General Review

Here we will review the main points and equations relating to the Einstein field equations under the set of symmetry conditions that we are to impose on them. Our initial development here will follow closely the one presented in [#!DiracGravity!#]. Under the conditions of time independence and of spherical symmetry around the origin of a spherical system of coordinates $(t,r,\theta,\phi)$, the Schwarzschild system of coordinates, the most general possible metric is given by the invariant interval, written in terms of this spherical system of coordinates,

$$ds^{2} = \,{\rm e}^{2\nu(r)} dt^{2} - \,{\rm e}^{2\lam... ... \left[ d\theta^{2} + \sin^{2}(\theta) d\phi^{2} \right],$$ (1)

where $\nu(r)$ and $\lambda(r)$ are two functions of only $r$. As one can see, in this work we will use the time-like signature $(+,-,-,-)$, following [#!DiracGravity!#]. Expressing the coefficients of $dt^{2}$ and $dr^{2}$ by the exponentials shown ensures that they have the physically required signs.

Let us now comment on the physical interpretation of these coordinates. In this coordinate system $r$ is such that a sphere centered at the origin $r=0$ and with its surface located at the position $r$ has total physical surface area equal to $4\pi r^{2}$. However, $r$ is not the true physical distance from the surface of the sphere to the origin, and furthermore $dr$ is not a variation of physical length in the radial direction. Note also that the time variable $t$ is not the true proper time at each spatial position. These two coordinates only recover their usual meanings in the $r\to\infty$ asymptotic limit, if that limit is accessible. In general the element of physical length in the radial direction is given by $\exp[\lambda(r)]dr$, and the element of proper time at each position is given by $\exp[\nu(r)]dt$. Finally note that, on the other hand, the quantities $rd\theta$ and $r\sin(\theta)d\phi$ are in fact true physical elements of arc length on the surface of the sphere, that is, on the spherical surface located at the position $r$.

From this invariant interval we can simply read out the metric tensor $g_{\mu\nu}$, that is, the metric tensor in its covariant form,

$$g_{\mu\nu} = \left[ \rule{0em}{8ex} \begin{array}{llll}... ...0 \hspace{4em} & -r^{2}\sin^{2}(\theta) \end{array} \right].$$ (2)

Note that we have for the determinant $g=\det(g_{\mu\nu})$ of the matrix $g_{\mu\nu}$ the value

$$\sqrt{-g} = \,{\rm e}^{\nu(r)+\lambda(r)}\,r^{2}\sin(\theta),$$ (3)

which is not zero for $r>0$ except at the two poles $\theta=0$ and $\theta=\pi$, which are just the usual singularities of the spherical system of coordinates, and can therefore be safely ignored. Recalling that $g_{\mu}^{\nu}=\delta_{\mu}^{\nu}$ is the identity matrix, we obtain at once the metric tensor $g^{\mu\nu}$, that is, the metric tensor in its contravariant form, which is the inverse matrix to the diagonal matrix $g_{\mu\nu}$ given above. The matrix $g_{\mu\nu}$ in invertible almost everywhere due to the fact that $g\neq 0$ almost everywhere, and the inverse is immediately found to be given by

$$g^{\mu\nu} = \left[ \rule{0em}{8ex} \begin{array}{llll}... ...\hspace{5em} & -r^{-2}\sin^{-2}(\theta) \end{array} \right].$$ (4)

The next step in the geometric development is to calculate from $g_{\mu\nu}$ and $g^{\mu\nu}$ the Christoffel symbol of the second kind $\Gamma^{\alpha}_{\;\;\mu\nu}$, which gives us the metric-compatible and torsion-free connection for the pseudo-Riemannian parallel transport in spacetime. It is given in terms of the metric tensor by

$$\Gamma^{\alpha}_{\;\;\mu\nu} = g^{\alpha\sigma} \Gamma_{\sigma\mu\nu}$$

$$= g^{\alpha\sigma} \frac{1}{2} \left( \rule{0em}{2.5ex} \partial_{\... ...\sigma\mu} + \partial_{\mu}g_{\nu\sigma} - \partial_{\sigma}g_{\mu\nu} \right).$$ (5)

Table: The values of the components of the connection $\Gamma^{\alpha}_{\;\;\mu\nu}$.

