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Variety of metrics

  1. Jul 23, 2004 #1
    How many metrics are there in special relativity and in general relativity?
    Can one metric be transformed into another?
    Can objects that don't have spherical symmetry have metrics?
    Is it possible for two different manifolds to intersect and for their
    metrics to be equal where they intersect?
    Do supermassive black holes have Schwarzschild metrics?
    If all the mass in the universe is at one point, space-time doesn't exist and nor does the concept of a metric.Do metrics become invalid at more normal mass-densities?
    Are metrics conserved:if a black hole evaporates, does a new black hole appear elsewhere in the universe?
    What sort of metric describes the space-time around a single proton?
    If a graviton has energy that curves space-time, what would its metric be?
    Last edited: Jul 23, 2004
  2. jcsd
  3. Jul 23, 2004 #2
    Do you mean metric or metric tensor? See


    re - " Can one metric be transformed into another?"

    To be 100% precise, not really. The metric tensor is a geometric object which is defines the magnitudes of vectors. I.e. that means that its defined as a bilinear map of two vectors into a scalar. I.e. the metric tensor is defined as a "machine" g(_, _) such that when you insert two vectors it outputs a scalar (i.e. a single number which is independant of any coordinate system used to evaluate the number). It is defined such that when you put in a vector you get the squared lenght of the vector or if you put in two different vectors you get the scalar product of the two vectors. Since its a geometrical object it can't be changed by any transformation. What you can transform are the components of the metric tensor.

    re - "Can objects that don't have spherical symmetry have metrics?"

    An object does not have a metric tensor. In GR there is a one-to-one association between a physical distribution of matter and the spacetime in which that matter is in. That association is given by Einstein's Field Equations (EFE). There is a metric associated with each such spacetime. The metric defines the geometry of the spacetime and is a solution to the EFE. The geometry of the spacetime is defined when the infinitesimal spacetime interval is defined between all neighboring events. That interval is given by the metric tensor.

    re - "Is it possible for two different manifolds to intersect and for their
    metrics to be equal where they intersect?"

    According to the definition of "manifold" they'd be the same manifold, not two different manifolds.

    re - "Do supermassive black holes have Schwarzschild metrics?"

    Its unlikely to happen in nature because its probable that all black holes have some angular momentum and as such are described by the Kerr metric and not the Schwarzschild metric.

  4. Jul 24, 2004 #3
    The metric defines the geometry of the spacetime and is a solution to the EFE


    How many solutions are there to the EFE- I gather there are 16 EFE's.
    Why can't EFEs be given solutions internal to a curvature generating mass?
    And why would 1kg of hydrogen curve space-time the same as 1 kg of electrons or photons (I'm thinking here of the Eotvos experiment which showed that the nature of a mass was irrelevant to how it accelerated)?
    Does dark energy curve space-time the same as other energy sources?

    I understand the use of the metric tensor but is it the metric that determines the curvature of a manifold? And does the metric in GR reflect how the basis vectors change from one point in space to another?
    Last edited: Jul 24, 2004
  5. Jul 24, 2004 #4


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    In Riemannian geometry, or the pseudo-Riemannian geometry in General Relativity, this is true. The components of the connection are then the Christoffel symbols, differential expressions of the components of the metric tensor with respect to the spactime coordinates. And a linear combination of these expressions constitutes the Riemann-Christoffel tensor, which is the complete description of the local curvature of spacetime; so that all comes straight out of the metric in this type of geometry. Einstein's tensor, the left side of his field equations, is formed by "contracting" the Riemann-Christoffel tensor, so again the left side of the equation is a differential expression based on the metric coefficients.

    Because the field equation is between second rank tensors in four dimensions, the naive number of component-by-component equations would be 4x4=16. But the tensors are symmetric, so that six of the equations are the same as another six, leaving only ten independent component-by-component equations.

    AFAIK, nobody has every done a complete characterization of the family of solutions of Einstein's equations. What we have instead is just a growing number of particular solutions, of which the Schwartzschild solution was the first.
    Last edited: Jul 24, 2004
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