Joel L. Weiner

Generalizing Birkhoff

University of Hawaii at Manoa, Department of Mathematics

Honolulu, HI USA

<EMAIL_ADDRESS>
Abstract

We generalize Birkhoff’s Theorem in the following fashion. We find necessary
and sufficient conditions for any spherically symmetric space-time to be
static in terms of the eigenvalues of the stress-energy tensor. In particular,
we generalize the Tolman-Oppenheimer-Volkoff equation and prove that
Birkhoff’s theorem holds under the weaker hypothesis of no pressure (with
respect to an appropriate frame.) We provide equations that show how the
coefficients of the metric relate to the eigenvalues of the stress-energy
tensor. These involve integrals that are simple functions of those
eigenvalues. We also determine among all static spherically symmetric space-
times those that are asymptotically flat. A few examples are presented taking
advantage of the results. The calculations are done by viewing the space-times
as warped products and the computations are done using Cartan’s moving frames
approach.

Key Words

spherically symmetric, static, asymptotically flat

# Generalizing Birkhoff

##  1\. Introduction

Birkhoff’s Theorem deals with spherically symmetric space-times. It asserts
that if the stress-energy tensor ${\bf T}=\mathbf{0}$, then such a space-time
is static. This further implies the existence of additional symmetries and, if
it makes sense to talk about space at infinity, that the space-time is
asymptotically flat. The most famous example of a spherically symmetric metric
is the Schwarzschild metric, which has a trivial stress-energy tensor and, of
course, is static. There are other examples of spherically symmetric space-
time metrics that are static such as the de Sitter space-time within an
appropriately restricted domain and the Reissner-Nordström space-time. The
former is not asymptotically flat but the latter is.

There have been many generalizations of Birkhoff’s Theorem, most from the
following viewpoint: “Given a space-time acted upon by a group of symmetries,
under what conditions do additional symmetries exist?” A number of papers, for
example Bokhari et al. [1] and Shanarii et al. [5], look for necessary
conditions on static spherically symmetric space-times in order that there
exist additional symmetries. Others, such as Szenthe [6] look for sufficient
conditions for the existence of a Killing vector field orthogonal to the
orbits of the action of a given symmetry group. In fact, the paper by Szenthe
gives a thorough review of the literature on that approach to the
generalization of Birkhoff’s Theorem.

Approaches such these just mentioned focus on the metric. Other approaches to
generalizing Birkhoff look for conditions on the stress-energy tensor ${\bf
T}$ of a spherically symmetric space-time in order for it to be static. Wald
[7], pp.125–128, examines that situation when the stress-energy tensor is that
of a perfect fluid and derives the Tolman-Oppenheimer-Volkoff (TOV) equation.
Moreover, in a paper by Cattoen et al. [2] an anisotropic version of the TOV
equation is derived. These TOV equations arise as necessary conditions
because, in both cases, it is assumed from the outset that the metric is
static.

Our goal is to find _necessary and sufficient_ conditions on ${\bf T}$ for a
spherically symmetric space-time to be static and show exactly how such
metrics relate to ${\bf T}$. Then, among those space-times that are static, we
further determine which are asymptotically flat. The paper is organized as
follows: In Section 2 we introduce the moving frame notation we use throughout
the paper and present some preliminary calculations; in particular, we get
formulas for the Ricci forms associated to a moving frame. In Section 3, we
use the formulas for the Ricci forms to prove the standard Birkhoff Theorem in
order to familiarize the reader to the techniques about to be employed in the
generalizations. Section 4 is devoted to our first generalization where we
consider stress-energy transformations which have at most 2 distinct
eigenvalues. Section 5 considers the most general situation where the stress-
energy transformation may have up to 3 distinct eigenvalues. It is in this
section that we obtain a generalization of the TOV equation and also prove, as
a corollary of the main theorem, the following result: If a spherically
symmetric space-time has no pressures with respect to what we call the
Birkhoff framing, then it has no energy density with respect to that framing,
and thus is a vacuum. Because of this result, the original Birkhoff Theorem
holds under a weaker assumption; this is a second corollary. A third corollary
to the main theorem asserts that spherically symmetric dust cannot be static.
Finally, Section 6 is devoted to determining which static spherically
symmetric space-times are asymptotically flat.

Lastly, we recommend the paper by Szenthe [6] for a through presentation of
differential topological aspects of the problem under consideration and hence
as a good jumping off point for what is to follow.

##  2\. Preliminary Calculations

Let us denote by $X$ a space-time which is spherically symmetric. As a
consequence of the spherical symmetry, there exists a fiber bundle $\beta:X\to
B$ whose fibers are the 2-spheres that are the orbits of the isometry group
that accounts for the spherically symmetric structure. We may assume, without
loss of generality that $B$ is diffeomorphic to an open ball and thus
$X=B\times\mathbf{S}^{2}$. This is certainly true locally. Moreover, due to
the fact that the symmetries are isometries and the orbits are spacial, the
metric $\bar{\mathbf{g}}$ on $X$ projects to a nonsingular metric $\mathbf{g}$
on $B$ which has index 1.

Through each event $\mathbf{x}\in X$ there is a round 2-sphere of curvature
$\frac{1}{r^{2}}$. This determines a function $r:X\to\mathbb{R}$ which we
assume is regular; this function projects to a regular function on $B$ which
we also denote by $r$. We choose $r$ to be a coordinate function on $B$.
Necessarily there exists a regular function $t$ on $B$ such that $dr$ and $dt$
are orthogonal with respect to $\mathbf{g}$. We then have a coordinate system
$t,r$ on $B$ and we finally assume that the image of $B$ under
$(t,r):B\to\mathbb{R}^{2}$ is a rectangle $(t_{0},t_{1})\times(r_{0},r_{1})$,
where $0<t_{0}<t_{1}\leq\infty$ and $0<r_{0}<r_{1}\leq\infty$. Again this is
always true locally, so we have not imposed any severe restrictions by these
assumptions. We finally observe that $X$ is a warped product; in fact

$X=B\times_{r}\mathbf{S}^{2},$

in the notation of O’Neill [4] . Thus, if we let $\mathbf{h}$ denote the
metric on the round 2-sphere of radius 1,

$\bar{\mathbf{g}}=\mathbf{g}+r^{2}\mathbf{h}.$

We intend to stay faithful to O’Neill’s terminology throughout this paper.

We will let $\bar{\bf{R}}$ denote the Ricci curvature tensor on $X$ viewed as
a symmetric bilinear form. This notation is convenient and will not lead to
any confusion since we can deal with the Riemann curvature tensor without
introducing a symbol to represent it. Will we use $\bf{R}$ to denote the Ricci
tensor for $B$, and, of course, the Ricci tensor for $\mathbb{S}^{2}$ is
$\bf{h}$. Our first goal is to obtain formulas for $\bar{\bf{R}}$ since that
is intimately connected to $\bf{T}$ by means of Einstein’s equation.

We will be working with moving frames. Thus we introduce all of the following,
all of which are defined locally. Let $\mathbf{a}_{\alpha},\alpha=0,1$, be an
orthonormal frame field on $B$ with $\mathbf{a}_{0}$ time-like. We denote its
dual frame field by $\theta^{\alpha}$. Let $\mathbf{b}_{\mu},\mu=2,3$, be an
orthonormal frame field on $\mathbb{S}^{2}$ with dual frame field
$\psi^{\mu}$. Then, there exist connection forms
${\theta^{\alpha}}\\!_{\beta}$ on $B$ such that

$d\theta^{\alpha}=-{\theta^{\alpha}}\\!_{\beta}\wedge\theta^{\beta}.$

If ${\Theta^{\alpha}}\\!_{\beta}$ are the curvature forms associated with
$\theta^{\alpha}$, then

${\Theta^{\alpha}}\\!_{\beta}=d{\theta^{\alpha}}\\!_{\beta}.$

In addition, there exist connection forms ${\psi^{\mu}}\\!_{\nu}$ on
$\mathbb{S}^{2}$ such that

$d\psi^{\mu}=-{\psi^{\mu}}\\!_{\nu}\wedge\psi^{\nu}$

and curvature forms ${\Psi^{\mu}}\\!_{\nu}$, where

${\Psi^{\mu}}\\!_{\nu}=d{\psi^{\mu}}\\!_{\nu}=\psi^{\mu}\wedge\psi_{\nu}.$

On $X$, we also let $\mathbf{a}_{\alpha}$ and $\mathbf{b}_{\mu}$ stand for the
horizontal and vertical vector fields that project to $\mathbf{a}_{\alpha}$ on
$B$ and $\mathbf{b}_{\mu}$ on $\mathbb{S}^{2}$, respectively. In addition, the
pullbacks of all the forms introduced above to $X$ by means of projection
mappings onto $B$ and $\mathbb{S}^{2}$ will be denoted by the same symbols.
Define vector fields $\bar{\mathbf{a}}_{\alpha}$ and $\bar{\mathbf{b}}_{\mu}$
on $X$ by

