3.3.3. Covariant Differentiation and Christoffel Symbol (2) - 3.3.3.3 Generalization of Covariant Differentiation & 3.3.3.4 Christoffel Transformation

3.3.3.3 Generalization of Covariant Differentiation (Moore의 17장 및 Problem 참조)

<Reference>

• https://physicspages.com/Moore%20Relativity.html Moore Problem 17.2 COVARIANT DERIVATIVE OF A GENERAL TENSOR”

■  앞장(chapter 3.3.3.2)의 공변 미분은 Tensor Type (1,0) 혹은 Tensor Type(0,1) 경우에 해당한다. 다음은 Mixed Tensor Type(1,2)의 공변 미분을 유도한다. 동 방법은 Moore 참조

$D_{bc}^aA_a^{}B_{}^bC_{}^c = scalar$ 이므로 ${\nabla _d}\left( {D_{bc}^aA_a^{}B_{}^bC_{}^c} \right) = {\partial _d}\left( {D_{bc}^aA_a^{}B_{}^bC_{}^c} \right)$가 성립

${\partial _d}\left( {D_{bc}^aA_a^{}B_{}^bC_{}^c} \right) = \left( {{\partial _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c + D_{bc}^a\left( {{\partial _d}A_a^{}} \right)B_{}^bC_{}^c + D_{bc}^aA_a^{}\left( {{\partial _d}B_{}^b} \right)C_{}^c + D_{bc}^aA_a^{}B_{}^b\left( {{\partial _d}C_{}^c} \right)$ ${\nabla _d}\left( {D_{bc}^aA_a^{}B_{}^bC_{}^c} \right) = \left( {{\nabla _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c + D_{bc}^a\left( {{\nabla _d}A_a^{}} \right)B_{}^bC_{}^c + D_{bc}^aA_a^{}\left( {{\nabla _d}B_{}^b} \right)C_{}^c + D_{bc}^aA_a^{}B_{}^b\left( {{\nabla _d}C_{}^c} \right)$ ${\nabla _d}\left( {D_{bc}^aA_a^{}B_{}^bC_{}^c} \right) = \left( {{\nabla _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c + D_{bc}^a\left( {{\nabla _d}A_a^{}} \right)B_{}^bC_{}^c + D_{bc}^aA_a^{}\left( {{\nabla _d}B_{}^b} \right)C_{}^c + D_{bc}^aA_a^{}B_{}^b\left( {{\nabla _d}C_{}^c} \right)$ $= \left( {{\nabla _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c + D_{bc}^a\left( {{\partial _d}A_a^{} - \;\Gamma _{ad}^m{A_m}} \right)B_{}^bC_{}^c + D_{bc}^aA_a^{}\left( {{\partial _d}{B^b} + \Gamma _{md}^b\;{B^m}} \right)C_{}^c$ $+ D_{bc}^aA_a^{}B_{}^b\left( {{\partial _d}{C^c} + \Gamma _{md}^c\;{C^m}} \right)$

${\nabla _d}\left( {D_{bc}^aA_a^{}B_{}^bC_{}^c} \right) = {\partial _d}\left( {D_{bc}^aA_a^{}B_{}^bC_{}^c} \right)$ 이므로 $\left( {{\nabla _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c + D_{bc}^a\left( {{\partial _d}A_a^{} - \;\Gamma _{ad}^m{A_m}} \right)B_{}^bC_{}^c + D_{bc}^aA_a^{}\left( {{\partial _d}{B^b} + \Gamma _{md}^b\;{B^m}} \right)C_{}^c + D_{bc}^aA_a^{}B_{}^b\left( {{\partial _d}{C^c} + \Gamma _{md}^c\;{C^m}} \right)$ $= \left( {{\partial _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c + D_{bc}^a\left( {{\partial _d}A_a^{}} \right)B_{}^bC_{}^c + D_{bc}^aA_a^{}\left( {{\partial _d}B_{}^b} \right)C_{}^c + D_{bc}^aA_a^{}B_{}^b\left( {{\partial _d}C_{}^c} \right)$

