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Stewart–Walker lemma
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Stewart–Walker lemma : ウィキペディア英語版
Stewart–Walker lemma
The Stewart–Walker lemma provides necessary and sufficient conditions for the linear perturbation of a tensor field to be gauge-invariant. \Delta \delta T = 0 if and only if one of the following holds
1. T_ = 0
2. T_ is a constant scalar field
3. T_ is a linear combination of products of delta functions \delta_^
== Derivation ==

A 1-parameter family of manifolds denoted by \mathcal_ with \mathcal_ = \mathcal^ has metric g_ = \eta_ + \epsilon h_. These manifolds can be put together to form a 5-manifold \mathcal. A smooth curve \gamma can be constructed through \mathcal with tangent 5-vector X, transverse to \mathcal_. If X is defined so that if h_ is the family of 1-parameter maps which map \mathcal \to \mathcal and p_ \in \mathcal_ then a point p_ \in \mathcal_ can be written as h_(p_). This also defines a pull back h_^ that maps a tensor field T_ \in \mathcal_ back onto \mathcal_. Given sufficient smoothness a Taylor expansion can be defined
:h_^(T_) = T_ + \epsilon \, h_^(\mathcal_T_) + O(\epsilon^)
\delta T = \epsilon h_^(\mathcal_T_) \equiv \epsilon (\mathcal_T_)_ 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 \delta T = \epsilon(\mathcal_T_)_0 - \epsilon(\mathcal_T_)_0 = \epsilon(\mathcal_T_\epsilon)_0. Picking a chart where X^ = (\xi^\mu,1) and Y^a = (0,1) then X^-Y^ = (\xi^,0) which is a well defined vector in any \mathcal_\epsilon and gives the result
:\Delta \delta T = \epsilon \mathcal_T_0.\,
The only three possible ways this can be satisfied are those of the lemma.

抄文引用元・出典: フリー百科事典『 ウィキペディア(Wikipedia)
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