Editing Apps/GoldreichWeber80 (section)

===Homologous Solution===
{{ GW80 }} discovered that the governing equations admit to an homologous, self-similar solution if they adopted a stream function of the form,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\psi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2}a \dot{a} \mathfrak{x}^2 \, ,</math>
  </td>
</tr>
</table>
</div>
which, when acted upon by the various relevant operators, gives,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\nabla_\mathfrak{x}\psi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~a \dot{a} \mathfrak{x} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\nabla^2_\mathfrak{x}\psi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{1}{2}a \dot{a} \biggr) \frac{1}{\mathfrak{x}^2} \frac{d}{d\mathfrak{x}} \biggl[\mathfrak{x}^2 \frac{d}{d\mathfrak{x}}  \mathfrak{x}^2 \biggr]
= 3 a \dot{a} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{d\psi}{dt}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\mathfrak{x}^2 \biggl[ \frac{1}{2}\dot{a}^2 + \frac{1}{2}a\ddot{a} \biggr] \, .</math>
  </td>
</tr>
</table>
</div>
Hence, the radial velocity profile is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~v_r = a^{-1}\nabla_\mathfrak{x} \psi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\dot{a}\mathfrak{x} \, ,
</math>
  </td>
</tr>
</table>
</div>
which, as foreshadowed above, exactly matches the radial velocity of the collapsing coordinate frame; the continuity equation gives,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\frac{d\ln \rho_c}{dt}  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>-~ \frac{3\dot{a}}{a}  </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>\Rightarrow~~~~\frac{d\ln \rho_c}{dt} + \frac{d\ln a^3}{dt} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, ,</math>
  </td>
</tr>
</table>
</div>
which means that, consistent with the expected relationship between the central density and the time-varying length scale [[#Length|established above]], the product, <math>~a^3 \rho_c</math>, is independent of time; and the Euler equation becomes,

<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~ - 3 f - \sigma </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ \frac{3}{4} \biggl( \frac{\pi G}{\kappa^3} \biggr)^{1/2} a(t) \biggr]  
\biggl\{ \mathfrak{x}^2 \biggl[ \frac{1}{2}\dot{a}^2 + \frac{1}{2}a\ddot{a} \biggr]  - 
\frac{1}{2} ( \dot{a} \mathfrak{x} )^2 \biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{3}{8} \biggl( \frac{\pi G}{\kappa^3} \biggr)^{1/2}   (a \mathfrak{x})^2 \ddot{a} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~~ \frac{(f + \sigma/3)}{\mathfrak{x}^2} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \frac{1}{8} \biggl( \frac{\pi G}{\kappa^3} \biggr)^{1/2}   a^2 \ddot{a} \, .
</math>
  </td>
</tr>

</table>
</div>
This matches equation (12) of {{ GW80 }}.

Because everything on the lefthand side of this scaled Euler equation depends only on the dimensionless spatial coordinate, <math>~\mathfrak{x}</math>, while everything on the righthand side depends only on time &#8212; via the parameter, <math>~a(t)</math> &#8212; both expressions must equal the same (dimensionless) constant.  {{ GW80 }} (see their equation 12) call this constant, <math>~\lambda/6</math>.  From the terms on the lefthand side, they conclude (see their equation 13) that the dimensionless gravitational potential is,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\sigma</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{2} \lambda ~\mathfrak{x}^2 - 3f \, .</math>
  </td>
</tr>
</table>
</div>
From the terms on the righthand side they conclude, furthermore, that the nonlinear differential equation governing the time-dependent variation of the scale length, <math>~a</math>, is,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
a^2 \ddot{a} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~\frac{4\lambda}{3} \biggl( \frac{\kappa^3}{\pi G} \biggr)^{1/2} \, .</math>
  </td>
</tr>
</table>
</div>



