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

Static vs Stationary?

  1. May 7, 2007 #1
    Just a quick question. What is the difference between a static vs stationary spacetime metric? does static imply stationary? is the converse true?
    What are a couple of example?

  2. jcsd
  3. May 7, 2007 #2
    A metric is stationary it has a timelike Killing vector field. If this Killing vector field is orthogonal to a family of spacelike surfaces of constant time, it is also considered static.

    Stationary, in non-technical terms, means basically that there exists a coordinate system where we can express the metric tensor independent of the time coordinate.

    Static is thus more restrictive than stationary.

    For example the Schwarzschild solution is static while the Kerr metric is stationary.
    Last edited: May 7, 2007
  4. May 7, 2007 #3
    cool. so basically we need the metric to have d/dt = 0 for all gmu,nu in order to be static and hence stationary?
    if d/dt does not equal zero it could be stationary but not static?
    static implies stationary but stationary does not imply static, correct?

    thank you
    Last edited: May 7, 2007
  5. May 7, 2007 #4

    Chris Hillman

    User Avatar
    Science Advisor

    Some definitions, examples, links, suggested reading

    A stationary spacetime admits a timelike Killing vector field. A static spacetime is a stationary spacetime in which the timelike Killing vector field has vanishing vorticity, or equivalently (by the Frobenius theorem) is hypersurface orthogonal.

    What does this mean? That a stationary spacetime is one in which you can find a family of observers who observe no changes in the gravitational field (or sources such as matter or electromagnetic fields) over time. A static spacetime is one which admits a slicing into spacelike hypersurfaces which are everywhere orthogonal to the world lines of our "bored observers".

    One can still imagine observers who move around an interact, or even light beams or other waves which are assumed to have neglible effect on the ambient gravitational field. So for example one can study scattering of waves by black holes.

    Yes. No.

    Some basic examples of static solutions:

    1. Any Weyl vacuum: see my post #12 in https://www.physicsforums.com/showthread.php?t=153765&highlight=Weyl+vacuum
    plus generalizations allowing for EM fields, scalar fields, etc.

    2. Any static spherically symmetric perfect fluid:

    Some basic examples of stationary solutions:

    1. Any Ernst vacuum (the family of all stationary axisymmetric vacuum solutions). This family includes the Kerr vacuum.

    2. The van Stockum dust http://en.wikipedia.org/w/index.php?title=Van_Stockum_dust&oldid=39874076

    3. The Meinel-Neugebauer disk (rigidly rotating dust surrounded by vacuum).

    4. The Bonnor beam (steady state cylindrical beam of incoherent electromagnetic radiation surrounded by vacuum):

    Some nonstationary solutions:

    1. Most cosmological models, including the FRW dusts.

    2. The monochromatic (single frequency) electromagnetic plane wave:
    and its gravitational wave analog.

    3. Colliding plane wave models.

    The standard reference on exact solutions of the EFE is the monograph by Kramer et al. cited at http://math.ucr.edu/home/baez/RelWWW/HTML/reading.html#advanced [Broken]
    Last edited by a moderator: May 2, 2017
  6. May 7, 2007 #5


    User Avatar
    Science Advisor

    Would this mean a spacetime containing only a nonspinning black hole would be stationary outside the event horizon but not inside?
  7. May 8, 2007 #6

    Chris Hillman

    User Avatar
    Science Advisor

    Yes, and that's a good point. Because no observers can exist between the inner and outer horizons of the Kerr vacuum solution who are not experiencing increasing curvature (tidal forces and so on), i.e. who are not observing rapid changes mitigating against their well-being.
  8. May 9, 2007 #7
    Is there a way to check if a spacetime given as metric in some arbitrary coordinates is static or stationary without having to obtain the Killing vector?

    For example, we know that a spacetime is Minkowski if all the components of Riemann tensor vanish at all points, no matter in what weird looking coordinates we work.

