Solution with Fluid Matter

We will now describe a method for obtaining the solution in the presence of fluid matter. We start with an informed guess about the form of the function $\lambda(r)$. The ansatz that we will present here is suggested by the well-known Jebsen-Birkhoff [#!JebsenTheorem!#,#!BirkhoffTheorem!#] theorem, which states that any spherically symmetric solution of the vacuum field equation must be both static and asymptotically flat, and therefore must be given by the Schwarzschild metric. It is to be noted, however, that this very statement implies, as a matter of course, that the solution at issue lies in a region that has continuous access to radial infinity.

It is usually stated that the theorem also implies that the geometry within the vacuous region between two spherically symmetric concentric shells, which do not need to be thin, is given by a radial section of the Schwarzschild solution, with the corresponding internal mass, between the corresponding two radii. However, this is not entirely correct, as one can see in the discussion presented in [#!JBTheoremError!#]. While it is correct for the spatial part of the geometry, there is a change in the temporal part. This can be seen in simple terms if one realizes that there must be a red-shift/blue-shift relationship between the internal bounded vacuous region and the external region at radial infinity. In order to see this it suffices to consider a monochromatic beam of light propagating from radial infinity towards the localized matter distribution, and passing through a thin radial hole made across the outer shell, into the bounded vacuous region. It is quite clear that the blue shift undergone by this beam of light is not the same that one would get in the absence of the outer shell, since it includes the blue-shift effect of the mass in the outer shell. Therefore, the coefficient $\exp[2\nu(r)]$ of the term $dt^{2}$ of the invariant interval, which gives this blue shift, must differ from the one in the Schwarzschild solution for the internal mass.

In short, we may safely assume that the Jebsen-Birkhoff theorem does have the interesting consequence that the geometry within a spherically symmetric empty shell of mass must be given by a flat Minkowski metric. In other words, spacetime is flat there, and thus the gravitational field must vanish inside an empty spherically symmetric shell, just as is the case for Newtonian gravitation. However, there is a difference between this bounded flat region and the flat space at radial infinity, because the relative rates of the passage of time differ between the two regions. In other words, while one should expect that, in the bounded vacuous region between two concentric shells, the metric should be such that the radial factor given by $\exp[2\lambda(r)]$ would be the one in the Schwarzschild solution, with the total mass that exists strictly within the external shell, one should not expect the same to be true for the temporal factor given by $\exp[2\nu(r)]$, which should display some significant difference with respect to the corresponding factor of the Schwarzschild solution for the internal mass. In fact, it is not difficult to see that the imposition of the condition $\nu(r)=-\lambda(r)$ on the component equations given in Equations (24) and (25) implies at once that $\mathfrak{T}(r)\equiv 0$, thus leading us back to the vacuum solution.

These facts strongly suggest that, in the case of the problem in the presence of fluid matter, which we are considering in this paper, the spatial part of the geometry at the position $r$ is that due only to the mass within the sphere whose surface is at the position $r$, and that at that location the solution should be given by the spatial part of the Schwarzschild metric with an appropriate value of the mass. If we consider the factor $\exp[2\lambda(r)]$ in the $dr^{2}$ term of the invariant interval, and its value in the case of the Schwarzschild solution, we are immediately led to consider writing this factor in the following way for the case of the solution in the presence of fluid matter,

$$\,{\rm e}^{2\lambda(r)} = \frac{r}{r-r_{M}\beta(r)},$$ (53)

where $r_{M}=2MG/c^{2}$ is the Schwarzschild radius associated to the total mass $M$ of the distribution of matter, and $\beta (r)$ is a dimensionless function, presumably with values in the interval $[0,1]$, so that $r_{M}\beta(r)$ effectively corresponds to a certain fraction of that total mass. This is the ansatz that we will use here. The case $\beta(r)\equiv 1$ corresponds, of course, to the value of the quantity $\exp[2\lambda(r)]$ for the case of the original Schwarzschild solution. Therefore, it is to be expected that in the $r\to\infty$ asymptotic limit we will get $\beta(r)\to 1$. Of course, the correctness of this ansatz will have to be tested by the successful imposition of the field equation, which is what we will go on to do right away.