$\left( {{\nabla _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c = \left( {{\partial _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c + D_{bc}^a\Gamma _{ad}^m{A_m}B_{}^bC_{}^c - D_{bc}^a\Gamma _{md}^bA_a^{}{B^m}C_{}^c - D_{bc}^a\Gamma _{md}^cA_a^{}B_{}^b\;{C^m}$
$D_{bc}^a\Gamma _{ad}^m{A_m}B_{}^bC_{}^c\;$ 항은 $m \to a,\;a \to m$ 이면, $D_{bc}^m\Gamma _{md}^a{A_a}B_{}^bC_{}^c$
$D_{bc}^a\Gamma _{md}^bA_a^{}{B^m}C_{}^c$ 항은 $m \to b,\;b \to m$ 이면, $D_{mc}^a\Gamma _{bd}^mA_a^{}{B^b}C_{}^c$.
$D_{bc}^a\Gamma _{md}^cA_a^{}B_{}^b\;{C^m}$ 항은 $m \to c,\;c \to m$ 이면, $D_{bm}^a\Gamma _{cd}^mA_a^{}B_{}^b\;{C^c}$

$\left( {{\nabla _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c = \left( {{\partial _d}D_{bc}^a} \right)A_a^{}B_{}^bC_{}^c + D_{bc}^m\Gamma _{md}^a{A_a}B_{}^bC_{}^c - D_{mc}^a\Gamma _{bd}^mA_a^{}{B^b}C_{}^c - D_{bm}^a\Gamma _{cd}^mA_a^{}B_{}^b\;{C^c}$ ${\nabla _d}D_{bc}^a = {\partial _d}D_{bc}^a + \Gamma _{md}^aD_{bc}^m - D_{mc}^a\Gamma _{bd}^m - D_{bm}^a\Gamma _{cd}^m$

■  공변 미분의 일반화

3.3.3.4 Christoffel Transformation

<Reference>

• Ryder, “Introduction to General Relativity 3.11. Some relations involving connection coefficients””

■  ${\nabla _{\rm{c}}}{\rm{A}}{{\rm{'}}_{\rm{a}}} = {\rm{\;A'}}_{{\rm{a\;\;}};{\rm{c}}}^{} = {\rm{\;\;A'}}_{{\rm{\;a\;}},{\rm{\;c}}}^{} - {\rm{\;\Gamma }}_{{\rm{ac}}}^{\bf{b}}{\rm{A}}{{\rm{'}}_{\bf{b}}}$

○ ${\rm{A'}}_{{\rm{a\;\;}};{\rm{c}}}^{}$와 ${\rm{A}}{{\rm{'}}_{\bf{b}}}$는 Tensor인 반면, ${\rm{A'}}_{{\rm{\;a\;}},{\rm{\;c}}}^{}$와 ${\rm{\Gamma }}_{{\rm{ac}}}^{\bf{b}}$는 Tensor가 아님

○ ${\rm{A'}}_{{\rm{a\;\;}};{\rm{c}}}^{}$와 ${\rm{A}}{{\rm{'}}_{\bf{b}}}$는 Tensor 이므로 각각 다음을 만족한다

