MA Emilio Schmidt
Introduction
This master's thesis investigates the dynamics and stability of filamentary gas structures using numerical simulations. These filaments are of physical interest because they are common sites of star formation.
As a theoretical reference, this work builds on the analytically derived magnetohydrodynamic models of Fiege and Pudritz (2000a). They solved the stationary magnetohydrodynamic equations for cylindrical, self-gravitating filaments with helical magnetic fields and obtained equilibrium solutions from them. Building on these equilibrium solutions, they performed in Fiege and Pudritz (2000b) a linear stability analysis to investigate instabilities that occur, associated growth rates and characteristic wavelengths.
The aim of this work is to numerically reproduce the equilibrium solutions described by Fiege and Pudritz (2000a). These equilibria are then disturbed in order to investigate the temporal development of the disturbances and compare the results with those of the linear stability analysis by Fiege and Pudritz (2000b).
Theoretical Framework
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 + r \frac{\partial_r^2 \rho}{\rho} = - a r \rho \end{align}
Since the density is positive by definition, the following substitution can be made without restriction:
\begin{align} u(r) = \ln\left(\frac{\rho(r)}{\rho_0}\right). \end{align}
Here, \(\rho_0 = \text{const.} > 0\) is a constant reference density. Its specific value is irrelevant for further consideration. It only serves to ensure that the argument of the logarithm is dimensionless. The first two derivatives of the new quantity \(u\) are then given by:
\begin{align} \partial_r u = \frac{\partial_r \rho}{\rho} \quad \text{and} \quad \partial_r^2 u = \frac{\partial_r^2 \rho}{\rho} - \left(\frac{\partial_r \rho}{\rho}\right)^2 \end{align}
Inserting this into the differential equation under consideration yields:
\begin{align} \frac{\partial_r u}{r} + \partial_r^2 u = - a \rho_0 \exp(u) \end{align}
To simplify this equation further, another substitution is performed:
\begin{align} z = u + \sqrt{2} t \, , \quad \text{with} \, t = \sqrt{2} \ln\left(\sqrt{a \rho_0} r\right) \end{align}
With \(\partial_r t = \frac{\sqrt{2}}{r}\), the following applies for the first two derivatives of \(u\) with respect to \(r\):
\begin{align} \partial_r u = \frac{\sqrt{2}}{r} \partial_t u \quad \text{and} \quad \partial_r^2 u = - \frac{\sqrt{2}}{r^2} \partial_t u + \frac{2}{r^2} \partial_t^2 u \end{align}
Inserting yields:
\begin{align} \phantom{\Rightarrow}\;& 2 \partial_t^2 u = - a \rho_0 r^2 \exp(u) \\[6pt] \Leftrightarrow\;& 2 \partial_t^2 u = - \exp(z) \end{align}
Furthermore, the following applies:
\begin{align} \phantom{\Rightarrow}\;& \partial_t^2 z = \partial_t^2 u + \sqrt{2} \partial_t^2 t \\[6pt] \Leftrightarrow\;& \partial_t^2 z = \partial_t^2 u \end{align}
Thus, it follows that:
\begin{align} \phantom{\Rightarrow}\;& 2 \partial_t^2 z = - \exp(z) \\[6pt] \Rightarrow\;& 2 \left(\partial_t z\right) \left(\partial_t^2 z\right) = - \left(\partial_t z\right) \exp(z) \\[6pt] \Leftrightarrow\;& \partial_t\left(\left(\partial_t z\right)^2 + \exp(z)\right) = 0 \\[6pt] \Rightarrow\;& \left(\partial_t z\right)^2 + \exp(z) = c_1 \, , \quad c_1 \in \mathbb{R} \\[6pt] \Leftrightarrow\;& \partial_t z = \pm \sqrt{c_1 - \exp(z)} \end{align}
At first, it might seem that both signs need to be treated separately. However, it can actually be shown that both equations lead to the same solutions[1]. In the following, therefore, only the equation with a positive sign will be considered. To solve this, the following substitution is used:
\begin{align} z = \ln(y) \end{align}
This results in the following for the differential equation under consideration:
\begin{align} \frac{\partial_t y}{y} = \sqrt{c_1 - y} \end{align}
Separation of variables yields:
\begin{align} \phantom{\Rightarrow}\;& \frac{\text{d}y}{y\sqrt{c_1 - y}} = \text{d}t \\[6pt] \Rightarrow\;& \frac{1}{\sqrt{c_1}} \ln\left(\bigg|\frac{\sqrt{c_1 - y} - \sqrt{c_1}}{\sqrt{c_1 - y} + \sqrt{c_1}}\bigg|\right) = t + C_2 \\[6pt] \Leftrightarrow\;& \bigg|\frac{\sqrt{c_1 - y} - \sqrt{c_1}}{\sqrt{c_1 - y} + \sqrt{c_1}}\bigg| = \exp\left(\sqrt{c_1} t + \sqrt{c_1} C_2\right) \, , \quad \text{with} \, c_2 = \exp\left(\sqrt{c_1} C_2\right) \\[6pt] \Leftrightarrow\;& \bigg|\frac{\sqrt{c_1 - y} - \sqrt{c_1}}{\sqrt{c_1 - y} + \sqrt{c_1}}\bigg| = c_2 \exp\left(\sqrt{c_1} t\right) \, , \quad \text{with} \, A = c_2 \exp\left(\sqrt{c_1} t\right) \\[6pt] \Leftrightarrow\;& \bigg|\frac{\sqrt{c_1 - y} - \sqrt{c_1}}{\sqrt{c_1 - y} + \sqrt{c_1}}\bigg| = A \, , \quad \text{with} \, y = \exp(z) \Rightarrow \sqrt{c_1 - y} > \sqrt{c_1} \; \forall y \\[6pt] \Leftrightarrow\;& \frac{\sqrt{c_1} - \sqrt{c_1 - y}}{\sqrt{c_1} + \sqrt{c_1 - y}} = A \\[6pt] \Leftrightarrow\;& y = 4 c_1 \frac{A}{(1 + A)^2} \\[6pt] \Leftrightarrow\;& \exp(u) = \frac{4 c_1}{a \rho_0 r^2} \frac{A}{(1 + A)^2} \\[6pt] \Leftrightarrow\;& \rho = \frac{4 c_1 \rho_0}{a \rho_0 r^2} \frac{A}{(1 + A)^2} \end{align}
The quantity \(A\) can be written as follows:
\begin{align} \phantom{\Rightarrow}\;& A = c_2 \exp\left(\sqrt{c_1} t\right) \\[6pt] \Leftrightarrow\;& A = c_2 \exp\left(\sqrt{2 c_1} \ln(\sqrt{a \rho_0} r)\right) \\[6pt] \Leftrightarrow\;& A = c_2 (\sqrt{a \rho_0} r)^{\sqrt{2 c_1}} \end{align}
Define additionally:
\begin{align} b := \sqrt{2 c_1} \quad \text{und} \quad r_0 := \left(c_2^\frac{1}{b} \sqrt{a \rho_0}\right)^{-1} \end{align}
This yields the following general solution for the differential equation under consideration:
