Difference between revisions of "User:Tohline/PGE/Euler"

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~~~\Rightarrow~~~ \frac{j^2}{\varpi^3} = \frac{(\varpi v_\varphi)^2}{\varpi^3} =  
~~~\Rightarrow~~~ \frac{j^2}{\varpi^3} = \frac{(\varpi v_\varphi)^2}{\varpi^3} =  
\frac{\omega_c^2}{\varpi} \biggl[ \frac{A^2\varpi}{A^2 + \varpi^2}\biggr]^2 =
\frac{\omega_c^2}{\varpi} \biggl[ \frac{A^2\varpi}{A^2 + \varpi^2}\biggr]^2 =
\omega_c^2 \biggl[ \frac{A^4\varpi}{(A^2 + \varpi^2)^2}\biggr]
\biggl[ \frac{\omega_c^2 A^4\varpi}{(A^2 + \varpi^2)^2}\biggr]
\, ,
\, ,
</math>
</math>
Line 608: Line 608:
<math>~
<math>~
\hat{e}_z~  \biggl[2\omega_c A^4 (A^2 + \varpi^2)^{-2} \biggr] \, .
\hat{e}_z~  \biggl[2\omega_c A^4 (A^2 + \varpi^2)^{-2} \biggr] \, .
</math>
  </td>
</tr>
</table>
[J] &nbsp; Hence,
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~(\vec{v} \cdot \nabla) \vec{v}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\hat{e}_\varpi \biggl[ - \frac{v_\varphi \cdot v_\varphi}{\varpi} \biggr] </math>
  </td>
</tr>
<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-~\hat{e}_\varpi \frac{\omega_c^2}{\varpi} \biggl[ \frac{A^2\varpi}{A^2 + \varpi^2}\biggr]^2
=
-~\hat{e}_\varpi \biggl[ \frac{\omega_c^2A^4 \varpi}{(A^2 + \varpi^2)^2} \biggr]
</math>
</math>
   </td>
   </td>

Revision as of 16:59, 7 August 2019

Euler Equation

Whitworth's (1981) Isothermal Free-Energy Surface
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Lagrangian Representation

in terms of velocity:

Among the principal governing equations we have included the

Lagrangian Representation
of the Euler Equation,

LSU Key.png

<math>\frac{d\vec{v}}{dt} = - \frac{1}{\rho} \nabla P - \nabla \Phi</math>

[BLRY07], p. 13, Eq. (1.55)

in terms of momentum density:

Multiplying this equation through by the mass density <math>~\rho</math> produces the relation,

<math>\rho\frac{d\vec{v}}{dt} = - \nabla P - \rho\nabla \Phi</math> ,

which may be rewritten as,

<math>\frac{d(\rho\vec{v})}{dt}- \vec{v}\frac{d\rho}{dt} = - \nabla P - \rho\nabla \Phi</math> .

Combining this with the Standard Lagrangian Representation of the Continuity Equation, we derive,

<math>\frac{d(\rho\vec{v})}{dt}+ (\rho\vec{v})\nabla\cdot\vec{v} = - \nabla P - \rho\nabla \Phi</math> .


Eulerian Representation

in terms of velocity:

By replacing the so-called Lagrangian (or "material") time derivative <math>d\vec{v}/dt</math> in the Lagrangian representation of the Euler equation by its Eulerian counterpart (see, for example, the wikipedia discussion titled, "Material_derivative", to understand how the Lagrangian and Eulerian descriptions of fluid motion differ from one another conceptually as well as how to mathematically transform from one description to the other), we directly obtain the

Eulerian Representation
of the Euler Equation,

<math>~\frac{\partial\vec{v}}{\partial t} + (\vec{v}\cdot \nabla) \vec{v}= - \frac{1}{\rho} \nabla P - \nabla \Phi</math>

in terms of momentum density:

As was done above in the context of the Lagrangian representation of the Euler equation, we can multiply this expression through by <math>~\rho</math> and combine it with the continuity equation to derive what is commonly referred to as the,

Conservative Form
of the Euler Equation,

<math>~\frac{\partial(\rho\vec{v})}{\partial t} + \nabla\cdot [(\rho\vec{v})\vec{v}]= - \nabla P - \rho \nabla \Phi</math>

[BLRY07], p. 8, Eq. (1.31)

The second term on the left-hand-side of this last expression represents the divergence of the "dyadic product" or "outer product" of the vector momentum density and the velocity vector, and is sometimes written as, <math>~\nabla\cdot [(\rho \vec{v}) \otimes \vec{v}]</math>.

in terms of the vorticity:

Drawing on one of the standard dot product rule vector identities, the nonlinear term on the left-hand-side of the Eulerian representation of the Euler equation can be rewritten as,

<math> (\vec{v}\cdot\nabla)\vec{v} = \frac{1}{2}\nabla(\vec{v}\cdot\vec{v}) - \vec{v}\times(\nabla\times\vec{v}) = \frac{1}{2}\nabla(v^2) + \vec{\zeta}\times \vec{v} , </math>

where,

<math> \vec\zeta \equiv \nabla\times\vec{v} </math>

is commonly referred to as the vorticity. Making this substitution leads to an expression for the,

Euler Equation
in terms of the Vorticity,

<math>~\frac{\partial\vec{v}}{\partial t} + \vec\zeta \times \vec{v}= - \frac{1}{\rho} \nabla P - \nabla \biggl[\Phi + \frac{1}{2}v^2 \biggr] </math>

Double Check Vector Identities

Let's plug a few different simple rotation profiles into the Euler equation, using a cylindrical-coordinate base to demonstrate that the three expressions are identical, namely, that

<math>~(\vec{v} \cdot \nabla) \vec{v}</math>

<math>~=</math>

<math>~\vec\zeta \times \vec{v} + \frac{1}{2}\nabla (v^2)</math>

<math>~=</math>

<math>~\nabla \Psi \, .</math>

Uniform Rotation

In the case of uniform rotation, we have,

<math>~\vec{v} = \hat{e}_\varphi (v_\varphi) = \hat{e}_\varphi (\varpi \omega_0) ~~~\Rightarrow~~~ \frac{j^2}{\varpi^3} = \frac{(\varpi v_\varphi)^2}{\varpi^3} = \frac{(\varpi^2\omega_0)^2}{\varpi^3} = \varpi \omega_0^2\, ,</math>

where, <math>~\omega_0</math> is independent of radial position. This also means that,

<math> \Psi \equiv - \int \frac{j^2(\varpi)}{\varpi^3} d\varpi = - \frac{1}{2} \varpi^2 \omega_0^2~; </math>

and,

<math>~\vec\zeta = \nabla \times \vec{v}</math>

<math>~=</math>

<math>~ \hat{e}_\varpi \biggl[ -\cancel{ \frac{\partial v_\varphi}{\partial z} }\biggr] + \hat{e}_z \biggl[ \frac{1}{\varpi} \frac{\partial (\varpi v_\varphi)}{\partial \varpi} \biggr] </math>

 

<math>~=</math>

<math>~ \hat{e}_z \biggl[ \frac{1}{\varpi} \frac{\partial (\varpi^2 \omega_0 )}{\partial \varpi} \biggr] </math>

 

<math>~=</math>

<math>~ \hat{e}_z ( 2\omega_0 ) </math>

[A]   Hence,

<math>~(\vec{v} \cdot \nabla) \vec{v}</math>

<math>~=</math>

<math>~\hat{e}_\varpi \biggl[ - \frac{v_\varphi \cdot v_\varphi}{\varpi} \biggr] </math>

 

<math>~=</math>

<math>~\hat{e}_\varpi \biggl[ - \frac{(\varpi \omega_0)\cdot (\varpi \omega_0)}{\varpi} \biggr] = - \hat{e}_\varpi (\varpi \omega_0^2) \, .</math>

[B}  Alternatively,

<math>~\vec\zeta \times \vec{v} + \frac{1}{2}\nabla (v^2)</math>

<math>~=</math>

<math>~\hat{e}_z ( 2\omega_0 ) \times \hat{e}_\varphi (\varpi \omega_0) + \hat{e}_\varpi \frac{1}{2} \biggl[ \frac{\partial}{\partial\varpi} (\varpi^2 \omega_0^2) \biggr]</math>

 

<math>~=</math>

<math>~ \hat{e}_\varpi \biggl\{ -( 2\omega_0 ) (\varpi \omega_0) + (\varpi \omega_0^2) \biggr\} = - \hat{e}_\varpi (\varpi \omega_0^2) \, .</math>

