Torque on the Accretion disc

In summary, the discussion revolved around the concept of torque and the equation for net torque acting on an accretion disc. It was noted that the net torque is calculated by taking the limit of dR approaching 0 and that the partial derivative is used because the torque is the difference in force between different radii. The equation for the derivative of specific angular momentum was also discussed and it was noted that the derivative with respect to time is 0 in a steady state, but the divergence of specific angular momentum is not 0.
  • #1
Jamipat
11
0
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

[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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  • #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.
 
  • #3
This sounds suspiciously like a variant on Lorentian aether, which has been thoroughly disproven.
 
  • #4
zhermes said:
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?
 
  • #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]
 
  • #6
zhermes said:
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]
 
  • #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.
 
  • #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.
 
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  • #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.
 

1. What is torque on the accretion disc?

Torque on the accretion disc refers to the twisting or rotational force that acts on the gas and dust particles as they spiral towards a central object, such as a black hole or a star.

2. How does torque affect the accretion disc?

Torque can transfer angular momentum from the accretion disc to the central object, causing the disc to spin faster and the central object to grow in mass. It can also create instabilities or warping within the disc, leading to changes in the flow of material.

3. What factors influence the amount of torque on the accretion disc?

The strength of the gravitational pull from the central object, the mass and velocity of the accreting material, and the structure and density of the disc all play a role in determining the torque on the accretion disc.

4. How is torque on the accretion disc measured?

Torque on the accretion disc can be measured indirectly by observing the changes in the disc's rotation and the behavior of the material within it. This can be done through techniques such as Doppler spectroscopy or monitoring the brightness of the accretion disc.

5. What are the implications of torque on the accretion disc?

The presence of torque on the accretion disc is crucial in understanding the formation and evolution of objects such as black holes and stars. It also plays a role in the transfer of material and energy within these systems, affecting their overall growth and behavior.

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