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August 5th, 2015, 02:38 PM   #1
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Why no r cancelation in divergence

hello

For cylindrical coords...

for a small area the flux in the radial direction sums up to approximately:

[F(r+Δr,θ ,z)-F(r,θ ,z)]r*Δθ*Δz/ΔV

but the 'r' multiplication is always then put inside the [ ] with the function so that:

[r*F(r+Δr,θ ,z)-r*F(r,θ ,z)]Δθ*Δz/ΔV


when the limit is taken it ends up being (1/r)*[d(r*F)/dr] instead of simply dF/dr.

Why?

Last edited by Kinroh; August 5th, 2015 at 02:45 PM.
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August 10th, 2015, 10:27 AM   #2
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I suspect that you have not gotten any responses because no one can determine exactly what question you are asking! What cancellation is it that you think should happen?
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August 10th, 2015, 11:54 AM   #3
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I see,

Well I am confused because isn't ΔV=r*Δθ*Δr*Δz

so if we are dividing: [F(r+Δr,θ ,z)-F(r,θ ,z)]r*Δθ*Δz/ΔV


that is: [F(r+Δr,θ ,z)-F(r,θ ,z)](r*Δθ*Δz)/(r*Δθ*Δr*Δz)

=[F(r+Δr,θ ,z)-F(r,θ ,z)]/Δr

everything cancels but the Δr

So then the small radial flux is approximately: [F(r+Δr,θ ,z)-F(r,θ ,z)]/Δr, which assume in the limit would be =dF/dr

But then then it is always given as (1/r)*d(r*F)/dr
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August 10th, 2015, 12:50 PM   #4
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I think part of my problem is that a polar area actually has another polar term that disappears in the limit, but is significant before the limit. But I am not sure which.

Last edited by Kinroh; August 10th, 2015 at 01:03 PM.
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August 10th, 2015, 01:47 PM   #5
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ok so MIGHT no what I'm thinking wrong:

So the function at the far radial end would be:

F(r+Δr,θ ,z)

But the small surface area would be:

ΔS=(r+Δr)*Δθ*ΔZ NOT ΔS=r*Δθ*ΔZ as this is the case for the inner radial end:
so the flux of outer end is :

F(r+Δr,θ ,z)*(r+Δr)*Δθ*ΔZ

and on the inner end is:

-F(r,θ ,z)*r*Δθ*ΔZ ( negative for opposite direction)

so the Sum of the two radial flux is:

[F(r+Δr,θ ,z)*(r+Δr)*Δθ*ΔZ]-[F(r,θ ,z)*r*Δθ*ΔZ]

In which case only the Δθ*ΔZ can factor out, and not a radial term so:

[F(r+Δr,θ ,z)*(r+Δr)]-[F(r,θ ,z)*r]*Δθ*ΔZ


the small Volume is r*Δθ*Δr*ΔZ

then the radial flux per volume is:

[F(r+Δr,θ ,z)*(r+Δr)]-[F(r,θ ,z)*r]*Δθ*ΔZ/r*Δθ*Δr*ΔZ

=F(r+Δr,θ ,z)*(r+Δr)]-[F(r,θ ,z)*r]*(1/r)*(1/Δr)

Which in the limit would perhaps be

(1/r)*d(r*F)/dr









r*Δθ*Δr*Δz)
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August 10th, 2015, 02:14 PM   #6
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since it the flux per volume, there are two flux's summed up in two different locations.. Maybe the volume should really be the average volume...

Meaning ΔV=(r+Δr/2)*Δθ*ΔZ*Δr

but in the limit the it simply becomes r*dr*dθ*dZ...and the answer is the same
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