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Relative acceleration of geodesics and spacetime curvature

  1. Mar 9, 2013 #1

    timmdeeg

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    Mass curves spacetime. The relative acceleration of nearby geodesics of free test particles indicates the sign of the spacetime curvature. Convergent geodesics mean positive, divergent negative curvature.

    But also the metric expansion of space curves spacetime. The geodesics may be convergent or divergent as well depending on whether the universe expands decelerating or accelerating. One could suspect that again convergent means positive and divergent negative curvature of spacetime. Is that true?

    I have some doubts whether this analogy holds, because Schwarzschild metric is one thing and FRW metric, which describes the expansion of the universe something much different.
     
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  3. Mar 9, 2013 #2

    bcrowell

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    The labels (1), (2), and (3) are mine. Statement 1 is a cause and effect statement. You seem to be thinking of 3 as analogous to 1, but it isn't; "metric expansion" is just a way of describing a certain type of spacetime, not a statement about why it curves.

    A more accurate restatement of 1 would be that mass couples to curvature, but curvature also couples to curvature. Just as an electromagnetic wave can exist without charges, spacetime curvature can exist without masses.

    I think the statement is a little too vaguely defined to be quite right. For example, it makes a difference whether you're talking about spacelike, lightlike, or timelike geodesics.

    They're both just metrics. The definition and interpretation of the Riemann tensor and other measures curvature isn't different for different metrics.

    The only generic description you can give of positive or negative curvature that would apply to all spacetimes would be that you could construct some invariant, and that could be positive or negative: http://en.wikipedia.org/wiki/Curvature_invariant_(general_relativity)

    When we talk about, e.g., the idea that a circle's circumference might be greater than 2pi times its radius, we're talking about *spatial* curvature, which is a type of sectional curvature: http://en.wikipedia.org/wiki/Sectional_curvature
     
  4. Mar 9, 2013 #3
    More precisely, matter-energy acts as the source of spacetime curvature in GR.
    This is a misleading simplification that only works as stated for the spatial projection of spacetime geodesics.
    Not exactly. If you are thinking of the FRW metric, there is not a way to pick a "curvature sign" just from the math, timelike geodesics can be either diverging from or contracting towards a singularity. Obviously observations lead us to chose the convention that has the timelike geodesics diverging (expansion), it is also purely conventional if you want to call that a positive or negative curvature of spacetime, I haven't seen it addressed in those terms anywhere. That kind of terminology (positive, flat or negative curvature) is normally reserved for the hypersurfaces since they do have constant sectional curvature in the FRW metric.
    You are more likely referring to constant sectional curvature.
     
  5. Mar 9, 2013 #4
    This might be a little misleading since, curvature couples to curvature, but light doesn't couple to light... I'll just say this a different way just in case: both electromagnetism and gravity can be free field theories.
     
  6. Mar 10, 2013 #5

    timmdeeg

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    I am talking about timelike geodesics. The idea was - may be naively - that test particles "don't know" whether they are freely falling (one behind the other) towards a mass (i) or are located somewhere in expanding FRW space (ii) while experiencing comparable relative acceleration in either case. Case (i) means negative spacetime curvature (as explained here), whereby the sign once was choosen by convention. As the particles don't distinguish between (i) and (ii) I suspected that their acceleration and deceleration resp. correlates to the same sign of the spacetime curvature in either case. Otherwise the sign of the curvature (ii) could be choosen arbitrarily and independent of that of (i). To me not the sign itself but their correlation is of interest. Oh, hopefully I could make myself understandable.

    Your statement
    seems to support that, if I don't misinterpret it. Do you say that the Riemann tensor doesn't distinguish between (i) and (ii)? Then said correlation should hold, right?

    Sorry this article is far beyond my scope. Does it mean that the sign of the mentioned invariant correlates to the sign of the spacetime curvature of (i) and (ii) as well?

    May I ask another question. The third case assumes a uniformly expanding universe,
    d²a/dt²=0 (scale factor a), with timelike geodesics still diverging, but linear. Would this mean flat spacetime? In contrast to relative acceleration/deceleration linear deviation of timelike geodesics means that there is no tidal force, like in flat (locally) not expanding space. Would you agree to that?

    I have checked Chapter 11 (my goodness, ha ha): Geodesic Deviation and Spacetime Curvature in "Gravitation" (MTW) and didn't find any hint either. They don't say anything about expanding space in this context.
     
  7. Mar 10, 2013 #6
    No, why? Vanishing Riemann tensor means flat spacetime, not vanishing second derivative of the scale factor.
    Here and in the rest of your post you seem to be mixing different solutions, if you stick to the FRW case there are no tidal forces to talk about , while if you want to talk about tidal forces like in Schwarzschld's vacuum exterior solution you are in a static spacetime so no expansion but still there is spacetime curvature.
     
