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General Relativity asymmetry identity

  1. Jan 9, 2017 #1
    1. The problem statement, all variables and given/known data

    I have ##R^{u}_{o b u } F^{o}_a - R^{u}_{oau}F^{o}_{b}##

    and I want to show that this equal to ##2R_{o[aF^{o}_b]}##

    where ## [ ] ## denotes antisymmetrization , and ##F_{uv} ## is a anitymstric tensor

    2. Relevant equations

    Since ##F_{uv} ##is antisymetric the antisymetrization ##2R_{o[aF^{o}_b}]## reduces to ## 3(R^{u}_{o b u } F^{o}_a - R^{u}_{oau}F^{o}_{b} + R_{oo}F^a_b)##

    3. The attempt at a solution

    Contracting to get the Ricci tensor and using that ##F_{uv} ## is antisymetric

    ##R^{u}_{o b u } F^{o}_a - R^{u}_{oau}F^{o}_{b}=R_{o b } F^{o}_a + R_{oa}F^{o}_{b}##
    ##=R{o b } F^{a}_o + R_{oa}F^{o}_{b}##

    I can see that there needs to be a ##R_{oo}F^a_b ## term added somewhere, which I cant see where this is going to come from unless this is zero ? which I cant see that it is, and even then Id get a factor of ##3## rather than the ##2## needed.

    Many thanks in advance.
     
  2. jcsd
  3. Jan 9, 2017 #2

    TSny

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    Go to this link and then scroll up into the previous section about 10 lines.
    https://en.wikipedia.org/wiki/Ricci_calculus#Differentiation

    It says,
    As with symmetrization, indices are not antisymmetrized when they are not on the same level, for example;
    $$ A_{[\alpha }B^{\beta }{}_{\gamma ]}={\dfrac {1}{2!}}\left(A_{\alpha }B^{\beta }{}_{\gamma }-A_{\gamma }B^{\beta }{}_{\alpha }\right)$$
    I don't know if you are meant to adopt this convention.
     
  4. Jan 10, 2017 #3
    ahh right thank you,
    and you can surely get this from the antisymmetrization of indices on the same level by raising an index?
     
  5. Jan 10, 2017 #4

    TSny

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    I don't think so unless I'm overlooking something.

    Suppose we have a tensor ##C_{\alpha}{} ^{\mu}{} _{\beta}##. Then we can antisymmetrize over ##\alpha## and ##\beta## to produce another tensor ##C_{[\alpha}{} ^{\mu}{} _{\beta]}##.

    But it would not generally be true that ##C_{[\alpha}{} ^{\mu}{} _{\beta]} = g^{\mu \tau} C_{[\alpha \tau \beta]}##.

    Instead, you would have to write ##C_{[\alpha}{} ^{\mu}{} _{\beta]} = g^{\mu \tau} C_{[\alpha |\tau |\beta]}## where we use another convention that indices located between vertical bars are to be ignored in the antisymmetrization.

    That's how I see it, anyway.
     
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