In Morris-Thorne Wormhole there is a quantity called proper radial distance given by:

$$\ell(r) = \pm \int^{r}_{r_{0}} \Biggr(\frac{r-b(r)}{r} \Biggr)^{-\frac{1}{2}}dr \tag{1}$$

This function must to be finite for every value of $r$.

The fact is, it seems that this is a strong condition for the mathematical viability of wormhole throat; in other words, we must to obey the finiteness of $(1)$.

But, considering now a specific shape function $b(r)$, given by $[1]$:

$$ b(r) = \frac{r}{\exp(r-r_{0})}, \tag{2}$$

we can calculate it's integral, which is (using the Integrate$[,r]$ function of Wolfram Mathematica) :

$$\ell(r) = \pm \Bigg(\frac{2\ln(e^{r/2} + \sqrt{e^r - e^{r_{0}}}) e^{-r/2} \sqrt{e^r - e^{r_{0}}} }{\sqrt{1 - e^{(-r + r_{0})}}}\Bigg)\Biggr]^{r}_{r_{0}} \tag{3}$$

The fact with $(3)$ is: when we calculate the definite integral it seems to be problematic; the part of $r_{0}$ becomes divergent, and therefore the condition of finitness required by $(1)$ becomes meaningless. Furthermore if we test the very notion of "$\ell(r)$ must to be finite everywhere", in particular it must to be finite on $r= r_{0}$, i.e., $\ell(r_{0})$; well, cleary $(3)$ isn't.

So my doubt is:

The proper radius $\ell(r)$ must to be finite on $r = r_{0}$, or we can just bypass this requirement in order to construct a well defined Morris-Thorne Wormhole?

$$ * * * $$

$[1]$ SAMANTA.G.C; et al. Traversable Wormholes with Exponential Shape Function in Modified Gravity and in General Relativity: A Comparative Study

  • $\begingroup$ Why do you think it blows up? It doesn't seem to blow up for me... $\endgroup$
    – Philip
    Commented Jun 18, 2020 at 20:32
  • $\begingroup$ @Philip I made a edit, try now. $\endgroup$
    – M.N.Raia
    Commented Jun 18, 2020 at 20:35
  • $\begingroup$ I assumed your edit, I calculated it myself and it doesn't blow up. Did you just plug $r = r_0$ in the denominator, or did you take the limit $r\to r_0$? $\endgroup$
    – Philip
    Commented Jun 18, 2020 at 20:35
  • $\begingroup$ @Philip I just plug on denominator. $\endgroup$
    – M.N.Raia
    Commented Jun 18, 2020 at 20:36
  • 1
    $\begingroup$ That's your problem. Take the limit carefully and you'll see what happens. It doesn't blow up. $\endgroup$
    – Philip
    Commented Jun 18, 2020 at 20:36

1 Answer 1


It's easy to show that your function $\ell(r)$ does not blow up as $r\to r_0$. You have that:

$$\ell(r) = \frac{2 e^{-r/2} \sqrt{e^r - e^{r_0}} \ln \left( e^{r/2} + \sqrt{e^r - e^{r_0}} \right)}{\sqrt{1 - e^{-r + r_0}}}. $$

Now, you might be tempted to naively replace $r = r_0$, but the denominator would seem to blow up, leading you to conclude (falsely) that the value diverges. It doesn't. Notice that there is also a term in the numerator which goes to $0$ as you set $r = r_0$. So this is a $\frac{0}{0}$ type situation, and needs to be treated with some care. A good rule of thumb is to take limits in general. (Mathematica allows you to do this, but frankly it's a good exercise to do it by hand too!)

In your case it's especially simple, since if you rewrite the denominator as $$\sqrt{1 - e^{-r + r_0}} = e^{-r/2} \sqrt{e^r - e^{r_0}},$$

then you simply get that

$$\ell(r) = 2 \ln \left( e^{r/2} + \sqrt{e^r - e^{r_0}} \right),$$

and now plugging in $r=r_0$ gives you:

$$\ell(r) = 2 \ln \left( e^{r_0/2}\right) = r_0,$$

which is certainly finite, so long as $r_0$ is finite.

  • $\begingroup$ Is it possible that $\ell$ be a negative number? $\endgroup$
    – M.N.Raia
    Commented Jun 19, 2020 at 3:34
  • $\begingroup$ I think this also has some problems, I think it $l(r)$ should be 0 at $r_0$, otherwise, it couldn't be smoothly extended to negative $r$. $\endgroup$
    – David Shaw
    Commented Mar 10, 2023 at 18:16

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