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Inverting a tensor equation

  1. Feb 18, 2008 #1
    1. The problem statement, all variables and given/known data
    I would like to know how to "invert" a tensor equation (see 2. Relevant equations).


    2. Relevant equations
    [tex]x^{a} = h^{a}{ } _{b} y^{b}[/tex].
    I just want to know how to get y-components in terms of x-components in general.



    3. The attempt at a solution
    I want to believe that [tex]h_{a}{ }^{b} h^{a}{ } _{c} = \delta^{b}{ } _{c}[/tex], but I'm not really sure that this is correct. Any help? Thanks.
     
    Last edited: Feb 18, 2008
  2. jcsd
  3. Feb 18, 2008 #2

    kdv

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    what is the definition of your h??
     
  4. Feb 18, 2008 #3
    [tex]h_{a b}=g_{a b}+k_{a} N_{b} + N_{a} k_{b}[/tex]

    where k and N are both null vectors and [tex]N_{a} k^{a} = -1[/tex]
    Here, h is called the transverse metric (i.e. transverse the the null vector fields k and N.
     
  5. Feb 18, 2008 #4

    kdv

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    Then you can calculate explicitly [tex] h_a^b h^a_c [/tex], no?
     
  6. Feb 18, 2008 #5
    I can expand it all out, but I don't know what to do with terms like [tex]g_{b}{}^{c} k_{c} N^{a}[/tex].
    I mean, I can't do the typical contraction-like operation with the metric g.
    Another thing: is [tex]g^{a}{}_{c} g_{b}{}^{c}=\delta^{a}{}_{b}[/tex] ?
    I'm pretty sure this is the case for the Minkowski metric, but what about a general metric?
     
    Last edited: Feb 18, 2008
  7. Feb 18, 2008 #6

    kdv

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    So [tex]h_{a }^b=g_{a}^b+k_{a} N^{b} + N_{a} k^{b}[/tex]
    and

    [tex]h^{a }_c=g^{a}_c+k^{a} N_{c} + N^{a} k_{c}[/tex]

    So then you may contract them over "a"
     
  8. Feb 18, 2008 #7

    kdv

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    [tex]g_{b}{}^{c} k_{c} N^{a}[/tex] is just [tex] k_b N^a [/tex]
     
  9. Feb 18, 2008 #8
    In that case, I'm getting [tex]h^{a}{}_{c} h_{b}{}^{c}=g^{a}{}_{c} g_{b}{}^{c}+2(k_{b} N^{a} + N_{b} k^{a} +1)=g^{a}{}_{c} g_{b}{}^{c} + 2(h^{a}{}_{b}-g^{a}{}_{b}+1)[/tex]. This doesn't look like the kronecker delta I want in order to verify that I have the inverse of h.
     
    Last edited: Feb 18, 2008
  10. Feb 18, 2008 #9

    kdv

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    Sorry, I just went back to your first post. I am not sure why you want o invert h. why not simply apply g_a^c to both sides of your first equation?
     
  11. Feb 18, 2008 #10
    [tex]g_{a}{}^{c} x^{a} =g_{a}{}^{c} h^{a}{ } _{b} y^{b}[/tex].
    [tex] x^{c} = h^{c}{ } _{b} y^{b}[/tex].
    I'm not getting it. Am I doing that wrong? I want to isolate the y-components in terms of the x-components. So my plan is to find the inverse of h, then apply it both sides of my first equation.
     
    Last edited: Feb 18, 2008
  12. Feb 18, 2008 #11
    The vectors [itex]k^\alpha,\,N^\alpha[/itex] are null eigenvectors of [itex]h_{\mu\nu}[/itex], i.e. [itex]h_{\mu\nu}\,N^\nu=0,\,h_{\mu\nu}\,k^\nu=0[/itex], thus [itex]h_{\mu\nu}\,h^{\nu\tau}=h_\mu{}^\tau[/itex].
    Now let [itex]x^\alpha=h^\alpha{}_\beta\,y^\beta \quad (\ast)[/itex] and contract with [itex]h^\gamma{}_\alpha[/itex] in order to get

    [tex]h^\gamma{}_\alpha\,x^\alpha=h^\gamma{}_\alpha\,h^\alpha{}_\beta\,y^\beta\Rightarrow h^\gamma{}_\alpha\,x^\alpha=h^\gamma{}_\beta\,y^\beta\overset{(\ast)}\Rightarrow h^\gamma{}_\alpha\,x^\alpha=x^\gamma \quad \forall x^\gamma[/tex]

    thus [itex](\ast)\Rightarrow y^\alpha=x^\alpha[/itex].
     
  13. Feb 18, 2008 #12

    kdv

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    Ah! I was not told they were null eigenvectors of h!
     
  14. Feb 18, 2008 #13
    Rainbow Child, your logic seems bulletproof, but the conclusion seems strange to me. If [tex]y^{a}=x^{a}[/tex], then doesn't that mean that h is just the identity (i.e. [tex]h^{a}{}_{b}=\delta^{a}{}_{b}[/tex])? I don't think that this can be since k is arbitrary.
     
    Last edited: Feb 18, 2008
  15. Feb 18, 2008 #14
    It should, if [itex]h^\alpha{}_\beta[/itex] has an inverse. But I don't think it has! :smile:
     
  16. Feb 18, 2008 #15
    Then that makes things much more difficult for me. I'll keep plugging away. Thank you kdv and Rainbow Child for your replies.
     
  17. Feb 18, 2008 #16
    Does every valid metric have an inverse? If so, then I would expect that h has an inverse because its interpretation is that it is the metric on a hypersurface that is orthogonal to the congruence defined by k.
     
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