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Stress-energy projected components:
Let $\rho = n^u n^v T_{uv}$ = the stress-energy density (2008 Alcubierre, eqn. 2.4.12)
Let $j_a = -n^u T_{uv} {\gamma_a}^v$ = the stress-energy spatial current (2008 Alcubierre, eqn. 2.4.12)
Let $S_{ab} = T_{uv} {\gamma_a}^u {\gamma_b}^v$ = the spatial stress tensor
Let $S = T_{uv} \gamma^{uv} = {S^a}_a$ = the spatial stress tensor trace
So $T_{uv} g^{uv} = T_{uv} (\gamma^{uv} - n^u n^v) = S - \rho$
$\perp j_a = {\gamma_a}^b j_b$
$= -{\gamma_a}^b n^u T_{uv} {\gamma^v}_b$
$= -n^u T_{uv} {\gamma^v}_a$
$= j_a$
so $j_a$ is spatial
$j_a n^a = -n^u T_{uv} {\gamma^v}_a n^a = 0$ so $j_a$ is spatial
So $T_{uv} = \delta_u^a \delta_v^b T_{ab}$
$= ({\gamma_u}^a - n_u n^a) ({\gamma_v}^b - n_v n^b) T_{ab}$
$= ({\gamma_u}^a {\gamma_v}^b - {\gamma_u}^a n_v n^b - n_u n^a {\gamma_v}^b + n_u n^a n_v n^b) T_{ab}$
$=
{\gamma_u}^a {\gamma_v}^b T_{ab}
- {\gamma_u}^a n_v n^b T_{ab}
- n_u n^a {\gamma_v}^b T_{ab}
+ n_u n^a n_v n^b T_{ab}
$
$= S_{uv} + n_v j_u + n_u j_v + n_u n_v \rho$
So $T_{uv} - \frac{1}{2} g_{uv} T_{ab} g^{ab}$
$= S_{uv} + n_v j_u + n_u j_v + n_u n_v \rho
- \frac{1}{2} (\gamma_{uv} - n_u n_v) (S_{ab} + n_b j_a + n_a j_b + n_a n_b \rho) (\gamma^{ab} - n^a n^b)
$
$= S_{uv}
+ n_v j_u
+ n_u j_v
+ n_u n_v \rho
- \frac{1}{2} (\gamma_{uv} - n_u n_v) (
S_{ab} \gamma^{ab}
- S_{ab} n^a n^b
+ n_b j_a \gamma^{ab}
- n_b j_a n^a n^b
+ n_a j_b \gamma^{ab}
- n_a j_b n^a n^b
+ n_a n_b \rho \gamma^{ab}
- n_a n_b \rho n^a n^b
)
$
$= S_{uv}
- \frac{1}{2} \gamma_{uv} (S - \rho)
+ n_v j_u
+ n_u j_v
+ n_u n_v \rho
+ \frac{1}{2} n_u n_v (S - \rho)
$
$= S_{uv}
+ n_v j_u
+ n_u j_v
+ n_u n_v \rho
- \frac{1}{2} (\gamma_{uv} - n_u n_v) (S - \rho)
$
$= S_{uv}
+ n_v j_u
+ n_u j_v
+ n_u n_v \rho
- \frac{1}{2} g_{uv} (S - \rho)
$
Einstein Field Equations in projected components:
Einstein Field Equations, contracting both indexes with the normal vector:
$n^u n^v G_{uv} = 8 \pi n^u n^v T_{uv}$
$n^u n^v G_{uv} = 8 \pi \rho$
using $2 n^u n^v G_{uv} = R^\perp + K^2 - K^{ab} K_{ab}$:
$R^\perp + K^2 - K^{ab} K_{ab} = 16 \pi \rho$ (2010 Baumgarte & Shapiro, eqn. 2.125)
$R^\perp + K^2 - K^{ab} K_{ab} - 16 \pi \rho = 0$
Hamiltonian constraint: Let...
$\mathcal{H} = n^u n^v G_{uv} - 8 \pi \rho$ (2008 Alcubierre, eqn. 2.5.10)
$\mathcal{H} = \frac{1}{2} (R^\perp + K^2 - K^{ab} K_{ab}) - 8 \pi \rho$ (2008 Alcubierre, eqn. 2.5.9)
Now when the Einstein Field Equations are satisfied, we will have $\mathcal{H} = 0$
Einstein Field Equations, contracting one index with normal and projecting the other:
$-n^u {\gamma_a}^v G_{uv} = -n^u {\gamma_a}^v 8 \pi T_{uv}$
using $j_a = -n^u {\gamma_a}^v T_{uv}$
$-n^u {\gamma_a}^v G_{uv} = 8 \pi j_a$
using ${\gamma_a}^u n^v G_{uv} = \nabla^\perp_a K - \nabla^\perp_b {K_a}^b$
$\nabla^\perp_b {K_a}^b - \nabla^\perp_a K = 8 \pi j_a$
$\nabla^\perp_b {K_a}^b - \nabla^\perp_a K - 8 \pi j_a = 0$
Momentum constraint: Let...
$\mathcal{M}_a = -n^u {\gamma_a}^v G_{uv} - 8 \pi j_a$ (2008 Alcubierre, eqn. 2.5.11)
$\mathcal{M}_a = \nabla^\perp_b {K_a}^b - \nabla^\perp_a K - 8 \pi j_a$
When the Einstein Field Equations are satisfied, we will have $\mathcal{M}_a = 0$
Einstein Field Equations, projecting both indexes.
