Ellis wormhole

Example

Use units where the gravitational constant and the speed of light equal $1$. Fix $n > 0$. On $\R \times \R \times S^2$ with coordinates $(t, \ell, \theta, \phi)$, the Ellis wormhole is the metric $$g = -dt^2 + d\ell^2 + \left(\ell^2 + n^2\right)\left(d\theta^2 + \sin^2\theta\, d\phi^2\right).$$

Check that it is a wormhole: the areal radius $r(\ell) = \sqrt{\ell^2 + n^2}$ has its minimum $n$ at $\ell = 0$, which is the throat. As $\ell \to \pm\infty$, $r(\ell)/\abs{\ell} \to 1$, so the two ends are asymptotically flat, and $g_{tt} = -1$ never vanishes, so there is no horizon.

Its Ricci curvature has one nonzero component, $$R_{\ell\ell} = -\frac{2n^2}{\left(\ell^2 + n^2\right)^2}.$$ For the null vector $k = \partial_t + \partial_\ell$ this gives $R_{ab}k^ak^b < 0$ everywhere, so the null energy condition fails. This Ricci tensor is the one produced by a massless scalar field $\varphi = \frac{1}{\sqrt{4\pi}}\arctan(\ell/n)$ whose kinetic energy has the opposite of the usual sign, through $R_{ab} = -8\pi\, \partial_a\varphi\, \partial_b\varphi$. Indeed $8\pi(\varphi')^2 = 2n^2/(\ell^2 + n^2)^2$.

The Ellis wormhole has zero mass at both ends: $g_{tt} = -1$ means that a particle at rest feels no gravitational pull. A free particle can cross the throat from one end to the other in finite time.