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Torque on the Accretion disc

by Jamipat
Tags: accretion, disc, torque
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Jamipat
#1
Feb27-12, 03:57 PM
P: 11
http://www.maths.qmul.ac.uk/~rpn/ASTM735/lecture3.pdf

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, the net torque is

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

Does anyone know why [itex]\frac{∂}{∂R}[/itex] is included in the expression?
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zhermes
#2
Feb27-12, 07:00 PM
P: 1,261
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.
Chronos
#3
Feb28-12, 12:45 AM
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P: 9,437
This sounds suspiciously like a variant on Lorentian aether, which has been thoroughly disproven.

Jamipat
#4
Feb29-12, 02:41 PM
P: 11
Torque on the Accretion disc

Quote Quote by zhermes View Post
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.
Why does dR [itex]\rightarrow[/itex] 0 makes it a correct expression?
zhermes
#5
Feb29-12, 02:53 PM
P: 1,261
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]
Jamipat
#6
Feb29-12, 03:40 PM
P: 11
Quote Quote by zhermes View Post
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]
If that's the definition of the derivative, shouldn't [itex]\frac{∂}{∂R}[/itex] on the expression be [itex]\frac{d}{dR}[/itex]
zhermes
#7
Feb29-12, 03:45 PM
P: 1,261
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.
Jamipat
#8
Feb29-12, 05:45 PM
P: 11
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]

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.
zhermes
#9
Feb29-12, 06:22 PM
P: 1,261
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.


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