This is the Christoffel symbol of the second kind, while the same quantity with all indices downstairs is the Christoffel symbol of the first kind. They are symmetric on the last two indices, as one can see, and they are non-tensors, because the derivative of a second-rank tensor is not a tensor. Note that the three terms within parenthesis correspond to cyclic permutations of the three indices. From now on the derivatives with respect to $r$ of $\nu(r)$, $\lambda(r)$, and any other quantities that depend only on $r$, will be denoted by primes.

In this calculation many of the components of $\Gamma^{\alpha}_{\;\;\mu\nu}$ turn out to be zero, as reported in [#!DiracGravity!#], and if one recalls that this quantity is symmetric on the pair of indices $(\mu,\nu)$, one gets the results, written in matrix form on the two lower indices, that are shown in Table 1. This matrix form is very useful as a basis for further calculations, such as that of the curvature tensor. Using these expressions in the equation defining the Ricci curvature tensor $R_{\mu\nu}$ in terms of the connection, which is given by

$$R_{\mu\nu} = \partial_{\nu}\Gamma^{\alpha}_{\;\;\mu\alpha... ...Gamma^{\alpha}_{\;\;\mu\beta} \Gamma^{\beta}_{\;\;\nu\alpha},$$ (6)

one gets a diagonal matrix for this curvature tensor, in its covariant form,

$$R_{\mu\nu} = \left[ \rule{0em}{8ex} \begin{array}{llll}... ...\hspace{2em} & 0 \hspace{2em} & R_{33} \end{array} \right],$$ (7)

with the four diagonal elements given in [#!DiracGravity!#], which are

$$R_{00} = \left\{ - \nu''(r) + \lambda'(r)\nu'(r) - \left[ \nu'(r) \right]^{2} - \frac{2\nu'(r)}{r} \right\} \,{\rm e}^{2\nu(r)-2\lambda(r)},$$

$$R_{11} = \left\{ \nu''(r) - \lambda'(r)\nu'(r) + \left[ \nu'(r) \right]^{2} - \frac{2\lambda'(r)}{r} \right\},$$

$$R_{22} = \left\{ \left[ 1 + r\nu'(r) - r\lambda'(r) \right] \,{\rm e}^{-2\lambda(r)} - 1 \right\},$$

$$R_{33} = \left\{ \left[ 1 + r\nu'(r) - r\lambda'(r) \right] \,{\rm e}^{-2\lambda(r)} - 1 \right\} \sin^{2}(\theta).$$ (8)

With the use of $g^{\mu\nu}$ the same quantities can be written, in mixed form, as

$$R_{0}^{0} = \left\{ - \nu''(r) + \lambda'(r)\nu'(r) - \left[ \nu'(r) \right]^{2} - \frac{2\nu'(r)}{r} \right\} \,{\rm e}^{-2\lambda(r)},$$

$$R_{1}^{1} = \left\{ - \nu''(r) + \lambda'(r)\nu'(r) - \left[ \nu'(r) \right]^{2} + \frac{2\lambda'(r)}{r} \right\} \,{\rm e}^{-2\lambda(r)},$$

$$R_{2}^{2} = - \left[ \frac{1}{r^{2}} + \frac{\nu'(r)}{r} - \frac{\lambda'(r)}{r} \right] \,{\rm e}^{-2\lambda(r)} + \frac{1}{r^{2}},$$

$$R_{3}^{3} = - \left[ \frac{1}{r^{2}} + \frac{\nu'(r)}{r} - \frac{\lambda'(r)}{r} \right] \,{\rm e}^{-2\lambda(r)} + \frac{1}{r^{2}}.$$ (9)

