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Mixed Einstein tensor

  1. Dec 22, 2011 #1
    1. The problem statement, all variables and given/known data
    Show that [itex]G_\alpha^0= g^{00}M_{0\alpha}+ g^{0\beta}M_{\alpha\beta}[/itex] 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 [itex]R_{\alpha\beta}=0[/itex], show that [itex]G_\alpha^0= g^{00}R_{0\alpha}[/itex] and [itex]G_0^\alpha= g^{0\alpha}R_{00}+ g^{\alpha\beta}R_{0\beta}[/itex]

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

    3. The attempt at a solution
    The problem also includes calculations for [itex]G_0^0[/itex] and [itex]G_\alpha^\beta[/itex], but I got those correct. For [itex]G_\alpha^0[/itex], this is what I'm getting:
    [itex]g^{0a}G_{\alpha a}= g^{0a}R_{\alpha a}- \frac{1}{2}g^{0a}g_{\alpha a}R[/itex], 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 [itex]G_0^0[/itex] and [itex]G_\alpha^\beta[/itex].

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

    [itex]=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})[/itex]
    [itex]=\frac{1}{2}(g^00R_{\alpha0}- g^{0\beta}R_{\alpha\beta})[/itex]
    [itex]=\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}[/itex]

    When [itex]R_{\alpha\beta}=0[/itex], [itex]G_\alpha^0= \frac{1}{2}g^{00}R_{\alpha0}[/itex]

    When I perform a similar calculation for [itex]G_0^\alpha[/itex], I come up with:
    [itex]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})[/itex]
    When [itex]R_{\alpha\beta}=0[/itex], [itex]G_0^\alpha= \frac{1}{2}g^{\alpha0}R_{00}[/itex]

    Could someone check my math?
     
  2. jcsd
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