\begin{align} \phantom{\Rightarrow}\;& \rho(r) = \frac{2 b^2}{a r^2} \frac{\left(\frac{r}{r_0}\right)^b} {\left(1 + \left(\frac{r}{r_0}\right)^b\right)^2} \\[6pt] \Leftrightarrow\;& \rho(r) = \frac{2 b^2}{a r_0^2} \frac{\left(\frac{r}{r_0}\right)^{b-2}} {\left(1 + \left(\frac{r}{r_0}\right)^b\right)^2} \end{align}
The integration constants \(r_0\) and \(b\) must then be determined using appropriate boundary conditions. For the problem considered, it makes sense to specify boundary conditions for the center of the cylinder, i.e., for \(r = 0\). In the considered case, the density at the center of the cylinder must be finite and nonzero, since both a vanishing and a diverging density at the origin would lead to unphysical behavior:
\begin{align} \rho(r=0) = \rho_c \, , \quad \text{with} \quad 0 < \rho_c < \infty \end{align}
Furthermore, rotational symmetry and the requirement for a smooth, regular solution imply that the density profile at the origin should not have a gradient. Consequently, \(\rho(r)\) has an extremum at \(r=0\), and the following applies:
\begin{align} \partial_r \rho(r) \big|_{r = 0} = 0 \end{align}
It should be noted that the substitution \(t=\sqrt{2}\ln\big(\sqrt{a\rho_0}r\big)\) is only defined for \(r>0\). The differential equation was thus initially solved in the domain \(r \in \mathbb{R}\). In order to formulate boundary conditions at \(r=0\), the solution must be extended continuously to \(r=0\). Accordingly, the central values are understood as right-sided limits:
\begin{align} \rho(r = 0) = \lim\limits_{r \rightarrow 0} \rho(r) \quad \text{and} \quad \partial_r \rho(r) \big|_{r = 0} = \lim\limits_{r \rightarrow 0} \partial_r \rho(r) \end{align}
With the first boundary condition, the following then applies:
\begin{align} \phantom{\Rightarrow}\;& \lim\limits_{r \rightarrow 0} \rho(r) = \rho_c \\[6pt] \Leftrightarrow\;& \lim\limits_{r \rightarrow 0} \frac{2 b^2}{a r_0^2} \frac{\left(\frac{r}{r_0}\right)^{b-2}} {\left(1 + \left(\frac{r}{r_0}\right)^b\right)^2} = \rho_c \end{align}
Since the denominator converges to \(1\) in the limit case \(r\rightarrow 0\), the local behavior of the density in the immediate vicinity of the origin is determined exclusively by the numerator. This, together with the fact that the density at the origin must not vanish or diverge, directly implies that the following must hold:
\begin{align} b = 2 \end{align}
Inserting into the solution for the density profile provides:
\begin{align} \rho(r) = \frac{8}{a r_0^2} \frac{1} {\left(1 + \left(\frac{r}{r_0}\right)^2\right)^2} \end{align}
Furthermore, the first boundary condition implies that:
\begin{align} \phantom{\Rightarrow}\;& \rho_c = \frac{8}{a r_0^2} \\[6pt] \Leftrightarrow\;& r_0^2 = \frac{8}{a \rho_c} \end{align}
Inserting this gives the stationary density profile of an infinitely long, isothermal, self-gravitating gas cylinder:
\begin{align} \rho(r) = \frac{\rho_c}{\left(1 + \frac{1}{8} a \rho_c r^2\right)^2} \end{align}
It is noticeable that the second boundary condition is not explicitly included in the determination of the integration constants \(r_0\) and \(b\). The reason for this is that the requirement for a finite central density different from zero already restricts the solution so much that only the case \(b=2\) remains. Thus, the density profile takes the above form and automatically satisfies the second boundary condition. It is therefore not redundant, but is already implicitly contained in the regularity and symmetry requirements.
In this context, the total mass per unit length \(\lambda_{\text{total}}\) of an infinitely long, isothermal gas cylinder held together by its own gravity can be introduced:
\begin{align} \phantom{\Rightarrow}\;& \lambda_{\text{total}} = 2 \pi \int_{0}^{\infty} \rho(r) r \text{d} r \\[6pt] \Leftrightarrow\;& \lambda_{\text{total}} = \frac{8 \pi}{a} \\[6pt] \Leftrightarrow\;& \lambda_{\text{total}} = \frac{2 R T}{G M} \end{align}
The solution for the density profile can therefore also be written as follows:
\begin{align} \rho(r) = \frac{\lambda_{\text{total}}}{\pi r_0^2} \frac{1}{\left(1 + \left(\frac{r}{r_0}\right)^2\right)^2} \end{align}
With the determined density distribution, the gravitational potential induced by the density distribution can now be determined using Poisson's equation. The Poisson equation, using the given symmetry, is given by:
\begin{align} \phantom{\Rightarrow}\;& \frac{1}{r} \partial_r \left(r \partial_r \phi_\text{ext}(r)\right) = 4 \pi G \rho(r) \\[6pt] \Leftrightarrow\;& \partial_r \left(r \partial_r \phi_\text{ext}(r)\right) = 4 \pi G \rho(r) r \end{align}
Integration yields:
\begin{align} \phantom{\Rightarrow}\;& r \partial_r \phi_\text{ext}(r) = 4 \pi G \int_{0}^{r} \rho(r^\prime) r^\prime \text{d} r \\[6pt] \Leftrightarrow\;& r \partial_r \phi_\text{ext}(r) = \frac{32 \pi G}{a r_0^2} \int_{0}^{r} \frac{r^\prime}{\left(1 + \left(\frac{r^\prime}{r_0}\right)^2\right)^2} \text{d} r \\[6pt] \Leftrightarrow\;& r \partial_r \phi_\text{ext}(r) = \frac{16 \pi G}{a} \frac{\left(\frac{r}{r_0}\right)^2} {1 + \left(\frac{r}{r_0}\right)^2} \\[6pt] \Leftrightarrow\;& \partial_r \phi_\text{ext}(r) = \frac{16 \pi G}{a r_0} \frac{\frac{r}{r_0}}{1 + \left(\frac{r}{r_0}\right)^2} \end{align}
Another integration yields the gravitational potential of of an infinitely long, isothermal, self-gravitating gas cylinder:
\begin{align} \phantom{\Rightarrow}\;& \phi_\text{ext}(r) = \frac{16 \pi G}{a r_0} \int_{0}^{r} \frac{\frac{r^\prime}{r_0}}{1 + \left(\frac{r^\prime}{r_0}\right)^2} \text{d} r \\[6pt] \Leftrightarrow\;& \phi_\text{ext}(r) = \frac{8 \pi G}{a} \ln\left(1 + \left(\frac{r}{r_0}\right)^2\right) \\[6pt] \Leftrightarrow\;& \phi_\text{ext}(r) = G \lambda_{\text{total}} \ln\left(1 + \left(\frac{r}{r_0}\right)^2\right) \end{align}
- ↑ Switching sign just flips the sign of \(b\), and the final \(\rho\) is invariant wrt \(b\to -b\).
