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D=11 supergravity equations of motion

  1. Sep 29, 2015 #1
    Quick question, I'm preparing to work on supergravity.
    For completeness I was deriving the equations of motion for the Bosonic sector of maximal sugra.

    The Action principle is ##S=\int \star R -\frac{1}{2}\star F_4\wedge F_4 + \frac{1}{6} F_4\wedge F_4\wedge A_3## with ##F_4 = dA_3##. The subscripts denote the degree of the forms, when I switch to index-notation these are omitted. The star is the hodge star operator.

    I succeeded in deriving the EFE for this theory of gravity ##R_{MN}-\frac{1}{2}Rg_{MN} = \frac{1}{12}\left( F_{MPQR}F_N^{PQR} - \frac{1}{8}F^2g_{MN}\right)## with ##F^2 = F_{MNOP}F^{MNOP}##.

    When I vary this action with respect to the 3-form potential ##A_3## I get
    [tex]d\star F_4 -\frac{1}{6}F_4\wedge F_4 = 0[/tex]

    The problem I have here is that in this PDF, the coefficient of the CS-term is different.

    Let me write out the relevant part of the variation below

    [tex]\delta S = -\frac{1}{2} \int \star d(\delta A_3)\wedge F_4 + \star F_4\wedge d(\delta A_3) + \frac{1}{6} \int d(\delta A_3)\wedge F_4 \wedge A_3 + F_4\wedge d(\delta A_3)\wedge A_3 + F_4\wedge F_4 \wedge \delta A_3[/tex]

    The first integral simplifies to ##-\int \star F_4 \wedge d(\delta A_3)## due to the identity ##\star A_p \wedge B_p = \star B_p\wedge A_p##.
    Partial integration immediately gives the first term of my result (positive sign).

    For the next integral I use that ##d(\delta A_3)\wedge F_4 \wedge A_3 = (-1)^{4\cdot 4} F_4 \wedge d(\delta A_3) \wedge A_3## which show the first 2 terms are the same. Moving the factor ##d(\delta A_3)## once more doesn't change anything.
    After partial integration the second integral becomes ##\frac{1}{6}\int -2 d(F_4 \wedge A_3) \wedge \delta A_3 + F_4 \wedge F_4 \wedge \delta A_3##.

    I figured I could exploit the fact that the Leibniz rule is modified, but because ##F_4## has degree 4 we don't pick up an extra sign. ##d(F_4\wedge A_3 = d(F_4)\wedge A_3 + (-1)^4 F_4\wedge dA_3 = 0+F_4\wedge F_4##

    In conclusion I get an apparent discrepancy with the resource I found.
    I haven't been able to confirm this resource though, the resources I found are mostly concerned with the transformation laws of the fields in the action (e.g. Supergravity by Freedman and Van Proeyen)

    Am I making this big of a mistake?

    Thanks,

    Joris
     
  2. jcsd
  3. Sep 29, 2015 #2

    fzero

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    Your mistake is small. You have to be very careful with signs when integrating by parts. For example,
    $$d [ ( \delta A_3 ) \wedge F_4 \wedge A_3 ] =
    d ( \delta A_3 )\wedge F_4 \wedge A_3 - (\delta A_3) \wedge F_4 \wedge F_4,$$
    because we have to move the ##d## past the odd form ##A_3## in the second term. The corresponding term in the variation of the action then comes with the opposite sign that you need to reinforce the last term in your expression. I'm sure that the sign of the next term can be explained in the same way.
     
  4. Sep 30, 2015 #3
    Thanks a million, I knew it was something like that.
    I couldn't find it unfortunately
     
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