{{SGFworkInProgress}}


<table border="1" cellpadding="10" align="center" width="75%">
<tr><td align="left">
As {{ GW80 }} point out, this nonlinear differential equation can be integrated twice to produce an algebraic relationship between <math>~a</math> and time, <math>~t</math>.   The required mathematical steps are identical to the steps used to analytically solve the [[ProjectsUnderway/CoreCollapseSupernovae#Nonrotating.2C_Spherically_Symmetric_Collapse|classic, spherically symmetric free-fall collapse problem]].  First, rewrite the equation as,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
\frac{d \dot{a} }{dt}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\frac{B}{2a^2} \, , </math>
  </td>
</tr>

</table>
where,
<div align="center">
<math>
~B \equiv \frac{8\lambda}{3} \biggl( \frac{\kappa^3}{\pi G} \biggr)^{1/2}  \, ,
</math>
</div>
has the same dimensions as the product, <math>~GM</math> (see the  [[ProjectsUnderway/CoreCollapseSupernovae#Nonrotating.2C_Spherically_Symmetric_Collapse|free-fall collapse problem]]), that is, the dimensions of "length-cubed per unit time-squared."   Then, multiply both sides by <math>~2\dot{a} = 2(da/dt)</math> to obtain,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~
2\dot{a} \frac{d\dot{a}}{dt}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-B \biggl( a^{-2} \frac{da}{dt} \biggr) </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~~
\frac{d\dot{a}^2}{dt}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~B \frac{d}{dt} \biggl( \frac{1}{a} \biggr) \, ,</math>
  </td>
</tr>

</table>
which integrates once to give,
<div align="center">
<math>
~\dot{a}^2 = \frac{B}{a} + C \, ,
</math>
</div>
or,
<div align="center">
<math>
~dt = \biggl( \frac{B}{a} + C \biggr)^{-1/2} da  \, .
</math>
</div>

For the case, <math>~C = 0</math>, this differential equation can be integrated straightforwardly to give (see equation 15 in {{ GW80 }}),

For the cases when <math>~C \ne 0</math>, [http://integrals.wolfram.com/index.jsp Wolfram Mathematica's online integrator] can be called upon to integrate this equation and provide the following closed-form solution,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~t</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\frac{a}{C} \biggl( \frac{B}{a} + C \biggr)^{1/2}
- \frac{B}{2C^{3/2}} \ln \biggl[2aC^{1/2}  \biggl( \frac{B}{a} + C \biggr)^{1/2}  + B + 2aC \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>


</td></tr>
</table>


As {{ GW80 }} point out, because all terms in this equation are inside the gradient operator, the sum of the terms inside the square brackets must equal a constant &#8212; that is, the sum must be independent of spatial position throughout the spherically symmetric configuration.  If, following the lead of {{ GW80 }}, we simply fold this integration constant into the potential, the Euler equation becomes (see their equation 8),
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial \psi}{\partial t}  - \biggl( \frac{\dot{a}}{a} \biggr)\psi +
H + \Phi + \frac{1}{2}\biggl(\frac{1}{a} \nabla_x\psi  \biggr)^2 </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0 \, .</math>
  </td>
</tr>
</table>
</div>

<div align="center">
<table border="0" cellpadding="5">

<tr>
  <td align="right">
<math>~\frac{\partial \rho}{\partial t} + \rho \nabla_r \cdot \vec{v} + \vec{v}\cdot \nabla_r \rho</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left" width="25%">
<math>~0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ \frac{1}{\rho} \frac{\partial \rho}{\partial t} + \nabla_r \cdot \vec{v} + \vec{v}\cdot \frac{\nabla_r \rho}{\rho}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left" width="25%">
<math>~0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~ \frac{1}{\rho} \frac{\partial \rho}{\partial t} + a^{-1} \nabla_x \cdot \biggl[  a^{-1} \nabla_x \psi \biggr] 
+ a^{-1} \nabla_x \psi  \cdot \frac{a^{-1}\nabla_x \rho}{\rho}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left" width="25%">
<math>~0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{1}{\rho} \frac{\partial \rho}{\partial t} + a^{-1}(a^{-1} \nabla_x\psi - \dot{a} \vec{x}) \cdot \frac{\nabla_x\rho}{\rho}
 + a^{-2} \nabla_x^2\psi </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left" width="25%">
<math>~0</math>
  </td>
</tr>
</table>
</div>