${\gamma_a}^u {\gamma_b}^v G_{uv} = 8 \pi {\gamma_a}^u {\gamma_b}^v T_{uv}$
using $S_{ab} = T_{uv} {\gamma_a}^u {\gamma_b}^v$:
${\gamma_a}^u {\gamma_b}^v G_{uv} = 8 \pi S_{uv}$
${\gamma_a}^u {\gamma_b}^v G_{uv} - 8 \pi S_{uv} = 0$
Energy constraint: Let...
$\mathcal{E}_{ab} = {\gamma_a}^u {\gamma_b}^v G_{uv} - 8 \pi S_{uv}$ (2008 Alcubierre, eqn. 2.5.12)
... this one is mentioned in 2008 Alcubierre, eqn. 2.5.12, but no mention of which values for $\perp G_{uv}$ are used ... is it the R definition? is it the K definition?
When the Einstein Field Equations are satisfied, we will have $\mathcal{E}_{ab} = 0$
starting with (from the "normal projections" worksheet):
$ {R^\perp}_{ab}
+ K K_{ab}
- K_{ac} {K^c}_b
=
{\gamma_a}^q {\gamma_b}^s R_{qs}
+ n^p n^r {\gamma_a}^q {\gamma_b}^s R_{pqrs}
$
$ n^p n^r {\gamma_a}^q {\gamma_b}^s R_{pqrs}
=
{R^\perp}_{ab}
+ K K_{ab}
- K_{ac} {K^c}_b
- {\gamma_a}^q {\gamma_b}^s R_{qs}
$
Substitute the trace-reversed Einstein field equations, $R_{ab} = 8 \pi (T_{ab} - \frac{1}{2} g_{ab} T)$
$
n^p n^r {\gamma_a}^q {\gamma_b}^s R_{pqrs}
=
{R^\perp}_{ab}
+ K K_{ab}
- K_{ac} {K^c}_b
- {\gamma_a}^u {\gamma_b}^v 8 \pi (T_{uv} - \frac{1}{2} g_{uv} T_{cd} g^{cd})
$ (2010 Baumgarte & Shapiro, 2.103)
$
n^p n^r {\gamma_a}^q {\gamma_b}^s R_{pqrs}
=
{R^\perp}_{ab}
+ K K_{ab}
- K_{ac} {K^c}_b
- 8 \pi (S_{ab} - \frac{1}{2} \gamma_{ab} (S - \rho))
$
... and this route is the one that turns into the time-derivative of the extrinsic curvature ...
... but how do we come up with the $\mathcal{E}_{ab}$ constraints from only projected, non-time-derivative terms?
We don't. The $\mathcal{E}_{ab}$ constraints give rise to the $K_{ab}$ ivp. So using ADM formalism this constraint is always satisfied.
$\mathcal{L}_\vec{n} K_{ab} =
n^u \partial_u K_{ab}
+ K_{ub} \partial_a n^u
+ K_{au} \partial_b n^u
$
$\mathcal{L}_\vec{n} K_{ab} =
n^u \nabla_u K_{ab}
+ K_{ub} \nabla_a n^u
+ K_{au} \nabla_b n^u
$
using $K_{ab} = -\nabla_a n_b - n_a a_b$
so $\nabla_a n_b = -K_{ab} - n_a a_b$
$\mathcal{L}_\vec{n} K_{ab} =
n^u \nabla_u (
-\nabla_a n_b - n_a a_b
)
+ {K^u}_b (
-K_{au} - n_a a_u
)
+ {K^u}_a (
-K_{bu} - n_b a_u
)
$
$\mathcal{L}_\vec{n} K_{ab} =
- n^u \nabla_u \nabla_a n_b
- a_b n^u \nabla_u n_a
- n_a n^u \nabla_u a_b
- {K^u}_b K_{au}
- {K^u}_a K_{bu}
- {K^u}_b n_a a_u
- {K^u}_a n_b a_u
$ (2010 Baumgarte & Shapiro, eqn. 2.74)
using $a_a = n^u \nabla_u n_a$
$\mathcal{L}_\vec{n} K_{ab} =
- n^u \nabla_u \nabla_a n_b
- a_a a_b
- n_a n^u \nabla_u a_b
- 2 K_{au} {K^u}_b
- {K^u}_b n_a a_u
- {K^u}_a n_b a_u
$
using $n_d {R^d}_{bua} = \nabla_u \nabla_a n_b - \nabla_a \nabla_u n_b$
$\mathcal{L}_\vec{n} K_{ab} =
- n_d n^u {R^d}_{bua}
- n^u \nabla_a \nabla_u n_b
- a_a a_b
- n_a n^u \nabla_u a_b
- 2 K_{au} {K^u}_b
- {K^u}_b n_a a_u
- {K^u}_a n_b a_u
$
TODO after a lot of work
$\mathcal{L}_\vec{n} K_{ab} =
-\frac{1}{\alpha} \nabla^\perp_a \nabla^\perp_b \alpha
- K_{ac} {K^c}_b
+ n^p n^r {\gamma_a}^q {\gamma_b}^s R_{pqrs}
$ (2010 Baumgarte & Shapiro, eqn. 2.82)
using the definition of $n^p n^r {\gamma_a}^q {\gamma_b}^s R_{pqrs}$ above ... without stress-energy:
$\mathcal{L}_\vec{n} K_{ab} =
-\frac{1}{\alpha} \nabla^\perp_a \nabla^\perp_b \alpha
- K_{ac} {K^c}_b
+ {R^\perp}_{ab}
+ K K_{ab}
- K_{ac} {K^c}_b
- {\gamma_a}^u {\gamma_b}^v R_{uv}
$
... and with stress-energy:
$\mathcal{L}_\vec{n} K_{ab} =
-\frac{1}{\alpha} \nabla^\perp_a \nabla^\perp_b \alpha
- K_{ac} {K^c}_b
+ {R^\perp}_{ab}
+ K K_{ab}
- K_{ac} {K^c}_b
- 8 \pi (S_{ab} - \frac{1}{2} \gamma_{ab} (S - \rho))
$
$\mathcal{L}_\vec{n} K_{ab} =
-\frac{1}{\alpha} \nabla^\perp_a \nabla^\perp_b \alpha
+ {R^\perp}_{ab}
+ K K_{ab}
- 2 K_{ac} {K^c}_b
- 8 \pi (S_{ab} - \frac{1}{2} \gamma_{ab} (S - \rho))
$
Extrinsic curvature initial value formulation:
using $\mathcal{L}_\vec{n} = \frac{1}{\alpha} (\mathcal{L}_\vec{t} - \mathcal{L}_\vec\beta)$
$\partial_t K_{ab} - \mathcal{L}_\vec\beta K_{ab} = \alpha \mathcal{L}_\vec{n} K_{ab}$
$\partial_t K_{ab}
=
- \nabla^\perp_a \nabla^\perp_b \alpha
+ \alpha (
{R^\perp}_{ab}
+ K K_{ab}
- 2 K_{ac} {K^c}_b
- 8 \pi (S_{ab} - \frac{1}{2} \gamma_{ab} (S - \rho))
)
+ \mathcal{L}_\vec\beta K_{ab}
$
Ricci tensor contracted with normal:
$n^a n^b R_{ab}$
$=n^a n^b {R^c}_{acb}$
$=n^b (\nabla_c \nabla_b n^c - \nabla_b \nabla_c n^c)$
$=n^b \nabla_c \nabla_b n^c - n^b \nabla_b \nabla_c n^c$
$=\nabla_c (n^b \nabla_b n^c) - (\nabla_c n^b)(\nabla_b n^c) - \nabla_b(n^b \nabla_c n^c) + (\nabla_b n^b) (\nabla_c n^c)$
$=\nabla_c (n^b \nabla_b n^c) - (-{K_c}^b + n_c a^b)(-{K_b}^c + n_b a^c) - \nabla_b(n^b \nabla_c n^c) + (-{K_b}^b + n_b a^b) (-{K_c}^c + n_c a^c)$
$=\nabla_c (n^b \nabla_b n^c) - ({K_c}^b {K_b}^c - n_c a^b {K_b}^c - n_b a^c {K_c}^b + n_b n_c a^b a^c) - \nabla_b(n^b \nabla_c n^c) + K^2$
$=\nabla_c (n^b \nabla_b n^c) - \nabla_b(n^b \nabla_c n^c) - K_{ab} K^{ab} + K^2$
$=\nabla_c (n^b \nabla_b n^c) - \nabla_c(n^c \nabla_b n^b) - K_{ab} K^{ab} + K^2$
$n^a n^b R_{ab}=K^2 - K_{ab} K^{ab} + \nabla_c (n^b \nabla_b n^c - n^c \nabla_b n^b)$ (2008 Alcubierre, eqn 2.7.3)
start with from the "normal projections" worksheet:
$R = R^\perp - K_{ab} K^{ab} + K^2 - 2 n^a n^b R_{ab}$
...substitute contracted Ricci tensor...
$R=R^\perp-K_{ab}K^{ab}+K^2-2(K^2 - K_{ab} K^{ab} + \nabla_c (n^b \nabla_b n^c - n^c \nabla_b n^b))$
$R=R^\perp-K_{ab}K^{ab}+K^2-2K^2 + 2K_{ab} K^{ab} - 2\nabla_c (n^b \nabla_b n^c - n^c \nabla_b n^b)$
$R=R^\perp+K_{ab}K^{ab}-K^2 - 2\nabla_c (n^b \nabla_b n^c - n^c \nabla_b n^b)$ (2008 Alcubierre, eqn 2.7.4)
Looks like now we disregard the $2\nabla_c (n^b \nabla_b n^c - n^c \nabla_b n^b)$
because it can be converted to a total integral via divergence theorem.
but this is only the case if we are integrating over a volume, i.e. true for $\int L dx^4$ and therefore $\int R dx^4$
So how can it still be true for $\mathcal{L} = \int \sqrt{-g} R dx^4$? Does $\int \nabla_c \sqrt{-g} dx^4$ disappear as well?
$R = R^\perp + K_{ab} K^{ab} - K^2$
Ok a suspicious thing about the "cancel total derivatives", that would imply:
$n^a n^b R_{ab} = K^2 - K_{ab} K^{ab}$
Is only true in the context of a spacetime-volume integral?
If so then how true is all of numerical relativity, where we are doing numerical spatial volume integration / spatial PDEs?
