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actually moving and so infer their three-dimensional motions. These observations
281
Appendix 11B: Supermassive black holes

imply that there exists a black hole of mass 2 5 — 106 M at the centre of our
Galaxy.
In a remarkable set of observations, a disc of H2 O masers has been detected in
the galaxy NGC 4258 using very long baseline interferometry (VLBI). The VLBI
observations measure the velocities of the masing clouds on scales of ∼ 0 3“2
parseconds and are well fitted by a thin (actually slightly warped) disc in circular
motion (see Figure 11.10). The mass of the central black hole is estimated to be
4 — 107 M .
Table 11.3 lists the masses of some potential supermassive black holes, with
a five-star rating. The masing disc of NGC 4258 gets a full five stars “ this is
the strongest observational evidence for a supermassive black hole. The stellar




Figure 11.10 The masing H2 O disc in the centre of NGC 4258. The lower
left-hand panel shows the variation in the line-of-sight velocity in km s’1 of
the material in the disc as a function of the distance along its major axis in
milliarcseconds. In the upper panel and lower right-hand panel the distance scales
are given in light years.
282 Schwarzschild black holes

Table 11.3 Potential supermassive black holes

Rating Source Evidence
Mbh /M

——— 2 — 109
M87 stars and optical disc
—— 1 — 109
NGC 3115 stars
—— 5 — 108
NGC 4594 (Sombrero) stars
—— 1 — 109
NGC 3377 stars
————— 4 — 107
NGC 4258 masing H2 O disc
—— 3 — 107
M31 (Andromeda) stars
—— 3 — 106
M32 stars
— — —— 2 5 — 106
Galactic centre stars and 3D motions


motions in the Galactic centre get four stars, though some astronomers might
argue that this evidence is so strong that it should rate five stars. Most of the other
observations are based on measurements of stellar-velocity dispersion. This is
fairly strong evidence but not completely convincing5 and so rates only two stars.


Appendix 11C: Conformal flatness of two-dimensional Riemannian
manifolds
Consider a general two-dimensional (pseudo-)Riemannian manifold in which the
points are labelled with some arbitrary coordinate system xa a = 1 2 . For any
such manifold to be conformally flat, we require that we can always find a
coordinate system x a in which the metric takes the form
gab x = 2
(11.33)
x ab

=
where 2 x is an arbitrary function of the new coordinates and ab
diag ±1 ±1 , the signs depending on the signature of the metric.
Suppose the primed coordinates are given by the transformation
1 2
x= x=
x1 x2 x1 x2
and
In order that (11.33) is satisfied, we thus require
12
= g ab = 0 (11.34)
g a b
11 22
“g = “ g ab = 0 (11.35)
g a b a b

where in the second equation the minus sign corresponds to the case where the
metric is positive- or negative-definite, and the plus sign corresponds to the case
where the metric is indeterminate.

5
The interpretation of velocity dispersion measurements requires some assumptions about the degree of velocity
anisotropy.
283
Exercises

It is straightforward to verify that (11.34) is satisfied identically if

= bc
ab g
a c

where is an arbitrary function of the coordinates and ab is the alternating
symbol, for which 11 = 22 = 0 and 12 = ’ 21 = 1. Moreover, substituting this
expression into (11.35), we find that
2
“1 g ab = 0
a b
g
where g = det gab . For a positive- or negative-definite metric the factor in square
brackets cannot be zero and, moreover, we can guarantee that g = 0. Thus, in this
case, we can satisfy our requirements by choosing 2 = g. For an indeterminate
metric, however, we must require that the above factor is zero, i.e. must not
be a null coordinate. In this case, we can again guarantee that g = 0, and so
we choose 2 = ’g. Thus we have shown explicitly that any two-dimensional
(pseudo-)Riemannian manifold is conformally flat.


Exercises
11.1 In the Schwarzschild geometry, we introduce the new coordinates

x = r sin cos y = r sin sin z = r cos

Find the form of the line element in these coordinates.
11.2 By introducing the new coordinate defined by
2
r= 1+
2
show that the line element for the Schwarzschild geometry can be written in the
isotropic form
’2
2 4
ds = c 1’ 1+ dt ’ 1 + + + sin2 d
2 2 2 2 2 2 2 2
d d
2 2 2
Show that g00 ≈ 1 ’ 2 / in the weak-field limit .
11.3 Show that the worldlines of radially moving photons in the Schwarzschild geometry
are given by
r
ct = r + 2 ln ’ 1 + constant (outgoing photon)
2
r
ct = ’r ’ 2 ln ’ 1 + constant (incoming photon)
2
284 Schwarzschild black holes

11.4 Show that, on introduction of the advanced Eddington“Finkelstein timelike coordi-
nate t = ct + 2 ln r/ 2 ’ 1 , the Schwarzschild line element takes the form

2 4c 2
2
ds2 = c2 1 ’ dt ’ dt dr ’ 1 + dr 2 ’ r 2 d + sin2 d
2 2
r r r
Hence show that the worldlines of radially moving photons in advanced Eddington“
Finkelstein coordinates are given by

r
ct = r + 4 ln ’ 1 + constant (outgoing photon)
2
ct = ’r + constant (incoming photon)

11.5 Show that, on introduction of the retarded Eddington“Finkelstein timelike coordinate
t— = ct + 2 ln r/ 2 ’ 1 , the Schwarzschild line element takes the form

2 4c — 2
dt— 2 +
ds2 = c2 1 ’ dt dr ’ 1 + dr 2 ’ r 2 d + sin2 d
2 2
r r r
Hence find the equations for the worldlines of radially moving photons in retarded
Eddington“Finkelstein coordinates. Use this result to sketch the spacetime diagram
showing the light-cone structure in this coordinate system.
11.6 A particle in the Schwarzschild geometry emits a radially outgoing photon at
coordinates tE rE , which is received by the distant fixed observer at tR rR .
Show that, if rE lies just outside the horizon r = 2 , the radial coordinate ˜seen™ by
the distant observer at the time tR is given by

rE tR ≈ 2 + 2 e’c tR ’rR /4



11.7 An observer sits on the surface of a star as it collapses to form a black hole. Once
an event horizon forms would the observer see any light from the star?
11.8 A spherical distribution of dust of coordinate radius R and total mass M collapses
from rest under its own gravity. Show that, as the collapse progresses, the coordinate
radius r of the star™s surface and the elapsed proper time of an observer sitting
on the surface are related by
1/2
r
1 r
r =’ dr
1 ’ r/R
1/2
2GM R


