Tetrad Formalism: Finding e^{a \mu}?

[ngkamsengpeter]

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

Thanks.

 

[Mentz114]

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

 

[ngkamsengpeter]

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

Thanks. Let see if I understand it correctly.

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

 

[Mentz114]

Thanks. Let see if I understand it correctly.

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

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.

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

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.

 

[ngkamsengpeter]

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.

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

Thanks.

 

[Mentz114]

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

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

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

 

[George Jones]

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

Thanks.

Are you familiar the coordinate basis [itex]\left\{ \partial / \partial x^\mu \right\}[/itex] associated with the coordinate system [itex]\left\{ x^\mu \right\}[/itex]?

 

[ngkamsengpeter]

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

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

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.