Symbol delta_ij

The passage around this formula

What must therefore vanish, identically, is not the slope but the correctly defined curvature-frame component built from it. For any static metric of the form d$s^2$ = -U($\mathbf x$)^2\,d$x_0^2$ + $\delta_{ij}$\,d$x^i$ d$x^j$ — flat spatial slices, all the position dependence carried by U — the only nonzero Christoffel symbols are $\Gamma^0_{0i}$ = $\partial_i$ $\ln$ U = $\phi_{,i}$ and $\Gamma^i_{00}$ = $U^2$$\phi_{,i}$ , both following directly from the metric’s definition. Substituting into the Riemann tensor’s defining combination for the mixed component $R^i${}_{0j0} , every term built from purely spatial Christoffels vanishes because the spatial slices are exactly flat, and the two surviving pieces are…

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