By making the substitution r = R cos2 /2 , or otherwise, show that the solution
can be expressed parametrically as
1/2
R R R
r= 1 + cos = + sin
2 2 2GM
Calculate the proper time experienced by the observer before the star collapses to
a point.
285
Exercises

11.9 A massive particle is released from rest at infinity in the Schwarzschild geometry.
Show that the covariant components of its subsequent 4-velocity at coordinate
radius r can be written as u = c2 T , where
’1
1/2
r 2 2
T = t+ 1’
1
dr
c
r r
Hence show that the line element of the Schwarzschild geometry in T r
coordinates is given by
2
ds2 = c2 dT 2 ’ dr + 2 c2 /r dT ’ r2 d + sin2 d
2 2



Is this new form singular at r = 2 ? What can you say about the hypersurface
T = constant? Show that observes infalling radially from rest at infinity have

T = 1 and hence give a physical interpretation of the T coordinate.
11.10 A massive particle is released from rest at coordinate radius r in the Schwarzschild
geometry. Show that a frame of orthonormal basis vectors defining the inertial
instantaneous rest frame of the particle may be taken as
’1/2 1/2
1 1 2 2
ˆ ˆ
= u= 1’ = 1’
e0 e1
0 1
c c r r
1 1
ˆ ˆ
= =
e2 e3
2 3
r sin
r
Hence show that the spatial components of the orthogonal connecting vector
between two such nearby particles satisfy
ˆ ˆ
ˆ
d2 2 c2 d2 c2 d2 c2
r
ˆ ˆ
ˆ
=+ 3 =’ 3 =’ 3
r
2 2 2
d r d r d r
11.11 Two compact masses, each of mass m, are connected by a light strong wire of
length l. The system is aligned in such a way that the two masses lie along a radial
line from a Schwarzschild black hole, and it is released from rest at coordinate
radius r. Obtain an expression for the tension in the wire immediately after the
system is released.
11.12 An astronaut, starting from rest at infinity, falls radially inwards towards a
Schwarzschild black hole with M = 105 M . Calculate the radial coordinate from
the centre of the black hole at which the astronaut first experiences a lateral tidal
force of 400 m s’2 m’1 and is therefore crushed. How does this radial coordinate
compare with the position of the event horizon?
11.13 An unpowered satellite is in radial free fall towards a Schwarzschild black hole.
Show that the principal stresses in the satellite are given by
2 c2 c2 c2
’ ’
r3 r3 r3
In what directions do these principal stresses act? Compare your answer with that
obtained in Exercise 11.10.
286 Schwarzschild black holes

11.14 An unpowered satellite follows a circular orbit of radius r around a Schwarzschild
black hole. Show that the principal stresses in the satellite are given by

c2 2r ’ 3 c2 r c2
’ ’
r3 r ’ 3 r3 r ’ 3 r3
In what directions do these principal stresses act?
11.15 Suppose that p and q are respectively the advanced and retarded Eddington“
Finkelstein time parameters, defined in terms of Schwarzschild coordinates by

r
p = ct + r + 2 ln ’1
2
r
q = ct ’ r ’ 2 ln ’1
2

A new set of (Kruskal) coordinates is defined by

ep/4 ’ e’q/4 ep/4 + e’q/4
v= u=
1 1
and
2 2

Show that these new coordinates are related to Schwarzschild coordinates for
r > 2 by
1/2
r r ct
v= ’1 exp sinh
2 4 4
1/2
r r ct
u= ’1 exp cosh
2 4 4

and for r < 2 by
1/2
r r ct
v = 1’ exp cosh
2 4 4
1/2
r r ct
u = 1’ exp sinh
2 4 4

11.16 Show that, in terms of the Kruskal coordinates u and v defined in Exercise 11.15,
the Schwarzschild line element takes the form
3
32 r
ds = exp ’ dv2 ’ du2 ’ r 2 d + sin2 d
2 2 2
2
r

where r is considered as a function of v and u and is defined implicitly by

r r
u2 ’ v 2 = ’ 1 exp
2 2

Show further that v is timelike and u is spacelike throughout the Schwarzschild
geometry.
287
Exercises

11.17 Perform an embedding into three-dimensional Euclidean space of the 2-space with
line element
= dr 2 + r 2 + a2 d
2 2
d

and hence show that the resulting 2-surface has a geometry reminiscent of a
wormhole.
11.18 By examining the paths of light rays in the Kruskal diagram, deduce that no
particle can pass through the Einstein“Rosen wormhole from region I to region I
or vice versa, before the throat of the wormhole pinches off.
11.19 A Schwarzschild black hole of mass M radiates as a blackbody of temperature
T = c3 / 8 kB GM . Show from first principles that the black hole has a lifetime
= M 3/ 3 , where = c4 / 15360 G2 . In its last second, calculate the total
energy radiated and estimate the typical energy of each radiated particle.
11.20 From observations of a compact binary system, one may calculate the mass
function
PK 3
fM =
2G
where P is the orbital period and K is the radial velocity amplitude. From Kepler™s
laws in Newtonian gravity, show that f M is related to the masses M1 and M2 of
the compact object and the companion star and the inclination angle i of the orbit
to the plane of the sky by
M1 sin3 i
3
fM =
M1 + M2 2
12
Further spherically symmetric geometries




In the preceding three chapters, we have considered in some detail the
Schwarzschild geometry, which represents the gravitational field outside a
static spherically symmetric object. We also considered the structure of the
Schwarzschild black hole, in which the empty-space field equations are satisfied
everywhere except at the central intrinsic singularity. In this chapter, we consider
solving the Einstein equations for a static spherically symmetric spacetime in
regions where the presence of other fields means that the energy“momentum
tensor is non-zero. In particular, we will concentrate on two physically inter-
esting situations. First, we discuss the relativistic gravitational equations for the
interior of a spherically symmetric matter distribution (or star); in this case the
energy“momentum tensor of the matter making up the star must be included in
the Einstein field equations. Second, we consider the spacetime geometry outside
a static spherically symmetric charged object; once again this is not a vacuum,
since it is filled with a static electric field whose energy“momentum must be
included in the field equations.


