1

I would like to better understand a hypothesis that Wald uses to derive the general local formula of a static spherically symmetric spacetime.

A spacetime is said to be spherically symmetric if its isometry group contains a subgroup isomorphic to the group $SO(3)$, and the orbits of this subgroup are two-dimensional spheres. One can show that the metric of such a spacetime can always be written in the form $$ds^2 = -f(t,r)dt^2 + h(t,r)dr^2 + r^2(d\theta^2 + \sin(\theta)^2 d\phi^2).$$ On the other hand, a spacetime is static if it has a timelike Killing vector field $\xi^a$ which is hypersurface-orthogonal. One can show that the metric of such a spacetime can always be written in the form $$ds^2 = -V^2(x^1,x^2,x^3) dt^2 + h_{ij}(x^1,x^2,x^3) dx^i dx^j,$$ with $i,j$ running from $1$ to $3$. When one combines both requirements, there is no a priori reason for the time coordinate $t$ in the first case be the same as the time coordinate $t$ in the second case.

On page 120 of his GR book, Wald shows that the orbit spheres lie wholly within the hypersurfaces orthogonal to $\xi^a$ if the static Killing field $\xi^a$ is unique. Then it is easy to show that one can write the metric in the form $$ds^2 = -f(r)dt^2 + h(r)dr^2 + r^2(d\theta^2 + \sin(\theta)^2 d\phi^2).$$

I would like to understand why it is reasonable to assume this uniqueness of $\xi^a$. Is there any relevant case for which this hypothesis does not hold and the metric cannot be written in that form? (Minkowski space is an obvious example of a static spherically symmetric spacetime for which $\xi^a$ is not unique, but with the metric still taking the given form.)

Qmechanic
  • 201,751
ram
  • 21
  • A good question. But your last sentence in brackets is in my understanding not true. You mean probably Minkowski's metric in spherical coordinates. It is static indeed but not spherically symmetric because it can be globally transformed into pseudo-euclidean form by corresponding coordinate transformation. – JanG Nov 04 '23 at 11:35
  • Thanks for the comment. Minkowski spacetime has the Poincaré group as its isometry group; this has the rotation group as a subgroup, with its orbits being perfect geometrical 2-spheres. Therefore, Minkowski spacetime is spherically symmetric. Notice that being a spherically symmetric spacetime is a coordinate independent property (see definition above). Notice also that if one can find coordinates such that the metric can be written in the form $ds^2 = -f(r)dt^2 + h(r)dr^2 + r^2(d\theta^2 + \sin(\theta)^2 d\phi^2)$, then the spacetime must be spherically symmetric. – ram Nov 04 '23 at 14:08
  • I understand your argument but lets take for example $R^2$ - a plane. I can foliate it completely by nested circles and one point. However, I would never say the plane is rotational symmetric. Please read my question Definition of spherically symmetric spacetime and tell me what you think about it. – JanG Nov 04 '23 at 15:12
  • By the way, in Minkowski metric $g_{00}$ and $g_{rr}$ are not functions of $r$, but constants equal $1$. I think that makes the difference to the spherically symmetric spacetime. – JanG Nov 04 '23 at 16:20
  • 1
    @JanG you are incorrect and ram is correct: Minkowski spacetime is of course spherically symmetric. Take the very simple example of Schwarzschild with $M \rightarrow 0$. The Poincare group includes rotations. – Eletie Nov 05 '23 at 18:42
  • @Eletie The exterior Schwarzschild metric for $M\rightarrow 0$ reads $ds^2=c^2 dt^2-dr^2-r^2 d\Omega^2$. That metric is nothing else as Minkowski metric written in spherically coordinate system. Both metrics are “flat”, i.e. all components of Riemann curvature tensor are zero . In contrast to this the spacetime described by similar metric ($ds^2=(1-p_1/4+p_1 r^2/R^2)^2 c^2 dt^2-dr^2-r^2 d\Omega^2$)[https://physics.stackexchange.com/a/679431/281096] (eq. 11) is spherically symmetric one. – JanG Nov 06 '23 at 09:22
  • 1
    @JanG you have misunderstood. I was simply claiming, as the OP did, that Minkowski spacetime possesses spherical symmetry, contrary to your comment. The specific cooordinate representation does not matter whatsoever. – Eletie Nov 06 '23 at 20:32
  • @Eletie Apparently I do not understand that notion properly. Flat spacetime metric is invariant under rotations about the 3-axis.. What is rotated? I prefer the definition of Crowell. However, does he mean with it foliation or fibration of spacetime? – JanG Nov 07 '23 at 11:45
  • @JanG The Wiki definition suffices for me: 'a spherically symmetric spacetime is a spacetime whose isometry group contains a subgroup which is isomorphic to the rotation group SO(3)'. You can formulate this with killing vectors if you wish. You'll also find the same result stated in most textbooks, e.g., Carroll says 'The simplest example is flat three-dimensional Euclidean space. If we pick an origin, then R3 is clearly spherically symmetric with respect to rotations around this origin.', etc. – Eletie Nov 07 '23 at 14:47
  • @Eletie I am of course aware of these definitions. For me they just say that such space is isotropic in every direction. To prove that $E^3$ is spherically symmetric one has to choose one point and show that rotation (of what?) about it doesn’t change anything (what?). In my understanding in Newton’s theory not space but matter fields are spherically symmetric and that in GR it is the other way around. The spherical symmetry of matter is due to spherical symmetry of spacetime. But it is only my thinking. – JanG Nov 07 '23 at 16:01
  • @Eletie One additional remark: 'The simplest example is flat three-dimensional Euclidean space. If we pick an origin, then $R^3$ is clearly spherically symmetric with respect to rotations around this origin”. Because it is valid for any point there such space is called spherically symmetric, right? However, what is in case of Schwarzschild spacetime? There exists only one point that fulfills the criterion of origin but we call the whole space spherically symmetric. – JanG Nov 07 '23 at 16:45
  • @JanG You seem to be using very non-standard definitions. This shouldn't be a contentious topic. No, I never said it had to be valid at every point. If you're aware of the definitions then you'll always find Minkowski spacetime is (almost trivially) spherically symmetric. Would you also argue Minkowski is not invariant under Poincare transformations because you can apply them at every point? You may also struggle with maximally symmetric spacetimes in this case.. – Eletie Nov 07 '23 at 17:22
  • @JanG if you want to use your own definitions of spherical symmetry that is fine, but be aware that this makes most of your statements incorrect to a general audience (and therefore unhelpful). – Eletie Nov 07 '23 at 17:25
  • @Eletie Thanks for your comments and advice. They are very valuable for me and hopefully for other user as well. – JanG Nov 07 '23 at 18:56

0 Answers0