Isn't the Lie derivative of the Minkowski metric zero?

Question

In the presentation of gauge invariance of weak gravitational fields, the Lie derivative of the Minkowski metric is involved. In many textbooks, it is written out in terms of a vector $\xi^\mu$ in the Minkowski background spacetime as ${\cal L}_\xi\eta_{\mu\nu}$. But since the metric has constant components everywhere in Minkowski spacetime, the difference of the pullback of the infinitesimal variation of the metric with the metric at the point of evaluation, should vanish. If that is the case, the texts should state this result is zero and that there is an isometry from the resulting Killing equation. Isn't the Lie derivative of the Minkowski metric zero everywhere?

Answer 1

If the components $\xi^\mu$ are constant (in the standard coordinates) then this is true, but not if $\xi$ is an arbitrary vector field - it could be squishing or stretching things. For example, the flow of the radial vector field $\xi^\mu(X) = X^\mu$ uniformly dilates lengths, so in this case $\mathcal L_\xi \eta = \eta$ and $\xi$ is not Killing.

Remember that the pullback $F_{t\xi}^* \eta$ isn't just translation of the base point, but involves the differential of the flow map $F_{t \xi}$; so the coordinate expression for $\mathcal L_\xi \eta$ involves derivatives of not only $\eta$, but also $\xi$.

Answer 2

Lie derivative of the Minkowski metric with respect to which vector field?

In coordinates, $\mathcal{L}_{\xi} \eta_{ab} = \xi_{a;b} + \xi_{b;a} = \xi_{a,b} + \xi_{b,a}$, if this equals zero, then $\xi$ is a Killing vector (the resulting equation is Killing's equation). If however, you have something like $\mathcal{L}_{\xi} \eta_{ab} = 2 c \eta_{ab}$, then, $\xi$ is a Homothetic vector field, for $c$ constant.