这是一份uwa西澳大学PHYS3011的成功案例

From the last equation, we get, by differentiation with respect to $\beta_{s}^{i}(t), i \geq 2$, and restricting the result to $\gamma$ (i.e., putting $t=t_{0}$ ), the equation
$$
\sum_{k=2}^{\operatorname{dim}{R} M} \Gamma{(j k)}^{i}(\gamma(s))\left[x^{k}(\gamma(s))-f^{k}(g(\xi))\right]=0
$$
for $i \geq 1, j \geq 2$ and every $\gamma(s)$. Hence
$$
\Gamma_{(j k)}^{i}(\gamma(s))=0 \quad \text { for } i \geq 1, j, k \geq 2 \text {, and } s \in J_{U^{*}}^{\prime}
$$
Moreover, by construction
$$
0=\left.\nabla_{\dot{\gamma}} E_{i}^{\prime}\right|{\gamma(s)}=\dot{\gamma}^{j}(s) \nabla{E_{j}^{\prime}} E_{i}^{\prime}=\left.\dot{\gamma}^{j}(s) \Gamma^{k}{ }{i j}(\gamma(s)) E{k}^{\prime}\right|_{\gamma(s)}
$$

PHYS3011 COURSE NOTES ：

$$
\begin{aligned}
&x^{1}(q)=x^{1}(p)+\ln \left(1+y^{1}(q)\right) \
&x^{i}(q)=x^{i}(p)+y^{i}(q) \quad \text { for } i \geq 2
\end{aligned}
$$
from which the Riemannian normal coordinates can be expressed as
$$
\begin{aligned}
&y^{1}(q)=\exp \left(x^{1}(q)-x^{1}(p)\right)-1 \
&y^{i}(q)=x^{i}(q)-x^{i}(p) \quad \text { for } i \geq 2 .
\end{aligned}
$$