Harvard Astronomy 201b

Ostriker’s 1964 Isothermal Cylinder model

In Uncategorized on April 14, 2013 at 4:43 pm

Particularly impressive is that, if you read the fine print, Jerry Ostriker was an NSF graduate fellow (meaning he was a grad student!) when he wrote this paper, which solves differential equations analytically at a level of technical virtuosity undoubtedly beyond anyone reared in the age of Mathematica.  Ostriker obtains solutions for polytropic cylinders, of which an isothermal cylinder is the case where n=\infty.  One has

P=K_n \rho ^{1+1/n},

which leads to the fundamental equation

K_n (n+1) \nabla ^2 \rho ^{1/n} = -4\pi G\rho.

Using the transformation

r \equiv \bigg[ \frac{(n+1)K_n}{4\pi G \lambda ^{1-1/n} } \bigg]^{1/2} \xi,

\rho \equiv \lambda \theta ^n,

one has a version of the Lane-Emden equation

\frac {1}{\xi}\frac{d}{d\xi} \big(\xi \frac{d\theta}{d\xi} \big) =\theta''+\frac{1}{\xi}\theta'=-\theta ^n.

For n=0, 1, and infinity there is a closed form solution, for other n a power series solution.  In the particular case of an isothermal cylinder, the EOS is that for an ideal gas, and one has

r\equiv \bigg[ \frac{K_I}{4\pi G \lambda} \big]^{1/2}\xi

and

\rho=\lambda e^{-\psi}.

Using the fundamental equation noted above and manipulating yields

\psi''+\frac{1}{\xi}\psi'=e^{-\psi}.

Letting z=-\psi +2\ln \xi and t=\sqrt{2}\ln\xi, we find

2\frac {d^2z}{dt^2}+e^z=0.

Letting z=\ln y the equation may be integrated to give

\psi(\xi)=2\ln\big(1+\frac{1}{8} \xi^2\big).

Using these results in expressions Ostriker provides in the paper that give general forms for density and mass, one has

\rho= \frac{\rho_0}{\big(1+\xi^2/8 \big)^2}

and M(\xi)=\frac{2 k_B T}{\mu m_0 G} \frac{1}{1+8/\xi^2}.

The expression for density should look familiar from the Pineda paper: it gets integrated to yield his eqn. (1) for the surface density of the cylinder.

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