<table border="1" cellpadding="5" align="center" width="75%">
<tr><td align="center" colspan="1">
Governing Equations from {{ GW80 }} After Initial ''Length'' Scaling (yet to be demonstrated)
</td></tr>

<tr><td align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{1}{\rho} \frac{\partial \rho}{\partial t} + a^{-1}(a^{-1} \nabla_x\psi - \dot{a} \vec{x}) \cdot \frac{\nabla_x\rho}{\rho}
 + a^{-2} \nabla_x^2\psi </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left" width="25%">
<math>~0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial \psi}{\partial t} - \frac{\dot{a}}{a} \vec{x}\cdot \nabla_x\psi + \frac{1}{2} a^{-2} | \nabla_x\psi|^2 
+ H + \Phi</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~
a^{-2} \nabla_x^2\Phi - 4\pi G \rho
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~0</math>
  </td>
</tr>

<tr><td align="left" colspan="3">
where, 
<div align="center">
<math>~\vec{x} \equiv \frac{\vec{r}}{a} \, ,</math>
</div>
and it is understood that derivatives in the <math>~\nabla_x</math> and <math>~\nabla_x^2</math> operators are taken with respect to the dimensionless radial coordinate, <math>~x</math>.
</td></tr>

</table>

</td></tr>
</table>

<!-- BEGIN PK2007 ASIDE

=PK2007 ASIDE= 

<div align="center">
<table border="1" width="90%" cellpadding="8">
<tr><td align="left">
<font color="red">'''ASIDE:'''</font>  It wasn't immediately obvious to me how the set of differential governing equations should be modified in order to accommodate a radially contracting (accelerating) coordinate system.  I did not understand the transformed set of equations presented by {{ GW80 }} as equations (7) and (8), for example.  I turned to {{ PK2007full }} &#8212; hereafter, {{ PK2007hereafter }} &#8212; for guidance.  {{ PK2007hereafter }} develop a set of governing equations that allows for coordinate rotation as well as expansion or contraction; here we will ignore any modifications due to rotation.

We note, first, that {{ PK2007hereafter }} (see their equation 4) adopt an accelerated radial coordinate of the same form as {{ GW80 }},
<div align="center">
<math>~\tilde{r} \equiv \biggl[ \frac{1}{a(t)} \biggr] \vec{r} \, ,</math>
</div>
but the {{ PK2007hereafter }} time-dependent scale factor is dimensionless, whereas the scale factor adopted by {{ GW80 }} &#8212; denoted here as <math>~a_{GW}(t)</math> &#8212; has units of length.  To transform from the KP07 notation, we ultimately will set,
<div align="center">
<math>~\mathfrak{x} = \frac{1}{a_0} \tilde{r}  ~~~~~\Rightarrow ~~~~~ a_{GW}(t) = a_0 a(t) \, ,</math>
</div>
where, <math>~a_0</math> is understood to be the {{ GW80 }} scale length at the onset of collapse, that is, at <math>~t = 0</math>.  According to {{ PK2007hereafter }}, this leads to a new "accelerated" time (see, again, their equation 4 with the exponent, <math>~\beta = 0</math>)
<div align="center">
<math>~\tau \equiv \int_0^t \frac{dt}{a(t)} \, .</math>
</div>
According to equation (7) of {{ PK2007hereafter }} &#8212; again, setting their exponent <math>~\beta=0</math> &#8212; the relationship between the fluid velocity in the inertial frame, <math>~\vec{v}</math>, to the fluid velocity measured in the accelerated frame, <math>~\tilde{v}</math>, is
<div align="center">
<math>~\vec{v} = \tilde{v} + \biggl[ \frac{d\ln a}{d\tau} \biggr] \tilde{r} \, .</math>
</div>
We note that, according to equation (8) of {{ PK2007hereafter }}, the first derivative of <math>~a(t)</math> with respect to ''physical'' time is,
<div align="center">
<math>~\dot{a} = \frac{d\ln a}{d\tau} \, ,</math>
</div>
so the transformation between velocities may equally well be written as,
<div align="center">
<math>~\vec{v} = \tilde{v} + \dot{a} \tilde{r} \, ;</math>
</div>
and we note that (see equation 9 of {{ PK2007hereafter }}),
<div align="center">
<math>~\ddot{a} = \frac{1}{a} \biggl[ \frac{d^2\ln a}{d\tau^2} \biggr] \, .</math>
</div>