♦ A a ;c' =∂ xd∂ x'a∂ xe∂ x'cAd ;e

♦ A'b=∂ xf∂ x'bAf

○ ${\rm{A'}}_{{\rm{\;a\;}},{\rm{\;c}}}^{} = \frac{\partial }{{\partial {x^{{\rm{'c}}}}}}\left( {{\rm{A'}}_{{\rm{\;a}}}^{}} \right) = \frac{\partial }{{\partial {x^{{\rm{'c}}}}}}\left( {\frac{{\partial {x^{\rm{f}}}}}{{\partial {x^{{\rm{'a}}}}}}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{}} \right) = \frac{{\partial {x^{\rm{f}}}}}{{\partial {x^{{\rm{'a}}}}}}{\rm{\;}}\frac{{\partial {\rm{A}}_{{\rm{\;}}{\bf{f}}}^{}}}{{\partial {x^{{\rm{'c}}}}}} + \frac{{{\partial ^2}{x^{\rm{f}}}}}{{\partial {x^{{\rm{'c}}}}\partial {x^{{\rm{'\;a}}}}}}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{} = \frac{{\partial {x^{\rm{f}}}}}{{\partial {x^{{\rm{'a}}}}}}{\rm{\;}}\frac{{\partial {x^{\bf{d}}}}}{{\partial {x^{{\rm{'c}}}}}}\frac{{\partial {\rm{A}}_{{\rm{\;}}{\bf{f}}}^{}}}{{\partial {x^{\bf{d}}}}} + \frac{{{\partial ^2}{x^{\rm{f}}}}}{{\partial {x^{{\rm{'c}}}}\partial {x^{{\rm{'\;a}}}}}}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{}{\rm{\;}}$ $= \frac{{\partial {x^{\rm{f}}}}}{{\partial {x^{{\rm{'a}}}}}}{\rm{\;}}\frac{{\partial {x^{\bf{d}}}}}{{\partial {x^{{\rm{'c}}}}}}\partial _{\rm{d}}^{}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{} + \frac{{{\partial ^2}{x^{\rm{f}}}}}{{\partial {x^{{\rm{'c}}}}\partial {x^{{\rm{'\;a}}}}}}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{}$

♦ Partial Derivative 과정과 동일하나 다음과 같이 변환함을 주의

♦ ${\rm{A}}{{\rm{'}}_{\bf{b}}} = \frac{{\partial {\rm{\;}}{x^{\rm{f}}}}}{{\partial {\rm{\;}}{x^{{\rm{'b}}}}}}{\rm{A}}_{\bf{f}}^{}$에서 ${\rm{A}}_{\bf{f}}^{}$와 동일하게 하기 위하여 ${\rm{A}}_{{\rm{\;a}}}^{} = \frac{{\partial {x^{\rm{f}}}}}{{\partial {x^{{\rm{'a}}}}}}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{}$로 공변 변환

♦ ${\rm{A}}_{{\rm{\;a\;\;}};{\rm{c}}}^{{\rm{'\;}}} = \frac{{\partial {\rm{\;}}{x^{\rm{d}}}}}{{\partial {\rm{\;}}{x^{{\rm{'a}}}}}}\frac{{\partial {\rm{\;}}{x^{\rm{e}}}}}{{\partial {\rm{\;}}{x^{{\rm{'c}}}}}}{\rm{A}}_{{\rm{d\;\;}};{\rm{e}}}^{}$의 d가 나타나도록 $\frac{{\partial {\rm{A}}_{{\rm{\;}}{\bf{f}}}^{}}}{{\partial {x^{{\rm{'c}}}}}} = \frac{{\partial {x^{\bf{d}}}}}{{\partial {x^{{\rm{'c}}}}}}\frac{{\partial {\rm{A}}_{{\rm{\;}}{\bf{f}}}^{}}}{{\partial {x^{\bf{d}}}}} = \frac{{\partial {x^{\bf{d}}}}}{{\partial {x^{{\rm{'c}}}}}}\partial _{\rm{d}}^{}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{}$게 변환

○ ${\rm{A}}_{{\rm{a\;\;}};{\rm{c}}}^{} = {\rm{\;\;A}}_{{\rm{\;a\;}},{\rm{\;c}}}^{} - {\rm{\;\Gamma }}_{{\rm{ac}}}^{\bf{b}}{{\rm{A}}_{\bf{b}}}$에 위에서 산출한 식들을 대입하면,

