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Spacetime curvature observer and/or coordinate dependent?

  1. Apr 13, 2012 #1
    Spacetime curvature observer and/or coordinate dependent?

    In another topic several people suggested that spacetime curvature is not absolute, it apparently depends on the observer and/or coordinate system. Apparently if someone goes fast (whatever that might mean in relativity) curvature is more in the time direction while if someone goes slow (again whatever that might mean) curvature is more in the spatial direction.

    [sarcasm on]Seems we got a way to determine absolute motion as we simply have to check the curvature for the two components to see how fast we are going[/sarcasm off]

    Any takers?
     
  2. jcsd
  3. Apr 13, 2012 #2

    Jonathan Scott

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    That sort of curvature is related to acceleration, and one person's accelerating frame of reference is another's free fall, so it does depend on the frame of reference.

    However, "going fast" is never a correct description from one's own point of view, so it's not quite clear what you mean.

    If you are describing different test objects moving through the same static fields, the curvature doesn't change but the effect changes with velocity, and it's the other way round from what you say, in that the curvature of space only affects moving things (as should be obvious).

    In a weak approximation, using an isotropic coordinate system (with one or more static sources), relativistic coordinate momentum per energy (not quite the same as velocity because the coordinate speed of light changes) changes at a rate g (1 + v2/c2) if the effective Newtonian field is g.

    If the test object is moving tangentially to the field, the same rule applies to the velocity, hence the effect that light is deflected by twice the Newtonian acceleration.
     
  4. Apr 13, 2012 #3

    Mentz114

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    The components of the Riemann tensor will change value in a local frame according to
    [tex]
    R_{abcd} = \Lambda^\alpha_a \Lambda^\beta_b \Lambda^\gamma_c \Lambda^\delta_d R_{\alpha\beta\gamma\delta}
    [/tex]
    where a,b,c,d are local frame indexes and α,β,γ,δ are coordinate indexes. The tetrad [itex]\Lambda[/itex] is defined so that
    [tex]
    \eta_{ab} = \Lambda^\alpha_a \Lambda^\beta_b g_{\alpha\beta}
    [/tex]
    where η is the Minkowski metric.

    Nobody said that.
    In the case of the Schwarzschild vacuum the velocities are relative to the source.
    What curvature and what components ? Velocities are relative to the source of the field.
     
  5. Apr 13, 2012 #4
    Yes the components change but curvature is absolute.

    Do you think we should talk about coordinate effects when we describe curvature or do you think that is wrong or misleading to say the least?
     
  6. Apr 13, 2012 #5

    Mentz114

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    Another example of curvature being dependent on an observer is the Langevin observer in the Born chart, where the 3D submanifold is curved ( non-Euclidean). That curvature depends on the angular velocity.

    If the Minkowski metric is written in spherical polar coords it acquires non-zero components of the Riemann tensor ( I can't remember how many - perhaps only one).

    Why did you phrase your question in the way that you did if you didn't want to talk about coordinate effects ?

    When I refer to curvature I mean the Riemann tensor. Do you think the only important number is the invariant Ricci scalar ?

    In answer to your question I would say that observed effects of curvature are observer dependent. For instance the Hagihara observer sees very different tidal effects compared to a Gull-Painleve observer, even though they are both moving in the same Schwarzschild spacetime.
     
  7. Apr 13, 2012 #6
    So then tell me how does an observer observe the effects of

    A) spatial curvature
    B) temporal curvature

    ?
     
  8. Apr 13, 2012 #7

    PAllen

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    Hmm. How can a tensor (all components, everywhere) vanish in one coordinate system but not in another? Seems to violate the tensor transform law...
     
  9. Apr 13, 2012 #8

    Dale

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    There are a bunch of different ways that you might use the word curvature. Several of them are tensor of higher order than 1, so I don't know if you want to call those absolute or not. One of them is a scalar field on spacetime, so that is clearly absolute.

    Then one other thing that you could talk about would be the curvature of a worldline. I think that this is the curvature refered to in the other discussion. That curvature is a scalar (the magnitude of the 4-acceleration), but it also depends on the worldline. So I don't know if you want to call that absolute or not.
     
  10. Apr 13, 2012 #9

    Mentz114

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    I just checked and I'm referring to an embedded spherical surface, which has one non-zero Riemann component. But probably not relevant to this topic.

    Off topic. It belongs in the other thread. Why don't you answer the points I made in post#5 ?
     
  11. Apr 13, 2012 #10

    PAllen

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    Right, that's no problem. A 2-sphere can be embedded in Euclidean 3-space, for example.
     
  12. Apr 13, 2012 #11
    All I want is to address statements that bodies are affected differently by space and time curvature depending on their velocity. These statements I consider wrong and very misleading.