The case with a core
We now consider an infinitely long cylinder with radius \(R_{\text{min}}\), surrounded by an isothermal gas that is held together by gravity from the core on the one hand and by its self-gravity on the other. It can be shown that this system follows the same differential equation for density distribution as the case without a core. The starting point here is also the hydrostatic Euler equation. For this, the external force density or force \(\text{d} \boldsymbol{F}_{\text{ext}}\) of the gas, including the core, is required on a mass element \(\rho(r) \text{d} V\), which is given by:
\begin{align} \text{d} \boldsymbol{F}_{\text{ext}} = - \frac{2 G \lambda(r) \rho(r) \text{d} V}{r} \boldsymbol{e}_r \, , \quad \text{mit} \quad r > R_{\text{min}} \end{align}
Here, \(\lambda(r)\) is the mass per unit length enclosed up to radius \(r\). It is given by:
\begin{align} \phantom{\Rightarrow}\;& \lambda(r) = 2 \pi \int_{0}^{r} \rho(r^\prime) r^\prime \text{d} r^\prime \\[6pt] \Leftrightarrow\;& \lambda(r) = \lambda_{\text{core}} + 2 \pi \int_{R_{\text{min}}}^{r} \rho(r^\prime) r^\prime \text{d} r^\prime \end{align}
Here, \(\lambda_{\text{core}}\) is the mass per unit length of the core. Substituting this into the hydrostatic Euler equation gives:
\begin{align} \phantom{\Rightarrow}\;& \boldsymbol{\nabla} P = - \frac{2 G \rho}{r} \left(\lambda_{\text{core}} + 2 \pi \int_{R_{\text{min}}}^{r} \rho(r^\prime) r^\prime \text{d} r^\prime\right) \boldsymbol{e}_r \\[6pt] \Leftrightarrow\;& \frac{R T}{M} \boldsymbol{\nabla} \rho = - \frac{2 G \rho}{r} \left(\lambda_{\text{core}} + 2 \pi \int_{R_{\text{min}}}^{r} \rho(r^\prime) r^\prime \text{d} r^\prime\right) \boldsymbol{e}_r \end{align}
Due to the given cylinder symmetry, the following then applies:
\begin{align} \phantom{\Rightarrow}\;& \frac{R T}{M} \partial_r \rho = - \frac{2 G \rho}{r} \left(\lambda_{\text{core}} + 2 \pi \int_{R_{\text{min}}}^{r} \rho(r^\prime) r^\prime \text{d} r^\prime\right) \\[6pt] \Leftrightarrow\;& \frac{R T}{2 G M} \frac{r \partial_r \rho}{\rho} = - \left(\lambda_{\text{core}} + 2 \pi \int_{R_{\text{min}}}^{r} \rho(r^\prime) r^\prime \text{d} r^\prime\right) \end{align}
Differentiating with respect to \(r\) then yields:
\begin{align} \phantom{\Rightarrow}\;& \frac{R T}{2 G M} \partial_r \left(\frac{r \partial_r \rho}{\rho}\right) = - 2 \pi \rho r \\[6pt] \Leftrightarrow\;& \frac{\partial_r \rho}{\rho} - r \left(\frac{\partial_r \rho}{\rho}\right)^2 + r \frac{\partial_r^2 \rho}{\rho} = - a r \rho \end{align}
Therefore, the general solution of the system under consideration is given by the general solution of the system without the core:
\begin{align} \phantom{\Rightarrow}\;& \rho(r) = \frac{2 b^2}{a r_0^2} \frac{\left(\frac{r}{r_0}\right)^{b-2}} {\left(1 + \left(\frac{r}{r_0}\right)^b\right)^2} \\[6pt] \Leftrightarrow\;& \rho(r) = \frac{b^2 \lambda_{\text{total}}}{4 \pi r_0^2} \frac{\left(\frac{r}{r_0}\right)^{b-2}} {\left(1 + \left(\frac{r}{r_0}\right)^b\right)^2} \end{align}
A condition for the existence of physically meaningful density profiles can be derived from Euler's hydrostatic equation. Under the assumed symmetry, the following then applies:
\begin{align} \phantom{\Rightarrow}\;& \frac{R T}{M} \text{d}_r \rho = - \frac{2 G \lambda \rho}{r} \\[6pt] \Leftrightarrow\;& \frac{R T}{M} \text{d}_r \ln(\rho) = - \frac{2 G \lambda}{r} \end{align}
Applying the general solution and then performing the derivations yields:
\begin{align} \frac{R T}{M} \left(\frac{b - 2}{r} - \frac{2 b}{r} \frac{\left(\frac{r}{r_0}\right)^b}{1 + \left(\frac{r}{r_0}\right)^b}\right) = - \frac{2 G \lambda}{r} \end{align}
Since this equation holds for all \(r \in \mathbb{R}_{> 0}\), it must be satisfied in particular on the surface of the core. With \(\lambda(r = R_{\text{min}}) = \lambda_{\text{core}}\), the following then follows:
\begin{align} \phantom{\Rightarrow}\;& \frac{R T}{M} \left(\frac{b - 2}{R_{\text{min}}} - \frac{2 b}{R_{\text{min}}} \frac{\left(\frac{R_{\text{min}}}{r_0}\right)^b}{1 + \left(\frac{R_{\text{min}}}{r_0}\right)^b}\right) = - \frac{2 G \lambda_{\text{core}}}{R_{\text{min}}} \\[6pt] \Leftrightarrow\;& \frac{r_0}{R_{\text{min}}} = \left(\frac{b + 2 - 4 \frac{\lambda_{\text{core}}}{\lambda_{\text{total}}}} {b - 2 + 4 \frac{\lambda_{\text{core}}}{\lambda_{\text{total}}}}\right)^{\frac{1}{b}} \end{align}
Since the left side of the equation is non-negative by definition, it follows that for the exponent \(b\), the following must apply:
\begin{align} b > \biggl|4 \frac{\lambda_{\text{core}}}{\lambda_{\text{total}}} - 2\biggr| \end{align}
The gas mass per unit length \(\lambda_{\text{gas}}\) can then be expressed in terms of the mass per unit length of the core and the exponent. The gas mass per unit length is generally given by:
\begin{align} \phantom{\Rightarrow}\;& \lambda_{\text{gas}} = 2 \pi \int_{R_{\text{min}}}^{\infty} \rho(r) r \text{d} r \\[6pt] \Leftrightarrow\;& \lambda_{\text{gas}} = \frac{b^2 \lambda_{\text{total}}}{2 r_0^b} \int_{R_{\text{min}}}^{\infty} \frac{r^{b-1}} {\left(1 + \left(\frac{r}{r_0}\right)^b\right)^2} \text{d} r \\[6pt] \Leftrightarrow\;& \lambda_{\text{gas}} = \frac{b \lambda_{\text{total}}}{2} \frac{1}{1 + \left(\frac{r}{r_0}\right)^b} \end{align}
Using the expression for \(\frac{r_0}{R_{\text{min}}}\) derived above, it follows:
\begin{align} \phantom{\Rightarrow}\;& \lambda_{\text{gas}} = \frac{b + 2}{4} \lambda_{\text{total}} - \lambda_{\text{core}} \\[6pt] \Leftrightarrow\;& \frac{\lambda_{\text{gas}} + \lambda_{\text{core}}}{\lambda_{\text{total}}} = \frac{b + 2}{4} \end{align}
If the core is a numerical necessity, it is only there to emulate the gas within the core radius. Therefore, the case \(b = 2\) continues to be considered for this case. For this case, the core mass per unit length must be selected as follows, for a given core radius:
\begin{align} \phantom{\Rightarrow}\;& \lambda_{\text{core}} = 2 \pi \int_{0}^{R_{\text{min}}} \rho(r,b=2) r \text{d} r \\[6pt] \Leftrightarrow\;& \lambda_{\text{core}} = \frac{2 \lambda_{\text{total}}}{r_0^2} \int_{0}^{R_{\text{min}}} \frac{r}{\left(1 + \left(\frac{r}{r_0}\right)^2\right)^2} \text{d} r \\[6pt] \Leftrightarrow\;& \lambda_{\text{core}} = \frac{R_{\text{min}}^2} {R_{\text{min}}^2 + r_0^2}\lambda_{\text{total}} \end{align}
This results in the following for the gas mass per unit length in the system:
\begin{align} \lambda_{\text{gas}} = \frac{r_0^2} {R_{\text{min}}^2 + r_0^2}\lambda_{\text{total}} \end{align}
The case with a core (Lothar)
The solution to the ODE reads
\begin{align} \rho &= \frac{\lambda b^2(r/r_0)^{b-2}}{4\pi r_0^2(1+(r/r_0)^b)^2}\\[2ex] \text{with}\quad \lambda &= \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, \(\lambda\) 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 \(\lambda_\text{c}\) 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+2-4\lambda_\text{c}/\lambda}{b-2+4\lambda_\text{c}/\lambda}\right)^{1/b} ~, \end{align} which in turn introduces the constraint \(b>\vert 4\lambda_\text{c}/\lambda-2\vert\).
The resulting gas mass is \begin{gather} \int_R^\infty 2\pi r\rho(r,r_0,b)\,\text dr = \lambda\frac{b+2}{4}-\lambda_\text{c}\\ \Leftrightarrow\quad \frac{\lambda_\text{c}+\lambda_\text{gas}}{\lambda}=\frac{b+2}{4} ~. \end{gather} That is, the amount of gas can be tuned via \(b\). Note that for \(\lambda_\text{c}<\lambda/2\), 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 \(\lambda_\text{c}=\lambda_\text{gas}=\lambda/2\) which yields \(r_0=R\).
Free-fall time of an an infinitely long gas cylinder
In the following, the free fall time \(t_{\text{ff}}\) of an infinitely long gas cylinder is calculated for the previously derived density profile, with the cylinder initially at rest. The calculation is performed in the pressureless limit \(P \rightarrow 0\). The dynamics of the system are then determined solely by its own gravity.
