Deriving the MTW version of the Einstein-Hilbert density

[TerryW]

Homework Statement

I've been working through a paper by Alexey Golovnev, title 'ADM and massive gravity' arXiv.1302.0687v4 [gr-qc] 26 March 2013. I am hoping to use his result for the Einstein-Hilbert density to achieve my aim of finding a way to derive Equation 21.90 in MTW. I have worked my way through the paper and can see that all the equations given are OK, with one crucial exception:

Homework Equations

The final link in the chain is this:

$-2 \sqrt{-g}\nabla_μ (K^i_in^μ) = -2∂_0(\sqrt{γ}K^i_i) + 2\sqrt{γ}^{(3)}\nabla_j(K^i_iN^j)$

My attempt to prove this identity ends up with an extra term. Can anyone tell me where I've gone wrong?

The Attempt at a Solution

Using the derivations in the paper:

$-2 \sqrt{-g}\nabla_μ (K^i_in^μ) = -2 \sqrt{-g}^{(4)}\nabla_0 (K^i_in^0) - 2 \sqrt{-g}^{(4)}\nabla_j (K^i_in^j)$

$= -2 ^{(4)}\nabla_0 (\sqrt{γ}NK^i_i\frac{1}{N}) - 2 \sqrt{-g}^{(4)}\nabla_j (K^i_in^j)$

$= -2 ^{(4)}\nabla_0 (\sqrt{γ}K^i_i) - 2 \sqrt{-g}[∂_j(K^i_in^j) + ^{(4)}Γ^j_{αj}(K^i_in^α)]$

$= -2 ∂_0 (\sqrt{γ}K^i_i) - 2 \sqrt{-g}∂_j(K^i_in^j) -2\sqrt{-g} ^{(4)}Γ^j_{0j}(K^i_in^0) -2\sqrt{-g} ^{(4)}Γ^j_{kj}(K^i_in^k)$

$= -2 ∂_0 (\sqrt{γ}K^i_i) + 2 \sqrt{-g}∂_j(K^i_i\frac{N^j}{N}) -2\sqrt{-g} ^{(4)}Γ^j_{0j}(K^i_i\frac{1}{N}) +2\sqrt{-g} ^{(4)}Γ^j_{kj}(K^i_i\frac{N^k}{N})$

$= -2 ∂_0 (\sqrt{γ}K^i_i) + 2 \sqrt{-g}∂_j(K^i_iN^j)(\frac{1}{N}) +2 \sqrt{-g}(K^i_iN^j)∂_j(\frac{1}{N})-2\sqrt{γ}N^{(4)}Γ^j_{0j}(K^i_i\frac{1}{N})$
$+2\sqrt{γ} N^{(4)}Γ^j_{kj}(K^i_i\frac{N^k}{N})$

$= -2 ∂_0 (\sqrt{γ}K^i_i) + 2 \sqrt{γ}∂_j(K^i_iN^j) +2\sqrt{γ}^{(4)}Γ^j_{kj}K^i_iN^k+2\sqrt{-g}(K^i_iN^j)(\frac{-1}{N^2})∂_jN -2\sqrt{γ} ^{(4)}Γ^j_{0j}(K^i_i)$

$= -2 ∂_0 (\sqrt{γ}K^i_i) + 2 \sqrt{γ}∂_j(K^i_iN^j) +2\sqrt{γ}(^{(3)}Γ^j_{kj} +\frac{N^j}{N}K_{jk})K^i_iN^k)-2\sqrt{γ}K^i_i(\frac{N^j}{N}∂_jN + ^{(4)}Γ^j_{0j})$

$= -2 ∂_0 (\sqrt{γ}K^i_i) + 2 \sqrt{γ}^{(3)}\nabla_j(K^i_iN^j) +2\sqrt{γ}\frac{N^j}{N}K_{jk}K^i_iN^k -2\sqrt{γ}K^i_i\frac{N^j}{N}∂_jN -2\sqrt{γ}K^i_{i}$ $^{(4)}Γ^j_{0j}$

The first two terms are what I was aiming for. The three remaining terms are 'surplus' to requirements!