$\bar{\mathbf{a}}_{\alpha}=\mathbf{a}_{\alpha}\quad{\rm
and}\quad\bar{\mathbf{b}}_{\mu}=\frac{1}{r}\mathbf{b}_{\mu}.$

Then $\bar{\mathbf{a}}_{\alpha},\bar{\mathbf{b}}_{\mu}$ is an orthonormal
frame on $X$ with dual frame $\bar{\theta}^{\alpha},\bar{\psi}^{\mu}$ where

$\bar{\theta}^{\alpha}=\theta^{\alpha}\quad{\rm
and}\quad\bar{\psi}^{\mu}=r\psi^{\mu}.$

We introduce connection forms
${\bar{\theta}^{\alpha}}\\!_{\beta},{\bar{\psi}^{\mu}}\\!_{\nu},{\bar{\omega}^{\alpha}}\\!_{\mu}$
and ${\bar{\omega}^{\mu}}\\!_{\alpha}$ where

$\displaystyle d\bar{\theta}^{\alpha}$ $\displaystyle=$
$\displaystyle-{\bar{\theta}^{\alpha}}\\!_{\beta}\wedge\bar{\theta}^{\beta}-{\bar{\omega}^{\alpha}}\\!_{\mu}\wedge\bar{\psi}^{\mu},$
$\displaystyle d\bar{\psi}^{\mu}$ $\displaystyle=$
$\displaystyle-{\bar{\omega}^{\mu}}\\!_{\alpha}\wedge\bar{\theta}^{\alpha}-{\bar{\psi}^{\mu}}\\!_{\nu}\wedge\bar{\psi}^{\nu}$

and note that
${\bar{\omega}^{\alpha}}\\!_{\mu}=\pm{\bar{\omega}^{\mu}}\\!_{\alpha}$, the
sign depending on the metrics $\bf{g}=\theta_{\alpha}\theta^{\alpha}$ on $B$
and $\bf{h}=\psi_{\mu}\psi^{\mu}$ on $\mathbb{S}^{2}$. We now compute
${\bar{\theta}^{\alpha}}\\!_{\beta},{\bar{\psi}^{\mu}}\\!_{\nu},{\bar{\omega}^{\mu}}\\!_{\alpha}$
and ${\bar{\omega}^{\alpha}}\\!_{\mu}$. We note that

$d\bar{\theta}^{\alpha}=d\theta^{\alpha}=-{\theta^{\alpha}}\\!_{\beta}\wedge\theta^{\beta}=-{\theta^{\alpha}}\\!_{\beta}\wedge\bar{\theta}^{\beta}.$

This implies that

${\bar{\omega}^{\alpha}}\\!_{\mu}\wedge\bar{\psi}^{\mu}=0.$

If we define $r_{\alpha}$ by $dr=r_{\alpha}\theta^{\alpha}$, then

$\displaystyle d\bar{\psi}^{\mu}$ $\displaystyle=$ $\displaystyle
d(r\psi^{\mu})=r_{\alpha}\theta^{\alpha}\wedge\psi^{\mu}-r{\psi^{\mu}}\\!_{\nu}\wedge\psi^{\nu}$
$\displaystyle=$ $\displaystyle-
r_{\alpha}\psi^{\mu}\wedge\bar{\theta}^{\alpha}-{\psi^{\mu}}\\!_{\nu}\wedge\bar{\psi}^{\nu}.$

From these last two equations we see that

${\bar{\theta}^{\alpha}}\\!_{\beta}={\theta^{\alpha}}\\!_{\beta},\quad{\bar{\psi}^{\mu}}\\!_{\nu}={\psi^{\mu}}\\!_{\nu}\quad{\rm
and}\quad{\bar{\omega}^{\mu}}\\!_{\alpha}=r_{\alpha}\psi^{\mu}=\frac{r_{\alpha}}{r}\bar{\psi}^{\mu}.$

It then follows that

${\bar{\omega}^{\alpha}}\\!_{\mu}=-\frac{r^{\alpha}}{r}\bar{\psi}_{\mu}$

since

$\displaystyle{\bar{\omega}^{\alpha}}\\!_{\mu}$ $\displaystyle=$
$\displaystyle\bar{g}^{\alpha\beta}\bar{\omega}_{\beta\mu}=-\bar{g}^{\alpha\beta}\bar{\omega}_{\mu\beta}=-\bar{g}^{\alpha\beta}\bar{g}_{\mu\nu}{\bar{\omega}^{\nu}}\\!_{\beta}$
$\displaystyle=$
$\displaystyle-\bar{g}^{\alpha\beta}\bar{g}_{\mu\nu}\frac{r_{\beta}}{r}\bar{\psi}^{\nu}=-\frac{r^{\alpha}}{r}\bar{\psi}_{\nu},$

keeping in mind that $\bar{g}_{\alpha\mu}=0$.

Before we proceed we discuss some notation. Let $\bf{}S$ be a symmetric
bilinear form on a vector space $V$ with frame $\mathbf{e}_{i}$ and dual frame
$\omega^{i}$. We may write

${\bf{S}}=S_{ij}\omega^{i}\omega^{j},\ {\rm with}\ S_{ij}=S_{ji},$

where we regard this as a the tensor product as opposed to a symmetric
product. We may introduce 1-forms $S_{i}$ by setting

$S_{i}=S_{ij}\omega^{j}$

and write

${\bf{S}}=S_{i}\omega^{i}.$

The 1-forms $S_{i}$ are characterized by the fact that

${\bf{S}}=S_{i}\omega^{i}\quad{\rm and}\quad
S_{i}(\mathbf{e}_{j})=S_{j}(\mathbf{e}_{i})$

and will be referred to as ${\bf{S}}$ 1-forms (without explicit mention of the
frame $\mathbf{e}_{i}$.)

We illustrate the use of this notation in two cases for tensor fields on a
Semi-Riemannian manifold $M$ with frame field $\mathbf{e}_{i}$ and dual frame
field $\omega^{i}$.

First, we consider a smooth function $f:M\to\mathbb{R}$. The Hessian of $f$,
denoted ${\bf{H}}f$, is defined by ${\bf{H}}f=f_{ij}\omega^{i}\omega^{j}$,
where $f_{ij}$ satisfies the following:

$f_{ij}\omega^{j}=df_{i}-f_{j}{\omega^{j}}\\!_{i}\quad{\rm and}\quad
f_{ij}=f_{ji}.$

Necessarily,

$({\bf{H}}f)_{i}=f_{ij}\omega^{j}.$

Second, we consider the Ricci tensor $\bf{R}$ on $M$. If ${\Omega^{i}}\\!_{j}$
are the curvature forms associated with the given frame field, then it is
straightforward to show that

$R_{j}=\mathbf{e}_{i}\lrcorner{\Omega^{i}}\\!_{j}.$

We now compute curvatures.