To determine the free fall time, an infinitesimal cylindrical shell with radius \(r_{\text{i}}\) and mass \(\rho(r_i) \text{d} V\) is considered, whose gas, according to Newton's shell theorem, is gravitationally accelerated radially inwards by the mass enclosed by the cylindrical shell. Due to symmetry, the problem is reduced to a one-dimensional, purely radial motion. The equation of motion to be solved is therefore:
\begin{align} \phantom{\Rightarrow}\;& \rho(r_{\text{i}}) \text{d} V \frac{\text{d}^2 r}{\text{d} t^2} = - \frac{2 G \lambda(r_{\text{i}}) \rho(r_{\text{i}}) \text{d} V}{r} \\[6pt] \Leftrightarrow\;& \frac{\text{d}^2 r}{\text{d} t^2} = - \frac{2 G \lambda(r_{\text{i}})}{r} \end{align}
Here, \(\lambda(r_{\text{i}})\) is the mass enclosed by the cylinder shell per unit length, which is given by the following equation for the given density distribution:
\begin{align} \phantom{\Rightarrow}\;& \lambda(r_{\text{i}}) = 2 \pi \int_{0}^{r_{\text{i}}} \rho(r) r \text{d} r \\[6pt] \Leftrightarrow\;& \lambda(r_{\text{i}}) = \frac{2 \lambda_{\text{total}}}{r_0^2} \int_{0}^{r_{\text{i}}} \frac{r}{\left(1 + \left(\frac{r}{r_0}\right)^2\right)^2} \text{d} r \\[6pt] \Leftrightarrow\;& \lambda(r_{\text{i}}) = \frac{r_{\text{i}}^2} {r_{\text{i}}^2 + r_0^2}\lambda_{\text{total}} \end{align}
Inserting yields:
\begin{align} \phantom{\Rightarrow}\;& \frac{\text{d}^2 r}{\text{d} t^2} = - \frac{2 G \lambda_{\text{total}}}{r} \frac{r_{\text{i}}^2}{r_{\text{i}}^2 + r_0^2} \, , \quad \text{def.:} \, K := 2 G \lambda_{\text{total}} \frac{r_{\text{i}}^2}{r_{\text{i}}^2 + r_0^2} \\[6pt] \Leftrightarrow\;& \frac{\text{d}^2 r}{\text{d} t^2} = - \frac{K}{r} \, , \quad \text{with} \ v = \frac{\text{d} r}{\text{d} t} \, , \, \frac{\text{d} v}{\text{d} t} = v \frac{\text{d} v}{\text{d} r} \\[6pt] \Leftrightarrow\;& v \frac{\text{d} v}{\text{d} r} = - \frac{K}{r} \\[6pt] \Leftrightarrow\;& v \text{d} v = - \frac{K}{r} \text{d} r \\[6pt] \Rightarrow\;& \frac{1}{2}v^2 = - K \ln(r) + C \end{align}
From the initial conditions \(r(t=0) = r_{\text{i}}\) and \(v(t=0) = 0\), the following applies to the integration constant:
\begin{align} C = K \ln(r_i) \end{align}
Thus follows:
\begin{align} \phantom{\Rightarrow}\;& \frac{1}{2}v^2 = - K \ln\left(\frac{r_{\text{i}}}{r}\right) \\[6pt] \Leftrightarrow\;& v = \pm \sqrt{2 K \ln\left(\frac{r_{\text{i}}}{r}\right)} \end{align}
Since the collapse is directed inwards, only the negative branch of the solution for the velocity is relevant, which is why only this branch will be considered in the following:
\begin{align} \phantom{\Rightarrow}\;& v = - \sqrt{2 K \ln\left(\frac{r_{\text{i}}}{r}\right)} \\[6pt] \Leftrightarrow\;& \frac{\text{d} r}{\text{d} t} = - \sqrt{2 K \ln\left(\frac{r_{\text{i}}}{r}\right)} \end{align}
By integrating this differential equation, the time \(t(r, r_i)\) required for a particle to fall from its starting position \(r_i\) to a radius \(r\), is obtained:
\begin{align} \phantom{\Rightarrow}\;& \int_{0}^{t} \text{d} t^\prime = - \int_{r_{\text{i}}}^{r} \frac{1}{\sqrt{2 K \ln\left(\frac{r_{\text{i}}}{r^\prime}\right)}} \text{d} r^\prime \\[6pt] \Leftrightarrow\;& t(r,r_i) = \sqrt{\frac{\pi}{2 K}} r_{\text{i}} \operatorname{erf}\left(\sqrt{\ln\left(\frac{r_i}{r}\right)}\right) \\[6pt] \Leftrightarrow\;& t(r,r_i) = \frac{1}{2} \sqrt{\frac{\pi\left(r_{\text{i}}^2 + r_0^2\right)}{G \lambda_{\text{total}}}} \operatorname{erf}\left(\sqrt{\ln\left(\frac{r_i}{r}\right)}\right) \end{align}
Here, \(\operatorname{erf}(x)\) is the Gaussian error function, which is defined as follows:
\begin{align} \operatorname{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2} \text{d} t \end{align}
The limit \(r \rightarrow 0\) then yields the free fall time \(t_{\text{ff}}(r_i)\) of a particle that is at its starting position at time \(t = 0\):
\begin{align} \phantom{\Rightarrow}\;& t_{\text{ff}}(r_i) = \lim\limits_{r \rightarrow 0} t(r,r_i) \, , \quad \text{with} \lim\limits_{x \rightarrow \infty} \operatorname{erf}(x) = 1 \\[6pt] \Leftrightarrow\;& t_{\text{ff}}(r_i) = \frac{1}{2} \sqrt{\frac{\pi\left(r_{\text{i}}^2 + r_0^2\right)}{G \lambda_{\text{total}}}} \end{align}
It follows that the free fall time does not approach zero when the starting position approaches zero, but converges to a finite limit value:
\begin{align} \lim_{r_i \rightarrow 0} t_{\text{ff}}(r_i) = \frac{r_0}{2}\sqrt{\frac{\pi}{G\lambda_{\text{total}}}} \end{align}
For arbitrarily small but finite starting positions \(r_i>0\), the free fall time cannot become arbitrarily small, but approaches this finite value.
Work Plan and Approach
The following section presents a rough work plan for this master's thesis, which aims to numerically reproduce the solutions developed by Fiege and Pudritz. The initial goal is to simulate the purely hydrostatic reference case, the infinitely long, isothermal, self-gravitating cylinder without magnetic fields, in PLUTO/belt. Due to the cylindrical symmetry, the first tests can be carried out in a one-dimensional setup in cylindrical coordinates.
Afterwards, the numerical setup will be gradually expanded to two and then three dimensions. Once the three-dimensional setup in cylindrical coordinates is stable and works reliably, the plan is to switch to a three-dimensional setup in Cartesian coordinates. Cartesian coordinates are needed for the considered problem because the instabilities to be investigated later are expected to break the cylindrical symmetry.
Once the three-dimensional setup is in Cartesian coordinates, the next step will be to extend it to magnetohydrodynamic calculations. The aim here is to implement the equilibrium solutions of Fiege and Pudritz (2000a) in PLUTO/belt. Finally, the equilibrium configuration should be disturbed, analogous to Fiege and Pudritz (2000b), in order to investigate the temporal development of the perturbations and compare them with the predictions of their linear stability analysis.