12.1 The form of the metric for a stellar interior
Most stars in the sky are nowhere near dense enough for general-relativistic effects
to be important in determining their structure. This is true for main sequence stars
(of which our Sun is an example), red giants and even such high-density objects
as white dwarfs. Thus, most stars will never even evolve into an object that is not
adequately described by the Newtonian theory of stellar structure.1 For neutron
stars, however, the extremely high densities involved (see Section 11.6) mean that
the internal gravitational forces will be very strong, and so we expect general-
relativistic effects to play a significant role in determining their structure and their
stability to collapse. As a result, it is of practical (as well as theoretical) interest

1
See, for example, S. Chandrasekhar, An Introduction to the Study of Stellar Structure, Dover, 1958.


288
289
12.1 The form of the metric for a stellar interior

to consider the relativistic equations governing the equilibrium of a centrally
symmetric self-gravitating distribution of matter.
Since we are assuming spherical symmetry and a static matter distribution, the
appropriate general form of the metric is that derived in Section 9.1, namely

ds2 = A r dt2 ’ B r dr 2 ’ r 2 d + sin2 d
2 2
(12.1)

As in our derivation of the Schwarzschild metric, the two functions A r and B r are
determined by solving the Einstein equations. For our present discussion, however,
we shall not solve the empty-space field equations R = 0, which are valid outside
the spherical object, but instead solve the full field equations that hold in the
interior of the object. These are most conveniently written in the form (8.15), namely

=’ ’ 2 Tg
1
(12.2)
R T

where T is the energy“momentum tensor of the matter of which the object is
composed, T ≡ T and = 8 G/c4 . For the discussion in this chapter we will
assume the matter to be described by a perfect fluid, so that
p
T = + 2 u u ’ pg (12.3)
c
where r is the proper mass density and p r is the isotropic pressure in the
instantaneous rest frame of the fluid, both of which may be taken as functions
only of the radial coordinate r for a static matter distribution. Using the fact that
u u = c2 , we find that
p2
T= + c ’p = c2 ’ 3p
c2
and so the field equations (12.2) read
p
=’ + u u ’2 c2 ’ p g
1
(12.4)
R 2
c
As shown in Section 9.2, the off-diagonal components of the Ricci tensor R
for the metric (12.1) are all zero and the diagonal components are given by
A A A B A
R00 = ’ + + ’ (12.5)
2B 4B A B rB
A A A B B
R11 = ’ + ’ (12.6)
2A 4A A B rB
1 r A B
R22 = ’1+ ’ (12.7)
2B
B A B
R33 = R22 sin2 (12.8)
290 Further spherically symmetric geometries

It is of interest first to determine the consequences of the vanishing off-diagonal
components of the Ricci tensor, R0i = 0 for i = 1 2 3. From the field equa-
tions (12.4), and using the fact that g0i = 0, we see immediately that we require
ui u0 = 0. Combining this with u u = c2 , we find that the covariant components
of the fluid 4-velocity are given by

=c A 1 0 0 0 (12.9)
u

and thus the spatial 3-velocity of the fluid must vanish everywhere. In particular,
we note that this conclusion holds without our assuming in advance that the
proper density and the pressure p are independent of t. Thus, the fact the metric
(12.1) is independent of t automatically implies that the matter distribution itself
is static and so the object is in a state of hydrostatic equilibrium. This is another
illustration how the equations of motion for matter follow directly from the field
equations (see Section 8.8).
= components of the field equations (12.2)
Let us now use the diagonal
to obtain the differential equations that the functions A r and B r must satisfy.
Inserting the expression (12.3) into the right-hand side of the field equations and
using the metric (12.1), we find that

R00 = ’ 2 c2 + 3p A
1
(12.10)
R11 = ’ 2 c2 ’ p B
1
(12.11)
R22 = ’ 2 c2 ’ p r 2
1
(12.12)
R33 = R22 sin2 (12.13)

From these equations, one quickly obtains

R00 R11 2R22
+ + 2 = ’2 c2
A B r

On substituting the expressions (12.5“12.8) for the Ricci tensor components and
simplifying, one finds that

1 rB
1’ + 2 = r 2 c2 (12.14)
B B

which can be rewritten in the form

1
d
r 1’ = r 2 c2
dr B
291
12.1 The form of the metric for a stellar interior

Integrating this expression with respect to r, and noting that the associated constant
of integration must be zero in order for B r to be non-zero at the origin (as
demanded by (12.14)), we find that the solution for B r is given by

’1
2Gm r
B r = 1’ (12.15)
c2 r

where we have defined the function
r
m r =4 ¯¯ r
r r 2 d¯ (12.16)
0
This function is worthy of further comment, since it has the appearance of being
the mass contained within a coordinate radius r. This interpretation is not quite
correct, however, since the proper spatial volume element for the metric (12.1) is
given by
d3 V = B r r 2 sin dr d d
Thus the proper integrated ˜mass™ (i.e. energy/c2 ) within a coordinate radius r is
’1/2
¯
r r 2Gm r
m r =4 ¯ B r r d¯ = 4
¯¯ r ¯ 1’ ¯r
2
r 2 d¯
r r
¯
c2 r
0 0