[C}  Or,

<math>~\nabla \Psi</math>

<math>~=</math>

<math>~\hat{e}_\varpi \biggl[- \frac{1}{2} \frac{\partial}{\partial\varpi} (\varpi^2 \omega_0^2) \biggr] = - \hat{e}_\varpi (\varpi \omega_0^2) \, .</math>

This demonstrates that, in the case of uniform angular velocity, all three expressions are identical.


Power Law

In the case of a power-law expression, we have,

<math>~\vec{v} = \hat{e}_\varphi (v_\varphi) = \hat{e}_\varphi \biggl[ \frac{j_0}{\varpi_0^2} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{(q-1)} \biggr] ~~~\Rightarrow~~~ \frac{j^2}{\varpi^3} = \biggl[ \frac{j_0^2}{\varpi_0^3} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{(2q-3)} \biggr] \, ,</math>

where, <math>~j_0</math> and <math>~\varpi_0</math> are both independent of radial position. This also means that,

<math> \Psi \equiv - \int \frac{j^2(\varpi)}{\varpi^3} d\varpi = - \frac{1}{2(q-1)} \biggl[ \frac{j_0^2}{\varpi_0^2} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{2(q-1)} \biggr]~; </math>

and,

<math>~\vec\zeta = \nabla \times \vec{v}</math>

<math>~=</math>

<math>~ \hat{e}_\varpi \biggl[ -\cancel{ \frac{\partial v_\varphi}{\partial z} }\biggr] + \hat{e}_z \biggl[ \frac{1}{\varpi} \frac{\partial (\varpi v_\varphi)}{\partial \varpi} \biggr] </math>

 

<math>~=</math>

<math>~ \hat{e}_z~ \frac{1}{\varpi} \frac{\partial }{\partial \varpi} \biggl[ \frac{j_0}{\varpi_0} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{q} \biggr] </math>

 

<math>~=</math>

<math>~ \hat{e}_z~ \frac{q}{\varpi} \biggl[ \frac{j_0}{\varpi_0^{q+1}} ( \varpi)^{q-1} \biggr] = \hat{e}_z~ q \biggl[ \frac{j_0}{\varpi_0^{3}} \biggl( \frac{\varpi}{\varpi_0} \biggr)^{q-2} \biggr]\, . </math>

[D]   Hence,

<math>~(\vec{v} \cdot \nabla) \vec{v}</math>

<math>~=</math>

<math>~\hat{e}_\varpi \biggl[ - \frac{v_\varphi \cdot v_\varphi}{\varpi} \biggr] </math>

 

<math>~=</math>

<math>~ - \hat{e}_\varpi \frac{1}{\varpi} \biggl[ \frac{j_0^2}{\varpi_0^4} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{2(q-1)} \biggr] = - \hat{e}_\varpi \biggl[ \frac{j_0^2}{\varpi_0^5} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{(2q-3)} \biggr]\, .</math>

[E}  Alternatively,

<math>~\vec\zeta \times \vec{v} + \frac{1}{2}\nabla (v^2)</math>

<math>~=</math>

<math>~\hat{e}_z~ q \biggl[ \frac{j_0}{\varpi_0^{3}} \biggl( \frac{\varpi}{\varpi_0} \biggr)^{q-2} \biggr] \times \hat{e}_\varphi \biggl[ \frac{j_0}{\varpi_0^2} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{(q-1)} \biggr] + \hat{e}_\varpi \frac{1}{2} \frac{\partial}{\partial\varpi} \biggl[ \frac{j_0^2}{\varpi_0^4} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{(2q-2)} \biggr]</math>

 

<math>~=</math>

