Hydrostatic Equilibrium in an Accretion Discby Jamipat Tags: accretion, disc, equilibrium, hydrostatic 

#1
Feb1612, 12:22 PM

P: 11

This is regarding an accretion disc orbiting a star. In the z (vertical) direction there is a hydrostatic equilibrium.
[itex]\frac{1}{ρ}[/itex][itex]\frac{∂P}{∂z}[/itex] = [itex]\frac{GMz}{(R^{2} + z^{2})^{3/2}}[/itex] The right hand side of the expression is the Gravitational potential energy and the left side is the pressure gradient. Can someone explain to me how the pressure gradient works in an accretion disc as I don't understand how the pressure gradient of an accretion disc is strong enough to equal the gravitational potential energy of the star? http://www.maths.qmul.ac.uk/~rpn/ASTM735/lecture3.pdf More information is on Section 3.3.1 in the lecture note. 



#2
Feb1612, 02:31 PM

P: 80

You're interpreting the equation in a wrong way. The pressure gradient DOES NOT balance the gravitational potential of the central object. The gravitational potential is actually balanced by the centrifugal forces, not the pressure gradient. And this balance is not described by the equation you wrote but rather the radial balance equation (the radial component of the NavierStokes equations)
The Hydrostatic equilibrium equation however is simply the NavierStokes equation in the vertical direction. It clearly does not say that the pressure gradient balances the gravitational force (notice that the gradient is wrt the vertical direction). If you replace ∂p by ∂rho Cs^2 and in the limit of a thin disc (z<<R) this gives you the density gradient as a function of the square of the keplerian angular velocity times the vertical distance from the disc plan. Which can then be solved to give you the vertical density profile (which turns out to be Gaussian). So simply what the vertical balance (hydrostatic equilibrium) is telling us is that the vertical density profile is Gaussian and for a disc to be thin the disc rotation has to be highly supersonic i.e Vk/Cs=R/H where Vk is the Keplerian rotation velocity, Cs is the sound speed and H is the disc thickness. 



#3
Feb1612, 03:00 PM

P: 1,262

If the pressure wasn't strong enough to resist gravity, the disk would collapse towards the midplane, and during the collapse the pressure increasesuntil it is strong enough to prevent further collapse. 



#4
Feb1612, 05:02 PM

P: 11

Hydrostatic Equilibrium in an Accretion Disc 



#5
Feb1612, 06:49 PM

P: 1,262

The force of gravity acting on a region of gas is directed towards the star. If the gas is exactly in the orbital plane, that force will be directed exactly radially inward (toward the star). If the gas is slightly above or below the orbital plane, there will be a strong radial component of the gravitational force (directed toward the star), but also a small component of the gravitational force in the vertical direction (directed towards the orbital plane).
The radial component is balanced by the centrifugal force. Another way of thinking about the same effect, is that the gas's angular momentum keeps it orbiting (and not being pulled inwards). The vertical component (the "z"component) of the force is balanced by the pressure of the disk. This is the equation that you gave in your original post. 



#6
Feb1612, 11:46 PM

P: 80

I had to mention the radial equilibrium because the OP was concerned about what force balances the gravitational potential (i.e prevent the material from falling onto the central object). 



#7
Feb1712, 01:02 PM

P: 11





#8
Feb1712, 01:17 PM

P: 1,262

The source of gravity is still the central star. Draw it outwhere R is the radial distance from the star to the location on the diskmidplane; and z is the distance above the midplane.
Use trigonometry to find the vertical component. 



#9
Feb1712, 01:24 PM

P: 80

the gravitational potential ∅ equals GM/r where r is the spherical radius i.e. r=√(R^2+z^2) Thus ∂∅/∂z is as stated: [itex]\frac{GMz}{(R^{2} + z^{2})^{3/2}}[/itex] 



#10
Feb1712, 03:11 PM

P: 11




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