1D Hydrodynamic Reference Test: Setup and Convergence
As explained above, the study starts with purely hydrodynamic, one-dimensional, isothermal simulations including self-gravity. The simulation setup is initialised according to an infinitely long, isothermal, self-gravitating cylinder with a core. Specifically, the density profile is set according to the previously derived density distribution, the velocity field is initialised to zero at the start, and the core mass is selected according to the relation given above. The temperature \(T\) and the characteristic radius \(r_0\) are selected based on typical values for observed filaments (cf. Seifried and Walch, 2015):
\begin{align} T = 10 \, \text{K} \quad \text{and} \quad r_0 = 0.033 \, \text{pc} \end{align}
For the initial simulations, a grid extending from \(R_{\min} = 10^{-4}\,\text{pc}\) to \(R_{\max} = 1\,\text{pc}\) is used. This ensures that the numerical domain completely encompasses the filament and that the outer region is sufficiently large to minimise boundary artefacts.
The grid is designed in such a way that the cells become exponentially larger towards the outside, according to:
\begin{align} \Delta r_i = r_{i-\frac{1}{2}} \left(10^{\Delta \xi} - 1\right) \, , \quad \text{with} \quad \Delta \xi = \frac{1}{N_r} \log_{10}\left(\frac{R_{\max}}{R_{\min}}\right) \end{align}
Here, \(\Delta r_i\) is the length of the \(i\)-th cell, \( r_{i-\frac{1}{2}}\) is the radius of the inner or left cell wall of the \(i\)-th cell, and \(N_r\) is the number of cells in the domain. This exponential growth of the domain ensures high resolution in the area of interest, i.e. in and around the filament, especially near the core, while a coarser resolution is sufficient further out, where in the case under consideration there are typically only very low densities or the filament transitions into an almost vacuum state. In this way, a good balance is achieved between accuracy in the relevant region and the highest possible numerical efficiency.
For the hydrodynamic variables, reflective boundary conditions are used at the inner edge of the domain. This is because the inner edge represents an impenetrable core and the reflective boundary condition prevents mass flow across the boundary surface. For the outer boundary, outflow–no-inflow boundary conditions are selected in order to model the domain as an open system. This allows matter, disturbances and waves to leave the domain, while preventing inflow from outside.
For the gravitational potential of the gas in the domain, a zero-gradient boundary condition is used at the inner edge. The reason for this is that the gravitational attraction of the core is considered separately in PLUTO/belt. Since gravitational potentials can be superimposed due to the linearity of Poisson's equation, the potential generated by the gas can then be linearly added to the potential generated by the core. At the outer boundary, the potential is specified by the analytical potential solution for an infinitely long, isothermal, self-gravitating cylinder without a core. This boundary condition is permissible in this setup because the core serves only as a numerical aid to emulate the case without a core.
The first simulation with this setup was performed with \(N_r = 1000\) cells in the radial direction. As a result, it can be seen that the maximum density in the system, which corresponds to the density in the innermost cell at the core due to the selected density profile, initially undergoes a damped oscillation. The maximum density then stabilises at a value of approximately \(\rho_{\max} = 3.36 \cdot 10^{-19}\,\frac{\text{g}}{\text{cm}^3}\). This corresponds to an upward deviation of about \(3\,\%\) from the initial and thus analytically expected value. This behaviour is illustrated in the following figure:
As a consistency check, both the radial density distribution \(\rho(r)\) and the gravitational potential \(\phi_{\text{ext}}(r)\) of the converged simulation are now considered and compared with the analytically expected solution. Both quantities are shown in the following figure, with the density distribution on the left and the corresponding gravitational potential on the right:
Due to conservation of mass, the increase in maximum density causes the density distribution to contract slightly. This is evident in the fact that the density peak increases while the characteristic radius \(r_0\) decreases. As a result, the profile becomes narrower overall and more concentrated towards the centre. The gravitational potential agrees very well with the analytically expected solution over the entire radial range. Minor differences can be explained by the fact that the gravitational potential is only uniquely determined up to an additive constant. A possible offset therefore corresponds merely to a choice of zero level and is physically uncritical. Overall, the comparison confirms that the numerically calculated potential resulting from the gas is consistent with the analytical reference solution.
It can be seen that the contraction of the density distribution is significantly weaker at higher resolutions. This can be seen in the following figure, which shows the density in the innermost cell, i.e. the first cell at the core, as a function of the number of cells \(N_r\) used:
Here, it can be seen that the density in the innermost cell converges towards the theoretically expected value as the number of cells increases. For \(N_r = 8000\), the maximum density deviates upwards by only about \(0.05\,\%\). In addition, the temporal evolution of the maximum density shows a rather monotonic increase with a single overshoot, followed by relaxation to the steady-state value without further significant oscillations. This behaviour is illustrated in the following figure. The left plot shows the density in the innermost cell as a function of time, and the middle plot shows the radial density distribution of the converged simulation as well as the theoretically expected density distribution used as the initial condition. For the sake of completeness, the right plot also shows the gravitational potential of the converged simulation together with the analytically expected potential solution. However, there is no qualitative or quantitative change in the potential compared to the case with \(N_r = 1000\) cells.
Free-fall time validation
Another consistency check is to verify whether the setup can reproduce the free fall time \(t_{\mathrm{ff}}\) in the two limiting cases \(r_i \rightarrow R_{\min}\) and \(r_i \rightarrow R_{\max}\). The analytical formula derived above yields the following expected values for these two limiting cases:
\begin{align} t_{\text{ff}}\left(r_i \rightarrow R_{\min}\right) = 1.07 \cdot 10^5\,\text{yr} \quad \text{and} \quad t_{\text{ff}}\left(r_i \rightarrow R_{\max}\right) = 3.26 \cdot 10^6\,\text{yr} \end{align}
To achieve this, the pressure must be switched off so that the dynamics of the system depend solely on its own gravity. To achieve this, the limiting case \(T \to 0\) is considered, because this then results in \(P \to 0\) from the isothermal equation of state. In the following simulations, a very low temperature of \(T = 10^{-3}\,\text{K}\) is used for this purpose, so that pressure forces are negligible compared to gravity. The total mass in the domain remains unchanged from the previous study.