♦ $\frac{{\partial {\rm{\;}}{x^{\rm{d}}}}}{{\partial {\rm{\;}}{x^{{\rm{'a}}}}}}\frac{{\partial {\rm{\;}}{x^{\rm{e}}}}}{{\partial {\rm{\;}}{x^{{\rm{'c}}}}}}{\rm{A}}_{{\rm{d\;\;}};{\rm{e}}}^{} = \frac{{\partial {x^{\rm{f}}}}}{{\partial {x^{{\rm{'a}}}}}}{\rm{\;}}\frac{{\partial {x^{\bf{d}}}}}{{\partial {x^{{\rm{'c}}}}}}\partial _{\rm{d}}^{}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{} + \frac{{{\partial ^2}{x^{\rm{f}}}}}{{\partial {x^{{\rm{'c}}}}\partial {x^{{\rm{'\;a}}}}}}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{} - {\rm{\Gamma }}_{{\rm{ac}}}^{\bf{b}}\frac{{\partial {\rm{\;}}{x^{\rm{f}}}}}{{\partial {\rm{\;}}{x^{{\rm{'b}}}}}}{\rm{A}}_{\bf{f}}^{}$

♦ 우변에서 A Vector가 f로 일치되어 있음을 알 수 있음

♦ 공변 미분(${\rm{A}}_{{\rm{a\;\;}};{\rm{c}}}^{} = {\rm{\;\;A}}_{{\rm{\;a\;}},{\rm{\;c}}}^{} - {\rm{\;\Gamma }}_{{\rm{ac}}}^{\bf{b}}{{\rm{A}}_{\bf{b}}}{\rm{\;}}$)을 이용하여 $\partial _{\rm{d}}^{}{\rm{A}}_{{\rm{\;}}{\bf{f}}}^{} = {\rm{A}}_{{\rm{\;f\;}},{\rm{\;d}}}^{}$를 표현하면, ${\rm{A}}_{{\rm{f\;\;}};{\rm{d}}}^{} = {\rm{\;\;A}}_{{\rm{\;f\;}},{\rm{\;d}}}^{} - {\rm{\;\Gamma }}_{{\rm{fd}}}^{\rm{e}}{{\rm{A}}_{\bf{e}}}$ 이므로 ${\rm{A}}_{{\rm{\;f\;}},{\rm{\;d}}}^{} = {\rm{A}}_{{\rm{f\;\;}};{\rm{d}}}^{} + {\rm{\;\Gamma }}_{{\rm{fd}}}^{\bf{e}}{{\rm{A}}_{\bf{e}}}$를 위 식에 대입
∂ xd∂ x'a∂ xe∂ x'cAd ;e=∂xf∂x'a ∂xd∂x'cAf ;d+ ΓfdeAe+∂2xf∂x'c∂x' aA f-Γacb∂ xf∂ x'bAf
=∂xf∂x'a ∂xd∂x'cAf ;d+∂xf∂x'a ∂xd∂x'cΓfdeAe+∂2xf∂x'c∂x' aA f-Γacb∂ xf∂ x'bAf
Γacb∂ xf∂ x'bAf=∂xf∂x'a ∂xd∂x'cΓfdeAe+∂2xf∂x'c∂x' aA f ∵∂ xd∂ x'a∂ xe∂ x'cAd ;e=∂xf∂x'a ∂xd∂x'cAf ;d , (d→f, e→d)

By relabelling (${\rm{A}}_{\bf{f}}^{}$ Vector가 임의의 Vector이므로 label 혹은 Index를 변경 가능)
Γacb∂ xf∂ x'bAf=∂xe∂x'a ∂xd∂x'cΓedfAf+∂2xf∂x'c∂x' aA f=∂xe∂x'a ∂xd∂x'cΓedfAf+∂2xf∂x'c∂x' aA f
Γacb∂ xf∂ x'b=∂xe∂x'a ∂xd∂x'cΓedf+∂2xf∂x'c∂x' a
양변에 $\frac{{\partial {\rm{\;}}{x^{{\rm{'b}}}}}}{{\partial {\rm{\;}}{x^{\rm{f}}}}}$를 곱하면,
$\therefore {\rm{\Gamma }}_{{\rm{ac}}}^{\bf{b}} = \frac{{\partial {\rm{\;}}{x^{{\rm{'b}}}}}}{{\partial {\rm{\;}}{x^{\rm{f}}}}}\frac{{\partial {x^{\bf{e}}}}}{{\partial {x^{{\rm{'a}}}}}}{\rm{\;}}\frac{{\partial {x^{\rm{d}}}}}{{\partial {x^{{\rm{'c}}}}}}{\rm{\Gamma }}_{{\rm{ed}}}^{\rm{f}} + \frac{{\partial {\rm{\;}}{x^{{\rm{'b}}}}}}{{\partial {\rm{\;}}{x^{\rm{f}}}}}\frac{{{\partial ^2}{x^{\rm{f}}}}}{{\partial {x^{{\rm{'c}}}}\partial {x^{{\rm{'\;a}}}}}}$

