ADM field Lagrangian for a source-free electromagnetic field

  • Thread starter TerryW
  • Start date
  • #1
TerryW
Gold Member
168
6

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??


Regards


TerryW
 

Answers and Replies

  • #2
TerryW
Gold Member
168
6
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
 

Related Threads on ADM field Lagrangian for a source-free electromagnetic field

  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
8
Views
5K
Replies
1
Views
2K
Replies
0
Views
3K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
2
Replies
36
Views
3K
  • Last Post
Replies
12
Views
3K
Top