I can reduce these three terms down to just one as follows

$2\sqrt{γ}\frac{N^j}{N}K_{jk}K^i_iN^k -2\sqrt{γ}K^i_i\frac{N^j}{N}∂_jN -2\sqrt{γ}K^i_{i}$ $^{(4)}Γ^j_{0j}$

$= 2\sqrt{γ}K^i_i[\frac{N^j}{N}K_{jk}N^k -\frac{N^j}{N}∂_jN - ^{(4)}Γ^j_{0j}]$

$= 2\sqrt{γ}K^i_i[\frac{N^j}{N}K_{jk}N^k -\frac{N^j}{N}∂_jN - [-\frac{N^j}{N}∂_jN -N(γ^{jk} - \frac{N^jN^k}{N^2})K_{jk} + ^{(3)}\nabla_j(N^j)]$

$= 2\sqrt{γ}K^i_i[Nγ^{jk}K_{jk} - ^{(3)}\nabla_j(N^j)]$

$= 2\sqrt{γ}K^i_i[Nγ^{jk}K_{jk} - γ^{jk}$ $^{(3)}\nabla_j(N_k)]$

$= -2\sqrt{γ}K^i_iγ^{jk}\,^{(4)}Γ_{k0j}$

I'd really appreciate it if someone can help me get rid of this last term!

 

[TerryW]

Homework Statement

I've been working through a paper by Alexey Golovnev, title 'ADM and massive gravity' arXiv.1302.0687v4 [gr-qc] 26 March 2013. I am hoping to use his result for the Einstein-Hilbert density to achieve my aim of finding a way to derive Equation 21.90 in MTW. I have worked my way through the paper and can see that all the equations given are OK, with one crucial exception:

Homework Equations

The final link in the chain is this:

$-2 \sqrt{-g}\nabla_μ (K^i_in^μ) = -2∂_0(\sqrt{γ}K^i_i) + 2\sqrt{γ}^{(3)}\nabla_j(K^i_iN^j)$

My attempt to prove this identity ends up with an extra term. Can anyone tell me where I've gone wrong?

The Attempt at a Solution

Using the derivations in the paper:

$-2 \sqrt{-g}\nabla_μ (K^i_in^μ) = -2 \sqrt{-g}^{(4)}\nabla_0 (K^i_in^0) - 2 \sqrt{-g}^{(4)}\nabla_j (K^i_in^j)$

$= -2 ^{(4)}\nabla_0 (\sqrt{γ}NK^i_i\frac{1}{N}) - 2 \sqrt{-g}^{(4)}\nabla_j (K^i_in^j)$

$= -2 ^{(4)}\nabla_0 (\sqrt{γ}K^i_i) - 2 \sqrt{-g}[∂_j(K^i_in^j) + ^{(4)}Γ^j_{αj}(K^i_in^α)]$

$= -2 ∂_0 (\sqrt{γ}K^i_i) - 2 \sqrt{-g}∂_j(K^i_in^j) -2\sqrt{-g} ^{(4)}Γ^j_{0j}(K^i_in^0) -2\sqrt{-g} ^{(4)}Γ^j_{kj}(K^i_in^k)$

$= -2 ∂_0 (\sqrt{γ}K^i_i) + 2 \sqrt{-g}∂_j(K^i_i\frac{N^j}{N}) -2\sqrt{-g} ^{(4)}Γ^j_{0j}(K^i_i\frac{1}{N}) +2\sqrt{-g} ^{(4)}Γ^j_{kj}(K^i_i\frac{N^k}{N})$

$= -2 ∂_0 (\sqrt{γ}K^i_i) + 2 \sqrt{-g}∂_j(K^i_iN^j)(\frac{1}{N}) +2 \sqrt{-g}(K^i_iN^j)∂_j(\frac{1}{N})-2\sqrt{γ}N^{(4)}Γ^j_{0j}(K^i_i\frac{1}{N})$
$+2\sqrt{γ} N^{(4)}Γ^j_{kj}(K^i_i\frac{N^k}{N})$

$= -2 ∂_0 (\sqrt{γ}K^i_i) + 2 \sqrt{γ}∂_j(K^i_iN^j) +2\sqrt{γ}^{(4)}Γ^j_{kj}K^i_iN^k+2\sqrt{-g}(K^i_iN^j)(\frac{-1}{N^2})∂_jN -2\sqrt{γ} ^{(4)}Γ^j_{0j}(K^i_i)$

