Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Metric expansion

  1. Jan 27, 2007 #1
    The pseudo metric of a Lorentzian space-time is an amalgamation of space and time.

    So then what expands? :confused:
  2. jcsd
  3. Jan 27, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    I assume you might be thinking of the so-callled FRW metrics (I apologize for the alphabet soup, but I think you know which ones I mean) thought to represent our universe in the large?

    Write down a FRW metric and look at it, and tell us what you think expands. (I think you're capable of this, and it would be instructive for you, I think).

    ps - the manifold of space-time is pseudo-Riemannian, but there's nothing "pseudo" about the metric. What separates a pseudo-Riemannian manifold from a Riemannian manifold is that the metric for the later must be positive definite. Because Lorentzian metrics have a -+++ signature they cannot be positive definite , due to the presence of the minus sign.
  4. Jan 27, 2007 #3
    As far as I understand it, it is definitly not a metric.
    For instance a metric must satisfy the triangle inequality, clearly the space-time pseudo-metric does not.

    Yes I am familiar with it.
    So the components are pure time and space and do not contain any cross terms? :confused:
    Last edited: Jan 28, 2007
  5. Jan 27, 2007 #4


    User Avatar
    Staff Emeritus
    Science Advisor

    AFAIK it's pseudo-Riemannian manifold and just-plain "metric" (not pseudo metric). Chris Hillman can probably comment on such mathematical nicities with more rigor than I can, however.

    As far as the FRW metric goes, only the space terms get multipled by the expansion scale factor a(t) and there are no cross space-time terms (i.e. no terms like dt dr).
  6. Jan 27, 2007 #5
    The components of the metric are a function of coordinates.
  7. Jan 27, 2007 #6
    The FRW manifolds are very special since by symmetry, there exists a
    time function (coordinate) [tex] t [/tex] acting as a notion of "global
    simultaneity". Moreover, there exists a set of distinguished observers,
    whose world lines are always normal to the "spatial hypersurfaces of
    simultaneity" (obtained by setting [tex] t= [/tex]constant).

    The expansion is described as an increase of distance (determined
    by an increasing scale factor) with [tex] t [/tex] between any pair
    of distinguished observers. Equivalently, the expansion is described
    by the fact that the scale of the "hypersurfaces of simultaneity"
    increases with [tex] t [/tex].
  8. Jan 27, 2007 #7
    Quite correct, so would that not imply that there must be one or more cross terms in the metric?
  9. Jan 27, 2007 #8
    Since the t curves are orthogonal to the hyper spatial surface, there is no cross terms.

    But I don't quite understand why the case of a hyperboloid can be homogeneous and isotropic (cosmological principle). I don't quite see it geometrically. Perhaps I don't really know what the axes are in the diagram of the hyperboloid.
  10. Jan 27, 2007 #9
    Actually I am talking about the cross terms in the FWR metric not about the hypersurface.
    Last edited: Jan 27, 2007
  11. Jan 28, 2007 #10

    Chris Hillman

    User Avatar
    Science Advisor

    Triangle inequality?

    Hi, MeJennifer,

    You asked about the triangle inequality in Lorentzian metrics ("metric" in the algebraic sense of indefinite bilinear forms, not in the general topology sense of "metric topology"!):

    Lorentzian metrics satisfy a reversed triangle inequality. See for example Barrett O'Neil, Semi-riemannian geometry.
    Last edited: Jan 28, 2007
  12. Jan 28, 2007 #11

    Chris Hillman

    User Avatar
    Science Advisor

    What expands?

    Hi, MeJennifer,

    I repeat a recommendation: Weinberg, The First Three Minutes. This should help you understand. A more technical alternative is the last chapter of D'Inverno, Understading Einstein's Relativity.
  13. Jan 28, 2007 #12
    Could you, in a nutshell, reason why you think one is a definitive example of space-time and the other definitively not?
  14. Jan 28, 2007 #13


    User Avatar
    Science Advisor

    I may be wrong but to me it seams more a matter of interpretation than only a matter of a mere mathematical derivation of the FRW metric, because in an FRW metric you can always perform a coordinate change to conformal coordinates that give you expanding space and expanding time (conformal time). I understand that with expanding you mean some coordinate that is multiplied by the scale factor. However, this conformal time does not correspond to our proper physical time that we measure locally with clocks.
    Last edited: Jan 28, 2007
  15. Jan 28, 2007 #14

    Chris Hillman

    User Avatar
    Science Advisor


    Hi again, MeJennifer,

    Like pervect, I am assuming that you are asking about FRW cosmological models when you ask:

    By "cross-terms" you probably mean nonzero off-diagonal components in the metric tensor as expressed in some specific coordinate chart for some specific spacetime model, such as the FRW dust with hyperslices orthogonal to the world lines of the dust particles which are locally isometric to [itex]E^3[/itex] (this mouthful is abbreviated in MTW and other books as "asymptotically open matter-dominated model", but it turns out that this name is seriously misleading, as per Cornish and Weeks).

    As it happens, it is possible to find charts in which the FRW models have diagonal metric tensors; such charts are often called "orthogonal coordinate charts", meaning that the coordinate basis vectors are mutually orthogonal. The exterior region Schwarzschild vacuum solution also has this property. Of course, in such cases, you can easily write down a coordinate transformation which destroys this property. Sometimes there are good reasons for doing this, as in passing to the Painleve chart for the Schwarzschild vacuum. However, most authors prefer to use an orthogonal chart, such as the Friedmann-Robertson-Walker chart, when discussing the FRW models.

    Note that not all Lorentzian spacetimes are "diagonalizable" in this sense. For example, exact solutions which model a circularly polarized gravitational wave propagating far from any massive objects are not "diagonalizable". Here's something for you to think about: by changing to a new coordinate chart, can we always "diagonalize" the metric tensor at a given event of a Lorentzian manifold? If not, what about Riemannian manifolds, which have positive definite metric tensors?

    So, three completely separate issues seem to be confusing you in this thread! I hope I have cleared up these confusions now!
    Last edited: Jan 28, 2007
  16. Jan 28, 2007 #15

    Chris Hillman

    User Avatar
    Science Advisor

    I said positive-definite, not "definitive"

    Hi, Jennifer,

    I said no such thing!! (If I hadn't seen previous posts by you, I'd think you were trolling, so in future, for all our sakes, please try harder not to misread standard technical terms as informal terms in natural language).

    Let me try to clarify. Mathematically speaking, bilinear forms can be:
    1. positive-definite (the only kind considered in general topology courses when one speaks of a metric topology),
    2. indefinite but non-degenerate, or
    3. degenerate.
    These all have physical interpretations in terms of
    1. Riemannian manifolds,
    2. Lorentzian manifolds, and
    3. Galilean manifolds (see "Newtonian spacetime" in MTW), or the induced metric on a light cone (less a point) in a Lorentzian manifold, and so on.
    In the planar case, these correspond to the three "flat" planar Kleinian geometries (which are two-dimensional homogeneous spaces):
    1. [itex]E^2[/itex], whose isometry group is [itex]E(2) = R^2 \, | \!\! \times SO(2)[/itex] with Lie algebra [itex]e(2)[/itex]
    2. [itex]E^{1,1}[/itex], whose isometry group is [itex]E(1,1) = R^2 \, | \!\! \times SO(1,1)[/itex] with Lie algebra [itex]e(1,1)[/itex]
    3. [itex]E^{1,0}[/itex], whose isometry group is [itex]E(1,0) = R^2 \, | \!\! \times SO(1,0)[/itex] with Lie algebra [itex]e(1,0)[/itex]
    These correspond in turn to the three kinds of planar trigonometries:
    1. elliptical trignometry ("high school trigonometry"),
    2. hyperbolic trigometry (careful!--- "hyperbolic geometry" uses euclidean trigonometry and the "hyperbolic plane" is a two-dimensional Riemannian manifold!),
    3. parabolic trigonometry
    and also to three (two dimensional) Cayley-Dickson algebras (special normed algebras having a multiplicative norm):
    1. complex numbers (which form a field in the sense of algebra),
    2. Minkowski numbers (as a ring, this algebra admits zero divisors, so the Minkowski numbers do not form a field!)
    3. Galilei numbers, aka "dual numbers", aka the exterior algebra on the real line (as a ring, this algebra admits idempotent elements.)