$\displaystyle{\bar{\Theta}^{\alpha}}\\!_{\beta}$ $\displaystyle=$
$\displaystyle
d{\bar{\theta}^{\alpha}}\\!_{\beta}+{\bar{\theta}^{\alpha}}\\!_{\gamma}\wedge{\bar{\theta}^{\gamma}}\\!_{\beta}+{\bar{\omega}^{\alpha}}\\!_{\mu}\wedge{\bar{\omega}^{\mu}}\\!_{\beta}$
$\displaystyle=$
$\displaystyle{\Theta^{\alpha}}\\!_{\beta}-\frac{r^{\alpha}}{r}\bar{\psi}_{\mu}\wedge\frac{r_{\beta}}{r}\bar{\psi}^{\mu}$
$\displaystyle=$ $\displaystyle{\Theta^{\alpha}}\\!_{\beta}.$
$\displaystyle{\bar{\Psi}^{\mu}}\\!_{\nu}$ $\displaystyle=$ $\displaystyle
d{\bar{\psi}^{\mu}}\\!_{\nu}+{\bar{\omega}^{\mu}}\\!_{\alpha}\wedge{\bar{\omega}^{\alpha}}\\!_{\nu}+{\bar{\psi}^{\mu}}\\!_{\lambda}\wedge{\bar{\psi}^{\lambda}}\\!_{\nu}$
$\displaystyle=$
$\displaystyle{\Psi^{\mu}}\\!_{\nu}-\frac{r_{\alpha}r^{\alpha}}{r^{2}}\bar{\psi}^{\mu}\wedge\bar{\psi}_{\nu}$
$\displaystyle=$ $\displaystyle{\Psi^{\mu}}\\!_{\nu}-\frac{||\nabla
r||^{2}}{r^{2}}\bar{\psi}^{\mu}\wedge\bar{\psi}_{\nu}.$
$\displaystyle{\bar{\Omega}^{\mu}}\\!_{\alpha}$ $\displaystyle=$
$\displaystyle
d{\bar{\omega}^{\mu}}\\!_{\alpha}+{\bar{\omega}^{\mu}}\\!_{\beta}\wedge{\bar{\theta}^{\beta}}\\!_{\alpha}+{\bar{\psi}^{\mu}}\\!_{\nu}\wedge{\bar{\omega}^{\nu}}\\!_{\alpha}$
$\displaystyle=$ $\displaystyle
d(r_{\alpha}\psi^{\mu})+\frac{r_{\beta}}{r}\bar{\psi}^{\mu}\wedge{\theta^{\beta}}\\!_{\alpha}+{\psi^{\mu}}\\!_{\nu}\wedge
r_{\alpha}\psi^{\nu}$ $\displaystyle=$ $\displaystyle
dr_{\alpha}\wedge\psi^{\mu}+r_{\alpha}d\psi^{\mu}+r_{\alpha}{\psi^{\mu}}\\!_{\nu}\wedge\psi^{\nu}-r_{\beta}{\theta^{\beta}}\\!_{\alpha}\wedge\psi^{\mu}$
$\displaystyle=$
$\displaystyle\frac{1}{r}r_{\alpha\beta}\bar{\theta}^{\beta}\wedge\bar{\psi}^{\mu}$
$\displaystyle=$
$\displaystyle-\bar{\psi}^{\mu}\wedge\frac{1}{r}({\bf{H}}r)_{\alpha}.$

Of course,

${\bar{\Omega}^{\alpha}}\\!_{\mu}=-\frac{1}{r}({\bf{H}}r)^{\alpha}\wedge\bar{\psi}_{\mu}.$

Next we compute the Ricci forms $\bar{\rho}_{\alpha}$ and $\bar{\rho}_{\mu}$
on $X$. Let $\rho_{\alpha}$ and $\rho_{\mu}=\psi_{\mu}$ denote the Ricci forms
on $B$ and $\mathbb{S}^{2}$, respectively.

$\displaystyle\bar{\rho}_{\alpha}$ $\displaystyle=$
$\displaystyle\bar{\mathbf{a}}_{\beta}\lrcorner{\bar{\Theta}^{\beta}}\\!_{\alpha}+\bar{\mathbf{b}}_{\mu}\lrcorner{\bar{\Omega}^{\mu}}\\!_{\alpha}$
$\displaystyle=$
$\displaystyle\mathbf{a}_{\beta}\lrcorner{\Theta^{\beta}}\\!_{\alpha}+\bar{\mathbf{b}}_{\mu}\lrcorner\left(\frac{1}{r}r_{\alpha\beta}\bar{\theta}^{\beta}\wedge\bar{\psi}^{\mu}\right)$
$\displaystyle=$
$\displaystyle\rho_{\alpha}-\frac{2}{r}r_{\alpha\beta}\bar{\theta}^{\beta}$
$\displaystyle=$ $\displaystyle\rho_{\alpha}-\frac{2}{r}({\bf{H}}r)_{\alpha}.$
$\displaystyle\bar{\rho}_{\mu}$ $\displaystyle=$
$\displaystyle\bar{\mathbf{a}}_{\alpha}\lrcorner{\bar{\Omega}^{\alpha}}\\!_{\mu}+\bar{\mathbf{b}}_{\nu}\lrcorner{\bar{\Psi}^{\nu}}\\!_{\mu}$
$\displaystyle=$
$\displaystyle\bar{\mathbf{a}}_{\alpha}\lrcorner\left(\frac{1}{r}\bar{\psi}_{\mu}\wedge({\bf{H}}r)^{\alpha}\right)+\bar{\mathbf{b}}_{\nu}\lrcorner\left({\Psi^{\nu}}\\!_{\mu}-\frac{||\nabla
r||^{2}}{r^{2}}\bar{\psi}^{\nu}\wedge\bar{\psi}_{\mu}\right).$
$\displaystyle=$
$\displaystyle-\frac{1}{r}\mathbf{a}_{\alpha}\lrcorner({\bf{Hr}})^{\alpha}\bar{\psi}_{\mu}+\frac{1}{r}\mathbf{b}_{\nu}\lrcorner{\Psi^{\nu}}\\!_{\mu}-\frac{||\nabla
r||^{2}}{r^{2}}(2\bar{\psi}_{\mu}-\bar{h}_{\mu\nu}\bar{\psi}^{\nu})$
$\displaystyle=$ $\displaystyle\frac{1}{r}\rho_{\mu}-\left(\frac{\Delta
r}{r}+\frac{||\nabla r||^{2}}{r^{2}}\right)\bar{\psi}_{\mu}$ $\displaystyle=$
$\displaystyle\left(\frac{1}{r^{2}}-\frac{\Delta r}{r}-\frac{||\nabla
r||^{2}}{r^{2}}\right)\bar{\psi}_{\mu}.$

To finish our computations of the Ricci forms we need to consider the metric
on $B$ in more detail. Thus we set

$\mathbf{g}=-e^{2}dt^{2}+g^{2}dr^{2},$

where we assume are that $e$ and $g$ are smooth and take on only positive
values. Clearly,

$\theta^{0}=edt,\theta_{0}=-edt\ {\rm and\ }\theta^{1}=\theta_{1}=gdr.$

From the assumptions just made we see that $dr=\frac{1}{g}\theta^{1}$ which
immediately implies that $r_{0}=0$ and $r_{1}=\frac{1}{g}$. Thus

$||\nabla r||^{2}=\frac{1}{g^{2}}.$

In addition,

$({\bf{H}}r)_{0}=-\frac{1}{g}{\theta^{1}}\\!_{0}\quad{\rm
and}\quad({\bf{H}}r)_{1}=d\left(\frac{1}{g}\right)=-\frac{1}{g^{2}}dg.$

We compute ${\theta^{0}}\\!_{1}$. Since
${\theta^{0}}\\!_{1}={\theta^{1}}\\!_{0}$, and

$\displaystyle d\theta^{0}$ $\displaystyle=$ $\displaystyle
d(edt)=e_{r}dr\wedge dt=-\frac{e_{r}}{eg}\theta^{0}\wedge\theta^{1},$
$\displaystyle d\theta^{1}$ $\displaystyle=$ $\displaystyle
d(gdr)=g_{t}dt\wedge dr=-\frac{g_{t}}{eg}\theta^{1}\wedge\theta^{0}$

we see that

${\theta^{0}}\\!_{1}=\frac{e_{r}}{eg}\theta^{0}+\frac{g_{t}}{eg}\theta^{1}.$

Noting that

$\displaystyle dg$ $\displaystyle=$ $\displaystyle
g_{t}dt+g_{r}dr=\frac{g_{t}}{e}\theta^{0}+\frac{g_{r}}{g}\theta^{1}$
$\displaystyle de$ $\displaystyle=$ $\displaystyle
e_{t}dt+e_{r}dr=\frac{e_{t}}{e}\theta^{0}+\frac{e_{r}}{g}\theta^{1}$

we obtain

$g_{0}=\frac{g_{t}}{e},g_{1}=\frac{g_{r}}{g},e_{0}=\frac{e_{t}}{e}{\rm\ and\
}e_{1}=\frac{e_{r}}{g}.$

Hence

${\theta^{0}}\\!_{1}=\frac{e_{1}}{e}\theta^{0}+\frac{g_{0}}{g}\theta^{1}=(\ln
e)_{1}\theta^{0}+(\ln g)_{0}\theta^{1}.$

As a consequence of what we just observed we have the following:

$({\bf{H}}r)_{0}=-\frac{1}{g}\left((\ln e)_{1}\theta^{0}+(\ln
g)_{0}\theta^{1}\right)\ {\rm and\ }({\bf{H}}r)_{1}=-\frac{1}{g}\left((\ln
g)_{0}\theta^{0}+(\ln g)_{1}\theta^{1}\right)$

and, moreover,

$\Delta r=-({\bf{H}}r)_{00}+({\bf{H}}r)_{11}=\frac{1}{g}\left[(\ln e)_{1}-(\ln
g)_{1}\right]=\frac{1}{g}\left(\ln\frac{e}{g}\right)_{1}.$

In terms of coordinates

$\Delta r=\frac{1}{g^{2}}\left(\ln\frac{e}{g}\right)_{r}.$

Since $\rho_{\alpha}=K_{B}\theta_{\alpha}$, where $K_{B}$ is the sectional
curvature of $B$, we now compute $K_{B}$. We do this by computing
${\Theta^{0}}\\!_{1}$.