$= -2 ∂_0 (\sqrt{γ}K^i_i) + 2 \sqrt{γ}∂_j(K^i_iN^j) +2\sqrt{γ}(^{(3)}Γ^j_{kj} +\frac{N^j}{N}K_{jk})K^i_iN^k)-2\sqrt{γ}K^i_i(\frac{N^j}{N}∂_jN + ^{(4)}Γ^j_{0j})$

$= -2 ∂_0 (\sqrt{γ}K^i_i) + 2 \sqrt{γ}^{(3)}\nabla_j(K^i_iN^j) +2\sqrt{γ}\frac{N^j}{N}K_{jk}K^i_iN^k -2\sqrt{γ}K^i_i\frac{N^j}{N}∂_jN -2\sqrt{γ}K^i_{i}$ $^{(4)}Γ^j_{0j}$

The first two terms are what I was aiming for. The three remaining terms are 'surplus' to requirements!

I can reduce these three terms down to just one as follows

$2\sqrt{γ}\frac{N^j}{N}K_{jk}K^i_iN^k -2\sqrt{γ}K^i_i\frac{N^j}{N}∂_jN -2\sqrt{γ}K^i_{i}$ $^{(4)}Γ^j_{0j}$

$= 2\sqrt{γ}K^i_i[\frac{N^j}{N}K_{jk}N^k -\frac{N^j}{N}∂_jN - ^{(4)}Γ^j_{0j}]$

$= 2\sqrt{γ}K^i_i[\frac{N^j}{N}K_{jk}N^k -\frac{N^j}{N}∂_jN - [-\frac{N^j}{N}∂_jN -N(γ^{jk} - \frac{N^jN^k}{N^2})K_{jk} + ^{(3)}\nabla_j(N^j)]$

$= 2\sqrt{γ}K^i_i[Nγ^{jk}K_{jk} - ^{(3)}\nabla_j(N^j)]$ (***)

$= 2\sqrt{γ}K^i_i[Nγ^{jk}K_{jk} - γ^{jk}$ $^{(3)}\nabla_j(N_k)]$

$= -2\sqrt{γ}K^i_iγ^{jk}\,^{(4)}Γ_{k0j}$

I'd really appreciate it if someone can help me get rid of this last term!

Resolution

If I use the general expression for the divergence of a vector (MTW 21.85 p 579):

$(A^α)_{;α} = \sqrt{-g}^{-½}(\sqrt{-g}A^α)_{,α}$

and then work on $K_{ij} = -^{(4)}\nabla_in_j$ (Golovnev (2)), I find that

$g^{ij}K_{ij} = K^i_i = -g^{ij}$ $^{(4)}\nabla_i n_j$ $= -^{(4)}\nabla_i n^i$

So $\sqrt{-g}K^i_i = \sqrt{-g}^{(4)}\nabla_i\frac{N^i}{N} = (\sqrt{-g}\frac{N^i}{N})_{,i} = (\sqrtγN^i)_{,i}$

Now use the general expression for the divergence of a vector in the 3-D space

$^{(3)}\nabla_iA^i = \sqrtγ^{-½}(\sqrtγA^i)_{,i}$ ...(1)

So $\sqrtγNK^i_i = (\sqrtγN^i)_{,i} = \sqrtγ^{(3)}\nabla_iN^i$

So $NK^i_i = ^{(3)}\nabla_iN^i$

which means that (***) above = 0, which achieves my aim of getting rid of the 'surplus to requirements' terms.This resolution was prompted by TSny, who sent me a much more elegant solution, again based on MTW 21.85 as follows;

$\sqrt{-g}^{(4)}\nabla_μ (K^i_in^μ) = (\sqrt{-g}(K^i_in^0)_{,0} - \sqrt{-g}K^i_in^j)_{j}$

$∴\sqrt{-g}^{(4)}\nabla_μ (K^i_in^μ) = (\frac {\sqrt{-g}}{N}K^i_i)_{,0} - (\frac {\sqrt{-g}}{N}K^i_iN^j)_{,j}$

Then using (1) and $\sqrt{-g} = \sqrtγN$

$\sqrt{-g}^{(4)}\nabla_μ (K^i_in^μ) = (\sqrtγK^i_i)_{,0} - \sqrtγ^{(3)}\nabla_j(K^i_iN^j)$

QED!