Interface Boundary Conditions

The condition of the continuity of $\lambda(r)$ at the interface $r_{1}$ implies that we must have that $\lambda_{i}(r_{1})=\lambda_{m}(r_{1})$, which from Equations (19) and (23) gives us the following relation between the parameters

  $$B-r_{\mu} = \frac{\kappa\rho_{0}}{3}\,r_{1}^{3}.$$ (28)

In addition to this, the condition of the continuity of $\lambda(r)$ at the interface $r_{2}$ implies that we must have $\lambda_{m}(r_{2})=\lambda_{s}(r_{2})$, which from Equations (17) and (23) gives us the following relation between the parameters

  $$B+r_{M} = \frac{\kappa\rho_{0}}{3}\,r_{2}^{3}.$$ (29)

This last condition already determines the integration constant $B$ in terms of the parameters of the problem,

  $$B = - r_{M} + \frac{\kappa\rho_{0}}{3}\,r_{2}^{3},$$ (30)

and the difference of the two conditions just obtained determines the integration parameter $r_{\mu}$ in terms of the parameters of the problem,

  $$r_{\mu} = - r_{M} + \frac{\kappa\rho_{0}}{3}\left(r_{2}^{3}-r_{1}^{3}\right).$$ (31)

We have therefore the solution for $\lambda(r)$ in the matter region, in terms of the parameters of the problem,

  $$\lambda_{m}(r) = -\, \frac{1}{2}\, \ln\! \left[ \frac {\kappa\rho_{0}\left(r_{2}^{3}-r^{3}\right)+3\left(r-r_{M}\right)} {3r} \right].$$ (32)

Let us point out that there is a consistency condition to be applied to this result, since we must have that the cubic polynomial appearing in the argument of the logarithm be strictly positive for all values of $r$ within the matter region, that is

  $$\kappa\rho_{0} \left(r_{2}^{3}-r^{3}\right)+3\left(r-r_{M}\right) > 0,$$ (33)

for all $r\in[r_{1},r_{2}]$. Note that the term with the cubes is necessarily non-negative, but that the other term may be negative, if $r_{M}$ is not smaller than $r_{1}$. Therefore, so long as $r_{M}<r_{1}$, this strict positivity condition is automatically satisfied. If, however, we have that $r_{1}<r_{M}<r_{2}$, then the condition must be actively verified for all $r\in[r_{M},r_{2}]$. If it fails, then there is no solution for that particular set of input parameters.

Since we have $\nu_{m}(r)$ written in terms of $P(r)$, and since we know the interface boundary conditions for $P(r)$ in limits from within the matter region, we are in a position to impose the boundary conditions on $\nu(r)$ across the interfaces, even without having available the complete solution for $\nu_{m}(r)$. To this end, let us note that from Equation (27) we have that $\nu_{m}(r_{1})=\nu_{m}(r_{2})=\nu_{1}$. At the interface $r_{1}$ the condition of the continuity of $\nu(r)$ implies that we must have $\nu_{i}(r_{1})=\nu_{m}(r_{1})$, which from Equations (20) and (27) gives us the following relation between the parameters,

  $$\nu_{1} = A + \frac{1}{2}\, \ln\!\left(\frac{r_{1}+r_{\mu}}{r_{1}}\right).$$ (34)

In addition to this, the condition of the continuity of $\nu(r)$ at the interface $r_{2}$ implies that we must have $\nu_{m}(r_{2})=\nu_{s}(r_{2})$, which from Equations (18) and (27) gives us the following relation between the parameters,

  $$\nu_{1} = \frac{1}{2}\, \ln\!\left(\frac{r_{2}-r_{M}}{r_{2}}\right).$$ (35)

This last condition gives us the integration constant $\nu_{1}$ in terms of the parameters of the problem, and its difference with the previous one determines the integration constant $A$,

  $$A = \frac{1}{2}\, \ln\! \left( \frac{1-r_{M}/r_{2}}{1+r_{\mu}/r_{1}} \right).$$ (36)

Note that we have that $A<0$ for any positive values of $r_{M}$ and $r_{\mu}$. This completes the determination of the solution for both $\nu(r)$ and $\lambda(r)$ in the inner vacuum region, for which we now have

$$\lambda_{i}(r) = -\, \frac{1}{2}\, \ln\!\left(\frac{r+r_{\mu}}{r}\right),$$ (37)

$$\nu_{i}(r) = \frac{1}{2}\, \ln\! \left( \frac{1-r_{M}/r_{2}}{1+r_{\mu}/r_{1}} \right) + \frac{1}{2}\, \ln\!\left(\frac{r+r_{\mu}}{r}\right),$$ (38)

with $r_{\mu}$ given by Equation (31). We also have the following form for the solution for $\nu(r)$ within the matter region, still in terms of $P(r)$,

  $$\nu_{m}(r) = \frac{1}{2}\, \ln\!\left(\frac{r_{2}-r_{M}}{r_{2}}\right) - \ln\!\left[\frac{\rho_{0}+P(r)}{\rho_{0}}\right].$$ (39)

At this point the situation is as follows, in regard to the complete solution of the problem. Given values of $r_{1}$, $r_{2}$ and $r_{M}$, which completely characterize the geometrical and physical nature of the object under study, we have the complete solution for both $\lambda(r)$ and $\nu(r)$ in the outer vacuum region. We also have the complete solution for both $\lambda(r)$ and $\nu(r)$ in the inner vacuum region, except for the determination of the parameter $\rho_{0}$. We have as well the complete solution for $\lambda(r)$ in the matter region, again up to the determination of the parameter $\rho_{0}$. The one element of the solution still missing is the complete solution for $\nu(r)$ in the matter region. However, since we have $\nu(r)$ determined in terms of $P(r)$ in this region, this can also be accomplished by the complete determination of $P(r)$ in this region, which is the task we tackle next. Let us emphasize that the parameter $\rho_{0}$ is not a free input parameter of the problem, since it must be chosen so that the given value of $r_{M}$ results, that is, the local value of the energy density must be chosen so that the given value of the asymptotic gravitational mass $M$ results at radial infinity.