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 + r_1 = r_0(1 + \xi)</math>. [For later reference, note that <math>~\xi</math> can be a function of <math>~r_0</math> as well as of <math>~t</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 (1+ \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 ( r_0 \xi )}{\partial t} = r_0 \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>. Finally, in maintaining a Lagrangian perspective, we will need to ensure that the same element of mass is being tracked as we "ride along" with the fluid element to its new position. For radial perturbations associated with a spherically symmetric configuration, this means that the differential mass in each spherical shell, <math>~dm = 4\pi r^2 \rho dr</math>, must remain constant; that is, <math>~4\pi r_0^2 \rho_0 dr_0</math> <math>~=</math> <math>~4\pi r^2 \rho dr</math>   <math>~=</math> <math>~4\pi [r_0(1+\xi)]^2 \rho_0(1+s_L) dr</math>   <math>~=</math> <math>~4\pi r_0^2 \rho_0 \biggl(1+2\xi + \cancelto{\scriptstyle\text{small}}{\xi^2} + \cdots \biggr) (1+s_L) dr</math>   <math>~\approx</math> <math>~4\pi r_0^2 \rho_0 \biggl(1+2\xi + s_L + 2\cancelto{\scriptstyle\mathrm{small}}{\xi s_L} \biggr) dr</math> <math>~\Rightarrow~~~ \frac{d}{dr}</math> <math>~\approx</math> <math>~(1+2\xi + s_L ) \frac{d}{dr_0} \, .</math>
<math>~\frac{d}{dt}\biggl[\rho_0(1+s_L)\biggr]</math> <math>~=</math> <math>~- \biggl\{ \frac{\rho_0(1+\cancelto{}{s_L})}{[r_0(1+\cancelto{}{\xi})]^2} \biggr\}(1+2\cancelto{}{\xi s_L}) \frac{\partial}{\partial r_0} \biggl\{ [r_0(1+\cancelto{}{\xi})]^2 v_r \biggr\}</math> <math>~\Rightarrow~~~ \frac{d s_L}{dt}</math> <math>~=</math> <math>~- \biggl[\frac{2v_r}{r_0} + \frac{\partial v_r}{\partial r_0} \biggr] - \frac{1}{\rho_0} \frac{d\rho_0}{dt}</math>	<math>~\frac{\partial}{\partial t}\biggl[\rho_0(1+s_E)\biggr]</math> <math>~=</math> <math>~- \frac{\rho_0(1+\cancelto{\mathrm{small}}{s_E})}{r_0^2} \frac{\partial}{\partial r_0} \biggl( r_0^2 v_r \biggr) - v_r \frac{\partial [\rho_0(1+\cancelto{\mathrm{small}}{s_E})] }{\partial r_0}</math> <math>~\Rightarrow~~~ \frac{\partial s_E}{\partial t}</math> <math>~=</math> <math>~- \biggl[\frac{2v_r}{r_0} + \frac{\partial v_r}{\partial r_0} \biggr] - \frac{v_r}{\rho_0} \frac{\partial \rho_0 }{\partial r_0}</math>
Note: The last term that appears on the righthand side of the two expressions appears to be different. But if, as we are assuming here, <math>~\rho_0</math> has no explicit time dependence but may be considered to be a function of the radial coordinate, <math>~r_0</math>, then the two terms are the same. This is because, quite generically for any scalar function <math>~q</math>, the total time-derivative (Lagrangian perspective) differs from the partial time-derivative (Eulerian perspective) via the expression, <math>dq/dt - \partial q /\partial t = \vec{v}\cdot \nabla q</math>. In our case, <math>~\partial \ln \rho_0/\partial t = 0</math>, so <math>~d\ln\rho_0/dt = \vec{v}\cdot \nabla \ln \rho_0</math>.
<math>~s_L ~~\rightarrow~~ \Delta_L(r_0) e^{i\omega t}</math>             … and …             <math>~s_E ~~\rightarrow~~ \Delta_E(r_0) e^{i\omega t}</math> <math>~\xi ~~\rightarrow~~ x(r_0) e^{i\omega t}</math>             <math>\Rightarrow</math>             <math>~v_r ~~\rightarrow~~ (i\omega)r_0 x(r_0) e^{i\omega t}</math>
<math>~e^{i\omega t} \biggl[ (i\omega)\Delta_L + \frac{d\Delta_L}{dt} \biggr]</math> <math>~=</math> <math>~- e^{i\omega t} \biggl[ 2(i\omega)x + (i\omega)x + (i\omega)r_0 \frac{\partial x}{\partial r_0} \biggr]</math> <math>~\Rightarrow ~~~r_0 \frac{\partial x}{\partial r_0} </math> <math>~=</math> <math>~- \Delta_L -3 x - \frac{1}{(i\omega)} \biggl[ v_r \frac{\partial\Delta_L}{\partial r_0} \biggr]</math>   <math>~=</math> <math>~- \Delta_L - x \biggl[3 + \cancelto{\scriptstyle\mathrm{small}}{\frac{\partial\Delta_L}{\partial \ln r_0} }\biggr]</math>	<math>~e^{i\omega t} (i\omega)\Delta_E </math> <math>~=</math> <math>~- e^{i\omega t} \biggl[ 2(i\omega)x + (i\omega)x + (i\omega)r_0 \frac{\partial x}{\partial r_0} \biggr] - \biggl[\frac{1}{\rho_0} \frac{\partial \rho_0}{\partial r_0} \biggr](i\omega)r_0 x e^{i\omega t}</math> <math>~\Rightarrow ~~~r_0 \frac{\partial x}{\partial r_0} </math> <math>~=</math> <math>~- \Delta_E - x \biggl[3 + \frac{\partial \ln\rho_0}{\partial \ln r_0} \biggr] </math>

See Also

• Eulerian Perspective
• Lagrangian Perspective