Active/Passive Diffeomorphisms – clarification on Rovelli’s

  • Thread starter albedo
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  • #26
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##(\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##)
 
  • #27
Fredrik
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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
$$\phi_{-t}(\phi_t(p))=(\phi_{-t}\circ\phi_t)(p)=\phi_{-t+t}(p)=\phi_0(p)=p.$$
 
  • #28
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Oh yes, thank you. The two notations in the definition tell exactly the same :smile:
 
  • #29
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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.
 

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