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ADM field Lagrangian for a source-free electromagnetic field

  1. Mar 17, 2019 at 8:12 AM #1

    TerryW

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    Gold Member

    1. The problem statement, all variables and given/known data

    I am trying to reproduce MTW's ADM version of the field Lagrangian for a source free electromagnetic field:

    ##4π\mathcal {L} = -\mathcal {E}^i∂A_i/∂t - ∅\mathcal {E}^i{}_{,i} - \frac{1}{2}Nγ^{-\frac{1}{2}}g_{ij}(\mathcal {E}^i\mathcal {E}^i + \mathcal {B}^i\mathcal {B}^i) + N^i[ijk]\mathcal {E}^j\mathcal {B}^k## .....(21.100)

    (I'm using γ instead of ##^{(3)}g## so ##(-^{4}g)^{\frac{1}{2}} = Nγ^{\frac{1}{2}}##)

    2. Relevant equations

    I have used as my start point "by what in flat spacetime would be"

    ##\quad\quad \frac{1}{4π}\big{[}A_{μ,ν}F^{μν} + \frac{1}{4} F_{μν}{}^{μν}\big{]} .....(21.99)##

    3. The attempt at a solution

    To begin, I recast (21.99) as:

    ##4π\mathcal {L} =\big{[}A_{μ,ν}g^{αμ}g^{βν}F^{αβ} + \frac{1}{4} F_{μν}g^{αμ}g^{βν}F_{αβ}\big{]}##

    I then worked on this to produce:

    ##4π\mathcal {L} =\frac{1}{N}\big{[}(-(γ^{\frac{1}{2}}γ^{ij}F_{i0}A_{j,0}) - A_0\frac{∂}{∂x^j}(γ^{\frac{1}{2}}γ^{ij}F_{i0})\big{]}\hspace{23mm}(A)##

    ##\quad\quad\quad\quad\quad+γ^{\frac{1}{2}}(A_{j,0} - A_{0,j})(\frac{γ^{ji}}{N})N^k(A_{k,i} - A_{i,k}) \hspace{21mm}(B)##

    ##\quad\quad\quad\quad\quad-\frac{1}{2}γ^{\frac{1}{2}}\big{[}(A_{i,0} - A_{0,i})(\frac{γ^{ij}}{N})(A_{j,0} - A_{0,j})\hspace{23mm}(C)##

    ##\quad\quad\quad\quad\quad-\frac{1}{4}γ^{\frac{1}{2}}(A_{j,i} - A_{i,j})Nγ^{ik}γ^{jl}(A_{l,k}-A_{k,l})\hspace{20mm}(D)##

    ##\quad\quad\quad\quad\quad+\frac{1}{2}γ^{\frac{1}{2}}(A_{j,i} - A_{i,j})γ^{jl}\frac{N^kN^i}{N}(A_{l,k}-A_{k,l})\hspace{20mm}(E)##

    From here on, I am working on assumptions which may not be entirely correct:

    If ##F_{i0} = E_i, γ^{\frac{1}{2}}γ^{ij}F_{i0} = \mathcal{E}^j##

    (A) becomes

    ##\frac{1}{N}(-\mathcal{E}^j\frac{∂A_j}{∂t} +φ\mathcal{E}^i{}_i)##

    If ##(A_{k,i}- A_{i,k}) = \frac{1}{2}[jki](A_{k,i}- A_{i,k})##

    (B) becomes

    ##\frac{1}{N}(\mathcal{E}^iN^k\mathcal{B}^j)[ijk]##
    Where ##[ijk] ## is needed because the i in ##\mathcal{E}^i ## and the k in ## N^k## are tied to the i,k in ##A_{i,k}##

    (C ) becomes

    ##-\frac{1}{2}(\frac{1}{N})γ^{-\frac{1}{2}}\mathcal{E}^i\mathcal{E}^jγ_{ij}##

    (D) becomes

    ##-\frac{N}{4}γ^{\frac{1}{2}}\mathcal{B}^m\mathcal{B}_n\frac{γ_{mn}γ^{mn}}{3}[mij][nlk]γ^{ik}γ^{jl}##

    which then becomes

    ##-\frac{N}{2}γ^{-\frac{1}{2}}\mathcal{B}^m\mathcal{B}_nγ_{mn}##

    (E) is a big problem because it is surplus to requirements and I can't see any way of making it disappear.

    So basically, I have produced a set of elements (A) to (D) which are almost the same as the elements in MTW (21.100) except for some annoying factors of 'N'.

    As I noted at the start, MTW make the point that

    ##\quad\quad \frac{1}{4π}\big{[}A_{μ,ν}F^{μν} + \frac{1}{4} F_{μν}{}^{μν}\big{]} .....(21.99)##

    is "what would in flat space-time be" and we are not in a flat spacetime, but I can't see a way of making a transformation which would be in any way useful. It would be really nice if ##E^i## in flat spacetime could become ##NE^i## as this would solve all the issues with (A) to (D), but that still leaves me with (E).


    Any ideas anyone??


    Regards


    TerryW
     
  2. jcsd
  3. Mar 22, 2019 at 7:00 PM #2
    Thanks for the thread! 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? The more details the better.
     
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