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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
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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]]

Latest revision as of 05:38, 21 August 2014

Wilber Berryhill is the title his parents gave him and he totally digs that title. The favorite hobby for him and his children is to play lacross and he'll be starting something else along with it. North Carolina is exactly where we've been living for many years and will by no means transfer. He is an info officer.

My site - love psychics (http://kjhkkb.net)