<math>~-\hat{e}_\varpi~ q \biggl[ \frac{j_0^2}{\varpi_0^{5}} \biggl( \frac{\varpi}{\varpi_0} \biggr)^{2q-3} \biggr] + \hat{e}_\varpi(q-1) \biggl[ \frac{j_0^2}{\varpi_0^5} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{(2q-3)} \biggr]</math>

 

<math>~=</math>

<math>~-\hat{e}_\varpi~ \biggl[ \frac{j_0^2}{\varpi_0^{5}} \biggl( \frac{\varpi}{\varpi_0} \biggr)^{2q-3} \biggr] \, . </math>

[F}  Or,

<math>~\nabla \Psi</math>

<math>~=</math>

<math>~\hat{e}_\varpi\frac{\partial}{\partial\varpi} \biggl\{- \frac{1}{2(q-1)} \biggl[ \frac{j_0^2}{\varpi_0^2} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{2(q-1)} \biggr] \biggr\}</math>

 

<math>~=</math>

<math>~- ~\hat{e}_\varpi\frac{\partial}{\partial\varpi} \biggl[ \frac{j_0^2}{\varpi_0^3} \biggl( \frac{\varpi}{\varpi_0}\biggr)^{2q-3} \biggr] </math>

This demonstrates that, in the case of power-law angular velocity profile, all three expressions are identical.


Uniform vφ

In the case of a uniform <math>~v_\varphi</math> (i.e., a flat rotation curve), we have,

<math>~\vec{v} = \hat{e}_\varphi (v_\varphi) = \hat{e}_\varphi v_0 ~~~\Rightarrow~~~ \frac{j^2}{\varpi^3} = \frac{v_0^2}{\varpi} \, ,</math>

where, <math>~v_0</math> is independent of radial position. This also means that,

<math> \Psi \equiv - \int \frac{j^2(\varpi)}{\varpi^3} d\varpi = - v_0^2 \ln \biggl( \frac{\varpi}{\varpi_0} \biggr)~; </math>

and,

<math>~\vec\zeta = \nabla \times \vec{v}</math>

<math>~=</math>

<math>~ \hat{e}_\varpi \biggl[ -\cancel{ \frac{\partial v_\varphi}{\partial z} }\biggr] + \hat{e}_z \biggl[ \frac{1}{\varpi} \frac{\partial (\varpi v_\varphi)}{\partial \varpi} \biggr] </math>

 

<math>~=</math>

<math>~ \hat{e}_z \biggl( \frac{v_0}{\varpi} \biggr) \, . </math>

[G]   Hence,

<math>~(\vec{v} \cdot \nabla) \vec{v}</math>

<math>~=</math>

<math>~\hat{e}_\varpi \biggl[ - \frac{v_\varphi \cdot v_\varphi}{\varpi} \biggr] = -~\hat{e}_\varpi \biggl[ \frac{v_0^2}{\varpi} \biggr] \, .</math>

[H}  Alternatively,

<math>~\vec\zeta \times \vec{v} + \frac{1}{2}\nabla (v^2)</math>

<math>~=</math>

<math>~ \hat{e}_z \biggl( \frac{v_0}{\varpi} \biggr) \times \hat{e}_\varphi v_0 + \hat{e}_\varpi~ \frac{1}{2} \frac{\partial}{\partial \varpi} (v_0^2) </math>

 

<math>~=</math>

<math>~ -~\hat{e}_\varpi \biggl( \frac{v_0^2}{\varpi} \biggr) \, . </math>

[I}  Or,

<math>~\nabla \Psi</math>

<math>~=</math>

<math>~\hat{e}_\varpi\frac{\partial}{\partial\varpi} \biggl\{- v_0^2 \ln \biggl( \frac{\varpi}{\varpi_0} \biggr)\biggr\}</math>

 

<math>~=</math>

<math>~-~ \hat{e}_\varpi v_0^2 \biggl(\frac{\varpi}{\varpi_0} \biggr)^{-1} \frac{1}{\varpi_0}</math>

 

<math>~=</math>

<math>~-~ \hat{e}_\varpi \biggl( \frac{v_0^2}{\varpi} \biggr) \, .</math>

This demonstrates that, in the case of a constant <math>~v_\varphi</math> profile, all three expressions are identical.