Back from our "normal projection" worksheet, start with our projected Einstein curvature:
$ {\gamma_a}^u {\gamma_b}^v G_{uv}
=
{R^\perp}_{ab}
+ K K_{ab}
- K_{ac} {K^c}_b
- \frac{1}{2} \gamma_{ab} R
- n^p n^r {\gamma_a}^q {\gamma_b}^s R_{pqrs}
$
Substitute $R = R^\perp + K_{ab} K^{ab} - K^2$
$ {\gamma_a}^u {\gamma_b}^v G_{uv}
=
{R^\perp}_{ab}
+ K K_{ab}
- K_{ac} {K^c}_b
- \frac{1}{2} \gamma_{ab} (
R^\perp + K_{ab} K^{ab} - K^2
)
- n^p n^r {\gamma_a}^q {\gamma_b}^s R_{pqrs}
$
$ {\gamma_a}^u {\gamma_b}^v G_{uv}
=
{R^\perp}_{ab}
- \frac{1}{2} \gamma_{ab} R^\perp
+ \frac{1}{2} \gamma_{ab} K^2
+ K K_{ab}
- \frac{1}{2} \gamma_{ab} K_{ab} K^{ab}
- K_{ac} {K^c}_b
- n^p n^r {\gamma_a}^q {\gamma_b}^s R_{pqrs}
$
... how to get rid of that last $ n^p n^r {\gamma_a}^q {\gamma_b}^s R_{pqrs}$ ?
Maybe you can't. As 2008 Alcubierre after eqn. 2.5.12 says, the $\mathcal{E}_{ab} = 0$ turn into the ADM equations.
Lagrangian is equal to scalar curvature:
$L=R$
Densitized Lagrangian:
$\mathcal{L}=\sqrt{-g}L$
$=\sqrt{-g}R$
$=\alpha\sqrt{\gamma}R$
...substitute Gauss-Codazzi equations...
$=\alpha\sqrt{\gamma}(R^\perp + K_{ab} K^{ab} - K^2$
$\mathcal{L}=\alpha\sqrt{\gamma}(R^\perp+K_{ij} K^{ij} - K^2)$ (2008 Alcubierre, eqn. 2.7.6)
Vacuum Action:
$ S = \int \sqrt{-g} R dx^4 $
$ S = \int \int \alpha \sqrt{\gamma} (R^{\perp} + K_{ij} K^{ij} - K^2) dx^3 dt $
$ \dot{\gamma}_{ij} = -2 \alpha K_{ij} + 2 \nabla^\perp_{(i} \beta_{j)} $
$ K_{ij} = -\frac{1}{2\alpha} \dot{\gamma}_{ij} + \frac{1}{\alpha} \nabla^\perp_{(i} \beta_{j)} $
$ \frac{\partial}{\partial \dot{\gamma}_{ij}} K_{kl} = -\frac{1}{2\alpha} \delta^i_k \delta^j_l $
conjugate momentum:
$ \pi_{ij} = \frac{\partial \mathcal{L}}{\partial \dot{\gamma}^{ij}} $
$ = \frac{\partial}{\partial \dot{\gamma}^{ij}} (\alpha \sqrt{\gamma} (R^{\perp} + K_{kl} K^{kl} - K^2)) $
$ = \frac{\partial}{\partial \dot{\gamma}^{ij}} \alpha \sqrt{\gamma} (R^{\perp} + \gamma_{km} \gamma_{ln} K^{kl} K^{mn} - K^{kl} \gamma_{kl} K^{mn} \gamma_{mn}) $
$ \pi_{ij} = \alpha \sqrt{\gamma} (\frac{\delta}{\delta\dot{\gamma}^{ij}} (\gamma_{km} \gamma_{ln} K^{kl} K^{mn}) - \frac{\delta}{\delta\dot{\gamma}^{ij}} (K^{kl} \gamma_{kl} K^{mn} \gamma_{mn})) $
$ \pi_{ij} = \alpha \sqrt{\gamma} (\gamma_{km} \gamma_{ln} \frac{\delta}{\delta\dot{\gamma}^{ij}} (K^{kl} K^{mn}) - \gamma_{kl} \gamma_{mn} \frac{\delta}{\delta\dot{\gamma}^{ij}} (K^{kl} K^{mn})) $
$ \pi_{ij} = \alpha \sqrt{\gamma} (\gamma_{km} \gamma_{ln} - \gamma_{kl} \gamma_{mn}) \frac{\delta}{\delta\dot{\gamma}^{ij}} (K^{kl} K^{mn}) $
$ \pi_{ij} = \alpha \sqrt{\gamma} (\gamma_{km} \gamma_{ln} - \gamma_{kl} \gamma_{mn}) (K^{kl} \frac{\delta}{\delta\dot{\gamma}^{ij}} K^{mn} + K^{mn} \frac{\delta}{\delta\dot{\gamma}^{ij}} K^{kl}) $
$ \pi_{ij} = -\alpha \sqrt{\gamma} (\gamma_{km} \gamma_{ln} - \gamma_{kl} \gamma_{mn}) (\frac{1}{2\alpha} K^{kl} \delta^m_i \delta^n_j + \frac{1}{2\alpha} K^{mn} \delta^k_i \delta^l_j) $
$ \pi_{ij} = -\alpha \frac{1}{2\alpha} \sqrt{\gamma} ((\gamma_{km} \gamma_{ln} - \gamma_{kl} \gamma_{mn}) K^{kl} \delta^m_i \delta^n_j + (\gamma_{km} \gamma_{ln} - \gamma_{kl} \gamma_{mn}) K^{mn} \delta^k_i \delta^l_j) $
$ \pi_{ij} = -\frac{1}{2} \sqrt{\gamma} ((\gamma_{km} \gamma_{ln} K^{kl} \delta^m_i \delta^n_j - \gamma_{kl} \gamma_{mn} K^{kl} \delta^m_i \delta^n_j) + (\gamma_{km} \gamma_{ln} K^{mn} \delta^k_i \delta^l_j - \gamma_{kl} \gamma_{mn} K^{mn} \delta^k_i \delta^l_j)) $
$ \pi_{ij} = -\frac{1}{2} \sqrt{\gamma} (K_{ij} - \gamma_{ij} K + K_{ij} - \gamma_{ij} K) $
$ \pi_{ij} = -\sqrt{\gamma} (K_{ij} - \gamma_{ij} K) $
$ \pi^{ij} = -\sqrt{\gamma} (K^{ij} - \gamma^{ij} K) $ (2008 Alcubierre, eqn. 2.7.8)
Notice that $\pi^{ij}$ is a tensor density with weight 1.