Therefore, let us consider each component equation in turn and thus obtain expressions for all the relevant quantities in terms of the single function $\beta (r)$. Starting from this ansatz, we may now use the $t$ component of the field equation, shown in Equation (24), in order to get the dimensionless quantity $\mathfrak{T}(r)$. That equation can be written in the form

$$\left[ r\,{\rm e}^{-2\lambda(r)} \right]' = 1 - \mathfrak{T}(r) \;\;\;\Rightarrow$$

$$\left[r-r_{M}\beta(r)\right]' = 1 - \mathfrak{T}(r) \;\;\;\Rightarrow$$

$$1-r_{M}\beta'(r) = 1 - \mathfrak{T}(r) \;\;\;\Rightarrow$$

$$\mathfrak{T}(r) = r_{M}\beta'(r),$$ (54)

where we used our ansatz, and which therefore determines $\mathfrak{T}(r)$ in terms of $\beta (r)$. Note that, since $\mathfrak{T}(r)$ is proportional to $r^{2}$ times the energy density, and must therefore be positive, we must have that $\beta'(r)\geq 0$ for all values of $r$ where the matter is located, and in fact everywhere. From the same $t$ component of the field equation, shown in Equation (24), we can get directly the quantity $r\lambda'(r)$, since that equation can be written as

$$1 - 2r\lambda'(r) = \,{\rm e}^{2\lambda(r)} \left[ 1 - \mathfrak{T}(r) \right]$$

$$= \frac{r-r_{M}\left[r\beta'(r)\right]}{r-r_{M}\beta(r)} \;\;\;\Rightarrow$$

$$2r\lambda'(r) = \frac {r-r_{M}\beta(r)-r+r_{M}r\beta'(r)} {r-r_{M}\beta(r)}$$

$$= \frac {-r_{M}\beta(r)+r_{M}r\beta'(r)} {r-r_{M}\beta(r)} \;\;\;\Rightarrow$$

$$\left[r\lambda'(r)\right] = -\, \frac{r_{M}}{2}\, \frac {\beta(r)-\left[r\beta'(r)\right]} {r-r_{M}\beta(r)},$$ (55)

where we again used our ansatz, as well as the solution for $\mathfrak{T}(r)$ given in Equation (54), and which therefore determines $\lambda'(r)$ in terms of $\beta (r)$. Using now, once more, that result for $\mathfrak{T}(r)$ and the $r$ component of the field equation, shown in Equation (25), we can get the quantity $r\nu'(r)$, since that equation can be written as

$$1 + 2r\nu'(r) = \,{\rm e}^{2\lambda(r)} \left[ 1 + \omega\mathfrak{T}(r) \right]$$

$$= \frac {r+\omega r_{M}\left[r\beta'(r)\right]} {r-r_{M}\beta(r)} \;\;\;\Rightarrow$$

$$2r\nu'(r) = \frac {r+\omega r_{M}r\beta'(r)-r+r_{M}\beta(r)} {r-r_{M}\beta(r)}$$

$$= \frac {\omega r_{M}r\beta'(r)+r_{M}\beta(r)} {r-r_{M}\beta(r)} \;\;\;\Rightarrow$$

$$\left[r\nu'(r)\right] = \frac{r_{M}}{2}\, \frac {\beta(r)+\omega\left[r\beta'(r)\right]} {r-r_{M}\beta(r)},$$ (56)

where we once again used our ansatz, and which therefore determines $\nu'(r)$ in terms of $\beta (r)$. Note that $\nu'(r)$ is not equal to $-\lambda'(r)$, as would be the case for the Schwarzschild solution. However, it is to be expected that $\nu'(r)$ has $-\lambda'(r)$ as its $r\to\infty$ asymptotic limit. With this we have two of the three quantities that appear in the left-hand side of the $\theta$ component of the field equation, shown in Equation (26). Since we have that

$$r \left[ r\nu'(r) \right]' = r^{2}\nu''(r) + r\nu'(r),$$ (57)

and using once again the result for $\mathfrak{T}(r)$ given in Equation (54), the $\theta$ component equation given in Equation (26) can now be written in the form

$$r \left[ r\nu'(r) \right]' + \left[ r\nu'(r) \right]... ...) \right] + \,{\rm e}^{2\lambda(r)} \omega r_{M}\beta'(r).$$ (58)

