Stewart–Walker lemma

The Stewart–Walker lemma provides necessary and sufficient conditions for the linear perturbation of a tensor field to be gauge-invariant. $\mathrm{\Delta }\delta T=0$ if and only if one of the following holds

1. $T_0=0$

2. $T_0$ is a constant scalar field

3. $T_0$ is a linear combination of products of delta functions ${\delta }_a^b$

Derivation

A 1-parameter family of manifolds denoted by ${{ℳ}}_{ϵ}$ with ${{ℳ}}_0={{ℳ}}^4$ has metric $g_{ik}={\eta }_{ik}+ϵh_{ik}$. These manifolds can be put together to form a 5-manifold ${𝒩}$. A smooth curve $\gamma$ can be constructed through ${𝒩}$ with tangent 5-vector $X$, transverse to ${{ℳ}}_{ϵ}$. If $X$ is defined so that if $h_t$ is the family of 1-parameter maps which map ${𝒩}\to {𝒩}$ and $p_0\in {{ℳ}}_0$ then a point $p_{ϵ}\in {{ℳ}}_{ϵ}$ can be written as $h_{ϵ}(p_0)$. This also defines a pull back $h_{ϵ}^{*}$ that maps a tensor field $T_{ϵ}\in {{ℳ}}_{ϵ}$ back onto ${{ℳ}}_0$. Given sufficient smoothness a Taylor expansion can be defined

$h_{ϵ}^{*}(T_{ϵ})=T_0+ϵh_{ϵ}^{*}({{ℒ}}_X{T}_{ϵ})+O({ϵ}^2)$

$\delta T=ϵh_{ϵ}^{*}({{ℒ}}_X{T}_{ϵ})\equiv ϵ({{ℒ}}_X{T}_{ϵ}{)}_0$ 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 $\mathrm{\Delta }\delta T=ϵ({{ℒ}}_X{T}_{ϵ}{)}_0-ϵ({{ℒ}}_Y{T}_{ϵ}{)}_0=ϵ({{ℒ}}_{X-Y}{T}_{ϵ}{)}_0$. Picking a chart where $X^a=({\xi }^{\mu },1)$ and $Y^a=(0,1)$ then $X^a-Y^a=({\xi }^{\mu },0)$ which is a well defined vector in any ${{ℳ}}_{ϵ}$ and gives the result

$\mathrm{\Delta }\delta T=ϵ{{ℒ}}_{\xi }{T}_0.$

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

Sources

• Describes derivation of result in section on Lie derivatives