Nevertheless, we note that it is m r , not m r , that appears in the radial metric
coefficient B r in (12.15). In particular, if the object extends to a coordinate
radius r = R, beyond which there is empty space, then the spacetime geometry
outside this radius is described by the Schwarzschild metric with mass parameter
M = m R , rather than M = m R . The difference E = M ’ M corresponds to the
gravitational binding energy of the object, which is the amount of energy required
to disperse the material of which the object consists to infinite spatial separation.
We now turn to determining the differential equation that must be satisfied by
the function A r in (12.1). In principle, this could be obtained by substituting for
B r using (12.15) in any of the equations (12.10“12.13). It is more convenient and
instructive, however, to use the conservation equation T = 0 directly, from
which the fluid equations of motion may be derived (as discussed in Section 8.3).
Using (12.3), we may write
p
T= + 2 uu ’ pg
c
1 p p
=√ + 2 uu + + 2 u u ’g p (12.17)
’g c c
where, in going from the first to the second line, the first term has been rewritten
using the expression (8.24) for the covariant divergence and the second term
g = 0 and that p is a scalar function.
has been manipulated by noting that
292 Further spherically symmetric geometries

From (12.9), however, we have u0 = c/ A and ui = 0, and since and p do not
depend on t the first term on the right-hand side of (12.17) must vanish. For the
same reason, the second term becomes equal to c2 + p 00 /A. From (3.21)
and (12.1), we have
= ’2g g00 = ’ 2 g
1 1
A
00

= 0 can be written as
and so the conservation equation T

c2 + p
A+g p=0
g
2A
Multiplying through by g and simplifying, one obtains
c2 + p
A+ p = 0 (12.18)
2A
Since A is a function only of r, the above equation is trivial for = 0, in which
case we recover the fact that p is independent of t. Similarly, for = 2 and
= 3 we find that the corresponding (tangential) derivatives of p also vanish,
as dictated by spherical symmetry. For = 1, however, the relation (12.18) is
non-trivial and reads (where primes denote d/dr)

2p
A
=’ 2 (12.19)
c +p
A

which gives a differential equation, in terms of r and p r , that A r must
satisfy in hydrostatic equilibrium.


12.2 The relativistic equations of stellar structure
The equations (12.15) and (12.19) show how to calculate the functions A r and
B r in the metric (12.1), given particular functions of r and p r . Specifying
these two functions does, however, imply an equation of state p = p by
elimination of r, and this is likely to be physically unrealistic for arbitrary choices
of r and p r . For astrophysical investigations, one is more interested in
building models of the density and pressure distribution inside a star under the
assumption of some (quasi-)realistic equation of state. Thus, it is usual to recast
the results obtained in the previous section into an alternative form.
In this approach, the first equation of stellar structure is taken simply from
(12.16) and written as
dm r
= 4 r2 r (12.20)
dr
293
12.2 The relativistic equations of stellar structure

which clearly relates the functions m r and r . The next step is to obtain an
equation linking m r and p r . This is most conveniently achieved by using
(12.7) and (12.12):
1 1
r A B
’1+ ’ =’ c2 ’ p r 2
2B 2
B A B
Eliminating the functions A and B using (12.19) and (12.15) and simplifying, one
obtains the second equation of stellar structure,

’1
1 4 G 3 Gm r 2Gm r
dp
=’ 2 c +p pr + 1’
2
(12.21)
c4 c2 c2 r
dr r

which is also known as the Oppenheimer“Volkoff equation. As mentioned above,
to obtain a closed system of equations we need to define the equation of state for
the matter, which gives the pressure in terms of the density, namely

p=p (12.22)

This provides the third (and final) equation of stellar structure. We note that, for
many astrophysical systems, the matter obeys a polytropic equation of state of
the form p = K , where K and are both constants. In the usual notation used
in this field, = 1 + 1/n, where n is known as the polytropic index.
The closed system of three equations (12.20“12.22) contains two coupled
first-order differential equations, and so to obtain a unique solution one must
specify two boundary conditions. The first is straightforward, since we must have
m 0 = 0, leaving just one further adjustable boundary condition to be specified.
It is most common to choose this adjustable parameter to be the central pressure
p 0 , or equivalently the central density 0 , which can be obtained immediately
from the equation of state (12.22). Very few exact solutions are known for real-
istic equations of state, and so in practice the system of equations is integrated
numerically on a computer. The procedure is to ˜integrate outwards™ from r = 0
(in practice in small radial steps of size r) until the pressure drops to zero. This
condition defines the surface r = R of the star, since otherwise there would be
an infinite pressure gradient, and hence an infinite force, on the material elements
constituting the outer layer of the star. For r > R r and p r are both zero and
m r = m R = M, and the spacetime geometry is described by the Schwarzschild
metric with mass parameter M.
Before looking for particular solutions to the set of equations (12.20“12.22), it is
worthwhile considering briefly their Newtonian limit. In fact, the forms of (12.20)
and (12.22) remain unchanged in this limit, and it is only the equation (12.21) for
the pressure gradient that is simplified. In the Newtonian limit we have p and
294 Further spherically symmetric geometries

therefore 4 r 3 p mc2 . Moreover, we require the metric (12.1) to be close to
Minkowski, and so we require 2Gm/ c2 r 1. Thus, the Oppenheimer“Volkoff
equation reduces to
dp Gm r r
=’ (12.23)
r2
dr
which is simply the Newtonian equation for hydrostatic equilibrium. Comparing
(12.23) with (12.21), we see that all the relativistic effects serve to steepen the
pressure gradient relative to the Newtonian case. Thus, for an object to remain
in hydrostatic equilibrium, the fluid of which it consists must experience stronger
internal forces when general-relativistic effects are taken into account.