  8. Mar 13, 2013 #7

    timmdeeg

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    Sorry for my time problem.
    Right, I should have mentioned, that this third case means the empty universe (Milne model). ä=0 is a consequence, not the cause.
    Why not? We talk about accelerated expansion, e.g. here, nearby particles see relative accelaration, like particles falling freely towards a mass.

    I am asking whether there is a relation and if yes which one between the signs of spacetime curvature if one compares the FRW and Schwarzschild cases. Such a relation could mean that said sign is negative in both cases, if test particles accelerate away from each other.

    I am not an expert. This question might well be meaningless. I don't know, how the sign of spacetime curvature in the FRW case is identified and whether or not it depends on the typ of expansion, accelerating - decelerating, as I was assuming.

    I am asking this question in this forum, because here are experts who usually lead people like my to a better understanding.
     
  9. Mar 13, 2013 #8

    Mentz114

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    The kinematic analysis of comoving geodesic congruences ( geodesic 'flow lines') provides a scalar called the expansion scalar. This gives a radially directed velocity field whose magnitude is independent of direction. The definition is
    [tex]\Theta=\nabla_\mu u^\mu[/tex]
    where ∇ is the covariant derivative. The covariant derivative depends only on first derivatives of the metric, but curvature on higher derivatives. So the connection between θ and the curvature components is not strong. In my opinion.
     
  10. Mar 13, 2013 #9
    So we are here dealing with the Ricci curvature, the part of the Riemann curvature that in simple terms would care about change in volume of the spatial hypersurface, this kind of relative motion of the dust's particles is usually not referred to as "tidal"; rather the relative acceleration of test particles that comes about due to the Weyl curvature part of the Riemann tensor is related to tidal forces.
    If you are talking about expansion you are clearly referring to the former (Ricci curvature) because the FRW solution has vanishing Weyl curvature.
    As I tried to explain the "curvature sign" you refer to is reserved for constant sectional curvatures, like that of the spatial hypersurfaces in the FRW metric or the spatial curvature of static solutions like Schwarzschild's, not to the "spacetime curvature".
     
  11. Mar 13, 2013 #10
    Op, I believe you are asking about the equation of geodesic deviation:
    [tex]
    \frac{D^2}{D\tau^2} \delta x^{\lambda} = R^{\lambda}_{\; \nu \mu\rho}\delta x^{\mu}\frac{d x^\nu}{d\tau}\frac{dx^\rho}{d\tau}
    [/tex]
    with
    [tex]
    \frac{DA^\mu}{D\tau} \equiv \frac{dA^{\mu}}{d\tau} + \Gamma^{\mu}_{\; \nu \lambda}\frac{dx^{\lambda}}{d\tau}A^\nu
    [/tex]
    Consider this equation for a metric with expansion, like FRW metric. This tells you how geodesics separate, in space, and time.
     
  12. Mar 13, 2013 #11

    Mentz114

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    That's a good candidate, and obviously coordinate independent like ##\Theta##. It is an acceleration, like Raychaudhuri's number ##d\Theta/d\tau##.

    This paper has a breakdown of expansion related tensors etc http://arxiv.org/abs/1012.4806
     
  13. Mar 19, 2013 #12

    timmdeeg

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    Yes, I was talking about excelerated expansion. So, in this case the stretching of things isn't called tidal stretching (perhaps by convention?), as I was assuming. My reasoning was that from a merely physical point of view - what happens to a loosely connected string of pearls - the two situations (falling radially towards a mass vs. isotropic expansion) are indistinguishable. The string is stretched in the same way, regardless whether or not there is Weyl curvature.

    Yes, you mentioned that before.

    I am not at all familiar with the meaning of "constant sectional curvature". What wikipedia writes here, reminds me of how spatial curvature depends on the relation energy density/critical density. I don't know if it is possible, but it would be great, if you could explain in simple words, what "constant sectional curvature" in the context of space curvature and spacetime curvature means.

    I think, I have some understanding in the the Weyl case. Here spacetime curvature and it's sign is linked to the change in shape of a ball of testparticles, to their relative acceleration, to their geodesics. But it seems there isn't something equivalent in the FRW case, right?. Any help is highly appreciated.
     
  14. Mar 19, 2013 #13
    But they are not indistinguishable, that's the key difference between Weyl (tidal) curvature and Ricci curvature, that the test particles have different behaviour: the former change shape without changing their volume, while the latter change volume (isotropic contraction/expansion).
    You are right is related to the critical density idea that talks about open, flat and closed spatial universes.
    The intuitive idea behind constant sectional curvature is quite simple (the formal explanation involves some knowledge about differential geometry, I guess at this point it is enough to know that sectional curvature determines completely the perhaps more familiar to you Riemann curvature) if you think that spatial hypersurfaces that have it are homogeneous and isotropic. Which happens to be what we expect of our spatial universe according to mainstream cosmology, wich uses the FRW model to come to this conclusion backed by what we observe with our telescopes.
    This is right except for the sign part, didn't we agree to leave that to spatial curvature discussions?
     
