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¯

= ’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