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I What is the source of curvature in an accelerating frame?

  1. Mar 25, 2016 #1
    I can understand that a free falling frame in a curved space with a non-zero Riemann tensor has a zero Ricci tensor but I have a doubt about the opposite; does an accelerating frame in a vacuum space have a non-zero Ricci or Riemann tensor? if so, where do components of energy momentum tensor come from then?
    Also what is the law of general transformation between the curved space-time and the flat space-time in a free-falling inertial frame?
     
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  3. Mar 25, 2016 #2

    PeterDonis

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    No. Both are zero.
     
  4. Mar 25, 2016 #3

    PeterDonis

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    You're going to have to be more specific about what you mean here. There are an infinite number of possible coordinates in curved spacetime.
     
  5. Mar 25, 2016 #4
    So if Ricci tensor is zero, the solution of the metric equation means a flat space. But the accelerating frame coordinates are not flat. Is it possible to have another solution to a zero Ricci?
     
  6. Mar 25, 2016 #5

    PeterDonis

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    "Flat" isn't a property of coordinates. It's a property of spacetime. The coordinates of an accelerating frame are not inertial, and don't have all the same properties as inertial coordinates; but that has nothing to do with the flatness or lack thereof of spacetime itself.

    If you want to check this, compute the Ricci tensor from the coordinates of an accelerating frame in flat spacetime--for example, from Rindler coordinates. You will find that it is zero.
     
  7. Mar 25, 2016 #6
    If it is zero, how does the observer in that accelerating frame explain the gravity which exists according to the equivalence principle.

    Also if the flattness is a property of a space, does this mean the space of an accelerating frame is also flat?
     
  8. Mar 26, 2016 #7

    A.T.

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    Gravitational coordinate acceleration doesn't require intrinsic spacetime curvature (which is related to tidal effects). For example, the cone spacetime in the below video is intrinsically flat (negligible tidal effect on small scale):



    See also this post:
     
  9. Mar 26, 2016 #8
    By operating any arbitrary coordinate transformation from a free falling inertial frame you get the geodesic equation - expressed in terms of the Christofflel symbol

    If the frame is free falling in a gravitational field, you find the generated inertial force to be the gravitational force in the new fixed referential (laboratory referential)
     
    Last edited by a moderator: Mar 27, 2016
  10. Mar 26, 2016 #9

    Nugatory

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    You are misunderstanding the equivalence principle. It doesn't say that acceleration is gravity, it says that the local effects of the two are equivalent. You can have acceleration without gravity.

    Saying "the space of an accelerating frame" makes no sense. The frame is just a rule we using for assigning coordinates to points on space, so different frames are still working with the same space. In the absence of gravity, space is flat no matter which frames we use describe it.
     
  11. Mar 26, 2016 #10

    A.T.

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    In the absence of gravity, spacetime is flat. As for space, it depends how you define it. For example, the spatial geometry measured with rulers at rest in a rotating frame is not flat.
     
  12. Mar 26, 2016 #11

    Nugatory

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    Thank you - yes - that was typing my reply too quickly. It's always spacetime curvature that we're talking about here, not space.
     
  13. Mar 26, 2016 #12
    So, my two concerns here in this thread can be demonstrated in those nice graphs;
    1) Graph C shows how a curved space-time (gravity) with non-vanishing Riemann tensor may have a vanishing Ricci tensor.
    2) Graph B shows how an accelerating frame with non-inertial coordinates can exit in a flat space-time (with no gravity).

    Regarding the first point, will the vanishing Ricci tensor imply an inertial frame, ignoring tidal effect? Can this be shown mathematically? For example, will vanishing all second derivatives of the metric imply automatically that we are ignoring the tidal effect?

    Regarding the second point, will be any corresponding vanishing Ricci tensor in 2D? For example, with plane polar coordinates in a flat space, the line elements;
    $$ds^2=dr^2+r^2 d\theta^2$$
    here the metric ##g_{22}## is not zero, and the second derivatives of it with respect to ##r## is not zero, will Ricci tensor be still zero?
     
    Last edited: Mar 26, 2016
  14. Mar 26, 2016 #13

    PeterDonis

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    \
    You can always construct a local inertial frame in any spacetime, whether the Ricci tensor vanishes or not.

