How to see the diffeomorphism invariance of a particular metric

Question

I understand how to show in general, that under the diffeomorphism $x^\mu\to x^\mu+\epsilon^\mu (x)$, the metric tensor changes as $$g'_{\mu\nu}(x')=g_{\mu\nu}(x)-\partial_\mu\epsilon_\nu(x)-\partial_\nu\epsilon_\mu(x),\tag{1}$$ and the differential $dx^\mu$ changes as $$dx'^\mu=\frac{\partial x^\mu}{\partial x^{\gamma}}dx^\gamma=dx^\mu+\partial_\gamma\epsilon^\mu dx^\gamma,$$ such that the line element $ds^2$ is invariant.

Now I'm trying to see this in a particular example, say, the Schwarzschild metric: $$ds^2=-\left(1-\frac{r_s}{r}\right)dt^2+\frac{dr^2}{\left(1-\frac{r_s}{r}\right)}+r^2\left(d\theta^2+\text{sin}^2\theta d\phi^2\right),$$ if I take a diffeomorphism in $r$, $$r\to r+\xi(r),$$ the $g_{tt}$ component of the metric for example then becomes $$g_{tt}=1-\frac{r_s}{r}+\frac{r_s}{r^2}\xi(r)+O(\xi^2),$$ which obviously doesn't transform as in $(1)$, and if $ds^2$ is to be invariant we would need another contribution from somewhere to cancel the term proportional to $\xi$. So what am I missing here?

Also, do diffeomorphisms in the angular coordinates also look the same, i.e. $$\theta\to\theta+\xi(\theta)?$$

I know I am probably missing something very basic, so apologies for my confusion.

Answer 1

The change in the metric is not what you say. The correct change is $$ \delta g_{\mu\nu} = ({\mathcal L}_\epsilon g )_{\mu\nu} \equiv \epsilon^\lambda\partial_\lambda g_{\mu\nu}+ g_{\lambda\nu} \partial_\mu \epsilon^\lambda + g_{\mu\lambda} \partial_\nu \epsilon^\lambda\\ = \nabla_\mu \epsilon_\nu+ \nabla_\nu \epsilon_\mu $$ i.e you need covariant derivatives rather than partials. A metric is unchanged by by an infinitesimal diffeomorphism only if ${\mathcal L}_\epsilon g =0$, in which case the vector field $\epsilon$ is a Killing vector.