I'm trying to find the metric describing the spacetime around an infinite cylinder of radius $\rho$ and mass density $m$. Since the spacetime is static and cylindrically symmetric, the metric must be of the following form:
$$\mathrm{d}s^2 = a(r) c^2 \mathrm{d}t^2 - b(r) \mathrm{d}r^2 - c(r) \mathrm{d}z^2 - r^2 \mathrm{d}\theta^2$$
Using these formulas I obtained the following Einstein tensor:
\begin{align} G_{tt} &= \frac{a(r) \left(c(r) \left(r b'(r) c'(r)-2 b(r) \left(r c''(r)+c'(r)\right)\right)+2 c(r)^2 b'(r)+r b(r) c'(r)^2\right)}{4 r b(r)^2 c(r)^2} \\ G_{rr} &= \frac{\left(r a'(r)+2 a(r)\right) c'(r)+2 c(r) a'(r)}{4 r a(r) c(r)} \\ G_{zz} &= -\frac{c(r) \left(r b(r) a'(r)^2+a(r) \left(r a'(r) b'(r)-2 b(r) \left(r a''(r)+a'(r)\right)\right)+2 a(r)^2 b'(r)\right)}{4 r a(r)^2 b(r)^2} \\ G_{\theta\theta} &= -\frac{r^2 \left(a(r) c(r) b'(r) \left(c(r) a'(r)+a(r) c'(r)\right)+b(r) \left(-a(r) c(r) \left(a'(r) c'(r)+2 a(r) c''(r)\right)+c(r)^2 \left(a'(r)^2-2 a(r) a''(r)\right)+a(r)^2 c'(r)^2\right)\right)}{4 a(r)^2 b(r)^2 c(r)^2} \end{align}
with all nondiagonal components equal to zero. From the Einstein field equations, the stress-energy tensor is $$T_{\mu\nu} = \frac{c^4}{8\pi G} G_{\mu\nu}$$
How can I proceed from here? Can the equations be simplified any further?
No comments:
Post a Comment