Active/Passive Diffeomorphisms – clarification on Rovelli’s

[albedo]

...
$(\mathcal L_XY)_p =\lim_{t\to 0}\frac{((\phi_{-t})_*Y)_p-Y_p}{t}$...

I stand with you with the demonstration but I think you should use the definition of the Lie derivative by including the flow map instead of p in the first term of the numerator (and similarly in the following lines):

$(\mathcal L_XY)_p =\lim_{t\to 0}\frac{((\phi_{-t})_*Y)_ {\phi_t(p)}-Y_p}{t}\\$

and make use of the definition of the flow map:

$\phi_{t}^{-1}=\phi_{-t}$ (together with $\phi_{t}\circ\phi_{s}=\phi_{t+s}$ and $\phi_{0}=id$)

 

[Fredrik]

I think you probably didn't understand the notation $((\phi_{-t})_*Y)_p$. (Not surprising since I didn't explain it, and I'm not sure how standard it is). For each smooth map $\phi:M\to M$ and each vector field $Y$, the vector field $\phi_*Y$ is defined by $(\phi_*Y)_p=\phi_* Y_{\phi^{-1}(p)}$ for all $p\in M$. This definition ensures that
$$((\phi_{-t})_*Y)_p = (\phi_{-t})_* Y_{(\phi_{-t})^{-1}(p)} =(\phi_{-t})_* Y_{\phi_t(p)}.$$ $Y_{\phi_t(p)}$ is pushed forward to the tangent space at
$$\phi_{-t}(\phi_t(p))=(\phi_{-t}\circ\phi_t)(p)=\phi_{-t+t}(p)=\phi_0(p)=p.$$

 

[albedo]

Oh yes, thank you. The two notations in the definition tell exactly the same

 

[albedo]

I think Haelfix's last post goes well along with the text from Thomas Thiemann, HERE, which probably grounds the argument more technically. As long as you change only $M$ under a coordinate transformation (passive diffeomorphism), the spatial volume $V_R(q)$ does change. However, if you consider the volume element over a region where a “physical” element $\rho$ is defined and built from the metric (and/or matter fields as he writes), e.g. the scalar $\rho=R$, then the transformation recovers the invariance for the volume element because both $q$ and $\rho$ transforms in the same way (pullbacks). I did not work out the demonstration but intuitively $\rho$ and $q$ when transformed together conjure so the integral does not change.