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Analogy between centrifugal force and gravity

  1. Dec 8, 2011 #1
    I understand the equivalence of gravity and uniform linear acceleration. But I'm less clear on the example involving circular motion (merry-go-round) and the equivalence of gravity and centrifugal force, because I'm unclear about the connections between SR & GR.

    The example does illuminate certain points, like the sense in which standing on the earth's surface is a matter of constant "acceleration" away from the spacetime geodesic (because centrifugal force is really constant acceleration away from the direction perpendicular to radial). And of course it shows a connection between time dilation due to relative motion (faster motion away from center) and due to gravity.

    But is this connection an identity or an analogy? How does one get from Lorentz transformations to a curved spacetime metric in this example?

    It seems to me that the motion-based length contraction would drop out of consideration because it would be perpendicular to the radial direction. Then do the tidal forces produced in the "centrifugal field" somehow match or account for the difference between coordinate ("reduced circumference") and proper length in curved spacetime?
     
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  3. Dec 9, 2011 #2

    atyy

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  4. Dec 9, 2011 #3
    Now I think I didn't understand the analogy between gravity and linear acceleration after all. There too we're comparing flat to curved spacetime. Also, there is no gradient of force/curvature varying with distance: in a constantly accelerating room, the inertial force is the same everywhere in the room. So my question about building a metric up out of Lorentz transformations was probably misguided. It's all just an analogy. Or is there a metric for a field of constant curvature?
     
  5. Dec 9, 2011 #4

    A.T.

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    This is about the uniform field:
    http://www.mathpages.com/home/kmath530/kmath530.htm

    This is about the rotating frame:
    http://www.projects.science.uu.nl/igg/dieks/rotation.pdf [Broken]
     
    Last edited by a moderator: May 5, 2017
  6. Dec 9, 2011 #5
    Thanks A.T.
    The first one is mostly over my head, but the second one set me straight on one basic thing: Lorentz contraction around the circumference of a rotating body already results in non-Euclidean geometry. (I had been focused on the gradient along the radial direction and so ignored contraction.)

    I had been thinking that one might derive a metric for the rotating frame in which a mass (or pseudo-mass) term (for an imaginary gravitational center) emerged out of the math, in something like the way it does in the Schwarzschild derivation. I don't know how to think about the fact that the field, while symmetric around the center, gets stronger rather than weaker away from it. But I imagined that the speed limit c would create some analog of an event horizon as a function of angular speed and radial distance. (It still isn't clear to me, either, as to whether we have to worry about issues involved in accelerating a very large physical disc, and what we're assuming if we just look at the rotating coordinate frame per se, existing, as it were, from eternity.)
     
  7. Dec 9, 2011 #6

    PeterDonis

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    I assume you mean non-Euclidean *spatial* geometry. The geometry of the spacetime as a whole is still flat.

    But actually, it's even more complicated than that. Strictly speaking, the term "the spatial geometry of a rotating disk" is not well-defined. (I don't know if you were imagining a rotating disk in the OP, or just observers all "rotating in unison" in some sense; imagining the actual disk makes it easier to see what's going on, IMO.)

    In order to properly define a "spatial geometry", we have to take a spacelike "slice" out of the spacetime--we have to pick out a specific 3-surface from the 4-dimensional spacetime and call that 3-surface "space", and then look at what geometry it has.

    For an inertial frame, this is easy: we just take any "surface of simultaneity" in that frame. This works because all observers at rest in the frame share the same surfaces of simultaneity, so the same "space" works for all of them.

    But for rotating observers, this doesn't work, because they have no surfaces of simultaneity in common! This is because, if you look closely, you will see that every single observer in the "rotating frame" has a different velocity vector (combination of speed and direction), as seen from a non-rotating inertial frame. This means no two of them are at rest relative to each other, so no two of them share the same definition of "simultaneity". And *that* means that there is no way to pick out any 3-surface as "space" that will naturally look like the "space" of more than one such observer.

    This does not mean we can't define "coordinates for the rotating disk" at all; we can (see below). But it does mean that no such coordinates can capture "the spatial geometry of the rotating disk", because there is no such thing. We can pick out a 3-surface by picking a particular definition of "simultaneity" and look at its geometry; the usual convention is to take the simultaneity of the observer at the center of the disk, who is not moving and who therefore has the same sense of simultaneity as the non-rotating inertial observer. But the geometry of such a 3-surface will not be "the" spatial geometry of the disk; the other rotating observers will see different spatial distances between each other.

    You can, but it has limitations (some of which you allude to). See here:

    http://en.wikipedia.org/wiki/Born_coordinates

    It does. The Born coordinates described in the Wikipedia article are only valid out to a certain radius, basically the radius at which the angular speed translates into a linear speed equal to the speed of light.

    You do. See here:

    http://en.wikipedia.org/wiki/Ehrenfest_paradox
     
  8. Dec 9, 2011 #7

    PeterDonis

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    After looking at the Dieks paper in more detail, I can expand some more on the issues I raised in my last post; I'm not even sure that saying the *spatial* "geometry" is non-Euclidean is a fruitful way to describe what's going on.

    Dieks' statement about the rotating frame having "non-Euclidean spatial geometry" is based on his equation (5), which is (I have rearranged terms slightly and changed to units where c = 1):

    [tex]dl^{2} = dz^{2} + dr^{2} + \frac{r^{2} d\phi^{2}}{1 - \omega^{2} r^{2}}[/tex]

    However, this so-called "spatial line element" is *not* the line element of *any* 3-dimensional spatial slice that can actually be "cut" out of the spacetime as a whole! At best, it is an "infinitesimal" line element that describes "spatial geometry" in the immediate vicinity of a single rotating observer. (Dieks says the line element describes the "spatial distance between two infinitesimally near points", but he then blithely goes on to talk about it as though it described the geometry of an entire 3-D spatial slice, when in fact the way he derives the expression clearly makes that impossible.)

    Let's look at the metric for the spacetime as a whole in the "rotating frame" to see what I mean (this is basically the Born chart from the Wikipedia page I linked to before; I have written [itex]d\tau[/itex] instead of [itex]ds[/itex] because Dieks uses the timelike sign convention for the metric signature):

    [tex]d\tau^{2} = \left( 1 - \omega^{2} r^{2} \right) dt^{2} - dz^{2} - dr^{2} - r^{2} d\phi^{2} - 2 \omega r^{2} dt d\phi[/tex]

    The obvious way to cut a 3-D spacelike slice is to set dt = 0 (this corresponds to the simultaneity convention of the observer at the center of the disk, who is not moving relative to an inertial frame); but this gives an induced metric on the slice that is flat (set dt = 0 in the above and switch signs to indicate a spacelike interval):

    [tex]dl^{2} = dz^{2} + dr^{2} + r^{2} d\phi^{2}[/tex]

    So that can't be the "non-Euclidean geometry" of the disk. And you will search in vain for any other way of cutting a 3-D slice that results in an induced spatial metric equal to Dieks' equation (5) above. In other words, there is *no* coordinate chart that can be constructed such that the equivalent of what I just did, setting "dt" to zero and seeing what's left of the metric, will yield Dieks' equation.
     
  9. Dec 10, 2011 #8

    A.T.

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    Here I disagree. To define a "spatial geometry" you don't need to know anything about 4D spacetime and slices trough it. You simply place rulers at rest in the rotating frame and you find that circumference > 2*pi*r.

    disk_at_rest.gif

    rotating_disk%20animated10fps.gif

    We had a very long discussion about this, where people brought the same arguments as you do here, about why this is not a sensible interpretation:

    You cannot define a slice through space time, for this space : It just means that the concept of "space as slice throuh space time" is not as general as some think.

    You cannot use Einstein synchronization of clocks (at r=const) : That is just a convention. You can use other conventions (signal from center). And you actually don't even need clocks to measure space geometry, just rulers.

    If you were building a giant rotating space station from prefabricated elements, you would have to produce more elements for the circumference, than for a non rotating space-station with the same radius. That is the physically relevant spatial geometry.

