Existence of the Singularity at the Origin

The existence of the singularity at the origin is equivalent to the statement that $r_{\mu}>0$, because the only way to avoid that singularity would be to have $r_{\mu}=0$. If we put $r_{\mu}=0$ and take the limit $r_{1}\rightarrow 0$ we no longer have a matter shell, and we obtain instead the Schwarzschild interior solution.

We start with a preliminary lemma, in which we will prove that the following combination of parameters

  $$\frac{1}{3}\, \Upsilon_{0}^{2}\left(r_{2}^{3}-r_{e}^{3}\right) - r_{M} > 0,$$ (58)

is strictly positive, where $r_{e}$ is the position of the maximum of the dimensionless pressure $p(r)$ within the interval $[r_{1},r_{2}]$. In order to do this, we consider the equation for $p(r)$ given in Equation (44). Applying that equation at $r_{2}$, since we have that $p(r_{2})=0$, we get for the derivative at the right end of the matter interval,

  $$p'(r_{2}) = -\, \frac{r_{M}}{2r_{2}\left(r_{2}-r_{M}\right)}.$$ (59)

Since by hypothesis we have that $r_{2}>r_{M}$ and that $r_{M}>0$, we conclude that the derivative $p'(r_{2})$ is strictly negative. In addition to this, since $p(r)$ is a positive function that is the solution of a first-order differential equation within $(r_{1},r_{2})$, it must be a continuous and differentiable function. Therefore, given that it is zero at both ends and always increases as we go to the interior of the interval, it must have a point of maximum $r_{e}$ somewhere in the interior of the interval, where we will have that $p'(r_{e})=0$. Using the differential equation for $p(r)$ given by Equation (44) at this point we thus obtain

  $$\frac{1}{2r_{e}} \left[1+p(r_{e})\right] \frac { \Upsilon_{0... ...0}^{2}\left(r_{2}^{3}-r_{e}^{3}\right) + 3\left(r_{e}-r_{M}\right) } = 0.$$ (60)

This can only be zero if the numerator is zero, so we have that

  $$\Upsilon_{0}^{2}r_{e}^{3}p(r_{e}) = \frac{1}{3}\, \Upsilon_{0}^{2} \left(r_{2}^{3}-r_{e}^{3}\right) - r_{M}.$$ (61)

Since $\Upsilon_{0}^{2}>0$ and at its maximum we must have $p(r_{e})>0$ for the dimensionless pressure, we conclude that our lemma holds,

  $$\frac{1}{3}\, \Upsilon_{0}^{2} \left(r_{2}^{3}-r_{e}^{3}\right) - r_{M} > 0.$$ (62)

Let us now consider the result for $r_{\mu}$ in terms of the given parameters of the problem, as shown in Equation (31), which we can write as

  $$r_{\mu} = \frac{1}{3}\, \Upsilon_{0}^{2} \left(r_{2}^{3}-r_{1}^{3}\right) - r_{M}.$$ (63)

By adding and subtracting terms to this equation, we can write it as

  $$r_{\mu} = \left[ \frac{1}{3}\, \Upsilon_{0}^{2} \left(r_{2}... ...\right] + \frac{1}{3}\, \Upsilon_{0}^{2} \left(r_{e}^{3}-r_{1}^{3}\right).$$ (64)

The quantity within square brackets is the one we just proved to be strictly positive in our lemma. The other term is also strictly positive because we certainly have that $r_{e}>r_{1}$. Therefore, we have our theorem,

  $$r_{\mu} > 0.$$ (65)

Therefore, every solution of the problem that exists at all is bound to have a singularity at the origin, which is characterized by the factor

  $$\ln\!\left(\frac{r+r_{\mu}}{r}\right),$$ (66)

that appears with a negative sign in $\lambda_{i}(r)$ and with a positive sign in $\nu_{i}(r)$. This implies that at this singular point we have that

$$\lim_{r\to 0}\lambda_{i}(r) = -\infty,$$ (67)

$$\lim_{r\to 0}\,{\rm e}^{\lambda_{i}(r)} = 0,$$ (68)

$$\lim_{r\to 0}\nu_{i}(r) = \infty,$$ (69)

$$\lim_{r\to 0}\,{\rm e}^{\nu_{i}(r)} = \infty.$$ (70)

Note that this singularity does not have any disastrous consequences, since it does not imply infinite concentrations of matter. In fact, we have $\rho(r)=0$ in the whole inner vacuum region, including at the origin. For the proper lengths in the radial direction, it just implies that they get progressively more contracted as we approach the origin, rather than being expanded with respect to the corresponding variations of the radial coordinate $r$, as is the case in the outer vacuum region. For the proper times it just means that we get progressively more severe red shifts as we approach the origin, rather than the blue shifts that we get as we approach the event horizon from the outer vacuum region.

As a corollary to the proof that $r_{\mu}>0$, note that this fact guarantees the positivity of the cubic polynomial in Equation (33). This is so because the second derivative of that polynomial is given by $-6\kappa\rho_{0}r$, being therefore negative for all $r\in[r_{1},r_{2}]$. This means that the graph of the cubic polynomial has a concavity turned downward throughout this interval. In addition to this, it is easy to see that at $r=r_{2}$ the polynomial is given by $3\left(r_{2}-r_{M}\right)$, which is strictly positive so long as $r_{2}>r_{M}$. Finally, at $r=r_{1}$ the polynomial is given by

  $$\kappa\rho_{0}\left(r_{2}^{3}-r_{1}^{3}\right) + 3\left(r_{1}-r_{M}\right) = 3\left(r_{1}+r_{\mu}\right),$$ (71)

where we used Equation (31), which is also strictly positive since $r_{\mu}>0$. As a consequence of this, we may conclude that, so long as the conditions $r_{2}>r_{M}$ and $r_{\mu}>0$ hold, as they must for physically sensible solutions, the polynomial is strictly positive for all $r\in[r_{1},r_{2}]$.