    Is there a similar test to determine if spacetime is static/stationary ?
  9. May 9, 2007 #8


    User Avatar
    Staff Emeritus
    Science Advisor

    Killing vectors can be described without reference to coordinates. If you imagine a one parameter group of isometries (i.e. diffeomorphisms that leave the metric unchanged), Killing vectors are the infinitesimal "generators" of that group.

    However, when you have a Killing vector, one usually takes advantage of the symmetry by a proper choice of coordinates.

    Note that Killing vectors can be time-like or space-like. It is the existence time-like Killing vectors that is necessary for a stationary system.
  10. May 9, 2007 #9
    I know that but imagine you are given a metric in some coordinates. Finding the Killing vectors require solving a system of differential equations and that is not trivial.

    Is it possible for check if the given metric specifies static/stationary spacetime without finding the killing vectors? Something similar to how we find out if a metric represents flat spacetime by computing the Riemann tensor components.

    A side question: If a Killing vector field is timelike at a point, is it timelike everywhere or it can flip between timelike, null and spacelike in different regions?
  11. May 9, 2007 #10


    User Avatar
    Staff Emeritus
    Science Advisor

    Not necessarily. If the metric tensor does not depend upon a particular coordinate, say k, then d/dk will be a Killing vector. Using a good choice of coordinate system, one can spot the Killing vectors directly from the metric tensor. For example, Schwarzschild metric does not depend on t or phi, and thus d/dt and d/dphi are Killing vectors.
  12. May 9, 2007 #11
    I am talking about absolute generic metric in generic coordinates, not a metric with obvious symmetries for which you can just read off the Killing vectors.
  13. May 9, 2007 #12

    Chris Hillman

    User Avatar
    Science Advisor

    Example of solving and interpreeting Killing equations

    Actually, the Killing equations are often much easier to solve than one might think. Indeed, GRTensorII often succeeds in solving them in under one second. Typical example: consider the EK6 vacuum pp wave in cylindrical Brinkmann chart:
    ds^2 = -4 m \, \log(R) \, dU^2 + 2 \, dU \, dV + dR^2 + R^2 \, d\Theta^2,
    -\infty < U, \, V < \infty, \; 0 < R < \infty, \; -\pi < \Theta < \pi
    or in GRTensorII speak
    Code (Text):

    Ndim_ :=  4:
    x1_ :=  U:
    x2_ :=  V:
    x3_ :=  R:
    x4_ :=  Theta:
    eta11_ := -1:
    eta22_ :=  1:
    eta33_ :=  1:
    eta44_ :=  1:
    bd11_ := -1/sqrt(2) - 2*m*log(R)/sqrt(2):
    bd12_ := -1/sqrt(2):
    bd21_ := -1/sqrt(2) + 2*m*log(R)/sqrt(2):
    bd22_ :=  1/sqrt(2):
    bd33_ :=  1:
    bd44_ :=  R:
    Info_ := `Ehlers-Kundt type 6 grav. beam PP wave (cylindrical Brinkmann chart)`:
    # A beam of gravitational radiation (at R = 0).
    # Chart covers -Infinity < U, Z < Infinity, 0 < R < Infinity, -Pi < Theta < Pi
    # Line element
    #       ds^2 = -4m log(R) dU^2 - 2 dU dV + dR^2 + R^2 dTheta^2
    #       -infty < U,V < infty, 0 < R < infty, -Pi < Theta < Pi
    # Three dimensional isometry group:
    #       null Killing vector field       @_V
    #       2nd K.V:                        @_U
    #       spacelike Killing vector field  @_Theta
    # Our observers are noninertial and spinning
    # Acceleration D_X X, where X = e_1 is m/R e_3
    # Vorticity vector is -m/(2*R) e_4
    # Expansion tensor
    #   H_(23) = H_(32) = -m/(2*R)
    # Electrogravitic tensor is
    #       E_(33) = E_(44) = m/R^2
    # Magnetogravitic tensor is
    #       B_(34) = B_(43) = m/R^2
    # Wave equation is
    #   *d*h =
    #   -2 h_(ZU) + 4m log(R) h_(ZZ) + h_(RR) + h_R/R + h_(Theta Theta)/R^2
    # Axially symmetric solution h(U,Z,R) = exp(a*U) f(Z,R) where
    #   -2a R f_Z + 4m R log(R) f_(ZZ) + R f_(RR) + f_R/R = 0
    # Cylindrically symmetric solution h(U,Z) = F(U) + G(U) log(R)
    # X = 1/sqrt(4*m*log(R)-2*m) [ @_U + sqrt(2*m)/R @_Theta ]
    # defines a timelike geodesic congruence with helical world lines.
    The commands
    Code (Text):

