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Continuing from the ADM worksheet...
${\Gamma^a}_{bc}$ is the Levi-Civita connection associated with the coordinate basis (whereas in previous worksheets it was ${{\tilde{\Gamma}}^a}_{bc}$).
Introduce new covariant derivative $\hat{\nabla}_a$ and associated connection ${\hat{\Gamma}^a}_{bc}$.
It should cancel the coordinate-basis $g_{\alpha\beta}$ metric, and be torsion-free.
Define the connection with respect to a non-coordinate locally-Minkowski basis $e_i$.
That leaves us with the equation from our "metric cancelling and torsion free connections" worksheet:
${\hat\Gamma^i}_{jk} = {e^i}_\alpha {e_j}^\beta {e_k}^\gamma {\tilde{\Gamma}^\alpha}_{\beta\gamma} + {e^i}_\alpha e_j ({e_k}^\alpha)$
${\hat\Gamma^\alpha}_{\beta\gamma} = {\tilde{\Gamma}^\alpha}_{\beta\gamma} + \partial_\beta ({e_k}^\alpha) {e^k}_\gamma$
On the Riemann worksheet we had:
${\hat{R}^\tilde{c}}_{\tilde{d}\tilde{a}\tilde{b}} = {\tilde{R}^\tilde{c}}_{\tilde{d}\tilde{a}\tilde{b}}$
...that one Riemann curvature derived from the covariant derivative with connection metric-cancelling torsion-free is the same as another Riemann curvature derived from metric-cancelling torsion-free connection of another basis.
Do we really need another connection on Lorentz coordinates?
Does Palatini action use an arbitrary connection, or does it use the affine connection extended to be compatible with the tetrad?
This is from Pullins' work, but why is it even here, especially if ${C_a}^{IJ} = 0$ in the end?
Is this to prove that any covariant Lorentz connection chosen that annihilates the Minkowski metric must be the corvariant Lorentz connection that annihilates the tetrad as well?
Same deal with the Wiki page on Tetradic Palatini action -- first it says "arbitrary connection" and then it says "that annihilates the tetrad."
If it annihilates the tetrad then it's not going to be arbitrary. It's going to be the one shown above.
Introduce a new covariant derivative and a new arbitrary connections on Lorentz and on curved coordinates:
$ \mathcal{D}_a {T_b}^I = \partial_a {T_b}^I - {\hat\Gamma^c}_{ba} {T_c}^I + {{\hat\omega_a}^I}_J K^J $
Require that it annihilates the Minkowski metric:
$ \hat\omega_{aIJ} = -\hat\omega_{aJI} $
covariant derivatives difference on Lorentz index:
$ (\mathcal{D}_a - \nabla_a) V^I $
$ = \partial_a V^I + {{\hat\omega_a}^I}_J V^J - \partial_a V^I - {{\omega_a}^I}_J V^J $
$ = ({{\hat\omega_a}^I}_J - {{\omega_a}^I}_J) V^J $
Let $ {{C_a}^I}_J = {{\hat\omega_a}^I}_J - {{\omega_a}^I}_J $
So $ {{C_a}^I}_J V^J = (\mathcal{D}_a - \nabla_a) V^I $