Extrinsic curvature in terms of conjugate momentum:
$ \gamma_{ij} \pi^{ij} = \gamma_{ij} (-\sqrt\gamma (K^{ij} - \gamma^{ij} K)) $
solve conjugate momentum trace:
$ \pi = -\sqrt\gamma (\gamma_{ij} K^{ij} - \gamma_{ij} \gamma^{ij} K) $
$ \pi = -\sqrt\gamma (K - 3 K) $
$ \pi = 2 \sqrt\gamma K $
$ K = \pi / (2 \sqrt\gamma) $
substitute into equation of conjugate momentum in terms of extrinsic curvature:
$ \pi^{ij} = -\sqrt{\gamma} (K^{ij} - \gamma^{ij} K) $
$ \pi^{ij} = -\sqrt{\gamma} K^{ij} + \sqrt\gamma \gamma^{ij} \pi / (2 \sqrt\gamma) $
$ \pi^{ij} = -\sqrt{\gamma} K^{ij} + \frac{1}{2} \gamma^{ij} \pi $
$ K^{ij} = -\frac{1}{\sqrt{\gamma}} (\pi^{ij} - \frac{1}{2} \gamma^{ij} \pi) $ (2008 Alcubierre, eqn. 2.7.9)
Extrinsic cuvrature derivative in terms of conjugate momentum:
$ \dot{\gamma}_{ij} = -2 \alpha K_{ij} + 2 \nabla^\perp_{(i} \beta_{j)} $
substitute $ K^{ij} = -\frac{1}{\sqrt{\gamma}} (\pi^{ij} - \frac{1}{2} \gamma^{ij} \pi) $:
$ \dot{\gamma}_{ij} = 2 \alpha \frac{1}{\sqrt\gamma} (\pi^{ij} - \frac{1}{2} \gamma^{ij} \pi) + 2 \nabla^\perp_{(i} \beta_{j)} $
useful identity:
$ K^2 - K_{ij} K^{ij} $
$ = (\pi / (2 \sqrt\gamma))^2 - \frac{1}{\gamma} (\pi_{ij} - \frac{1}{2} \gamma_{ij} \pi) (\pi^{ij} - \frac{1}{2} \gamma^{ij} \pi) $
$ = \frac{1}{4\gamma} \pi^2 - \frac{1}{\gamma} (\pi_{ij} \pi^{ij} - \frac{1}{2} \pi \gamma^{ij} \pi_{ij} - \frac{1}{2} \pi \gamma_{ij} \pi^{ij} + \frac{1}{4} \pi \gamma_{ij} \gamma^{ij} \pi) $
$ = \frac{1}{\gamma} ( \frac{1}{4} \pi^2 - \pi_{ij} \pi^{ij} + \frac{1}{2} \pi^2 + \frac{1}{2} \pi^2 - \frac{3}{4} \pi^2 ) $
$ = \frac{1}{\gamma} (\frac{1}{2} \pi^2 - \pi_{ij} \pi^{ij}) $
Hamiltonian density:
$ \mathscr{H} = \pi^{ij} \dot{\gamma}_{ij} - \mathcal{L} $
$ = -\alpha \sqrt{\gamma} (R^{\perp} + K_{ij} K^{ij} - K^2) - \sqrt{\gamma} (K^{ij} - \gamma^{ij} K) (-2 \alpha K_{ij} + 2 \nabla^\perp_{(i} \beta_{j)}) $
$ = -\alpha \sqrt{\gamma} (R^{\perp} + K_{ij} K^{ij} - K^2) + 2 \alpha \sqrt{\gamma} (K^{ij} - \gamma^{ij} K) K_{ij} - 2 \sqrt{\gamma} (K^{ij} - \gamma^{ij} K) \nabla^\perp_{(i} \beta_{j)} $
$ = -\sqrt{\gamma} (\alpha (R^{\perp} + K_{ij} K^{ij} - K^2 - 2 (K^{ij} - \gamma^{ij} K) K_{ij}) + 2 (K^{ij} - \gamma^{ij} K) \nabla^\perp_{(i} \beta_{j)} ) $
$ = -\sqrt{\gamma} (\alpha (R^{\perp} + K_{ij} K^{ij} - K^2 - 2 K^{ij} K_{ij} + 2 K^2) + 2 (K^{ij} - \gamma^{ij} K) \nabla^\perp_{(i} \beta_{j)} ) $
$ = -\sqrt{\gamma} (\alpha (R^{\perp} - K_{ij} K^{ij} + K^2) + 2 (K^{ij} - \gamma^{ij} K) \nabla^\perp_{(i} \beta_{j)} ) $
$ = -\sqrt{\gamma} (\alpha (R^{\perp} - K_{ij} K^{ij} + K^2) - 2 \pi^{ij} \frac{1}{\sqrt{\gamma}} \nabla^\perp_{(i} \beta_{j)} ) $
$ = -\sqrt{\gamma} (\alpha (R^{\perp} + \frac{1}{\gamma} (\frac{1}{2} \pi^2 - \pi_{ij} \pi^{ij})) - 2 \pi^{ij} \frac{1}{\sqrt{\gamma}} \nabla^\perp_{(i} \beta_{j)} ) $
$ = \alpha (-\sqrt{\gamma} R^{\perp} + \frac{1}{\sqrt{\gamma}} (\pi_{ij} \pi^{ij} - \frac{1}{2} \pi^2)) + 2 \pi^{ij} \nabla^\perp_i \beta_j $
using integration by parts: $ 2 \pi^{ij} \nabla^\perp_j \beta_i = -2 \beta_i \nabla^\perp_j \pi^{ij} $
(TODO prove that integration-by-parts works with the spatial projected covariant derivative)
