# Similarity Solution

Much of the material in this chapter has been drawn from §4.1 of a review article by Tohline (1982) titled, Hydrodynamic Collapse.

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Several authors (references given, below) have shown that when isothermal pressure gradients are important during a gas cloud's collapse, the equations governing the collapse admit a set of similarity solutions. Certain properties of these solutions can be described analytically and are instructive models for comparison with more detailed, numerical collapse calculations.

## Establishing Set of Governing Equations

Drawing from an accompanying chapter's introductory discussion, we begin with the set of governing equations that describe the collapse of isothermal spheres from an Eulerian frame of reference.

Eulerian Frame
 $~\frac{\partial M_r}{\partial r}$ $~=$ $~4\pi r^2 \rho \, ,$ $~\frac{\partial M_r}{\partial t}$ $~=$ $~- 4\pi r^2 \rho v_r \, ,$ $~\frac{\partial v_r}{\partial t} + v_r \frac{\partial v_r}{\partial r}$ $~=$ $~- c_s^2 \biggl( \frac{\partial \ln \rho}{\partial r}\biggr) - \frac{GM_r}{r^2} \, .$

Notice that, following Larson's (1969) lead, we have replaced the standard continuity equation with the following equivalent statement of mass conservation:

 $~\frac{dM_r}{dt}$ $~=$ $~0$ $~\Rightarrow ~~~ 0$ $~=$ $~\frac{\partial M_r}{\partial t} + v_r ~\frac{\partial M_r}{\partial r}$ $~=$ $~\frac{\partial M_r}{\partial t} +4\pi r^2 \rho v_r \, .$

## Solution

### Summary

A similarity solution becomes possible for these equations when the single independent variable,

$~\zeta = \frac{c_s t}{r} \, ,$

is used to replace both $~r$ and $~t$. Then, if $~M_r$, $~\rho$, and $~v_r$ assume the following forms,

 $~M_r(r,t)$ $~=$ $~\biggl(\frac{c_s^3 t}{G}\biggr) m(\zeta) \, ,$ $~\rho(r,t)$ $~=$ $~\biggl(\frac{c_s^2 }{4\pi G r^2}\biggr) \Rho (\zeta) \, ,$ $~v_r(r,t)$ $~=$ $~- c_s U(\zeta) \, ,$

the three coupled partial differential equations reduce to two coupled ordinary differential equations for the functions, $~\Rho (\zeta)$ and $~U(\zeta)$, namely,

 $~\frac{dU}{d\zeta}$ $~=$ $~ \frac{(\zeta U +1) [\Rho (\zeta U +1) -2)]}{[ (\zeta U +1)^2 - \zeta^2]} \, ,$ $~\frac{dP}{d\zeta}$ $~=$ $~\frac{\zeta \Rho [2-\Rho (\zeta U +1)]}{[ (\zeta U +1)^2 - \zeta^2]} \, ,$

and a single equation defining $~m(\zeta)$,

 $~m(\zeta)$ $~=$ $~\Rho \biggl[ U + \frac{1}{\zeta} \biggr] \, .$

The parameters $~\zeta, m, \Rho$, and $~U$, and this summary set of equations are exactly those used by Hunter (1977) in his analysis of this problem. But they differ in form from the relations used by Larson (1969), Penston (1969), and Shu (1977) primarily because these authors chose to use a similarity variable,

$~x = \pm \frac{1}{\zeta} \, ,$

instead of $~\zeta$. Hunter's analysis is the most complete and his relations will be used here, but a transformation between his presentation and those of the other authors can be easily obtained from Table 1 of Hunter (1977) which, for convenience, is reproduced here.

Analogous to Table 1 from Hunter (1977)
Relations Between the Variables Used by Different Authors

Physical
Quantity
Herein Larson (1969) Penston (1969) Shu (1977)
$~\frac{c_s t}{r}$ $~\zeta$ $~- \frac{1}{x}$ $~- \frac{1}{x}$ $~+ \frac{1}{x}$
$~- \frac{v_r}{c_s}$ $~U$ $~\xi$ $~- V$ $~-v$
$~\frac{4\pi G\rho r^2}{c_s^2}$ $~\Rho$ $~x^2\eta$ $~x^2 e^Q$ $~x^2\alpha$
$~\frac{GM_r}{c_s^3 t}$ $~m$ $~-N$ $~m$
$~\ln(4\pi G\rho t^2)$ $~Q$ $~\ln\eta$ $~Q$ $~\ln\alpha$
$~\frac{r}{(- c_s t)}$ $~y$ $~x$ $~x$ $~-x$

Adopting Hunter's notation, this dimensionless variable name, $~\Rho$, is the capital Greek letter, $~\rho$, and should not be confused with the variable name, $~P$, that represents herein the ideal gas pressure.

