Deriving Einstein Eq. from Variational Principle

[rezkyputra]

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

 

[TSny]

Welcome to PF!

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}

First term on left is not written correctly. You have $a$ as an upper index twice.

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 \\

\end{align}

You have $a$ as an upper index on the left but lower on the right.

 

[rezkyputra]

Okay, it was a typo, i got it correct indices on my notes

so let me retype it

\begin{align}
\delta \Gamma_{ab}^d &= \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_{ab}^d &= \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_{ab}^d &= \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 \\
\label{eq:8}
&= - \nabla_c \, \delta g^{dc} + \frac{1}{2} \, g_{ab} \, \nabla^d \, \delta g^{ab}
\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 \\
\label{eq:9}
&= \frac{1}{2} g_{qc} \, \nabla^d \, \delta g^{cq}
\end{align}

they both still doesn't prove this eq

\begin{equation}
\label{eq:10}
g^{ab} \delta \Gamma_{ab}^d - g^{ad}\delta \Gamma_{ac}^c = g_{ab} \, \nabla^d \, \delta g^{ab} - \nabla_c \, \delta g^{dc}
\end{equation}

They still doesn't add up!

 

[TSny]

I believe eq's 8 and 9 do add to give eq. 10.