Next, we note that {{ GW80 }} introduce a variable to track the dimensionless density,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~f^3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{\rho}{\rho_c} \biggr) = \biggl( \frac{\pi G}{\kappa} \biggr)^{3/2} [a_{GW}(t)]^3 \rho \, .</math>
  </td>
</tr>
</table>
Comparing this to equation (10) of {{ PK2007hereafter }}, which introduces a density field, <math>~\tilde\rho</math>, as viewed in the accelerated frame of reference of the form,
<div align="center">
<math>~\tilde\rho = [a(t)]^\alpha \rho \, ,</math>
</div>
we see that, by setting the exponent <math>~\alpha = 3</math>, the {{ GW80 }} dimensionless density can be retrieved from the {{ PK2007hereafter }} work by setting,
<div align="center">
<math>~f^3= \frac{\tilde\rho}{\rho_0} \, ,</math>
</div>
where,
<div align="center">
<math>~\rho_0 \equiv \biggl( \frac{\kappa}{\pi G a_0^2} \biggr)^{3/2} \, .</math>
</div>
{{ PK2007hereafter }} then claim that, in the accelerating reference frame, the continuity equation and Euler equation become, respectively,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial \tilde\rho}{\partial \tau} + \tilde{\nabla}\cdot(\tilde\rho \tilde{v})</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(3-\nu)\biggl[ \frac{d\ln a}{d\tau} \biggr] \tilde\rho \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial \tilde\rho \tilde{v} }{\partial \tau} + \tilde{\nabla} \cdot(\tilde\rho \tilde{v} \tilde{v})  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(2-\nu)\biggl[ \frac{d\ln a}{d\tau} \biggr] \tilde\rho \tilde{v} - 
\biggl[ \frac{d^2\ln a}{d\tau^2} \biggr] \tilde\rho \tilde{r} - \tilde{\nabla}\tilde{P} \, ,</math>
  </td>
</tr>
</table>
</div>
where {{ PK2007hereafter }} have introduced <math>~\nu</math> as a "dimensionality parameter of the problem."  In an effort to rewrite the left-hand-side of their Euler equation in a form that matches the {{ GW80 }} Euler equation, we note that,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\nabla\cdot [(\tilde\rho \tilde{v}) \tilde{v}]</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tilde\rho(\tilde{v}\cdot \tilde\nabla) \tilde{v} + \tilde{v}[\tilde\nabla \cdot (\tilde\rho \tilde{v})] \, ,</math>
  </td>
</tr>
</table>
</div>
and, with the help of the {{ PK2007hereafter }} continuity equation,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial (\tilde\rho \tilde{v})}{\partial\tau}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tilde\rho \frac{\partial \tilde{v}}{\partial\tau} + \tilde{v} \frac{\partial \tilde\rho}{\partial\tau} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\tilde\rho \frac{\partial \tilde{v}}{\partial\tau} + \tilde{v} \biggl[ 
(3-\nu)\biggl( \frac{d\ln a}{d\tau} \biggr) \tilde\rho
- \tilde{\nabla}\cdot(\tilde\rho \tilde{v})
\biggr] \, .</math>
  </td>
</tr>
</table>
</div>
Hence, the Euler equation becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ \tilde\rho \frac{\partial \tilde{v}}{\partial\tau}  
+ \tilde\rho(\tilde{v}\cdot \tilde\nabla) \tilde{v} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~\biggl[ \frac{d\ln a}{d\tau} \biggr] \tilde\rho \tilde{v} - 
\biggl[ \frac{d^2\ln a}{d\tau^2} \biggr] \tilde\rho \tilde{r} - \tilde{\nabla}\tilde{P} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~ \Rightarrow ~~~ \frac{\partial \tilde{v}}{\partial\tau}  + (\tilde{v}\cdot \tilde\nabla) \tilde{v} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~\dot{a} \tilde{v} - 
a \ddot{a} \tilde{r} - \tilde{\nabla}\tilde{H} \, .</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~ \Rightarrow ~~~ \frac{\partial \tilde{v}}{\partial\tau}  + \frac{1}{2} \tilde\nabla({\tilde{v}} \cdot \tilde{v} ) + \tilde{\zeta}\times \tilde{v}  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~\dot{a} \tilde{v} - 
a \ddot{a} \tilde{r} - \tilde{\nabla}\tilde{H} \, ,</math>
  </td>
</tr>
</table>
</div>
where the vector identity that has been used to obtain this last expression has been drawn from our [[PGE/Euler#in_terms_of_the_vorticity:|separate presentation of the Euler equation written in terms of the fluid vorticity]], <math>~\tilde\zeta \equiv \tilde\nabla \times \tilde{v}</math>.  

