# 4 of Einstein equations without 2nd order time derivative

This question is related to my previous one and it was a homework problem and was due two weeks ago.

Problem：prove that four of Einsteins' equations $$G_{0\nu} = 8\pi T_{0\nu}$$ have to 2nd order time derivative and thus serve as constraint equations.

I found a nice argument in Carroll's lecture notes, $$\partial_0 G^{0\nu} =-\partial_i G^{i\nu} -\Gamma^{\mu}_{\mu\lambda} G^{\lambda \nu} -\Gamma^{\nu}_{\mu\lambda} G^{\mu \lambda}$$ since there are at most 2nd order time derivative on the RHS, $G^{0\nu}$ has at most 1st order time derivative.

But $G_{0\nu}$ is not proportional to $G^{0\nu}$ for a general metric. In fact $G_{0\nu} = G^{\mu\lambda}g_{0\mu} g_{\lambda \mu}$ contains ingredients other than $G^{0\mu}$.

Of course we can use $G_{\mu\nu}{}^{;\nu}=0$ to prove that there is not 2nd order $\partial^0$ in $G_{0\nu}$. But now "spatial derivative" here is a mixed derivative, $\partial^i = g^{i\mu} \partial_\mu$. Two such spatial derivative could contain 2nd order time derivative.

I suspect that only the $G^{0\nu}$ version is true. However a talk here use $G_{0\nu}$ instead of $G^{0\nu}$ for the constraint equations, where they didn't provide a proof.

Now let me compute it explicitly in the explicit form, while terms containing at most 1st order time derivative will be dropped. So the sign $\sim$ really means both sides have the same term of 2nd order time derivative.

$$R_{0\nu} = R^{\alpha}{}_{0\alpha\nu} \sim \Gamma^{\alpha}{}_{0\nu,\alpha}-\Gamma^{\alpha}{}_{0\alpha,\nu} \sim \Gamma^{0}{}_{0\nu,0} -\Gamma^{\alpha}{}_{0\alpha,\nu}$$

$$R \sim \Gamma^{\alpha\beta}{}_{\alpha,\beta} - \Gamma^{\alpha\beta}{}_{\beta,\alpha} \sim \Gamma^{\alpha 0}{}_{\alpha,0} - \Gamma^{0\alpha}{}_{\alpha,0}$$ $$G_{0\nu} =\Gamma^{0}{}_{0\nu,0} -\Gamma^{\alpha}{}_{0\alpha,\nu} - \frac{1}{2}g_{0\nu}(\Gamma^{\alpha 0}{}_{\alpha,0} - \Gamma^{0\alpha}{}_{\alpha,0})$$

First evaluate things in the parenthesis, namely the scalar curvature,

$$\Gamma^{\alpha 0}{}_{\alpha 0} = (g^{0\beta} \Gamma^{\alpha}{}_{\beta\alpha})_{,0} = (g^{0\beta} \ln(|g|^{\frac{1}{2}})_{,\beta})_{,0} \sim g^{00} \ln(|g|^{\frac{1}{2}})_{,00}$$

$$\Gamma^{\alpha}{}_{0\alpha} = g^{\alpha\beta} \Gamma^{0}{}_{\alpha\beta}=\frac{1}{2}g^{\alpha\beta} g^{0\mu}(g_{\mu\alpha,\beta} + g_{\mu\beta,\alpha} - g_{\alpha\beta, \mu} ) = -g^{\alpha\beta} g^{0\mu}{}_{,\beta}g_{\mu\alpha} - \frac{1}{2} g^{\alpha\beta} g^{0\mu} g_{\alpha\beta,\mu} = -g^{0\mu}{}_{,\mu} -g^{0\mu}\ln(|g|^{\frac{1}{2}})_{,\mu} \sim -g^{00}_{,0} - g^{00}\ln(|g|^{\frac{1}{2}})_{,0}$$

and thus, $$R \sim 2g^{00}\ln(|g|^{\frac{1}{2}})_{,00} + g^{00}{}_{,00}$$

Now for Ricci tensor, $$\Gamma^{0}{}_{0\nu} = \frac{1}{2}g^{0\mu}(g_{\mu 0, \nu} + g_{\mu \nu, 0} - g_{0\nu, \mu} ) \sim - \frac{1}{2}g^{0\mu}{}_{,0}( g_{\mu 0} \delta_{\nu 0} + g_{\mu\nu} ) -\frac{1}{2} g^{00}g_{0\nu,0}$$ $$\Gamma^{\alpha}{}_{0\alpha,\nu} \sim \delta_{0\nu} \ln(|g|^{\frac{1}{2}})_{,00}$$

Hence, $$R_{0\nu} \sim - \frac{1}{2}g^{0\mu}{}_{,00}( g_{\mu 0} \delta_{\nu 0} + g_{\mu\nu} ) -\frac{1}{2} g^{00}g_{0\nu,00} - \delta_{0\nu} \ln(|g|^{\frac{1}{2}})_{,00}$$

Finally the Einstein tensor $$G_{0\nu} \sim - \frac{1}{2}g^{0\mu}{}_{,00}( g_{\mu 0} \delta_{\nu 0} + g_{\mu\nu} ) -\frac{1}{2} g^{00}g_{0\nu,00} - \delta_{0\nu} \ln(|g|^{\frac{1}{2}})_{,00} - g_{0\nu} g^{00}\ln(|g|^{\frac{1}{2}})_{,00} -\frac{1}{2}g_{0\nu} g^{00}{}_{,00}$$

Maybe there are some subtle mistakes in the calculation(please point them out), but I can't simply it any further and at along show it $\sim 0$.

My questions

1. Is the "$G_{0\nu}$" version of the statement right?

2. Is it possible to prove it using the trick that similar to that of Carroll's?

3. Any mistake I made in the explicit calculation? Or any magic identity could make it $\sim 0$