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Hydrostatic Equilibrium in an Accretion Disc

  1. Feb 16, 2012 #1
    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.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 16, 2012 #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.
  4. Feb 16, 2012 #3
    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.

    You're talking about the radial-equilibrium equation. The OP is concerned with the vertical equilibrium condition---which is unaffected by the centrifugal force.
  5. Feb 16, 2012 #4
    For an accretion disk, I thought its angular momentum was preventing the star's gravity from pulling it.
  6. Feb 16, 2012 #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.
  7. Feb 16, 2012 #6
    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).
  8. Feb 17, 2012 #7
    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.
  9. Feb 17, 2012 #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.
  10. Feb 17, 2012 #9

    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]
  11. Feb 17, 2012 #10
    Thanks for your help guys.
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