----
Now, let's shift to ''physical'' parameters &#8212; or example,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\tilde{v}</math>
  </td>
  <td align="center">
<math>~~~\rightarrow~~~</math>
  </td>
  <td align="left">
<math>~\vec{v} - \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} = \vec{v} - \dot{a} \tilde{r} ~~ \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial}{\partial\tau}</math>
  </td>
  <td align="center">
<math>~~~\rightarrow~~~</math>
  </td>
  <td align="left">
<math>~\frac{\partial t}{\partial\tau} \frac{\partial}{\partial t} = a \frac{\partial}{\partial t} </math>
  </td>
</tr>
</table>
</div>
&#8212; and, following {{ GW80 }}, set the vorticity to zero.  The Euler equation becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial }{\partial t}  \biggl[ \vec{v} - \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \biggr]
+ \frac{1}{2}a^{-1} \tilde\nabla \biggl[ \biggl( \vec{v} - \dot{a} \tilde{r} \biggr) \cdot \biggl( \vec{v} - \dot{a} \tilde{r} \biggr) \biggr]  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~\frac{\dot{a}}{a} \biggl( \vec{v} - \dot{a} \tilde{r} \biggr)  - 
\ddot{a} \tilde{r} - a^{-1}\tilde{\nabla}\tilde{H} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~
\frac{\partial \vec{v} }{\partial t}  - \biggl[ \biggl(\frac{\dot{a}}{a} \biggr)\frac{\partial\vec{r}}{\partial t} + \frac{\ddot{a}}{a} \vec{r} - \biggl( \frac{\dot{a}}{a}\biggr)^2 \vec{r} \biggr]
+ \frac{1}{2}a^{-1} \tilde\nabla \biggl[ \vec{v} \cdot \vec{v} -2\dot{a} \vec{v} \tilde{r} + (\dot{a} \tilde{r} )^2 \biggr]  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-~\frac{\dot{a}}{a} \biggl( \vec{v} - \frac{\dot{a}}{a} \vec{r} \biggr)  - 
\frac{\ddot{a}}{a} \vec{r} - a^{-1}\tilde{\nabla}\tilde{H} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~
\frac{\partial \vec{v} }{\partial t}  
+ \frac{1}{2}a^{-1} \tilde\nabla \biggl[ \vec{v} \cdot \vec{v} -2\dot{a} \vec{v} \tilde{r} + (\dot{a} \tilde{r} )^2 \biggr]  
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\frac{\dot{a}}{a} \biggl(\frac{\partial\vec{r}}{\partial t} - \vec{v} \biggr)  - a^{-1}\tilde{\nabla}\tilde{H}</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~~
\frac{\partial \vec{v} }{\partial t}  
+ a^{-1} \tilde\nabla \biggl[ \frac{1}{2}(\vec{v} \cdot \vec{v}) - \dot{a} \vec{v} \tilde{r} + \tilde{H} \biggr]  + \biggl( \frac{\dot{a}}{a} \biggr)^2 \vec{r} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>\frac{\dot{a}}{a} \biggl(\frac{\partial\vec{r}}{\partial t} - \vec{v} \biggr)  \, .</math>
  </td>
</tr>
</table>
</div>

