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Jamipat
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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?
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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