We will tend to write all relevant tensor quantities in this mixed form, which is the most useful for our purposes here. Note that the exponential $\exp[2\nu(r)]$ has vanished from these expressions, which therefore contain only the functions $\lambda(r)$, $\lambda'(r)$, $\nu'(r)$ and $\nu''(r)$. Note also that it turns out that $R_{2}^{2}=R_{3}^{3}$, as a consequence of the symmetries that we imposed. The last geometric element that we need to discuss here is the scalar curvature $R=R_{\mu}^{\mu}$, which can be written as

$$R = R_{0}^{0} + R_{1}^{1} + R_{2}^{2} + R_{3}^{3}.$$ (10)

We therefore can write our result for the scalar curvature as

$$\frac{1}{2}\, R = \left\{ - \nu''(r) + \lambda'(r)\n... ...^{2}} \right\} \,{\rm e}^{-2\lambda(r)} + \frac{1}{r^{2}}.$$ (11)

In the theory of General Relativity the equation determining the gravitational field is written in terms of the Ricci curvature tensor $R_{\mu\nu}$. The equation also involves the matter energy-momentum tensor $T_{\mu\nu}$, which plays the role of the source of the gravitational field. The Einstein gravitational field equation is a tensor equation which, written in its mixed form, using our notation here, with the signature $(+,-,-,-)$, following [#!DiracGravity!#], is given by

$$R_{\mu}^{\nu} - \frac{1}{2}\, R\,g_{\mu}^{\nu} = -\kappa\, T_{\mu}^{\nu},$$ (12)

where $\kappa=8\pi G/c^{4}$, $G$ is the universal gravitational constant and $c$ is the speed of light.

Note that the imposition of spherical symmetry and time independence on the solution of the Einstein field equation reduces the problem of finding that solution to a much simpler one-dimensional one, on the variable $r$. While one can take any metric at all, so long as the functions involved in it, such as $\nu(r)$ and $\lambda(r)$, are differentiable to the second order, and just calculate $R_{\mu}^{\nu}$ and $R$ in order to simply verify whatever results for $T_{\mu}^{\nu}$, with such a deeply non-linear equation one is not free to choose the matter energy-momentum tensor $T_{\mu}^{\nu}$ in an arbitrary way. Both the general structure of the theory and the symmetry conditions will impose restrictions on the possible values of this tensor. For example, since we have, due to the imposition of the symmetries, that $R_{2}^{2}=R_{3}^{3}$, and since we also have that $g_{2}^{2}=g_{3}^{3}$, it follows at once that $T_{2}^{2}=T_{3}^{3}$. Also, since $R_{\mu}^{\nu}$ and $g_{\mu}^{\nu}$ are symmetric tensors, so must be $T_{\mu}^{\nu}$.

At this point we must pause in order to consider what information we have obtained so far about $T_{\mu}^{\nu}$. First of all, since under the current hypotheses the left-hand side of the Einstein field equation turns out to be diagonal, and since we have also the additional fact that the expressions of the last two component equations turn out to be identical, the same must be true for the matter energy-momentum tensor $T_{\mu}^{\nu}$ on the right-hand side of the equation, which must therefore be diagonal,

$$T_{\mu}^{\nu} = \left[ \rule{0em}{8ex} \begin{array}{ll... ...e{2em} & 0 \hspace{2em} & T_{3}^{3}(r) \end{array} \right],$$ (13)

and which must also satisfy $T_{2}^{2}(r)=T_{3}^{3}(r)$. In addition to this, since there are no dependencies on $t$, $\theta$ or $\phi$, it follows that $T_{\mu}^{\nu}$ can depend only on $r$. Note, however, that we still do not have any further information about any possible relations between $T_{0}^{0}(r)$, $T_{1}^{1}(r)$ and $T_{2}^{2}(r)$. From this point on, in order to simplify the notation, we will use the variable names $T_{0}(r)$, $T_{1}(r)$, $T_{2}(r)$ and $T_{3}(r)$ for the diagonal elements $T_{0}^{0}(r)$, $T_{1}^{1}(r)$, $T_{2}^{2}(r)$ and $T_{3}^{3}(r)$, respectively, of the energy-momentum tensor $T_{\mu}^{\nu}$ in its mixed form.