Now, let's tackle the continuity equation:
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial \tilde\rho}{\partial \tau} + \tilde\rho \tilde{\nabla}\cdot \tilde{v} + \tilde{v} \cdot \tilde\nabla \tilde\rho  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(3-\nu)\biggl[ \frac{d\ln a}{d\tau} \biggr] \tilde\rho </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~~\frac{a}{\tilde\rho}\frac{\partial \tilde\rho}{\partial t} + \tilde{\nabla}\cdot \tilde{v} + \tilde{v} \cdot \frac{\tilde\nabla \tilde\rho}{\tilde\rho}  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(3-\nu) \dot{a} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~~
\frac{a}{\tilde\rho}\frac{\partial \tilde\rho}{\partial t} 
+ (\vec{v} - \dot{a}\tilde{r})\cdot \frac{\tilde\nabla \tilde\rho}{\tilde\rho}  
+ \tilde{\nabla}\cdot (\vec{v} - \dot{a}\tilde{r}) 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(3-\nu) \dot{a} </math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~~
\frac{1}{a^3\rho}\frac{\partial (a^3\rho)}{\partial t} 
+ a^{-1}(\vec{v} - \dot{a}\tilde{r})\cdot \frac{\tilde\nabla \rho}{\rho}  
+ a^{-1}\tilde{\nabla}\cdot \vec{v} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(3-\nu) \frac{\dot{a}}{a}  + a^{-1}\tilde{\nabla}\cdot (\dot{a}\tilde{r}) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~~
\frac{1}{\rho}\frac{\partial \rho}{\partial t} 
+ a^{-1}(\vec{v} - \dot{a}\tilde{r})\cdot \frac{\tilde\nabla \rho}{\rho}  
+ a^{-1}\tilde{\nabla}\cdot \vec{v} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(3-\nu) \frac{\dot{a}}{a}  + a^{-1}\tilde{\nabla}\cdot (\dot{a}\tilde{r}) -3 \frac{\dot{a}}{a}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\dot{a}}{a} (\tilde{\nabla}\cdot \tilde{r}-\nu)  
</math>
  </td>
</tr>

</table>
</div>
If we set <math>~\nu = 3</math>, this last expression appears to match equation (7) of {{ GW80 }}.


----
With the aid of the continuity equation, the left-hand-side of the Euler equation can be rewritten as,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial \tilde\rho \tilde{v} }{\partial \tau} + \tilde{\nabla} \cdot(\tilde\rho \tilde{v} \tilde{v})  </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl[ \tilde\rho \frac{\partial \tilde{v} }{\partial \tau} + \tilde{v} \frac{\partial \tilde\rho }{\partial \tau} \biggr] + 
\biggl[ (\tilde{v} \cdot \tilde{\nabla} ) \tilde\rho \tilde{v} + (\tilde\rho \tilde{v} \cdot \tilde{\nabla})\tilde{v} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\tilde\rho \frac{\partial \tilde{v} }{\partial \tau} + 
\biggl[(3-\nu)\biggl( \frac{d\ln a}{d\tau} \biggr) \tilde\rho \tilde{v}  - (\tilde{v} \cdot \tilde{\nabla} ) \tilde\rho \tilde{v} \biggr] + 
\biggl[ (\tilde{v} \cdot \tilde{\nabla} ) \tilde\rho \tilde{v} + (\tilde\rho \tilde{v} \cdot \tilde{\nabla})\tilde{v} \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\tilde\rho \biggl[ \frac{\partial \tilde{v} }{\partial \tau} + 
(3-\nu)\biggl( \frac{d\ln a}{d\tau} \biggr) \tilde{v}  + 
(\tilde{v} \cdot \tilde{\nabla})\tilde{v} \biggr] \, .
</math>
  </td>
</tr>
</table>
</div>
Hence, the Euler equation becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>
\frac{\partial \tilde{v} }{\partial \tau} 
+ (\tilde{v} \cdot \tilde{\nabla})\tilde{v} 
+ \biggl( \frac{d\ln a}{d\tau} \biggr) \tilde{v}  
+ \biggl( \frac{d^2\ln a}{d\tau^2} \biggr) \tilde{r}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ - \frac{\tilde{\nabla}\tilde{P}}{\tilde\rho} \, .</math>
  </td>
</tr>
</table>
</div>

