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Lorentz Contraction Circular Motion

  1. Jan 7, 2010 #1
    Does Acceleration affect Lorentz contraction?

    Suppose their was a Circle spinning around its center then its outer edges would decrease in length.

    looking somewhat similar to a saw blade or something.

    This doesn't make sense though because then then object would change shape depending on what frame of reference you are looking at.

    However all the examples of Lorentz contraction i have seen involve objects moving in a straight line.

    Circular motion could be viewed as an object constantly accelerating around a point, which is what lead me to the question, does acceleration affect Lorentz contraction.
     
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  3. Jan 7, 2010 #2

    A.T.

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    Last edited by a moderator: May 4, 2017
  4. Jan 7, 2010 #3

    bcrowell

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    Here's my own way of understanding it: http://www.lightandmatter.com/html_books/genrel/ch03/ch03.html#Section3.4 [Broken]

    It occurs to me for the first time that some aspects of this problem are very similar to Bell's spaceship paradox, http://math.ucr.edu/home/baez/physics/Relativity/SR/spaceship_puzzle.html [Broken] . Essentially Ehrenfest's paradox is just the same paradox, but wrapped around into a circle.
     
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  5. Jan 7, 2010 #4

    Fredrik

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    Short answer: It isn't possible to increase the angular velocity of a solid disc without forcefully stretching it. If the circumference remains the same, it's because the material has been forcefully streched exactly by a factor of [itex]\gamma[/itex], so that it exactly compensates for the Lorentz contraction.

    I would be surprised if the radius remains the same. The rotating disc would try to restore itself from the forceful stretching. (It's like it consists of a bunch of rubber bands of different sizes, all stretched to longer lengths that their equilibrium lengths). The sum of those forces on an (arbitrary) atom in the disc would be toward the center. But there's also a centrifugal force pulling the atom away from the center. These two forces would have to exactly cancel for the disc to keep its shape.

    I agree that it's essentially the same as Bell's spaceship paradox.

    I wouldn't say that the geometry is non-euclidean in the rotating frame, since the rotating frame agrees with the non-rotating frame about which hypersurface of spacetime to call "space, at time t". That hypersurface is flat, and flatness is a coordinate independent property.
     
  6. Jan 8, 2010 #5

    A.T.

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    Do you agree that a ruler at rest in the rotating frame will measure a circumference greater than 2*PI ?
    http://img688.imageshack.us/img688/4590/circleruler.png [Broken]

    If yes, how can the spatial geometry be euclidean in the rotating frame? A circumference greater than 2*PI usually implies negative curvature.
     
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  7. Jan 8, 2010 #6

    Ich

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    But you would certainly expect that space is orthogonal to time, which is a bit of a problem in the rotating frame.
     
  8. Jan 8, 2010 #7

    Fredrik

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    Yes.

    The spatial geometry of the rotating frame depends only on which subset of spacetime the rotating frame considers "space" at a given time t. This is the set of points that are assigned time coordinate t by the rotating frame, and it's the same set of points that are assigned time coordinate t by an inertial frame that's co-moving with the point at the center.

    The measurements you describe don't have anything to do with the spatial geometry in the rotating frame. Those measurements will just tell you the lengths of the segments in a bunch of different inertial frames. (Someone should slap Brian Greene with a fish :wink:)
     
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  9. Jan 8, 2010 #8

    bcrowell

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    Yes, this is a little subtle. Correct me if I'm wrong, but here is what I think is the situation.

    In the nonrotating frame, we have
    [tex]
    d s^2=d t^2 - d r^2 - r^2d \theta^2
    [/tex]
    With the transformation [itex]\theta'=\theta-\omega t[/itex], we get
    [tex]
    d s^2=(1-\omega^2 r^2)d t^2 - d r^2 - r^2d \theta'^2 - 2\omega r^2d \theta'd t
    [/tex]
    The presence of the cross-term means that there is no clean separation between the time and spatial coordinates, so there's no clean way to separate the metric into parts that you can identify as temporal and spatial. You can complete the square,
    [tex]
    d s^2=(1-\omega^2 r^2)\left[d t+\frac{\omega r^2}{1-\omega^2 r^2}d \theta'\right]^2 - d r^2 - \frac{r^2}{1-\omega^2r^2}d \theta'^2
    [/tex]
    The thing in square brackets is not quite the total differential of [itex]T = t + \frac{\omega r^2}{1-\omega^2 r^2}\theta'[/itex]. I need to think about this a little more. I think you can handle this with Rindler coordinates or something...? Anyway, if you restrict yourself to synchronizing clocks at the circumference of the disk, where r is constant, then you don't have to worry about the fact that the the thing in square brackets isn't the total differential of T. Admittedly there is something not quite right here, but passing over that point:
    [tex]
    d s^2=(1-\omega^2 r^2)dT^2 - d r^2 - \frac{r^2}{1-\omega^2r^2}d \theta'^2
    [/tex]
    This separates cleanly into parts that can be identified as temporal and spatial. The coordinate T isn't quite what you want as a synchronized time coordinate, because it has a discontinuity as a function of [itex]\theta'[/itex]. You can't synchronize clocks in a rotating coordinate system. This metric is still flat, because you got it by doing a coordinate transformation on a flat metric. However, if you now write down just the spatial part,
    [tex]
    d s^2= d r^2 + \frac{r^2}{1-\omega^2r^2}d \theta'^2
    [/tex]
    it's non-Euclidean.

