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Stationary axisymmetric spacetime

  1. Dec 23, 2007 #1
    I am trying to understand the definition of stationary axisymmetric spacetimes and I am confused!
    Following Wald's definition in his book "General Relativity" a spacetime is called stationary axisymmetic when it exists a timilike Killing field [itex] \xi^\alpha [/itex] and a spacelike Killing field [itex] \psi^\alpha [/itex] whose integral curves are closed. In this case we can choose a coordinate system [itex] (x^0=t,x^1=\phi,x^2,x^3) [/itex] where the Killing fields are

    [tex]\xi^\alpha= \frac{\partial}{\partial\,t}, & \psi^\alpha=\frac{\partial}{\partial\,\phi} [/tex]

    Calculating now the integral curves of [itex] \psi^\alpha [/itex] we have

    [tex] \frac{d\,t}{d\,\lambda}=0, \quad \frac{d\,\phi}{d\,\lambda}=1, \quad \frac{d\,x^2}{d\,\lambda}=0, \quad \frac{d\,x^3}{d\,\lambda}=0 \Rightarrow[/tex]
    [tex] t(\lambda)=c_t, \quad \phi(\lambda)=\lambda+c_\phi, \quad x^2(\lambda)=c_2, \quad x^3(\lambda)=c_3 [/tex]

    and here is the problem. The above curve is not closed for any interval [itex] \lambda \in (\alpha,\beta) [/itex].
    The only answer I can think is that someone must make somekind of identification, say

    [tex] \phi(\lambda)=\lambda+c_\phi, \quad \lambda \in [0,2\,\pi) [/tex]
    [tex] \phi(\lambda+2\,\pi)=\phi(\lambda), \quad \lambda \geq 2\,\pi [/tex]

    in order to produce a closed curve.
    Is this the correct answer?
  2. jcsd
  3. Dec 23, 2007 #2


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    AFAIK, yes - the point is that [itex]\phi[/itex] is an angular coordinate, so it is periodic. Wald really only uses this result to show that if one choses coordinates t and [itex]\phi[/itex] properly, being stationary and axissymmetric solution is equivalent to saying that none the metric coefficients are functions of either t (time) or [tex]\phi[/itex] (the angle) - so that if you rotate the metric around the axis of symmetry, it doesn't change.
    Last edited: Dec 23, 2007
  4. Dec 23, 2007 #3
    Thanks pervect!

    So if I have an albitrary specetime with a Killing field [tex] \xi^\alpha =\frac{\partial}{\partial\,z}[/tex] where the coordinate [itex] z [/itex] is not an angular coordinate, I can make the space time axisymmetric by imposing periodical conditions on [itex] z [/itex].
    I assume that can be done by the cost of altering the topology of the spacetime,

    i.e. if for the original spacetime one chart could desribe the whole manifold, after imposing the periodical conditions we need infinity charts to cover the manifold.
  5. Dec 23, 2007 #4


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    I'm not so sure I'd go that far!
  6. Dec 23, 2007 #5

    George Jones

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    Is exterior Schwarzschild stationary and axisymmetric?
  7. Dec 23, 2007 #6
    I hope this is not a ...tricky question! :smile:

    Schwarzschild' s exterior solution is not only stationary but also static. As far as I can understand it is also axisymmetric because it admits the Killing field [tex]\xi^\alpha=\partial_\phi [/tex] which corresponds to the angular coordinate [tex] \phi [/tex].

    Am I loosing something?
  8. Dec 24, 2007 #7

    George Jones

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    No, I misread what you wrote in post #3 and thought that Schwarzschild was a counterexample. What I had in mind is captured better by what follows.

    Consider [itex]\mathbb{R}^2[/itex] covered by the standard single chart. The infinite cylinder can be realized as a quotient manifold obtained from the equivalence relation [itex]\left( x , y \right) \sim \left( x+n , y \right)[/itex] for [itex]n \in \mathbb{N}[/itex].