----
Now, let's shift to ''physical'' parameters.  For example,
<div align="center">
<table border="0" cellpadding="5" align="center">
<tr>
  <td align="right">
<math>~\tilde{v}</math>
  </td>
  <td align="center">
<math>~~~\rightarrow~~~</math>
  </td>
  <td align="left">
<math>~\vec{v} - \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial}{\partial\tau}</math>
  </td>
  <td align="center">
<math>~~~\rightarrow~~~</math>
  </td>
  <td align="left">
<math>~\frac{\partial t}{\partial\tau} \frac{\partial}{\partial t} = a \frac{\partial}{\partial t} \, .</math>
  </td>
</tr>
</table>
</div>

Hence, the Euler equation becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~ - \tilde{\nabla}\tilde{H} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
a\frac{\partial}{\partial t} \biggl[ \vec{v} - \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \biggr]
+ (\tilde{v} \cdot \tilde{\nabla})\tilde{v} 
+ \dot{a} \biggl[ \vec{v} - \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \biggr]  
+ \ddot{a} \vec{r}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>a\frac{\partial \vec{v} }{\partial t} - 
a \biggl[ \biggl(\frac{\ddot{a}}{a} \biggr) \vec{r} - \biggl(\frac{\dot{a}}{a} \biggr)^2 \vec{r} + \biggl(\frac{\dot{a}}{a} \biggr) \frac{\partial \vec{r} }{\partial t} \biggr]
+ (\tilde{v} \cdot \tilde{\nabla})\tilde{v} 
+ \dot{a} \biggl[ \vec{v} - \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \biggr]  
+ \ddot{a} \vec{r}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>a\frac{\partial \vec{v} }{\partial t} + (\tilde{v} \cdot \tilde{\nabla})\tilde{v} 
+ \dot{a} \biggl[ \vec{v} - \frac{\partial \vec{r} }{\partial t} \biggr]  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>a\frac{\partial \vec{v} }{\partial t} + \dot{a} \biggl[ \vec{v} - \frac{\partial \vec{r} }{\partial t} \biggr]  
+ \biggl\{\biggl[ \vec{v} - \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \biggr] \cdot \tilde{\nabla} \biggr\} \biggl[ \vec{v} - \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>a\frac{\partial \vec{v} }{\partial t} + \dot{a} \biggl[ \vec{v} - \frac{\partial \vec{r} }{\partial t} \biggr]  
+ (\vec{v} \cdot \tilde\nabla)\biggl[ \vec{v} - \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \biggr]
- \biggl(\frac{\dot{a}}{a} \biggr) \vec{r}  \cdot \tilde{\nabla} 
\biggl[ \vec{v} - \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl\{ a\frac{\partial \vec{v} }{\partial t} + \dot{a} \biggl[ \vec{v} - \frac{\partial \vec{r} }{\partial t} \biggr]  
+ (\vec{v} \cdot \tilde\nabla)\vec{v} 
- \biggl(\frac{\dot{a}}{a} \biggr) \vec{r}  \cdot \tilde{\nabla} \vec{v} \biggr\}
- (\vec{v} \cdot \tilde\nabla)\biggl[ \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \biggr]
+ \biggl(\frac{\dot{a}}{a} \biggr) \vec{r}  \cdot \tilde{\nabla} \biggl[\biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl\{ a\frac{\partial \vec{v} }{\partial t} + \dot{a} \biggl[ \vec{v} - \frac{\partial \vec{r} }{\partial t} \biggr]  
+ (\vec{v} \cdot \tilde\nabla)\vec{v} 
- \biggl(\frac{\dot{a}}{a} \biggr) \vec{r}  \cdot \tilde{\nabla} \vec{v} \biggr\}
- \biggl[ \vec{v} - \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \biggr] \cdot \tilde{\nabla} \biggl[\dot{a} \tilde{r}  \biggr] 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>
\biggl\{ a\frac{\partial \vec{v} }{\partial t}  + (\vec{v} \cdot \tilde\nabla)\vec{v} 
- \biggl(\frac{\dot{a}}{a} \biggr) \vec{r}  \cdot \tilde{\nabla} \vec{v} \biggr\}
+ \dot{a} \biggl\{ \biggl[ \vec{v} - \frac{\partial \vec{r} }{\partial t} \biggr] 
- \biggl[ \vec{v} - \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \biggr] \cdot \tilde{\nabla} \biggl[\tilde{r}  \biggr] \biggr\}
</math>
  </td>
</tr>
</table>
</div>