12.3 The Schwarzschild constant-density interior solution
The simplest analytic interior solution for a relativistic star is obtained by making
the assumption that, throughout the star,

= constant

which constitutes an equation of state. There is no physical justification for this
assumption, but it is on the borderline of being realistic. It corresponds to an
ultra-stiff equation of state that represents an incompressible fluid. Consequently,
the speed of sound in the fluid, which is proportional to dp/d 1/2 , is infinite
(which is clearly not allowed relativistically). Nevertheless, it is believed that the
interiors of dense neutron stars are of nearly uniform density, and so this simple
case is of some practical interest.
Equation (12.20) immediately integrates to give

for r ¤ R
4
r3
mr = 3 (12.24)
R3 ≡ M
4
for r > R
3


where R is the radius of the star, as yet undetermined, and M is the mass parameter
for the Schwarzschild metric describing the spacetime geometry outside the star.
Moreover, the Oppenheimer“Volkoff equation (12.21) becomes
’1
4G 8G 2
dp
= ’ 4 r c2 + p c + 3p 1’
2
r
3c2
3c
dr
This equation is separable and we may write
¯ ¯r
pr r
4G
dp r d¯
=’
c2 + p
¯ c2 + 3p
¯ 1 ’ 8 G r 2 / 3c2
¯
3c4 0
p0
295
12.3 The Schwarzschild constant-density interior solution

where p0 = p 0 is the central pressure of the star. Performing these standard
integrals, one finds that
1/2
c2 + 3p c2 + 3p0 8G 2
= 1’ (12.25)
r
c2 + p c2 + p0 3c2
At the surface r = R of the star, the pressure p is zero and so the left-hand side
of the above equation equals unity. Thus, we obtain
2
c2 + p0
3c2
R= 1’
2
c2 + 3p0
8G

which gives the radius of a star of uniform density with a central pressure p0 .
Alternatively, we may rearrange this result and use (12.24) to obtain a useful
expression for the central pressure,

1 ’ 1 ’ 2 /R 1/2
p0 = c2
(12.26)
3 1 ’ 2 /R 1/2 ’ 1

= GM/c2 . Using this expression to replace p0 in (12.25) gives
where

1 ’ 2 r 2 /R3 1/2 ’ 1 ’ 2 /R 1/2
pr = c for r ¤ R
2
(12.27)
3 1 ’ 2 /R 1/2 ’ 1 ’ 2 r 2 /R3 1/2

To obtain the complete solution to the problem, it remains to determine the
functions A r and B r in the metric (12.1). From (12.15) and (12.24), we
immediately find that
’1
2 r2
B r = 1’ 3 (12.28)
R
In particular, we note that at the star™s surface, where r = R, the above solution
matches with the corresponding expression from the Schwarzschild metric for
the exterior solution. The function A r is obtained from (12.19), (12.24) and
(12.27). One may fix the integration constant arising from (12.19) by imposing
the boundary condition that A r matches the corresponding expression in the
Schwarzschild metric at r = R. One then finds
1/2 2
1/2
c2 2 r2
2
Ar = 3 1’ ’ 1’ 3 (12.29)
4 R R

The expressions (12.28) and (12.29) constitute Schwarzschild™s interior solution
for a constant-density object.
296 Further spherically symmetric geometries

12.4 Buchdahl™s theorem
The most important feature of the Schwarzschild constant-density solution
discussed above is that it imposes a constraint connecting the star™s ˜mass™ M
and its (coordinate) radius R. To derive this constraint, one notices that (12.26)
implies that p0 ’ as /R ’ 4/9. Since pressure is a general scalar, this
infinity will persist in any coordinate system, and so one can only avoid this
behaviour by demanding that

4
GM
(12.30)
<
c2 R 9

Although we have only shown that this constraint holds for an object of constant
density, Buchdahl™s theorem states that (12.30) is in fact valid for any equation
of state. This theorem can be proved directly from the Einstein equations but
requires considerable care and lies outside the scope of our discussion.
Equation (12.30) can be regarded as providing an upper limit on the mass of
a star for a fixed radius. If one attempts to pack more mass inside R than is
allowed by (12.30), general relativity admits no static solution: the hydrostatic
equilibrium is destroyed by the increased gravitational attraction. Such a star must
therefore collapse inwards without stopping. Throughout the collapse, the exterior
geometry is described by the Schwarzschild metric, and so eventually one obtains
a Schwarzschild black hole. The limit (12.30) is, in fact, quite easily reached. For
example, the density of a neutron star is around 1016 kg m’3 and, assuming it to
be of uniform density, we find from (12.30) and (12.24) that M < 7 — 1031 kg.
This is approximately 35 solar masses, which is of same order as the most massive
stars in our Galaxy.



12.5 The metric outside a spherically symmetric charged mass
We now turn to our second physical application, namely the form of the metric
outside a static spherically symmetric charged body. The exterior of such an
object is not a vacuum, since it is filled with a static electric field. We must
therefore once again solve the Einstein field equations for a static spherically
symmetric spacetime in the presence of a non-zero energy“momentum tensor,
this time representing the electromagnetic field of the object.
Since we are assuming spherical symmetry and a static object, the general form
of the metric is once more given by (12.1). The two functions A r and B r
are determined by solving the full Einstein equations outside the spherical object;
these equations are again most conveniently written in the form (12.2). In this
297
12.5 The metric outside a spherically symmetric charged mass

case, however, T is the energy“momentum tensor of the electromagnetic field
of the charged object, which from Exercise 8.3 has the general form

’1
=’ ’ 4g F F
1
(12.31)
T FF
0


where F = A ’ A is the electromagnetic field strength tensor and A is the
electromagnetic 4-potential. The first point to note about this energy“momentum
tensor is that it has zero trace,

’1
T ≡T =’ ’4 =0
1
FF FF
0


Thus, in this case, the Einstein field equations (12.2) take the simplified form

=’ T (12.32)
R

In addition to the Einstein field equations, our solution must also satisfy the
Maxwell equations. In the region outside the charged object, the 4-current density
j is zero and so the Maxwell equations read

=0 (12.33)
F
+ + =0 (12.34)
F F F

The Einstein and Maxwell equations are coupled together, since F enters the
gravitational field equations through the energy“momentum tensor (12.31) and
the metric g enters the electromagnetic field equations through the covariant
derivative.
The constraint imposed on the metric coefficients g (or gravitational fields)
by requiring the solution to be spherically symmetric and static is embodied in the
choice of line element (12.1). We thus begin by considering the corresponding
consequences of these symmetry constraints for the form of the electromagnetic
field. In this case, the electromagnetic 4-potential in t r coordinates takes
the form
r
= ar 00 (12.35)
A
c2

where r and a r depend only on r and may be interpreted respectively as
the electrostatic potential and the radial component of the 3-vector potential as
r’ (the extra factor of 1/c multiplying r in (12.35), as compared with
the usual form in Minkowski coordinates, is a result of taking x0 = t rather than
x0 = ct; also, note that the 3-vector potential a r should not be confused with
298 Further spherically symmetric geometries

the function A r in the metric (12.1)). From (12.35), the field-strength tensor has
the form
⎛ ⎞
0 ’1 0 0
⎜1 0 0 0⎟
⎜ ⎟
F =E r ⎜ ⎟ (12.36)
⎝0 0 0 0⎠
0 000
where E r is an arbitrary function of r only and may be interpreted as the radial
component of the static electric field as r ’ . Thus our task is to use the
Einstein equations (12.32) and the Maxwell equations (12.33“12.34) to determine
the three unknown functions A r , B r and E r .
Let us begin by using the Maxwell equations. As discussed in Section 6.6, the
equations (12.34) are automatically satisfied by the definition of F . Moreover,
from Exercise 4.10, since F is antisymmetric we may rewrite the covariant
divergence in the first Maxwell equation (12.33) to obtain

1
=√ ’gF =0 (12.37)
F
’g
where g is the determinant of the metric. For a diagonal line element such as
(12.1), the determinant is simply the product of the diagonal elements, so that
g = ’A r B r r 4 sin2 . Given the form of F in (12.36), the expression (12.37)
yields the single equation

ABr 2 F 10 = 0
1

Writing the required contravariant component as F 10 = g 1 g 0 F = g 11 g 00 F10 =
’E/ AB , we thus obtain the equation
r 2E
d
=0

dr AB
This integrates to give
k ArBr
Er = (12.38)
r2
where k is a constant of integration. If we make the assumption that the metric is
asymptotically flat then A r ’ c2 and B r ’ 1 as r ’ . Identifying E r with
the radial electric field component at infinity, we thus require k = Q/ 4 0 c ,
where Q is the total charge of the object.
We now turn to the Einstein equations (12.32). The Ricci tensor components
for the metric (12.1) are given in (12.5“12.8), and the form of the electromagnetic
field energy“momentum tensor T may be found by substituting the form (12.36)
for F into the expression (12.31). On performing this substitution, one quickly
299
12.5 The metric outside a spherically symmetric charged mass

finds that the off-diagonal components of T are zero, and so the Einstein
equations for = are satisfied identically. For the diagonal components of the
Einstein equations, one finds

R00 = ’ 2 c2 0 E 2 /B
1
(12.39)
R11 = c2 0 E 2 /A
1
(12.40)
2

R22 = ’ 2 c2 0 r 2 E 2 / AB
1
(12.41)
R33 = R22 sin2 (12.42)

where we have used the facts that F0 1 = g 11 F01 = E/B and F1 0 = g 00 F10 = E/A;
we have also made use of the relation 0 0 = 1/c2 . From (12.39) and (12.40),
we immediately obtain

BR00 + AR11 = 0

On substituting the expressions (12.5, 12.6) for the Ricci tensor coefficients and
rearranging, this yields

A B+B A = 0

which implies that AB = constant. We may fix this constant from the requirement
that the metric is asymptotically flat as r ’ , and so we have

A r B r = c2 (12.43)

A further independent equation may be obtained from the 22-component (12.41)
of the Einstein equations. Inserting the expression (12.7) for the Ricci tensor
component and using (12.38), one finds that

GQ2
A + rA = c 1’
2
4 0 c4 r 2

Noting that A + rA = rA and integrating, one thus obtains

GQ2
2GM
A r =c 1’ 2 +
2
4 0 c4 r 2
cr

where we have identified the integration constant as ’2GM/c2 , M being the mass
of the object, since the line element must reduce to the Schwarzschild case when
Q = 0. The solutions for B r and E r are then found immediately from (12.43)
and (12.38) respectively.
300 Further spherically symmetric geometries

Thus, collecting our results together and defining the constants = GM/c2 and
q 2 = GQ2 / 4 0 c4 , the line element for the spacetime outside a static spherically
symmetric body of mass M and charge Q has the form

’1
q2 q2
2 2
ds = c 1’ + 2 dt ’ 1 ’ +2 dr 2 ’ r 2 d + sin2 d
2 2 2 2 2
r r r r

(12.44)

from which one may read off the metric coefficients g that determine the
gravitational field of the object. The resulting solution is known as the Reissner“
Nordström geometry. The electromagnetic F of the field of the object is given
by (12.36) with

Q
Er = 2
4 0r




12.6 The Reissner“Nordström geometry: charged black holes
The Reissner“Nordström (RN) metric (12.44) is only valid down to the surface of
the charged object. As in our discussion of the Schwarzschild solution, however,
it is of interest to consider the structure of the full RN geometry, namely the
solution to the coupled Einstein“Maxwell field equations for a charged point mass
located at the origin r = 0, in which case the RN metric is valid for all positive r.
Calculation of the invariant curvature scalar R shows that the only
R
intrinsic singularity in the RN metric occurs at r = 0. In the ˜Schwarzschild-
like™ coordinates t r , however, the RN metric also possesses a coordinate
singularity wherever r satisfies

q2
2
r ≡ 1’ + 2 =0 (12.45)
r r
with r = ’1/g11 r = g00 r /c2 . Multiplying (12.45) through by r 2 and solving
the resulting quadratic equation, we find that the coordinate singularities occur
on the surfaces r = r± , where

r± = ± ’ q2
2 1/2
(12.46)

It is clear that there exist three distinct cases, depending on the relative values of
2 and q 2 ; we now discuss these in turn.
301
12.6 The Reissner“Nordström geometry: charged black holes

• Case 1: 2 < q 2 In this case r± are both imaginary, and so no coordinate singularities
exist. The metric is therefore regular for all positive values of r. Since the function r
always remains positive, the coordinate t is always timelike and r is always spacelike.
Thus, the intrinsic singularity at r = 0 is a timelike line, as opposed to a spacelike line
in the Schwarzschild case. This means that the singularity does not necessarily lie in
the future of timelike trajectories and so, in principle, can be avoided. In the absence
of any event horizons, however, r = 0 is a naked singularity, which is visible to the
outside world. The physical consequences of a naked singularity, such as the existence
of closed timelike curves, appear so extreme that Penrose has suggested the existence
of a cosmic censorship hypothesis, which would only allow singularities that are hidden
behind an event horizon. As a result, the case 2 < q 2 is not considered physically
realistic.

• Case 2: 2 > q 2 In this case, r± are both real and so there exist two coordinate
singularities, occurring on the surfaces r = r± . The situation at r = r+ is very similar to
the Schwarzschild case at r = 2 . For r > r+ , the function r is positive and so the
coordinates t and r are timelike and spacelike respectively. In the region r’ < r < r+ ,
however, r becomes negative and so the physical natures of the coordinates t and
r are interchanged. Thus, a massive particle or photon that enters the surface r = r+
from outside must necessarily move in the direction of decreasing r, and thus r = r+
is an event horizon. The major difference from the Schwarzschild geometry is that
the irreversible infall of the particle need only continue to the surface r = r’ , since
for r < r’ the function r is again positive and so t and r recover their timelike
and spacelike properties. Within r = r’ , one may (with a rocket engine) move in
the direction of either positive or negative r, or stand still. Thus, one may avoid the
intrinsic singularity at r = 0, which is consistent with the fact that r = 0 is a timelike
line. Perhaps even more astonishing is what happens if one then chooses to travel
back in the direction of positive r in the region r < r’ . On performing a maximal
analytic extension of the RN geometry, in analogy with the Kruskal extension for the
Schwarzschild geometry discussed in Section 11.9, one finds that one may re-cross
the surface r = r’ , but this time from the inside. Once again one is moving from a
region in which r is spacelike to a region in which it is timelike, but this time the
sense is reversed and one is forced to move in the direction of increasing r. Thus
r = r’ acts as an ˜inside-out™ event horizon. Moreover, one is eventually forceably
ejected from the surface r = r+ but, according to the maximum analytic extension, the
particle emerges into a asymptotically flat spacetime different from that from which it
first entered the black hole. As discussed in Section 11.9, however, such matters are
at best highly speculative, and we shall not pursue them further here.

• Case 3: 2 = q 2 In this case, called the extreme Reissner“Nordström black hole, the
function r is positive everywhere except at r = , where it equals zero. Thus, the
coordinate r is everywhere spacelike except at r = , where it becomes null, and hence
r = is an event horizon. The extreme case is basically the same as that considered
in case 2, but with the region r’ < r < r+ removed.
302 Further spherically symmetric geometries

We may illustrate the properties of the RN spacetime in more detail by considering
the paths of photons and massive particles in the geometry, which we now go on
to discuss. Since the case 2 > q 2 is the most physically reasonable RN spacetime,
we shall restrict our discussion to this situation.



12.7 Radial photon trajectories in the RN geometry
Let us begin by investigating the paths of radially incoming and outgoing photons
in the RN metric for the case 2 > q 2 . Since ds = d = d = 0 for a radially
moving photon, we have immediately from (12.44) that

’1
q2 r2
1 2 1
dt
= ± = 1’ +2 =± (12.47)
c r ’ r’ r ’ r+
dr c r r

where, in the second equality, we have used the result (12.46); the plus sign
corresponds to an outgoing photon and the minus sign to an incoming photon. On
integrating, we obtain

2
2 r+
r’ r r
ct = r ’ ’1 + ’ 1 + constant
ln ln outgoing
r+ ’ r’ r+ ’ r’
r’ r+
2
2 r+
r’ r r
ct = ’r + ’1 ’ ’ 1 + constant
ln ln ingoing
r+ ’ r’ r+ ’ r’
r’ r+

We will concentrate in particular on the ingoing radial photons. To develop a
better description of infalling particles in general, we may construct the equivalent
of the advanced Eddington“Finkelstein coordinates derived for the Schwarzschild
metric in Section 11.5. Once again this coordinate system is based on radially
infalling photons, and the trick is to use the integration constant as the new
coordinate, which we denote by p. As before, p is a null coordinate and it is more
convenient to work instead with the timelike coordinate t defined by ct = p ’ r.
Thus, we have
2
2 r+
r’ r r
ct = ct ’ ’1 + ’1
ln ln (12.48)
r+ ’ r’ r+ ’ r’
r’ r+

On differentiating, or from (12.47) directly, one obtains

1
c dt = dp ’ dr = c dt + ’ 1 dr (12.49)
r
303
12.7 Radial photon trajectories in the RN geometry

where r is defined in (12.45). Using the above expression to substitute for c
in (12.44), one quickly finds that

ds2 = c2 dt 2 ’ 2 1 ’ dt dr ’ 2 ’ dr 2 ’ r 2 d + sin2 d
2 2



which is the RN metric in advanced Eddington“Finkelstein coordinates. In partic-
ular, we note that this form is regular for all positive values of r and has an
instrinsic singularity at r = 0.
From (12.47) and (12.49), one finds that, in advanced Eddington“Finkelstein
coordinates, the equation for ingoing radial photon trajectories is

ct + r = constant (12.50)

whereas the trajectories for outgoing radial photons satisfy the differential equation
2’
dt
= (12.51)
c
dr


Event horizon Event horizon

r = r“ r = r+


III II I




r=0




Figure 12.1 Spacetime diagram of the Reissner“Nordström solution in advanced
Eddington“Finkelstein coordinates. The straight diagonal lines are ingoing
photon worldlines whereas the curved lines correspond to outgoing photon world-
lines.
304 Further spherically symmetric geometries

We may use these equations to determine the light-cone structure of the RN metric
in these coordinates. For ingoing radial photons, the trajectories (12.50) are simply
straight lines at 45 in a spacetime diagram. For outgoing radial photons, (12.51)
gives the gradient of the trajectory at any point in the spacetime diagram, and so
one may sketch these without solving (12.51) explicitly. This resulting spacetime
diagram is shown in Figure 12.1. It is worth noting that the light-cone structure
depicted confirms the nature of the event horizon at r = r+ . Moreover, the light-
cones remain tilted over in the region r’ < r < r+ , indicating that any particle
falling into this region must move inwards until it reaches r = r’ . Once in the
region r < r’ , the lightcones are no longer tilted and so particles need not fall into
the singularity r = 0. As was the case in Section 11.5 for the Schwarzschild metric,
however, this spacetime diagram may be somewhat misleading. For an outward-
moving particle in the region r < r’ , Figure 12.1 suggests that it can only reach
r = r’ asymptotically, but by peforming an analytic extension of the RN solution
one can show that the particle can cross the surface r = r’ in finite proper time.


12.8 Radial massive particle trajectories in the RN geometry
We now consider the trajectories of radially moving massive particles for the
case 2 > q 2 . To simplify our discussion, we will assume that the particles are
electrically neutral. In this case, the particles will follow geodesics. In the more
general case of an electrically charged particle, one must also take into account
the Lorenz force on the particle produced by the electromagnetic field of the black
hole. The equation of motion for the particle is then given by (6.13).
For a radially moving particle, the 4-velocity has the form
™™
= u0 u1 0 0 = t r 0 0
u
where the dots denote differentiation with respect to the proper time of the
particle. The geodesic equations of motion, obeyed by neutral particles in the RN
metric, are most conveniently written in the form (3.56):
u=
™ 1
g uu
2

Since the metric coefficients in the RN line element (12.44) do not depend on t,
we immediately obtain

u0 = g00 t = constant

The radial equation of motion may then be obtained using the normalisation
condition g u u = c2 , which gives

g00 u0 2 + g11 u1 = c2
2
305
Exercises

c2∆(r)




u0 > c2


u0 = c2
c2

u0 < c2
r
r“ r+




Figure 12.2 The limits of radial motion for a neutral massive particle in the
Reissner“Nordström geometry.


r = g00 /c2 = ’1/g11 , one finds that
Using the fact, from (12.45), that

u2
r +c
™ r= 2
2 2 0
(12.52)
c

This clearly has the form of an ˜energy™ equation, in which c2 r plays the role
of a potential. Qualitative information on the properties of the radial trajectories
can be obtained directly from (12.52) by simply plotting the function c2 r ;
this plot is shown in Figure 12.2. The radial limits of the motion depend on
the choice of the constant u0 , as indicated. The case u0 = c2 corresponds to the
particle being released from rest at infinity. In all cases, there exists an inner radial
limit that is greater than zero. This indicates that a neutral particle moving freely
under gravity cannot reach the central intrinisic singularity at r = 0 but is instead
repelled once it has approached to within some finite distance. As mentioned in
Section 12.6. performing a maximum analytic extension the RN metric suggests
that the particle passes back through r = r’ and r = r+ and ultimately emerges
in a different asymptotically flat spacetime.


Exercises
12.1 For a general static diagonal metric, show that the 4-velocity of a perfect fluid in
the spacetime must have the form
c
u =√ 1000
g00
306 Further spherically symmetric geometries

12.2 Calculate the gravitational binding energy E = M ’ M of a spherical star of constant
density and coordinate radius R. Compare your answer with the corresponding
Newtonian result and interpret your findings physically.
12.3 Derive the Oppenheimer“Volkoff equation from the Einstein equations for a static
spherically symmetric perfect-fluid distribution, and show that it reduces to the
standard equation for hydrostatic equilibrium in the Newtonian limit.
12.4 In Newtonian gravity, show directly that the equation for hydrostatic equilibrium is

dp r Gm r r
=’
r2
dr
12.5 Show that, in the Newtonian limit, the equation before (12.15) reduces to

d r Gm r
=
dr r
where r is the Newtonian gravitational potential.
12.6 For a spherical star of uniform density and central pressure p0 , verify that the
Oppenheimer“Volkoff equation requires p r to satisfy
1/2
c2 + 3p r c2 + 3p0 8G 2
= 1’ r
c2 + p r c 2 + p0 3c2

and hence show that
1 ’ 2 r 2 /R3 1/2 ’ 1 ’ 2 /R 1/2
pr = c 2
3 1 ’ 2 /R 1/2 ’ 1 ’ 2 r 2 /R3 1/2

where R is the coordinate radius of the star.
12.7 In Newtonian gravity, obtain the expression for p r for a spherical star of uniform
density , central pressure p0 and radius R. Compare your result with that obtained
in Exercise 12.6.
12.8 Show that, for a spherical star of uniform density ,

c2 16c6
2 2
and
R< M<
243 G3
3G
If a photon is emitted from the star™s surface and received by a stationary observer
at infinity, show that the observed redshift must obey the constraint z < 2. Show
also, however, that the observed redshift for a photon emitted from the star™s centre
can be arbitrarily large.
12.9 For a spherical star of uniform density , show that in order for the star not to lie
within its own Schwarzschild radius, one requires

3c6
2
M<
32 G3
Compare this limit with that derived in Exercise 12.8.
307
Exercises

12.10 For a spherical uniform-density star of mass M and coordinate radius R, show that
the line element of spatial sections with t = constant can be written in the form

Rc2
= + sin2 + sin2 d
2 2 2 2
d d d
2GM
12.11 Consider a static infinitely long cylindrical configuration of matter that is invariant
to translations and Lorentz boosts along the axis of symmetry (a cosmic string).
Adopting ˜cylindrical polar™ coordinates ct r z , show that a self-consistent
solution to the Einstein field equations may be obtained if the stress-energy tensor

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