    So I think the story here is that the geometry in the rotating frame is non-Euclidean, in the sense that if you do the best job of clock synchronization that you can, in the most natural way, then the hypersurfaces of constant time have curvature. The time coordinate T is natural because it corresponds to putting n clocks around the circumference of the disk and synchronizing clock j with clock j+1 in the sense that an inertial observer who is momentarily coincides with the midpoint between the clocks will receive flashes of light from them simultaneously. You synchronize 1 with 2, 2 with 3, ... and end up with perfect synchronization except that there's a big discontinuity between clocks n and 1.
     
  10. Jan 8, 2010 #9

    A.T.

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    You don't need to consider spacetime to measure spatial geometry in a frame. You just need a bunch of rulers at rest in that frame. And if you lay out that bunch of resting rulers in the rotating frame, you will see that the spatial geometry is non-euclidean there. The rulers are at rest, so time doesn't matter.
    The measurements I describe to determine spatial geometry is: using rulers a rest. What is wrong with that? You could use the same method to measure the non-euclidean spatial geometry in a Schwarzschild-spacetime.
    A rotating frame is a "bunch of different inertial frames". When you lay out rulers in a Schwarzschild-spacetime. you also get just "lengths of the segments in a bunch of different inertial frames", because inertial frames are only local there. But you still call the measurement of these resting rulers which stretch across different inertial frames: "the spatial geometry".
    Maybe that is the core of the disagreement here: Fredrik is talking about the space-time metric, while I mean the purely spatial geometry.
     
    Last edited: Jan 8, 2010
  11. Jan 8, 2010 #10

    Fredrik

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    The frame is a function from spacetime to [itex]\mathbb R^4[/itex], and until you have considered its definition, you don't even know what the word "spatial" refers to.

    What's right with it? I don't know why you think this has anything to do with the spatial geometry in the rotating frame. (I haven't completely ruled out that I have misunderstood something).

    I know what you mean, but the metric of space is induced by the metric of spacetime. More specifically, it's the inclusion function pullback of the spactime metric: Define [itex]I:S_t\rightarrow M[/itex], where St is space at time t and M is spacetime, by I(x)=x for all x in St. The metric on St is [itex]I^*g[/itex] (where g is the metric on M). We clearly can't talk about the geometry of St until we know what St is.

    So how do we find St? It's defined as the set of all points that are assigned time t by the rotating frame, so we need to know how the rotating frame is defined. It's defined by the substitution [itex]\varphi\rightarrow\varphi-\omega t[/itex] in the usual polar coordinates

    [tex]x=r\cos\varphi[/tex]
    [tex]y=r\sin\varphi[/tex]

    So the relationship between the inertial coordinates and the rotating coordinates is

    [tex]t=t'[/tex]
    [tex]x=r'\cos(\varphi'-\omega t')[/tex]
    [tex]y=r'\sin(\varphi'-\omega t)[/tex]
    [tex]z=z'[/tex]

    where I have put primes on all the coordinates of the rotating frame to clearly distinguish them from the coordinates of the inertial frame. Since the rotating frame assigns time t to exactly those points that are assigned time t by the inertial frame, St is just the hypersurface that's assigned time t by the inertial frame. The inclusion function pullback of the Minkowski metric to such a hypersurface is clearly Euclidean.
     
    Last edited: Jan 8, 2010
  12. Jan 8, 2010 #11

    Ich

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    Fredrik, you may of course call any 3-D-slice "space". But usually, you would like to have space orthogonal to the bundle of observer worldlines that defines it. That is not possible if rotation is involved, you run into the problems bcrowell describes: you have to slice the infinite helix into pieces, and those pieces are not euclidean, of course.
    If you borrow the synchronization from the inertial frame - as it is done in the real world to define a global time - that's ok, but arguably not "space in the rotating frame".
     
  13. Jan 8, 2010 #12

    Fredrik

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    I think my edit (which I made before I saw your post) answers that. How else would you define the rotating frame? Each segment of the world line of any tiny piece of the disc defines a local inertial frame, but I don't see why anyone would say that those world lines have anything to do with a "rotating frame".
     
