Axial symmetry

And finally, let's break the spherical symmetry we have been dealing with thus far down to axial one. (This is the case already in the Kerr-Newman solution, but there the breaking is due to rotation only: the mass and charge are spherically symmetrically distributed.)

Are there any static axially symmetrical solutions? Yes! Plenty!

In the spherically symmetrical case for pure $g_{\mu\nu}$ (or $g_{\mu\nu} \oplus A_\mu$) the Schwarzschild (respectively Reissner-Nordstrom) solution is the only solution, i.e. even if you solve time-dependent problem, a solution you obtain can be put in the above written (``static'') form by an appropriate change of coordinates (the Birkhoff theorem). This is not true already for the scalar solutions. There the static solution is unique, but time-dependent problem is in no way reducible to it (and no time-dependent solutions are known; this problem is already for numerical investigations only).

For the axial symmetry, even time-independent problem has infinite number of inequivalent solutions, but nevertheless they all can be classified, because each of them turns out to be determined by a solution of a single 2-dimensional Laplace equation.

Here is one of them, for pure gravity and which contains Schwarzschild solution for a particular value of the parameter ( $\alpha = 1$):

It has two symmetries: $t \rightarrow t+const$ and $\phi \rightarrow \phi + const$. There is an infinitely redshifted surface at $r = \frac{2M}{\alpha}$(where $dtdt$-component of the metric becomes zero) and this surface is singular, i.e. curvature invariants blow up there.

One can attempt to analyze it further, but it is not even clear how to proceed. You get stuck already trying to choose the ``right coordinates''. (In [3] it is claimed that the right coordinates for this solution are spheroidal ones, in which metric looks even more complicated, but the arguments there are not very convincing). By analogy with areas of symmetry orbits in the spherically symmetrical case, one can calculate circumference lengths (which are coordinate independent quantities) for circles around the symmetry axis on the $r=2k$ surface and find that they are finite or infinite depending on $\theta$. Thus, going ``from our Minkowski infinity down there'' you would arrive not to something ``on the axis of symmetry'', but to a region which is ``infinitely large'' (see [5]). The $r=2k$ (hyper)surface is not singular around the poles, so that the non-singular region is not simply connected; there are also some fancy ways of continuing this solution past this surface which gives rise to ``double-sheeted topology'' (see [4]).

And so on.