    The question if you can reconcile this with the idea that "space must be a 3D slice trough a 4D space-time" is of zero physical relevance. If you cannot, then you can always go back to the basics defined by Einstein, before 4D space-time was introduced into Relativty by Minkowski:

    Time is what clocks measure.
    Space is what rulers measure.
     
    Last edited: Dec 10, 2011
  10. Dec 10, 2011 #9

    DrGreg

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    When I first encountered this a year or two ago, this was something I found hard to grasp at first, but in fact it isn't always the case that 3D space must correspond to a 3D surface-of-simultaneity in 4D spacetime. That is something that seems natural in "static" coordinate systems such as Minkowski coordinates, or Rindler coordinates, or Schwarzschild coordinates, but in this case we are dealing with a lattice of observers who are "stationary" but not "static" in the technical GR sense of those terms. We have a congruence of worldlines for observers on a rotating disk who determine themselves to be a fixed distance from each other (e.g. by a radar signal that undergoes no red-shift or blue-shift on the round trip). The metric for the 3D space of these observers needs to determine a distance between neighbouring worldlines, which is the formula you quoted above, and is the constant orthogonal separation of two infinitesimally close worldlines. That is sufficient to determine the distance between points in 3D space that are at rest with respect to each other.

    It's true you cannot "glue together" the small separations to form a consistent 3D surface in 4D spacetime, but that doesn't matter because no such surface is required to define distance in 3D space. To put it technically, the 3D space we are constructing here is not the orthogonal complement of the worldlines; it is a quotient space. For static worldlines, the two concepts coincide, but for stationary non-static worldlines they do not.

    Another way of looking at this is that, for a stationary lattice you do not need a definition of simultaneity to measure distance between lattice points because the distance is constant over time.

    This topic is covered in more detail in Rindler W (2006, 2nd ed), Relativity: Special, General, and Cosmological, Oxford University Press, ISBN 978-0-19-856732-5 Section 9.7 p.198. Earlier in that chapter, Rindler proves a more general result to show an arbitrary stationary spacetime's metric can be expressed in the form[tex]
    ds^2 = e^{2\Phi/c^2}\left( c\,dt - \frac{w_i}{c^2}dx^i\right)^2 - k_{ij}\,dx^i\,dx^j
    [/tex]where [itex]\Phi,\,w_i,\,k_{ij}[/itex] are all independent of t. Here, [itex]k_{ij}\,dx^i\,dx^j[/itex]is the time-independent metric of 3D space, but [itex]c\,dt - w_i\,dx^i/c^2[/itex] isn't the differential of any coordinate, except in the static case when [itex]\textbf{w} = \textbf{0}[/itex].
     
  11. Dec 10, 2011 #10

    PeterDonis

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    Ah, you're right, I wasn't taking into account the fact that the "rotating" worldlines are stationary. So the line element Dieks gives is the (non-Euclidean) metric on the quotient space, and that space can be thought of as the "space of the rotating frame", because it is time independent. Got it.

    I agree on the geometry (see my response to DrGreg just above), but the actual physical process of building the station gets into the Ehrenfest paradox: how do you spin up the station? If you build it in its non-rotating state, there's nowhere to put the extra elements for the circumference; or, alternatively, there will be missing elements in the radius if you build the circumference with the extra elements included.

    Instead, I would say the elements have to have some elasticity to them, so they can expand or compress as needed to adjust to the changing stresses as the station spins up. (For example, they could be a souped-up version of the sound-absorbing mounts used on submarines to mount equipment to the hull; the mounts expand and compress as needed to cancel the vibration of the equipment, so no sound is emitted by the hull.) Basically, the circumferential elements will experience compressive stress and the radial elements will experience tensile stress (stretching). What state the station finally settles into when its rotation frequency is established will depend on the detailed material properties; in principle there could be any combination of compressing the circumference and stretching the radius.
     
  12. Dec 10, 2011 #11

    A.T.

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    I build it already spinning. The elements are basically rulers, with rockets attached. But that's an engineering problem. Physics doesn't care how the rulers arrive at rest in a common rotating frame. It just says, that if they do, they will measure a non-Euclidean geometry.
     
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