勾配

まず $\mathop{\mathop{\emph ▽}\nolimits }\nolimits f$ を求めてみよう。

$$\mathop{\mathop{\emph ▽}\nolimits }\nolimits f = \sum_i(\mathop{\mathop{\emph ▽}\nolimits }\nolimits f)_i\emph u^... ...h u^i =\sum_i\sum_\mu\alpha_i^\mu\partial_\mu f \sum_\nu\alpha^i_\nu\emph u^\nu$$

$$= \sum_\mu\sum_\nu\delta^\mu_\nu\emph u^\nu\partial_\mu f =\sum_\mu\emph u^\mu\partial_\mu f$$

$$= \sum_\mu\frac{\emph e^\mu}{\sqrt{g_{\mu\mu}}}\partial_\mu f$$ (C.64)

これが $\sum_\mu(\mathop{\mathop{\emph ▽}\nolimits }\nolimits f)_\mu\emph e^\mu$ に等しいはずなので、 次式が得られる。

$$(\mathop{\mathop{\emph ▽}\nolimits }\nolimits f)_\mu=\frac1{\sqrt{g_{\mu\mu}}}\D f{x^\mu}$$ (C.65)