User:Tohline/SSC/Perspective Reconciliation

Reconciling Eulerian versus Lagrangian Perspectives

Linearizing the Key Relations

Continuity Equation
Lagrangian Perspective	Eulerian Perspective
<math>~\frac{d\rho}{dt}</math> <math>~=</math> <math>~- \rho \nabla\cdot\vec{v}</math>	<math>~\frac{\partial\rho}{\partial t}</math> <math>~=</math> <math>~- \rho \nabla\cdot\vec{v} - \vec{v}\cdot \nabla\rho</math>
Spherically Symmetric Initial Configurations & Purely Radial Perturbations
<math>~\frac{d\rho}{dt}</math> <math>~=</math> <math>~- \frac{\rho}{r^2} \frac{\partial}{\partial r} \biggl( r^2 v_r \biggr)</math>	<math>~\frac{\partial\rho}{\partial t}</math> <math>~=</math> <math>~- \frac{\rho}{r^2} \frac{\partial}{\partial r} \biggl( r^2 v_r \biggr) - v_r \frac{\partial \rho}{\partial r}</math>
In an interval of time, <math>~dt = \partial t</math>, a fluid element initially at position <math>~r_0</math> moves to position, <math>~r = r_0 + \xi</math>. On the righthand side of the expression, the radial coordinate will be handled as follows: From the Lagrangian perspective, <math>~r \rightarrow (r_0 + \xi)</math>, while from the Eulerian perspective, we want to stay at the original coordinate location, so <math>~r \rightarrow r_0</math>. From both perspectives, <math>~v_r = \frac{\partial\xi}{\partial t} \, .</math> Riding with the fluid element (Lagrangian perspective), <math>~\rho \rightarrow (\rho_0 + \rho_L) = \rho_0(1+s_L)</math>, while at a fixed coordinate location (Eulerian perspective), <math>~\rho \rightarrow (\rho_0 + \rho_E) + \rho_0(1 + s_E)</math>.
<math>~\frac{d}{dt}\biggl[\rho_0(1+s_L)\biggr]</math> <math>~=</math> <math>~- \frac{\rho_0(1+s_L)}{r^2} \frac{\partial}{\partial r} \biggl( r^2 v_r \biggr)</math>	<math>~\frac{\partial}{\partial t}\biggl[\rho_0(1+s_E)\biggr]</math> <math>~=</math> <math>~- \frac{\rho_0(1+s_E)}{r^2} \frac{\partial}{\partial r} \biggl( r^2 v_r \biggr) - v_r \frac{\partial \rho_0(1+s_E)}{\partial r}</math>

See Also

• Eulerian Perspective
• Lagrangian Perspective