Spacetime around a massive cylinder

Question

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?

Answer 1

In the case that $T_{\mu\nu}=0$ (i.e. outside the cylinder $ r>\rho$), these equations reduce to: $$ f'(r) = 0 \quad \text{and}\quad h'(r) = 0$$.

Consequently, both are constants. The constant value of $f$ can be absorbed by a rescaling of $t$, which leaves (with $h(r) = A^{-2}$)

$$ \mathrm{d}s^2 = c^2 \mathrm{d}t^2 - A^{-2} \mathrm{d}r^2 - \mathrm{d}z^2 - r^2 \mathrm{d}\theta^2 $$

or equivalently (after a rescaling of $r$)

$$ \mathrm{d}s^2 = c^2 \mathrm{d}t^2 - \mathrm{d}r^2 - \mathrm{d}z^2 - A^2 r^2 \mathrm{d}\theta^2 $$

This metric is locally flat (i.e. it Riemann tensor vanishes). However, it is not identical to the Minkowski metric globally. This can be seen by parallel transporting a vector along a closed curve that goes around the origin. Doing so will reveal the vector to be rotated by $\alpha = 2\pi (A-1)$ radians.

You can repeat this calculation inside the cylinder, choosing an (cylindrically symmetric) energy momentum distribution inside the cylinder. By matching the two solution at the boundary (see e.g. ArXiv:1111.6468 section 2.1) you find that $\alpha = \frac{8\pi G}{c^4}\mu$, where $\mu$ is the mass per unit length of the cylinder. This result is independent of the radial mass distribution in the cylinder, as long as it is static and cylinder symmetric.

EDIT: The above case is specific to the case to that of a straight cosmic string (i.e. a static cylinder that only features a tension along its length (equal to its energy density in natural units) and no tension/pressure in the radial or azimuthal directions), which corresponds to the original question asked.

In the more general case the equations still simplify significantly in the vacuum exterior. With some straight forward algebra one can show that the vacuum equations reduce to:

\begin{align} 0 &= a(r a''+a')-r a'^2\\ 0 &= c(r c''+c')-r c'^2\\ 0 &= a' b c - a b' c + a b c'\\ 0 &= r a' c' + 2(a'c+a c') \end{align}

This has the general solution (after some rescalings of $t$ and $z$ to eliminate some overall constants): \begin{align} a(r) &= r^\alpha\\ b(r) &= \beta r^{\frac{\alpha^2}{2+\alpha}}\\ c(r) &= r^{-\frac{2\alpha}{2+\alpha}} \end{align}

where $\alpha$ and $\beta$ are constants that depend on the details of the cylinder. (When $\alpha =0$ it reduces to the cosmic string case.)