2005 #11
    The red statements are conflicting. If you measure a smaller circumference than someone riding on top of the disc, then you would also have a smaller [tex]\pi[/tex]. Look at [tex]c={\pi}d[/tex]. If d stays the same and c increases, then [tex]\pi[/tex] must increase too.

    I have attached a picture to help you visualize it (you may have to wait for it to be approved by a mentor). If we take the formula for the circumference of a circle drawn on the surface of this sphere to be [tex]c=\pi'd[/tex] then [tex]\pi'[/tex] must be less than the Euclidean [tex]\pi=3.14159[/tex]. For the case of the specific circle drawn, you can easily see that the diameter (red line) makes a half circle across the surface of the sphere, which if extended to be a full circle would be the same size as the circumference of the blue circle. Since the diameter is half the size of the circumference of the circle, then in this case [tex]\pi'=2[/tex], which is less than 3.14159. As you can see, if the diameter is reduced, [tex]\pi'[/tex] will increase, but if the diameter is increased [tex]\pi'[/tex] will decrease. As the diameter approaches 0, [tex]\pi'[/tex] will approach 3.14159, and as the diameter approaches the circumference of the sphere, [tex]\pi'[/tex] will approach 0.

    Attached Files:

    Last edited: Aug 10, 2005
  13. Aug 10, 2005 #12
    Yes, I understand that part you corrected that was just a stupid mistake on my part, I was mostly wondering about these two parts inparticular here. The ratio wouldn't be Pi and it would infact seem to satisfy a positive curvature instead of a negative curvature, as c/d is smaller than in euclidean geometry unless I am being an idiot and making more dumb mistakes(which is very possible :yuck:). Thanks for the help
    Last edited: Aug 10, 2005
  14. Aug 10, 2005 #13
    I may have not been explicit enough in my wording, so I'll clarify just so I know we're even talking about the same situation. When I said "If you measure the diameter and the circumference, you will still find the ratio between them to be [tex]\pi[/tex]", I meant if you are observing the disc spin from a non-spinning inertial reference frame. In this reference frame your geometry will not change. There is no reason for a non-spinning inertial reference frame to take on a new geometry just because you observe a spinning disc in it. From this reference frame space is still Euclidean, so any disc has to have a circumference of [tex]c={\pi}d[/tex] where [tex]\pi=3.14159[/tex]. Although if you observed the individual atoms they would appear shrunken due to their relative speed, there is no possible way to construct a disc in Euclidean space which does not follow [tex]c={\pi}d; \pi=3.14159[/tex]. So what has to happen is the atoms will be pulled apart from each other, and depending on the rigidity of the disc, the disc will either have to break or stretch. In the case of a perfectly rigid disc, it is assumed that the atoms are just not as close together along the circumference if the disc is spinning close to the speed of light (the disc will become less dense but will maintain its shape). This means even for a spinning disc you will measure [tex]\pi=3.14159[/tex] in a non-spinning inertial reference frame. (For convenience, you can also imagine the disc as having been constructed on a platform which was already rotating, thus the disc won't have any problems accepting its non-Euclidean geometry and won't have to stretch or break to fit.)

    But if I am riding along on the disc, and you are watching me measure the disc, you will notice that my meter stick will be the same size as yours when I am measuring the radius but smaller than yours when I am measuring the circumference. From your reference frame, it is easy to see that I will get a larger value for the circumference of the disc than you did because my meter stick will be smaller but I will still have to measure the same length (as judged by you), thus my [tex]\pi>3.14159[/tex]. From my reference frame, I would say the disc is in a gravitational field, thus non-Euclidean, so I will also not be surprised when my [tex]\pi>3.14159[/tex].

    But for circles on the surface of a sphere, [tex]\pi {\leq} 3.14159[/tex].
    Last edited: Aug 10, 2005
  15. Aug 11, 2005 #14


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    I think I probably need to amplify the point I made a bit earlier with an example.

    Suppose you take a rotating disk, and mark your starting point with a red dot, or some other mark.

    Now, you very carefully take a ruler, and start mesauring the circumference of disk. In order to measure the circumference, you use a local inertial frame. Because the disk is accelerating, nearby points will not be moving at the same velocity as you are, but it turns out that this is not an insurmountable problem, *IF* you only measure the distance to a nearby points. WHen you get to the nearby point, you then use the inertial frame for that point to measure the distance to the next. You take the limit, where you make a very large number of measurements to points that become closer and closer.

    This is no more problematical than measuring lengths while living in a gravitational field, something we do all the time. Note that I've assumed that you have a ruler that is rigid enough not to stretch appreciably under the forces it experiences - while it can't be ideally rigid, it can be rigid enough if the distance is short enough.

    WHen you reach the red dot, you stop. You then proudly claim "Look, I've measured the circumference of the disk". You also note that it (the circumference) is not pi times the radius.

    But, there is a flaw in the procedure. This flaw will be revealed if you also plant stopwatches around the circumference of the disk as you measure the distances, and you carefully synchronize each stopwatch to the nearby ones in the appropriate inertial frame (the one in which you perform the length measurement).

    When you get all the way round the disk, and stop at the red dot, you'll see that your stopwatch does NOT match the stopwatch on the red dot!

    This means that you are not really measuring the length of a closed curve in space-time, the length of a a set of points that are all "at the same time". What you have measured is an open arc, whose ending time isn't the same as the starting time.

    For more details see the links I quoted (the sci.physics.faq, and the paper by Tartaglia).
    Last edited: Aug 11, 2005
  16. Aug 12, 2005 #15
    So are you saying the geometry of a rotating disc will not have negative curvature after all or just that it'd be harder to prove in the thought experiment than it seems at first sight or that there is no way to describe the geometry for a rotating disc?
  17. Aug 12, 2005 #16


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    To borrow some words from Tartaglia, I'm saying that the geometry of a rotating disk is "conventional", i.e. it depends on convention. Different authors have used different conventions to describe it. Tartaglia also calls the circumerence "not well defined" for the same reason.

    And I'm also saying that it's _definitely_ wrong to compute the travel time of a light beam around the circumference as circumference/velocity.
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