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Does an observer's proper frame give a unique foliation?

  1. Mar 11, 2015 #1

    andrewkirk

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    By 'proper frame' of observer O, I mean any reference frame (coordinate system) in which

    (Condition A:) The worldline of O is always at the spatial origin for every time coordinate.

    Clearly such a frame is not unique because spatial rotations do not invalidate (A). What I am interested in is whether it determines a unique foliation of spacetime, so that the time coordinate is unique, modulo a change of time units.

    My guess is that the answer is No, but that a proper frameof O is 'approximately unique' near O, meaning something like that for any two frames C1 and C2 that are proper relative to O, the time coordinates of a point x under frames C1 and C2 are the same, to first order, for x near O.

    I can't recall seeing a theorem about this. The closest I can remember is about the existence of Normal Coordinates, but that's a somewhat different issue, as Normal Coordinates are for a locally inertial frame, and the proper frames we are talking about here need not be inertial.

    Thank you.
     
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  3. Mar 11, 2015 #2

    PeterDonis

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    A more precise way to say this is that, for any observer O, we can construct a "proper frame" for O that assigns unique coordinates to events in some neighborhood of O's worldline, but it will not assign unique coordinates to events outside that neighborhood. Such a coordinate chart will define a unique foliation of spacetime within that same neighborhood of O's worldline, but not outside it. (The size of the neighborhood will be of order ##1 / a##, where ##a## is the proper acceleration of O's worldline.)

    This basically means C1 and C2 are related by a spatial rotation only. If the rotation is not time-dependent, then yes, I think what you say is true. But if, for example, C2 is rotating at some angular velocity ##\omega## relative to C1, then the two frames will not define the same foliation of spacetime within the same neighborhood of O's worldline; the time coordinates they assign can be made to be the same only on O's worldline, not throughout the neighborhood.

    More precisely, Riemann normal coordinates are for a local inertial frame. But no local inertial frame can be a "proper frame" in your sense for any observer, because even if the observer is inertial, his worldline will diverge from the "time axis" of the local inertial frame once it gets far enough from the origin of the LIF for tidal gravity to become non-negligible.

    What you are describing are Fermi normal coordinates, which can be constructed around the worldline of any observer at all, inertial or with arbitrary (and possibly time-dependent) proper acceleration. These coordinates define a "proper frame" in your sense, because the observer's worldline is always at the spatial origin. But the coordinate values this frame assigns will not, in general, be the same as for Riemann normal coordinates, even for an inertial observer within a small patch of spacetime on which the Riemann normal coordinates are defined.
     
  4. Mar 13, 2015 #3

    andrewkirk

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    Thank you Peter. If we add the constraint that the 'proper' coordinate system must everywhere have a time axis that is orthogonal to the constant-time hypersurfaces (isotemps), does that make the foliation it generates unique?

    Formally, given spacetime [itex]M[/itex] and a time-like worldline [itex]\gamma:(0,1)\to M[/itex] of an observer, with image Im [itex]\gamma=W[/itex], is it the case that, for any two coordinate systems [itex]C1,C2:M\to \mathbb{R}^4[/itex] that satisfy the [itex]W[/itex]-orthogonality criterion below, the isotemps that intersect [itex]W[/itex] will agree wherever they are both defined (ie for any two points [itex]u,v\in M[/itex] that are [itex]C1[/itex]-simultaneous with each other and with a point [itex]p1\in W[/itex], at least one of which, say [itex]u[/itex], is also [itex]C2[/itex]-simultaneous with a point [itex]p2\in W[/itex], we must have [itex]p1=p2[/itex] and [itex]u,v[/itex] must also be [itex]C2[/itex]-simultaneous).

    [itex]W[/itex]-Orthogonality Criterion
    We say a global coordinate system [itex]C:M\to \mathbb{R}^4[/itex] satisfies this criterion if, for every point [itex]p\in M[/itex] that has a point [itex]q_p\in W[/itex] with which it is [itex]C[/itex]-simultaneous, and for all [itex]x,y,z\in\mathbb{R}[/itex], it is the case that [itex]\langle\vec{e}_0(p),\vec{v}\rangle=0[/itex], where [itex]\vec{e}_0(p)\in T_pM[/itex] is the time coordinate vector of coordinate system [itex]C[/itex] at [itex]p[/itex] and [itex]\vec{v}\in T_pM[/itex] is the vector with components [itex](0,x,y,z)[/itex] in the basis of [itex]T_pM[/itex] derived from coordinate system [itex]C[/itex] (ie the time axis is everywhere orthogonal to the isotemps).

    This criterion is weaker than a global orthogonality criterion because it only requires the orthogonality to apply at points that are simultaneous with points in the worldline [itex]W[/itex].

    Example
    Consider a 2D spacetime plotted on the number plane with time axis vertical, so that isotemps under the natural, Cartesian coordinate system (call that [itex]C1[/itex]) are horizontal lines. Let [itex]C2[/itex] be the coordinate system whose time axes are still vertical but spatial axes (isotemps) have a constant gradient of 10%. Both of these coordinate systems are in a sense proper coordinate systems of an observer whose worldline heads vertically upwards along the [itex]C1[/itex] time axis, and they define different isotemps. But if we impose the Orthogonality Criterion, it rules out [itex]C2[/itex], whose isotemps are not orthogonal to the time axes.
     
    Last edited: Mar 13, 2015
  5. Mar 13, 2015 #4

    PeterDonis

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    Fermi normal coordinates already obey this constraint. As I said, they give a unique foliation in a neighborhood of O's worldline, but not outside it. There's no way to do any better; the non-uniqueness is because of spacetime curvature and (if applicable) the proper acceleration of the worldline. You can't eliminate those by changing coordinates.
     
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