证明： ∂ ln g ∂ u α = Γ 1 α 1 + Γ 2 α 2 \frac {\partial \ln \sqrt {g}}{\partial u^{\alpha }}= \Gamma _{1\alpha }^{1}+ \Gamma _{2\alpha }^{2} ∂uα∂lng=Γ1α1+Γ2α2
证：
左式计算为
∂ ln g ∂ u α = 1 2 g ∂ g ∂ u α \frac{\partial\ln\sqrt{g}}{\partial u^\alpha}=\frac1{2g}\frac{\partial g}{\partial u^\alpha} ∂uα∂lng=2g1∂uα∂g
注意 g = g 11 g 22 − g 12 g 21 g=g_{11}g_{22}-g_{12}g_{21} g=g11g22−g12g21 , g 12 = g 21 g_{12}=g_{21} g12=g21 ,求导有
∂ g ∂ u α = ∂ g 11 ∂ u α g 22 + ∂ g 22 ∂ u α g 11 − 2 ∂ g 12 ∂ u α g 12 \frac{\partial g}{\partial u^\alpha}=\frac{\partial g_{11}}{\partial u^\alpha}g_{22}+\frac{\partial g_{22}}{\partial u^\alpha}g_{11}-2\frac{\partial g_{12}}{\partial u^\alpha}g_{12} ∂uα∂g=∂uα∂g11g22+∂uα∂g22g11−2∂uα∂g12g12
再计算右式的Christoffel符号，
Γ α β γ = 1 2 g γ ξ ( ∂ g α ξ ∂ u β + ∂ g β ξ ∂ u α − ∂ g α β ∂ u ξ ) {\Gamma_{\alpha\beta}^{\gamma}=\frac12g^{\gamma\xi}(\frac{\partial g_{\alpha\xi}}{\partial u^\beta}+\frac{\partial g_{\beta\xi}}{\partial u^\alpha}-\frac{\partial g_{\alpha\beta}}{\partial u^\xi})} Γαβγ=21gγξ(∂uβ∂gαξ+∂uα∂gβξ−∂uξ∂gαβ)
在上式中令 α = γ , β = α \alpha=\gamma,\beta=\alpha α=γ,β=α,有
Γ γ α γ = 1 2 g γ ξ ( ∂ g γ ξ ∂ u α + ∂ g α ξ ∂ u γ − ∂ g γ α ∂ u ξ ) \Gamma_{\gamma\alpha}^\gamma=\frac12g^{\gamma\xi}(\frac{\partial g_{\gamma\xi}}{\partial u^\alpha}+\frac{\partial g_{\alpha\xi}}{\partial u^\gamma}-\frac{\partial g_{\gamma\alpha}}{\partial u^\xi}) Γγαγ=21gγξ(∂uα∂gγξ+∂uγ∂gαξ−∂uξ∂gγα)
分别代入 γ = 1 , 2 \gamma=1,2 γ=1,2 有
Γ 1 α 1 = 1 2 g 1 ξ ( ∂ g 1 ξ ∂ u α + ∂ g α ξ ∂ u 1 − ∂ g 1 α ∂ u ξ ) Γ 2 α 2 = 1 2 g 2 ξ ( ∂ g 2 ξ ∂ u α + ∂ g α ξ ∂ u 2 − ∂ g 2 α ∂ u ξ ) \Gamma_{1\alpha}^{1}=\frac12g^{1\xi}(\frac{\partial g_{1\xi}}{\partial u^\alpha}+\frac{\partial g_{\alpha\xi}}{\partial u^1}-\frac{\partial g_{1\alpha}}{\partial u^\xi})\\\Gamma_{2\alpha}^{2}=\frac12g^{2\xi}(\frac{\partial g_{2\xi}}{\partial u^\alpha}+\frac{\partial g_{\alpha\xi}}{\partial u^2}-\frac{\partial g_{2\alpha}}{\partial u^\xi}) Γ1α1=21g1ξ(∂uα∂g1ξ+∂u1∂gαξ−∂uξ∂g1α)Γ2α2=21g2ξ(∂uα∂g2ξ+∂u2∂gαξ−∂uξ∂g2α)
求和有
Γ 1 α 1 + Γ 2 α 2 = 1 2 g 11 ( ∂ g 11 ∂ u α ) + 1 2 g 12 ( ∂ g 12 ∂ u α + ∂ g α 2 ∂ u 1 − ∂ g 12 ∂ u 2 ) + 1 2 g 21 ( ∂ g 21 ∂ u α + ∂ g α 1 ∂ u 2 − ∂ g 2 α ∂ u 1 ) + 1 2 g 22 ( ∂ g 22 ∂ u α ) \begin{aligned}\Gamma_{1\alpha}^{1}+\Gamma_{2\alpha}^{2}&=\frac12g^{11}(\frac{\partial g_{11}}{\partial u^\alpha})+\frac12g^{12}(\frac{\partial g_{12}}{\partial u^\alpha}+\frac{\partial g_{\alpha2}}{\partial u^1}-\frac{\partial g_{12}}{\partial u^2})\\&+\frac12g^{21}(\frac{\partial g_{21}}{\partial u^\alpha}+\frac{\partial g_{\alpha1}}{\partial u^2}-\frac{\partial g_{2\alpha}}{\partial u^1})+\frac12g^{22}(\frac{\partial g_{22}}{\partial u^\alpha})\end{aligned} Γ1α1+Γ2α2=21g11(∂uα∂g11)+21g12(∂uα∂g12+∂u1∂gα2−∂u2∂g12)+21g21(∂uα∂g21+∂u2∂gα1−∂u1∂g2α)+21g22(∂uα∂g22)
因为 g 21 = g 12 g^{21}=g^{12} g21=g12 ,上式其实是
Γ 1 α 1 + Γ 2 α 2 = 1 2 g 11 ( ∂ g 11 ∂ u α ) + 1 2 ( g 12 + g 21 ) ( ∂ g 12 ∂ u α ) + 1 2 g 22 ( ∂ g 22 ∂ u α ) \Gamma_{1\alpha}^1+\Gamma_{2\alpha}^2=\frac12g^{11}(\frac{\partial g_{11}}{\partial u^\alpha})+\frac12(g^{12}+g^{21})(\frac{\partial g_{12}}{\partial u^\alpha})+\frac12g^{22}(\frac{\partial g_{22}}{\partial u^\alpha}) Γ1α1+Γ2α2=21g11(∂uα∂g11)+21(g12+g21)(∂uα∂g12)+21g22(∂uα∂g22)
再依据二阶情形逆矩阵的等价关系
( g 11 g 12 g 21 g 22 ) = 1 g ( g 22 − g 12 − g 21 g 11 ) \begin{pmatrix}g^{11}&g^{12}\\g^{21}&g^{22}\end{pmatrix}=\frac{1}{g}\begin{pmatrix}g_{22}&-g_{12}\\-g_{21}&g_{11}\end{pmatrix} (g11g21g12g22)=g1(g22−g21−g12g11)
修改系数就得到结论。