The main general consistency condition imposed by the structure of the theory is that the covariant divergence of $T_{\mu}^{\nu}$ must vanish, that is, the condition that we must have that

$$D_{\nu}T_{\mu}^{\nu} = 0.$$ (14)

This is due to the fact that the combination of tensors that constitutes the left-hand side of the Einstein field equation satisfies the Bianci identity of the Ricci curvature tensor,

$$D_{\nu}\! \left( R^{\mu\nu} - \frac{1}{2}\, R\,g^{\mu\nu} \right) = 0,$$ (15)

which therefore implies the requirement that the covariant divergence of $T^{\mu\nu}$ must vanish,

$$D_{\nu} T^{\mu\nu} = 0.$$ (16)

Since $g_{\mu\nu}$ and $g^{\mu\nu}$ behave as constants under covariant differentiation, we may then write this condition as the requirement on $T_{\mu}^{\nu}$ given in Equation (14). At this point we must calculate this consistency condition for the specific case of time independence and spherical symmetry. In other words, we must now write explicitly, under these conditions, the expression shown in Equation (14). This rather long calculation is done in detail in Section A.1 of Appendix A. As one can see there, three of the four conditions in Equation (14) are automatically satisfied, since it results from that calculation that

$$D_{\nu}T_{0}^{\nu}(r) \equiv 0,$$

$$D_{\nu}T_{2}^{\nu}(r) \equiv 0,$$

$$D_{\nu}T_{3}^{\nu}(r) \equiv 0.$$ (17)

Therefore, the only non-trivial condition is that given by $D_{\nu}T_{1}^{\nu}(r)=0$, which results in

$$\left[r\nu'(r)\right] [T_{0}(r)-T_{1}(r)] = \left[rT'_{1}(r)\right] + [2T_{1}(r)-T_{2}(r)-T_{3}(r)],$$ (18)

and which can also be written as

$$\left[r\nu'(r)\right] = \frac{\left[rT'_{1}(r)\right]}{T_... ...r)} + \frac{2T_{1}(r)-T_{2}(r)-T_{3}(r)}{T_{0}(r)-T_{1}(r)}.$$ (19)

Note that this consistency condition on $T_{\mu}^{\nu}(r)$ ends up involving only the element of the metric given by the function $\nu'(r)$.

We now have at hand all the elements needed in order to write the Einstein gravitational field equation under our hypotheses about the geometry, as well as the relevant consistency condition. Using the elements $R_{\mu}^{\nu}$ and $R$, as well as the fact that $g_{\mu}^{\nu}=\delta_{\mu}^{\nu}$, we may now write the left-hand side of the components of the Einstein field equation, in mixed form, as

$$R_{0}^{0} - \frac{1}{2}\, R\,g_{0}^{0} = \left[ \frac{1}{r^{2}} - \frac{2\lambda'(r)}{r} \right] \,{\rm e}^{-2\lambda(r)} - \frac{1}{r^{2}},$$

$$R_{1}^{1} - \frac{1}{2}\, R\,g_{1}^{1} = \left[ \frac{1}{r^{2}} + \frac{2\nu'(r)}{r} \right] \,{\rm e}^{-2\lambda(r)} - \frac{1}{r^{2}},$$

$$R_{2}^{2} - \frac{1}{2}\, R\,g_{2}^{2} = - \left\{ - \nu''(r) + \lambda'(r)\nu'(r) - \left[ \nu'(r) \right... ... - \frac{\nu'(r)}{r} + \frac{\lambda'(r)}{r} \right\} \,{\rm e}^{-2\lambda(r)},$$

$$R_{3}^{3} - \frac{1}{2}\, R\,g_{3}^{3} = - \left\{ - \nu''(r) + \lambda'(r)\nu'(r) - \left[ \nu'(r) \right... ... - \frac{\nu'(r)}{r} + \frac{\lambda'(r)}{r} \right\} \,{\rm e}^{-2\lambda(r)}.$$ (20)