$\displaystyle{\bar{\Theta}^{0}}\\!_{1}$ $\displaystyle=$ $\displaystyle
d{\bar{\theta}^{0}}\\!_{1}=d\left((\ln e)_{1}\theta^{0}+(\ln
g)_{0}\theta^{1}\right)$ $\displaystyle=$ $\displaystyle(\ln
e)_{11}\theta^{1}\wedge\theta^{0}-(\ln
e)_{1}{\theta^{0}}\\!_{1}\wedge\theta^{1}+(\ln
g)_{00}\theta^{0}\wedge\theta^{1}-(\ln
g)_{0}{\theta^{1}}\\!_{0}\wedge\theta^{0}$ $\displaystyle=$
$\displaystyle\left(-(\ln e)_{11}-(\ln e)_{1}\>^{2}+(\ln g)_{00}+(\ln
g)_{0}\,^{2}\right)\theta^{0}\wedge\theta^{1}$ $\displaystyle=$
$\displaystyle\left(-\frac{e_{11}}{e}+\frac{g_{00}}{g}\right)\theta^{0}\wedge\theta_{1}.$

Thus

$K_{B}=-\frac{e_{11}}{e}+\frac{g_{00}}{g}=-\frac{1}{eg}\left[\left(\frac{e_{r}}{g}\right)_{r}-\left(\frac{g_{t}}{e}\right)_{t}\right].$

Substituting formulas for
$\rho_{\alpha}=K_{B}\theta_{\alpha},(Hr)_{\alpha},\Delta r$ and $||\nabla
r||^{2}$ into the formulas for the Ricci forms we get the following:

$\displaystyle\bar{\rho}_{0}$ $\displaystyle=$
$\displaystyle\left(-K_{B}+\frac{2}{rg}(\ln
e)_{1}\right)\bar{\theta}^{0}+\frac{2}{rg}(\ln g)_{0}\bar{\theta}^{1}$
$\displaystyle\bar{\rho}_{1}$ $\displaystyle=$ $\displaystyle\frac{2}{rg}(\ln
g)_{0}\bar{\theta}^{0}+\left(K_{B}+\frac{2}{rg}(\ln
g)_{1}\right)\bar{\theta}^{1}$ $\displaystyle\bar{\rho}_{\mu}$
$\displaystyle=$
$\displaystyle\left(\frac{1}{r^{2}}\left(1-\frac{1}{g^{2}}\right)-\frac{1}{rg}\left(\ln\frac{e}{g}\right)_{1}\right)\bar{\psi}_{\mu}$

##  3\. Birkhoff’s Theorem

Since we have the machinery to prove Birkhoff’s Theorem, we will. As is well
known, if ${\bf{T}}=0$, then the Ricci tensor $\bar{\bf{R}}$ on $X$ is
trivial. This implies that the Ricci forms must be trivial as well. This gives
us the following four equations:

(1) $\displaystyle K_{B}=\frac{2}{rg}(\ln e)_{1}$ (2) $\displaystyle
K_{B}=-\frac{2}{rg}(\ln g)_{1}$ (3) $\displaystyle\frac{2}{rg}(\ln g)_{0}=0$
(4)
$\displaystyle\frac{1}{r^{2}}\left(1-\frac{1}{g^{2}}\right)-\frac{1}{rg}\left(\ln\frac{e}{g}\right)_{1}=0$

From equation (3) we immediately see that $g=g(r)$, that is, $g$ is a function
of $r$ only. Equations (1) and (2) imply that $(\ln eg)_{1}=0$. Thus,
$eg(r)=f(t)$ and we conclude that $e=\frac{f(t)}{g(r)}$, where $f(t)>0$, for
all $t$. Thus we introduce a new time variable $\tau$ where $d\tau=f(t)dt$ and
the metric on $B$ takes the form

$-\frac{1}{g^{2}}d\tau^{2}+g^{2}dr^{2}.$

Noting that $e=\frac{1}{g}$ in the new $\tau,r$ coordinates and introducing
the function $h=e^{2}$, equation (4) becomes

$\frac{1}{r^{2}}(1-h)-\frac{h}{r}(\ln h)_{r}=0,$

that is,

(5) $rh_{r}+h=1.$

This is equivalent to $(rh)_{r}=1$ and leads immediately to

$h=1+\frac{c}{r},$

where $c$ is the constant of integration.

Note however that we have not yet established that equations (1) and (2) are
satisfied by our solution

$\bar{\mathbf{g}}=-hd\tau^{2}+h^{-1}dr^{2}+r^{2}d\sigma^{2},$

where we have introduced $d\sigma^{2}$ for the metric on $\mathbb{S}^{2}$.

Since we know that the difference of these two equations is satisfied, it will
suffice to show that equation (1) is. First, we express $K_{B}$ in terms of
$h$.

$K_{B}=-\left(\frac{e_{r}}{g}\right)_{r}=-\frac{1}{2}h_{rr}.$

Since

$\frac{2}{rg}(\ln e)_{1}=\frac{1}{rg^{2}}(\ln e^{2})_{r}=\frac{1}{r}h_{r}.$

equation (1) may be written

$rh_{rr}+2h_{r}=0.$

This follows immediately from equation (5).

Thus under the assumptions made we find the metric has the following
representation:

$\bar{\mathbf{g}}=-\frac{r+c}{r}d\tau^{2}+\frac{r}{r+c}dr^{2}+r^{2}d\sigma^{2}.$

This is clearly a static metric since $\frac{\partial}{\partial\tau}$ is an
irrotational Killing vector field. If $r_{1}=\infty$ then the metric is also
asymptotically flat. (We will precisely define what we mean by “asymptotically
flat” in Section 6.) We also note that such a spherically symmetric space-time
must have trivial Ricci curvature and hence trivial stress-energy tensor.

If $c>0$ then the metric $\bar{\mathbf{g}}$ is defined for all real $r$ and
becomes singular as $r$ approaches $0$. If $c=0$, $X$ is flat; in fact, it is
open submanifold of Minkowski space-time. If $c<0$, then one customarily sets
$c=-2M$ (where $M>0$) and notes that the metric is defined only for $r>2M$. In
this last case we have obtained the Schwarzchild metric.

We close with a statement of what we have proved.

Birkhoff’s Theorem If a space-time has spherical spatial symmetry and
vanishing Ricci curvature, then is must be static and (assuming
$r_{1}=\infty$) it must also be asymptotically flat. If the metic is singular
as $r\to r_{0}>0$ (and $r_{1}=\infty$) then the space-time metric is the
Schwarzchild metric.

##  4\. Generalizing Birkhoff

Let $T_{S}X$ be the vector bundle of 2-dimensional vector spaces tangent to
the spatial symmetry spheres and $T_{R}X$ be the vector bundle that is the
orthogonal complement of $T_{S}X$. We let $\bar{\bf{g}}_{S}$ and $\bf{T}_{S}$
denote the restrictions of $\bar{\bf{g}}$ and $\bf{T}$ to $T_{S}X$; similarly,
let $\bar{\bf{g}}_{R}$ and $\bf{T}_{R}$ denote the restriction of
$\bar{\bf{g}}$ and $\bf{T}$ to $T_{R}X$. Because of the symmetry assumption,
there exists a function $\lambda:X\to\mathbb{R}$ (that is the pullback of a
function on $B$) such that $\bf{T}_{S}=\lambda\bar{\bf{g}}_{S}$. As an
operator this means that $\bf{T}^{\\#}$ (the linear transformation associated
to $\bf{T}$ by means of the metric) has any fiber of $T_{S}X$ as an invariant
subspace. Thus, the fibers of $T_{R}X$, being the orthogonal complements of
the fibers of $T_{S}M$, must be invariant subspaces of $\bf{T}^{\\#}$ as well.
We will assume that $\bf{T}_{R}=-\mu\bar{\bf{g}}_{R}$, where
$\mu:X\to\mathbb{R}$ (which also must be the pullback of a function defined on
B.) As a consequence of that assumption,

$\bf{T}=-\mu\bar{\bf{g}}_{R}+\lambda\bar{\bf{g}}_{S}.$

This section is devoted to studying space-times which have spherical spatial
symmetry and stress-energy tensors of this sort. The discussion given below
mimics the proof of Birkhoff’s Theorem.

