- #1
rezkyputra
- 5
- 2
Homework Statement
Okay, in Carrol's Intro to Spacetime and Geometry, Chapter 4, Eq. 4.63 to 4.65 require a derivation of a difference between Christoffel Symbol. I did the calculation and found my answer to be somewhat correct in form, but the indices doesn't match up
Homework Equations
So we need someway to change this prove this eq.
\begin{equation}
g^{ab} \delta \Gamma_{bc}^a - g^{ad}\delta \Gamma_{ac}^c = g_{ab} \, \nabla^d \, \delta g^{ab} - \nabla_c \, \delta g^{dc}
\end{equation}
also the variation of Chirstoffel Symbol is:
\begin{equation}
\delta \Gamma_{bc}^a = -\frac{1}{2} \left(g_{qc} \, \nabla_b \, \delta g^{aq} + g_{qb} \, \nabla_c \, \delta g^{aq} - g_{rb} \, g_{sc} \, \nabla^a \, \delta g^{rs} \right)
\end{equation}
The Attempt at a Solution
So first I'll break that into the first and second term
First Term
\begin{align}
\delta \Gamma_{bc}^a &= \frac{1}{2} \left( - g_{qb} \, \nabla_a \, \delta g^{dq} - g_{qa} \, \nabla_b \, \delta g^{dq} + g_{ra} \, g_{sb} \, \nabla^d \, \delta g^{rs} \right) \nonumber \\
g^{ab} \delta \Gamma_{bc}^a &= \frac{1}{2} \left( - g_{qb} \, g^{ab} \, \nabla_a \, \delta g^{dq} - g_{qa} \, g^{ab} \, \nabla_b \, \delta g^{dq} + g_{ra} \, g_{sb} \, g^{ab} \, \nabla^d \, \delta g^{rs} \right) \nonumber \\
g^{ab} \delta \Gamma_{bc}^a &= \frac{1}{2} \left( - g_{qb} \, g^{ab} \, \nabla_a \, \delta g^{dq} - g_{qa} \, g^{ab} \, \nabla_b \, \delta g^{dq} + g_{ra} \, g_{sb} \, \frac{1}{2} \left[g^{ab} + g^{ab} \right] \, \nabla^d \, \delta g^{rs} \right) \nonumber \\
&= \frac{1}{2} \left( - \delta_q^a \, \nabla_a \, \delta g^{dq} - \delta_q^b \, \nabla_b \, \delta g^{dq} + \frac{1}{2} \left[\delta_r^b \, g_{sb} + \delta_s^a \, g_{ra}\right] \, \nabla^d \, \delta g^{rs} \right) \nonumber \\
&= \frac{1}{2} \left(- \nabla_q \, \delta g^{dq} - \nabla_q \, \delta g^{dq} + \, g_{ab} \, \nabla^d \, \delta g^{ab} \right) \nonumber \\
&= - \nabla_q \, \delta g^{dq} + \frac{1}{2} \, g_{ab} \, \nabla^d \, \delta g^{ab} \nonumber \\
&q \ is \ "loose" \ so \ just \ set \ it \ equal \ to \ c \\
&= - \nabla_c \, \delta g^{dc} + \frac{1}{2} \, g_{ab} \, \nabla^d \, \delta g^{ab} \nonumber \\
\end{align}
Second Term
\begin{align}
- \delta \Gamma_{ac}^c &= \frac{1}{2} \left(g_{qc} \, \nabla_a \, \delta g^{cq} + g_{qa} \, \nabla_c \, \delta g^{cq} - g_{ra} \, g_{sc} \, \nabla^c \, \delta g^{rs} \right) \nonumber \\
- g^{ad}\delta \Gamma_{ac}^c &= \frac{1}{2} \left(g_{qc} \, g^{ad} \, \nabla_a \, \delta g^{cq} + g_{qa} \, g^{ad} \, \nabla_c \, \delta g^{cq} - g_{ra} \, g_{sc} \, g^{ad} \, \nabla^c \, \delta g^{rs} \right) \nonumber \\
&= \frac{1}{2} \left(g_{qc} \, \nabla^d \, \delta g^{cq} + \delta_q^d \, \nabla_c \, \delta g^{cq} -\delta_r^d \, \nabla_s \, \delta g^{rs} \right) \nonumber \\
&= \frac{1}{2} \left(g_{qc} \, \nabla^d \, \delta g^{cq} + \nabla_c \, \delta g^{cd} - \nabla_s \, \delta g^{ds} \right) \nonumber \\
&s \ is \ "loose" \ so \ just \ set \ it \ equal \ to \ c \\
&= \frac{1}{2} g_{qc} \, \nabla^d \, \delta g^{cq}
\end{align}
Now those two term adds up into
\begin{align}
&= \frac{1}{2} g_{qc} \, \nabla^d \, \delta g^{cq} + \frac{1}{2} \, g_{ab} \, \nabla^d \, \delta g^{ab} - \nabla_c \, \delta g^{dc}
\end{align}
Soo the last term is correct, but the first two, whila having the same form as the answer, their indices doesn't add up (i think)
so did i miss something? or are there anything wrong in my calculation?
Thanks in advance