Hydrostatic Equilibrium in an Accretion Disc

In summary: OP wrote.In summary, an accretion disc orbiting a star is in hydrostatic equilibrium in the vertical direction, with the pressure gradient balancing the vertical gravitational force. This is described by the hydrostatic equilibrium equation, which is simply the Navier-Stokes equation in the vertical direction. The pressure gradient does not balance the gravitational potential of the star, which is actually balanced by the centrifugal forces. However, the pressure gradient is crucial in maintaining the vertical stability of the disk and preventing it from collapsing towards the mid-plane. This can be seen in the equation where the pressure gradient is equal to the gravitational potential energy of the star
  • #1
Jamipat
11
0
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 [Broken]

More information is on Section 3.3.1 in the lecture note.
 
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  • #2
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 Navier-Stokes equations)

The Hydrostatic equilibrium equation however is simply the Navier-Stokes 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
Jamipat said:
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?
If you have an accretion disk that is in hydrostatic equilibrium, the vertical pressure must balance the vertical gravitational force---as the equation says. This will generally occur by establishing an appropriate density-profile in the vertical direction (same ideas as what HossamCFD says at one point).

If the pressure wasn't strong enough to resist gravity, the disk would collapse towards the mid-plane, and during the collapse the pressure increases---until it is strong enough to prevent further collapse.

HossamCFD said:
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 Navier-Stokes equations)
You're talking about the radial-equilibrium equation. The OP is concerned with the vertical equilibrium condition---which is unaffected by the centrifugal force.
 
  • #4
zhermes said:
If you have an accretion disk that is in hydrostatic equilibrium, the vertical pressure must balance the vertical gravitational force---as the equation says. This will generally occur by establishing an appropriate density-profile in the vertical direction (same ideas as what HossamCFD says at one point).

If the pressure wasn't strong enough to resist gravity, the disk would collapse towards the mid-plane, and during the collapse the pressure increases---until it is strong enough to prevent further collapse.

For an accretion disk, I thought its angular momentum was preventing the star's gravity from pulling it.
 
  • #5
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
zhermes said:
You're talking about the radial-equilibrium equation. The OP is concerned with the vertical equilibrium condition---which is unaffected by the centrifugal force.

I say so explicitly in the very same paragraph you quoted.


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
zhermes said:
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.

At the vertical direction, shoudn't the gravitational force be -[itex]\frac{GM}{z^{2}}[/itex] instead of [itex]\frac{GMz}{(R^{2} + z^{2})^{3/2}}[/itex] as the gravitational force is acting in the z direction.
 
  • #8
The source of gravity is still the central star. Draw it out---where R is the radial distance from the star to the location on the disk-midplane; and z is the distance above the mid-plane.
Use trigonometry to find the vertical component.
 
  • #9
Jamipat said:
At the vertical direction, shoudn't the gravitational force be -[itex]\frac{GM}{z^{2}}[/itex] instead of [itex]\frac{GMz}{(R^{2} + z^{2})^{3/2}}[/itex] as the gravitational force is acting in the z direction.


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
zhermes said:
The source of gravity is still the central star. Draw it out---where R is the radial distance from the star to the location on the disk-midplane; and z is the distance above the mid-plane.
Use trigonometry to find the vertical component.

HossamCFD said:
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]

Thanks for your help guys.
 

1. What is hydrostatic equilibrium in an accretion disc?

Hydrostatic equilibrium in an accretion disc refers to the balance between the inward gravitational force and the outward pressure force within the disc. This balance is necessary for the disc to maintain its shape and structure.

2. How is hydrostatic equilibrium achieved in an accretion disc?

Hydrostatic equilibrium is achieved through the combination of two forces: gravity, which pulls material inward, and pressure, which pushes material outward. These forces must be equal for the disc to remain stable.

3. What role does angular momentum play in hydrostatic equilibrium?

Angular momentum is a crucial factor in maintaining hydrostatic equilibrium in an accretion disc. As the disc rotates, material closer to the center has a higher angular velocity and thus a higher centrifugal force, which helps to balance the inward pull of gravity.

4. How does the mass of the central object affect hydrostatic equilibrium in an accretion disc?

The mass of the central object has a direct impact on hydrostatic equilibrium in an accretion disc. A more massive central object will have a stronger gravitational pull, requiring a higher outward pressure force to maintain equilibrium.

5. What are the implications of hydrostatic equilibrium for accretion disc dynamics?

The balance of forces in hydrostatic equilibrium has significant implications for the dynamics of accretion discs. It determines the size and shape of the disc, the speed of material moving through it, and the overall stability of the disc's structure.

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