A straightforward calculation, using Einstein’s equation, implies

$\bar{\bf{R}}=-8\pi\lambda\bar{\bf{g}}_{R}+8\pi\mu\bar{\bf{g}}_{S}.$

This last equation is equivalent to
$\bar{\rho}_{\alpha}=-8\pi\lambda\,\bar{\theta}_{\alpha}$ and
$\bar{\rho}_{\mu}=8\pi\mu\,\bar{\psi}_{\mu}$. Hence

(6) $\displaystyle K_{B}-\frac{2}{rg}(\ln e)_{1}=-8\pi\lambda\qquad$ (7)
$\displaystyle K_{B}+\frac{2}{rg}(\ln g)_{1}=-8\pi\lambda\qquad$ (8)
$\displaystyle\frac{2}{rg}(\ln g)_{0}=0\qquad\quad\quad$ (9)
$\displaystyle\frac{1}{r^{2}}\left(1-\frac{1}{g^{2}}\right)-\frac{1}{rg}\left(\ln\frac{e}{g}\right)_{1}=8\pi\mu$

Half the sum and difference of equations (6) and (7) will be more useful. They
follow.

(10) $\displaystyle\frac{1}{rg}\left(\ln eg\right)_{1}=0\qquad$ (11)
$\displaystyle\quad
K_{B}+\frac{1}{rg}\left(\ln\frac{g}{e}\right)_{1}=-8\pi\lambda\ $

We begin by introducing the function $h:X\to\mathbb{R}$, where
$h=\frac{1}{g^{2}}$. We first observe that equation (8) is equivalent to
asserting $h=h(r)$. In light of that fact, equation (10) is equivalent to the
following: there exists a time coordinate $\tau$ so that relative to the
$\tau,r$ coordinate system $e^{2}=h$ and $g^{2}=h^{-1}$, that is, there exists
coordinates $\tau,r$ such that

(12) $\bar{\bf{g}}=-h(r)d\tau^{2}+h(r)^{-1}dr^{2}+r^{2}d\sigma^{2}.$

Clearly the metric in equation (12) is that of a static space-time. In fact,
it is the form of all the metrics mentioned in the introduction.

Equation (9) written in terms of $h$ becomes

$\frac{1}{r^{2}}(1-h)-\frac{1}{r}h_{r}=8\pi\mu,$

which can be rewritten as

(13) $(rh)_{r}=1-8\pi r^{2}\mu.$

Given that $h=h(r)>0$, the preceding equation is equivalent to stating that
$\mu=\mu(r)$, there exist $c$ and $r_{*}\in(r_{0},r_{1})$ such that

(14) $h(r)=\frac{r+c}{r}-\frac{2}{r}\int_{r_{*}}^{r}4\pi x^{2}\mu(x)dx$

and

(15) $\int_{r_{*}}^{r}4\pi x^{2}\mu(x)dx<\frac{r+c}{2},\quad{\rm for\
}r\in(r_{0},r_{1}).$

We only need to consider equation (11). It clearly corresponds to a condition
on $\mu$ and $\lambda$. In terms of $h$, equation (11) becomes

$-\frac{1}{2}h_{rr}-\frac{1}{r}h_{r}=-8\pi\lambda,$

which can be rewritten

$(rh)_{rr}=16\pi r\lambda.$

Because of equation (13), this last equation is equivalent to

(16) $\lambda=-\mu-\frac{r}{2}\mu_{r}.$

As a consequence of the preceding discussion we have our first generalization
of Birkhoff’s Theorem.

######  Theorem 1.

If $X$ is a spherically symmetric space-time with metric $\bar{\bf{g}}$ and
stress-energy tensor $\bf{T}=-\mu\bar{\bf{g}}_{R}+\lambda\bar{\bf{g}}_{S}$,
then $\mu=\mu(r)$, there exist $c$ and $r_{*}\in(r_{0},r_{1})$ such that $\mu$
satisfies inequality (15), equation (16) holds and there exists a time
coordinate $\tau$ so that the metric $\bar{\mathbf{g}}$ is given by equation
(12) with $h$ defined by (14). Conversely, if $X$ is a space-time with metric
${\rm\bar{\bf{g}}}$ given by equation (12) where $h$ is defined by equation
(14), with $\mu=\mu(r)$ necessarily satisfying inequality (15) and $\lambda$
is given by equation (16), then $X$ with metric $\bar{\bf{g}}$ is spherically
symmetric and has stress-energy tensor
$\bf{T}=-\mu\bar{\bf{g}}_{R}+\lambda\bar{\bf{g}}_{S}$.

Obviously, a metric $\bar{\bf{g}}$ given by equation (12) where $h$ satisfies
equation (14) is static and any $\lambda$ given by equation (16) satisfies
$\lambda=\lambda(r)$ when $\mu=\mu(r)$. We see we have such metrics for every
$\mu=\mu(r)$ which satisfies (15). We therefore consider some examples.

######  Example 1.

Suppose $\bf{T}$ is a multiple of the metric, that is, $\lambda=-\mu$
throughout $X$. It follows from Theorem 1 that $\mu$ and $\lambda$ are
constants. If we suppose that $r_{0}=0$, $c=0$ and $r_{*}=0$ (which does not
present any problems given the form of the integrand) in order that the metric
extends to the world line of the center of symmetry, then inequality (15)
implies $r_{1}<\sqrt{\frac{3}{8\pi\mu}}$. We get a de-Sitter metric if
$\mu>0$, a Minkowski space if $\mu=0$ and an anti-de Sitter space if $\mu<0$.

######  Example 2.

Suppose $r_{0}=0,r_{1}=\infty$ and assume $\rho$ and $b$ are positive
constants. Set

$\mu=\rho\ {\rm for\ }0<r<b<\sqrt{\frac{3}{8\pi\rho}}\quad{\rm and}\quad\mu=0\
{\rm for\ }r\geq b.$

We choose $c=r_{*}=0$, and then note $r_{1}=\infty$ is allowed in this
situation since inequality (15) is satisfied for all $r>0$. In addition, by
equation (16)

$\lambda=-\mu\ {\rm for\ }r\neq b\quad{\rm
and}\quad\lambda(b)=\frac{b\rho}{2}\delta_{r-b}.$

Introducing $m(r)=\frac{4}{3}\pi r^{3}\rho$, which can be regarded as some
sort of mass of the body bounded by the 2-sphere of curvature
$\frac{1}{r^{2}}$ with density $\rho$, then

$h(r)=1-\frac{2m(r)}{r}\ {\rm for\ }r<b\quad{\rm and}\quad
h(r)=1-\frac{2m(b)}{r}\ {\rm for\ }r\geq b.$

This metric is the static approximation[3] of the simplest GEneric Object of
Dark Energy (GEODE).

######  Example 3.

Suppose $\mu$ is constant and $r_{0}=r_{*}>0$. Let
$c-\frac{8\pi\mu}{3}r_{0}^{3}=R_{S}$, which we regard as the Schwarzchild
radius, and $\Lambda=8\pi\mu$, which we regard as the cosmological constant.
We get the Kottler-Weyl-Trefftz metric for which

$h(r)=1-\frac{R_{S}}{r}-\frac{\Lambda r^{2}}{3}.$

##  5\. Further Generalization

We have been using special frames throughout this paper and at this point we
want to give those frames a name. Recall that $r$ and the direction of
$\frac{\partial}{\partial t}$ have geometric meaning and thus physical
meaning.

######  Definition 1.

We call an orthonormal frame a Birkhoff frame if the time-like vector of the
frame is a multiple of $\frac{\partial}{\partial t}$ and the first spacial
vector of the frame is a multiple of $\frac{\partial}{\partial r}$. The
coframe dual to this frame is referred to as a Birkhoff coframe.

