Understanding the displacement vector
Posted: Sat May 28, 2022 2:18 pm
The physics behind the displacement vector, specifically the radial part (\xi_r), confuse me a bit,
Using the relation between lagrangian and eulerian perturbations for scalar quantities, it quickly becomes apparent that
\delta r = r' + \xi_r
However, I have some nagging feeling in the back of my mind, that r' might be zero (although I haven't been able to find explicit arguments for it anywhere).
Am I completely off the mark? Or am I just overcomplicating things?
Kind regards
- Peter
Using the relation between lagrangian and eulerian perturbations for scalar quantities, it quickly becomes apparent that
\delta r = r' + \xi_r
However, I have some nagging feeling in the back of my mind, that r' might be zero (although I haven't been able to find explicit arguments for it anywhere).
Am I completely off the mark? Or am I just overcomplicating things?
Kind regards
- Peter