2
$\begingroup$

In this reference $[1]$ the author created a Inflating Morris-Throne Wormhole (IMTW) given by:

$$ds^2=-e^{\Phi(r)}dt^2+e^{2\xi t}\Biggr\{\frac{1}{1-\frac{b(r)}{r}}dr^2+r^2d\theta^2+r^2sin^2\theta d\phi^2\Biggr\} \tag{1}$$

Which is slightly different from canonical Morris-Thorne Wormhole (MTW):

$$ds^2=-e^{\Phi(r)}dt^2+\frac{1}{1-\frac{b(r)}{r}}dr^2+r^2d\theta^2+r^2sin^2\theta d\phi^2 \tag{2}$$

Mt doubt lies on physical interpretation of that exponential factor. Again from $[1]$, the function $\xi$ is in fact a constant function given by:

$$\xi = \sqrt{\frac{\Lambda}{3}} \tag{3}$$

Where $\Lambda$ is, interrestingly, the Cosmological constant. My doubt is then:

What is the physical interpretation of a scale factor when this factor is constructed using a physical constant? Or, in other words, what is the main physical motivation that drives someone to modify the metric $(2)$ into the form given by $(1)$?

$$ * * * $$

$[1]$ https://demonstrations.wolfram.com/ToyModelOfAnInflatingWormhole/

$\endgroup$
7
  • $\begingroup$ Have you tried chugging this metric down in Einstein field equation and see what the Einstein tensor implies in the modified case? $\endgroup$
    – aitfel
    Jun 6, 2020 at 3:01
  • $\begingroup$ No. I should do this of course. Also, A metric with cosmological constant need to be a solution of EFE with cosmological constant, right? $\endgroup$
    – M.N.Raia
    Jun 6, 2020 at 3:46
  • $\begingroup$ [Here is the Einstein tensor without $\Lambda g_{\mu\nu}$ term][1] [1]: i.stack.imgur.com/9F4uF.jpg $\endgroup$
    – aitfel
    Jun 6, 2020 at 5:00
  • $\begingroup$ @aitfel Please, could you give me your whole .nb? $\endgroup$
    – M.N.Raia
    Jun 6, 2020 at 5:06
  • $\begingroup$ Sure but it consists of just 3 lines since I used ccgrg package. But how do I share it? $\endgroup$
    – aitfel
    Jun 6, 2020 at 5:10

1 Answer 1

1
$\begingroup$

Note that in both spacetimes, $t$ has similar interpretations: it is related (through a red-shift factor $e^{\Phi(r)/2}$) to the proper-time of observers "at rest" in the coordinate system (for $r>b(r)$). Moreover, the hypersurfaces $t = const$ in both spacetimes are conformally related by a constant factor -- which basically means they are the "same" 3d space, only rescaled (differently for each different $t$, by $e^{\xi t}$). So, I would suppose that the main motivation from going from (2) to (1) is having a spacetime in which, now, there is a family of observers who see the same "kind" spatial sections as the static observers in (2), but now the physical distance between any two of them is exponentially increasing (according to themselves).

Regarding which equation (1) solves, you just have to calculate the Einstein tensor $G_{\mu \nu}$ related to this line-element. Whether you interpret that as solution of Einstein equation with or without a cosmological constant is a matter of taste; depends simply if you infer the energy-momentum tensor (which has not been given beforehand) as equal to $G_{\mu \nu}$ or to $G_{\mu\nu}+\Lambda g_{\mu \nu}$.

$\endgroup$

Your Answer

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

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