    However it seems that quite a few folks on this forum are actively supporting these statements.

    So now I wonder why I am even bothering......
     
  13. Apr 13, 2012 #12

    Mentz114

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    The geodesic equations show clearly that coordinate acceleration depends on velocity. They contain the Christoffel symbols which, along with their derivatives appear in the curvature tensors.

    Your question about measuring temporal curvature brought to mind gravitational time dilation, which depends only on the time parts of the metric. The acceleration needed to remain at constant r in the Schwarzschild vacuum is the Newtonian value divided by the square root of g00. But I'm not sure if that is relevant either.

    Anyhow it is late in my timezone so I'm clocking off.
     
  14. Apr 13, 2012 #13

    Jonathan Scott

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    If you move in a straight line in local space which is curved (as described by the metric) relative to the coordinate system, then that obviously adds to the acceleration relative to the coordinate system. Just as in ordinary mechanics, if the linear curvature is 1/r, then the resulting acceleration is v2/r. In this case, r = c2/g, so this causes an acceleration of g times v2/c2 (for motion perpendicular to the direction of the field).
     
  15. Apr 13, 2012 #14

    Jonathan Scott

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    If you're bothered by the implications for the principle of equivalence of different accelerations for different velocities, that's not actually a problem if you think about it more carefully.

    The principle of equivalence holds within a local coordinate system which is small enough to approximate SR. If you consider a local box at rest in the field and view it from an isotropic coordinate system, it appears to be curved, along with local space, towards the source. A light ray crossing the box is twice as curved (at 2g). This means that relative to a local horizontal line across the box, which follows the curvature of space, the light actually appears to be accelerating downwards at only 1g, the same as any material object at rest or moving at any other speed. It's all consistent.
     
  16. Apr 13, 2012 #15
    I agree based on prior explanations from a number of people here in the forums:

    The curvature or flatness of space (as opposed to spacetime) IS a coordinate dependent idea.

    Overall, can be a confusing issue due in large part to language.....

    From my [physics forums] notes of posts:

    [one perspective]
    https://www.physicsforums.com/showthread.php?p=3661242&posted=1#post3661242


    And a further perspective:
    Those simple perspectives clinch the issue for me;there was no dissent in the thread(s).

    Other [supporting] perspectives:

    [I don't have the original thread(s)]

    DrGreg:

    "...... let's restrict our attention to 2D spacetime, i.e. 1 space dimension and 1 time dimension, i.e. motion along a straight line. …
    In the absence of gravitation, an inertial frame corresponds to a flat sheet of graph paper with a square grid. If we switch to a different inertial frame we "rotate" to a different square grid, but it is the same flat sheet of paper. (The words "rotation" and "square" here are relative to the Minkowski geometry of spacetime, which doesn't look quite like rotation to our Euclidean eyes, but nevertheless it preserves the Minkowski equivalents of "length" (spacetime interval) and "angle" (rapidity).)

    If we switch to a non-inertial frame [an accelerated observer] but still in the absence of gravitation), we are now drawing a curved grid, but still on the same flat sheet of paper. Thus, relative to a non-inertial observer, an inertial object seems to follow a curved trajectory through spacetime, but this is due to the curvature of the grid lines, not the curvature of the paper which is still flat.

    When we introduce gravitation, the paper itself becomes curved. (I am talking now of the sort of curvature that cannot be "flattened" without distortion. The curvature of a cylinder or cone doesn't count as "curvature" in this sense.) Now we find that it is impossible to draw a square grid to cover the whole of the curved surface. The best we can do is draw a grid that is approximately square over a small region, but which is forced to either curve or stretch or squash at larger distances. This grid defines a local inertial frame, where it is square, but that same frame cannot be inertial across the whole of spacetime.

    So, to summarize, "spacetime curvature" refers to the curvature of the graph paper, regardless of observer, whereas visible curvature in space is related to the distorted, non-square grid lines drawn on the graph paper, and depends on the choice of observer...."


    DrGreg:
    When we talk of curvature in spacetime (either curvature of a worldline, or curvature of spacetime itself) we don't mean the kind of curves that result from using a non-inertial coordinate system, i.e, non-square graph paper in my analogy.

    Four acceleration:
    [both from pervect I suspect]

    In Minkowski coordinates in Special Relativity, 4-acceleration is just the coordinate derivative of 4-velocity with respect to arc-length (proper time), and the 4-velocity is the unit tangent vector of worldline. As the 4-velocity has a constant length its derivative must be orthogonal to it. The 4-acceleration is the curvature vector; orthogonal to the worldline and its length is the reciprocal of the worldline's "radius of curvature".