And $ \mathcal{D}_a V^I = {{C_a}^I}_J V^J + \nabla_a V^I $
covariant derivatives difference on curved index:
$ (\mathcal{D}_a - \nabla_a) V^b $
$ = \partial_a V^b + {\hat\Gamma^b}_{ca} V^c - \partial_a V^b - {\Gamma^b}_{ca} V^c $
$ = {\hat\Gamma^b}_{ca} V^c - {\Gamma^b}_{ca} V^c $
Let $ {\bar\Gamma^b}_{ca} = {\hat\Gamma^b}_{ca} - {\Gamma^b}_{ca} $
So $ {\bar\Gamma^b}_{ca} V^a = (\mathcal{D}_a - \nabla_a) V^b $
And $ \mathcal{D}_a V^b = \nabla_a V^b + {\bar\Gamma^b}_{ca} V^c $
covariant derivatives difference on both:
$ (\mathcal{D}_a - \nabla_a) T^{bI} $
$ = {\bar\Gamma^b}_{ca} T^{cI} + {{C_a}^I}_J T^{bJ} $
So $ \mathcal{D}_a T^{bI} = \nabla_a T^{bI} + {\bar\Gamma^b}_{ca} T^{cI} + {{C_a}^I}_J T^{bJ} $
Curvature of the new connections in terms of covariant derivative differences:
$ {{\hat\Omega_{ab}}^I}_J V^J = \mathcal{D}_a \mathcal{D}_b V^I - \mathcal{D}_b \mathcal{D}_a V^I $
$ = \mathcal{D}_a ({{C_b}^I}_J V^J + \nabla_b V^I)
- \mathcal{D}_b ({{C_a}^I}_J V^J + \nabla_a V^I) $
$ = \nabla_a ({{C_b}^I}_J V^J + \nabla_b V^I)
- {\bar\Gamma^c}_{ba} ({{C_c}^I}_J V^J + \nabla_c V^I)
+ {{C_a}^I}_K ({{C_b}^K}_J V^J + \nabla_b V^K)
- \nabla_b ({{C_a}^I}_J V^J + \nabla_a V^I)
+ {\bar\Gamma^c}_{ab} ({{C_c}^I}_J V^J + \nabla_c V^I)
- {{C_b}^I}_K ({{C_a}^K}_J V^J + \nabla_a V^K) $
$ = \nabla_a ({{C_b}^I}_J V^J + \nabla_b V^I)
- \nabla_b ({{C_a}^I}_J V^J + \nabla_a V^I)
+ {{C_a}^I}_K ({{C_b}^K}_J V^J + \nabla_b V^K)
- {{C_b}^I}_K ({{C_a}^K}_J V^J + \nabla_a V^K)
+ ({\bar\Gamma^c}_{ab} - {\bar\Gamma^c}_{ba}) ({{C_c}^I}_J V^J + \nabla_c V^I)
$
assume ${\bar\Gamma^c}_{ab} = {\bar\Gamma^c}_{ba} $
$ = \nabla_a ({{C_b}^I}_J V^J + \nabla_b V^I)
- \nabla_b ({{C_a}^I}_J V^J + \nabla_a V^I)
+ {{C_a}^I}_K ({{C_b}^K}_J V^J + \nabla_b V^K)
- {{C_b}^I}_K ({{C_a}^K}_J V^J + \nabla_a V^K) $
$ = \nabla_a {{C_b}^I}_J V^J + \nabla_a \nabla_b V^I
- \nabla_b {{C_a}^I}_J V^J - \nabla_b \nabla_a V^I
+ {{C_a}^I}_K {{C_b}^K}_J V^J + {{C_a}^I}_K \nabla_b V^K
- {{C_b}^I}_K {{C_a}^K}_J V^J - {{C_b}^I}_K \nabla_a V^K $
$ = \nabla_a \nabla_b V^I - \nabla_b \nabla_a V^I
+ \nabla_a {{C_b}^I}_J V^J - \nabla_b {{C_a}^I}_J V^J
+ {{C_a}^I}_K {{C_b}^K}_J V^J - {{C_b}^I}_K {{C_a}^K}_J V^J
+ {{C_a}^I}_K \nabla_b V^K - {{C_b}^I}_K \nabla_a V^K
$
$ = {{R_{ab}}^I}_J V^J + 2 ( \nabla_{[a} {{C_{b]}}^I}_J + {C_{[a}}^{IK} C_{b]KJ} ) $
$ {\hat\Omega_{ab}}^{IJ} - {R_{ab}}^{IJ} = 2 ( \nabla_{[a} {{C_{b]}}^I}_J + {C_{[a}}^{IK} C_{b]KJ} ) $
TODO show $ R = {e^a}_I {e^b}_J {R_{ab}}^{IJ} = {e^a}_I {e^b}_J {\hat\Omega_{ab}}^{IJ} $
action on curvature of Lorentz covariant derivative, with Riemann curvature substituted:
$ S = \int \sqrt{-g} R dx^4 $
$ = \int e R dx^4 $