  15. Mar 19, 2013 #14

    timmdeeg

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    Well, is there a misunderstanding? My references are:

    http://books.google.de/books?id=ybh...avitational waves light ray triangles&f=false
    Gravitational waves: According to Fig.1.1 the spacetime curvature oszillates from convex via flat to concave. And J.A.Wheeler in "A Journey into Gravity and Spacetime" connects diverging trajectories with negative spacetime curvature and vice versa, discussing a gravitational wave passing by.

    http://www.pitt.edu/~jdnorton/teaching/HPS_0410/chapters/general_relativity/index.html
    Uniqueness of Free Fall
    Same connection of trajectories with the sign of the spacetime curvature.

    So we have Weyl curvature in either case. In fact this made me thinking whether there exists a somehow complementary situation in the expanding FRW case. But this reasoning seems too simple.

    Thanks for explaining constant sectional curvature, I might come back to that later.

    This was curt, sorry.
     
    Last edited: Mar 19, 2013
  16. Mar 19, 2013 #15

    pervect

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    As some others have indirectly mentioned, this isn't clear. The sort of curvature measured by the relative acceleration of geodesics isn't a single number. There isn't any way to give this colleciton of numbers a unique sign. So in this sense, there isn't any "sign of the space-time curvature".

    If you use the expansion scalar, which is rather closely related to geodesic deviation, instead of geodesic deviation. it is a single number, and it has a sign. I'm not sure how to refer to it informally to avoid confusion, though.

    I think that what you are saying here follows from Raychurdi's equation if you use the expansion scalar.

    Note that the "expansion of space" isn't a cause, it's more of an effect or result of the matter distribution.

    http://en.wikipedia.org/w/index.php?title=Raychaudhuri_equation&oldid=544131115

     
  17. Mar 19, 2013 #16

    Mentz114

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    Last edited: Mar 19, 2013
  18. Mar 20, 2013 #17

    timmdeeg

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    Oh, sorry, I don't remember why.

    Thanks for this link. So it seems that the expansion scalar though being related to said behaviour of geodesics, it's sign should not be interpreted as the sign of space-time curvature.

    Coming back to Weyl curvature, would you agree that there are definitve signs (signs based on math) of space-time curvature, as indicated in post 14? These authors talk about geodesic deviation, but as you say "the relative acceleration of geodesics isn't a single number". Or perhaps, as there is stretching and squashing inextricably linked to each other, signs have been adopted just to make things easier? I think TrickyDicky has mentioned that such signs are due to convention. And perhaps this motivation lacks in the expanding universe, because there is stretching only. How do you think about that?

    One more question about the geodesic equation. Does it distinguish beween Weyl and Ricci curvature?

    Any help appreciated.
     
  19. Mar 20, 2013 #18
    Consider the equation of geodesic deviation again. Notice that the relative acceleration of the difference between geodesics is proportional to the Riemann curvature tensor. Also note that the Riemann curvature tensor decomposes as follows:
    [tex]
    R_{abcd} = S_{abcd}+E_{abcd}+C_{abcd}
    [/tex]
    where [itex]C_{abcd}[/itex] is the Weyl curvature tensor. Then the equation of geodesic deviation knows about the the Weyl curvature, for off diagonal pieces, naturally. The other parts of the decomposition involve the Ricci tensor and Ricci scalar and other contractions. Here is the wiki page for that decomposition:Ricci decomposition
     
  20. Mar 20, 2013 #19

    Mentz114

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    In the online version of his book "Riemannian Manifolds : An introduction to curvature", Lee says ( page 9)
    I could be taking this out of context, but it seems pretty unequivocal.

    The expansion scalar is a bit subtle, and probably it only has local significance.
     
  21. Mar 21, 2013 #20

    timmdeeg

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    The relative acceration between geodesics is a number, which has a sign. Could you give me an idea, how to understand that such a number can be proportional to a tensor, which is a matrix. Wikipedia writes: "The curvature tensor represents the tidal force experienced by a rigid body moving along a geodesic in a sense made precise by the Jacobi equation."
    Does this mean that the local tidal force, which the tensor represents, determines and is proportional to the relative accereratiion of (timelike) geodesics?

    Yes, that's a good hint. Here is a little more:
    I guess, regarding converge/diverge relative acceration is meant. Also on page 9 in my words: The sectional curvature of a Riemannian metric is given by a number K, whch is positive (sphere), negative (hyperboloid) or zero (euclidean).
    I am not sure, does the sign of the sectional curvature K then correspond to the sign of the curvature, that causes the geodesics to converge and to diverge, resp.?
     
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