In order to be able to write out this equation, we must now calculate in terms of $\beta (r)$ the quantity $r\left[r\nu'(r)\right]'$. We can do this by simply differentiating the quantity $\left[r\nu'(r)\right]$. The detailed calculation can be found in Subsection A.2.1 of Appendix A, and the result is

$$r\left[r\nu'(r)\right]' = r_{M}\, \frac { \left( \beg... ...r \beta(r) \end{array} \right) } {2[r-r_{M}\beta(r)]^{2}},$$ (59)

which therefore determines $\nu''(r)$ in terms of $\beta (r)$. We now have all the quantities that appear in the left-hand side of the $\theta$ component of the field equation, shown in Equation (26). In fact, just as in the case of the vacuum solution, the solution in the presence of fluid matter is characterized by the following set of quantities, this time written in terms of $\beta (r)$,

$$\mathfrak{T}(r) = r_{M}\beta'(r),$$ (60)

$$\,{\rm e}^{2\lambda(r)} = \rule{0em}{5ex} \frac{r}{r-r_{M}\beta(r)},$$ (61)

$$r\nu'(r) = \frac{r_{M}}{2}\, \frac {\beta(r)+\omega\left[r\beta'(r)\right]} {r-r_{M}\beta(r)},$$ (62)

$$r\lambda'(r) = -\, \frac{r_{M}}{2}\, \frac {\beta(r)-\left[r\beta'(r)\right]} {r-r_{M}\beta(r)},$$ (63)

$$r\left[r\nu'(r)\right]'(r) = \frac{r_{M}}{2}\, \frac { ... ... r \beta(r) \end{array} \right) } {[r-r_{M}\beta(r)]^{2}}.$$ (64)

Note that the quantity $\exp[2\nu(r)]$ remains undetermined, which does not really present a problem, since it does not appear in the components of the field equation. Once $\beta (r)$ is determined in each particular case, $\nu(r)$ can be obtained from $\nu'(r)$ by straightforward integration. These expressions are to be used in what follows in order to define the asymptotic conditions in the $r\to\infty$ asymptotic limit. This is what we will do next. Later on we will return to the discussion of the single equation yet to be satisfied, the $\theta$ component equation in the form shown in Equation (58).

On the one hand, in Section 3 we calculated from the vacuum solution the values of the $r\to\infty$ asymptotic limits, for the various corresponding quantities involved in the solution in the presence of fluid mater, asymptotic values that were given in Equations (47)--(51). On the other hand, in this section we calculated the solution in the presence of fluid matter, and listed what is essentially the same set of relevant quantities in Equations (60)--(64). We are now ready to discuss the corresponding asymptotic conditions. It is to be expected, of course, that they will result in corresponding asymptotic conditions on $\beta (r)$ and its derivatives. We start by discussing the asymptotic condition on $\lambda(r)$. As we already discussed before, the expression in Equation (48) can only be the asymptotic limit of the expression in Equation (61) if we have that

$$\lim_{r\to\infty} \beta(r) = 1.$$ (65)

Note that we do not have to worry directly about the asymptotic condition on $\nu(r)$, because $\nu(r)$ itself does not appear in the component equations, which contain only its derivatives. Therefore, next we discuss the asymptotic condition on $\lambda'(r)$. Given that $\beta(r)\to 1$, the expression in Equation (50) can only be the asymptotic limit of the expression in Equation (63) if we have that

$$\lim_{r\to\infty} \left[r\beta'(r)\right] = 0.$$ (66)

Next we discuss the asymptotic condition on $\nu'(r)$. It is quite clear that the conditions above on $\beta (r)$ and $\beta'(r)$ are sufficient to ensure that the expression in Equation (49) will be the asymptotic limit of the expression in Equation (62). Let us recall that, once we have $\nu'(r)$ in terms of a known function $\beta (r)$, we can obtain $\nu(r)$ from $\nu'(r)$ by straightforward integration. When doing this integration, the integration constant must be chosen so that $\nu(r)$, just like $\lambda(r)$, goes to zero for $r\to\infty$, of course. In that limit we also expect that $\nu(r)=-\lambda(r)$. Finally, we discuss the asymptotic condition on $\nu''(r)$. Given that $\beta(r)\to 1$ and that $\left[r\beta'(r)\right]\to 0$, the expression in Equation (51) can only be the asymptotic limit of the expression in Equation (64) if we have that

$$\lim_{r\to\infty} \left[r^{2}\beta''(r)\right] = 0.$$ (67)

Therefore, we have the complete set of asymptotic conditions to be satisfied by $\beta (r)$ and its derivatives,

$$\lim_{r\to\infty}\beta(r) = 1,$$

$$\lim_{r\to\infty}\left[r\beta'(r)\right] = 0,$$

$$\lim_{r\to\infty}\left[r^{2}\beta''(r)\right] = 0.$$ (68)

