ADM field Lagrangian for a source-free electromagnetic field

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TerryW
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Homework Statement



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}}##)

Homework 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)##

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??RegardsTerryW
 
on Phys.org
Hi anyone who is following my ramblings.

I've sorted this now! I should have looked further down the page and spotted that MTW give a definition for
##\mathcal{E}^i## in 21.103.

Using this in my recast of 21.99 introduces a whole lot of extra terms but they all cancel out, including my problematical (E) term and leaving me with just the terms I need for 21.100 with no bits left over and no problems with unwanted 'N's.

:smile::smile:

TerryW