Stewart–Walker lemma explained

The Stewart–Walker lemma provides necessary and sufficient conditions for the linear perturbation of a tensor field to be gauge-invariant.

\Delta\deltaT=0

if and only if one of the following holds

1.

T0=0

2.

T0

is a constant scalar field

3.

T0

is a linear combination of products of delta functions
b
\delta
a

Derivation

A 1-parameter family of manifolds denoted by

l{M}\epsilon

with

l{M}0=l{M}4

has metric

gik=ηik+\epsilonhik

. These manifolds can be put together to form a 5-manifold

l{N}

. A smooth curve

\gamma

can be constructed through

l{N}

with tangent 5-vector

X

, transverse to

l{M}\epsilon

. If

X

is defined so that if

ht

is the family of 1-parameter maps which map

l{N}\tol{N}

and

p0\inl{M}0

then a point

p\epsilon\inl{M}\epsilon

can be written as

h\epsilon(p0)

. This also defines a pull back
*
h
\epsilon
that maps a tensor field

T\epsilon\inl{M}\epsilon

back onto

l{M}0

. Given sufficient smoothness a Taylor expansion can be defined
*
h
\epsilon

(T\epsilon)=T0+\epsilon

*
h
\epsilon

(l{L}XT\epsilon)+O(\epsilon2)

\deltaT=\epsilon

*
h
\epsilon

(l{L}XT\epsilon)\equiv\epsilon(l{L}XT\epsilon)0

is the linear perturbation of

T

. However, since the choice of

X

is dependent on the choice of gauge another gauge can be taken. Therefore the differences in gauge become

\Delta\deltaT=\epsilon(l{L}XT\epsilon)0-\epsilon(l{L}YT\epsilon)0=\epsilon(l{L}X-YT\epsilon)0

. Picking a chart where

Xa=(\xi\mu,1)

and

Ya=(0,1)

then

Xa-Ya=(\xi\mu,0)

which is a well defined vector in any

l{M}\epsilon

and gives the result

\Delta\deltaT=\epsilonl{L}\xiT0.

The only three possible ways this can be satisfied are those of the lemma.

Sources