- #1

synoe

- 23

- 0

[tex]

g^\prime_{\mu\nu}(x)=g_{\mu\nu}(x)

[/tex]

under the transformation [itex]x^\mu\to x^{\prime\mu}(x)[/itex]. This means under infinitesimal transformation

[tex]

x^\mu\to x^\mu+\epsilon \xi^\mu(x)

[/tex]

where [itex]\epsilon[/itex] is any infinitesimal constant, the vector field [itex]\xi^\mu(x)[/itex] satisfies Killing equation:

[tex]

\nabla_\mu\xi_\nu(x)+\nabla_\nu\xi_\mu(x)=0.

[/tex]

I have questions about above description.

1. Why the infinitesimal parameter [itex]\epsilon[/itex] is constant and [itex]\xi^\mu[/itex] is depend on [itex]x[/itex]? The most general form seems to be [itex]\epsilon(x)\xi^\mu(x)[/itex].

2.Is this transformation global or local? It's global symmetry because [itex]\epsilon[/itex] is constant?

3.I have seen the description like "isometry means the metric is independent on the coordinate of the isometry direction." That is, if the isometry transformation is [itex]x^0\to x^0+\epsilon\xi(x)[/itex], [itex]g_{\mu\nu}=g_{\mu\nu}(x^1,\dots)[/itex] or [itex]\partial_0g_{\mu\nu}=0[/itex]. But I think this is true only when the Killing equation have a constant solution. The statement is true in more general?