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Orthonormal basis of 1 forms for the rotating c metric

  1. Mar 28, 2016 #1
    1. The problem statement, all variables and given/known data

    Write down an orthonormal basis of 1 forms for the rotating C-metric

    Use the result to find the corresponding dual basis of vectors


    See attached file for metric and appropriate equations


    The two equations on the left are for our vectors. the equations on the right are for our 1-forms/dual-vectors.


    3. The attempt at a solution
    g^μν= inverse metric on the manifold
    η^μν=inverse minkowski metric=diag(-1,1,1,1)= minkowski m etric
    E_a=non coordinate basis vectors for metric
    Θ_a=non coordinate 1-forms for metric

    I'm confused how we read off the g_μν from the metric above. Will this be equal to g^μν?

    Do we have to expand the brackets or are the coordinates

    1. dt-αx^2dφ
    2.dy
    3.dx
    4.dφ+αx^2

    and the g_μν components just the factors in front of these square rooted? What are the vierbeins?
    Any help would be appreciated.

    Thanks
     

    Attached Files:

  2. jcsd
  3. Apr 2, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Apr 4, 2016 #3

    stevendaryl

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    The metric components [itex]g_{\mu \nu}[/itex] are just the coefficients in the expansion of [itex]ds^2[/itex]. You write:

    [itex]ds^2 = g_{tt} dt^2 + 2 g_{tx} dt dx + 2 g_{ty} dt dy + 2 g_{t\varphi} dt d\varphi + g_{xx} dx^2 + 2 g_{xy} dx dy + 2 g_{x\varphi} dx d\varphi + g_{yy} dy^2 + 2 g_{y\varphi} dy d\varphi + g_{\varphi \varphi} d\varphi^2[/itex]

    (The reason for the factors of 2 is because it really should be [itex]g_{xy} dx dy + g_{yx} dy dx[/itex], but those two terms are equal, so I just wrote [itex]2 g_{xy} dx dy[/itex]). So if you expand your expression for [itex]ds^2[/itex], you can just read off the components [itex]g_{\mu \nu}[/itex].

    As for the second question: No, [itex]g^{\mu \nu} [/itex] is not equal to [itex]g_{\mu \nu}[/itex], in general. Viewed as 4x4 matrices, [itex]g^{\mu \nu}[/itex] is the inverse of [itex]g_{\mu \nu}[/itex]. Or in terms of components:

    [itex]\sum_{\alpha} g^{\mu \alpha} g_{\alpha \nu} = \delta^\mu_\nu[/itex], where [itex]\delta^\mu_\nu[/itex] is 1 if [itex]\mu = \nu[/itex] and zero otherwise.
     
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