### Proof2

First, note that,

 $~\frac{\partial \zeta}{\partial t}$ $~=$ $~\frac{c_s}{r} = \frac{\zeta}{t} \, ;$ $~\frac{\partial \zeta}{\partial r}$ $~=$ $~- \frac{c_s t}{r^2} = -\frac{\zeta^2}{c_st} \, .$

Next, let's take partial derivatives, with respect to both $~r$ and $~t$, of the three primary physical variables, $~M_r$, $~\rho$, and $~v_r$.

 $~\frac{\partial M_r}{\partial r}$ $~=$ $~ \biggl( \frac{c_s^3 t}{G} \biggr)\frac{dm}{d\zeta} \biggl( \frac{\partial \zeta}{\partial r} \biggr) = \biggl( \frac{c_s^3 t}{G} \biggr)\frac{dm}{d\zeta} \biggl(- \frac{\zeta^2}{c_s t} \biggr) = -\biggl( \frac{c_s^2 \zeta^2}{G} \biggr)\frac{dm}{d\zeta} \, ;$ $~\frac{\partial M_r}{\partial t}$ $~=$ $~ \biggl( \frac{c_s^3 t}{G} \biggr)\frac{dm}{d\zeta} \biggl( \frac{\partial \zeta}{\partial t} \biggr) + \biggl( \frac{c_s^3}{G}\biggr) m = \biggl( \frac{c_s^3 t}{G} \biggr)\frac{dm}{d\zeta} \biggl( \frac{\zeta}{t} \biggr) + \biggl( \frac{c_s^3}{G}\biggr) m = \biggl( \frac{c_s^3 }{G} \biggr) \biggl[ \zeta ~\frac{dm}{d\zeta} + m \biggr] \, ;$ $~\frac{\partial v_r}{\partial r}$ $~=$ $~ -c_s \frac{dU}{d\zeta} \biggl( \frac{\partial \zeta}{\partial r} \biggr) = -c_s \frac{dU}{d\zeta} \biggl( - \frac{\zeta^2}{c_s t} \biggr) = \frac{\zeta^2}{t} \frac{dU}{d\zeta} \, ;$

### Proof

Plugging the similarity solution expressions for $~M_r$ and $~\rho$ into the first of the three governing equations gives,

 $~\frac{\partial}{\partial r} \biggl[ \biggl(\frac{c_s^3 t}{G}\biggr) m(\zeta) \biggr]$ $~=$ $~4\pi r^2 \biggl[ \biggl(\frac{c_s^2 }{4\pi G r^2}\biggr) \Rho (\zeta) \biggr]$ $~\Rightarrow ~~~ (c_s t ) \frac{\partial}{\partial r} \biggl[ m(\zeta) \biggr]$ $~=$ $~\Rho (\zeta) \, .$

Plugging the similarity solution expressions for $~M_r$, $~\rho$, and $~v_r$ into the second of the three governing equations gives,

 $~\frac{\partial}{\partial t} \biggl[ \biggl(\frac{c_s^3 t}{G}\biggr) m(\zeta) \biggr]$ $~=$ $~- 4\pi r^2 \biggl[ \biggl(\frac{c_s^2 }{4\pi G r^2}\biggr) \Rho (\zeta) \biggr] \biggl[ -c_s U(\zeta)\biggr]$ $~\Rightarrow ~~~ \frac{\partial}{\partial t} \biggl[ t m(\zeta) \biggr]$ $~=$ $~\Rho (\zeta) U(\zeta)$ $~\Rightarrow ~~~ m(\zeta) + t \biggl[ \frac{\partial m(\zeta)}{\partial t} \biggr]$ $~=$ $~\Rho (\zeta) U(\zeta) \, .$

And, plugging the similarity solution expressions for $~M_r$, $~\rho$, and $~v_r$ into the third of the three governing equations gives,

