Difference between revisions of "User:Tohline/SSC/Perspective Reconciliation"

From VistrailsWiki
Jump to navigation Jump to search
(→‎Reconciling Eulerian versus Lagrangian Perspectives: Rewrite both formulations in terms of similar variables)
Line 86: Line 86:
</math><br />
</math><br />
</div>
</div>
Note that we are convinced that this expression is error-free because, for example, when the structural properties of an equilibrium, <math>~n=1</math> polytrope are plugged into it, we obtain exactly the same 2<sup>nd</sup>-order ODE as published by Murphy.


==Eulerian Reformulation==
==Eulerian Reformulation==

Revision as of 17:50, 22 June 2015

Reconciling Eulerian versus Lagrangian Perspectives

Whitworth's (1981) Isothermal Free-Energy Surface
|   Tiled Menu   |   Tables of Content   |  Banner Video   |  Tohline Home Page   |


Comment by J. E. Tohline on 22 June 2015: Last night I realized that a key to understanding how to reconcile the Eulerian and Lagrangian perspectives was an analysis of the eigenvalue problem for the homogeneous sphere.

In an accompanying discussion, we have reviewed T. E. Sterne's (1937, MNRAS, 97, 582) study of radial pulsation modes in the homogeneous sphere. He solved the eigenvalue problem as defined by the

Adiabatic Wave (or Radial Pulsation) Equation

LSU Key.png

<math>~ \frac{d^2x}{dr_0^2} + \biggl[\frac{4}{r_0} - \biggl(\frac{g_0 \rho_0}{P_0}\biggr) \biggr] \frac{dx}{dr_0} + \biggl(\frac{\rho_0}{\gamma_\mathrm{g} P_0} \biggr)\biggl[\omega^2 + (4 - 3\gamma_\mathrm{g})\frac{g_0}{r_0} \biggr] x = 0 </math>

and, hence, as established via a Lagrangian formulation of the problem. After finishing that review, we became aware that a separate study of radial pulsation modes in the homogeneous sphere has been published by S. Rosseland (1969) In his book titled, The Pulsation Theory of Variable Stars (see, specifically his § 3.2, beginning on p. 27). Rosseland solved an eigenvalue problem as defined by the relation (see his equation 2.23 on p. 20, with the adiabatic condition being enforced by setting the right-hand-side equal to zero),

<math>~\frac{\partial}{\partial r} \biggl( \gamma P_0 \nabla\cdot \vec{\xi}\biggr) + \biggl( \omega^2 + \frac{4g_0}{r}\biggr) \rho_0 \xi</math>

<math>~=</math>

<math>~0 \, ,</math>

where,

<math>~\vec\xi = \mathbf{\hat{e}}_r \xi(r) \, ,</math>

which he derived in an earlier section of his book via an Eulerian formulation of the problem. Realizing that, for a spherically symmetric system,

<math>\nabla\cdot \vec\xi = \frac{1}{r^2}\frac{\partial}{\partial r}\biggl(r^2 \xi\biggr) = \frac{\partial \xi}{\partial r} + \frac{2\xi}{r} \, ,</math>

as is demonstrated in and accompanying discussion, we can rewrite this relation in the more familiar form of a 2nd-order ODE, namely,

<math>~P_0 \frac{\partial^2 \xi}{\partial r^2} + \biggl[ \frac{2P_0}{r}- \rho_0 g_0 \biggr] \frac{\partial \xi}{\partial r} + \biggl[ \biggl( \frac{\omega^2\rho_c}{\gamma} + \frac{4\rho_c g_0}{\gamma r}\biggr) \biggl(\frac{\rho_0}{\rho_c}\biggr) - \biggl(\frac{2\rho_c g_0 }{r}\biggr)\biggl(\frac{\rho_0}{\rho_c}\biggr) - \frac{2P_0}{r^2} \biggr] \xi </math>

<math>~=</math>

<math>~0 \, .</math>

Lagrangian Reformulation

Defining the characteristic time for dynamical oscillations in spherically symmetric configurations (SSC) as,

<math> \tau_\mathrm{SSC} \equiv \biggl[ \frac{R^2 \rho_c}{P_c} \biggr]^{1/2} , </math>

and the characteristic gravitational acceleration as,

<math> g_\mathrm{SSC} \equiv \frac{P_c}{R \rho_c} \, , </math>

we can rewrite the Lagrangian-formulated wave equation as,

<math> \biggl(\frac{P_0}{P_c}\biggr)\frac{d^2x}{d\chi_0^2} + \biggl[\frac{4}{\chi_0}\biggl(\frac{P_0}{P_c}\biggr) - \biggl(\frac{\rho_0}{\rho_c}\biggr) \biggl(\frac{g_0}{g_\mathrm{SSC}}\biggr) \biggr] \frac{dx}{d\chi_0} + \biggl(\frac{\rho_0}{\rho_c}\biggr) \biggl(\frac{1}{\gamma_\mathrm{g}} \biggr)\biggl[\tau_\mathrm{SSC}^2 \omega^2 + (4 - 3\gamma_\mathrm{g})\biggl(\frac{g_0}{g_\mathrm{SSC}}\biggr) \frac{1}{\chi_0} \biggr] x = 0 . </math>

Note that we are convinced that this expression is error-free because, for example, when the structural properties of an equilibrium, <math>~n=1</math> polytrope are plugged into it, we obtain exactly the same 2nd-order ODE as published by Murphy.

Eulerian Reformulation

Using the same characteristic time scale and gravitational acceleration, we can similarly rewrite the Eulerian-formulated expression as,

<math>~\biggl(\frac{P_0}{P_c}\biggr) \frac{\partial^2 \xi}{\partial \chi_0^2} + \biggl[ \frac{2}{\chi_0}\biggl(\frac{P_0}{P_c}\biggr) - \frac{g_0 }{g_\mathrm{SSC}}\biggl(\frac{\rho_0}{\rho_c}\biggr) \biggr] \frac{\partial \xi}{\partial \chi_0} + \biggl\{ \biggl[\frac{\omega^2\tau_\mathrm{SSC}^2}{\gamma} + \frac{2}{\chi_0 } \biggl(\frac{2}{\gamma } - 1\biggr)\frac{g_0}{g_\mathrm{SSC}}\biggr] \biggl(\frac{\rho_0}{\rho_c}\biggr) - \frac{2}{\chi_0^2} \biggl(\frac{P_0}{P_c}\biggr) \biggr\} \xi </math>

<math>~=</math>

<math>~0 \, .</math>


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