General Relativity – Step-by-Step Guide to Calculating the Ricci Tensor

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I am currently working through an exercise to calculate the component $R_{22}$ of the Ricci tensor for the line element $ds^2=a^2dt^2 -a^2dx^2 – \frac{a^2e^{2x}}{2}dy^2 +2a^2e^xdydt -a^2dz^2$. The question first asks for the value of $\Gamma^{0}_{12}$, which I calculate to be $\frac{e^{x}}{2}$. I am told to assume the following values for the connection coefficients: $\Gamma^{0}_{12}=\Gamma^{0}_{21}=\Gamma^{0}_{10}=\Gamma^{0}_{01}=1$, $\Gamma^{1}_{22}=\frac{e^{2x}}{2}$, $\Gamma^{2}_{10}=-e^{-x}$, $\Gamma^{1}_{02}=\Gamma^{1}_{20}=\frac{e^{x}}{2}$ and all others are zero.

Using the relation for the Ricci tensor, I find that the only non-zero components are: $R_{22}=\partial_1(\Gamma^{0}_{12})+\Gamma^{0}_{10}\Gamma^{1}_{22}-\Gamma^{0}_{21}\Gamma^{1}_{02}-\Gamma^{1}_{20}\Gamma^{0}_{12}$. This is where the problem arises: using the assumed values for the connection coefficients (with $\Gamma^{0}_{12}=\Gamma^{0}_{21}=1$) I find that $R_{22}=e^{2x}-e^{x}$, while using the values $\Gamma^{0}_{12}=\Gamma^{0}_{21}=\frac{e^{x}}{2}$ (the rest being those assumed) I find that $R_{22}=e^{2x}$. I am told that the second result is correct. It seems to be the case that the assumed value for $\Gamma^{0}_{12}$ is incorrect.

Could someone provide clarification as to whether there is indeed a mistake in the question? As a beginner in GR, I find myself questioning the basics.

EDIT:

There was a mistype in the above metric the correct one should be: $ds^2=a^2dt^2 -a^2dx^2 + \frac{a^2e^{2x}}{2}dy^2 +2a^2e^xdydt -a^2dz^2$

This then gives:

$$g_{ab} = \left[\begin{matrix}a^{2} & 0 & a^{2} e^{x} & 0\\0 & – a^{2} & 0 & 0\\a^{2} e^{x} & 0 & + \frac{a^{2} e^{2 x}}{2} & 0\\0 & 0 & 0 & – a^{2}\end{matrix}\right]$$

and $$g^{ab} = \frac{1}{a^2}\left[\begin{matrix}-1 & 0 & 2 e^{-x} & 0\\0 & – 1 & 0 & 0\\ 2e^{-x} & 0 & -2e^{-2x} & 0\\0 & 0 & 0 & – 1\end{matrix}\right]$$

Apologies for the mistake.

Best Answer

According to the below python code the metric $$ \left[\begin{matrix}a^{2} & 0 & a^{2} e^{x} & 0\\0 & - a^{2} & 0 & 0\\a^{2} e^{x} & 0 & \frac{a^{2} e^{2 x}}{2} & 0\\0 & 0 & 0 & - a^{2}\end{matrix}\right] $$ has the following Christoffel symbols ${\Gamma^\mu}_{\nu\rho}$ of the second kind ${\Gamma^0}_{\nu\rho},{\Gamma^1}_{\nu\rho},{\Gamma^2}_{\nu\rho},{\Gamma^3}_{\nu\rho}$: \begin{align} \left[\begin{matrix}0 & 1 & 0 & 0\\1 & 0 & \frac{e^{x}}{2} & 0\\0 & \frac{e^{x}}{2} & 0 & 0\\0 & 0 & 0 & 0\end{matrix}\right], \left[\begin{matrix}0 & 0 & \frac{e^{x}}{2} & 0\\0 & 0 & 0 & 0\\\frac{e^{x}}{2} & 0 & \frac{e^{2 x}}{2} & 0\\0 & 0 & 0 & 0\end{matrix}\right], \left[\begin{matrix}0 & - e^{- x} & 0 & 0\\- e^{- x} & 0 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{matrix}\right], \left[\begin{matrix}0 & 0 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\\0 & 0 & 0 & 0\end{matrix}\right] \end{align} The Ricci tensor is \begin{align} R_{\mu\nu}=\left[\begin{matrix}1 & 0 & e^{x} & 0\\0 & 0 & 0 & 0\\e^{x} & 0 & e^{2 x} & 0\\0 & 0 & 0 & 0\end{matrix}\right] \end{align}

# https://itp.uni-frankfurt.de/~hanauske/VARTC/VARTCorona/python/GraviPy-tutorial.html

from gravipy.tensorial import *
from sympy import init_printing, pprint, print_latex, exp
import inspect
init_printing()


a, t, x, y, z, M = symbols('a t, x, y, z, M')

# create a coordinate four-vector object instantiating 
# the Coordinates class
C = Coordinates('\chi', [t, x, y, z])
# define a matrix of a metric tensor components
Metric = Matrix([[a**2,0,a**2*exp(x),0],[0,-a**2,0,0],[a**2*exp(x),0,a**2*exp(2*x)/2,0],[0,0,0,-a**2]])
print_latex( Metric )
# create a metric tensor object instantiating the MetricTensor class
g = MetricTensor('g', C, Metric)
Ga = Christoffel('Ga', g)
#for i in [1,2,3,4]: # first kind
#    print_latex( Ga(i,All,All) )
for i in [-1,-2,-3,-4]: # second kind
    print_latex( Ga(i,All,All) )
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