In the following simulations, a grid with \(N_r = 8000\) cells is used to keep numerical resolution effects as small as possible. The only change compared to the previous study concerns the boundary condition at the inner boundary for the hydrodynamic variables. Outflow-no-inflow boundary conditions are used there instead of reflective ones. This is necessary because in this test, the inner boundary is supposed to act as a sink into which gas can flow, and not as an impenetrable wall as before. Inflow into the domain is prevented because such an inflow does not exist physically and otherwise mass could be artificially introduced into the domain due to numerical effects. All other boundary conditions for the hydrodynamic variables and for the gravitational potential remain unchanged.
In order to determine the free fall time in the two boundary cases from the simulation, suitable measurement variables are required that can be used to clearly identify the corresponding times. For this purpose, it is useful to consider the temporal course of the cumulative mass loss over the inner edge \(\Delta M_{R_{\min}}(t)\). This is defined by:
\begin{align} \Delta M_{R_{\min}}(t):= \int_{0}^{t} \frac{\mathrm{d} M_{R_{\min}}}{\mathrm{d} t^\prime}\mathrm{d} t^\prime \end{align}
Here, \(\frac{\mathrm{d} M_{R_{\min}}}{\mathrm{d} t}\) denotes the mass flow across the inner edge at time \(t\), i.e., the mass flowing out of the domain per unit time across \(R_{\min}\).
To determine the two times, the ratio of the cumulative mass loss \(\Delta M_{R_{\min}}(t)\) to the final cumulative mass loss \(\Delta M_{R_{\min}}(t_{\mathrm{end}})\) is considered. Here, \(t_{\mathrm{end}}\) denotes the time at which the simulation ends. The simulation is run for a sufficiently long time so that approximately the entire mass has flowed out of the domain via the inner boundary. Specifically, the two limiting cases \(r_i \rightarrow R_{\min}\) and \(r_i \rightarrow R_{\max}\) are defined by the first times at which the normalized cumulative mass loss exceeds a corresponding threshold. Ideally, the times could be defined as follows:
\begin{align} t_{\text{ff}}\left(r_i \rightarrow R_{\min}\right) = \min\left\{t : \frac{\Delta M_{R_{\min}}(t)}{\Delta M_{R_{\min}}(t_{\text{end}})} \geq 0\right\} \quad \text{and} \quad t_{\text{ff}}\left(r_i \rightarrow R_{\max}\right) = \min\left\{t : \frac{\Delta M_{R_{\min}}(t)}{\Delta M_{R_{\min}}(t_{\text{end}})} = 1\right\} \end{align}
However, these definitions are not numerically practical. They would require that in the time interval \([0,\,t_{\text{ff}}(r_i \rightarrow R_{\min})]\), in which no mass outflow is expected analytically, there is also no mass loss numerically, and that from \(t = t_{\text{ff}}(r_i \rightarrow R_{\max})\) onwards, the entire mass has completely flowed out. However, due to numerical effects, very small numerical outflows or residual masses typically occur. Therefore, the two times are defined using a small threshold \(\varepsilon \ll 1\):
\begin{align} t_{\text{ff}}\left(r_i \rightarrow R_{\min}\right) = \min\left\{t : \frac{\Delta M_{R_{\min}}(t)}{\Delta M_{R_{\min}}(t_{\text{end}})} \geq \varepsilon\right\} \quad \text{and} \quad t_{\text{ff}}\left(r_i \rightarrow R_{\max}\right) = \min\left\{t : \frac{\Delta M_{R_{\min}}(t)}{\Delta M_{R_{\min}}(t_{\text{end}})} \geq 1-\varepsilon\right\} \end{align}
Thus, \(t_{\text{ff}}(r_i \rightarrow R_{\min})\) is interpreted as the point in time at which mass outflow clearly becomes distinguishable from numerical noise, while \(t_{\text{ff}}(r_i \rightarrow R_{\max})\) marks the point in time at which the outflow is practically over.
In the following, the tolerance \(\varepsilon = 10^{-4}\) is chosen. This choice ensures that the numerically determined times describe, with good approximation, the point in time at which a measurable mass outflow begins, or the point in time at which the outflow is practically complete. The following figure shows the temporal evolution of the cumulative mass loss. In addition, vertical lines are drawn to mark the free fall times determined in this way for the two limiting cases.
The figure shows that the free fall times determined from the simulation for the two boundary cases considered correspond very well with the analytically expected times, especially considering that \(\varepsilon\) is a parameter that can be freely selected within certain limits. For the limiting case \(r_i \rightarrow R_{\min}\), the simulation yields \(t_{\text{ff}}\left(r_i \rightarrow R_{\min}\right) = 9.45 \cdot 10^4\,\text{yr}\), which corresponds to a deviation of about \(12\,\%\) from the analytical solution. For the limiting case \(r_i \rightarrow R_{\max}\), the simulation yields \(t_{\text{ff}}\left(r_i \rightarrow R_{\max}\right) = 3.12 \cdot 10^6\,\text{yr}\), which corresponds to a deviation of about \(4.2\,\%\) from the analytical value.
Overall, the results of this section, in combination with the results from the previous study, show that the hydrodynamic setup, including self-gravity, is numerically stable with the boundary conditions used. With a sufficiently high resolution, both the stationary profiles and the analytically expected free fall times are accurately reproduced in the boundary cases considered. This creates a reliable reference that serves as a starting point for the subsequent extension to two dimensions.
2D Hydrodynamic Reference Test: Stability
Next, the one-dimensional setup is expanded to a two-dimensional one. Since the previous study showed that the one-dimensional setup can reliably reproduce the analytical solution at sufficiently high resolution, the coarsest possible resolution, at which the one-dimensional simulation still runs stably and does not produce any obviously unphysical results, is deliberately chosen for the investigation now. This increases the numerical efficiency for the following tests. The aim of this study is not to reproduce the analytical solution again with maximum accuracy, but to verify the extension to two dimensions. In particular, the aim is to check whether the setup is fundamentally stable in 2D and whether or which numerical or implementation-related problems occur during the transition from one to two dimensions.
It appears that the coarsest grid at which the one-dimensional simulation still runs stably and does not produce any obviously unphysical results has \(N_r = 250\) cells in the radial direction. The figure below shows the density in the innermost cell as a function of time on the left. The middle plot shows the radial density distribution of the converged simulation and the theoretically expected density distribution. The right-hand plot additionally shows the gravitational potential of the converged simulation together with the analytically expected potential solution.
