Deriving conjugate momenta from the Einstein-Hilbert density

[TerryW]

Homework Statement

This post contains the answer to my thread of 10th August...
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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) $Γ^0{}_{ij} = -\frac{1}{N}K_{ij}$

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

 

[member 553083]

$= \sqrt {^{(4)}g} ( -\frac{1}{N}K_{pq} - g_{pq} ^{(4)}(-\frac{1}{N}K_{rs}) g^{rs}) g^{pq} g^{jq}$$= \sqrt {^{(4)}g} \frac{1}{N^2}K_{pq} (δ^p_iδ^q_j- g_{pq} g^{rs}δ^p_iδ^q_j)$$= \sqrt {^{(4)}g} \frac{1}{N^2}K_{ij} (γ^{ij}- g_{ij} g^{rs}γ^{rs})$$= \sqrt {^{(4)}g} \frac{1}{N^2}K_{ij} (γ^{ij}- γ^{ij})$$= \sqrt {^{(4)}g} \frac{1}{N^2}K_{ij} (K^{ij}- γ^{ij}K^k_k)$$= \sqrt {^{(4)}g} \frac{1}{N^2}(K^{kl}-γ^{kl}K^i_i)$which is the same as the result derived from Lagrangian.