For a spherically symmetric space-time, we have shown that

${\bf{T}}=-\mu_{0}\bar{\theta}_{0}\bar{\theta}^{0}+\kappa(\bar{\theta}_{1}\bar{\theta}^{0}-\bar{\theta}_{0}\bar{\theta}^{1})+\mu_{1}\bar{\theta}_{1}\bar{\theta}^{1}+\lambda\bar{\psi}_{\nu}\bar{\psi}^{\nu}$

with respect to a Birkhoff coframe. The components $\mu_{\alpha},\lambda$ and
$\kappa$ are pullbacks of functions on $B$, i.e., they depend only on $t$ and
$r$. We will understand that unless specifically restricted to be functions of
$r$ these components are to be regarded as functions of $t$ and $r.$

It follows from Einstein’s equation that $\bar{\bf{R}}$ has the same
eigenvectors with respect to $\bar{\bf{g}}$ as $\bf{T}$ does. By examining the
Ricci forms of a spacially symmetric space-time we see that
$\bar{\bf{R}}(\bar{\mathbf{a}}_{0}.\bar{\mathbf{a}}_{1})=\frac{2}{rg}(\ln
g)_{0}$, where $\bar{\mathbf{a}}_{0},\bar{\mathbf{a}}_{1}$ are the first two
vectors of a Birkhoff frame. If the metric on $X$ is spherically symmetric and
static, then $\bar{\bf{R}}(\bar{\mathbf{a}}_{0}.\bar{\mathbf{a}}_{1})=0$ and
consequently $\bar{\mathbf{a}}_{0}$ and $\bar{\mathbf{a}}_{1}$ are
eigenvectors of $\bar{\bf{R}}$ with respect to $\bar{\bf{g}}$. Hence,
$\bar{\mathbf{a}}_{0}$ and $\bar{\mathbf{a}}_{1}$ are eigenvectors of $\bf{T}$
with respect to $\bar{\bf{g}}$ as well. Obviously, the functions
$\mu_{\alpha}$ and $\lambda$ depend only on $r$ and $\kappa=0$. The following
has been shown.

Observation. _If $X$ is a static spherically symmetric space-time, then
${\bf{T}}$ is diagonalized by a Birkhoff frame with components
$\mu_{\alpha}=\mu_{\alpha}(r)$ and $\lambda=\lambda(r)$, i.e., the stress-
energy tensor _

(17)
${\bf{T}}=-\mu_{0}(r)\bar{\theta}_{0}\bar{\theta}^{0}+\mu_{1}(r)\bar{\theta}_{1}\bar{\theta}^{1}+\lambda(r)\bar{\psi}_{\nu}\bar{\psi}^{\nu}.$

Based on the Observation, a reasonable starting point for any further
generalizations would require that $\bf{T}$ satisfy equation (17). However,
are we going to demand anything of the $\mu_{\alpha}$ and $\lambda$? We will
only impose a condition on $\mu_{1}$. In fact, our starting point for the
derivation of our main theorem is the imposition of the following conditions:

$\displaystyle(i)$ $\displaystyle\quad X\ {\rm is\ a\ spherically\ symmetric\
space\\!-\\!time}.$ $\displaystyle(ii)$ $\displaystyle\quad\bf{T}{\rm\ has\
the\ form\ given\ in\ equation\ (\ref{eqn:T}}).$ $\displaystyle(iii)$
$\displaystyle\quad\mu_{1}=\mu_{1}(r).$

Because of $(i)$ and $(ii)$, Einstein’s equation yields the following:

(18) $\displaystyle\frac{1}{rg}\left(\ln eg\right)_{1}=4\pi(\mu_{0}+\mu_{1})$
(19) $\displaystyle
K_{B}+\frac{1}{rg}\left(\ln\frac{g}{e}\right)_{1}=-8\pi\lambda$ (20)
$\displaystyle\frac{2}{rg}(\ln g)_{0}=0$ (21)
$\displaystyle\frac{1}{r^{2}}\left(1-\frac{1}{g^{2}}\right)-\frac{1}{rg}\left(\ln\frac{e}{g}\right)_{1}=4\pi(\mu_{0}-\mu_{1})$

As before, we let $h=\frac{1}{g^{2}}$ but also introduce $p=eg$. (The function
$p$ stands for product not pressure.) Just as we noticed in earlier sections,
equation (20) is equivalent to asserting $h=h(r)$. Rewriting equation (18) in
terms of $h$ and $p$ gives

(22) $h\left(\ln p\right)_{r}=4\pi r(\mu_{0}+\mu_{1}).$

Moreover, rewriting (21) using $h$ and $p$, we get the following:

$\displaystyle\frac{1}{r^{2}}(1-h)-\frac{h}{r}(\ln
hp)_{r}=4\pi(\mu_{0}-\mu_{1})$
$\displaystyle\frac{1}{r^{2}}(1-h)-\frac{1}{r}{h_{r}}-\frac{h}{r}(\ln
p)_{r}=4\pi(\mu_{0}-\mu_{1})$

Using equation (22) to substitute into the third term, this last equation
becomes after some simplification

(23) $(rh)_{r}=1-8\pi r^{2}\mu_{0}.$

The preceding statement is equivalent to asserting that $\mu_{0}=\mu_{0}(r)$,
there exists $c$ and $r_{*}\in(r_{0},r_{1})$ such that

(24) $h(r)=\frac{r+c}{r}-\frac{2}{r}\int_{r_{*}}^{r}4\pi x^{2}\mu_{0}(x)dx.$

and

(25) $\int_{r_{*}}^{r}4\pi x^{2}\mu_{0}(x)dx<\frac{r+c}{2},\quad{\rm for\
}r\in(r_{0},r_{1}).$

Taking into account $(iii)$, we can now solve equation (22) for $p$. We find
that there exists a function of $t$, $f(t)>0$, so that

(26) $p=f(t)\exp\left[\int_{r_{*}}^{r}\frac{4\pi
x^{2}(\mu_{0}(x)+\mu_{1}(x))}{x+c-2\int_{r_{*}}^{x}4\pi
y^{2}\mu_{0}(y)dy}dx\right].$

We introduce a new time variable $\tau$ so that $f(t)dt=d\tau$ and note that
in $\tau,r$ coordinates $p=p(r)$. The function $p$ is then given by equation
(26) with $f(t)$ removed. The metric has the following form:

(27) $\bar{\bf{g}}=-p^{2}hd\tau^{2}+h^{-1}dr^{2}+r^{2}d\sigma^{2}.$

What we have observed so far is that, because of $(iii)$, equations (18), (20)
and (21) are equivalent to asserting the following: $\mu_{0}=\mu_{0}(r)$,
which is subject to equation (25), and there exists coordinates $\tau,r$ so
that $\bar{\mathbf{g}}$ is given by equation (27) with $h$ and $p$ given by
equations (24) and (26), with $f(t)$ removed, respectively.

We still have to deal with equation (19); this should correspond to a
condition on $\mu_{0},\mu_{1}$ and $\lambda$. Written in terms of $h$ and $p$,
equation (19) becomes

$\frac{1}{2p}\left(\frac{(p^{2}h)_{r}}{p}\right)_{r}+\frac{1}{rp}\left(ph\right)_{r}=8\pi\lambda.$

We note this implies that $\lambda=\lambda(r)$ and proceed with a calculation.
After carrying out the indicated differentiations, we get

$(\ln p)^{2}_{r}h+(\ln p)_{rr}h+(\ln p)_{r}h_{r}+\frac{1}{2}(\ln
p)_{r}h_{r}+\frac{1}{2}h_{rr}+\frac{1}{r}h_{r}+\frac{1}{r}(\ln
p)_{r}h=8\pi\lambda.$

Regrouping, we can write

$(rh)_{rr}+2r[(\ln p)_{r}h]_{r}+(\ln p)_{r}h+(\ln p)_{r}[2r(\ln
p)_{r}h+(rh_{r})_{r}]=16\pi r\lambda.$

Using equations (22) and (23), we get

(28)
$\lambda=\frac{1}{4}(3\mu_{1}-\mu_{0})+\frac{r}{2}(\mu_{1})_{r}+\frac{(\mu_{0}+\mu_{1})(1+8\pi
r^{2}\mu_{1})}{4h}.$

(It easily follows that if $\mu_{1}=-\mu_{0}$ then this equation reduces to
equation (16).)

The forgoing has proved the following generalization of Birkhoff’s theorem.

######  Theorem 2.

If $X$ is a spherically symmetric space-time with

${\bf{T}}=-\mu_{0}\bar{\theta}_{0}\bar{\theta}^{0}+\mu_{1}\bar{\theta}_{1}\bar{\theta}^{1}+\lambda\bar{\psi}_{\nu}\bar{\psi}^{\nu}$

and $\mu_{1}=\mu_{1}(r)$, then $\mu_{0}=\mu_{0}(r)$ and there exist $c$ and
$r_{*}\in(r_{0},r_{1})$ such that $\mu_{0}$ satisfies inequality (25),
equation (28) holds and there exists a time coordinate $\tau$ such that the
metric $\bar{\bf{g}}$ is given by equation (27), where $h$ is given by
equation (24) and

(29) $p=\exp\left[\int_{r_{*}}^{r}\frac{4\pi
x^{2}(\mu_{0}(x)+\mu_{1}(x))}{x+c-2\int_{r_{*}}^{x}4\pi
y^{2}\mu_{0}(y)dy}dx\right].$

Conversely, if $X$ is a space-time with metric $\bar{\bf{g}}$ given by
equation (27) where $h$ is given by (24) and $p$ is given by equation (29),
with $\mu_{\alpha}=\mu_{\alpha}(r)$, and $\mu_{0}$ necessarily satisfying
equation (25), and $\lambda$ given by equation (28), then $X$ is a spherically
symmetric space-time and has stress-energy tensor

${\bf{T}}=-\mu_{0}\bar{\theta}_{0}\bar{\theta}^{0}+\mu_{1}\bar{\theta}_{1}\bar{\theta}^{1}+\lambda\bar{\psi}_{\nu}\bar{\psi}^{\nu}.$

It is clear from Observation that the metrics described in Theorem 2 account
for all static spherically symmetric space-times. We now consider some
examples.

