Understanding the displacement vector

[pjuelsgaard]

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

[rhtownsend]

Good question, Peter! Because Eulerian perturbations are fixed position, by definition r' = 0 and so delta r = xi_r