To define a traversable wormhole, there should be some conditions on the metric components, such as:

I) No event horizon,

II) Minimum value for the shape function (considering a spherical symmetric solution, we should have $b(r_t)=r_t$ ),

III) The flare--out condition ($b'(r_t)<1)$ ans ...

If this condition satisfies besides the null energy condition, we have a traversable wormhole solution with nonexotic matter.

My question is that, could we describe the spacetime through the throat with the wormhole metric or it is another patch? It seems that on the above conditions, it is not possible to use the mentioned wormhole solution in the throat. But, then, why in the literature this metric is used for throat passage description?

  • 1
    $\begingroup$ There are geometries like the Ellis wormhole that are traversible and a single patch covers the wormhole and the two spacetimes either side of it. $\endgroup$ Mar 5, 2020 at 17:35
  • $\begingroup$ @JohnRennie Do you mean that I understand the issue correctly? Do we need two patches to describe a spacetime like Morris-Thorne wormhole? $\endgroup$
    – Astrolabe
    Mar 5, 2020 at 17:57
  • $\begingroup$ I can't remember the Morris-Thorne metric offhand, but I don't recall needing two patches for it, $\endgroup$ Mar 5, 2020 at 18:03

1 Answer 1


There's a variety of metrics used for the Morris-Thorne metric. The two most common are the Schwarzschild coordinates and the proper radial coordinateS.

The Schwarzschild coordinates are just done using the classic spherical symmetric coordinates,

$$ds^2 = -e^{2\phi_{\pm}(r)}dt^2 + \frac{dr^2}{1 - b_\pm(r) / r} + r^2 d\Omega^2$$

This is done on two different patches glued together at $r_0$, with the functions $+$ and $-$ depending on which part you are on. But you can otherwise switch to proper radial coordinates, via the coordinate trnsform

$$l = \pm \int_{r_0}^r \frac{dr'}{\sqrt{1 - b_\pm(r') / r'}}$$

In which case you get the new metric

$$ds^2 = -e^{2\psi(l)} dt^2 + dl^2 + r^2(l) d\Omega^2$$

which is defined on the entire manifold.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.