4.2. Classification of metric perturbations
Now we consider small perturbations of the spacetime metric away from the Robertson-Walker form:
![]() |
(4.11) |
We have introduced two 3-scalar fields
(x,
) and
(x,
), one 3-vector field
w(x,
) = wi
ei, and
one symmetric, traceless second-rank 3-tensor field
h(x,
) = hij
ei
ej. No generality is lost by making
hij traceless since any trace part can be put into
. The
factors of 2 and signs have been chosen to simplify later expressions.
Equation (4.11) is completely general:
gµ has 10
independent components and we have introduced 10 independent fields
(1 + 1 + 3 + 5 for
+
+
w + h). In fact, only 6 of these
fields can represent physical degrees of freedom because we are free
to transform our 4 coordinates
(
, xi)
without changing any
physical quantities. Infinitesimal coordinate transformations, called
gauge transformations, result in changes of the fields
(
,
,
w, h) because the spacetime scalar
ds2 =
gµ
dxµ
dx
must
be invariant under general coordinate transformations. We shall explore
the consequences of this invariance later. Coordinate invariance
complicates general relativity compared with other gauge theories (e.g.,
electromagnetism) in which the spacetime coordinates are fixed while other
variables change under the appropriate gauge transformations.
Unless stated explicitly to the contrary, in the following we shall
treat the perturbation variables
(,
,
wi, hij) exclusively as
3-tensors (of rank 0, 1, or 2 according to the number of indices) with
components raised and lowered using
ij
and
ij.
In doing this we choose to use
ij
as the 3-metric in the perturbed hypersurface of constant
despite the fact that the
spatial part of the 4-metric (divided by a2) is given
by (1 -
2
)
ij +
2hij. This treatment is satisfactory because we will
assume that the metric perturbations are small and we will neglect all
terms quadratic in them. However, we will use
gµ
to raise
4-vector components:
Gµ
=
gµ
G
.
Do take care to distinguish Latin from Greek!
We have introduced 3-scalar, 3-vector, and 3-tensor perturbations. (From now on we will drop the prefix 3- since it should be clear from the context whether 3- or 4- is implied.) Are these the famous scalar, vector, and tensor metric perturbations? Not quite! Recall the decomposition of a vector into longitudinal and transverse parts:
![]() |
(4.12) |
Since w|| =
-w for some
scalar w, how can it be
called a vector perturbation? By definition, only the transverse
component
w
represents a vector perturbation.
There is a similar decomposition theorem for tensor fields: Any differentiable traceless symmetric 3-tensor field hij(x) may be decomposed into a sum of parts, called longitudinal, solenoidal, and transverse:
![]() |
(4.13) |
The various parts are defined in terms of a scalar field h(x) and transverse (or solenoidal) vector field h(x) such that
![]() |
(4.14) |
where we have denoted symmetrization with parentheses and have employed the traceless symmetric double gradient operator:
![]() |
(4.15) |
Note that the divergences of h|| and
h are
longitudinal and transverse vectors, respectively (it doesn't matter which
index is contracted on the divergence since h is symmetric):
![]() |
(4.16) |
where 2
h
(
2
hi) ei. (We do not call
h
the
transverse part, as we would by extension from
w
,
because "transverse" is conventionally used to refer to the
tensor part.) The longitudinal tensor h|| is also called
the scalar part of h, the solenoidal part
h
is also
called the vector part, and the transverse-traceless part
hT
is also called the tensor part. This classification of the spatial metric
perturbations hij was first performed by
Lifshitz (1946).
The purpose of this decomposition is to separate hij into parts that can be obtained from scalars, vectors, and tensors. Is the decomposition unique? Not quite. It is clear, first of all, that h and hi are defined only up to a constant. But there may be additional freedom (Stewart 1990).
First, the vector h is defined only up to solutions of
Killing's equation
i
hj +
j
hi = 0, called Killing vectors
(Misner et al. 1973).
The reader can easily verify that one such solution (using the
quasi-Cartesian coordinates of eq. 4.3) is
(hx, hy, hz) =
(y, - x, 0).
In an open space (K
0) this solution would be excluded because it is
unbounded -- our perturbations should not diverge! - but in a closed
space (K > 0) the coordinates have a bounded range. This
Killing vector,
and its obvious cousins, correspond to global rotations of the spatial
coordinates and not to physical perturbations.
Next, there may also be non-uniqueness associated with the tensor (and scalar) component:
![]() |
(4.17) |
where is
some scalar field. From eqs. (4.5) and (4.6) one can show
2(
i
) =
i(
2 +
2K)
so
that
i
ji =
0 as required for the tensor component. However, we also require
hij, T to be traceless, implying
(
2 +
3K)
=
0. Thus, the
tensor mode is defined only up to eq. (4.17) with bounded solutions of
(
2 +
3K)
=
0. In fact, this condition also implies
ij =
Dij
,
so we may equally well attribute
ij
to the scalar mode hij, || . Thus, we are free to add any
multiple of
to h (the scalar mode) provided we subtract Dij
from the tensor mode. In an open space (K
0) there are no nontrivial
bounded solutions to
(
2 +
3K)
=
0 but in a
closed space (K > 0) there are four linearly independent solutions
(Stewart 1990).
Once again, these solutions correspond
to redefinitions of the coordinates with no physical significance.
Kodama & Sasaki
(1984,
Appendix B)
gave a proof of the tensor decomposition theorem,
but they missed the additional vector and scalar/tensor mode solutions
present in a closed space. In practice, it is easy to exclude these modes,
and so we shall ignore them hereafter.
Thus, we conclude that the most general perturbations of the
Robertson-Walker metric may be decomposed at each point in space into four
scalar parts each having 1 degree of freedom
(,
,
w||, h||), two vector parts each
having 2 degrees of freedom
(w
,
h
), and one
tensor part having 2 degrees of
freedom (hT, which lost 3 degrees of freedom to the
transversality condition). The total number of degrees of freedom is 10.
Why do we bother with this mathematical classification? First and foremost,
the different metric components represent distinct physical phenomena.
(By way of comparison, in previous lectures we have already seen that
v|| and
v
play very different roles in fluid
motion.) Ordinary Newtonian gravity obviously is a scalar phenomenon
(the Newtonian potential is a 3-scalar), while gravitomagnetism and
gravitational radiation - both of which are absent from Newton's laws,
and will be discussed below - are vector and tensor phenomena,
respectively. Moreover, this spatial decomposition can also be applied
to the Einstein and stress-energy tensors, allowing us to see clearly
(at least in some coordinate systems) the physical sources for each type
of gravity. Finally, the classification will help us to eliminate
unphysical gauge degrees of freedom. There are at least four of them,
corresponding to two of the scalar fields and one transverse vector field.
We will not write the weak-field Einstein equations for the general metric of eq. (4.11). Instead, we will consider only two particular gauge choices, each of which allows for all physical degrees of freedom (and more, in the case of synchronous gauge). First, however, we must examine the stress-energy tensor.