MA Emilio Schmidt

The isothermal self-gravitating cylinder

Sought is the stationary density profile \(\rho\) of an infinitely long, isothermal, self-gravitating gas cylinder. To do this, the hydrostatic Euler equation has to be solved, which is given by:

\begin{align} \boldsymbol{\nabla} P = \boldsymbol{f}_\text{ext} \end{align}

Here, \(P\) is the pressure field and \(\boldsymbol{f}_\text{ext}\) is the force density acting on the gas. In the considered case, this is the gas's own gravity. Since the gravitational field is a conservative force field, it can be expressed by a potential \(\phi_\text{ext}\):

\begin{align} \boldsymbol{f}_\text{ext} = - \rho \boldsymbol{\nabla} \phi_\text{ext} \end{align}

Inserting yields:

\begin{align} \boldsymbol{\nabla} P = - \rho \boldsymbol{\nabla} \phi_\text{ext} \end{align}

The pressure field of the gas under consideration can be described by the equation of state of an isothermal gas, which is given by:

\begin{align} P = \frac{R T}{M} \rho \end{align}

Here, \(R\) denotes the universal gas constant, \(T\) the spatially homogeneous temperature distribution, and \(M\) the molar mass of the gas. Inserting into the hydrostatic Euler equation yields:

\begin{align} \phantom{\Rightarrow}\;& \frac{R T}{M} \boldsymbol{\nabla} \rho = - \rho \boldsymbol{\nabla} \phi_\text{ext} \\[6pt] \Leftrightarrow\;& \frac{1}{\rho} \boldsymbol{\nabla} \rho = - \frac{M}{R T} \boldsymbol{\nabla} \phi_\text{ext} \\[6pt] \Leftrightarrow\;& \boldsymbol{\nabla} \ln(\rho) = - \frac{M}{R T} \boldsymbol{\nabla} \phi_\text{ext} \\[6pt] \Rightarrow\;& \Delta \ln(\rho) = - \frac{M}{R T} \Delta \phi_\text{ext} \end{align}

It proves useful to use the Poisson equation of Newtonian gravity, which is given by:

\begin{align} \Delta \phi_\text{ext} = 4 \pi G \rho \end{align}

Inserting yields:

\begin{align} \phantom{\Rightarrow}\;& \Delta \ln(\rho) = - \frac{4 \pi G M}{R T} \rho \, , \quad \text{def.:} \, a := \frac{4 \pi G M}{R T} \\[6pt] \Leftrightarrow\;& \Delta \ln(\rho) = - a \rho \end{align}

For an infinitely long, straight, self-gravitating gas cylinder without an externally imposed azimuthal or axial structure, there is symmetry with respect to rotation about the cylinder axis and translation along the axis. The underlying equations share this symmetry, which is why corresponding derivatives disappear in the considered equation:

\begin{align} \phantom{\Rightarrow}\;& \frac{1}{r} \partial_r \big(r \partial_r \ln(\rho)\big) = - a \rho \\[6pt] \Leftrightarrow\;& \partial_r \bigg(r \frac{\partial_r \rho}{\rho}\bigg) = - a r \rho \\[6pt] \Leftrightarrow\;& \frac{\partial_r \rho}{\rho} - r \bigg(\frac{\partial_r \rho}{\rho}\bigg)^2 + \frac{\partial_r^2 \rho}{\rho} = - a r \rho \\[6pt] \Leftrightarrow\;& \rho \partial_r \rho - r (\partial_r \rho)^2 + r \rho \partial_r^2 \rho = - a r \rho^3 \end{align}

To be continued ...

The case with a core

The solution to the ODE reads

\begin{align} \rho &= \frac{\tilde M b^2(r/r_0)^{b-2}}{4\pi r_0^2(1+(r/r_0)^b)^2}\\[2ex] \text{with}\quad \tilde M &= \frac{2k_\text{B}T}{Gm} = \int_0^\infty 2\pi r\rho(r,r_0,b{=}2)\,\text dr \end{align} being the only possible total mass/length in the case of gas only, where \(b=2\) is necessary for zero force at \(r=0\). Notably, \(\tilde M\) does not depend on \(r_0\), i.e. every width of \(\rho\) is equally allowed. The reason is that no length scale can be constructed from the input parameters \(k_\text{B}T\), \(G\) and \(m\).

This changes in the presence of a core with mass/length \(M_\text{c}\) and radius \(R\). Vanishing force at \(r=0\) is replaced by force balance at \(r=R\), which leads to the condition \begin{align} \frac{r_0}{R} &= \left(\frac{b+2-4M_\text{c}/\tilde M}{b-2+4M_\text{c}/\tilde M}\right)^{1/b} ~, \end{align} which in turn introduces the constraint \(b>\vert 4M_\text{c}/\tilde M-2\vert\).

The resulting gas mass is \begin{gather} \int_R^\infty 2\pi r\rho(r,r_0,b)\,\text dr = \tilde M\frac{b+2}{4}-M_\text{c}\\ \Leftrightarrow\quad \frac{M_\text{c}+M_\text{gas}}{\tilde M}=\frac{b+2}{4} ~. \end{gather} That is, the amount of gas can be tuned via \(b\). Note that for \(M_\text{c}<\tilde M/2\), it cannot be chosen arbitrarily small.

If the core is actually a numerical necessity and only meant to emulate the gas contained inside, it's \(b=2\) again, and the most convenient choice is \(M_\text{c}=M_\text{gas}=\tilde M/2\) which yields \(r_0=R\).