j-Constant Rotation

In the case of so-called j-constant rotation, we have,

<math>~\vec{v} = \hat{e}_\varphi (v_\varphi) = \hat{e}_\varphi ~\omega_c \biggl[ \frac{A^2\varpi}{A^2 + \varpi^2}\biggr] ~~~\Rightarrow~~~ \frac{j^2}{\varpi^3} = \frac{(\varpi v_\varphi)^2}{\varpi^3} = \frac{\omega_c^2}{\varpi} \biggl[ \frac{A^2\varpi}{A^2 + \varpi^2}\biggr]^2 = \biggl[ \frac{\omega_c^2 A^4\varpi}{(A^2 + \varpi^2)^2}\biggr] \, , </math>

where, <math>~\omega_c</math>, and the characteristic length, <math>~A</math>, are both independent of radial position. This also means that,

<math> \Psi \equiv - \int \frac{j^2(\varpi)}{\varpi^3} d\varpi = +\frac{1}{2}\biggl[ \frac{\omega_c^2 A^4}{(A^2 + \varpi^2)}\biggr]~; </math>

and,

<math>~\vec\zeta = \nabla \times \vec{v}</math>

<math>~=</math>

<math>~ \hat{e}_\varpi \biggl[ -\cancel{ \frac{\partial v_\varphi}{\partial z} }\biggr] + \hat{e}_z \biggl[ \frac{1}{\varpi} \frac{\partial (\varpi v_\varphi)}{\partial \varpi} \biggr] </math>

 

<math>~=</math>

<math>~ \hat{e}_z \biggl\{ \frac{\omega_c}{\varpi} \frac{\partial }{\partial \varpi} \biggl[ \frac{A^2\varpi^2}{A^2 + \varpi^2}\biggr]\biggr\} </math>

 

<math>~=</math>

<math>~ \hat{e}_z~ \frac{\omega_c}{\varpi} \biggl\{ \biggl[ 2A^2\varpi(A^2 + \varpi^2)^{-1} \biggr] - \biggl[ 2A^2\varpi^3(A^2 + \varpi^2)^{-2} \biggr]\biggr\} </math>

 

<math>~=</math>

<math>~ \hat{e}_z~ \biggl[2\omega_c A^4 (A^2 + \varpi^2)^{-2} \biggr] \, . </math>

[J]   Hence,

<math>~(\vec{v} \cdot \nabla) \vec{v}</math>

<math>~=</math>

<math>~\hat{e}_\varpi \biggl[ - \frac{v_\varphi \cdot v_\varphi}{\varpi} \biggr] </math>

 

<math>~=</math>

<math>~ -~\hat{e}_\varpi \frac{\omega_c^2}{\varpi} \biggl[ \frac{A^2\varpi}{A^2 + \varpi^2}\biggr]^2 = -~\hat{e}_\varpi \biggl[ \frac{\omega_c^2A^4 \varpi}{(A^2 + \varpi^2)^2} \biggr] </math>

Related Discussions


Whitworth's (1981) Isothermal Free-Energy Surface

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