How Does Torque Affect the Dynamics of an Accretion Disc?

[Jamipat]

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$\pi$R³$\nu$Ʃ$\frac{dΩ}{dR}$

The torque acting on the outer edge of the ring is

RF_out = [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?

 

[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.

 

[Chronos]

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

 

[Jamipat]

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?

 

[zhermes]

Because that's the definition of a derivative
$$f'(x) \equiv \lim_{dx \to 0} \frac{ f(x+dx) - f(x) }{ dx } = \lim_{dx \to 0} \frac{ f|_{x+dx} - f|_x }{ dx }$$

 

[Jamipat]

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 $\frac{d}{dR}$

 

[zhermes]

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]

T_m = $\frac{dj}{dt}$

where T_m 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)

= v_R$\frac{∂j}{∂R}$ [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 v_R$\frac{∂j}{∂R}$ equal to 0.

I don't understand why $\frac{dj}{dt}$ isn't 0.

 

[zhermes]

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.