# Deriving Einstein eq. from Variational Principle: DIfference Between Chistoffel Symbols

1. Jul 11, 2016

### rezkyputra

1. The problem statement, all variables and given/known data
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 doesnt match up

2. Relevant equations
So we need someway to change this prove this eq.

g^{ab} \delta \Gamma_{bc}^a - g^{ad}\delta \Gamma_{ac}^c = g_{ab} \, \nabla^d \, \delta g^{ab} - \nabla_c \, \delta g^{dc}

also the variation of Chirstoffel Symbol is:

\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)

3. 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 doesnt add up (i think)

so did i miss something? or are there anything wrong in my calculation?

2. Jul 11, 2016

### TSny

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

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

3. Jul 11, 2016

### 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 doesnt prove this eq

\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}