######  Example 4.

The purpose of this example is to present a non-trivial closed form solution
satisfying the conditions given in Theorem 2. Suppose there exists constants
$m_{0}$ and $m_{1}$ such that $\mu_{\alpha}=\frac{m_{\alpha}}{r^{2}},\ {\rm
for\ }r_{0}<r<r_{1}.$ Because of the form of the integrands with which we are
dealing, we may take $r_{*}=r_{0}$. The calculations are straightforward; we
get the following, where $C=c+8\pi m_{0}r_{0}$.

$h(r)=1-8\pi m_{0}+\frac{C}{r}\quad{\rm and}\quad
p=\left(\frac{h(r)}{h(r_{0})}\right)^{\frac{4\pi(m_{0}+m_{1})}{1-8\pi m_{0}}}$

so that

$\bar{\bf{g}}=-h(r_{0})^{\frac{8\pi(m_{0}+m_{1})}{8\pi
m_{0}-1}}h(r)^{\frac{1-8\pi m_{1}}{1-8\pi
m_{0}}}d\tau^{2}+\frac{1}{h(r)}dr^{2}+r^{2}d\sigma^{2}$

and

$\lambda=\frac{(m_{0}+m_{1})[8\pi(m_{0}+m_{1})r-C]}{4[(1-8\pi
m_{0})r+C]}\frac{1}{r^{2}}.$

For physical reasons we want $m_{0}\geq 0$. To get a Lorentzian metric, with
$\frac{\partial}{\partial\tau}$ time-like, one of the following will do:

$\displaystyle{\rm If\ }m_{0}<\frac{1}{8\pi}\ {\rm and\ }C\geq 0,\ {\rm then\
}r_{0}\ {\ }r_{1}\ {\rm are\ unrestricted}.$ $\displaystyle{\rm If\
}m_{0}<\frac{1}{8\pi}\ {\rm and\ }C<0,\ {\rm then\ }r_{0}\geq\frac{C}{8\pi
m_{0}-1}.$ $\displaystyle{\rm If\ }m_{0}>\frac{1}{8\pi},\ {\rm then\ }C>0\
{\rm and\ }r_{1}\leq\frac{C}{8\pi m_{0}-1}.$

An interesting case occurs if $m_{0}+m_{1}=0$. Then

$\bar{\bf{g}}=-h(r)d\tau^{2}+h(r)^{-1}dr^{2}+r^{2}d\sigma^{2}\quad{\rm
and}\quad\lambda=0.$

This is a collection of metrics which includes the Schwartzchild metric, which
occurs when $m_{0}=0$, that have no transversal pressure with respect to the
Birkhoff frame.

######  Example 5.

We suppose all of the following: $r_{0}=c=r_{*}=0,r_{1}=\infty$ and $\rho$ is
a non-negative valued function on $(0,\infty)$. We define $m(r)$ and place an
additional restriction on $\rho$ by asserting

$m(r)=\int_{0}^{r}4\pi x^{2}\rho(x)dx<\frac{r}{2},\ {\rm for\ all\ }r>0.$

Moreover, suppose that $\bf{T}$ is the stress-energy tensor of a perfect fluid
and set $\rho=\mu_{0}\geq 0$ and $P=\mu_{1}=\lambda$. Substituting into
equation (28) yields the TOV equation

$P_{r}=-(P+\rho)\frac{m(r)+4\pi r^{3}P}{r[r-2m(r)]}.$

In order for the metric to exist at $r=0$ it is necessary and sufficient that

$\left|\int_{0}^{r}x(\rho(x)+P(x))dx\right|<\infty,\ {\rm for\ any\ }r>0.$

Remark. If we write $h(r)=1-\frac{2m(r)}{r}$ so that
$m(r)=c+\int_{r_{*}}^{r}4\pi x^{2}\mu_{0}(x)dx$, then equation (28) can be
written as follows:

$\lambda-\mu_{1}=\frac{r}{2}\left[(\mu_{1})_{r}+\frac{(\mu_{0}+\mu_{1})(m(r)+4\pi
r^{3}\mu_{1})}{r[r-2m(r)]}\right].$

We notice its close relation to TOV; hence we call it or equation (28) the
generalized TOV equation.

We end this section by applying Theorem 2 to the study of stress-energy
tensors $\bf{T}$ which are “close to $\bf{0}$.” What we mean by that will be
made clear in what is to follow.

######  Definition 2.

Let $\bf{B}:V\times V\to\mathbb{R}$ be a symmetric bilinear form on the vector
space $V$. The null space of $\bf{B}$, $N(\bf{B})$, is defined as follows:

$N(\bf{B})=\\{\bf{w}\in V:\bf{B}(\bf{w},\bf{v})=0,\ {\rm for\ all\ }\bf{v}\in
V\\}.$

The next lemma is a straightforward result from linear algebra.

######  Lemma 1.

Suppose $V$ is an $n$-dimensional vector space with inner product. If
$\dim(N({\bf{B}}))=n-1$, then the 1-dimensional subspace of $V$ orthogonal to
$N({\bf{B}})$ is the eigenspace of $\bf{B}$ with respect to that metric
associated to the nonzero eigenvalue of $\bf{B}$ with respect to that metric.

######  Corollary 1.

Let $X$ be a spherically symmetric space-time such that $N({\bf{T}}_{x})$
contains the spacial vectors of the Birkhoff frame, for every $x\in X$, then
$\bf{T}=0$

###### Proof.

Suppose the hypothesis are true. From the definition of the Birkhoff frame and
Lemma 1, we see that

$\bf{T}=-\mu_{0}\bar{\theta}_{0}\bar{\theta}^{0}.$

All of the conditions $(i),(ii)$ and $(iii)$ are satisfied since $\mu_{1}$ is
a constant function. Hence, $\mu_{0}=\mu_{0}(r)$ and (28) holds. Substituting
$0$ for $\mu_{1}$ and $\lambda$ in equation (28), we get

$-\frac{\mu_{0}}{4}+\frac{\mu_{0}}{4h}=0.$

Thus, on any open interval where $\mu_{0}\neq 0$, we must have $h=1$; that is

$c=2\int_{r_{*}}^{r}4\pi x^{2}\mu_{0}(x)dx.$

This is only possible if $\mu_{0}=0$. ∎

As a consequence of this corollary we have the following.

######  Corollary 2.

Birkhoff’s Theorem is still true for a spherically symmetric space-time $X$ if
we replace the condition that the space-time is a vacuum by the condition that
$N({\bf{T}}_{x})$ contains the spacial vectors of the Birkhoff frame, for
every $x\in X$.

We have yet one more corollary, but first a definition.

######  Definition 3.

Let $X$ be a space-time. We say that $X$ is dust, if $N({\bf{T}}_{x})$ is the
orthogonal complement of a time-like vector, for all $x\in X$.

An immediate consequence of the Observation and Corollary 1 is the well-known
result.

######  Corollary 3.

Suppose $X$ is a spherically symmetric space time. If $X$ is dust, then $X$
cannot be static.

##  6\. Asymptotics

For the space-times we are considering where the coordinate $r$ has geometric
meaning, we have the following definition.

Definition of Asymptotically Flat. A space-time $X$ which is spherically
symmetric is said to be asymptotically flat if the metric $\bar{\bf{g}}$
approaches a limiting value and the Riemann curvature tensor of $\bar{\bf{g}}$
approaches 0 as $r$ approaches $r_{1}$.

These conditions are certainly met by the metrics that arise in Birkhoff’s
Theorem.

