Magnetohydrostatic Configuration

In section 4.2, we obtained the maximum mass which is supported against the self-gravity using the virial analysis. In this section, we will survey result of more realistic calculation.

Formalism was obtained by Mouschovias (1976a,6), which was extended by Tomisaka, Ikeuchi, & Nakamura (1988) to include the effect of rotation. Magnetohydrostatic equlibrium is achived on a balance between the Lorentz force, gravity, thermal pressure force, and the centrifugal force as

\frac{1}{4\pi}\mbox{\boldmath${(\nabla\times B)\times B}$}
...p - \rho \mbox{\boldmath${\omega\times (\omega \times r)}$}=0.
\end{displaymath} (C.13)

In the axisymmetric case, the poloidal magnetic fields is obtained by the magnetic flux function, $\Phi$, or the $\phi$-component of the vector potential as
$\displaystyle B_z$ $\textstyle =$ $\displaystyle \frac{1}{r}\frac{\partial \Phi}{\partial r},$  
$\displaystyle B_r$ $\textstyle =$ $\displaystyle -\frac{1}{r}\frac{\partial \Phi}{\partial z},$  
$\displaystyle B_\phi$ $\textstyle =$ $\displaystyle \frac{b_\phi}{r}.$ (C.14)

Equation (C.13) leads to
$\displaystyle -\frac{1}{4\pi r^2}\Delta_1\Phi \frac{\partial \Phi}{\partial z}
-\frac{b_\phi}{4\pi r^2}\frac{\partial b_\phi}{\partial z}$ $\textstyle =$ $\displaystyle \rho \frac{\partial \psi}{\partial z}+\frac{\partial p}{\partial z},$ (C.15)
$\displaystyle -\frac{1}{4\pi r^2}\Delta_1\Phi \frac{\partial \Phi}{\partial r}
-\frac{b_\phi}{4\pi r^2}\frac{\partial b_\phi}{\partial r}$ $\textstyle =$ $\displaystyle \rho\frac{\partial \psi}{\partial r}+\frac{\partial p}{\partial r}-\rho \omega^2 r,$ (C.16)
$\displaystyle -\frac{\partial b_\phi}{\partial r}\frac{\partial \Phi}{\partial z}+\frac{\partial b_\phi}{\partial z}\frac{\partial \Phi}{\partial r}$ $\textstyle =$ $\displaystyle 0,$ (C.17)

\frac{\partial^2 \Phi}{\partial z^2}+r\f... r}\left(\frac{1}{r}\frac{\partial \Phi}{\partial r}\right).
\end{displaymath} (C.18)

Equation (C.17) indicates $b_\phi$ is a function of $\Phi$ as $b_\phi(\Phi)$, which is constant along one magnetic field line. Ferraro's isorotation law demands, that is, to satisfy the stead-state induction equation $\omega$ is constant along a magnetic field. This means $\omega$ is also constant along one magnetic field line, $\omega(\Phi)$. From this, the density distribution in one flux tube is written
\end{displaymath} (C.19)

This means $q$ is also constant along one magnetic field line, $q(\Phi)$ . Since the forces are expressed by the defrivative of function $q$
(\nabla q)\exp \left[ -\left(\psi-\frac{1}{2}r^2\omega^2\rig...
... \nabla \psi - \rho r \omega^2 -\rho r^2 \omega \nabla \omega,
\end{displaymath} (C.20)

where equation (C.19) is used, equation (C.15) and (C.16) are rewritten as
$\displaystyle \Delta_1\Phi \frac{\partial \Phi}{\partial z}+\frac{\partial (b_\phi^2/2)}{\partial z}$ $\textstyle =$ $\displaystyle -4\pi r^2\left\{\frac{\partial q}{\partial z}\exp\left[ -\left(\p...
...right)/c_s^2\right]+\rho r^2 \omega \frac{\partial \omega}{\partial z}\right\},$ (C.21)
$\displaystyle \Delta_1\Phi \frac{\partial \Phi}{\partial r}+\frac{\partial (b_\phi^2/2)}{\partial r}$ $\textstyle =$ $\displaystyle -4\pi r^2\left\{\frac{\partial q}{\partial r}\exp\left[ -\left(\p...
...right)/c_s^2\right]+\rho r^2 \omega \frac{\partial \omega}{\partial r}\right\}.$ (C.22)

Finally, using the fact that $b$, $q$, and $\omega$ are functions of $\Phi$, these two equations are reduced to
\Delta_1 \Phi=-\frac{d (b_\phi^2/2)}{d \Phi}
-4\pi r^2 \lef...
.../c_s^2\right]+\rho r^2 \omega \frac{d \omega}{d \Phi}\right\}.
\end{displaymath} (C.23)

Another equation to be coupled is the Poisson equation as
\Delta \psi=4\pi G \frac{q}{c_s^2}\exp\left[ -\left(\psi-\frac{1}{2}r^2\omega^2\right)/c_s^2\right].
\end{displaymath} (C.24)

The source terms of equations (C.23) and (C.24) are given by determining the mass $\delta m(\Phi)$ and the angular momentum $\delta L(\Phi)$ contained in a flux tube $\Phi$- $\Phi+\delta \Phi$. Mass and angular momentum distribution of
$\displaystyle \delta m(\Phi)$ $\textstyle =$ $\displaystyle 2 \int_0^{z_s(\Phi)} dz \int_{r(z,\Phi)}^{r(z,\Phi+\delta\Phi)}dr 2\pi r \rho(r,z),$ (C.25)
$\displaystyle \delta L(\Phi)$ $\textstyle =$ $\displaystyle 2 \int_0^{z_s(\Phi)} dz \int_{r(z,\Phi)}^{r(z,\Phi+\delta\Phi)}dr 2\pi r \rho(r,z)r^2\omega(\Phi)$ (C.26)

is chosen artitrary in nature, where $z_s(\Phi)$ is the the height of the cloud surface where the magnetic potential is equal to $\Phi$. For example, $m(\Phi)$ and $L(\Phi)$ are chosen as a uniformly rorating uniform-density spherical cloud threaded by uniform magnetic field. Since $\int_{r(\Phi)}^{r(\Phi+\delta\Phi)} 2\pi r dr=\delta \Phi \pi \partial r^2/\partial \Phi $
$\displaystyle q(\Phi)$ $\textstyle =$ $\displaystyle {\left(d m/d \Phi \right)}/{\int_0^{z_s(\Phi)} dz 2\pi \left(\par...{1}{c_s^2}\exp\left[ -\left(\psi-\frac{1}{2}r^2\omega^2\right)/c_s^2\right]},$ (C.27)
$\displaystyle \omega(\Phi)$ $\textstyle =$ $\displaystyle {\left(d L/d \Phi \right)}/{\int_0^{z_s(\Phi)} dz 2\pi \left(\par...
...2}q(\Phi)r^2\exp\left[ -\left(\psi-\frac{1}{2}r^2\omega^2\right)/c_s^2\right]},$ (C.28)

The source terms of PDEs [eqs (C.23) and (C.24)] are given from equations (C.27) and (C.28). While the functons $q(\Phi)$ and $\omega(\Phi)$ are determined from the solution of these PDEs after $m(\Phi)$ and $L(\Phi)$ are chosen. This can be solved by a self-consistent field method.

Kohji Tomisaka 2012-10-03