Exercise 16, chapter 3 (Tetrad) in Carroll

[Pouramat]

Homework Statement

Using Tetrad formalism, I am trying to solve exercise 16 carroll.
In final steps I cannot read elements of Riemann tensor correctly. please help.

Relevant Equations

$$
ds^2 = d \psi^2 + \sin^2 \theta (d \theta^2 + \sin^2 \theta d \phi^2)
$$

My attempt at solution:
in tetrad formalism:
$$ds^2=e^1e^1+e^2e^2+e^3e^3≡e^ae^a$$
so we can read vielbeins as following:
$$
\begin{align}
e^1 &=d \psi;\\
e^2 &= \sin \psi \, d\theta;\\
e^3 &= \sin⁡ \psi \,\sin⁡ \theta \, d\phi
\end{align}
$$
componets of spin connection could be written by using $de^a+{\omega^a}_b∧e^b=0$ as:
$$
\begin{align}
de^1+ {\omega^1}_2 \wedge e^2 + {\omega^1}_3 \wedge e^3 = 0 \\
de^2+ {\omega^2}_1 \wedge e^1 + {\omega^2}_3 \wedge e^3 = 0 \\
de^3+ {\omega^3}_1 \wedge e^1 + {\omega^3}_2 \wedge e^2 = 0
\end{align}
$$
Now we can lower the indices of $\omega$s and write:
$$
\begin{align}
{\omega^1}_2 = \omega_{12} = -\omega_{21} = \omega_1 \\
{\omega^1}_3 = \omega_{13} = -\omega_{31} = \omega_3 \\
{\omega^2}_3 = \omega_{23} = -\omega_{32} = \omega_2
\end{align}
$$
Now I can rewrite torsionless equations using :
$$
\begin{align}
de^1+ \omega_1 \wedge e^2 + \omega_3 \wedge e^3 = 0 \\
de^2 - \omega_1 \wedge e^1 + \omega_2 \wedge e^3 = 0 \\
de^3 - \omega_3 \wedge e^1 - \omega_2 \wedge e^2 = 0
\end{align}
$$
Now we can calculate exterior derivative of e's:
$$
\begin{align*}
de^1 &= d(d \psi) = 0 \\
de^2 &= \cos \psi \, d\psi \wedge d\theta = \cot \psi \,e^1 \wedge e^2 \\
de^3 &= \cos \psi \, \sin \theta \, d\psi \wedge d \phi+ \sin \psi \, \cos \theta \, d\theta \wedge d \phi = \cot \psi \, e^1 \wedge e^3 + \frac{\cot \theta}{\sin \psi} e^2 \wedge e^3
\end{align*}
$$
We can read ωs from the equations:
$$
\begin{align}
\omega_1 &= \cot \psi \, d \psi \\
\omega_2 &= -\cos \theta \, d \phi \\
\omega_3 &= \cos \psi \, \sin \theta \, d \psi
\end{align}
$$
Using Cartan's structure equation for ${R^a}_b=d{\omega^a}_b+{\omega^a}_c{\omega^c}_b$:
$$
\begin{align}
{R^1}_2 &= d{\omega^1}_2 + {\omega^1}_3 \wedge{\omega^3}_2 \\
{R^1}_3 &= d{\omega^1}_3 + {\omega^1}_2 \wedge{\omega^2}_3 \\
{R^2}_3 &= d{\omega^2}_3 + {\omega^2}_1 \wedge{\omega^1}_3
\end{align}
$$$$
\begin{align}
{R^1}_2 &= 0 \\
{R^1}_3 &= \sin \theta \, d \theta \wedge d\phi - \cot \psi \, \cos \psi \, \sin \theta \, d \psi \wedge d\phi \\
{R^2}_3 &= - \sin \psi \, \sin \theta \, d \psi \wedge d\phi + \cos \psi \, \cos \theta \, d\theta \wedge d\phi - \cot \psi \, \cos \theta \, d\psi \wedge d\phi
\end{align}
$$
The Problem is now begining, I cannot read elements of Reimann tensor using these equations. Can anyone help?

 

[TSny]

I think everything is OK until you get to the following:

We can read ωs from the equations:

$\omega_1 = \cot \psi \, d \psi$
$\omega_2 = -\cos \theta \, d \phi$
$\omega_3 = \cos \psi \, \sin \theta \, d \psi$

I get different results for $\omega_1$ and $\omega_3$. Can you show how you got these?

The Problem is now begining, I cannot read elements of Reimann tensor using these equations. Can anyone help?

You can follow Carroll's example that starts on page 490. In particular, equation (J.49) is used to get the components $R{^\rho}{_{\sigma \mu \nu}}$ of the Riemann tensor. Here the indices denote $\psi$, $\theta$, or $\phi$. For example, you could work out $R{^\psi}{_{\theta \psi \theta}}$.