Let’s elaborate on the conditions specified in the definition. That
$\bar{\bf{g}}$ has a limit as $r\to r_{1}$ means that $\lim_{r\to r_{1}}e^{2}$
and $\lim_{r\to r_{1}}g^{2}$ are positive reals. To say that the Riemann
curvature tensor approaches 0 means that all of its components with respect to
some orthonormal moving frame approach 0; thus the Riemann curvature tensor
approaches 0 as $r\to r_{1}$ if and only if

${\bar{\Theta}^{0}}\\!_{1}\to 0,\quad{\bar{\Psi}}^{2}\\!_{3}\to
0,\quad{\bar{\Omega}^{\alpha}}\\!_{\nu}\to 0\quad{\rm as}\quad r\to r_{1}.$

From the formulas for these curvature forms we see that these last limits take
the following form:

$K_{B}\to 0,\ \frac{1}{r^{2}}\left(1-\frac{1}{g^{2}}\right)\to 0,\
\frac{1}{rg^{2}}(\ln e)_{r}\to 0,\ \frac{1}{rg^{2}}(\ln g)_{r}\to 0\ {\rm as}\
r\to r_{1}.$

We will now characterize those static spherically symmetric space-times that
are asymptotically flat. However, we will do so under the physically
reasonable assumption that $\mu_{0}\geq 0$.

######  Theorem 3.

Assume $X$ is a static space-time with spherical spatial symmetry and

${\bf{T}}=-\mu_{0}\bar{\theta}_{0}\bar{\theta}^{0}+\mu_{1}\bar{\theta}_{1}\bar{\theta}^{1}+\lambda\bar{\psi}_{\nu}\bar{\psi}^{\nu},$

where $\mu_{0}\geq 0$. Then $X$ is asymptotically flat if and only if one of
the following occurs:

$\displaystyle{\rm(i)}$ $\displaystyle\ r_{1}<\infty,\
h(r)=1+\frac{2}{r}\int_{r}^{r_{1}}4\pi x^{2}\mu_{0}(x)dx,\ {\rm and}$
$\displaystyle\qquad\qquad\lim_{r\to r_{1}}\mu_{0}=\lim_{r\to
r_{1}}\mu_{1}=\lim_{r\to r_{1}}(\mu_{1})_{r}=0.$ $\displaystyle{\rm(ii)}$
$\displaystyle\ r_{1}=\infty,\lim_{r\to\infty}r^{2}\mu_{0}(r)=l,\ {\rm where\
}0\leq l<\frac{1}{8\pi},\ {\rm and}$
$\displaystyle\qquad\qquad\int_{r_{*}}^{\infty}x(\mu_{0}(x)+\mu_{1}(x))dx\
{\rm exists}.$

###### Proof.

First, let’s write the four limits given above in terms of $h$, $\mu$ andf
$\lambda$. We can do this by taking advantage of Einstein’s equation, more
explicitly, equations (18), (19) and (21). We get that the above limits hold,
and thus the Riemann curvature tensor approaches 0 as $r\to r_{1}$, if and
only if the following do:

(30) $\lim_{r\to r_{1}}\frac{1-h}{r^{2}}=0,\ \lim_{r\to r_{1}}\mu_{0}=0,\
\lim_{r\to r_{1}}\mu_{1}=0,\ \lim_{r\to r_{1}}\lambda=0$

Second, as a consequence of equation (28),

(31) $\lim_{r\to r_{1}}{\bf{T}}=0\Leftrightarrow\lim_{r\to
r_{1}}\mu_{\alpha}=\lim_{r\to r_{1}}\lambda=0\Leftrightarrow\lim_{r\to
r_{1}}\mu_{\alpha}=\lim_{r\to r_{1}}(\mu_{1})_{r}=0.$

Assume $X$ is asymptotically flat. Suppose $r_{1}<\infty$. Since the limit of
the metric exists, $\lim_{r\to r_{1}}h$ is positive real. As a consequence of
the first limit appearing in (30), $r_{1}<\infty$ implies $\lim_{r\to
r_{1}}h(r)=1$. Using equation (24), this implies

$c=2\int_{r_{*}}^{r_{1}}4\pi x^{2}\mu_{0}(x)dx.$

Consequently

$h(r)=1+\frac{2}{r}\int_{r}^{r_{1}}4\pi x^{2}\mu_{0}(x)dx.$

Since the Riemann curvature tensor approaches 0 as $r\to r_{1}$, it follows,
from (30) and (31), that the remainder to the conditions in (i) hold.

Now suppose that $r_{1}=\infty$. Thus, it must be true that
$\lim_{r\to\infty}h=n$, where $n>0$. According to equation (24) this implies

$1-8\pi\lim_{r\to\infty}\frac{1}{r}\int_{r_{*}}^{r}x^{2}\mu_{0}(x)dx=n.$

If we let

$l=\lim_{r\to\infty}\frac{1}{r}\int_{t_{*}}^{r}x^{2}\mu_{0}(x)dx,$

then $0\leq l<\frac{1}{8\pi}$, since we have assumed that $\mu_{0}\geq 0$. The
integral appearing in the above limit is an nondecreasing function of $r$.
Thus the integral is bounded for all $r$ or approaches $\infty$ as
$r\to\infty$. If it is bounded for all $r$ then
$\lim_{r\to\infty}r^{2}\mu_{0}(r)=0$. If the integral approaches $\infty$ as
$r\to\infty$, then we apply l’Hospital’s Rule to the limit and obtain
$\lim_{r\to\infty}r^{2}\mu_{0}(r)=l$. Either way,
$\lim_{r\to\infty}r^{2}\mu_{0}(r)=l$, where $0\leq l<\frac{1}{8\pi}$.

It must also be true that $\lim_{r\to\infty}p^{2}h$ is a positive real. Since
the $\lim_{r\to\infty}h$ exists and $p$ is an exponential

$\lim_{r\to\infty}\ln p=\lim_{r\to\infty}\int_{r_{*}}^{r}\frac{4\pi
x(\mu_{0}(x)+\mu_{1}(x))}{h}dx\ {\rm exists}.$

Again, since $\lim_{r\to\infty}h=n$,
$\int_{r_{*}}^{\infty}x(\mu_{0}(x)+\mu_{1}(x))dx$ must exist.

Now assume that (i) holds. Clearly $\lim_{r\to r_{1}}h=1$. Thus the first
limit in (30) holds. By (31) the remainder of the limits in (30) holds. Since
the limits of $h,\mu_{0}$ and $\mu_{1}$ all exist as $r\to r_{1}$, the
integral in the definition of $p$ is a proper integral even when the upper
limit is $r_{1}$. Hence, $\lim_{r\to r_{1}}p^{2}h$ exists. Indeed, $X$ is
asymptotically flat.

Finally, assume what is asserted in (ii) is true. It is not hard to show that
$\lim_{r\to\infty}h(r)$ is a positive real from the assertion that
$\lim_{r\to\infty}r^{2}\mu_{0}(r)=l$, where $0\leq l<\frac{1}{8\pi}$. It also
follows from this limit that the second limit in (30) holds. Since
$\int_{r_{*}}^{\infty}x(\mu_{0}(x)+\mu_{1}(x))dx$ exists and
$\lim_{x\to\infty}h$ exists, $\lim_{r\to\infty}\ln p$ exists and thus
$\lim_{r\to\infty}p^{2}h$ is a positive real. Moveover, that improper integral
given in (ii) exists implies that
$\lim_{r\to\infty}r(\mu_{0}(r)+\mu_{1}(r))=0$. But the existence of the first
limit in (ii) implies $\lim_{r\to\infty}r\mu_{0}(r)=0$. Thus
$\lim_{r\to\infty}r\mu_{1}(r)=0$. Hence, the third limit in (30) holds.
However, we can apply l’Hospital’s Rule to
$\lim_{r\to\infty}\frac{\mu_{1}(r)}{r^{-1}}=0$ to show that
$\lim_{r\to\infty}(\mu_{1})_{r}=0$. By (31), the fourth limit in (30) holds.
Thus $X$ is asymptotically flat. ∎

We consider the examples presented above. In Example 1, only the Minkowski
space-time is asymptotically flat. All the metrics in Example 2 are
asymptoticaly flat. The only space-times in Example 4 that is asymptotically
flat are those for which $m_{0}+m_{1}=0$.

For the space-times in Example 5 to be asymptotically flat we must require
that $\lim_{r\to\infty}r^{2}\rho(r)<\frac{1}{8\pi}$. (The assumption that
$m(r)<\frac{r}{2}$, for all $r>0$, would only imply
$\lim_{r\to\infty}r^{2}\rho(r)\leq\frac{1}{8\pi}$.) We must also require that
$\int_{r_{*}}^{\infty}x(\rho(x)+P(x))dx$ exist.

##  7\. Acknowledgement

I would like to thank Kevin Croker for his assistance revising the
introduction to this paper and his suggestions regarding the readability of
the paper in general.

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