$ = \int e {\hat\Omega_{ab}}^{ab} dx^4 $
$ = \int e {e^a}_I {e^b}_J {\hat\Omega_{ab}}^{IJ} dx^4 $
$ S = \int e {e^a}_I {e^b}_J ({R_{ab}}^{IJ} + 2 \nabla_{[a} {C_{b]}}^{IJ} + 2 {C_{[a}}^{IK} {C_{b]K}}^J) dx^4 $
TODO show $ \delta / \delta {C_a}^{IJ} (\nabla_b {C_c}^{KL}) = 0 $
$ \delta S / \delta {C_a}^{IJ} = \delta / \delta {C_a}^{IJ} \int e {e^b}_K {e^c}_L ({R_{bc}}^{KL} + 2 \nabla_{[b} {C_{c]}}^{KL} + 2 {C_{[b}}^{KM} {C_{c]M}}^L) dx^4 $
$ = \int 2 e {e^b}_K {e^c}_L \delta / \delta {C_a}^{IJ} {C_{[b}}^{KM} {C_{c]M}}^L dx^4 $
$ = \int e {e^b}_K {e^c}_L ({C_{cM}}^L \delta / \delta {C_a}^{IJ} {C_b}^{KM} + {C_b}^{KM} \delta / \delta {C_a}^{IJ} {C_{cM}}^L - {C_{bM}}^L \delta / \delta {C_a}^{IJ} {C_c}^{KM} - {C_c}^{KM} \delta / \delta {C_a}^{IJ} {C_{bM}}^L) dx^4 $
$ = \int e {e^b}_K {e^c}_L (
{C_{cM}}^L \delta^a_b \delta^K_I \delta^M_J
+ {C_b}^{KM} \delta^a_c \delta^N_I \delta^L_J \eta_{MN}
- {C_{bM}}^L \delta^a_c \delta^K_I \delta^M_J
- {C_c}^{KM} \delta^a_b \delta^N_I \delta^L_J \eta_{MN}
) dx^4 $
$ = \int e {e^a}_K {e^b}_L (
\delta^K_I \delta^M_J {C_{bM}}^L
- \delta^K_J \delta^M_I {C_{bM}}^L
- \delta^L_I \delta^M_J {C_{bM}}^K
+ \delta^L_J \delta^M_I {C_{bM}}^K
) dx^4 $
$ = \int e {e^a}_K {e^b}_L (\delta^K_I \delta^M_J {C_{bM}}^L - \delta^K_J \delta^M_I {C_{bM}}^L - \delta^L_I \delta^M_J {C_{bM}}^K + \delta^L_J \delta^M_I {C_{bM}}^K) $
$ = \int e ({e^a}_K {e^b}_L (\delta^K_I \delta^M_J {C_{bM}}^L - \delta^K_J \delta^M_I {C_{bM}}^L) - {e^b}_K {e^a}_L (\delta^K_I \delta^M_J {C_{bM}}^L - \delta^K_J \delta^M_I {C_{bM}}^L)) $
$ = \int e ({e^a}_K {e^b}_L - {e^b}_K {e^a}_L) (\delta^K_I \delta^M_J {C_{bM}}^L - \delta^K_J \delta^M_I {C_{bM}}^L) $
$ = \int e ({e^a}_K {e^b}_L - {e^b}_K {e^a}_L) (\delta^K_I \delta^M_J - \delta^K_J \delta^M_I) {C_{bM}}^L $
$ = \int 2 e ({e^a}_K {e^b}_L - {e^b}_K {e^a}_L) \delta^K_{[I} \delta^M_{J]} {C_{bM}}^L $
$ = \int 4 e {e^{[a}}_K {e^{b]}}_L \delta^K_{[I} \delta^M_{J]} {C_{bM}}^L $
$ \delta S / \delta {C_a}^{IJ} = 0 $ for $ \delta S / \delta {C_a}^{IJ} = 4 e {e^{[a}}_K {e^{b]}}_L \delta^K_{[I} \delta^M_{J]} {C_{bM}}^L $
Therefore either $ e = 0 $ or ${e^{[a}}_K {e^{b]}}_L \delta^K_{[I} \delta^M_{J]} {C_{bM}}^L = 0$
For $ {e^{[a}}_K {e^{b]}}_L \delta^K_{[I} \delta^M_{J]} $ full rank (TODO show) this implies
$ {C_{bM}}^L = 0$
TODO:
$ S = \int e {e^a}_I {e^b}_J {\hat\Omega_{ab}}^{IJ} dx^4 $
$ \delta S / \delta {e^a}_I = \delta / \delta {e^a}_I \int e {e^b}_J {e^c}_K {\hat\Omega_{bc}}^{JK} dx^4 $
$ = \int (
\delta / \delta {e^a}_I e {e^b}_J {e^c}_K {\hat\Omega_{bc}}^{JK}
+ e \delta / \delta {e^a}_I {e^b}_J {e^c}_K {\hat\Omega_{bc}}^{JK}
+ e {e^b}_J \delta / \delta {e^a}_I {e^c}_K {\hat\Omega_{bc}}^{JK}