 

[Pouramat]

Dear TSny;
I revised my notes and got the mistake in it, as you mentioned:
$$
\begin{align}
{\omega^2}_1 &= -{\omega^1}_2 = -\cot \psi \, e^2\\
{\omega^3}_1 &= -{\omega^1}_3 = \frac{cos \psi}{\sin^2 \psi} \, e^3 \\
{\omega^3}_2 &= -{\omega^2}_3 = \frac{\cot \theta}{\sin \psi} \, e^3
\end{align}
$$

 

[romsofia]

The Problem is now begining, I cannot read elements of Reimann tensor using these equations. Can anyone help?

http://sites.science.oregonstate.edu/physics/coursewikis/GDF/book/gdf/components.html
and $\Omega^i_j$ is defined here:http://sites.science.oregonstate.edu/physics/coursewikis/GDF/book/gdf/curvature.html

 

[TSny]

Dear TSny;
I revised my notes and got the mistake in it, as you mentioned:
$$
\begin{align}
{\omega^2}_1 &= -{\omega^1}_2 = -\cot \psi \, e^2\\
{\omega^3}_1 &= -{\omega^1}_3 = \frac{cos \psi}{\sin^2 \psi} \, e^3 \\
{\omega^3}_2 &= -{\omega^2}_3 = \frac{\cot \theta}{\sin \psi} \, e^3
\end{align}
$$

These look good except that I believe you have the wrong sign for ${\omega^2}_1$.

I think you should express these in terms of $d\psi$, $d\theta$, and $d\phi$ so that you can get the ${R^a}_b$ in terms of $d\psi$, $d\theta$, and $d\phi$ as you did in your first post.

I am far from being an expert in GR. This is the first example of using the tetrad formalism that I have ever worked through. So, we are going through this together.

Once we get the ${R^a}_b$, we can finally get the various ${R^\rho}_{\sigma \mu \nu}$.

 

[Pouramat]

Merry X-mas
Ok, let's start:
$$
\begin{align}
{\omega^1}_2 &= -{\omega^2}_1 = -\cot \psi \, e^2 = - \cos \psi \, d\theta\\
{\omega^3}_1 &= -{\omega^1}_3 = \frac{cos \psi}{\sin^2 \psi} \, e^3 = \cot \psi \, \sin \theta \, d\phi\\
{\omega^3}_2 &= -{\omega^2}_3 = \frac{\cot \theta}{\sin \psi} \, e^3 = \cos \theta \, \sin \theta \, d\phi
\end{align}
$$
taking exterior derivative:
$$
\begin{align}
d{\omega^1}_2 &= \sin \psi \, d\psi \wedge d\theta \\
d{\omega^1}_3 &= -\cot \psi \, \cos \theta \, d\theta \wedge d\phi + \frac{\sin \theta}{\sin^2 \psi} \, d \psi \wedge d\phi \\
d{\omega^2}_3 &= (-sin^2 \theta + \cos^2 \theta)\, d\theta \wedge d\phi
\end{align}
$$
now we can evaluate Riemann tensor:
$$
\begin{align}
{R^1}_2 &= d {\omega^1}_2 + {\omega^1}_3 \wedge {\omega^3}_2 = \sin \psi \, d\psi \wedge d\theta +(\cot \psi \, \sin \theta \, d\phi) \wedge (\cos \theta \, \sin \theta \, d\phi) = \sin \psi \, d\psi \wedge d\theta \\

{R^1}_3 &= d {\omega^1}_3 + {\omega^1}_2 \wedge {\omega^2}_3 = -\cot \psi \, \cos \theta \, d\theta \wedge d\phi + \frac{\sin \theta}{\sin^2 \psi} \, d \psi \wedge d\phi + ( \cos \psi \, d\theta) \wedge (\cos \theta \, \sin \theta \, d\phi) = -\cot \psi \, \cos \theta \, d\theta \wedge d\phi + \frac{\sin \theta}{\sin^2 \psi} \, d \psi \wedge d\phi + \cos \psi \, \cos \theta \, \sin \theta \, d\theta \wedge d\phi \\

{R^2}_3 &= d {\omega^2}_3 + {\omega^2}_1 \wedge {\omega^1}_3 =
(-sin^2 \theta + \cos^2 \theta)\, d\theta \wedge d\phi - (\cos \psi \, d\theta) \wedge (\cot \psi \, \sin \theta \, d\phi) = (-sin^2 \theta + \cos^2 \theta)\, d\theta \wedge d\phi - \sin \psi \, \sin \theta \, d\theta \wedge d\phi
\end{align}
$$
ok?

 

[TSny]

Merry X-mas

Thank you!

Ok, let's start:
$$
\begin{align}
{\omega^1}_2 &= -{\omega^2}_1 = -\cot \psi \, e^2 = - \cos \psi \, d\theta\\
{\omega^3}_1 &= -{\omega^1}_3 = \frac{cos \psi}{\sin^2 \psi} \, e^3 = \cot \psi \, \sin \theta \, d\phi\\
{\omega^3}_2 &= -{\omega^2}_3 = \frac{\cot \theta}{\sin \psi} \, e^3 = \cos \theta \, \sin \theta \, d\phi
\end{align}
$$

For ${\omega^3}_1$, I overlooked an error you had back in post #3. There you had ${\omega^3}_1 = \frac{\cos \theta}{\sin^2 \theta} e^3$. The $\sin \theta$ should not be squared.

For ${\omega^3}_2$, I don't believe the $\sin \theta$ should be in the final expression.

My results are:
$$
\begin{align}
{\omega^1}_2 &= -{\omega^2}_1 = - \cos \psi \, d\theta\\
{\omega^3}_1 &= -{\omega^1}_3 = \cos\psi \, \sin \theta \, d\phi\\
{\omega^3}_2 &= -{\omega^2}_3 = \cos \theta \, d\phi
\end{align}
$$