# Mixed Einstein tensor

1. Dec 22, 2011

### PhyPsy

1. The problem statement, all variables and given/known data
Show that $G_\alpha^0= g^{00}M_{0\alpha}+ g^{0\beta}M_{\alpha\beta}$ where Greek letters indicate summation over 3 space coordinates (M consists of terms involving the metric and its first derivatives; see equations below)
Also, when Ricci tensor $R_{\alpha\beta}=0$, show that $G_\alpha^0= g^{00}R_{0\alpha}$ and $G_0^\alpha= g^{0\alpha}R_{00}+ g^{\alpha\beta}R_{0\beta}$

2. Relevant equations
$R_{00}= -\frac{1}{2}g^{\alpha\beta}g_{\alpha\beta,00}+ M_{00}$
$R_{0\alpha}= \frac{1}{2}g^{0\beta}g_{\alpha\beta,00}+ M_{0\alpha}$
$R_{\alpha\beta}= -\frac{1}{2}g^{00}g_{\alpha\beta,00}+ M_{\alpha\beta}$
$G_{ab}= R_{ab}- \frac{1}{2}g_{ab}R$

3. The attempt at a solution
The problem also includes calculations for $G_0^0$ and $G_\alpha^\beta$, but I got those correct. For $G_\alpha^0$, this is what I'm getting:
$g^{0a}G_{\alpha a}= g^{0a}R_{\alpha a}- \frac{1}{2}g^{0a}g_{\alpha a}R$, where English letters indicate summation over time and space coordinates (4 dimensions); I'm not entirely sure when changing the tensor to covariant form that the new index should be a summation over all four dimensions, but it gave me the right answer for $G_0^0$ and $G_\alpha^\beta$.

$G_\alpha^0= g^{0\beta}R_{\alpha\beta}+ g^{00}R_{\alpha0}- \frac{1}{2}\delta^0_\alpha g^{ab}R_{ab}$
$=g^{0\beta}R_{\alpha\beta}+ g^{00}R_{\alpha0}- \frac{1}{2}\delta^0_\alpha(g^{00}R_{00}+ 2g^{0\beta}R_{0\beta}+ g^{\alpha\beta}R_{\alpha\beta})$; breaking out the Ricci scalar into terms involving the metric tensor and the Ricci tensor is another spot where I'm not totally confident.

$=g^{0\beta}R_{\alpha\beta}+ g^{00}R_{\alpha0}- \frac{1}{2}(g^{00}R_{\alpha0}+ 2g^{0\beta}R_{\alpha\beta}+ g^{0\beta}R_{\alpha\beta})$
$=\frac{1}{2}(g^00R_{\alpha0}- g^{0\beta}R_{\alpha\beta})$
$=\frac{1}{4}g^{00}g^{0\beta}g_{\alpha\beta,00}+ \frac{1}{2}g^{00}M_{0\alpha}+ \frac{1}{4}g^{0\beta}g^{00}g_{\alpha\beta,00}- \frac{1}{2}g^{0\beta}M_{\alpha\beta}= \frac{1}{2}g^{00}g^{0\beta}g_{\alpha\beta,00}+ \frac{1}{2}g^{00}M_{0\alpha}- \frac{1}{2}g^{0\beta}M_{\alpha\beta}$

When $R_{\alpha\beta}=0$, $G_\alpha^0= \frac{1}{2}g^{00}R_{\alpha0}$

When I perform a similar calculation for $G_0^\alpha$, I come up with:
$g^{\alpha a}R_{0a}= g^{\alpha0}R_{00}+ g^{\alpha\beta}R_{0\beta}- \frac{1}{2}\delta^\alpha_0(g^{00}R_{00}+ 2g^{\beta0}R_{\beta0}+ g^{\alpha\beta}R_{\alpha\beta})$
When $R_{\alpha\beta}=0$, $G_0^\alpha= \frac{1}{2}g^{\alpha0}R_{00}$

Could someone check my math?