# A What is right expression for momentum constraint in GR?

Tags:
1. Jul 30, 2016

### Sergei65

Momentum constraint in GR in ADM formalism is written in the form

$$\mathcal M_i=\gamma_{ij}D_k\pi^{kj},~~~~~~~~~~(1a)$$ or equivalently

$$\mathcal M_i=D_k\pi^{k}_i,~~~~~~~~~~(1b)$$ where
$\pi^{ij}=-\gamma^{1/2}\left(K^{ij}-\gamma^{ij}K\right)~$, $K=\gamma^{ij}K_{ij}~$, $\gamma=\det \gamma_{ij}~$ and $D_i~$ is covariant derivative. This is from DeWitt1967 parer and original ADM parer.

However, those who deal with numerical relativity uses $$\mathcal M_i=D_jK^j_i-D_iK.~~~~~~~~~~~~~~~(2)$$

What formula is right? (they coincides only if $\gamma$ does not depend on spatial coordinates, which is evidently not the case.

2. Jul 30, 2016

### Ravi Mohan

I think the equations 1a, 1b and 2 are all same (besides a factor of $-\sqrt{\gamma}$ in equation 2). To establish the equality you need to use the fact that the intrinsic covariant derivative is (pullback) metric compatible. Also $D_i\sqrt{\gamma}=\frac{1}{2}\sqrt{\gamma}\gamma^{ab}D_i\gamma_{ab}=0$

3. Jul 30, 2016

### Sergei65

Let me ask, why we could not write $D_i\sqrt {\gamma}=\partial_i \sqrt {\gamma}\sim\gamma^{ab}\partial_i\gamma_{ab}\ne0$? I ask this because it is well known that $d\gamma\sim\gamma^{ab}d\gamma_{ab}$, where $d$ is usual differencial. From the other hand it seems that $D_i\gamma=\partial_i \gamma$.

Last edited: Jul 30, 2016
4. Jul 30, 2016

### Sergei65

Last edited: Jul 30, 2016