■  ∇cAa= A' ;ca= ∂cA'a+Γbca A'b

○ ${\rm{A}}_{{\rm{\;\;}};{\rm{c}}}^{\rm{a}}$와 ${{\rm{A}}^{\rm{b}}}$는 Tensor인 반면, ${\rm{A}}_{{\rm{\;\;}},{\rm{\;c}}}^{\rm{a}}$와 ${\rm{\Gamma }}_{{\rm{bc}}}^{\rm{a}}$는 Tensor가 아님

○ ${\rm{A}}_{{\rm{\;\;}};{\rm{c}}}^{\rm{a}}$와 ${{\rm{A}}^{\rm{b}}}$는 Tensor 이므로 각각 다음을 만족한다

♦ A ;c' a=∂ x'a∂ xd∂ xe∂ x'cA ;ed

♦ A' b=∂ x'b∂ xfAf

○ A' , ca=∂∂x'cA' a=∂∂x'c∂x'a∂xfA f=∂2x'a∂x'c∂xfA f+∂x'a∂xf ∂A f∂x'c=∂2x'a∂x'c∂xfA f+∂x'a∂xf∂xd∂x'c∂A f∂xd
=∂x'a∂xf∂xd∂x'c∂dA f+∂2x'a∂x'c∂xfA f

♦ Partial Derivative 과정과 동일하나 다음과 같이 변환함을 주의

♦ ${A^{{\rm{'\;}}{\bf{b}}}} = \frac{{\partial {\rm{\;}}{x^{{\rm{'b}}}}}}{{\partial {\rm{\;}}{x^{\rm{f}}}}}{\rm{A}}_{}^{\rm{f}}$에서 ${\rm{A}}_{}^{\bf{f}}$와 동일하게 하기 위하여 ${\rm{A'}}_{}^{\rm{a}} = \frac{{\partial {x^{{\rm{'a}}}}}}{{\partial {x^{\rm{f}}}}}{\rm{A}}_{\rm{\;}}^{\bf{f}}$로 반변 변환

♦ ${\rm{A}}_{{\rm{\;\;}};{\rm{c}}}^{{\rm{'\;a}}} = \frac{{\partial {\rm{\;}}{x^{{\rm{'a}}}}}}{{\partial {\rm{\;}}{x^{\rm{d}}}}}\frac{{\partial {\rm{\;}}{x^{\rm{e}}}}}{{\partial {\rm{\;}}{x^{{\rm{'c}}}}}}{\rm{A}}_{{\rm{\;\;}};{\rm{e}}}^{\rm{d}}$의 d가 나타나도록 $\frac{{\partial {\rm{A}}_{\rm{\;}}^{\bf{f}}}}{{\partial {x^{{\rm{'c}}}}}} = \frac{{\partial {x^{\bf{d}}}}}{{\partial {x^{{\rm{'c}}}}}}\frac{{\partial {\rm{A}}_{\rm{\;}}^{\bf{f}}}}{{\partial {x^{\bf{d}}}}} = \frac{{\partial {x^{\bf{d}}}}}{{\partial {x^{{\rm{'c}}}}}}\partial _{\rm{d}}^{}{\rm{A}}_{\rm{\;}}^{\bf{f}}$게 변환

○ ${\rm{A}}_{{\rm{\;\;}};{\rm{c}}}^{\rm{a}} = {\rm{\;\;A}}_{{\rm{\;\;}},{\rm{\;c}}}^{\rm{a}} - {\rm{\;\Gamma }}_{{\bf{b}}{\rm{c}}}^{\rm{a}}{\rm{A}}_{}^{\bf{b}}$에 위에서 산출한 식들을 대입하면,