 $~\frac{\partial }{\partial t} \biggl[ - c_s U(\zeta) \biggr] + \biggl[ - c_s U(\zeta) \biggr] \frac{\partial }{\partial r} \biggl[ - c_s U(\zeta) \biggr]$ $~=$ $~- c_s^2 \biggl[\biggl(\frac{c_s^2 }{4\pi G r^2}\biggr) \Rho (\zeta) \biggr]^{-1} \frac{\partial }{\partial r}\biggl[ \biggl(\frac{c_s^2 }{4\pi G r^2}\biggr) \Rho (\zeta) \biggr] - \frac{G}{r^2}\biggl[ \biggl(\frac{c_s^3 t}{G}\biggr) m(\zeta) \biggr]$ $~\Rightarrow ~~~ \frac{\partial }{\partial t} \biggl[ U(\zeta) \biggr] - c_s U(\zeta) \frac{\partial }{\partial r} \biggl[ U(\zeta) \biggr]$ $~=$ $~\biggl[ \frac{c_s r^2}{\Rho (\zeta)} \biggr]\frac{\partial }{\partial r}\biggl[ \biggl(\frac{\Rho (\zeta)}{r^2}\biggr) \biggr] + \biggl[ \frac{c_s^2 t}{r^2} \biggr] m(\zeta)$ $~\Rightarrow ~~~ \frac{\partial U}{\partial t} - (c_s U) \frac{\partial U}{\partial r}$ $~=$ $~\frac{c_s}{\Rho} \biggl[ \biggl( \frac{\partial \Rho}{\partial r}\biggr) -\frac{2\Rho}{r} \biggr] + \biggl[ \frac{c_s^2 t}{r^2} \biggr] m(\zeta) \, .$

Now, from the functional dependence of $~m(\zeta)$ on $~\Rho(\zeta)$ and $~U(\zeta)$, we have,

 $~\frac{\partial m}{\partial r}$ $~=$ $~ \biggl[ U + \frac{1}{\zeta} \biggr] \frac{\partial\Rho}{\partial r} + \Rho \biggl[ \frac{\partial U}{\partial r} - \frac{1}{\zeta^2} \frac{\partial \zeta}{\partial r}\biggr]$ $~=$ $~ \biggl[ U + \frac{1}{\zeta} \biggr] \frac{\partial\Rho}{\partial r} + \Rho \biggl[ \frac{\partial U}{\partial r} + \frac{1}{r \zeta} \biggr] \, ,$

and,

 $~\frac{\partial m}{\partial t}$ $~=$ $~ \biggl[ U + \frac{1}{\zeta} \biggr] \frac{\partial\Rho}{\partial t} + \Rho \biggl[ \frac{\partial U}{\partial t} - \frac{1}{\zeta^2} \frac{\partial \zeta}{\partial t}\biggr]$ $~=$ $~ \biggl[ U + \frac{1}{\zeta} \biggr] \frac{\partial\Rho}{\partial t} + \Rho \biggl[ \frac{\partial U}{\partial t} - \frac{1}{t \zeta} \biggr] \, .$

Hence, the first two governing equations become, respectively,

 $~\Rho$ $~=$ $~(r\zeta) \biggl\{ \biggl[ U + \frac{1}{\zeta} \biggr] \frac{\partial\Rho}{\partial r} + \Rho \biggl[ \frac{\partial U}{\partial r} + \frac{1}{r \zeta} \biggr] \biggr\}$ $~=$ $~ \biggl[ r\zeta U + r \biggr] \frac{\partial\Rho}{\partial r} + (r\zeta \Rho ) \frac{\partial U}{\partial r} + \Rho$ $~\Rightarrow~~~0$ $~=$ $~ \biggl[ \zeta U + 1\biggr] \frac{\partial\Rho}{\partial r} + (\zeta \Rho ) \frac{\partial U}{\partial r} \, ;$ $~\Rho (\zeta) U(\zeta)$ $~=$ $~ \Rho\biggl[ U + \frac{1}{\zeta}\biggr] + t \biggl\{ \biggl[ U + \frac{1}{\zeta} \biggr] \frac{\partial\Rho}{\partial t} + \Rho \biggl[ \frac{\partial U}{\partial t} - \frac{1}{t \zeta} \biggr] \biggr\}$ $~\Rightarrow ~~~0$ $~=$ $~ t \biggl\{ \biggl[ U + \frac{1}{\zeta} \biggr] \frac{\partial\Rho}{\partial t} + \Rho \biggl[ \frac{\partial U}{\partial t} \biggr] \biggr\}$ $~\Rightarrow ~~~0$ $~=$ $~ \biggl[ \zeta U + 1 \biggr] \frac{\partial\Rho}{\partial t} + (\zeta \Rho) \frac{\partial U}{\partial t} \, .$