It can be seen that the simulation with \(N_r = 250\) cells in the radial direction converges faster in simulated physical time than the simulation with \(N_r = 1000\) cells. However, this is due to numerical effects. In this case, too, an increase in central density and, due to mass conservation, a decrease in the characteristic radius can be observed. As expected, this contraction is more pronounced than in the case of \(N_r = 1000\). In addition, the coarse resolution shows a slight deviation in the gravitational potential. For small radii, the numerical potential is slightly below the analytical solution. This is consistent with the contracted density distribution, as it means that more mass is located at small radii and the potential in the centre is correspondingly deeper.
For the two-dimensional simulations, a domain is considered that is identical to the previous study in the radial direction and extends in the azimuthal direction from \(\varphi_{\min}=0\) to \(\varphi_{\max}=2\pi\). The domain is discretised by an \(N_r \times N_{\varphi}\) grid. In the radial direction, as in the previous study, a grid with \(N_r = 250\) cells is used, with the cells increasing exponentially in size towards the outside. The number of cells in the azimuthal direction \(N_{\varphi}\) is chosen so that the cells are approximately square. This achieves a comparable resolution in the radial and azimuthal directions, which reduces anisotropic numerical errors and stronger dissipation in one direction. For locally square cells, the following must approximately apply:
\begin{align} r_i \Delta \varphi_i = \Delta r_i \end{align}
Here, \(r_i\) is the radius of the \(i\)-th cell centre, \(\Delta r_i\) is the radial width of the \(i\)-th cell, and \(\Delta\varphi\) is the azimuthal cell width. A uniform cell size is used for this in the azimuthal direction. The following therefore applies:
\begin{align} \phantom{\Rightarrow}\;& \Delta \varphi_i = \Delta \varphi = \frac{\varphi_{\max} - \varphi_{\min}}{N_{\varphi}} \\[6pt] \Leftrightarrow\;& \Delta \varphi = \frac{2 \pi}{N_{\varphi}} \end{align}
Inserting yields:
\begin{align} \frac{2 \pi r_i}{N_{\varphi}} = r_{i-\frac{1}{2}} \left(10^{\Delta \xi} - 1\right) \end{align}
Since \(r_i\) describes the radius of the cell centre and \(r_{i-\frac12}\) describes the radius of the inner cell edge, the following applies:
\begin{align} r_i = r_{i-\frac12} + \frac{\Delta r_i}{2}\, . \end{align}
If the grid has a sufficiently fine resolution, i.e. \(\Delta r_i \ll r_i\), then the difference between \(r_i\) and \(r_{i-\frac12}\) is only a small fraction of \(r_i\). Therefore, as a first approximation, we can assume that the following applies:
\begin{align} r_{i-\frac12} \approx r_i \, . \end{align}
Thus follows:
\begin{align} \phantom{\Rightarrow}\;& \frac{2 \pi}{N_{\varphi}} = 10^{\Delta \xi} - 1 \\[6pt] \Leftrightarrow\;& N_{\varphi} = \frac{2 \pi}{10^{\Delta \xi} - 1} \end{align}
With \(N_r = 250\), \(R_{\min} = 10^{-4}\,\text{pc}\) and \(R_{\max} = 1\,\text{pc}\), the following applies:
\begin{align} N_{\varphi} = \frac{2 \pi}{10^{\frac{2}{125}} - 1} \approx 167 \end{align}
Therefore, a \(250 \times 167\) grid will be used in the following.
For the hydrodynamic quantities and the gravitational potential, the same boundary conditions are used in the radial direction as in the previous study. In the azimuthal direction, periodic boundary conditions are chosen for both the hydrodynamic quantities and the gravitational potential. This is justified by the angular geometry, since \(\varphi=0\) and \(\varphi=2\pi\) describe the same physical location.
It can be seen that the simulation runs stable without any new numerical or implementation-related problems occurring. The following figure shows the time evolution of the maximum density of the two-dimensional simulation in the domain. For comparison, the time evolution of the one-dimensional simulation with \(N_r = 250\) cells is also shown.
The figure shows that the temporal evolution of the maximum density in the two-dimensional simulation is practically identical to that in the one-dimensional simulation. In particular, it initially shows a damped oscillation and then approaches a steady state value. From about \(t \sim 3 \cdot 10^8\,\text{yr}\) onwards, the oscillation has completely damped out.
The following figure shows both the density and the gravitational potential of the converged simulation as a function of the Cartesian coordinates \(x = r \sin(\varphi)\) and \(y = r \cos(\varphi)\). For the density representation, the central area of the domain was zoomed in on more than for the gravitational potential, since the density is essentially concentrated in the filament and no longer exhibits any relevant structures at larger radii.
The figure confirms that the two-dimensional simulation maintains the expected cylindrical symmetry. Both the density and the gravitational potential are purely radially dependent and show no systematic \(\varphi\)-dependence. In particular, the density peak remains centred and the isolines run circularly, indicating that the periodic boundary conditions are consistently implemented in the azimuthal direction and no artefacts arise at \(\varphi = 0\) or \(\varphi = 2\pi\).
This observation can also be confirmed by the following figure. On the left-hand side, the radial profiles are shown for five different azimuth angles. The density is shown at the top and the gravitational potential at the bottom. On the right-hand side, the corresponding profiles are shown for an azimuth angle close to \(\varphi \approx 2\pi\), together with a fit and the analytically expected solution, with the density distribution at the top and the gravitational potential at the bottom.
The figure confirms that the converged two-dimensional simulation remains cylindrically symmetric. On the left, it can be seen that the radial profiles for the different azimuth angles completely overlap, both for density and for gravitational potential. This means that there are no systematic \(\varphi\)-dependencies, and in particular, even close to \(\varphi \approx 2\pi\), there are no signs of artefacts caused by the periodic boundary conditions.
As in the one-dimensional case with coarse resolution, the density profile continues to show a contraction. The potential agrees very well with the analytical expectation over large radii. The small deviations in the inner region are consistent with the changed mass distribution. It should also be noted that the potential is basically only determined up to an additive constant, so that a possible offset is physically uncritical. Overall, both the temporal evolution of the maximum density and the fit parameters agree with the corresponding one-dimensional reference run at the same radial resolution, so that the extension to two dimensions does not introduce any additional systematic effects.
Overall, this two-dimensional reference test shows that the hydrodynamic setup, including self-gravity and periodic boundary conditions in the \(\varphi\) direction, is stable and consistently reproduces the corresponding one-dimensional case, thus providing the basis for subsequent extension to higher dimensions.