Deriving conjugate momenta from the Einstein-Hilbert density

  • Thread starter TerryW
  • Start date
  • #1
TerryW
Gold Member
173
8

Homework Statement



This post contains the answer to my thread of 10th August...
[/B]
in which I asked if anyone could point out how to derive

##\pi^{ij} = \sqrt {^{(4)}g} (^{(4)} \Gamma ^0 \,_{pq} - g_{pq} ^{(4)} \Gamma ^0\, _{rs} g^{rs}) g^{pq} g^{jq}##

from

##\mathfrak {L}## = (4)R ##\sqrt{-^{(4)}g}##


Homework Equations



I've finally come up with a solution, again using equations derived from Golovnev's ArXiv paper. The useful equations are:

##K_{ij} = -\frac{1}{2N}(γ_{ij,0} - ^{(3)}∇_{i}N_{j} - ^{(3)}∇_{j}N_{i}) \quad\quad## Golovnev (3),

##\sqrt{-^{(4)}g}^{(4)}R = \sqrt{γ}N(^{(3)}R + K^i_iK^j_j - K^{ij}K_{ij} )\quad\quad γ^{ik}γ^{jl}##*Golovnev (13) and

##Γ^0{}_{ij} = -\frac{1}{N}K_{ij}\quad\quad## Golovnev (11)



3. The solution

##K^i_iK^j_j = γ^{ij}K_{ij}γ^{kl}K_{kl}\quad##and ##K^{ij}K_{ij} = K_{ij}γ^{ik}γ^{jl}K_{kl}##

∴ ##\sqrt{-^{(4)}g}^{(4)}R = \sqrt{γ}N(^{(3)}R + γ^{ij}γ^{kl}(\frac{1}{4N^2})(γ_{ij,0} - 2^{(3)}∇_{(i}N_{j)})(γ_{kl,0} - 2^{(3)}∇_{(k}N_{l)}))##
##\quad\quad γ^{ik}γ^{jl}(\frac{1}{4N^2})(γ_{ij,0} - 2^{(3)}∇_{(i}N_{j)})(γ_{kl,0} - 2^{(3)}∇_{(k}N_{l)}))##

∴ ##\sqrt{-^{(4)}g}^{(4)}R = \sqrt{γ}N(^{(3)}R + (\frac{1}{4N^2})(γ^{ij}γ^{kl}-γ^{ik}γ^{jl})(γ_{ij,0} - 2^{(3)}∇_{(i}N_{j)})(γ_{kl,0} - 2^{(3)}∇_{(k}N_{l)}))##

∴ ##\sqrt{-^{(4)}g}^{(4)}R = \sqrt{γ}N(^{(3)}R + (\frac{1}{4N^2})(γ^{ij}γ^{kl}-γ^{ik}γ^{jl})(γ_{ij,0}γ_{kl,0} -γ_{ij,0}(2^{(3)}∇_{(k}N_{l)}) - γ_{kl,0}(2^{(3)}∇_{(i}N_{j)}) + (2^{(3)}∇_{(i}N_{j)})(2^{(3)}∇_{(k}N_{l)})))##

∴ ##\sqrt{-^{(4)}g}^{(4)}R = \sqrt{γ}N(^{(3)}R + (\frac{1}{4N^2})(γ^{ij}γ^{kl}-γ^{ik}γ^{jl})(δ^k_iδ^l_jγ_{kl,0}γ_{kl,0} -δ^k_iδ^l_jγ_{kl,0}(2^{(3)}∇_{(k}N_{l)}) - γ_{kl,0}(2^{(3)}∇_{(i}N_{j)}) + (2^{(3)}∇_{(i}N_{j)})(2^{(3)}∇_{(k}N_{l)})))##

I now depart from MTW by deriving ##π^{kl} = \frac{∂(action)}{∂(γ_{kl,0})}## and note that ##(^{(3)}R)## and ##(2^{(3)}∇_{(i}N_{j)})(2^{(3)}∇_{(k}N_{l)})## are both independent of ##γ_{kl,0}## so:

##π^{kl} = \frac{∂(\sqrt{-^{(4)}g}^{(4)}R)}{∂(γ_{kl,0})}= \frac{\sqrt{γ}N}{4N^2}(γ^{ij}γ^{kl}-γ^{ik}γ^{jl})(2δ^k_iδ^l_jγ_{kl,0} - δ^k_iδ^l_j(2^{(3)}∇_{(k}N_{l)}) - (2^{(3)}∇_{(i}N_{j)}))##

##= \frac{\sqrt{γ}N}{4N^2}(γ^{ij}γ^{kl}-γ^{ik}γ^{jl})(2γ_{ij,0} - (2^{(3)}∇_{(i}N_{j)}) - (2^{(3)}∇_{(i}N_{j)}))##

##= \frac{\sqrt{γ}N}{4N^2}(γ^{ij}γ^{kl}-γ^{ik}γ^{jl})(-4NK_{ij}))## from Golovnev's (3) above

##= \frac{\sqrt{γ}N}{4N^2}(K^{kl}-γ^{kl}K^i_i)##

then, using Golovnev's equation (11) ##Γ^0{}_{ij} = -\frac{1}{N}K_{ij}##

##\pi^{ij} = \sqrt {^{(4)}g} (^{(4)} \Gamma ^0 \,_{pq} - g_{pq} ^{(4)} \Gamma ^0\, _{rs} g^{rs}) g^{pq} g^{jq}##
 

Answers and Replies

Related Threads on Deriving conjugate momenta from the Einstein-Hilbert density

Replies
1
Views
760
Replies
0
Views
683
  • Last Post
Replies
6
Views
2K
Replies
6
Views
4K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
2K
Replies
7
Views
2K
Replies
4
Views
997
Top