Now, we can use these two relations to replace derivatives of $~\Rho$ with derivatives of $~U$ — or visa versa — in the third governing relation. In the first case, we obtain,

 $~ \frac{\Rho}{c_s} \biggl[\frac{\partial U}{\partial t} - (c_s U) \frac{\partial U}{\partial r} \biggr]$ $~=$ $~\biggl( \frac{\partial \Rho}{\partial r}\biggr) -\frac{2\Rho}{r} + \frac{\Rho^2}{r} \biggl[\zeta U + 1\biggr]$ $~=$ $~ \frac{\Rho^2(\zeta U + 1)}{r} -\frac{2\Rho}{r} - \biggl( \frac{\partial U}{\partial r}\biggr) \biggl[ \frac{\zeta \Rho}{(\zeta U + 1)} \biggr]$ $~ \Rightarrow ~~~ \frac{1}{r} \biggl[ \Rho^2(\zeta U + 1) - 2\Rho \biggr]$ $~=$ $~ \frac{\Rho}{c_s} \biggl[\frac{\partial U}{\partial t}\biggr] - (\Rho U) \frac{\partial U}{\partial r} + \biggl( \frac{\partial U}{\partial r}\biggr) \biggl[ \frac{\zeta \Rho}{(\zeta U + 1)} \biggr]$ $~ \Rightarrow ~~~ \biggl[ \Rho(\zeta U + 1) - 2 \biggr]$ $~=$ $~ \frac{r}{c_s} \biggl[\frac{\partial U}{\partial t}\biggr] + \biggl( \frac{\partial U}{\partial r}\biggr) \biggl[ \frac{r \zeta }{(\zeta U + 1)} - (rU) \biggr]$ $~=$ $~ \frac{t}{\zeta} \biggl[\frac{\partial U}{\partial t}\biggr] + r \biggl( \frac{\partial U}{\partial r}\biggr) \biggl[ \frac{\zeta - U (\zeta U + 1)}{(\zeta U + 1)} \biggr] \, .$

And, given that,

 $~\frac{\partial U}{\partial t}$ $~=$ $~\biggl( \frac{dU}{d\zeta} \biggr) \frac{\partial \zeta}{\partial t} = \biggl( \frac{dU}{d\zeta} \biggr)\frac{c_s}{r} = \biggl( \frac{dU}{d\zeta} \biggr)\frac{\zeta}{t} \, ;$ and $~\frac{\partial U}{\partial r}$ $~=$ $~ \biggl( \frac{dU}{d \zeta} \biggr) \frac{\partial \zeta}{\partial r} = - \frac{c_s t}{r^2} \biggl( \frac{dU}{d \zeta} \biggr) = -\frac{\zeta^2}{c_st} \biggl( \frac{dU}{d \zeta} \biggr) \, ,$

we can rewrite this as an ODE of the form,

 $~ \biggl[ \Rho(\zeta U + 1) - 2 \biggr]$ $~=$ $~ \biggl(\frac{d U}{d\zeta}\biggr) -\zeta \biggl( \frac{d U}{d\zeta }\biggr) \biggl[ \frac{\zeta - U (\zeta U + 1)}{(\zeta U + 1)} \biggr]$ $~\Rightarrow ~~~ [ \Rho(\zeta U + 1) - 2 ](\zeta U + 1)$ $~=$ $~ \biggl(\frac{d U}{d\zeta}\biggr) \biggl\{(\zeta U + 1) -\zeta \biggl[ \zeta - U (\zeta U + 1) \biggr]\biggr\}$ $~=$ $~ \biggl(\frac{d U}{d\zeta}\biggr) \biggl[ \zeta^2U^2 + 2\zeta U + 1 - \zeta^2 \biggr]$ $~ \Rightarrow ~~~ \frac{d U}{d\zeta}$ $~=$ $~ \frac{ [\Rho(\zeta U + 1) - 2 ](\zeta U + 1)}{ [ (\zeta U + 1)^2 - \zeta^2 ] } \, .$

