# 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. – Michael Seifert Jul 12 '19 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 '19 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. – Michael Seifert Jul 12 '19 at 15:52