2 Adde comment on physical interpretation

Recall that the definition of the Lie derivative of a tensor field $T$ with respect to a vector field $X$ is given by "dragging" $T$ with respect to the one-parameter (quasi) group $\phi_t$ generated by $X$, i.e., computing $\phi_t^*(T)$, and differentiating wrt $t$ at $t = 0$. But to say that $X$ is a Killing field means that the $\phi_t$ are (partial) isometries, and so not only preserve the metric tensor but also the Riemann curvature tensor and its contraction the Ricci tensor or any other tensor field that is defined canonically from the metric tensor and so preserved by isometries. Thus any such tensor field is preserved by dragging, i.e., $\phi_t^*(T)$ is constant in $t$ and so has a zero derivative.

Regarding the physical interpretation, let me try to answer a slightly different question. Recall that the Ricci tensor comes up as the Euler-Lagrange expression for the Einstein-Hilbert functional, and that the latter is invariant under the group of ALL diffeomorphisms. So it is natural to ask what the Noether Theorem (connecting one-parameter groups that preserve a Lagrangian to constants of the motion of the corresponding Euler-Lagrange equations) leads to in this case. The answer is that it gives the contracted Bianchi identity for the Ricci tensor. Perhaps this is what your question about physical significance was aiming at.

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Recall that the definition of the Lie derivative of a tensor field $T$ with respect to a vector field $X$ is given by "dragging" $T$ with respect to the one-parameter (quasi) group $\phi_t$ generated by $X$, i.e., computing $\phi_t^*(T)$, and differentiating wrt $t$ at $t = 0$. But to say that $X$ is a Killing field means that the $\phi_t$ are (partial) isometries, and so not only preserve the metric tensor but also the Riemann curvature tensor and its contraction the Ricci tensor or any other tensor field that is defined canonically from the metric tensor and so preserved by isometries. Thus any such tensor field is preserved by dragging, i.e., $\phi_t^*(T)$ is constant in $t$ and so has a zero derivative.