In the second case, we obtain,

 $~\frac{c_s}{\Rho} \biggl( \frac{\partial \Rho}{\partial r}\biggr) + \frac{\zeta}{t} \biggl[ \Rho (\zeta U + 1 ) - 2\biggr]$ $~=$ $~ \frac{\partial U}{\partial t} - (c_s U) \frac{\partial U}{\partial r}$ $~=$ $~ -\biggl[ \frac{\zeta U +1}{\zeta \Rho} \biggr] \frac{\partial \Rho}{\partial t} - (c_s U) \biggl[ \frac{\zeta U + 1}{\zeta\Rho} \biggr] \frac{\partial \Rho}{\partial r}$ $~\Rightarrow ~~~ \frac{\zeta}{t} \biggl[ 2- \Rho (\zeta U + 1 ) \biggr]$ $~=$ $~ \biggl\{ \frac{c_s}{\Rho} + (c_s U) \biggl[ \frac{\zeta U + 1}{\zeta\Rho} \biggr] \biggr\}\frac{\partial \Rho}{\partial r} + \biggl[ \frac{\zeta U +1}{\zeta \Rho} \biggr] \frac{\partial \Rho}{\partial t}$ $~\Rightarrow ~~~ \frac{\zeta^2 \Rho}{c_s t} \biggl[ 2- \Rho (\zeta U + 1 ) \biggr]$ $~=$ $~ \biggl[ \zeta + U (\zeta U + 1 ) \biggr] \frac{\partial \Rho}{\partial r} + \frac{1}{c_s}\biggl[ \zeta U +1 \biggr] \frac{\partial \Rho}{\partial t} \, .$

And, given that,

 $~\frac{\partial \Rho}{\partial t}$ $~=$ $~\biggl( \frac{d\Rho}{d\zeta} \biggr) \frac{\partial \zeta}{\partial t} = \biggl( \frac{d\Rho}{d\zeta} \biggr)\frac{c_s}{r} = \biggl( \frac{d\Rho}{d\zeta} \biggr)\frac{\zeta}{t} \, ;$ and $~\frac{\partial \Rho}{\partial r}$ $~=$ $~ \biggl( \frac{d\Rho}{d \zeta} \biggr) \frac{\partial \zeta}{\partial r} = - \frac{c_s t}{r^2} \biggl( \frac{d\Rho}{d \zeta} \biggr) = -\frac{\zeta^2}{c_st} \biggl( \frac{d\Rho}{d \zeta} \biggr) \, ,$

we can rewrite this as an ODE of the form,

 $~ \frac{\zeta^2 \Rho}{c_s t} \biggl[ 2- \Rho (\zeta U + 1 ) \biggr]$ $~=$ $~ - \frac{\zeta^2}{c_s t} \biggl[ \zeta + U (\zeta U + 1 ) \biggr] \frac{\partial \Rho}{\partial \zeta} + \frac{\zeta}{c_s t}\biggl[ \zeta U +1 \biggr] \frac{\partial \Rho}{\partial \zeta}$ $~\Rightarrow~~~ \zeta\Rho [ 2- \Rho (\zeta U + 1 ) ]$ $~=$ $~ - \zeta\biggl[ \zeta + U (\zeta U + 1 ) \biggr] \frac{\partial \Rho}{\partial \zeta} + \biggl[ \zeta U +1 \biggr] \frac{\partial \Rho}{\partial \zeta}$ $~=$ $~ \biggl\{( \zeta U +1 ) - \zeta [ \zeta + U (\zeta U + 1 ) ] \biggr\}\frac{\partial \Rho}{\partial \zeta}$ $~=$ $~ \biggl\{\zeta U +1 -\zeta^2 - \zeta U (\zeta U + 1 ) \biggr\}\frac{\partial \Rho}{\partial \zeta}$

# See Especially

 © 2014 - 2021 by Joel E. Tohline |   H_Book Home   |   YouTube   | Appendices: | Equations | Variables | References | Ramblings | Images | myphys.lsu | ADS | Recommended citation:   Tohline, Joel E. (2021), The Structure, Stability, & Dynamics of Self-Gravitating Fluids, a (MediaWiki-based) Vistrails.org publication, https://www.vistrails.org/index.php/User:Tohline/citation