♦ ∂ x'a∂ xd∂ xe∂ x'cA ;ed=∂x'a∂xf∂xd∂x'c∂dA f+∂2x'a∂x'c∂xfA f+Γbca∂ x'b∂ xfAf

♦ 우변에서 A Vector가 f로 일치되어 있음을 알 수 있음

♦ 공변 미분(${\rm{A'}}_{{\rm{\;\;}};{\rm{c}}}^{\rm{a}} = {\rm{\;\;A'}}_{{\rm{\;\;}},{\rm{\;c}}}^{\rm{a}} - {\rm{\;\Gamma }}_{{\bf{b}}{\rm{c}}}^{\rm{a}}{\rm{A'}}_{}^{\bf{b}}{\rm{\;}}$)을 이용하여 $\partial _{\rm{d}}^{}{\rm{A}}_{\rm{\;}}^{\bf{f}} = {\rm{A}}_{{\rm{\;\;}},{\rm{\;d}}}^{\rm{f}}$를 표현하면, ${\rm{A}}_{{\rm{\;\;}};{\rm{d}}}^{\rm{f}} = {\rm{\;\;A}}_{{\rm{\;\;}},{\rm{\;d}}}^{\rm{f}} + {\rm{\;\Gamma }}_{{\bf{e}}{\rm{d}}}^{\rm{f}}{\rm{A}}_{}^{\bf{e}}$ 이므로 $\partial _{\rm{d}}^{}{\rm{A}}_{\rm{\;}}^{\bf{f}} = {\rm{A}}_{{\rm{\;\;}},{\rm{\;d}}}^{\rm{f}} = {\rm{A}}_{{\rm{\;\;}};{\rm{d}}}^{\rm{f}} - {\rm{\;\Gamma }}_{{\bf{e}}{\rm{d}}}^{\rm{f}}{\rm{A}}_{}^{\bf{e}}$를 위 식에 대입
∂ x'a∂ xd∂ xe∂ x'cA ;ed=∂x'a∂xf∂xd∂x'cA ;df- ΓedfAe+∂2x'a∂x'c∂xfA f+Γbca∂ x'b∂ xfAf
=∂x'a∂xf∂xd∂x'cA ;df-∂x'a∂xf∂xd∂x'cΓedfAe+∂2x'a∂x'c∂xfA f+Γbca∂ x'b∂ xfAf
Γbca∂ x'b∂ xfAf=∂x'a∂xf∂xd∂x'cΓedfAe-∂2x'a∂x'c∂xfA f ∵∂ x'a∂ xd∂ xe∂ x'cA ;ed=∂x'a∂xf∂xd∂x'cA ;df , (d→f, e→d)

By relabelling (${\rm{A}}_{\bf{f}}^{}$ Vector가 임의의 Vector이므로 label 혹은 Index를 변경 가능)
Γbca∂ x'b∂ xfAf=∂x'a∂xe∂xd∂x'cΓfdeAf-∂2x'a∂x'c∂xfA f=∂x'a∂xe∂xd∂x'cΓfde-∂2x'a∂x'c∂xfA f (e→f, f→e)
Γbca∂ x'b∂ xf=∂x'a∂xe∂xd∂x'cΓfde-∂2x'a∂x'c∂xf

양변에 $\frac{{\partial {\rm{\;}}{x^{\rm{f}}}}}{{\partial {\rm{\;}}{x^{{\rm{'b}}}}}}$. 를 곱하면,
∴Γbca=∂xd∂x'cΓfde-∂ xf∂ x'b∂2x'a∂x'c∂xf=∂x'a∂xe∂ xf∂ x'b∂xd∂x'cΓfde-∂ xf∂x'b∂ xe∂x'c∂2x'a∂ xe∂xf =∂x'a∂xe∂ xf∂ x'b∂xd∂x'cΓfde-∂ xf∂x'b∂ xd∂x'c∂2x'a∂ xd∂xf

■  Ryder 책 연습문제에서는 쉽게 풀린다고 하는데…. 난 왜 이렇게 힘들게? 다른 책은?