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## Homework Statement

Show that [itex]R_{00}= -\frac{1}{2}g^{\alpha\beta}g_{\alpha\beta,00}+ M_{00}[/itex], where M can be expressed solely in terms of the initial data, which is the metric tensor and its first derivatives.

## Homework Equations

[itex]R_{ab}= R^c_{acb}= \Gamma^a_{bd,c}- \Gamma^a_{bc,d}[/itex] in geodesic coordinates

[itex]\Gamma^a_{bc}= \frac{1}{2}g^{ad}(g_{dc,b}+ g_{db,c}- g_{bc,d})[/itex]

## The Attempt at a Solution

[itex]R_{00}= R^a_{0a0}= \Gamma^a_{00,a}- \Gamma^a_{0a,0}[/itex]

[itex]R_{00}= (g_{b0,0}- \frac{1}{2}g_{00,b})g^{ab}_{,a}+ g^{ab}g_{b0,a0}- \frac{1}{2}g^{ab}g_{00,ab}- \frac{1}{2}(g_{ba,0}+ g_{b0,a}- g_0a,b)g^{ab}_{,0}- \frac{1}{2}g^{ab}g_{ba,00}- \frac{1}{2}g^{ab}g_{b0,0a}+ \frac{1}{2}g^{ab}g_{0a,0b}[/itex]

a and b are summations from 0 to 3, while [itex]\alpha[/itex] and [itex]\beta[/itex] are summations for space coordinates, so they are 1 to 3. Since the one second-derivative term in the expression I am trying to reach is summed over the space coordinates, I need to split one of the terms above. I split [itex]-\frac{1}{2}g^{ab}g_{ba,00}= -\frac{1}{2}(g^{\alpha\beta}g_{\beta\alpha,00}+ g^{a0}g_{0a,00}+ g^{0\beta}g_{\beta0,00})[/itex].

So I define [itex]M_{00}= (g_{b0,0}- \frac{1}{2}g_{00,b})g^{ab}_{,a}- \frac{1}{2}(g_{ba,0}+ g_{b0,a}- g_0a,b)g^{ab}_{,0}[/itex], as all the rest of the terms have second derivatives of the metric tensor, which are not in the initial data.

[itex]R_{00}= -\frac{1}{2}(g^{\alpha\beta}g_{\beta\alpha,00}+ g^{a0}g_{0a,00}+ g^{0\beta}g_{\beta0,00}- g^{ab}g_{b0,a0}+ g^{ab}g_{00,ab}- g^{ab}g_{0a,0b})+ M_{00}[/itex]

So now I have to show that [itex]-\frac{1}{2}(g^{a0}g_{0a,00}+ g^{0\beta}g_{\beta0,00}- g^{ab}g_{b0,a0}+ g^{ab}g_{00,ab}- g^{ab}g_{0a,0b})=0[/itex]

I have simplified those five terms down to three: [itex]\frac{1}{2}(g^{a\beta}g_{0a,0\beta}- g^{a\beta}g_{00,a\beta}+ g^{\alpha\beta}g_{\beta0,\alpha0})[/itex]

I do not know how to simplify that any further. Can someone help?

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