This Wikipedia article says that the metric in polar Riemann normal coordinates ($\xi_1,...,\xi_n$, with $\xi_1$ as the 'radial' coordinate), satisfies $g_{1i}=\delta_{1i}$. By definition of normal coordinates I can see easily why $g_{11}=1$. But I don't see why $g_{1i\neq 1}=0$.
Asked
Active
Viewed 185 times
1 Answers
1
I think the argument goes as follows: it is clear from the definition of normal coordinates that $\Gamma_{11}^i=0$ (which says that the radial vector doesn't change under parallel transport along a radial path). Writing this in terms of the metric gives $$ 2\partial_1 g_{1i}-\partial_i g_{11}=0. $$ Setting $i=1$ gives $\partial_1g_{11}=0$, which means that $g_{11}$ is constant everywhere -- therefore $\partial_i g_{11}=0$. Substituting this into the equation above gives $\partial_1 g_{1i}=0$ -- which means that $g_{1i}$ is constant everywhere. Since $g_{1i}=\delta_{1i}$ at the origin, $g_{1i}=\delta_{1i}$ everywhere.
I should add that $g_{1i}=\delta_{1i}$ implies $\Gamma^{1}_{1i}=0$ (which isn't obvious from the definition of normal coordinates at all), which roughly says that the the radial component of a vector doesn't change under parallel transport along a radial path, or utilising $\Gamma^{1}_{i1}=\Gamma^1_{1i}=0$, that the radial component of the radial vector doesn't change under parallel transport along an arbitrary path.
If you prefer to work in the cartesian system, $x_1,...,x_n$, (instead of polar coordinates) then the argument is: $x^j x^k\Gamma^{i}_{jk}=0$ implies $$ 2x^j\partial_jF_i=\partial_i(x^jF_j) $$ where $F_i\equiv x^kg_{ki}$. This has the (unique) solution $F_i=x^j\delta_{ij}$, given the small $x$ behaviour $F_i=x^j\delta_{ij}+O(|x|^2)$ (which follows from $g_{ij}(0)=\delta_{ij}$).
dennis
- 473