1. Apr 11, 2013

### ngkamsengpeter

I am new to tetrad formalism in general relativity. I understand that $e^{a}_{\mu}$ is the component of a tetrad basis but what is meaning of $e^{a \mu}$ and how do i find it? For example, $e^{a}_{\mu}$ is a diagonal matrix (a,b,c,d), how do I find $e^{a \mu}$? Just raise the index using metric tensor $g^{\nu \mu}$?

Thanks.

2. Apr 11, 2013

### Mentz114

There is a good explanation in section 5.8 of the attached notes.

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3. Apr 11, 2013

### ngkamsengpeter

Thanks. Let see if I understand it correctly.

Let say my metric tensor $g_{\mu \nu}$ is a diagonal matrix (a^2,b^2,c^2,d^2) and we choose the tetrad $e^{m}_{\mu}$ as (a,b,c,d), then the $\gamma_{mn}$ is equal to (1,1,1,1) right? And we use this $\gamma_{mn}$ to raise and lower the tetrad indices m and n right?

Besides, if my metric tensor is not a diagonal matrix, is there a simple way to find the corresponding tetrad?

Thanks.

4. Apr 11, 2013

### Mentz114

According to equ(3), for an orthonormal tetrad, $\gamma_m \cdot\gamma_n = \eta_{mn}$, but in general $\gamma_m \cdot\gamma_n = \gamma_{mn}$. I assume the dot is the tensor product.

If you can write the line element as a sum of squares, it's possible to read off the static frame field. But for the frame field of a general worldline, I don't know a method that works every time.
I have had success finding a frame cobasis for a 4-velocity $u^\mu$ by setting the components of the cobasis vectors so ${\vec{e}}_0=u_\mu$, which points the time-like cobasis vector in the right direction. One has to guess the structure of the other ${\vec{e}}_i$, while keeping the vectors independent, and put in unknowns for the components. Then calculate $g_{\mu\nu} = -{\vec{e}}_0 \otimes {\vec{e}}_0 + \sum_i {\vec{e}}_i \otimes {\vec{e}}_i$ and iteratively eliminate the unknowns. There could be better ways to do this, but I don't know them.

( The tensor product above corresponds to equ(12) in the notes, with $\gamma=\eta$, if the rows of the tetrad ${e^m}_\mu$ are the cobasis vectors)).

If that isn't clear I can show you an example.

Last edited: Apr 11, 2013
5. Apr 11, 2013

### ngkamsengpeter

Can you show me an example so that it is much clearer?

Thanks.

6. Apr 11, 2013

### Mentz114

I'm going to use matrices to show this because it will be quicker. Given this metric
$\left[ \begin{array}{cccc} g00 & g01 & 0 & 0\\\ g01 & g11 & 0 & 0\\\ 0 & 0 & g22 & 0\\\ 0 & 0 & 0 & g33\end{array} \right]$
and making a guess at the cobasis vectors,
$\mathbf{e_0}=\left[ \begin{array}{cccc} a & b & 0 & 0\end{array} \right],\ \mathbf{e_1}=\left[ \begin{array}{cccc} c & d & 0 & 0\end{array} \right],\ \mathbf{e_2}=\left[ \begin{array}{cccc} 0 & 0 & \sqrt{g22} & 0\end{array} \right],\ \mathbf{e_3}=\left[ \begin{array}{cccc} 0 & 0 & 0 & \sqrt{g33}\end{array} \right]$
where a,b,c,d are unknown. Now we calculate the tensor product of the covector ( see my previous) and get
$\left[ \begin{array}{cccc} {c}^{2}-{a}^{2} & c\,d-a\,b & 0 & 0\\\ c\,d-a\,b & d^2-b^2 & 0 & 0\\\ 0 & 0 & g22 & 0\\\ 0 & 0 & 0 & g33\end{array} \right]$
Now we can start eliminating a,b,c and d. For instance, we have the equation
$c\,d-a\,b=g01$ or $c=\frac{g01+a\,b}{d}$. Now substitute this value into the cobasis vectors and recalculate the tensor product and repeat.

To give some physical meaning to this choose ${\mathbf{e}}_0$ to be the 4-velocity covector of some worldline. For a static observer b will be zero.

This is a basic, brute-force method ( I have software to do this ) and maybe someone will come up with something elegant.

Over to you, have fun

I get $b=0$, $a=\sqrt{\frac{{g01}^{2}}{g11}-g00}$, $c=\frac{g01}{\sqrt{g11}}$ and $d=\sqrt{g11}$

Last edited: Apr 11, 2013
7. Apr 11, 2013

### George Jones

Staff Emeritus
Are you familiar the coordinate basis $\left\{ \partial / \partial x^\mu \right\}$ associated with the coordinate system $\left\{ x^\mu \right\}$?

8. Apr 18, 2013

### ngkamsengpeter

Thanks for the example.

Actually I am trying to study the dirac equation in curved spacetime. Do you have example how to calculate the spin connection?

Thanks.