# Power-Law Density Distributions

Here we begin with the same second-order, one-dimensional ODE that governs the structure of polytropic spheres, namely, the

Lane-Emden Equation
$\frac{1}{\xi^2} \frac{d}{d\xi}\biggl( \xi^2 \frac{d\Theta_H}{d\xi} \biggr) = - \Theta_H^n$ ,

and examine whether or not this governing relation can be satisfied by a power-law enthalpy distribution of the form,

$\Theta_H = A \xi^{-\alpha} ,$

where $A$ and $\alpha$ are assumed to be constants. We note, up front, that such a solution will not satisfy the boundary conditions that are imposed on polytropic spheres. But the simplistic form of a power-law solution can nevertheless sometimes be instructive.

## Derivation

Plugging the power-law expression for the dimensionless enthalpy into both sides of the Lane-Emden equation gives,

$-\alpha (1 -\alpha) A \xi^{-(2 +\alpha)} = - A^n \xi^{-\alpha n} . $

Hence, the power-law enthalpy distribution works as long as,

$\alpha = \frac{2}{n-1} ~~~~~~\mathrm{and}~~~~~~ A = [\alpha (1 -\alpha)]^{1/(n-1)} = \biggl[ \frac{2(n-3)}{(n-1)^2} \biggr]^{1/(n-1)}.$

This means that hydrostatic balance can be established at all radial positions within a spherically symmetric configuration for power-law density distributions of the form,

$\frac{\rho}{\rho_c} = \biggl[ \frac{2(n-3)}{(n-1)^2} \biggr]^{n/(n-1)} \xi^{- 2n/(n-1)}.$

(Note that, in this case, the subscript c should not represent the central conditions but, rather, conditions at some characteristic radial position within the configuration.)

## Examples

It looks like the derived solution makes some physical sense only for polytropic indices $n > 3$. For $n=4$, the relevant power-law density distribution is,

$\frac{\rho}{\rho_c} = \biggl[ \frac{2}{9} \biggr]^{4/3} \xi^{- 8/3}.$

For $n=(3+\epsilon)$ and $\epsilon \ll 1$,

$\frac{\rho}{\rho_c} \approx \biggl[ \frac{\epsilon}{2} \biggr] \xi^{- 3}.$

For $n \gg 1$,

$\frac{\rho}{\rho_c} \approx \biggl[ \frac{2}{n} \biggr] \xi^{- 2}.$

Hence, for polytropic indices in the range $\infty > n > 3$, the relevant power-law density distribution lies between $\rho \propto \xi^{-2}$ and $\rho \propto \xi^{-3}$.

## Isothermal Equation of State

Suppose the gas is isothermal so that the relevant equation of state is,

$P = c_s^2 \rho ,$

where $c_s$ is the sound speed. To determine what power-law density distribution will satisfy hydrostatic equilibrium in this case, it is better to return to the original statement of hydrostatic balance for spherically symmetric configurations,

$\frac{1}{\rho} \frac{dP}{dr} = -\frac{d\Phi}{dr} .$

Plugging in the isothermal equation of state and assuming a radial density distribution of the form,

$\rho(r) = \rho_0 \biggl( \frac{r}{r_0} \biggr)^{-\beta} ,$

we obtain,

$\frac{d\Phi}{dr} = \beta \biggl(\frac{c_s^2}{r}\biggr) .$

Therefore, the Poisson equation gives,

$\frac{1}{r^2}\frac{d}{dr}\biggl[r^2 \frac{d\Phi}{dr}\biggr] = \beta \biggl(\frac{c_s^2}{r^2}\biggr)= 4\pi G \rho_0 \biggl( \frac{r}{r_0} \biggr)^{-\beta} .$

This relation can be satisfied only if,

$\beta = 2 ~~~~~\mathrm{and}~~~~~ \rho_0 = \frac{c_s^2}{2\pi G r_0^2} .$

Hence, hydrostatic balance can be achieved for an isothermal gas with a power-law density distribution of the form, $\rho \propto r^{-2}$. Because an isothermal $P(\rho)$ equation of state is obtained by setting $n = \infty$ in the more general polytropic equation of state, the result just derived is consistent with the above, more general analysis which showed that, for values of the polytropic index $n \gg 1$, the equilibrium power-law density distribution tends toward a $\rho \propto r^{-2}$ distribution.

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