    No; ignoring all second derivatives of the metric implies that we are ignoring all tidal effects. This is what we do in a local inertial frame: we choose a particular event of interest, and we construct coordinates such that the metric ##g_{\mu \nu}## at the chosen event is equal to the Minkowski metric ##\eta_{\mu \nu}##, and all first derivatives of the metric at the chosen event are zero. This can always be done. But we cannot make all of the second derivatives of the metric vanish, unless the spacetime as a whole is flat. The best we can do in a curved spacetime is to restrict attention to a patch of spacetime around the chosen event that is small enough that the second derivatives of the metric, within that patch, can be ignored, to whatever level of accuracy of measurement we are using. This amounts to choosing the patch of spacetime such that within it, tidal effects are small enough that we can ignore them.

    But the Riemann tensor is zero; the flat plane is still a flat plane. (Try computing the Riemann tensor explicitly; it's not as simple as just computing the second derivatives of the metric.) So the Ricci tensor is obviously zero; but that's because the space is flat.
     
  15. Mar 26, 2016 #14
    I like the bold very much.
     
  16. Mar 26, 2016 #15

    vanhees71

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    In this case, it's very easy, because we know how we can find new coordinates, where the metric components become constant, namly Cartesian coordinates
    $$x=r \cos \vartheta, \quad y=r \sin \vartheta.$$
    We also know that in the new coordinates the metric components are just given by ##\delta_{jk}##. This is easy to prove, because
    $$g_{jk}' \mathrm{d} {q'}^j \mathrm{d} {q'}^k=\delta_{jk} \mathrm{d} {q'}^j \mathrm{d} {q'}^k =\mathrm{d} x^2 + \mathrm{d} y^2.$$
    Indeed, it's easy to prove. We have
    $$\mathrm{d}x = \mathrm{d}r \cos \vartheta - r \sin \vartheta \mathrm{d} \vartheta, \quad \mathrm{d}y = \mathrm{d}r \sin \vartheta + r \cos \vartheta \mathrm{d} \vartheta,$$
    and thus indeed
    $$\mathrm{d} x^2 + \mathrm{d} y^2 = \mathrm{d} r^2 + r^2 \mathrm{d} \vartheta^2.$$
    Since in the new coordinates the metric components are constant the curvature tensor vanishes and does so in any coordinate system.
     
  17. Mar 27, 2016 #16
    Nice trick, so the lesson is; wherever the space is flat, Ricci tensor will be always zero no matter how the metric is a function of coordinates because we can always find a new coordinates-system in the same space where the metric is constant.
    But this is an indirect proof right? the complete proof is to calculate the components of Ricci tensor in the former coordinates.
     
  18. Mar 27, 2016 #17

    Nugatory

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    If the tensor is zero in one coordinate system, then it will be zero in all coordinate systems. Thus, if you can show that it's zero in any coordinate system, your proof is complete.
     
  19. Mar 27, 2016 #18
    I am still confused about one thing, having said that Ricci tensor is nill does not help in calculating the equation of motion or geodescics, right? I mean to calculate the trajectory of particle in graph C which represents the gravity, Ricci must not be zero because the motion of the particle is a function of ##T_{ij}##. So having that a local inertial frame with a vanishing Ricci tensor is just to show the correctness of the equivalence principle in a small scale but it is not related to the equation of motion.
     
  20. Mar 28, 2016 #19

    PeterDonis

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    I'm not sure what you mean, but as far as I can tell, this is incorrect. See below.

    No, it isn't. We're talking here about the motion of a test particle, not the motion of the source. The motion of a test particle, assuming it is geodesic, is determined entirely by the spacetime geometry, which in the case under discussion is a vacuum solution, with zero Ricci tensor. "Vacuum" means the stress-energy tensor is zero.

    If the Ricci tensor vanishes in one frame, it vanishes in all frames. You can't use a zero Ricci tensor in one frame and a nonzero one in another; that's inconsistent.
     
  21. Mar 28, 2016 #20
    I will try to be more clear. Einstein field equation equates the left hand side which is represented by Einstein tensor, or Ricci tensor, with the right hand side which is represented by energy momentum tensor.
    When Einstein in his famous paper, spoke about vanishing Ricci tensor in a matter-free field, in what context did he mention that? Was that in the context of explaining an accelerating frame in a flat space with no matter? If so, what does this have to do with derivation of geodesics in a field with matter that requiring non-vanishing energy momentum tensor and consequently non-vanishing Ricci tensor?
     
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