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Shan K
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I was studying about Accretion disks and found some difficulties regarding some concepts. My questions are
1. For an accretion disk, ## v_r<<c_s ## Why?
2. The conservation of mass equation is derived as,
The mass in an annulus of radius ##\Delta r## is ##2\pi r \Delta r\Sigma ## and it is changing due to the mismatch of mass outflow and inflow which is written as,
$$ \frac{\partial}{\partial t}(2πr\Delta r\Sigma) = v_r(r,t)2πr\Sigma(r,t)−v_r(r+\Delta r,t)2π(r+\Delta r)\Sigma(r+\Delta r,t)\\ \approx-2\pi\Delta r\frac{\partial (r\Sigma v_r)}{\partial r} $$
up to this I have no problem but it further says that this equation in the ## \Delta r\rightarrow0 ## limit gives,
$$r\frac{\partial\Sigma}{\partial t}+\frac{\partial (r\Sigma v_r)}{\partial r}=0$$
and I can't able to derive how that limit gives that equation.
3. The angular momentum conservation equation is derived from the Navier-Stokes equation as,
$$ \frac{\partial \vec{v}}{\partial t}+(\vec{v}.\vec{\nabla})\vec{v}=-\frac{1}{\rho}(\vec{\nabla}P-\vec{\nabla}.\sigma)-\vec{\nabla}\Phi $$
In the axisymmetry assumption and no momentum loss or gain from the ## z ## direction one gets,
$$\Sigma\Big(\frac{\partial v_\phi}{\partial t}+\frac{v_rv_\phi}{r}+v_r\frac{\partial v_\phi}{\partial r}\Big)=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 T_{r\phi}) $$
Now my question is, shouldn't the RHS be ## \frac{1}{r}\frac{\partial}{\partial r}(r T_{r\phi}) ## instead of ## \frac{1}{r^2}\frac{\partial}{\partial r}(r^2 T_{r\phi}) ## because the ## r ## part of ## \vec{\nabla}.\vec{f}## in the cylindrical coordinate is ## \frac{1}{r}\frac{\partial}{\partial r}(r f_r) ##.
Thanks in advance..
1. For an accretion disk, ## v_r<<c_s ## Why?
2. The conservation of mass equation is derived as,
The mass in an annulus of radius ##\Delta r## is ##2\pi r \Delta r\Sigma ## and it is changing due to the mismatch of mass outflow and inflow which is written as,
$$ \frac{\partial}{\partial t}(2πr\Delta r\Sigma) = v_r(r,t)2πr\Sigma(r,t)−v_r(r+\Delta r,t)2π(r+\Delta r)\Sigma(r+\Delta r,t)\\ \approx-2\pi\Delta r\frac{\partial (r\Sigma v_r)}{\partial r} $$
up to this I have no problem but it further says that this equation in the ## \Delta r\rightarrow0 ## limit gives,
$$r\frac{\partial\Sigma}{\partial t}+\frac{\partial (r\Sigma v_r)}{\partial r}=0$$
and I can't able to derive how that limit gives that equation.
3. The angular momentum conservation equation is derived from the Navier-Stokes equation as,
$$ \frac{\partial \vec{v}}{\partial t}+(\vec{v}.\vec{\nabla})\vec{v}=-\frac{1}{\rho}(\vec{\nabla}P-\vec{\nabla}.\sigma)-\vec{\nabla}\Phi $$
In the axisymmetry assumption and no momentum loss or gain from the ## z ## direction one gets,
$$\Sigma\Big(\frac{\partial v_\phi}{\partial t}+\frac{v_rv_\phi}{r}+v_r\frac{\partial v_\phi}{\partial r}\Big)=\frac{1}{r^2}\frac{\partial}{\partial r}(r^2 T_{r\phi}) $$
Now my question is, shouldn't the RHS be ## \frac{1}{r}\frac{\partial}{\partial r}(r T_{r\phi}) ## instead of ## \frac{1}{r^2}\frac{\partial}{\partial r}(r^2 T_{r\phi}) ## because the ## r ## part of ## \vec{\nabla}.\vec{f}## in the cylindrical coordinate is ## \frac{1}{r}\frac{\partial}{\partial r}(r f_r) ##.
Thanks in advance..
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