       grcalc> uu(U,V,R,Theta);
       grcalc> vv(U,V,R,Theta);
       grcalc> rr(U,V,R,Theta);
       grcalc> tt(U,V,R,Theta);
    keqs := [seq(seq(grcomponent(LieD[kv0,g(dn,dn)],
    for guy in keqs do value(guy); od;
    casesplit(keqs); op(%)[1]; pdsolve(%);
       {tt(U, V, R, Theta) = _C3, vv(U, V, R, Theta) = _C1,
       uu(U, V, R, Theta) = _C2, rr(U, V, R, Theta) = 0}
    shows that the Killing vectors are
    [tex] \partial_U, \; \partial_V, \; \partial_\Theta [/tex]
    Here [itex]\partial_V[/itex] is the wave vector (a null translation), [itex]\partial_\Theta[/itex] is rotation about the axis of spatial symmetry, and [itex]\partial_U[/itex] is timelike where [itex]R>1[/itex], null at [itex]R=1[/itex], and spacelike on [itex]0 <R<1[/itex]. This vacuum solution can be matched to a null dust solution, incidently, to form the Bonnor beam. We just verified that the vacuum exterior is stationary outside [itex]R=1[/itex]. Next, computing normalizing and computing the vorticity tensor shows this doesn't vanish, so this Killing vector field is timelike but not hypersurface orthogonal. In my post #3, I listed the Bonnor beam as stationary but nonstatic. Now you know why.

    Well, in the case of the Rindler chart on Minkowski spacetime
    [tex]ds^2 = -x^2 \, dt^2 + dx^2 + dy^2 + dz^2, \;
    -\infty < t, \, y \, z < \infty, \; 0 < x < \infty [/tex]
    the metric components depend only on [itex]x[/itex], so the other three coordinate vectors are Killing vectors. Since [itex]\partial_t[/itex] is timelike, you can tell at a glance that this metric is stationary. Computing the vorticity tensor will tell you whether or not it is in fact static!

    [EDIT: just noticed that cristo already pointed this out.]

    Yes, this happens in the Kerr vacuum at the horizon. Indeed, if you draw the Killing flow in the Kruskal-Szekeres chart for the special case of the Schwarzschild vacuum, you obtain nested hyperbolas plus the "X" representing the event horizon of the "eternal black hole".

    Something similar happens in the case of the EK6 pp vacuum gravitational wave studied above, as you see.

    In the 5 part arXiv eprint by Vacaru (which I assume everyone here is studying because of its obvious interest), some of you might have noticed the Delphic comment in the introduction that near the horizon, the Schwarzschild vacuum resembles a pp-wave. This refers to a construction of Penrose, the Penrose limit, according to which near any null geodesic, any Lorentzian spacetime resembles a pp-wave. But the world sheet of the horizon consists of the world lines of special null geodesics, so...

    There are various tensors which can recognize various geometrically interesting (locally defined!) circumstances besides local flatness, such as local conformal flatness. In some circumstances the vanishing of particular tensors or certain scalar invariants can enable one to recognize particular spacetimes and thus certain symmetries, but this is exceptional.
    Last edited: May 9, 2007
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook