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i ij
x’y 2!
r r r
3xi xj ’ r 2
1 xi ij
= + yi 3 + yi yj +···
r5
r r

where r ≡ x is the spatial distance from the origin to the field point and ≡
i
/ xi . One may then write the solution (17.31) as
4G 1 xi
¯ ct x = ’ ctr y d3 y + ctr y yi d3 y
h T T
c4 r r3
3xi xj ’ r 2 ij
+ ctr y yi yj d3 y + · · · (17.36)
T
r5
481
17.8 The compact-source approximation

where ctr = ct ’ x ’ y . This multipole expansion may be written in a particularly
compact form:
’1
4G 1
¯ i1 i2 ···i
ct x = ’ ···
h M ctr i1 i2 i
!
c4 r
=0

where the multipole moments of the source distribution at any time t are given by
i1 i2 ···i
ct = ct y yi1 yi2 · · · yi d3 y
M T

Since the fall-off with distance of the term associated with the th multipole
moment goes as 1/r +1 , the gravitational field at large distances from the source
is well approximated by only the first few terms of the multipole expansion.


17.8 The compact-source approximation
Let us suppose that the source is some matter distribution localised near the origin
O of our coordinate system. If we take our field point x to be a distance r from
O that is large compared with the spatial extent of the source, we need consider
only the first term in the multipole expansion (17.36). Moreover, we assume that
the source particles have speeds that are sufficiently small compared with c for us
to take ctr = ct ’ r in the argument of the stress“energy tensor. Thus, the solution
in the compact-source approximation is given by

4G
¯ ct x = ’ ct ’ r y d3 y (17.37)
h T
c4 r

In this approximation, we are thus considering only the far-field solution to the
linearised gravitational equations, which varies as 1/r.
From (17.37), we see that calculating the gravitational field has been reduced
to integrating T over the source at a fixed retarded time ct ’ r. The physical
interpretation of the various components of this integral is as follows:

T 00 d3 y, total energy of source particles (including rest mass energy) ≡ Mc2 ;
T 0i d3 y c— total momentum of source particles in the xi -direction ≡ P i c;
T ij d3 y, integrated internal stresses in the source.

For an isolated source, the quantities M and P i are constants in the linear
theory (this is easily proved directly from the conservation equation T = 0).6
Moreover, without loss of generality, we may take our spatial coordinates xi to

6
We shall see later that a source does in fact lose energy via the emission of gravitational radiation, but the
energy“momentum carried away by the gravitational field is quadratic in h and hence neglected in the
linear theory.
482 Linearised general relativity

correspond to the ˜centre-of-momentum™ frame of the source particles, in which
case P i = 0. Thus, from (17.37), in centre-of-momentum coordinates we have

4GM
¯ ¯ ¯
h00 = ’ 2 hi0 = h0i = 0 (17.38)
cr

The remaining components of the gravitational field are then given by the inte-
grated stress within the source,
4G
¯
hij ct x = ’ 4 T ij ct y d3 y (17.39)
cr r

where r denotes that the expression in the brackets is evaluated at ct = ct ’ r.
The integral in (17.39) is surprisingly troublesome to evaluate directly. Fortu-
nately, there exists an alternative route that leads to a very neat expression for this
quantity. We first recall that T = 0 (where, for consistency with (17.39), we
are considering T as a function of the coordinates ct y and so 0 = / ct
and k = / yk ). From this result, we may write
+ =0
00 0k
(17.40)
0T kT

+ =0
i0 ik
(17.41)
0T kT

Let us now consider the integral

T ik y j d3 y = y j d3 y + T ij d3 y
ik
kT
k

where the integral is taken over a region of space enclosing the source, so that
T = 0 on the boundary surface S of the region. Using Gauss™ theorem to convert
the integral on the left-hand side to an integral over the surface S, we find that
its value is zero. Hence, on using (17.41), we can write
1d
T ij d3 y = ’ y j d3 y = y j d3 y = T i0 y j d3 y
ik i0
kT 0T
c dt
For later convenience, interchanging i and j and adding gives
1d
T ij d3 y = T i0 y j + T j0 yi d3 y (17.42)
2c dy0
We must now consider the integral

T 0k yi y j d3 y = y i y j d3 y + T 0i y j + T 0j yi d3 y
0k
kT
k

where, once again, we may use Gauss™ theorem to show that the left-hand side is
zero. Using (17.40), we thus have
1d
T 0i y j + T 0j yi d3 y = T 00 yi y j d3 y (17.43)
c dt
483
17.9 Stationary sources

Combining (17.42) and (17.43) yields

1 d2
T d y= 2 2
3
T 00 yi y j d3 y
ij
2c dt
Inserting this expression into (17.39), we finally obtain the quadrupole formula

2 ij
¯ ij ct x = ’ 2G d I ct (17.44)
h
dt 2
c6 r r


where we have defined the quadrupole-moment tensor of the energy density of
the source,

I ij ct = T 00 ct y yi y j d3 y (17.45)

which is a constant tensor on each hypersurface of constant time. In the next
chapter, we will use this formula to determine the far-field gravitational radiation
generated by a time-varying matter source.


17.9 Stationary sources
Let us return to the general solution (17.31) to the linearised field equation. In the
previous section we confined our attention to the far-field solution for a compact
source. This behaves like 1/r as a function of distance and depends only on
the mass and inertia tensor of the source. As shown in the multipole expansion
(17.36), other properties of the source generate a field that falls off more rapidly
with distance. In general, it is often impossible to obtain a simple expression
for the solution (17.31). Nevertheless, the solution simplifies somewhat when the
source is stationary.
A stationary source has 0 T = 0, i.e. the energy“momentum tensor is constant
in time. Note that this does not necessarily imply that the source is static (so that
its constituent particles are not moving), which would additionally require the
form of T to be invariant to the transformation t ’ ’t. A typical example of a
stationary, but non-static, source is a uniform rigid sphere rotating with constant
angular velocity. The main advantage of the stationary-source limit is that the
time dependence vanishes and thus retardation is irrelevant. Hence, the general
solution (17.31) to the linearised field equations reduces to

4G Ty3
¯ x =’ (17.46)
h dy
x’y
c4
484 Linearised general relativity

One can perform a multipole expansion of this solution identical to that given
in (17.36) but for which all time dependence is omitted. Indeed, in this case, it
becomes somewhat simpler to interpret the various multipole moments physically.
A particularly interesting special case is the non-relativistic stationary source.
Consider a source having a well-defined spatial velocity field ui x , where the
speed u of any constituent particle is small enough compared with c that we
can neglect terms of order u2 /c2 and higher in its energy“momentum tensor. In
particular, we will take u = 1 ’ u2 /c2 ’1/2 ≈ 1. Moreover, the pressure p within
the source is everywhere much smaller than the energy density and may thus be
neglected. From the discussion of energy“momentum tensors in Section 8.1 we
see that, for such a source,
T 00 = c2 T 0i = c ui T ij = ui uj
where x is the proper-density distribution of the source. We see that
ij / T 00 ∼ u2 /c2 and so we should take T ij ≈ 0 to the order of our approxi-
T
mation. The corresponding solution (17.46) to the linearised field equations can
then be written as
i
¯ 00 = 4 ¯ 0i = A ¯
hij = 0 (17.47)
h h
2
c c
where we have defined the gravitational scalar potential and gravitational spatial
vector potential Ai by
y
x ≡ ’G d3 y (17.48)
x’y
y ui y 3
4G
A x ≡’ 2
i
(17.49)
dy
x’y
c
¯ ¯
The corresponding components of h are given by h = h ’ 2 1
h. The
¯ ¯
result (17.47) implies that h = h00 and, on lowering indices, we find that the
non-zero components are

2 Ai
h00 = h11 = h22 = h33 = h0i = (17.50)
c2 c

It should be remembered that raising or lowering a spatial (roman) index intro-
duces a minus sign. Thus the numerical value of Ai is minus that of Ai , the latter
being the ith component of the spatial vector A. The obvious analogy between the
equations (17.48, 17.49) and their counterparts in the theory of electromagnetism
will be discussed in detail in Section 17.11.
For the most part, in this chapter we adopt the viewpoint that h is simply a
rank-2 tensor field defined in Cartesian coordinates on a background Minkowski
485
17.10 Static sources and the Newtonian limit

spacetime. At this point, however, it is useful to revert to the viewpoint in which
g= + h defines the metric of a (slightly) curved spacetime. From (17.50),
we may therefore write the line element, in the limit of the non-relativistic source
considered here and in quasi-Minkowski coordinates, as

2 2Ai 2
ds2 = 1 + c2 dt2 + c dt dxi ’ 1 ’ 2 2
dxi (17.51)
c2 c c

in which it is worth noting that Ai dxi = ’ ij Ai dxj = ’A · dx. Determining the
geodesics of this line element provides a straightforward means of calculating the
trajectories of test particles in the gravitational field of a non-relativistic source
(in the weak-field limit). In particular, we note that we need not assume that the
test particles are slow-moving, and so the trajectories of photons in this limit may
also be found by determining the null geodesics of the line element (17.51).


17.10 Static sources and the Newtonian limit
A special case of stationary sources are static sources, for which the constituent
particles are not moving. In this case the only non-zero component of the source
energy“momentum tensor is the rest energy T 00 = c2 , where x is the proper
density distribution of the source. Indeed, this Newtonian source limit is clearly
equivalent to a stationary source with a vanishing velocity field ui x = 0. Thus,
from (17.50), we immediately find that in this case the non-zero elements of
h are

2
h00 = h11 = h22 = h33 = (17.52)
c2

In fact, the above solution remains valid to a good approximation even if the
source particles are moving, provided that the source energy“momentum tensor is
still dominated by the rest energy of the matter distribution, so that T 00 T 0i
and T 00 T ij .
The line element corresponding to (17.52) is given by

2 2
ds2 = 1 + c2 dt2 ’ 1 ’ 2
(17.53)
d
c2 c2

where d 2 = dx2 + dy2 + dz2 ; (17.53) is often referred to as the line element in
the Newtonian limit. Moreover, this line element is easily adapted to allow for
arbitrary spatial coordinate transformations, since d 2 is simply the line element
of three-dimensional Euclidean space. Thus if, for example, we adopt spatial
486 Linearised general relativity

spherical polar coordinates then one need only rewrite the spatial line element as
d 2 = dr 2 + r 2 d 2 + r 2 sin2 d 2 .
It is interesting to compare (17.53) with our discussion of the Newtonian limit
in Chapter 7, where we considered weak gravitational fields, static sources and
slowly moving test particles. Under these assumptions, we found that we recovered
the Newtonian equation of motion for a test particle provided that we made the
identification h00 = 2 /c2 , where is the Newtonian gravitational potential. In
the solution (17.53), we have arrived at the Newtonian limit without making any
restriction on the velocity of the test particle. This generalisation is important,
as previously we needed to consider only the effects of the g00 -component of
the metric, but, as the above solution shows, the trajectories of relativistic test
particles and photons also depend on the metric spatial components.
As an example of the line element (17.53), let us consider the simple case of
a static spherical object of mass M, so that the Newtonian gravitational potential
is given by = ’GM/r, where r is a radial coordinate. In this case, adopting
spherical polar spatial coordinates, the line element in the Newtonian limit is
given by
2GM 2GM
ds2 = c2 1 ’ dt2 ’ 1 + dr 2 + r 2 d + r 2 sin2 d
2 2
c2 r c2 r
which is straightforwardly shown to be identical to the Schwarzschild solution,
to first order in M. In the Solar System, this approximation is sufficiently accu-
rate to determine correctly the bending of light and gravitational redshifts (the
Shapiro effect) induced by the Sun, giving identical results to those discussed in
Chapter 10. The accuracy of the above approximation is, however, insufficient to
predict perihelion shifts correctly. This is not surprising, since perihelion shift is
a cumulative effect.


17.11 The energy“momentum of the gravitational field
Physically, one would expect the gravitational field to carry energy“momentum
just as, for example, the electromagnetic field does. Unfortunately, the task of
assigning an energy density to a gravitational field is famously difficult, both
technically and in principle. From our discussion of the equivalence principle in
Chapter 7, we know that transforming coordinates to a freely falling frame can
always eliminate gravitational effects at any one event. As a result, there is no local
notion of gravitational energy density in general relativity. Moreover, in a general
spacetime there is no reason why energy and momentum should be conserved.
In electromagnetism, for example, the conservation of energy and momentum
for the field is a direct consequence of the symmetries of the Minkowski space-
time assumed in the theory. In a general spacetime, however, there are no such
487
17.11 The energy“momentum of the gravitational field

symmetries. Even in the linearised gravitational theory developed in this chapter,
the field h represents a weak distortion of Minkowski space and so the Lorentz
symmetry properties are lost.
Nevertheless, as we have remarked several times, one can also regard the
linearised theory as describing a simple rank-2 tensor field h in Cartesian
inertial coordinates propagating in a fixed Minkowski spacetime background. We
might therefore hope to assign an energy“momentum tensor to this field just as
we do for electromagnetism, or any other field theory in Minkowski spacetime.
As was discussed in Section 17.4, the linearised gravitational theory ignores the
energy“momentum associated with the gravitational field itself (i.e. the ˜gravity
of gravity™). To include this contribution, and thereby go beyond the linearised
theory, one must modify the linearised field equations to read
8G
G1 =’ +t
T
c4
1
where G is the linearised Einstein tensor, T is the energy“momentum tensor
of any matter present and t is the energy“momentum tensor of the gravitational
field itself. Trivially rearranging this equation gives
8G 8G
G1 + t =’ 4 T
c4 c
Returning to the exact Einstein equations, however, we may expand beyond first
order to obtain
8G
G ≡ G 1 +G 2 +··· = ’ 4 T (17.54)
c
where superscripts in parentheses indicate the order of the expansion in h . This
suggests that, to a good approximation, we should make the identification
c4
t≡ G2 (17.55)
8G
This is also in keeping with our experience of other field theories in Minkowski
spacetime, such as electromagnetism, in which the energy“momentum tensor is
quadratic in the field variable. One should not, however, be too firmly guided by
the analogy with electromagnetism. The reason why the electromagnetic energy“
momentum tensor is quadratic in the field variable is that the electromagnetic
field (constituted by photons in the quantum description) does not carry charge
and so cannot act as its own source. Indeed, this is the physical reason why
electromagnetism is a linear theory. In the gravitational case, however, one could
in fact include the higher-order terms in (17.54) in the definition of t ; these
terms correspond to the contribution to the total energy“momentum arising from
the gravitational interaction of the gravitational field with itself. Nevertheless,
when the gravitational field is weak these higher-order terms may be neglected.
488 Linearised general relativity

As one might expect from such an heuristic approach, however, there are some
shortcomings of the identification (17.55), which we now outline. The terms in
the Einstein tensor that are second-order in h are given by

G2 =R2 ’2 R 2 ’ 2h R 1 + 2 h R1
1 1 1
(17.56)

2
where R denotes the terms in the Ricci tensor that are second-order in h and
R 1 and R 2 denote the terms in the Ricci scalar that are first- and second-order
in h respectively. Although (17.56), and hence t , is covariant under global
Lorentz transformations (although not under general coordinate transformations,
as one might expect), it may be may shown, after considerable algebra, that
it is not invariant under the gauge transformation (17.5) (or equivalently the
infinitesimal coordinate transformation (17.4)). One way of circumventing this
problem is to take seriously the fact that the energy“momentum of a gravitational
field at a point in spacetime has no real meaning in general relativity, since at
any particular event one can always transform to a free-falling frame in which
gravitational effects disappear. This suggests that, at each point in spacetime, one
2
should average G over a small region in order to probe the physical curvature
of the spacetime, which gives a gauge-invariant measure of the gravitational
field strength. Denoting this averaging process by · · · , one should thus replace
(17.55) by

c4
≡ G2 (17.57)
t
8G

Having made this identification, our task is now an algebraic one of determining
2
the form of G as a function of h . This is rather a cumbersome calculation,
but the job is made somewhat easier by averaging over small spacetime regions.
Since we are averaging over all directions at each point, first derivatives average
a = 0. This has the
to zero. Thus, for any function of position a x , we have
ab = a b + a b = 0, and hence we
important consequence that
may swap derivatives in products and inherit only a minus sign, i.e.

a b =’ a (17.58)
b

Let us begin by considering the last two terms on the right-hand side of (17.56),
which depend on the first-order Ricci tensor and Ricci scalar. It will prove most
convenient to express these in terms of the energy“momentum tensor T of any
matter present. The first-order (linearised) field equation (17.11) can be written as

R1 =’ T ’2
1
T
489
17.11 The energy“momentum of the gravitational field

where T ≡ T and ≡ 8 G/c4 . We also note from this equation that R 1 = T .
Thus, we may write (17.56) as

¯
G2 =R2 ’2 R2 ’2 ’
1 1
(17.59)
h hT

2
It therefore remains only to find the form of R , from which R 2 may
be obtained by contraction. The standard expression for the full Ricci tensor
is obtained by contracting (17.7) on its first and last indices. Thus, the terms
second-order in h are given by


R2 = ’ + ’
2 2 1 1 1 1
(17.60)


where, on the right-hand side, the superscripts in parentheses denote the order of
expansion in h for the connection coefficients. The connection coefficients to
first order were calculated in (17.6), and now including the second-order terms
we have


= + +···
1 2


= h+ h’ ’ 2h + h’ +···
1 1
h h h
2


Inserting these expressions into (17.60) and simplifying, one finds after a little
algebra


R 2 = ’4 + 2h h+ ’ ’
1 1
h h h h h
+2 ’ +2 ’2 h+ ’
1 1 1
h h h h h h h
(17.61)


Although the third group of terms on the right-hand side is not manifestly symmet-
ric in and , this symmetry is easy to verify. In fact, in subsequent calculations
it is convenient to maintain manifest symmetry by writing out this term again
with and reversed and multiplying both terms by one-half.
2
To evaluate the averaged expression (17.57), we must now calculate R .
One first makes use of the result (17.58) to rewrite products of first derivatives in
(17.61) in terms of second derivatives. Using the first-order field equation (17.10)
to substitute for terms of the form 2 h , and then applying (17.58) once more
490 Linearised general relativity

to rewrite terms containing second derivatives as products of first derivatives, one
finally obtains

R2 = ’2 +2 h’
1
h h h h h h h
4

+ 2h T + 2hT ’ hT ’ 4h (17.62)
T

where we have made use of the symmetrisation notation discussed in Chapter 4.
Contracting this expression, and once again making use of the result (17.58) and
the first-order field equation (17.10), one quickly finds that

R 2 = ’2
1
(17.63)
hT

Combining the expressions (17.59), (17.62) and (17.63) and writing the result
¯
(mostly) in terms of the trace reverse field h = h ’ 21
h, we thus find that
the energy“momentum tensor (17.57) of the gravitational field is given by

¯ ¯ ¯ ¯ ¯ ¯
’1
=4 ’2 ’2
1 1
t h h h h h h
(17.64)
¯
’ T+
4h hT

It may be verified by direct substitution that this expression is indeed invariant
under the gauge transformation (17.5), as required. We shall use this tensor in the
next chapter to determine the energy carried by gravitational waves.


Appendix 17A: The Einstein“Maxwell formulation of linearised gravity
In our discussion of non-relativistic stationary sources in Section 17.9, we found
that the expressions for the gravitational field exhibited a remarkable similarity to
the corresponding results in electromagnetism. We now pursue further the analogy
between linearised general relativity and electromagnetism for non-relativistic
stationary sources.
As discussed in Section 17.9, for such a source we may write

Ai
2g g
=h =h =h = 2 h= hij = 0
00 11 22 33 0i
(17.65)
h
c c
here we denote the gravitational scalar and vector potentials by and Ag
g
respectively. The linearised field equations may then be written as
16 G
=4 G Ag =
2 2
and (17.66)
j
g
c2
491
Appendix 17A: The Einstein“Maxwell formulation of linearised gravity

where we have defined the momentum density (or matter current density) j ≡ v.
These equations have the solutions (17.48, 17.49), which we write as

4G
y jy
x = ’G Ag x = ’
d3 y d3 y
and
g
x’y x’y
c2
Comparing the above results with the corresponding equations in electromag-
netism for the electric potential and the magnetic vector potential in the absence
of time-varying fields, there is a direct analogy on making the identifications
1 16 G
”’ ”’
and
0 0
c2
4G
The minus signs in these relations are a result of the fact that the electric force
repels like charges, whereas the gravitational force attracts (like) masses. Clearly,
in the electromagnetic case, and j correspond to the charge and current densities
respectively, rather than the matter and momentum densities. We can take the
analogy further by defining the gravitoelectric and gravitomagnetic fields

Eg = ’ Bg = — Ag
and (17.67)
g

Using the equations (17.66), it is straightforward to verify that the fields Eg and
Bg satisfy the gravitational Maxwell equations


· Eg = ’4 G · Bg = 0
16 G
— Eg = 0 — Bg = ’ j
c2

The equations for Eg describe the standard gravitational field produced by a static
mass distribution, whereas the equations for Bg provide a notationally familiar
means of determining the ˜extra™ gravitational field produced by moving masses
in a stationary non-relativistic source.
Although the gravitational Maxwell equations completely determine the gravi-
tational fields produced by a non-relativistic stationary source, they do not deter-
mine the effect of such fields on the motion of a test particle. In electromagnetism
one must, in addition, postulate the Lorentz force law. From our discussion in
Section 8.8, however, one might suspect that in the case of gravitation the corre-
sponding force law could be derived rather than postulated. The equation of
motion for a test particle in a gravitation field is the geodesic equation

x+
¨ x x =0
™™ (17.68)
492 Linearised general relativity

where the dots denote differentiation with respect to the proper time of the
particle. Let us assume that the test particle is slow-moving, i.e. its speed v
is sufficiently small compared with c that we may neglect terms in v2 /c2 and
higher. Hence we may take v = 1 ’ v2 /c2 ’1/2 ≈ 1. Writing x = ct x , the
4-velocity of the particle may thus be written

™ = cv≈cv
x v

This immediately implies that x = 0 and, moreover, that dt/d = 1, so we may
¨
replace dots with derivatives with respect to t. Thus, the spatial components of
(17.68) may be written as

d 2 xi
≈ ’ c2 00 + 2c + v ≈ ’ c2 00 + 2c
i i j i ij i i j
(17.69)
0j v ij v 0j v
dt2
where in the first approximate equality we have expanded the summation in
(17.68) into terms containing respectively two time components, one time and one
spatial component, and two spatial components. In the second approximation, we
have neglected the purely spatial terms since their ratio with respect to the purely
temporal term c2 i 00 is of order v2 /c2 . To first order in the gravitational field
h , the connection coefficients are given by (17.6). Inserting this expression into
(17.69) and remembering that for a stationary field 0 h = 0, one obtains

d 2 xi 1 2 i
≈ 2 c h00 + c h0j ’ j hi vj = ’ 2 c2 j h00 ’ c ’ j h0k vj
1
i ij ik
k h0j
0
dt2
Substituting the expressions (17.65) and remembering that one inherits a minus
sign on raising or lower a spatial (roman) index, the equation of motion may be
written as
d2 x
≈’ g +v— — Ag
dt2
Thus, using (17.67), one obtains the gravitational Lorenz force law

d2 x
≈ Eg + v — Bg
dt2

for slow-moving particles in the gravitational field of a stationary non-relativistic
source. The first term on the right-hand side gives the standard Newtonian result
for the motion of a test particle in the field of a static non-relativistic source,
whereas the second term gives a notationally familiar result for the ˜extra™ force
felt by a moving test particle in the presence of the ˜extra™ field produced by
moving masses in a stationary non-relativistic source.
493
Exercises

Exercises
17.1 In a region of spacetime with a weak gravitational field, there exist coordinates in
which the metric takes the form g = + h . Show that h is not a tensor
under a general coordinate transformation. Show further that, to first order in h ,

= ’h
g

where h = h.
17.2 For an infinitesimal general coordinate transformation x = x + x , show that
to first order in the inverse transformation is given by
x
= ’
x
= +h
17.3 If g with h 1, verify that, to first order in h ,

= h+ h’ h’
1
R h
2

= h+ h’ h’
2
1
R h
2

R= h’
2
h

Hence show that the linearised Einstein field equations are given by

h+ h’ h’ h’ h’ = ’2 T
2 2
h

17.4 The trace reverse of h is defined by

¯ ≡h ’2
1
h h

¯
¯ ¯
Show that h = ’h and h = h . Hence show that the linearised Einstein field
equations in Exercise 17.3 can be written as
2¯ ¯ ¯ ¯
h+ ’ h’ h = ’2 T
h

17.5 Obtain an expression for the covariant components R of the linearised Riemann
tensor in Exercise 17.3 and show that it is invariant under a gauge transformation
of the form (17.5). Hence show that the linearised Einstein field equations are also
invariant under such a gauge transformation.
17.6 From the linearised Einstein field equations, show that T = 0.
17.7 For a plane gravity wave of the form h = A exp ik x , show that the linearised
Riemann tensor is given by

= k k h +k k h ’k k h ’k k h
1
R 2

Hence show that the linearised Ricci tensor is given by

= k w + k w ’ k2 h
1
R 2
494 Linearised general relativity

¯
where k2 = k k and w = k h . Hence show that the linearised Einstein field
equations require that
= k w +k w
k2 h

From your answer to Exercise 17.7, show that for k2 = 0 one requires R = 0.
17.8
Hence show that this case does not correspond to a physical wave but merely a
periodic oscillation of the coordinate system.
¯
17.9 From your answer to Exercise 17.7, show that for k2 = 0 one requires k h = 0.
Hence show that the wavevector k is an eigenvector of the Riemann tensor in the
k = 0.
sense that R
17.10 Show explicitly that

¯ ¯
=h0 x ’2 G x ’y T d4 y
h x y

is a solution of the linearised Einstein field equations in the Lorenz gauge if
¯
h 0 x is any solution of the linearised field equations in vacuo and G x ’ y
satisfies
x ’y = x ’y
2 4
xG


17.11 The Green™s function G x ’ y satisfies the equation

x ’y = x ’y
2 4
xG


Show that the four-dimensional Dirac delta function can be written as
1
x ’y = exp ik x ’ y
4
d4 k
4
2

Hence, by writing the Green™s function in terms of its Fourier transform G k ,
show that
1 1
G x ’y = ’ exp ik x ’ y d4 k
4 2
2 k

where k2 = k k .
17.12 Verify that the solution in Exercise 17.10 satisfies the Lorenz gauge condition.
17.13 Prove the results (17.32, 17.33) for the derivatives of a function of retarded time.
17.14 By writing r ≡ x = ij xi xj ’1/2 , show that

1 xi
=’
i
r r

Hence show that
3xi xj ’ r 2
1 ij
=
ij
r5
r
495
Exercises

17.15 In Newtonian gravity, the gravitational potential x produced by some density
distribution x is given by
y
x = ’G d3 y
x’y
V

where the integral extends over the volume of the distribution. Show that
x·y
1 1 1
= + 3+
x’y x3
x x
Hence show that the gravitational potential can be written as

GM G d · x 1
x =’ ’ +
x3 x3
x
where
M= d=
y d3 y y y d3 y
and
V V

= 0, show that
17.16 From the conservation equation T

+ k T 0k = 0 + k T ik = 0
00 i0
and
0T 0T

By integrating each equation over a spatial volume V whose bounding surface S
encloses the energy“momentum source and using the three-dimensional divergence
theorem, show that the quantities

Mc2 ≡ Pi ≡
T 00 d3 y T i0 d3 y
and
V

are constants and give a physical interpretation of them.
17.17 For a stationary source, show that 0 T 0i = 0. Hence show that

T 0i y j + T 0j yi d3 y = 0
V

where the spatial volume V encloses the source.
17.18 For a non-relativistic stationary source, show that, in centre-of-momentum
coordinates,
4GM 1
¯
h00 x = ’ 2 +
x2
cx
2G 1
¯
h0i x = 3 3 xj J ij +
x3
cx
¯
hij x = 0

where the quantities M and J ij are given by

M= J ij = y i p j y ’ y j pi y
y d3 y d3 y
and
V V
496 Linearised general relativity

in which y is the proper density distribution of the source and pi y = y ui y
is the momentum density distribution of the source. Give a physical interpretation
of J ij .
Hint: You will find your answer to Exercise 17.17 useful.
17.19 Use your answer to Exercise 17.18 to show that, for a stationary non-relativistic
source, the gravitational scalar and vector potentials respectively are given to
leading order in 1/ x by
2G
GM
x =’ Ag x = ’ J —x
and
g
c2 x 3
x

where J = y — p d3 y is the total angular momentum vector of the source.
Show further that these expressions are exact in the linear theory for a spherically
symmetric source.
17.20 Use your answer to Exercise 17.19 to show that, in the linear theory, the line
element outside a spherically symmetric matter distribution rotating about the
z-axis at a steady rate is given by
2GM 4GJ 2GM
ds2 = c2 1 ’ dt2 + x dy ’y dx dt ’ 1 + 2 dx2 +dy2 +dz2
c2 r c2 r 3 cr
where r = x . Show that this is equal to the Kerr line element to first order in M
and J .
Hint: Ai dxi = ’ ij Aj dxi = ’A · dx and J — x · dx = J · x — dx .
17.21 If g = + h , show that the terms in the Einstein tensor that are second order
in h are given by

G2 =R2 ’2 R 2 ’ 2h R 1 + 2 h R1
1 1 1


where R 2 denotes the terms in the Ricci tensor that are second order in h , and
R 1 and R 2 denote the terms in the Ricci scalar that are first and second order
in h respectively. Show further that this quantity is not invariant under a gauge
transformation of the form (17.5).
17.22 Verify that the energy“momentum tensor of the linearised gravitational field is
given by (17.64). Show further that this tensor is invariant under a gauge trans-
formation of the form (17.5).
17.23 Use your answer to Exercise 17.19 to show that, in the linear theory, a spherically
symmetric body of mass M rotating steadily with angular momentum J produces
gravitoelectric and gravitomagnetic fields given respectively by
2G
GM ˆ ˆˆ
Eg x = ’ Bg x = J ’3 J ·x x
and
x
x2 2x3
c
ˆ
where x is a unit vector in the x- direction.
a=
Hint: For any scalar field and spatial vector fields a and b one has
—a+ — a and — a — b = a · b ’ b · a + b · a ’ a · b. Also,
1/ x = ’3x/ x 5 .
3
497
Exercises

17.24 Consider a particle moving under gravity at speed v in a circular orbit of radius
r in the equatorial plane of the body in Exercise 17.23. Show that the vector
acceleration of the particle is given by
GM ˆ 2GJv ˆ
a=’ r± 2 3 r
r2 cr
where r is the position vector of the orbiting particle and the plus and minus signs
corresponding to prograde and retrograde orbits respectively. Hence show that the
angular velocity of the particle is given to first order in J by
GM 2GJ GM
= “ 24
2
r3 cr r
where the minus and plus signs now correspond to prograde and retrograde orbits
respectively. Thus show that the retrograde orbit has a shorter period than the
prograde orbit.
17.25 In electromagnetism, the magnetic dipole moment of a current density distribution
j y is defined by m = 2 y — j d3 y, and the force and torque on the dipole
1

in a magnetic field B are given by F = m · B and T = m — B respectively.
Hence deduce that, in linearised gravity, the force and torque exerted by the
gravitomagnetic field Bg on a spinning body with spin angular momentum s are
given respectively by

Fg = s· Tg = s — Bg
1 1
and
Bg
2 2

Thus show that the spin angular momentum of the body will evolve as
ds
= s — Bg
1
2
dt
and therefore that s precesses about Bg with angular velocity = ’ 2 Bg (i.e. in
1

the negative sense). This is called the Lens“Thirring precession.
17.26 A gyroscope is in orbit about the massive rotating body in Exercise 17.23. Use
your answer to Exercise 17.25 to show that the precessional angular velocity vector
of the gyroscope is given by
G ˆˆ
= 3 J ·x x’J
2x3
c
where x is the position vector of the gyroscope relative to the centre of the massive
body. Show that this result agrees with that derived in Section 13.20 when x points
along J .
18
Gravitational waves




In the previous chapter, we saw that the linearised field equations of general
relativity could be written in the form of a wave equation

= ’2 T (18.1)
h

¯
provided that the h satisfy the Lorenz gauge condition

¯ =0 (18.2)
h

This suggests the existence of gravitational waves in an analogous manner to that
in which Maxwell™s equations predict electromagnetic waves. In this chapter, we
discuss in detail the propagation, generation and detection of such gravitational
radiation. As in the previous chapter, we will adopt the viewpoint that h is
simply a symmetric tensor field (under global Lorentz transformations) defined
on a flat Minkowski background spacetime.



18.1 Plane gravitational waves and polarisation states
In Section 17.5, we showed that the general solution of the linearised field
equations in vacuo may be written as the superposition of plane-wave solutions
of the form
¯ =A exp ik x (18.3)
h

where the A are constant (and, in general, complex) components of a symmetric
tensor and k are the constant (real) components of a vector. The Lorenz gauge
condition is satisfied provided that the additional constraint

A k =0 (18.4)

498
499
18.1 Plane gravitational waves and polarisation states

is obeyed. Physical solutions corresponding to propagating plane gravitational
waves in empty space may be obtained by taking the real part of (18.3):
¯ = exp ik x
h A

= 2A exp ik x + 2 A exp ’ik x
1 1


which is clearly just a superposition of two plane waves of the form (18.3).
The constants A are the components of the amplitude tensor, and the k ≡
k are the components of the 4-wavevector. It is conventional to denote
the components of the 4-wavevector by k = /c k , where k is the spatial
3-wavevector in the direction of propagation and is the angular frequency of
the wave. The nullity of k implies that 2 = c2 k 2 , and so both the group and
phase velocity of a gravitational wave are equal to the speed of light.
Since A = A , the amplitude tensor has 10 different (complex) components,
but the four Lorenz gauge conditions (18.4) reduce the number of independent
components to six. Moreover, we still have the freedom to make a further gauge
transformation of the form (17.5), which will preserve the Lorenz gauge provided
x so that they satisfy 2 = 0. As we show
that we choose the four functions
below, this may be used to reduce the number of independent components in the
amplitude matrix from six to just two. This results in two possible polarisations
for plane gravitational waves.
It is convenient to consider the concrete example of a plane gravitational
wave propagating in the x3 -direction, in which case the components of the 4-
wavevector are
=k00k (18.5)
k

where k = /c. The Lorenz gauge condition (18.4) then immediately gives
A 3 = A 0 . Together with the symmetry of the amplitude tensor, this implies
that all the components A can be expressed in terms of the six quantities
A00 A01 A02 A11 A12 A22 :
⎛ 00 ⎞
A01 A02 A00
A
⎜A01 A11 A12 A01 ⎟
⎜ ⎟
A = ⎜ 02 02 ⎟
⎝A A⎠
12 22
A A
A00 A01 A02 A00

We may now perform a gauge transformation of the form (17.5) to simplify the
amplitude tensor still further. To preserve the Lorenz gauge condition we must
ensure that 2 = 0. A suitable transformation, which satisfies this condition, is
given by
= exp ik x
500 Gravitational waves

where the are constants. Substituting this expression into the transformation
¯
law (17.12) for the trace reverse tensor h , which we assume to be of the form
(18.3), one quickly finds that the amplitude tensor transforms as
=A ’i k ’i k +i (18.6)
A k
Using the expression (18.5) for the 4-wavevector and the result (18.6), we obtain
00 11
= A00 ’ ik + = A11 ’ ik ’
0 3 0 3
A A
01 12
= A01 ’ ik = A12
1
A A
02
= A02 ’ ik A 22 = A22 ’ ik ’
2 0 3
A
Now, by choosing the constants as follows,
= ’i 2A00 + A11 + A22 / 4k = ’iA01 /k
0 1


= ’iA02 /k = ’i 2A00 ’ A11 ’ A22 / 4k
2 3


we obtain
A 00 = A 01 = A 02 = 0 A 11 = ’A 22
and
On dropping primes, the first condition means that only A11 A12 and A22 are
non-zero. Moreover, the second condition means that only two of these can be
specified independently. Choosing A11 ≡ a and A12 ≡ b as the two independent
(in general, complex) components in our new gauge, we thus have
⎛ ⎞
00 00
⎜0 a b 0⎟
⎜ ⎟
ATT = ⎜ ⎟ (18.7)
⎝ 0 b ’a 0 ⎠
00 00
for a wave travelling in the x3 -direction. As indicated, the new gauge we have
adopted is known as the transverse-traceless gauge (or TT gauge), which we will
discuss in more detail in Section 18.3. For now we simply note that (18.7) implies
¯ ¯
hTT = 0 = hTT (hence the term traceless) and hTT = hTT for our plane wave.
It is also convenient to introduce the two linear polarisation tensors e1 and
e2 , the components of which are obtained by setting a = 1 b = 0 and a = 0 b = 1
respectively in (18.7). The general amplitude tensor in the TT gauge for a wave
travelling in the x3 -direction can then be written as
ATT = ae1 + be2
It follows that all possible polarisations of the gravitational wave may be obtained
by superposing just two polarisations, with arbitrary amplitudes and relative
phases.
501
18.2 Analogy between gravitational and electromagnetic waves

18.2 Analogy between gravitational and electromagnetic waves
Before going on to discuss gravitational waves in more detail, it is instructive to
illustrate the close analogy with electromagnetic waves. By adopting the Lorenz
gauge condition A = 0, the electromagnetic field equations in free space take
the form 2 A = 0. These admit plane-wave solutions of the form

A= Q exp ik x

where the Q are the constant components of the amplitude vector. The field
equations again imply that the 4-wavevector k is null and the Lorenz gauge
condition requires that Q k = 0, thereby reducing the number of independent
components in the amplitude vector to three. In particular, if we again consider
a wave propagating in the x3 -direction then k = k 0 0 k and the Lorenz
gauge condition implies that Q0 = Q3 , so that

= Q0 Q1 Q2 Q0
Q

The Lorenz gauge condition is preserved by any further gauge transformation
of the form A ’ A + , provided that 2 = 0. An appropriate gauge
transformation that satisfies this condition is

= exp ik x

is a constant. This yields Q = Q + i k , and so
where

Q 0 = Q0 + i k Q 1 = Q1 Q 2 = Q2

By choosing = ’iQ0 /k, on dropping primes we have Q0 = 0. In the new
gauge, the amplitude vector has just two independent components, Q1 and Q2 ,
and the electromagnetic fields are transverse to the direction of propagation. By
introducing the two linear polarisation vectors

e1 = 0 1 0 0 e2 = 0 0 1 0
and

we may write the general amplitude vector as

Q = ae1 + be2

where a and b are arbitrary (in general, complex) constants.
If b = 0 then as the electromagnetic wave passes a free positive test charge this
will oscillate in the x1 -direction with a magnitude that varies sinusoidally with
time. Similarly, if a = 0 then the test charge will oscillate in the x2 -direction. The
particular combinations of linear polarisations given by b = ±ia give circularly
polarised waves, in which the mutually orthogonal linear oscillations combine in
such a way that the test charge moves in a circle.
502 Gravitational waves

18.3 Transforming to the transverse-traceless gauge
In Section 18.1, we considered only the transformation into the TT gauge of a
plane gravitational wave travelling in the x3 -direction. We now consider a general
¯
gravitational perturbation h satisfying the empty-space linearised field equation
and the Lorenz gauge condition. As discussed previously, a gauge transformation
of the form (17.5) will preserve the Lorenz gauge condition provided that the four
x satisfy 2 = 0. From (17.12), the trace-reverse field tensor
functions
transforms as
¯ ¯
=h ’ ’ +
h
¯ ¯
Since the components h also satisfy the in vacuo wave equation 2 h = 0,
this gauge transformation may be used to set any four linear combinations of the
¯
h to zero. The TT gauge is defined by choosing

¯ TT ¯
h0i ≡ 0 hTT ≡ 0
and (18.8)

¯
This last condition means that hTT = hTT , and these quantities may therefore
be used interchangeably. Moreover, setting = 0 and = j respectively in the
¯
Lorenz gauge condition hTT = 0, and using (18.8), gives the constraints

¯ 00 ¯ ij
=0 =0
and (18.9)
0 hTT i hTT


We note that, if the gravitational field perturbation is non-stationary (i.e. it depends
on t), as for a general gravitational wave disturbance, the first constraint in (18.9)
0
implies that h00 also vanishes and so hTT = 0 for all . In other words, in this
TT
ij
case only the spatial components hTT are non-zero.
Let us now consider the particular case of an arbitrary plane gravitational wave
of the form (18.3) and satisfying the Lorenz gauge condition. The conditions
(18.4) immediately imply that

A0i = 0 =0
and ATT
TT

Moreover, the conditions (18.9) also require that
ij
A00 = 0 ATT kj = 0
and
TT

These last conditions ensure that, quite generally, a plane gravitational wave is
transverse, like electromagnetic waves.
The above conditions tell us the constraints on the form of ATT . We must
now consider how to construct this tensor for a plane wave with a given spatial
wavevector k and amplitude matrix A . First, it is clear that we need consider
503
18.3 Transforming to the transverse-traceless gauge
ij
only the spatial components ATT , since the remaining components are all zero.
Moreover, from the above conditions, this spatial tensor must be orthogonal to k
and traceless. We therefore introduce the spatial projection tensor

Pij ≡ ’ ni nj
ij

which projects spatial tensor components onto the surface orthogonal to the unit
spatial vector with components ni . The action of the projection tensor is easily
illustrated by applying it to an arbitrary spatial vector vi . One quickly finds that
ni Pj vj = 0 and Pk Pj vj = Pj vj , as required. In the case of our plane gravitational
i ik i

ˆ
wave, we choose ni to lie in the direction of the spatial wavevector, so that ni = ki ,
and thus obtain the components of the spatial amplitude tensor that are transverse
to the direction of propagation, namely
ij j
AT = Pk Pl Akl
i


The trace of this tensor is given by AT i = Pkl Akl , which in general does not
i
vanish. Using the fact that Pii = 3 ’ 1 = 2, we may however construct a traceless
tensor that still remains transverse to k; this is given by

ij j
ATT = Pk Pl ’ 2 P ij Pkl Akl
1
i
(18.10)

For a plane gravitational wave travelling in the x3 -direction, so that k =
k 0 0 k , it is a simple matter to verify that (18.10) produces an amplitude
matrix of the form
⎛ ⎞
0 0 0 0
⎜0 1 A11 ’ A22 0⎟
A12
⎜ ⎟
ATT = ⎜ ⎟
2 (18.11)
⎝0 0⎠
22 ’ A11
1
12
A A
2
0 0 0 0

which agrees with that given in (18.7). In fact this result illustrates that there is
a quick and simple algorithm for transforming a plane wave travelling along one
of the coordinate directions into the TT gauge. We see that the transformation
(18.11) corresponds to setting to zero all components that are not transverse to
the direction of wave propagation and subtracting one-half the resulting trace
from the remaining diagonal elements, to make the final tensor traceless. There
is, however, nothing special about our choice of x3 -direction and so the above
prescription must be true for a plane wave travelling in any of the three coordinate
directions.
504 Gravitational waves

18.4 The effect of a gravitational wave on free particles
Let us now consider the motion of a set of test particles, initially at rest, in the
presence of a gravitational wave. In fact, in the latter case it is not enough to
consider the trajectory of just a single test particle, as we discuss below. To obtain
a coordinate-independent measure of the effects of the wave, it is necessary to
consider the relative motion of a set of nearby particles.
First consider a single free test particle, whose 4-velocity u must satisfy the
geodesic equation
du
+ u u =0
d
Suppose that the particle is initially at rest in our chosen coordinate system, so
that u = c 1 0 0 0 . The geodesic equation then reads
du
= ’c2 = ’ 2 c2 0 h 0 + 0 h0 ’
1
h00
00
d
where in the last equality we have used (17.6) to obtain the connection coefficients
to first order in terms of the derivatives of h . Let us now adopt the TT gauge,
which we may do for any general gravitational wave disturbance in vacuo. From
the discussion in Section 18.3, we know that hTT = 0 for all values of . Thus,
0
initially, du /d = 0 and so the particle will still be at rest a moment later.
The argument may then be repeated, showing that the particle remains at rest
forever, regardless of the passing of the gravitational wave. In other words u =
c 1 0 0 0 is a solution of the geodesic equation in this case, as may readily be
verified by direct substitution.
What has gone wrong here? The key point is that ˜at rest™ in this context means
simply that the particle has constant spatial coordinates. What we have uncovered
is that by choosing the TT gauge we have found a coordinate system that stays
attached to individual particles. This has no coordinate-invariant physical meaning.
To obtain a proper physical interpretation of the effect of a passing gravitational
wave, we must consider a set of nearby particles.
Let us therefore consider a cloud of non-interacting free test particles. From
the above discussion, the worldlines of the particles are curves having constant
= 0 1 2 3 giving
spatial coordinates. Thus the small spacelike vector
the coordinate separation between any two nearby particles is constant (this may
also be shown explicitly by demonstrating that the equation of geodesic deviation
= constant as a solution in this case). Although the coordinate
(7.24) has
separation of the particles is constant, this does not mean that their physical spatial
separation l is constant. The latter is given by

l2 = ’gij = ’ hij
ij ij
ij
505

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