    The single chart that covers [itex]\mathbb{R}^2[/itex] does give rise to an infinite number of charts on the cylinder, but an infinite number of charts aren't needed to cover the cylinder, since other charts are possible..
    Last edited: Dec 24, 2007
  9. Dec 24, 2007 #8


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    A simple example of what's bothering me about the question. Suppose we have a plane Then we know that in polar coordinates, that one coordinate chart of the plane is:

    ds^2 = dr^2 + r^2 d[itex]\theta^2[/itex]

    where 0<=r<[itex]\infty[/itex] and 0<=r<2[itex]\pi[/itex]

    (not really sure about all the equal signs here in the domain here, but let's not sweat that right now). Now the question is, can we imagine some other topology (i.e. not a plane) that has the line element

    ds^2 = dr^2 + r^2 d[itex]\theta^2[/itex]

    defined over a different domain? I think that maybe there could be. What comes to mind is the surface initially postulated by Riemann to handle the logarithmic function on the complex plane, which has an infinite number of sheets. (And there are obvious subsets generated by cutting and pasting).

    I think that we have to exclude the origin though - there shouldn't be any real problem including the origin of the plane (it's just a coordinate singularity), but there is a real problem with the manifold at R=0 for the Riemann log surface.

    But I don't think getting into this really sheds much light on Wald's intent - it's getting more into the math end of the things than the physics end.
    Last edited: Dec 24, 2007
  10. Dec 25, 2007 #9
    Ok, let' s return to physics! :smile:

    There is a very nice article of Gott and Alpert, GRG 16, 243-247 (1984) about (2+1) Spacetime.
    In there terminology having the line element:

    [tex] d\,s^2=-d\,t'^2+B_o\,d\,r^2+r^2\,d\,\phi^2, \quad 0\leq\phi<2\pi [/tex]

    which decribes a flat Minkowski space and performing the coordinate transormation

    [tex] r'=\sqrt{B_o}\,r,\quad \phi'=\frac{\phi}{\sqrt{B_o}} [/tex]

    we end up with flat (2 +1)-Minkowski metric

    [tex] d\,s^2=-d\,t'^2+d\,r'^2+r'^2\,d\,\phi'^2[/tex]

    written in polar coordinates. But now the coordinate [tex]\phi'[/tex] has range

    [tex] 0\leq\phi'<\frac{2\pi}{\sqrt{B_o}} [/tex]

    producing a conical singurality. As they mention:

    "However, the singularity is focusing, in that particle trajectories which are initially
    parallel and pass on opposite sides of r = 0 will eventually intersect each other at
    an angle of [tex] 2\pi(1-\frac{1}{\sqrt{B_o}}[/tex]."

    So the physics is altered (!)

    Thus we come again in the first question.
    The former line element is axisymmetric, but the latter is not, if we do not demand the periodical conditions.
    And as far as I can understand this is an counterexample for George Jones' s post, because even though we can still have one chart [tex](t',r',\phi')[/tex] to cover the whole manifold, if we insist on axisymmetric demand we must use infinity charts.

    Thus I think that the range of the coordinates are crucial. (The other possibility is that I am are mixed up :smile: )

    Merry Christmas to everybody!
  11. Dec 28, 2007 #10

    George Jones

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    Consider a slightly altered version of my example in post #7 that is the strip of the plane given given by [itex]\left\{\left(x,y\right) | -1 \leq x \leq 1, -\infty < y < \infty\right\}[/itex]. In this strip, polar coordinates still work in the patch [itex]\left\{\left(r,\theta\right) | 0 < r < 1, 0< \theta < 2\pi\right\}[/itex].

    Now roll the paper into into a cylinder and glue, i.e., identify the lines [itex]x = -1[/itex] and [itex]x = 1[/itex]. This results in a space that is not simply connected, but that still is intrinsically flat. The components of the metric in the polar coordinate patch (a small part of the space) with domain as in the previous paragraph are just as you gave.
    Last edited: Dec 28, 2007
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