+ e {e^b}_J {e^c}_K \delta / \delta {e^a}_I {\hat\Omega_{bc}}^{JK}
) dx^4 $
using $ \delta / \delta {e^a}_I {\hat\Omega_{bc}}^{JK} = 0$
and $ \delta e = e {e^a}_I \delta {e_a}^I = -e {e_a}^I \delta {e^a}_I $
so $ \delta / \delta {e^a}_I e = -e {e_a}^I $
$ = \int (
-e {e_a}^I {e^b}_J {e^c}_K {\hat\Omega_{bc}}^{JK}
+ e \delta / \delta {e^a}_I {e^b}_J {e^c}_K {\hat\Omega_{bc}}^{JK}
+ e {e^b}_J \delta / \delta {e^a}_I {e^c}_K {\hat\Omega_{bc}}^{JK}
) dx^4 $
$ = \int e (
-{e_a}^I {e^b}_J {e^c}_K {\hat\Omega_{bc}}^{JK}
+ \delta^I_J \delta^b_a {e^c}_K {\hat\Omega_{bc}}^{JK}
+ {e^b}_J \delta^I_K \delta^c_a {\hat\Omega_{bc}}^{JK}
) dx^4 $
$ = \int e (
-{e_a}^I {e^b}_J {e^c}_K {\hat\Omega_{bc}}^{JK}
+ {e^c}_K {\hat\Omega_{ac}}^{IK}
+ {e^b}_J {\hat\Omega_{ba}}^{JI}
) dx^4 $
$ = \int e (
-{e_a}^I {e^b}_J {e^c}_K {\hat\Omega_{bc}}^{JK}
+ {e^b}_J {\hat\Omega_{ab}}^{IJ}
+ {e^b}_J {\hat\Omega_{ba}}^{JI}
) dx^4 $
using ${\hat\Omega_{ab}}^{IJ} = -{\hat\Omega_{ba}}^{IJ} = {\hat\Omega_{ba}}^{JI}$
$ = \int 2 e (
{e^b}_J {\hat\Omega_{ab}}^{IJ}
- {1\over 2} {e_a}^I {e^b}_J {e^c}_K {\hat\Omega_{bc}}^{JK}
) dx^4 $
$ \delta S / \delta {e^a}_I = 0$ for either $e = 0$ or
$ {e^b}_J {\hat\Omega_{ab}}^{IJ} - {1\over 2} {e_a}^I {e^b}_J {e^c}_K {\hat\Omega_{bc}}^{JK} = 0$
replace ${\hat\Omega_{ab}}^{IJ}$ with ${R_{ab}}^{IJ}$ to get...
$ {e^b}_J {R_{ab}}^{IJ} - {1\over 2} {e_a}^I {e^b}_J {e^c}_K {R_{bc}}^{JK} = 0$
$ {R_{ab}}^{Ib} - {1\over 2} {e_a}^I {R_{bc}}^{bc} = 0$
$ {R_a}^I - {1\over 2} {e_a}^I R = 0$
Transform both sides by $e_{bI}$ to get
$ R_{ab} - {1\over 2} g_{ab} R = 0$
and there is the Einstein Field Equations
Spin connection and extrinsic curvature relation
$ {{\omega_\alpha}^I}_\beta = {\Gamma^I}_{\beta\alpha} - \partial_\alpha {e_\beta}^I $
Using $\omega_{\alpha IJ} = -\omega_{\alpha JI}$...
$ -{\omega_{\alpha\beta}}^I = {\Gamma^I}_{\beta\alpha} - \partial_\alpha {e_\beta}^I $
solve for time direction, and in spatial frame:
$ -{\omega_{ab}}^\hat{t} = {\Gamma^\hat{t}}_{ba} - \partial_a {e_b}^\hat{t} $
$ -{\omega_{ab}}^\hat{t} = {e_\mu}^\hat{t} {\Gamma^\mu}_{ba} - \partial_a {e_b}^\hat{t} $
$ -{\omega_{ab}}^\hat{t} = {e_t}^\hat{t} {\Gamma^t}_{ba} + {e_c}^\hat{t} {\Gamma^c}_{ba} - \partial_a {e_b}^\hat{t} $
use the ADM tetrad definition that ${e_t}^{\hat{t}} = \alpha$ and ${e_b}^\hat{t} = 0$
$ -{\omega_{ab}}^\hat{t} = \alpha {\Gamma^t}_{ba} $
using the extrinsic curvature spatial components in ADM metric: $K_{ab} = -\alpha {\Gamma^t}_{ba}$
$ -{\omega_{ab}}^\hat{t} = -{1\over\alpha} \alpha K_{ab} $
$ {\omega_{ab}}^\hat{t} = K_{ab} $
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