About Stewart Walker: American musician (1974-) | Biography, Facts, Career, Life
peoplepill id: stewart-walker
SW
1 views today
1 views this week
American musician

# Stewart Walker

Stewart Walker
The basics

## Quick Facts

 Intro American musician Is Musician Record producer Composer From United States of America Field Business Music Gender male Birth 1 January 1974, Charlotte, Mecklenburg County, North Carolina, U.S.A. Age 48 years Profiles The details (from wikipedia)

## Biography

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. $T_{0}=0$

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 ${\mathcal {M}}_{\epsilon }$ with ${\mathcal {M}}_{0}={\mathcal {M}}^{4}$ has metric $g_{ik}=\eta _{ik}+\epsilon h_{ik}$ . These manifolds can be put together to form a 5-manifold ${\mathcal {N}}$ . A smooth curve $\gamma$ can be constructed through ${\mathcal {N}}$ with tangent 5-vector $X$ , transverse to ${\mathcal {M}}_{\epsilon }$ . If $X$ is defined so that if $h_{t}$ is the family of 1-parameter maps which map ${\mathcal {N}}\to {\mathcal {N}}$ and $p_{0}\in {\mathcal {M}}_{0}$ then a point $p_{\epsilon }\in {\mathcal {M}}_{\epsilon }$ can be written as $h_{\epsilon }(p_{0})$ . This also defines a pull back $h_{\epsilon }^{*}$ that maps a tensor field $T_{\epsilon }\in {\mathcal {M}}_{\epsilon }$ back onto ${\mathcal {M}}_{0}$ . Given sufficient smoothness a Taylor expansion can be defined

$\delta T=\epsilon h_{\epsilon }^{*}({\mathcal {L}}_{X}T_{\epsilon })\equiv \epsilon ({\mathcal {L}}_{X}T_{\epsilon })_{0}$ 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 $\Delta \delta T=\epsilon ({\mathcal {L}}_{X}T_{\epsilon })_{0}-\epsilon ({\mathcal {L}}_{Y}T_{\epsilon })_{0}=\epsilon ({\mathcal {L}}_{X-Y}T_{\epsilon })_{0}$ . 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 ${\mathcal {M}}_{\epsilon }$ and gives the result

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

The contents of this page are sourced from Wikipedia article. The contents are available under the CC BY-SA 4.0 license.