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Kerr Metric

  1. Apr 1, 2004 #1
    How can I obtain a Kerr metric by using the Einstein equation?
     
  2. jcsd
  3. Apr 5, 2004 #2

    DW

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    The stress energy tensor for the Kerr Newman spacetime is such a mess I wish you luck. The T^00 term by itelf has 33 terms in the numerator alone. Normally instead of going that rout certain symmetries and boundary conditions are assumed from which the metric is worked out.
     
  4. Apr 5, 2004 #3

    Stingray

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    Its not simple. The most elegant way is by using the Kerr-Schild construction. Let the metric be of the form g=eta+k.k, where eta is the minkowski metric, and k is a null vector. There are various nice theorems that can be proven with this (even though it looks perturbative, the results are exact) that eventually lead to Kerr. I believe Chandrasekhar's black hole book does this, although it may not be the best place to start.
     
  5. Apr 9, 2004 #4
    Was trying to follow the construction by way of complex transformations of the Schwartzschild metric (in advanced Eddington-Finkelstein coordinates) through the null tetrad formalism in Ray D'Inverno's Introducing Einstein's Relativity. The result is supposed to be the Boyer-Lindquist form of the Kerr metric. But there are gaps in the complex transformation process that I don't understand. For example:- it suddenly comes up with a choice of the null tetrad for the Schwartzschild metric (in advanced Eddington-Finkelstein coordinates) and asks the reader to check that [itex]g^{ab}=l^a n^b + l^b n^a - m^a \bar{m}^b - m^b \bar{m}^a[/itex], and I'm wondering whether there is a typo inside that choice [itex]n^a=(-1,-\frac{1}{2} (1-\frac{2m}{r}),0,0)[/itex] (shouldn't it be [itex]n^a=(-1,-\frac{1}{2} (1-\frac{2m}{r})^{-1},0,0)[/itex]?), etc. Has anyone out there followed the construction? Am I missing some more fundamental math that's able to generate that choice of the null tetrad? Are such algebraic stuff (construction of the Kerr metric) more suited for computer programs to work out? Is it really necessary to learn how to do it by hand? :confused:
     
  6. Nov 28, 2006 #5

    Chris Hillman

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    Deriving the Kerr vacuum solution?

    Hi, Vu,

    Well there are dozens of papers in the literature which offer "elementary" or "fast" or otherwise distinguished "derivations", for example http://www.arxiv.org/abs/gr-qc/0305035 and innumerable books.

    I was about to tell Vu that tastes differ, and indeed, while the Kerr-Schild Ansatz is certainly important, I happen to prefer the approach based on the Ernst equation because of its generality (applies to all axisymmetric stationary electrovacuum solutions) and its "obvious connection" with mathematical techniques which are valuable in studying other nonlinear partial differential equations, such as symmetry analysis, as well as its "unexpected connections" with other important topics in gtr such as colliding plane waves.

    Before I say anything else, I should point out that the basic reference for exact solutions in gtr is the monograph Exact solutions of Einstein's field equations, by Hans Stephani et al., 2nd ed., Cambridge University Press, 2003. Of all the textbooks on gtr, not suprisingly, the one by Stephani has the nicest discussion of exact solutions. However, not even Stephani attempts to even mention all the "usual suspects" in the world of exact solutions, which has in any case been augmented by a relative newcomer, the Neugebauer-Meinel rigidly rotating disk of dust model (1995), which may be seen twenty years from now as fully important as Kerr's discovery in 1963; see http://www.arxiv.org/abs/gr-qc/0301107

    I was about the mention D'Inverno's slick description of the so-called "Janis trick", which is a very fast (but rather mysterious) way of obtaining the Kerr vacuum, when I noticed this:

    Probably not; the discussion is perhaps a bit too sketchy.

    First, the NP tetrad he gives is indeed a valid NP tetrad for the Schwarzschild vacuum (written in the ingoing Eddington chart), but there are many possible choices of NP tetrad. In particular, in (19.4), (19.5) and (19.10) he gives a standard recipe for obtaining a tetrad from a real frame field.

    The most common frame field for the ingoing Eddington chart is the "slowfall frame" corresponding to observers who accelerate radially outward with with an acceleration vector of magnitude [tex]m/r^2[/tex], which of course would be just enough to maintain their position in Newtonian gravity (modulo the obvious question of choice of interpretation of the Schwarzschild radial coordinate), but who slowly fall in gtr due, if you like, to the fact that the gravitational field carries energy, which increases in density as you get closer, so the gravitational field is a bit stronger than Newton would lead you to expect (roughly speaking!).

    However, applying the recipe he mentions to this frame field gives a different choice of NP tetrad, and I agree that omitting to point this out seems sure to cause confusion. However, in my reading, D'Inverno is not implying that he expects his readers to guess where this tetrad comes from, only to check that it does work (give the right metric tensor).

    Second, in (19.28), it looks like D'Inverno forgot to mention something important: the radial function r(x,y,z) is given implicitly by
    [tex] \frac{x^2+y^2}{r^2 + a^2} + \frac{z^2}{r^2} = 1[/tex]
    and thus, the surfaces of constant Boyer-Lindquist "radius" should be thought of as something like oblate spheroids (rougly what you would expect from the equipotentials of a rotating fluid drop treated in Newtonian gravitation), even though they plot in the Boyer-Lindquest chart as prolate spheroids.

    Yes and no, respectively. See http://www.math.ucr.edu/home/baez/RelWWW/software.html for some suggestions. GRTensorII is particularly well suited for computations with specific exact or approximate solutions (or families of solutions, possibly defined using some constraint equations whose solutions are regarded as "known"). This is free and fairly well debugged, but it runs under Maple which is definitely NOT free, unless you are a registered university student, in which case I urge you to immediately obtain Maple, at literally 1/20-th the list price.

    If you think the Janis trick is slick, your mouth may water when I say that the derivation of the Kerr vacuum via the Ernst equation starts with the linear polynomial
    [itex]\frac{1}{p(\xi, \eta)} = \alpha \, \xi + \beta \, \eta [/itex]
    which is a solution of the Ernst equation written in prolate spheroidal coordinates on FLAT spacetime. Prolate spheroidal coordinates are related to cylindrical coordinates by the transformation
    [itex] z = A \, \xi \, \eta, \; \; r = A \, \sqrt{\xi^2-1} \, \sqrt{1-\eta^2}, [/itex]
    [itex] 1 < \xi < \infty, \; -1 < \eta < 1 [/itex]
    with inverse transformation
    [itex] 2 \, A \, \xi = \sqrt{(z+A)^2+r^2} + \sqrt{(z-A)^2+r^2}, \; \; 2 \, A \, \eta = \sqrt{(z+A)^2+r^2} - \sqrt{(z-A)^2+r^2}, [/itex]
    [itex] -\infty < z < \infty, \; 0 < r < \infty [/itex]

    The (upper half plane form of the) Ernst equation is (in standard vector calculus notation)
    [itex] p \, \Box p = | \nabla p |^2 - | \nabla q |^2 [/itex]
    [itex] p \, \Box q = 2 \, \nabla p \cdot \nabla q [/itex]
    This has a large and interesting symmetry group, which suggests an attack based on symmetries. It also has interesting Baecklund automorphisms which have been deftly exploited by Chandrasekhar and others.

    Chris Hillman
     
    Last edited: Nov 28, 2006
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