Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Torque on the Accretion disc

  1. Feb 27, 2012 #1
    http://www.maths.qmul.ac.uk/~rpn/ASTM735/lecture3.pdf [Broken]

    The diagram on page 26 is the accretion disc.

    The torque acting on the inner edge of the ring (the one that has a thickness of dR in the diagram) is

    RFin = -2[itex]\pi[/itex]R3[itex]\nu[/itex]Ʃ[itex]\frac{dΩ}{dR}[/itex]

    The torque acting on the outer edge of the ring is

    RFout = [2[itex]\pi[/itex]R3[itex]\nu[/itex]Ʃ[itex]\frac{dΩ}{dR}[/itex]]R+dR

    I would think that the net torque acting would be

    T = [2[itex]\pi[/itex]R3[itex]\nu[/itex]Ʃ[itex]\frac{dΩ}{dR}[/itex]]R+dR - [2[itex]\pi[/itex]R3[itex]\nu[/itex]Ʃ[itex]\frac{dΩ}{dR}[/itex]]R

    = [2[itex]\pi[/itex]R3[itex]\nu[/itex]Ʃ[itex]\frac{dΩ}{dR}[/itex]]dR

    but according to equation (66) on page 27 of http://www.maths.qmul.ac.uk/~rpn/ASTM735/lecture3.pdf [Broken], the net torque is


    Does anyone know why [itex]\frac{∂}{∂R}[/itex] is included in the expression?
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Feb 27, 2012 #2
    Assume you take the limit that [itex] dR \rightarrow 0[/itex] then that is the correct expression, just based on the definition of the derivative.
    Conceptually its because the torque is the difference in force between different radii.
  4. Feb 28, 2012 #3


    User Avatar
    Science Advisor
    Gold Member

    This sounds suspiciously like a variant on Lorentian aether, which has been thoroughly disproven.
  5. Feb 29, 2012 #4
    Why does dR [itex]\rightarrow[/itex] 0 makes it a correct expression?
  6. Feb 29, 2012 #5
    Because that's the definition of a derivative
    [tex] f'(x) \equiv \lim_{dx \to 0} \frac{ f(x+dx) - f(x) }{ dx } = \lim_{dx \to 0} \frac{ f|_{x+dx} - f|_x }{ dx } [/tex]
  7. Feb 29, 2012 #6
    If that's the definition of the derivative, shouldn't [itex]\frac{∂}{∂R}[/itex] on the expression be [itex]\frac{d}{dR}[/itex]
  8. Feb 29, 2012 #7
    No, because you're only looking at the explicit R dependence---in your equation f(x) would be just the explicitly R dependent part. So you're just doing the partial derivative.
  9. Feb 29, 2012 #8
    Tm = [itex]\frac{dj}{dt}[/itex]

    where Tm is the torque per unit mass, and j = R2Ω(R) which is the specific angular momentum (angular momentum per unit mass)

    It may be shown that

    [itex]\frac{dj}{dt}[/itex] = [itex]\frac{∂j}{∂t}[/itex] + (v.[itex]\nabla[/itex])j (convective derivative)

    = vR[itex]\frac{∂j}{∂R}[/itex] [equation (68) of page 27 http://www.maths.qmul.ac.uk/~rpn/ASTM735/lecture3.pdf] [Broken]

    Looking at the convective derivative, I'm guessing that [itex]\frac{∂j}{∂t}[/itex] = 0 due to the derivative of j = R2Ω(R) with respect to t. If that's the case, then surely

    [itex]\frac{dj}{dt}[/itex] should also be 0 which makes vR[itex]\frac{∂j}{∂R}[/itex] equal to 0.

    I don't understand why [itex]\frac{dj}{dt}[/itex] isn't 0.
    Last edited by a moderator: May 5, 2017
  10. Feb 29, 2012 #9
    The change in specific angular momentum would be zero if there were no torque. If there is torque, its not zero.

    The equation you gave, which (at least) looks right---says that [itex] \frac{dj}{dt} = (v \dot \Del ) j = v_R \frac{ \partial j}{\partial R}[/itex] (i think you're right about the partial w.r.t. time being zero---at least if we assume a steady state). But the divergence of the specific angular momentum isn't zero---so neither is the R.H.S.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook