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What metric from given manifold?

  1. Jun 28, 2011 #1
    Given a manifold as algebraic variety, say sphere, how do we obtain possible metrics? how do we classify them?
    If spcaetime manifold is n-sphere, Einstein's vacuum (for now) equation would be some special metric among many other possible metrics?

    i'm curious what role Einstein's equation plays. It is not determining spacetime manifold itself. It determines some metric as its solution. Then why do we say matter (source term to EInstein equation) curve the spacetime? It changes metric, but what about spacetime itself?
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  3. Jun 29, 2011 #2
    The class of possible metrics on a manifold is too wide - we have everything from a discrete metric to standard Riemannian metrics. We can't reject the possibility of another metric for spacetime only on mathematical grounds.However, the metric of general relativity can be shown to be,in a way, the only metric compatible with newtonian gravity. The ultimate test of Einstein's equations is experimental evidence, as with every physical theory. I don't know how to address the issue the best.
  4. Jun 29, 2011 #3


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    The best way we can. Meaning there is no general procedure that generates all the metrics. But what is a metric? It is the specification of a symmetric nongenerate bilinear form of a given signature (p,q) on each tangent space T_xM, such that these bilinear forms vary smoothly with x.

    One possible way to put a metric on a manifold M is to embed said manifold in a manifold N already equipped with a metric g, then restrict g to M. For instance, if g is riemannian (i.e. signature (n,0)), than g restricts to a riemannian metric on M. If R^{n+1} is equipped with the Minkowski metric [itex]g=(dx^1)^2+\ldots+(dx^n)^2-(dz)^2[/itex] (signature (n,1)), and if we restrict it to the hyperboloid H given by [itex](x^1)^2+\ldots+(x^n)^2-z^2=-1[/itex], z>0, then this is a riemannian metric on H, and in fact, (H,g|) is a model of hyperbolic n-space.

    Note however that in the non-riemannian case, this restriction of the metric business does not always work. For it is possible that the restriction of such a nondegenerate bilinear form to a subspace be degenerate.

    That depends on what kind of classification you're interested in. Usually one classifies pseudo-riemmnian manifold up to isometries.. i.e. diffeomorphisms which preserve the metric. But one might be content to have a classificatioin up to conformal equivalence... i.e. diffeomorphisms which preserves the metric up to a (variable) constant.

    Yes, but be aware that there are relations between the topology of a manifold and the kind of metrics it can support. Usually, these theorem are stated in terms of a relation between the curvature and some topological invariant. One well-known example is the Gauss-Bonnet theorem. It states that for a 2-dimensional closed riemannian manifold of Gaussian curvature K and Euler caracteristic [itex]\chi(M)[/itex], the following equation hold:

    For instance, one concludes from this that the 2-sphere does not support a metric of negative or zero curvature, since [itex]\chi(S^2)=2>0[/itex].

    Curvature (R) is a property of the metric: if you know the metric, you can compute the curvature. It is a certain expression involving partials derivatives of the metric (g). Hence, the Einstein's equations are a system of partial differential equations that basically say: if you know the distribution of "matter" (stress-energy tensor, T) in the universe at a given time t, then solving the resulting equations give the metric at that time, and in particular, the curvature.

    Now the curvature (more precisely, the metric) determines the trajectories of the stuff in the universe as a function of time by means of the geodesic equations. So at time t+dt, the stress-enery tensor is not what it was at time t because stuff has moved! So the metric has changed also, according the Einstein's equations. Hence the trajectory of the stuff is gonna change also, etc. So in that way, the Einstein's equation + the geodesic equation work together to determine the evolution of the stuff in the universe. (This is sometimes explained in laymen terms by saying that the position of matter determines curvature and curvature determines the position of matter.)
  5. Jul 10, 2011 #4
    I found this interesting post and it triggered several questions in my mind:

    Can you elaborate on this way to assign a metric to a manifold?
    Wouldn't be there ways to make restrictions that assured nondegeneracy of the subspace?
    The Hilbert action is usually regarded as a 4 instead of 2 dim version of Gauss Bonnet theorem, how does this translate to the Einstein field equations?

    I have a question regarding this. In a manifold without a metric, that has an affine connection, curvature is a property of the connection, right?. So it seems you don't need a metric to have the notion of curvature, Could you clarify this? I'm confused about the dependence of curvature on the metric vs. the connection.
  6. Jul 10, 2011 #5


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    Do you have something in particular in mind?

    It's the first time I hear something like that. In my mind, the Hilbert-Einstein action functional is a real-valued map F from the space of lorentz metric on spacetime whose critical points correspond to solutions of the Einstein equations. I.e. F'(g)=0 is the variational formulation of the Einstein equations.

    Could you elaborate on the interpretation of the action as a Gauss-Bonnet analogue?

    It is indeed as you tell it. I only meant that given a metric, there is a "preferred" connection associated to it (Levi-Civita) and hence a "preferred curvature".
  7. Jul 10, 2011 #6
    Not really, just was curious as to what kind of restrictions could give us a nondegenerate metric in the case one wanted to put a metric in the submanifold of a Lorentzian manifold.

    I got this comparison from a commentary about the Gauss-Bonnet theorem in the book "Vector calculus 5th ed." from Marsden and Tromba (page 484). It is just mentioned as an invitation to go deeper into the study of differential geometry and GR since the book is just about vector calculus, but it made me think it was some usual knowledge.
    I checked it and indeed there is a resemblance between the first term of the left side of the Gauss-Bonnet theorem and the Hilbert action RHS.

    [tex]\frac{1}{2}\int_M{R\sqrt{g} d^2x}-\int_{\partial M}{K\sqrt{g} d^1x}=2\pi\chi[/tex]

    [tex]S=-\frac{1}{2k}\int{R\sqrt{-g} d^4x}[/tex]

    R= Ricci curvature scalar, here the fact that the Ricci scalar in 2 dimensions is twice the Gaussian curvature is used.

    With the main difference being the obvious fact that in the 4 dim case the signature is not positive-definite.
  8. Jul 10, 2011 #7

    George Jones

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    Metrics on manifolds are not unique, but a metric obtained by a restriction is unique.
    Consider a 4-dimensional vector space [itex]V[/itex] and bilinear form [itex]g[/itex] such there is a basis [itex]\left\{e_1 , e_2 , e_3 , e_4 \right\}[/itex] with [itex] 1 = g \left(e_1, e_1 \right) = g \left(e_2, e_2 \right) = -g \left(e_3, e_3 \right) = -g \left(e_4, e_4 \right)[/itex] and [itex]0 = g \left(e_\mu, e_\nu \right)[/itex] when [itex]\mu \neq\nu[/itex].

    Let [itex]U = \mathbf{span} \left\{ e_1 + e_3 , e_2 + e_4 \right\}[/itex]. What is [itex]g[/itex] restricted to [itex]U[/itex]?
  9. Jul 10, 2011 #8


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    Oh I see! Thanks for pointing that out..

    Notice that if one considers Gauss-Bonnet (without boundary) as a function of g... so a "Gauss-Bonnet action" if you will,


    Then Gauss-Bonnet formula implies GB'(g)=0 for all g.
  10. Oct 3, 2011 #9
    what i meant in my question was, even if the metric changes due to matter, it doesn't change spacetime manifold itself. n-sphere in out example. presence of matter on the sphere could deform metric field on it, but not sphere itself. (changes only up to diffeomorphism)

    i don't know what meaning exactly this underlying manifold physically has. But in my "classical" notion of spacetime, there might be some properties of manifold not contained in metric. (are they all non physical?)
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