It thus results that we get some fairly simple expressions for the left-hand sides of the four components of the field equation in mixed form. Note once more that the exponential $\exp[2\nu(r)]$ is absent from these expressions, and also that the expressions of the last two component equations are identical, and therefore not independent from each other. We must now write the right-hand side of these equations, and thus introduce the energy-momentum tensor $T_{\mu}^{\nu}(r)$ that was discussed before.

Let us recall that, for an homogeneous and isotropic cosmological spacetime, containing equally homogeneous and isotropic fluid matter, in the co-moving system of coordinates, in which the matter is locally at rest, we would have that $T_{0}(r)=\rho(r)$, where $\rho(r)$ is the energy density, and that $T_{1}(r)=T_{2}(r)=T_{3}(r)=-P(r)$, where $P(r)$ is the pressure. We can expect a similar situation in our case here, for the values of the components of the energy-momentum tensor. Note that the pure radiation condition given by the equation of state $\rho(r)=3P(r)$ translates here to the simple invariant condition $T(r)=0$ on the trace of $T_{\mu}^{\nu}(r)$. Note also that this condition holds for any type of massless matter field. It is important to observe that up to this point we have assumed no more about these quantities than what is implied by the structure of the field equation itself. Since we must have that $T_{2}(r)=T_{3}(r)$ as a consequence of the symmetries that we imposed, we have only three independent components of the field equation, to which we now add the consistency condition as an ancillary condition,

$$\left[ \frac{1}{r^{2}} - \frac{2\lambda'(r)}{r} \right] \,{\rm e}^{-2\lambda(r)} - \frac{1}{r^{2}} = -\kappa\,T_{0}(r),$$

$$\left[ \frac{1}{r^{2}} + \frac{2\nu'(r)}{r} \right] \,{\rm e}^{-2\lambda(r)} - \frac{1}{r^{2}} = -\kappa\,T_{1}(r),$$

$$\left\{ \rule{0em}{3ex} \nu''(r) - \lambda'(r)\nu'(r) + \right. \hspace{7em}$$

$$\left. + \left[ \nu'(r) \right]^{2} + \frac{\nu'(r)}{r} - \frac{\lambda'(r)}{r} \rule{0em}{3ex} \right\} \,{\rm e}^{-2\lambda(r)} = -\kappa\,T_{2}(r),$$

$$\left[r\nu'(r)\right] = \frac{\left[rT'_{1}(r)\right]}{T_... ...r)} + \frac{2T_{1}(r)-T_{2}(r)-T_{3}(r)}{T_{0}(r)-T_{1}(r)},$$ (21)

where we recall that $T_{3}(r)=T_{2}(r)$. We will now impose on the components of $T_{\mu}^{\nu}(r)$ the equation of state for fluid matter. Since the equation of state determines the nature of the fluid, and assuming that no phase transitions occur within the volume occupied by the matter, we must have the same equation of state throughout the volume of the fluid matter. In other words, the equation of state must not dependent on the position $r$. Both the energy density $\rho(r)$ and the pressure $P(r)$ may depend on $r$, but the relations between them may not. This means that we are assuming, in this simplest case, that there is a certain homogeneity regarding the state of the matter, which is assumed not to undergo a phase transition along the possible values of $r$. This implies that we should have the relations

$$T_{1}(r) = -\omega T_{0}(r),$$

$$T_{2}(r) = -\omega T_{0}(r),$$

$$T_{3}(r) = -\omega T_{0}(r),$$ (22)

which automatically satisfy the condition that $T_{3}(r)=T_{2}(r)$, and where $\omega$ is a positive real number in the interval $(0,1/3]$. The value $\omega=0$ corresponds to pressureless dust and the value $\omega=1/3$ corresponds to pure relativistic radiation. The value of $\omega$ in this range determines what fraction of the energy is bound in the form or rest mass and what fraction is in the form of relativistic radiation. Multiplying the first three component equations in Equation (21) by $r^{2}$ and making the replacements indicated above, which result in the fluid matter being described by the single function $T_{0}(r)$, we get

$$\left\{ 1 - 2\left[r\lambda'(r)\right] \right\} \,{\rm e}^{-2\lambda(r)} = 1 - \kappa r^{2}T_{0}(r),$$

$$\left\{ 1 + 2\left[r\nu'(r)\right] \right\} \,{\rm e}^{-2\lambda(r)} = 1 + \omega\kappa r^{2}T_{0}(r),$$

$$\left\{ \rule{0em}{3ex} r^{2}\nu''(r) - \left[ r\lambda'(r) \right] \left[ r\nu'(r) \right] + \right. \hspace{6em}$$

$$\left. + \left[ r\nu'(r) \right]^{2} + \left[ r\nu'(r) \right] - \left[ r\lambda'(r) \right] \rule{0em}{3ex} \right\} \,{\rm e}^{-2\lambda(r)} = \omega\kappa r^{2}T_{0}(r),$$

$$\left[r\nu'(r)\right] = -\, \frac{\omega}{1+\omega}\, \frac{\left[rT'_{0}(r)\right]}{T_{0}(r)}.$$ (23)

At this point it is convenient to write all the equations in terms of the single function given by $\mathfrak{T}(r)=\kappa r^{2}T_{0}(r)$, which also happens to be dimensionless, because due to the dimensions of the Einstein field equation the quantity $\kappa T_{0}(r)$ has the dimensions of $[\mbox{length}]^{-2}$,

$$\left\{ 1 - \left[ 2r\lambda'(r) \right] \right\} \,{\rm e}^{-2\lambda(r)} = 1 - \mathfrak{T}(r),$$ (24)

$$\left\{ 1 + \left[ 2r\nu'(r) \right] \right\} \,{\rm e}^{-2\lambda(r)} = 1 + \omega\mathfrak{T}(r),$$ (25)

$$\left\{ \rule{0em}{3ex} r^{2}\nu''(r) - \left[ r\lambda'(r) \right] \left[ r\nu'(r) \right] + \right. \hspace{6em}$$

$$\left. + \left[ r\nu'(r) \right]^{2} + \left[ r\nu'(r) \right] - \left[ r\lambda'(r) \right] \rule{0em}{3ex} \right\} \,{\rm e}^{-2\lambda(r)} = \omega\mathfrak{T}(r),$$ (26)

$$\left[r\nu'(r)\right] = \frac{2\omega}{1+\omega} - \fra... ...ga}\, \frac{\left[r\mathfrak{T}'(r)\right]}{\mathfrak{T}(r)}.$$ (27)

These are the four equations that must be satisfied by the functions $\nu(r)$, $\lambda(r)$ and $\mathfrak{T}(r)$ that represent a solution of the Einstein gravitational field equation in the presence of fluid matter, under the hypotheses of time independence, spherical symmetry and a simple homogeneous local equation of state for the fluid matter. In the sequence we will first recover the solution for empty space, that is, we will derive from these equations the Schwarzschild solution, and then we will establish the solution in the presence of the fluid matter. Ultimately, the Schwarzschild solution will play the role of being the $r\to\infty$ asymptotic limit of the solution in the presence of localized fluid matter.