    The metric tensors of the three (homogeneous) plane Kleinian geometries are, when represented in a Cartesian coordinate chart, diagonal of form [itex](1,1), \; (-1,1), \; (1,0)[/itex] respectively (see O'Neill for details). The two possibly unfamiliar Cayley-Dickson algebras, once reformulated as differential rings, permit "factoring" of analogues of the Laplace equation, a la Dirac, and to a limited extent supports analogues of the notion of a holomorphic function from a standard complex analysis course, but the obvious analogues of the crucial Cauchy integral theorems do not hold, so one should be cautious about concluding that there are two straightforward analogues of complex analysis formulated in terms of Minkowski or Galilei numbers (in fact, there are no such straightforward analogues). And exterior algebra provides the algebraic underpinning of Cartan's notion of exterior calculus; see for example Spivak's textbook, Calculus on Manifolds.

    EDIT: troll alert! In my opinion, a poster in this thread other than MeJennifer is attempting to troll us in this board. Therefore, I will depart PF until this situation is resolved. Good luck, Jennifer, and keep trying! I can see that you have been making progress in your understanding, but a little temporary if painful "back-sliding" might be part of the learning process.
    Last edited: Jan 28, 2007
  17. Jan 28, 2007 #16
  18. Jan 28, 2007 #17

    Chris Hillman

    User Avatar
    Science Advisor

    Orthogonal coordinate charts are distinct from conformal coordinate charts

    Hi, hellfire,

    This is a fourth issue which will probably confuse Jennifer: another general type of coordinate chart are conformal charts on Lorentzian manifolds, in which the metric tensor is diagonal, with the form of a scalar function times the usual line element of Minkowski spacetime.

    Note well: conformal coordinate charts are a special case of orthogonal coordinate charts, but not every orthogonal coordinate chart is a conformal chart!
  19. Jan 28, 2007 #18
    Well to me "general topologyl" and Riemannian manifolds are not identical. Riemannian manifolds are simply a subset. I had, perhaps falsely, the impression that you exclude the possibility that space-time can be modeled topologically. That was why I asked the particular question. But if you did not then I apologize.

    I did not misread and I really meant definitively.
    Last edited: Jan 28, 2007
  20. Jan 28, 2007 #19

    Chris Hillman

    User Avatar
    Science Advisor

    Apology accepted, MeJennifer, but I will take a break to await moderator intervention

    I didn't say that either! You appear to be deliberately misreading what I write, although I am trying to avoid drawing that conclusion. Please read carefully--- physics is confusing enough without confusion due to careless reading!

    Riemannian manifolds have positive-definite metrics. However, these are not metric spaces in the sense of general topology! This is the exact same issue which keeps coming up, regarding the multiplicity of notions of "distance in the large".

    No, no, no! I never said that, and of course it is not true!

    I said that the topology which is used in say [itex]E^{1,1}[/itex] is the euclidean topology; because the metric tensor of this homogeneous space (one of the three "flat" Kleinian geometries which can be defined on the euclidean plane) is indefinite, it does not define a metric topology.

    You did, but it is understandable--- the terminology really is confusing and all the misconceptions which you have mentioned in this thread are common ones.

    Apology accepted, but because of rampant trolling/harrassment by another poster in this board, which I have reported to the moderators, I will duck out now and wait for the moderators to resolve the issue.

    Good luck, Jennifer, and keep trying!

    EDIT: I have requested that the moderator immediately delete a new post by the suspected troll which contained the email address of a well known physicist. As I assume everyone here knows very well, posting an email address or other personal information in a public forum is highly deprecated and generally considered to constitute harrassment, so everyone, please avoid this in future.
    Last edited: Jan 28, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook