
#1
Feb2712, 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 RF_{in} = 2[itex]\pi[/itex]R^{3}[itex]\nu[/itex]Ʃ[itex]\frac{dΩ}{dR}[/itex] The torque acting on the outer edge of the ring is RF_{out} = [2[itex]\pi[/itex]R^{3}[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]R^{3}[itex]\nu[/itex]Ʃ[itex]\frac{dΩ}{dR}[/itex]]_{R+dR}  [2[itex]\pi[/itex]R^{3}[itex]\nu[/itex]Ʃ[itex]\frac{dΩ}{dR}[/itex]]_{R} = [2[itex]\pi[/itex]R^{3}[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]R^{3}[itex]\nu[/itex]Ʃ[itex]\frac{dΩ}{dR}[/itex]]dR Does anyone know why [itex]\frac{∂}{∂R}[/itex] is included in the expression? 



#2
Feb2712, 07:00 PM

P: 1,262

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
Feb2812, 12:45 AM

Sci Advisor
PF Gold
P: 9,182

This sounds suspiciously like a variant on Lorentian aether, which has been thoroughly disproven.




#4
Feb2912, 02:41 PM

P: 11

Torque on the Accretion disc 



#5
Feb2912, 02:53 PM

P: 1,262

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
Feb2912, 03:40 PM

P: 11





#7
Feb2912, 03:45 PM

P: 1,262

No, because you're only looking at the explicit R dependencein your equation f(x) would be just the explicitly R dependent part. So you're just doing the partial derivative.




#8
Feb2912, 05:45 PM

P: 11

T_{m} = [itex]\frac{dj}{dt}[/itex]
where T_{m} is the torque per unit mass, and j = R^{2}Ω(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) = v_{R}[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 = R^{2}Ω(R) with respect to t. If that's the case, then surely [itex]\frac{dj}{dt}[/itex] should also be 0 which makes v_{R}[itex]\frac{∂j}{∂R}[/itex] equal to 0. I don't understand why [itex]\frac{dj}{dt}[/itex] isn't 0. 



#9
Feb2912, 06:22 PM

P: 1,262

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 rightsays 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 zeroat least if we assume a steady state). But the divergence of the specific angular momentum isn't zeroso neither is the R.H.S. 


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