Chapter 4. Spherically Symmetric Flow

Basics

Let us consider systems in steady state (Parker's wind & Bondi accretion)

The EoC requires that $\dot M=4\pi r^2\rho u$ is a constant, which gives out the relation

$$\frac1\rho\frac{\text d\rho}{\text dr}+\frac2r+\frac1v\frac{\text dv}{\text dr}=0$$

The EoM gives

$$\begin{align*} v\frac{\text dv}{\text dr}&=-\frac1\rho\frac{\text dP}{\text dr}-\frac{GM_*}{r^2}\\ &=-\frac{c_s^2}\rho\frac{\text d\rho}{\text dr}-\frac{GM_*}{r^2} \end{align*}$$

Combine EoC and EoM to eliminate the density gradient $\text d\rho/\text dr$, we find

$$v\frac{\text dv}{\text dr}=\frac{2c_s^2}r+\frac{c_s^2}v\frac{\text dv}{\text dr}-\frac{GM_*}{r^2}$$

$$\Rightarrow \frac12\left(1-\frac{c_s^2}{u^2}\right)\frac{\text dv^2}{\text dr}=-\frac{GM_*}{r^2}\left(1-\frac{2c_s^2r}{GM_*}\right)$$

To solve this ODE, we apply the so-called trans-sonic condition, that is, $v=cs$ at $r=r\text c\equiv GM_*/2c_s^2$.

Isothermal Solution

In isothermal flow, $c_s$ is a constant, the solution is thus

$$\mathcal M^2-\ln \mathcal M^2=4\ln\xi+\frac4\xi-3$$

where $\mathcal M$ is the Mach number, and $\xi\equiv r/r_\text c$. The solutions are shown as below.

• I & II - unphysical

• III - supersonic

• IV - subsonic

• V - wind solution (Parker's wind)

• The other trans-sonic solution - accretion solution (Bondi accretion)

Accretion Solution

If the flow is not isothermal, we assume a polytropic equation of state $P=K\rho^\gamma$. Bernoulli's theorem for steady flow ensures that $\frac12u^2+h+\phi$ is a constant along each streamline, where $h$ is the specific enthalpy

$$h(p)\equiv\int\frac{\text dp}{\rho(p)}=\frac K{\gamma-1}\gamma\rho^{\gamma-1}=\frac{c_s^2}{\gamma-1}$$

and $\phi$ is the gravitational potential. Thus at radius $r$,

$$\frac12v^2+\frac{c_s^2}{\gamma-1}-\frac{GM_*}{r}=\frac{c_\infty^2}{\gamma-1}$$

where $c_\infty$ is the adiabatic sound speed at infinity. Here we have applied the infinite boundary condition, that is, $v\to0$ as $r\to\infty$.

Setting $v=c_s$ at

$$r=r_c\equiv\frac{GM_*}{2c_{s,\text c}},\quad c_{s,\text c}\equiv c_s(r_\text c)$$

as we are only interested in the trans-sonic solution. Consequently,

$$\begin{align*} r_\text c&=\frac{5-3\gamma}2\frac{GM_*}{2c_\infty^2}\\ c_{s,\text c}&=\left(\frac2{5-3\gamma}\right)^{1/2}c_\infty\\ \rho_\text c&=\left(\frac2{5-3\gamma}\right)^{1/(\gamma-1)}\rho_\infty \end{align*}$$

$\rho_\text c$ is calculated from the EoS.

Obviously, for $\gamma\ge5/3$, $r_\text c\le0$, so there is no trans-sonic solution for adiabatic (no radiative cooling) EoS. $\gamma=5/3$ is critical for spherical, point-mass accretion.

Mass Accretion Rate

$$\begin{align*} \dot M_B&=4\pi\rho r^2v=4\pi\rho_\text c r_\text c^2v_{s,\text c}\\ &=4\pi\left(\frac2{5-3\gamma}\right)^{\frac{5-3\gamma}{2(\gamma-1)}}\left(\frac{GM_*}{2c_\infty^2}\right)^2\rho_\infty c_\infty\quad\left(\propto \rho M_*^2T^{-3/2}\right)\\ &\equiv4\pi q(\gamma)R_B^2\rho_\infty c_\infty \end{align*}$$

where we define

$$q(\gamma)\equiv\frac14\left(\frac2{5-3\gamma}\right)^{\frac{5-3\gamma}{2(\gamma-1)}}$$

and the Bondi radius

$$R_B\equiv\frac{GM_*}{c_\infty^2}$$

• For diffusive ISM accretion onto a sun-like object

  $$\rho_\infty\sim10^{-24}\text{ g/cm}^3,\quad T\sim10^4\text{ K},\quad \gamma\sim1.4$$

  $$\Rightarrow \dot M_B\sim10^{-15}\left(\frac{M_*}{M_\odot}\right)^2M_\odot/\text{yr}$$

  This rate is extremely small. Even if we neglect the accretion feedback, a sun-like object can only accrete $\sim10^{-5}M_\odot$ in Hubble time.

• For denser and colder molecular cloud

  $$\rho_\infty\sim10^{-20}\text{ g/cm}^3,\quad T\sim10\text{ K},\quad \gamma\sim1.4$$

  $$\Rightarrow \dot M_B\sim3\times10^{-7}\left(\frac{M_*}{M_\odot}\right)^2M_\odot/\text{yr}$$

  In this way, the accretion timescale is quite interesting, as it only takes several million years for a sun-like object to double its mass. For a SMBH $\sim10^6M_\odot$, the timescale to reach double mass is only 0.3 yr.

We note that in reality, BH feedback suppresses the accretion somehow. Thus, $\dot M_B$ is the upper limit of the actual accretion rate. In this case,

$$\frac{\text dM_*}{\text dt}=AM_*^2$$

$$\Rightarrow\frac1{M_*}-\frac1{M_0}=-A(t-t_0)\Rightarrow M_*=\frac{M_0}{1-AM_0(t-t_0)}$$

The growth of $M*$ is so rapid that before $t$ goes to infinity, $M$ diverges within *Bondi timescale $t_B$

$$t_B\equiv\frac1{AM_0}$$

This is known as the super-exponential growth.

Accretion Flows

Radiative Luminosity

The luminosity caused by radiation can be estimated as

$$L_\text{acc}\sim\frac{\text dE_\text{g}}{\text dt}\sim\frac{\text d}{\text dt}\left(\frac{GM^2}{R}\right)\sim\frac{GM\dot M}{R}-\frac{GM^2}{R^2}\dot R$$

The second term is important only when there is significant orbital shrink in the system, such as binary spiral-in or stellar contraction. For steady accretion onto a point mass, it is negligable, so we can focus on the first term only.

• Neutron star

  $$M_*\sim M_\odot,\quad R_*\sim10\text{ km}\Rightarrow L_\text{acc}\sim\frac{GM_*}{R_*}\dot M\sim10^{38}\left(\frac{\dot M}{10^{-8}M_\odot/\text{yr}}\right)\text{ erg/s}$$
• Black hole

  $$L_\text{acc}\sim\frac{GM}{R_\text{ISCO}}\dot M$$

  where $R_\text{ISCO}$ is the innermost stable circular orbit. For a Schwarzschild black hole,

  $$R_\text{ISCO}=3R_{Sch}=\frac{6GM}{c^2}$$

  so the accretion rate is approximately

  $$L_\text{acc}\sim\frac16\dot Mc^2$$

We define the radiation efficiency $\eta$ so that

$$L_\text{acc}\equiv\eta\dot Mc^2$$

$\eta$ can be of the order $10^{-1}$ for black holes, but is negligible for less compact objects, such as normal stars.

Eddington Accretion Rate

For any astrophysical system, for materials to accrete, the gravity should exceed the radiation force, that is

$$\frac{GM}{r^2}\ge\frac{L}{4\pi r^2}\frac{\rho\kappa}{c}$$

Immediately, we notice that $L$ cannot exceed a critical luminosity $L_{Edd}$,

$$L\le L_{Edd}\equiv\frac{4\pi cGM}{\kappa}$$

$L_{Edd}$ is the famous Eddington luminosity. Assuming $\kappa$ is solely given by the electron scattering, the typical Eddington luminosity is

$$L_{Edd}\sim1.3\times10^{38}\left(\frac{M}{M_\odot}\right)\text{ erg/s}$$

Since $\dot M=Mc^2$, there is also a critical $\dot M_{Edd}$, known as the Eddington accretion rate, given as

$$\dot M\le\dot M_{Edd}\equiv\frac{4\pi GM}{\eta c\kappa}\propto M$$

Note that

$$\frac{\text dM}{\text dt}\propto BM\Rightarrow M=M_0e^{B(t-t_0)}$$

So Eddington accretion undergoes exponential growth.