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What class of metric is this?

  1. Nov 22, 2014 #1
    Hello,

    this is the metric I am talking about:

    $$ ds^2= (dt - A_idx^i)^2 - a^2(t)\delta_{ij}dx^idx^j $$

    I never see one like this. How the metric tensor matrix would be?
     
  2. jcsd
  3. Nov 22, 2014 #2

    Matterwave

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    How do you think you can obtain the metric tensor from the space-time interval that you gave?
     
  4. Nov 22, 2014 #3
    I think the matrix would be:
    $$ g_{\mu\nu} =
    \left( \begin{array}{ccc}
    1 & 2A_1 & 2A_2 & 2A_3 \\
    0 & A_1^2 - a^2 & 0 & 0 \\
    0 & 0 & A_2^2 - a^2 & 0 \\
    0 & 0 & 0 & A_3^2 - a^2 \end{array} \right) $$
     
  5. Nov 22, 2014 #4

    Matterwave

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    I think you are missing some more off-diagonal terms. The first parenthesis looks like it will also give terms that look like ##dx^1 dx^2## etc. Make sure to foil correctly. :)

    Also note that the metric must be a symmetric matrix, and your matrix is definitely not symmetric.
     
  6. Nov 22, 2014 #5
    Do you mean:

    $$ (A_i dx^i)^2 = A_i^2 dx^idx^j $$ or $$ 2 . A_i dt . dx^i \longrightarrow 2A_i dtdx^i + 2A_i dx^i dt $$

    So with the latter I get:

    $$ g_{\mu\nu} =
    \left( \begin{array}{ccc}
    1 & 2A_1 & 2A_2 & 2A_3 \\
    2A_1 & A_1^2 - a^2 & 0 & 0 \\
    2A_2 & 0 & A_2^2 - a^2 & 0 \\
    2A_3 & 0 & 0 & A_3^2 - a^2 \end{array} \right) $$
     
  7. Nov 22, 2014 #6

    Matterwave

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    Notice that ##A_i dx^i=A_1 dx^1+A_2 dx^2+A_3 dx^3##, then you want to take ##(dt-A_1 dx^1-A_2 dx^2-A_3 dx^3)^2## what do you get? You can see that there will be not only terms like ##dt dx^1## etc in there, there will be terms like ##dx^1 dx^2##, which are all 0 in your matrix.
     
  8. Nov 23, 2014 #7

    Mentz114

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    Are you sure ? I think ##a^2(t)\delta_{ij}dx^idx^j## means diagonal spatial elements only.

    The metric an FLRW type expanding cosmology with anisotropic matter flow.
     
  9. Nov 23, 2014 #8

    Matterwave

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    I don't see how you can avoid the off diagonal terms when clearly the squaring of the first term will give you terms like ##dx^1 dx^2##. There's a sum inside the square. I'm talking about the term ##(dt-A_i dx^i)^2##
     
  10. Nov 23, 2014 #9
    Mmm!! nice! I just only have a doubt, as I know the metric tensor is symmetric, when I obtain, for example, -2A2A3dx²dx³ should I write also in the ds² formula the symmetric term -2A3A2dx³dx² ? i think it is not necesary, right?

    Well the matrix I obtained now is:

    $$
    \left( \begin{array}{ccc}
    1 & -2A_1 & -2A_2 & -2A_3 \\
    -2A_1 & A_1^2 - a^2 & 2A_1 A_2 & 2A_1 A_3 \\
    -2A_2 & 2A_1 A_2 & A_2^2 - a^2 & 2A_2 A_3 \\
    -2A_3 & 2A_1 A_3 & 2A_2 A_3 & A_3^2 - a^2 \end{array} \right)
    $$
     
  11. Nov 23, 2014 #10

    Mentz114

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    Yes, I missed that sum. Very weird. Otherwise the metric is not unusual.
     
  12. Nov 23, 2014 #11

    Matterwave

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    To check your answer, try to obtain your original ##ds^2=(dt-A_i dx^i)^2-a^2 \delta_{ij} dx^i dx^j## by using ##ds^2=g_{\mu\nu}dx^\mu dx^\nu##. Match the two sides to see if they give you the same expression. :)
     
  13. Nov 23, 2014 #12
    So I must write explicitly the 16 terms in the ds² expression and the matrix seems right :)
     
  14. Nov 23, 2014 #13

    Matterwave

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    To check to make sure. But notice that since ##g_{\mu\nu}=g_{\nu\mu}## when you take the sum ##g_{\mu\nu}dx^\mu dx^\nu## you will get terms like ##dx^1 dx^2## and then another term like ##dx^2 dx^1## repeated, with the same factor in front.

    What I'm getting at is I think you maybe have a factor of 2 off on some of your off diagonal terms, so you might want to double check.
     
  15. Nov 23, 2014 #14
    Oh, I did not notice. If my intuition does not fail, the off-diagonal terms in the ds² equation when you obtain something like: -2A2A3dx²dx³ must be splitted in two terms dividing by two? so you would have: -A2A3dx²dx³ - A3A2dx³dx² ?
     
  16. Nov 23, 2014 #15

    Matterwave

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    Yeah, basically.
     
  17. Nov 23, 2014 #16
    :D

    $$
    \left( \begin{array}{ccc}
    1 & -A_1 & -A_2 & -A_3 \\
    -A_1 & A_1^2 - a^2 & A_1 A_2 & A_1 A_3 \\
    -A_2 & A_1 A_2 & A_2^2 - a^2 & A_2 A_3 \\
    -A_3 & A_1 A_3 & A_2 A_3 & A_3^2 - a^2 \end{array} \right)
    $$
     
  18. Nov 23, 2014 #17
    So now in order to define a natural vierbein I must diagonalize this matrix:

    $$ g_{\mu\nu} = e^{\alpha}_{\mu}\eta_{\alpha\beta}e^{\beta}_{\nu} $$

    right?
     
  19. Nov 23, 2014 #18

    Matterwave

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    You must diagonalize the metric into the form diag(-1,1,1,1) or diag(1,-1,-1,-1) depending on the signature.
     
  20. Nov 23, 2014 #19

    DrGreg

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    Yes. (In the sense that Matterwave just said.)

    But here is a hint. You should find it easier to work with the original form of the metric in post #1, rather than the matrix you have just calculated.
     
  21. Nov 23, 2014 #20
    Mmm interesting.

    I must find an analytical tranform to obtain something like ## (\alpha dt^2 +\beta_i (dx^i)²) ## from ##(dt - A_i dx^i)^2 ##... I am wondering how. Maybe second grade equations... roots... ?
     
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