Christoffel identity

  • Thread starter etotheipi
  • Start date
  • #1
etotheipi
Homework Statement:
We need to show that ##\Gamma_{cb}^e = \frac{1}{2}g^{ea}(\partial_c g_{ab} + \partial_b g_{ac} - \partial_a g_{cb})##
Relevant Equations:
N/A
We use ##g_{\alpha \beta} = \vec{e}_{\alpha} \cdot \vec{e}_{\beta}## to show that$$\partial_c g_{ab} = \partial_c (\vec{e}_a \cdot \vec{e}_b) = \vec{e}_a \cdot \partial_c \vec{e}_b + \vec{e}_b \cdot \partial_c \vec{e}_a$$then because ##\partial_{\alpha} \vec{e}_{\beta} := \Gamma_{\alpha \beta}^{\gamma} \vec{e}_{\gamma}##, we get$$\begin{align*}\partial_c g_{ab} = \vec{e}_a \cdot \Gamma_{cb}^{d} \vec{e}_{d} + \vec{e}_b \cdot \Gamma_{ca}^{d} \vec{e}_{d}

&= \Gamma_{cb}^{d} \vec{e}_a \cdot \vec{e}_d + \Gamma_{ca}^{d} \vec{e}_b \cdot \vec{e}_d \\

&= \Gamma_{cb}^{d} g_{ad} + \Gamma_{ca}^{d} g_{bd}

\end{align*}$$Now I am uncertain as to how to proceed. I wonder if someone can give me a hint? I try contracting both sides with ##g^{ea}## but I can't see how that helps. Thanks!
 

Answers and Replies

  • #2
Ibix
Science Advisor
Insights Author
2020 Award
8,320
7,742
I'm not certain what you know about Christoffel symbols at this point. If you are allowed to use that ##\Gamma^a_{bc}=\Gamma^a_{cb}## then you can permute the indices in your last expression and use that symmetry to get three equations in three terms like ##\Gamma^\cdot_{\cdot\cdot}g_{\cdot\cdot}##. That ought to get you where you're going.
 
  • #3
etotheipi
Thanks! I'll write out my solution then, using this fact about symmetry in the downstairs indices of the Christoffel symbols,$$\begin{align*}
\partial_c g_{ab} &= \Gamma_{bc}^d g_{ad} + \Gamma_{ca}^d g_{bd} \\
\partial_b g_{ca} &= \Gamma_{ab}^d g_{cd} + \Gamma_{bc}^d g_{ad} \\
\partial_a g_{bc} &= \Gamma_{ca}^d g_{bd} + \Gamma_{ab}^d g_{cd}
\end{align*}$$It follows that$$\partial_c g_{ab} + \partial_b g_{ca} - \partial_a g_{bc} = 2\Gamma_{bc}^d g_{ad}$$Now I halve both sides, and contract both sides with ##g^{ea}##, making use of the identity ##g^{\alpha \beta}g_{\beta \gamma} = \delta^{\alpha}_{\gamma}##,$$\frac{1}{2} g^{ea} \left(\partial_c g_{ab} + \partial_b g_{ca} - \partial_a g_{bc} \right) = \Gamma_{bc}^d g^{ea} g_{ad} = \Gamma^e_{bc}$$Nice! Muchas gracias señor Ibix!
 
  • Like
Likes PhDeezNutz and Ibix

Related Threads on Christoffel identity

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
8
Views
329
E
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
509
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
6
Views
2K
Top