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The '''Stewart–Walker lemma''' provides necessary and sufficient conditions for the [[linear]] [[wiktionary:perturbation|perturbation]] of a [[tensor]] field to be [[gauge theory|gauge]]-invariant. <math>\Delta \delta T = 0</math> [[if and only if]] one of the following holds
 
1. <math>T_{0} = 0</math>
 
2. <math>T_{0}</math> is a constant scalar field
 
3. <math>T_{0}</math> is a linear combination of products of delta functions <math>\delta_{a}^{b}</math>
 
== Derivation ==
 
A 1-parameter family of manifolds denoted by <math>\mathcal{M}_{\epsilon}</math> with <math>\mathcal{M}_{0} = \mathcal{M}^{4}</math> has [[Metric (mathematics)|metric]] <math>g_{ik} = \eta_{ik} + \epsilon h_{ik}</math>. These manifolds can be put together to form a 5-manifold <math>\mathcal{N}</math>. A smooth curve <math>\gamma</math> can be constructed through <math>\mathcal{N}</math> with tangent 5-vector <math>X</math>, transverse to <math>\mathcal{M}_{\epsilon}</math>. If <math>X</math> is defined so that  if <math>h_{t}</math> is the family of 1-parameter maps which map <math>\mathcal{N} \to \mathcal{N}</math> and <math>p_{0} \in \mathcal{M}_{0}</math> then a point <math>p_{\epsilon} \in \mathcal{M}_{\epsilon}</math> can be written as <math>h_{\epsilon}(p_{0})</math>. This also defines a [[pullback (differential geometry)|pull back]] <math>h_{\epsilon}^{*}</math> that maps a tensor field <math>T_{\epsilon} \in \mathcal{M}_{\epsilon} </math> back onto <math>\mathcal{M}_{0}</math>. Given sufficient smoothness a Taylor expansion can be defined
 
:<math>h_{\epsilon}^{*}(T_{\epsilon}) = T_{0} + \epsilon \, h_{\epsilon}^{*}(\mathcal{L}_{X}T_{\epsilon}) + O(\epsilon^{2})</math>
 
<math>\delta T = \epsilon h_{\epsilon}^{*}(\mathcal{L}_{X}T_{\epsilon}) \equiv \epsilon (\mathcal{L}_{X}T_{\epsilon})_{0}</math> is the linear perturbation of <math>T</math>. However, since the choice of <math>X</math> is dependent on the choice of [[Gauge theory|gauge]] another gauge can be taken. Therefore the differences in gauge become <math>\Delta \delta T = \epsilon(\mathcal{L}_{X}T_{\epsilon})_0 - \epsilon(\mathcal{L}_{Y}T_{\epsilon})_0 = \epsilon(\mathcal{L}_{X-Y}T_\epsilon)_0</math>. Picking a [[Chart (topology)|chart]] where <math>X^{a} = (\xi^\mu,1)</math> and <math>Y^a = (0,1)</math> then <math>X^{a}-Y^{a} = (\xi^{\mu},0)</math> which is a well defined vector in any <math>\mathcal{M}_\epsilon</math> and gives the result
 
:<math>\Delta \delta T = \epsilon \mathcal{L}_{\xi}T_0.\,</math>
 
The only three possible ways this can be satisfied are those of the lemma.
 
== Sources ==
 
{{refbegin}}
* {{cite book | author=Stewart J. | title=Advanced General Relativity | location=Cambridge | publisher=Cambridge University Press | year=1991 | isbn=0-521-44946-4}}  Describes derivation of result in section on Lie derivatives
{{refend}}
 
{{DEFAULTSORT:Stewart-Walker lemma}}
[[Category:Tensors]]
[[Category:Lemmas]]

Revision as of 18:58, 1 June 2013

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

Derivation

A 1-parameter family of manifolds denoted by ϵ with 0=4 has metric gik=ηik+ϵhik. These manifolds can be put together to form a 5-manifold 𝒩. A smooth curve γ can be constructed through 𝒩 with tangent 5-vector X, transverse to ϵ. If X is defined so that if ht is the family of 1-parameter maps which map 𝒩𝒩 and p00 then a point pϵϵ can be written as hϵ(p0). This also defines a pull back hϵ* that maps a tensor field Tϵϵ back onto 0. Given sufficient smoothness a Taylor expansion can be defined

hϵ*(Tϵ)=T0+ϵhϵ*(XTϵ)+O(ϵ2)

δT=ϵhϵ*(XTϵ)ϵ(XTϵ)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 ΔδT=ϵ(XTϵ)0ϵ(YTϵ)0=ϵ(XYTϵ)0. Picking a chart where Xa=(ξμ,1) and Ya=(0,1) then XaYa=(ξμ,0) which is a well defined vector in any ϵ and gives the result

ΔδT=ϵξT0.

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

Sources

Template:Refbegin

  • 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.

    My blog: http://www.primaboinca.com/view_profile.php?userid=5889534 Describes derivation of result in section on Lie derivatives

Template:Refend