$ = \alpha (-\sqrt{\gamma} R^{\perp} + \frac{1}{\sqrt{\gamma}} (\pi_{ij} \pi^{ij} - \frac{1}{2} \pi^2)) - 2 \beta^i \gamma_{ij} \nabla^\perp_k \pi^{jk} $
Somewhere in the middle of this we had:
$\mathscr{H} = -\sqrt{\gamma} (\alpha (R^{\perp} - K_{ij} K^{ij} + K^2) - 2 \pi^{ij} \frac{1}{\sqrt{\gamma}} \nabla^\perp_{(i} \beta_{j)} ) $
...TODO should look like $\mathscr{H} = -2 \sqrt{\gamma} ( \alpha \mathcal{H} + \beta_i \mathcal{M}^i )$
...TODO that means defining earlier ... $\mathcal{H} = R^{\perp} - K_{ij} K^{ij} + K^2$ ... but next we define it as the densitized curvature ... which is it?
Constraints:
Let $\mathcal{H} = -\sqrt\gamma R^\perp + \frac{1}{\sqrt{\gamma}} (\pi_{ij} \pi^{ij} - \frac{1}{2}\pi^2) $
So $\mathcal{H} = -\sqrt\gamma (R^\perp + K^2 - {K^i}_j {K^j}_i) $
So $\mathcal{H} = -\sqrt\gamma (R^\perp + \epsilon_{ijm} \epsilon^{klm} {K^i}_k {K^j}_l) $
Let $ \mathcal{M}^i = 2 \nabla^\perp_j \pi^{ij} $
$ \mathscr{H} = \alpha \mathcal{H} - \beta_i \mathcal{M}^i $
(I'm using Alcubierre's symbols from the numerical relativity community. In the LQG world this usually looks like $H = C N + C^i N_i$.)
Assume $\mathcal{H} = 0, \mathcal{M}^i = 0$.
(TODO Alcubierre's book, eqn 2.7.11, defines Hamiltonian as non-densitized and puts the density into this equation. Maybe I should too? Where did I get the densitized version from anyways?)
Let $ \textbf{H} = \int \mathscr{H} dx^3 $
$ S = \int \textbf{H} dt $
Derivatives of momentum constraint:
$ \beta_i \mathcal{M}^i = -2 \beta^k \gamma_{ki} \nabla^\perp_j \pi^{ij} $
$ = -2 \beta_i \nabla^\perp_j \pi^{ij} $
using integration-by-parts of spatial covariant derivative (TODO verify you can do this)
$ = 2 \pi^{ij} \nabla^\perp_i \beta_j $
$ = \pi^{ij} \cdot 2 \nabla^\perp_{(i} \beta_{j)} $
using $ \mathcal{L}_\vec\beta \gamma_{ij} = 2 \nabla^\perp_{(i} \beta_{j)} $
$ = \pi^{ij} \mathcal{L}_\vec\beta \gamma_{ij} $
using integration by parts of Lie derivative (TODO verify you can do this)
$ = -\gamma_{ij} \mathcal{L}_\vec\beta \pi^{ij} $
so $ \frac{\partial}{\partial \gamma_{ij}} ( \beta_k \mathcal{M}^k ) = -\mathcal{L}_\vec\beta \pi^{ij} $
and $ \frac{\partial}{\partial \pi^{ij}} ( \beta_k \mathcal{M}^k ) = \mathcal{L}_\vec\beta \gamma_{ij} $
Let $ \{f,g\} $ represent the Poisson bracket of $f$ and $g$, defined as:
$ \{f,g\} = \frac{\partial f}{\partial \gamma_{ij}} \frac{\partial g }{\partial \pi^{ij}} - \frac{\partial f}{\partial \pi^{ij}} \frac{\partial g}{\partial \gamma_{ij}} $
There is also the option to define an extended phase space: $\{\gamma_{ij}, \beta_i, \alpha\}$ and $\{\pi^{ij}, p^i, p\}$. (think of some letters for those).
The Poisson bracket can then be defined as:
$\{f, g\} =
\frac{\partial f}{\partial \gamma_{ij}} \frac{\partial g}{\partial \pi^{ij}}
- \frac{\partial f}{\partial \pi^{ij}} \frac{\partial g}{\partial \gamma_{ij}}
+ \frac{\partial f}{\partial \beta_i} \frac{\partial g}{\partial p^i}
- \frac{\partial f}{\partial p^i} \frac{\partial g}{\partial \beta_i}
+ \frac{\partial f}{\partial \alpha} \frac{\partial g}{\partial p}
- \frac{\partial f}{\partial p} \frac{\partial g}{\partial \alpha}
$.
