Einstein Tensor - Explicit Form

Explicit Form

The Ricci tensor depends only on the metric tensor, so the Einstein tensor can be defined directly with just the metric tensor. However, this expression is complex and rarely quoted in textbooks. The complexity of this expression can be shown using the formula for the Ricci tensor in terms of Christoffel symbols:

\begin{align} G_{\alpha\beta} &= R_{\alpha\beta} - \frac{1}{2} g_{\alpha\beta} R \\ &= R_{\alpha\beta} - \frac{1}{2} g_{\alpha\beta} g^{\gamma\zeta} R_{\gamma\zeta} \\ &= (\delta^\gamma_\alpha \delta^\zeta_\beta - \frac{1}{2} g_{\alpha\beta}g^{\gamma\zeta}) R_{\gamma\zeta} \\ &= (\delta^\gamma_\alpha \delta^\zeta_\beta - \frac{1}{2} g_{\alpha\beta}g^{\gamma\zeta})(\Gamma^\epsilon_{\gamma\zeta,\epsilon} - \Gamma^\epsilon_{\gamma\epsilon,\zeta} + \Gamma^\epsilon_{\epsilon\sigma} \Gamma^\sigma_{\gamma\zeta} - \Gamma^\epsilon_{\zeta\sigma} \Gamma^\sigma_{\epsilon\gamma}), \end{align}

where is the Kronecker tensor and the Christoffel symbol is defined as

Before cancellations, this formula results in individual terms. Cancellations bring this number down somewhat.

In the special case of a locally inertial reference frame near a point, the first derivatives of the metric tensor vanish and the component form of the Einstein tensor is considerably simplified:

where square brackets conventionally denote antisymmetrization over bracketed indices, i.e.

Read more about this topic:  Einstein Tensor

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