We must now discuss what happens near $r=0$. At this point we will have conditions associated to the regularity of the energy density. If we consider that, according to the definitions in Section 2, the energy density is given by

$$T_{0}(r) = \frac{\mathfrak{T}(r)}{\kappa r^{2}},$$ (69)

where we also have that $\mathfrak{T}(r)=r_{M}\beta'(r)$, as shown in Equation (54), it follows that we have

$$T_{0}(r) = \frac{r_{M}}{\kappa}\, \frac{\beta'(r)}{r^{2}}.$$ (70)

If the energy density $T_{0}(r)$ is to be non-singular at $r=0$, and have a limited integral around that point, then we must have that the limit

$$\lim_{r\to 0}\frac{\beta'(r)}{r^{2}}$$ (71)

exists and is finite. This implies that at least for the first derivative of $\beta (r)$ we must have that $\beta'(0)=0$. In fact, since we must also have that $\beta'(r)/r\to 0$ when $r\to 0$, one can show that the same has to be true for the second derivative as well, that is, we also have that $\beta''(0)=0$. This leads us to a picture of a function $\beta (r)$ that has both zero derivative and zero second derivative at both ends of the real $r$ semi-axis.

One may also argue that it is necessary that $\beta(0)$ not be a strictly positive number. The argument leading to this condition is as follows. According to the motivation leading to the construction of our solution, the quantity $r_{M}\beta(r)$, where $r_{M}=2MG/c^{2}$ and $\beta(r)\in[0,1]$, is effectively a certain fraction of the total mass $M$. If $\beta(0)>0$ then this quantity has a finite and non-zero positive limit when we make $r\to 0$. What this means is that there is a certain finite and non-zero mass, given by $r_{M}\beta(0)$, which is inside a sphere whose surface is at the radial position $r$, and that this holds for all $r$. However, this means that for some value of $r$ this finite and non-zero mass will be inside the Schwarzschild radius associated to itself, thus leading to the existence of an event horizon within the matter distribution, which contradicts our hypotheses here. In order to avoid this, and assuming that $\beta(r)\in[0,1]$, we might be tempted to conclude that we must have that

$$\lim_{r\to 0}\beta(r) = 0,$$ (72)

that is, that we must have $\beta(0)=0$. However, we should not take this heuristic motivation too seriously, and while it seems inevitably true that we cannot have $\beta(0)>0$, we also cannot definitely assert that we must have $\beta(0)=0$, because there is really no reason why $\beta (r)$ cannot be negative. Instead, once the ansatz given in Equation (53) is assumed, we must then follow wherever the field equation takes us. In fact, contrary to what the heuristic intuition may seem to indicate, $\beta(0)$ is indeed always negative. When $\beta (r)$ becomes negative it becomes impossible for the factor $\exp[2\lambda(r)]$ to diverge, as it does at the event horizon of the Schwarzschild solution, and this makes the solution regular and avoids the existence of event horizons within the matter distribution. Considering the fact that, due to the positivity of the energy, we must have $\beta'(r)\geq 0$ for all $r$, and taking into account the asymptotic conditions derived here for $\beta (r)$ and its derivatives, as well as the conditions at $r=0$, we can state that the function $\beta (r)$ must have a very simple qualitative behavior, going monotonically from some negative value at $r=0$ to the value $1$ for $r\to\infty$. We have therefore the complete set of relevant conditions at the two ends of the real $r$ semi-axis,

$$\lim_{r\to 0} \beta'(r) = 0,$$

$$\lim_{r\to 0} \beta''(r) = 0,$$

$$\lim_{r\to\infty}\beta(r) = 1,$$

$$\lim_{r\to\infty}\left[r\beta'(r)\right] = 0,$$

$$\lim_{r\to\infty}\left[r^{2}\beta''(r)\right] = 0.$$ (73)

Note that, since we have here a second-order field equation, which will give origin to a second-order equation for $\beta (r)$, we in fact can satisfy only two independent conditions using the corresponding integration constants. Therefore any additional conditions must arise as automatic consequences of the first two. As we will see, it is possible to reduce the equation for $\beta (r)$ to a form in which one has only two free parameters to deal with, both physically meaningful, one of which will be given by the total mass $M$ present, leading to the parameter $r_{M}$, and only one, the parameter $\omega$, which directly affects the integration process itself. Since we have fewer parameters to adjust than conditions to meet, the problem can only be solved if there are internal consistency structures within the system that guarantee that most conditions are automatically met when we adjust the parameters in order to satisfy what conditions we can. We will see that this is indeed the case.