## Torque on the Accretion disc

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$\pi$R3$\nu$Ʃ$\frac{dΩ}{dR}$

The torque acting on the outer edge of the ring is

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

I would think that the net torque acting would be

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

= [2$\pi$R3$\nu$Ʃ$\frac{dΩ}{dR}$]dR

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

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

Does anyone know why $\frac{∂}{∂R}$ is included in the expression?
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 Assume you take the limit that $dR \rightarrow 0$ 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.
 Recognitions: Gold Member Science Advisor This sounds suspiciously like a variant on Lorentian aether, which has been thoroughly disproven.

## Torque on the Accretion disc

 Quote by zhermes Assume you take the limit that $dR \rightarrow 0$ 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 $\rightarrow$ 0 makes it a correct expression?
 Because that's the definition of a derivative $$f'(x) \equiv \lim_{dx \to 0} \frac{ f(x+dx) - f(x) }{ dx } = \lim_{dx \to 0} \frac{ f|_{x+dx} - f|_x }{ dx }$$

 Quote by zhermes Because that's the definition of a derivative $$f'(x) \equiv \lim_{dx \to 0} \frac{ f(x+dx) - f(x) }{ dx } = \lim_{dx \to 0} \frac{ f|_{x+dx} - f|_x }{ dx }$$
If that's the definition of the derivative, shouldn't $\frac{∂}{∂R}$ on the expression be $\frac{d}{dR}$
 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.
 Tm = $\frac{dj}{dt}$ 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 $\frac{dj}{dt}$ = $\frac{∂j}{∂t}$ + (v.$\nabla$)j (convective derivative) = vR$\frac{∂j}{∂R}$ [equation (68) of page 27 http://www.maths.qmul.ac.uk/~rpn/ASTM735/lecture3.pdf] Looking at the convective derivative, I'm guessing that $\frac{∂j}{∂t}$ = 0 due to the derivative of j = R2Ω(R) with respect to t. If that's the case, then surely $\frac{dj}{dt}$ should also be 0 which makes vR$\frac{∂j}{∂R}$ equal to 0. I don't understand why $\frac{dj}{dt}$ isn't 0.
 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 $\frac{dj}{dt} = (v \dot \Del ) j = v_R \frac{ \partial j}{\partial R}$ (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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