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In Nakaharas book (Geometry, Topology, Physics) he states the isometry condition and then derives the Killing equation ("by a simple calculation") on page 279:

$$\begin{equation}\frac{\partial(x^\kappa+\epsilon X^\kappa)}{\partial x^\mu}\frac{\partial(x^\lambda+\epsilon X^\lambda)}{\partial x^\nu}\end{equation}g_{\kappa\lambda}(x+\epsilon X) = g_{\mu\nu}(x) \tag{7.96b}$$

$$\begin{equation}X^\xi\partial_\xi g_{\mu\nu}+\partial_\mu X^\kappa g_{\kappa \nu}+\partial_\nu X^\lambda g_{\mu\lambda} = 0\tag{7.120a}\end{equation}$$

How does one derive the second from the first?

blablu
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1 Answers1

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The steps are to note that: $$\frac{\partial(x^\kappa+\epsilon X^\kappa)}{\partial x^\mu} = \delta^\kappa_\mu+\epsilon \partial_\mu X^\kappa,\qquad g_{\kappa\lambda}(x+\epsilon X) = g_{\kappa\lambda}(x)+\epsilon X^\xi\partial_\xi g_{\kappa\lambda}(x)+O(\epsilon^2)$$ Now combine to get: $$\frac{\partial(x^\kappa+\epsilon X^\kappa)}{\partial x^\mu}\frac{\partial(x^\lambda+\epsilon X^\lambda)}{\partial x^\nu}g_{\kappa\lambda}(x+\epsilon X) = (\delta^\kappa_\mu+\epsilon \partial_\mu X^\kappa)(\delta^\lambda_\nu+\epsilon \partial_\nu X^\lambda) (g_{\kappa\lambda}+\epsilon X^\xi\partial_x g_{\kappa\lambda} +O(\epsilon^2))$$ multplying it out gives: $$g_{\mu\nu}+\epsilon (\partial_\nu X^\kappa g_{\kappa\mu}+\partial_\mu X^\lambda g_{\lambda\nu}+\partial_\xi X^\xi g_{\mu\nu})+O(\epsilon^2)$$ by setting this equal to $g_{\mu\nu}$ you get the equation desired.

s.harp
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