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Question about Tetrad

  1. Apr 11, 2013 #1
    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.
     
  2. jcsd
  3. Apr 11, 2013 #2

    Mentz114

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

    Attached Files:

  4. Apr 11, 2013 #3
    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.
     
  5. Apr 11, 2013 #4

    Mentz114

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    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
  6. Apr 11, 2013 #5
    Can you show me an example so that it is much clearer?

    Thanks.
     
  7. Apr 11, 2013 #6

    Mentz114

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    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 :wink:

    [edit]
    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
  8. Apr 11, 2013 #7

    George Jones

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    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]?
     
  9. Apr 18, 2013 #8
    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.
     
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