Then choose the phase space hypersurface/subset such that $p = 0, p^i = 0$. I guess you could pick any values for $p$ and $p^i$ so long as they are constant, such that $\frac{\partial}{\partial p} = 0$ and $\frac{\partial}{\partial p^i} = 0$.
This causes your Poisson bracket value to reduce to the original definition of
$\{f, g\} =\frac{\partial f}{\partial \gamma_{ij}} \frac{\partial g}{\partial \pi^{ij}}
- \frac{\partial f}{\partial \pi^{ij}} \frac{\partial g}{\partial \gamma_{ij}}
$. (This doesn't sit right. If a coordinate variable remains constaint, this does not imply that the derivative of another variable wrt that coordinate variable will be constant.)
Poisson bracket of momentum constraint and arbitrary function:
$ \{f(\pi, \gamma), \beta_k \mathcal{M}^k\} $
$ = \frac{\partial f}{\partial \gamma_{ij}} \frac{\partial}{\partial \pi^{ij}} (\beta_k \mathcal{M}^k) - \frac{\partial f}{\partial \pi^{ij}} \frac{\partial}{\partial \gamma_{ij}} (\beta_k \mathcal{M}^k) $
$ = \frac{\partial f}{\partial \gamma_{ij}} \mathcal{L}_\vec\beta \gamma_{ij} + \frac{\partial f}{\partial \pi^{ij}} \mathcal{L}_\vec\beta \pi^{ij} $
$ = \mathcal{L}_\vec{\beta} f $
Derivatives of Hamiltonian constraint:
$ \frac{\partial}{\partial \gamma_{ij}} \mathcal{H} $
$ = \frac{\partial}{\partial \gamma_{ij}} (-\sqrt\gamma R^\perp + \frac{1}{\sqrt{\gamma}} (\gamma_{km} \gamma_{ln} \pi^{kl} \pi^{mn} - \frac{1}{2} \gamma_{kl} \gamma_{mn} \pi^{kl} \pi^{mn})) $
$ = -(
\frac{1}{2} \frac{1}{\sqrt{\gamma}} R^\perp \frac{\partial}{\partial \gamma_{ij}} (\gamma)
+ \sqrt\gamma \frac{\partial}{\partial \gamma_{ij}} (R^\perp)
)
-\frac{1}{2} \frac{1}{\gamma^{3/2}} \frac{\partial}{\partial \gamma_{ij}} (\gamma) (\gamma_{km} \gamma_{ln} \pi^{kl} \pi^{mn} - \frac{1}{2} \gamma_{kl} \gamma_{mn} \pi^{kl} \pi^{mn})
+ \frac{2}{\sqrt\gamma} ( {\pi^i}_k \pi^{kj} - \frac{1}{2} \pi \pi^{ij}
) $
$ = - \frac{1}{2} \frac{1}{\sqrt{\gamma}} R^\perp \gamma \gamma^{ij}
- \sqrt\gamma \frac{\partial}{\partial \gamma_{ij}} (R^\perp)
- \frac{1}{2} \frac{1}{\gamma^{3/2}} \gamma \gamma^{ij} ( \pi_{kl} \pi^{kl} - \frac{1}{2} \pi^2)
+ \frac{2}{\sqrt\gamma} ( {\pi^i}_k \pi^{kj} - \frac{1}{2} \pi \pi^{ij}
) $
$ =
- \frac{1}{2} \sqrt\gamma R^\perp \gamma^{ij}
- \sqrt\gamma \frac{\partial}{\partial \gamma_{ij}} (R^\perp)
- \frac{1}{2} \frac{1}{\sqrt{\gamma}} \gamma^{ij} ( \pi_{kl} \pi^{kl} - \frac{1}{2} \pi^2)
+ \frac{2}{\sqrt{\gamma}} ( {\pi^i}_k \pi^{kj} - \frac{1}{2} \pi \pi^{ij})
$
$ \frac{\partial}{\partial \pi^{ij}} \mathcal{H} $
$ = \frac{\partial}{\partial \pi^{ij}} (-\sqrt\gamma R^\perp + \frac{1}{\sqrt{\gamma}} (\gamma_{km} \gamma_{ln} \pi^{kl} \pi^{mn} - \frac{1}{2} \gamma_{kl} \gamma_{mn} \pi^{kl} \pi^{mn})) $
$ = \frac{1}{\sqrt{\gamma}} (
\gamma_{im} \gamma_{jn} \pi^{mn}
+ \gamma_{ki} \gamma_{lj} \pi^{kl}
- \frac{1}{2} (
\gamma_{ij} \gamma_{mn} \pi^{mn}
+ \gamma_{kl} \gamma_{ij} \pi^{kl}
)
) $
$ = \frac{2}{\sqrt{\gamma}} ( \pi_{ij} - \frac{1}{2} \gamma_{ij} \pi ) $
Derivatives of the Hamiltonian:
$\frac{\partial}{\partial \gamma_{ij}} \mathscr{H} = \frac{\partial}{\partial \gamma_{ij}} (\alpha \mathcal{H} + \beta_i \mathcal{M}^i)$
$= \alpha \frac{\partial}{\partial \gamma_{ij}} \mathcal{H}$
$= \alpha (
- \frac{1}{2} \sqrt\gamma R^\perp \gamma^{ij}
- \sqrt\gamma \frac{\partial}{\partial \gamma_{ij}} (R^\perp)
- \frac{1}{2} \frac{1}{\sqrt{\gamma}} \gamma^{ij} ( \pi_{kl} \pi^{kl} - \frac{1}{2} \pi^2)
+ \frac{2}{\sqrt{\gamma}} ( {\pi^i}_k \pi^{kj} - \frac{1}{2} \pi \pi^{ij})
)$
$\frac{\partial}{\partial \pi^{ij}} \mathscr{H} = \frac{\partial}{\partial \pi^{ij}} (\alpha \mathcal{H} + \beta_i \mathcal{M}^i)$
$= \alpha \frac{\partial}{\partial \pi^{ij}} \mathcal{H} + \beta_i \frac{\partial}{\partial \pi^{ij}} \mathcal{M}^i$
...