    In non-inertial coordinates in GR, the 4-acceleration is defined as a covariant derivative. This takes into account (and removes) any curvature of spacetime or "apparent curvature" due to using a "non-square grid", and leaves us with curvature that is a property of the worldline itself, not the spacetime or the choice of coordinates. Then everything else I said in the last paragraph is still true in a coordinate-independent sense.

    A geodesic has zero 4-acceleration and zero curvature.



    Pervect:
    If you draw a largish parallelogram on a curved surface, you'll find that the opposite sides might not necessarily be equal in length, even though the sides are parallel. Now, imagine such a parallelogram, but that one of the sides of this parallelogram is the time axis on a space-time diagram. What you'll see is something that looks pretty much like gravitational time dilation. One timelike worldline will be shorter than another, even though they are connected by parallel geodesics.

    from another discussion:


    The Stress energy Tensor

    The "amount of gravity produced" by the object is not a function of its energy alone, it's a function of its stress-energy tensor [ SET] of which energy is only one component. In a frame in which the object is moving, there will be other non-zero components of the SET as well as the energy, and their effects will offset the apparent "effect" of the increase in energy, so the final result will be the same [gravitation] as it is for a frame in which the object is at rest.

    [A way to think about this is that in the frame of the object, the 'other' mass is the one that is moving....so the moving object cannot have 'greater gravity.....]
     
  17. Apr 13, 2012 #16
    Jonathan Scott post #2:

    is entirely correct, but that is not the 'spacetime curvature' as described in my prior post by Dr Greg. These 'acceleration effects' are the curved grids lines as described by Dr Greg, not the 'spacetime' graph paper.
     
    Last edited: Apr 13, 2012
  18. Apr 13, 2012 #17
    Dalespam:
    This too is curved grid lines on gravitational spacetime....so as noted, you have to define to what type of curvature you refer. But for a given worldine seems like it is a scalar invarient.
     
  19. Apr 14, 2012 #18
    passionflower...I think your original question deserves a lot more discussion....It has taken me a long time to develop a perspective with help from several experts here who agree with you....this discussion has so far not caused me to change my mind.

    post #15:

    I 'wanna' reneg on my agreement, but agree with passionflower's original implied statement:
    that GRAVITATIONAL spacetime curvature [curved graph paper as described by docAl] is invarient, but acceleration overlays [curved grid lines drawn on the curved graph paper] are
    not.

    How can you experts agree on such an issue when no one entity in the Einstein Field Equations can claim preeminence in describing 'gravity'.....

    Also, if no matter how fast you see a particle whizzing by, it will never become a black hole, how can it change 'gravitational curvature'? Such speed does not add to gravitational curvature yet you generally PERCEIVE the spacetime as being curved .....the gravitational spacetime curvature component [curved graph paper] must be considered frame independent.
     
  20. Apr 14, 2012 #19

    Mentz114

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    Re: Spacetime curvature observer dependent?

    Is spacetime curvature observer dependent ?

    If we take 'spacetime curvature' to mean gravity then the answer must be 'yes'.

    1. A freely falling observer feels no gravity, but a stationary one does. This is because the elements of curvature that depend on the first derivatives of the metric can be transformed away for a point on a geodesic worldline. See for instance Fermi coordinates here http://en.wikipedia.org/wiki/Fermi_coordinates or Riemann coordinates here http://users.monash.edu.au/~leo/research/papers/files/lcb96-01.pdf. The latter is a lot better than the Fermi page.

    2. Tidal effects ( also gravity ) can be different for different observers in the same spacetime. The tidal tensor in the coordinate basis is defined as
    [tex]
    T_{\mu\nu}=R_{\mu\rho\nu\sigma}U^\rho U^\sigma
    [/tex]
    (which already contains an observer dependent vector U) and being a tensor can be transformed to a local frame ( which may be geodesic or not) by the usual tetrad
    [tex]
    \hat{T}_{ab}=\Lambda^\mu_a \Lambda^\nu_b T_{\mu\nu}
    [/tex]
    and since the tidal effects are components of T, they can change without violating the tensorial properties of T. I gave an example in an earlier post of two observers who see different tidal effects in the Schwarzschild vacuum.

    There could be something wrong with my logic, in which case, please politely point it out.
     
    Last edited: Apr 14, 2012
  21. Apr 14, 2012 #20
    Naty1 quoting someone else:
    This should probably be qualified with the bolded words as:
    and there probably should be a few more qualifying statement in there. Anyway, without any qualification it is not always true. For example a stationary observer in the field of a rotating gravitational body is not equivalent to a moving observer in the field of a non rotating gravitational body.
     
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