# Cylindrical universe cosmology in general relativity

Is there a compact cylindrical universe solution to the Einstein equation, with space homogeneity, without using "artificial" periodic boundaries? I'm expecting a metric of the following shape: $$\begin{equation}\tag{1} ds^2 = f^2(t, r) \, dt^2 - g^2(t, r)(dx^2 + dy^2) - b^2(t) \, dz^2, \end{equation}$$ where $$f(t, r)$$, $$g(t, r)$$ and $$b(t)$$ are some functions and $$r = \sqrt{x^2 + y^2}$$. I'm also expecting $$f(t, r) = 1$$ and $$g(t, r) = a(t) \, g(r)$$. The cylindrical planar space section (coordinates $$x$$ and $$y$$) should be homogeneous and compact.

I'm thinking of something similar to a 2D version of the spherical closed universe in cosmology ($$k = +1$$), with an extra space dimension, of FLRW metric $$\begin{equation}\tag{2} ds^2 = dt^2 - \frac{a^2(t)}{(1 + r^2 / 4)^2}(dx^2 + dy^2 + dz^2) - b^2(t) \, du^2. \end{equation}$$

The metric (1) I'm looking for would have an homogeneous 3D space, would be 2D isotropic in the cylindrical plane (in coordinates $$x$$ and $$y$$), but would not be isotropic in 3D.

EDIT 2 (I've fixed some important mistakes): If I assume $$f(t, r) = 1$$ and $$g(t, r) = a(t) \, g(r)$$, I could compute the Christoffel symbols and get these (all the other symbols are trivially 0. Here: $$i, j, k = 1, 2$$ for coordinates $$x$$ and $$y$$): \begin{align} \Gamma_{ij}^0 &= a \dot{a} g^2 \, \delta_{ij}, \\[12pt] \Gamma_{33}^0 &= b \dot{b}, \\[12pt] \Gamma_{0 j}^i &= \frac{\dot{a}}{a} \, \delta_{ij}, \\[12pt] \Gamma_{jk}^i &= \frac{1}{g}(\partial_j g \, \delta_{ik} + \partial_k g \, \delta_{ij} - \partial_i \, g \, \delta_{jk}), \\[12pt] \Gamma_{03}^3 &= \frac{\dot{b}}{b}. \end{align} Then, I compute the Riemann components (all other components are 0): \begin{align} R^0_{i 0 j} &= a \ddot{a} \, g^2 \, \delta_{ij}, \\[12pt] R^0_{303} &= b \ddot{b}, \\[12pt] R^i_{33j} &= -\, \frac{\dot{a} \dot{b} b}{a} \, \delta_{ij}, \\[12pt] R^i_{jkl} &= (\dot{a}^2 g^2 + \mathcal{Q}(r))(\delta_{ki} \, \delta_{lj} - \delta_{li} \, \delta_{kj}) + \text{anisotropic terms}. \end{align} The anisotropic terms cancel if $$\begin{equation}\tag{3} \frac{g^{\prime \prime}}{g} = \frac{g^{\prime}}{g \, r} + 2 \frac{g^{\prime 2}}{g^2}, \end{equation}$$ which implies the solution $$\begin{equation}\tag{4} g(r) = \frac{A}{1 + B \, r^2}, \end{equation}$$ where $$A$$ and $$B$$ are constants (they can be absorbed into the coordinates, so $$A = 1$$ and $$B = -1, 0, 1$$ like the parameter $$k$$ in cosmology). Then $$\begin{equation}\tag{5} \mathcal{Q}(r) = 4 B \, g^2. \end{equation}$$ I then compute the Einstein tensor (again: $$i, j = 1, 2$$ only): \begin{align} G_{00} &= -\, \Big( 2 \frac{\dot{a} \dot{b}}{a b} + \frac{\dot{a}^2}{a^2} + \frac{4 B}{a^2} \Big), \tag{6} \\[12pt] G_{ij} &= -\, \Big( \frac{\ddot{a}}{a} + \frac{\ddot{b}}{b} + \frac{\dot{a} \dot{b}}{a b} \Big) \, g_{ij}, \tag{7} \\[12pt] G_{33} &= -\, \Big( 2 \frac{\ddot{a}}{a} + \frac{\dot{a}^2}{a^2} + \frac{4 B}{a^2} \Big) \, g_{33}. \tag{8} \end{align} These components are similar (but have some differences) with what we find in the usual spherical FLRW cosmology.

I guess that I need to introduce some anisotropic pressure ($$\sigma = p_{\perp}$$ in the $$x$$-$$y$$ plane and $$p = p_{\parallel}$$ along the $$z$$ axis). Is there any interesting paper on anisotropic-cylindrical cosmologies?

• Would having a conical singularity at $r = 0$ count as not having "artificial" periodic boundaries? Depending on the behaviour of $g$ as $r \to 0$, this is easy to accomplish. However, a conical singularity can also be obtained by removing a wedge from a well-behaved spacetime and identifying the "cuts", so it may not be what you're looking for. Jul 12, 2019 at 15:33
• @MichaelSeifert, I'm interested in a regular metric at $r = 0$ (so no conical singularity). Space should be homogeneous.
– Cham
Jul 12, 2019 at 15:38
• Ah, OK. If it's regular at $r = 0$, then it would seem to be topologically trivial. But you used the topology tag and wanted the spacetime to be "cylindrical", which is why I'm confused. Jul 12, 2019 at 15:52