Lie bracket of momentum constraint and momentum constraint:
(TODO was this $\mathcal{H}$ the Hamiltonian constraint scalar, coupled with $\alpha$, or was it $\mathscr{H}$ the Hamiltonian?).
$ \{ \beta_k \mathcal{M}^k, b_k \mathcal{M}^k \} $
$ = \frac{\partial}{\partial \gamma_{ij}} (\beta_k \mathcal{M}^k ) \frac{\partial}{\partial \pi^{ij}} (b_l \mathcal{M}^l) - \frac{\partial}{\partial \pi^{ij}} ( \beta_k \mathcal{M}^k ) \frac{\partial}{\partial \gamma_{ij}} ( b_l \mathcal{M}^l ) $
$ = -\mathcal{L}_\vec\beta \pi^{ij} \frac{\partial}{\partial \pi^{ij}} (b_l \mathcal{M}^l) - \mathcal{L}_\vec\beta \gamma_{ij} \frac{\partial}{\partial \gamma^{ij}} ( b_l \mathcal{M}^l ) $
$ = -\mathcal{L}_\vec\beta (b_k \mathcal{M}^k ) $
$ = [b_k \mathcal{M}^k, \vec\beta] $
TODO show this is equal to $ = [\vec{b}, \vec\beta]_k \mathcal{M}^k $
(i.e. show that $\mathcal{L}_\vec\beta M^k = -2 \mathcal{L}_\vec\beta \nabla^\perp_j \pi^{jk} = 0$)
Lie bracket of momentum constraint and Hamiltonian constraint:
(TODO was this $\mathcal{H}$ the Hamiltonian constraint scalar, coupled with $\alpha$, or was it $\mathscr{H}$ the Hamiltonian?).
$ \{ \beta_k \mathcal{M}^k, \alpha \mathcal{H} \} $
$ = \frac{\partial}{\partial \gamma_{ij}} (\beta_k \mathcal{M}^k ) \frac{\partial}{\partial \pi^{ij}} (\alpha \mathcal{H}) - \frac{\partial}{\partial \pi^{ij}} ( \beta_k \mathcal{M}^k ) \frac{\partial}{\partial \gamma_{ij}} ( \alpha \mathcal{H} ) $
$ = -\mathcal{L}_\vec\beta \pi^{ij} \frac{\partial}{\partial \pi^{ij}} (\alpha \mathcal{H}) - \mathcal{L}_\vec\beta \gamma_{ij} \frac{\partial}{\partial \gamma^{ij}} ( \alpha \mathcal{H} ) $
$ = -\mathcal{L}_\vec\beta (\alpha \mathcal{H}) $
TODO show this is equal to $-\mathcal{H} \mathcal{L}_\vec\beta \alpha $
i.e. show $\mathcal{L}_\vec\beta \mathcal{H} = \mathcal{L}_\vec\beta (-\sqrt\gamma R^\perp + \frac{1}{\sqrt{\gamma}} (\pi_{ij} \pi^{ij} - \frac{1}{2} \pi^2)) = 0$
Lie bracket of Hamiltonian constraint and Hamiltonian constraint:
(TODO was this $\mathcal{H}$ the Hamiltonian constraint scalar, coupled with $\alpha$, or was it $\mathscr{H}$ the Hamiltonian?).
$ \{ \alpha \mathcal{H}, a \mathcal{H} \} $
$ =
\frac{\partial}{\partial \gamma_{ij}} (\alpha \mathcal{H})
\frac{\partial}{\partial \pi^{ij}} (a \mathcal{H})
-
\frac{\partial}{\partial \gamma_{ij}} (a \mathcal{H})
\frac{\partial}{\partial \pi^{ij}} (\alpha \mathcal{H})
$
$ =
\frac{\partial}{\partial \gamma_{ij}} (\alpha \mathcal{H})
\frac{\partial}{\partial \pi^{ij}} (a \mathcal{H})
-
\frac{\partial}{\partial \gamma_{ij}} (a \mathcal{H})
\frac{\partial}{\partial \pi^{ij}} (\alpha \mathcal{H})
$
TODO show this is equal to $(\alpha \partial^k a - a \partial^k \alpha) \mathcal{M}_k $
TODO:
$\dot{f} =? \frac{d}{dt} f = \{f, \mathscr{H}\}$, and how does this relate to the Lie derivative, and to the total derivative definition $\frac{d}{dt} = \frac{\partial}{\partial t} - \mathcal{L}_{\vec{\beta}}$ and the Lie deriative statements above?
implied by $\mathcal{H} = \mathcal{M}^i = 0$:
$\dot{\mathcal{H}} = \{ \mathcal{H}, \mathscr{H} \} = 0$
$\dot{\mathcal{M}}^i = \{ \mathcal{M}^i, \mathscr{H} \} = 0$
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