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title: Priya Project

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# Priya Project
Accretion Disk: Accretion disk plays a very vital role in the growth and evolution of SMBH. In the accretion disk model the matter around the SMBH.

Accretion Disk is a Flatter astronomical object, made of rapidly rotating gas into a spiral onto a central gravitating body.

In the accretion disk, the high angular momentum of the rotating matter is gradually transported outward by stress.
This gradual loss of angular momentum allows matter to progressively move inward, toward the center of gravity.

Accretion disk physics is governed by a non-linear combination of many processes including gravity, hydrodynamics, viscosity, radiation, and magnetic fields.
accretion disk models
*Analytic Models
Depends upon 3 main components 
1 Geometry (vertically thin or thick)
2 Mass Supply Rate (sub-Eddington or Super-Eddington accretion rate)
3 Optical depth  (opaque or transparent)


{Standard Disk Model}


\{Geometrically thin and Optically thick}

Accretion is stationary and Axially symmetric.\
Vertical extension across the disc plane is much smaller than its radial (across the plane )extension.\

Disc structure depends on radial coordinates or may be described by ordinary differential equations. \
Accretion disk is supported by thermal pressure(gas+radiation).\

The accretion power is dissipated in the optically thick part of the disk .\

All disk quantities are just power laws as a function of distance to a black hole.\
The accretion rate is very sub-Eddington.\

Opacity is very high dominated by electron scattering. Gas goes down on tight spirals approximated by circular, free orbitals.\
High luminosity, and high efficiency of radiative cooling.


\{Basic Disk Parameters}
Optically thick, geometrically thin accretion disks; have conditions when large amounts of cold gas are falling into the center. the parameters which govern the properties of such systems are the accretion rate, the BH mass, and the BH spin. \

These parameters determine the geometry, gas temp, luminosity, and the emitted spectrum.\
We calculated the Accretion rate in units of $ \frac{l}{l_{edd}} $ Which refers to the normalized accretion rate. \

Consider the ring; with radius $r$ and radial thickness $dr$.\
Total mass,  $$ dm = 2\pi r \Sigma dr$$
$$ \Sigma = surface density per square centimeters $$
$$ \Sigma(r) = \rho(r) H(r) $$
$$ H(r) = disk height at radius r $$
\textbf{Continuity equation }
$$\frac{d\rho}{dt}+\nabla{.(\rho \Vec{V}}) = 0$$
in steady-state $\frac{\partial(\rho r)}{\partial t} = \frac{\partial H(r)}{\partial t } = \frac{\Sigma (r)}{\partial t} = 0$
$$\rho(r) = \frac{\Sigma(r)}{H(r)}$$
$$\frac{d}{dt}[\frac{\Sigma(r)}{H(r)}]+\frac{1}{r}\frac{\partial}{\partial r}[\frac{\Sigma(r)}{H(r)} V_{\Vec{r}}]= 0 $$
$$\frac{d \Sigma(t)}{dt} + \frac{1}{r} \frac{\partial}{\partial r}[\Sigma(r) V_{\Vec{(r)}}]=0$$
$$ 2\pi r \frac{d\Sigma}{dt} + \frac{\partial ( 2\pi r \Sigma v(r))}{\partial t}  = 0 $$
$$ r v(r) \rho H = constant = \frac{-\dot{M}}{4\pi}$$
Accretion rate , $$ \dot{M} = -2\pi r \rho v(r) 2 H$$
$$ \dot{M}= -2\pi r \Sigma v(r)= constant $$
Keplarian Angular Velocity =
$$ \Omega = \Omega_k = \frac{v_k}{r} = {(\frac{GM}{r^{3}})}^{1/2}$$
A major assumption in standard disk models is that the accreted gas moves inward slowly while retaining in circular orbital motion. 
$$ V_\phi(r) \approx V_k(r) $$
$$ V_r (r) << V_\phi(r) $$
Angular Momentum 
$$ S = 2 \pi r dr \Sigma r^{2}\Omega $$ Angular momentum of the gas in a ring of the unit radial thickness. 

\{Assumption} - The motion of particles of a ring is coupled to the motion of the gas particles just outside and just inside its location through the same kind of friction and viscosity.\\

The net radial change of the torque across the ring must be equal to the rate of change of the angular momentum of the ring.\\
$$\frac{\partial N}{\partial r}= \frac{\partial (2 \pi r \Sigma r^2 \Omega)}{\partial t} = \frac{\partial S}{\partial t} $$
$$=2\pi r V_r \Sigma \frac{\partial(r^2 \Omega)}{\partial r} $$ 
$$= -\dot{M}\frac{\partial (r^{2}r)}{\partial r}$$
Integration of dN along $N_2 -N_1$, where $N_2 = N(r+dr), N_1 = N(r)$
$$ \Omega = {(\frac{GM}{r^{3}})}^{1/2}$$
We get , $$=-\dot{M}[\frac{1}{2}{GM}^{1/2}r^{-1/2}]$$
$$N(r) = -\dot{M}{GMr}^{1/2}+ const$$

We carry the calculations inward over all rings down to the innermost radius of the disk.\\
There is no viscosity and torque from inside, $$N_1(r_{in})=0$$
$$N(r)= \dot{M}[[{GMr}^{1/2}]-[{GMr_{in}}^{1/2}]]f(r)$$
$$f(r) = 1 -(\frac{r_{in}}{r})^{1/2}$$
Gas inside $r_{in}$ is falling .\

For fast Rotating Kerr Black hole $a=0.998$ $r_{in} \approx 1.2 r_g$
For non-rotating  Kerr Black  hole $r_{in}=6r_g$
$r_{out}$ is difficult to define as it is a function of pressure and gravity at a large distance.\

\{Luminosity, Emissivity, and Temperature}


The local energy at the radius r is determined by the loss of gravitational energy of the inward-going material (+ve) and the work done by the torque on the exterior disk (+ve or -ve).\

Mechanical energy ${E_G}$
$$dE_g = \frac{1}{2}V^{2}dm-\frac{GMdm}{r}=-\frac{1}{2}\frac{GMdm}{r}$$
$V^{2}=\frac{GM}{r}$
$$dE_g = -\frac{1}{2}\frac{GM\dot{M}dt}{r}$$
$$L_G=\frac{dE_G}{dt} = -\frac{GM\dot{M}}{r} = -\frac{\dot{M}GM}{2r}$$
Energy-related to torque N as $$ dE_N = N d\theta$$
$$L_N = -\frac{dE_N}{dt}=-N\Omega = -\frac{\dot{M}GMf(r)}{r}$$
$$L_r = L_G+L_N$$
We get 
$$\frac{dL_r}{dr} = \frac{3GM\dot{M}f(r)}{2r^{2}}$$
Therefore $$L_r =-\frac{3GM\dot{M}}{2r} $$

\{Emissitivity per unit area }


The measure of the material's strength to emit infrared energy.
$$D(r) = \frac{1}{2}\frac{1}{2\pi r}\frac{dL_r}{dr} = \frac{1}{4\pi r}\frac{3GM\dot{M}f(r)}{2r^{2}}$$
$$ D(r) = \frac{3GM\dot{M}f(r)}{8\pi r^{3}}$$
$dL_r$ is released over the area of 2$\pi rdr$ where r = $1.36r_{in}$
\subsection{Radius Dependent disk Temperature }
$$ D(r) = \sigma T(r)^{4}$$
$$T(r) = (\frac{3Gm\dot{M}f(r)}{8\sigma \pi r^{3}})^{1/4}$$
Radius in units of $r_g$ & accretion rate in $\frac{\dot{M}}{\dot{M_{edd}}}$
\section{Luminosity, Emissivity, and Temperature}
The local energy at the radius r is determined by the loss of gravitational energy of the inward-going material (+ve) and the work done by the torque on the exterior disk (+ve or -ve).\\
Mechanical energy ${E_G}$
$$dE_g = \frac{1}{2}V^{2}dm-\frac{GMdm}{r}=-\frac{1}{2}\frac{GMdm}{r}$$
$V^{2}=\frac{GM}{r}$
$$dE_g = -\frac{1}{2}\frac{GM\dot{M}dt}{r}$$
$$L_G=\frac{dE_G}{dt} = -\frac{GM\dot{M}}{r} = -\frac{\dot{M}GM}{2r}$$
Energy-related to torque N as $$ dE_N = N d\theta$$
$$L_N = -\frac{dE_N}{dt}=-N\Omega = -\frac{\dot{M}GMf(r)}{r}$$
$$L_r = L_G+L_N$$
We get 
$$\frac{dL_r}{dr} = \frac{3GM\dot{M}f(r)}{2r^{2}}$$
Therefore $$L_r =-\frac{3GM\dot{M}}{2r} $$
\newpage
\{Emissitivity per unit area }
The measure of the material's strength to emit infrared energy.
$$D(r) = \frac{1}{2}\frac{1}{2\pi r}\frac{dL_r}{dr} = \frac{1}{4\pi r}\frac{3GM\dot{M}f(r)}{2r^{2}}$$
$$ D(r) = \frac{3GM\dot{M}f(r)}{8\pi r^{3}}$$
$dL_r$ is released over the area of 2$\pi rdr$ where r = $1.36r_{in}$

{Radius Dependent disk Temperature }
$$ D(r) = \sigma T(r)^{4}$$
$$T(r) = (\frac{3Gm\dot{M}f(r)}{8\sigma \pi r^{3}})^{1/4}$$
Radius in units of $r_g$ \& accretion rate in $\frac{\dot{M}}{\dot{M_{edd}}}$
\subsection{Expression for Normalised Temperature}
$$T (\frac{r}{r_g}) \propto M_g^{-1/4}[\frac{\dot{M}}{\dot{M_{edd}}}]^{1/4}[\frac{r}{r_g}]^{-3/4}f(r) $$

Newtonian approximation to the disk effective temp is 

$$ T(r) \approx 8.6 \times 10^{5} \dot{M_1}^{1/4} M_g^{-1/2} f(r) [\frac{r}{r_g}] ^{-3/4}K$$
 
\{Viscosity}
The Radial velocity $v_r(r)$ is must depend on the friction between adjacent rings, which is determined by the viscosity in the disk. 
\subsection{Rate of shearing(A)}
$$ A = r \frac{d\Omega}{dr} = r \frac{d[\sqrt{GM}r^{-3/2}]}{dr} = \frac{-3}{2}\frac{\sqrt{GM}}{\sqrt{r^{3}}}= \frac{-3\Omega}{2}$$
\subsection{ Absolute value of the viscous stress}
$$ V\rho A = V\frac{\Sigma}{H}A = \frac{3}{2} V \frac{\Sigma}{H} \Omega$$
Viscosity force per unit length $V\Sigma A$,
Torque, $$ N(r) = (2\pi r) (V\Sigma A) r = \dot{M}[GMr]^{1/2}f(r)$$
then ,  $$\dot{M}= \frac{-3\pi V\Sigma}{f(r)}$$
And , {$$ V(r) = \frac{3\nu}{2rf(r)}$$
Higher the viscosity , the larger is the radial velocity of the in-flowing gas. For a given accretion rate, a higher viscosity .\
Viscosity depends on a typical length (H) and a typical velocity (the sound speed ,$V_s$) in the disk. \
The assumption is that $$ \nu = {\alpha }\prime ` \nu_s H$$