  14. Jan 8, 2010 #13

    pervect

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    This is *the* key point, in my opinion. One might add "using the Einstein convention" to the above remark.

    There are enough papers on this problem to fill a book. "Relativity in rotating frames: relativistic physics in rotating reference frames", Guido Rizzi, Matteo Luca Ruggiero for instance. While this book is good in that it shows a large number of viewpoints, it might be less useful for teaching or learning.

    One can't really talk about the geometry of space until one defines how to slice space from space-time, by determining the mechanism that defines the points of space-time that compromise the 3 dimensional "space" corresponding to some specific time instant t. The problem becomes even more acute when one tries to explain this point to a more general audience - I'm not convinced any attempt I've seen has really succeeded :-(.
     
  15. Jan 8, 2010 #14

    Ich

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    Have a look at http://books.google.com/books?id=IyJhCHAryuUC&pg=PA90#v=onepage&q=&f=false". (I hope the link works)
    I'm sure I've read a more exhaustive paper by Gron, but I can't find it right now.
    --But I see that pervect joined the discussion, IIRC he gave me the link back then.
     
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  16. Jan 8, 2010 #15
    I question if a tiny piece of a rotating disc can be defined as a local inertial frame. An observer located at any point would feel and measure acceleration and even locally it would seem difficult to transform that fact away. Perhaps you mean local inertial frames in the sense of momentarily co-moving frames or perhaps I misunderstanding what you are saying?

    For example can we claim somebody standing on the surface of the Earth or in the more extreme standing on a neutron star is in a local inertial frame?
     
  17. Jan 8, 2010 #16

    A.T.

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    For me "spatial geometry" in a frame means the spatial metric in that frame, determined by spatial distances which are measured with rulers at rest in that frame.

    Do you agree that the spatial metric around a big mass is non-Euclidean, because the circumference at a distance r is different from 2*PI*r, when both lengths are measured with rulers at rest? The same applies to rulers at rest in a rotating frame.
     
  18. Jan 8, 2010 #17
    You can synchronise clocks in rotating frame using the Einstein convention as long as you restrict yourself to a small portion of the circumference. It is only impossible to synchronise clocks if you try to synchronise the clocks all the way around.
     
  19. Jan 8, 2010 #18

    A.T.

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    How is that preventing you from measuring spatial geometry with rulers? You can't synchronize clocks around a massive object either, but you can still say that the spatial geometry is non-Euclidean there. What is different in a rotating frame?

    Why do you have to consider the time dimension at all? You have rulers placed at rest in the rotating frame. What they measure doesn't change with time.
     
  20. Jan 8, 2010 #19

    bcrowell

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    Gron, Relativistic description of a rotating disk, Am. J. Phys. 43 (1975) 869

    The Gron paper discusses this in excruciating detail. He considers three different observers: S is at rest with respect to the disk's axis; Sk is an observer in an inertial frame whose position and velocity instantaneously coincide with a point on the edge of the disk; S' is an observer rotating with the disk.

    In SR, there is a notion of an inertial frame. S and Sk are inertial in the SR sense.

    In GR the notion of an inertial frame is not useful. What's useful is the notion of a free-falling frame, i.e., the frame of an observer who is not subject to any nongravitational forces.

    The historical interest of the example comes from the fact that Einstein used it as a bridge from SR to GR. He figured out that the spatial geometry was non-Euclidean, as measured by rulers at rest with respect to the disk (Gron's observer S'). By the equivalence principle, S' can describe the motion of Sk as arising from a gravitational field. S' says, in agreement with all the other observers, that the local four-dimensional spacetime is flat. However, you can have a flat spacetime that is still permeated with a gravitational field, according to one observer's coordinates.

    S' says, "There's an outward gravitational field in this region of flat space. By hanging on to the disk, I can keep from falling. Sk is falling. Sk's frame is a free-falling frame, and he agrees with me on that. I consider my frame to be inertial and Sk's to be noninertial, but Sk says it's the other way around."

    Sk says, "There's no gravitational field in this region of flat space. S' is accelerating due to the force from the disk. My frame is a free-falling frame, and S' agrees with me on that. I consider my frame to be inertial and the frame of S' to be noninertial, but S' says it's the other way around."
     
  21. Jan 8, 2010 #20

    bcrowell

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    In the four-dimensional space around a massive object, you can measure the Riemann tensor, which is local and coordinate-independent. Therefore the issue doesn't arise.

    Here we're talking about measuring the curvature of a three-dimensional subspace defined by a constant t. Unless you can define t, you can't define what subspace you're talking about. Once the subspace is defined, then you can measure the three-dimensional Riemann tensor within that subspace.
     
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