And the continuity equation becomes,
<div align="center">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~(3-\nu) \dot{a} </math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{a}{\tilde\rho} \frac{\partial \tilde\rho}{\partial t} 
+ \tilde{\nabla}\cdot \biggl[  \vec{v} - \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \biggr]
+ \biggl[ \vec{v} - \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \biggr] \cdot \frac{\tilde{\nabla} \tilde\rho}{\tilde\rho} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{a}{\tilde\rho} \frac{\partial \tilde\rho}{\partial t} 
+ \tilde{\nabla}\cdot \vec{v} 
- \tilde{\nabla}\cdot \biggl[  \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \biggr]
+ \biggl[ \vec{v} 
- \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \biggr] \cdot \frac{\tilde{\nabla} \tilde\rho}{\tilde\rho} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ (3-\nu) \dot{a} 
+ \tilde{\nabla}\cdot \biggl[  \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \biggr]
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{a^2 \rho} \frac{\partial (a^3\rho)}{\partial t} 
+ \tilde{\nabla}\cdot \vec{v} 
+ \biggl[ \vec{v} 
- \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \biggr] \cdot \frac{\tilde{\nabla} \rho}{\rho} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{a}{\rho} \frac{\partial \rho}{\partial t} + 3\dot{a} 
+ \tilde{\nabla}\cdot \vec{v} 
+ \biggl[ \vec{v} 
- \biggl(\frac{\dot{a}}{a} \biggr) \vec{r} \biggr] \cdot \frac{\tilde{\nabla} \rho}{\rho} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{1}{\rho} \frac{\partial \rho}{\partial t} 
+ a^{-1}\tilde{\nabla}\cdot \vec{v} 
+ \biggl[ \vec{v} 
- \dot{a} \biggl( \frac{\vec{r}}{a}\biggr) \biggr] \cdot \frac{a^{-1}\tilde{\nabla} \rho}{\rho} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math> 
\frac{\dot{a}}{a} \biggl[\tilde{\nabla}\cdot \biggl( \frac{\vec{r}}{a} \biggr) -\nu \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{1}{\rho} \frac{\partial \rho}{\partial t} 
+ a^{-1}\tilde{\nabla}\cdot \vec{v} 
+ a^{-1} \biggl[ \vec{v} 
- \dot{a} \vec{\mathfrak{x}} \biggr] \cdot \frac{\tilde{\nabla} \rho}{\rho} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math> 
\frac{\dot{a}}{a} \biggl[\tilde{\nabla}\cdot \vec{\mathfrak{x}} -\nu \biggr]
</math>
  </td>
</tr>
</table>
</div>

</td></tr>
</table>
</div>

END PK2007 ASIDE -->