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Post by mondo on Apr 15, 2022 6:56:03 GMT
According to a figure [8] and path $K$ author says: "Clearly $\overline{H(z)} = z$ has no circulation along either of the arcs". Is that because the vector field points radially outwards in a direction that is normal to these arc therefore there is curl round $z$ instead the field only moves away from $z$?
Next, I have no idea how in the equation below: (fluid out) - (fluid in) = $2[\frac{1}{2}a^2\phi - \frac{1}{2}b^2\phi]$. Why $\frac{1}{2}$ and why the square braces is said to be the shaded area inside $K$?
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Post by Admin on Apr 15, 2022 10:39:33 GMT
According to a figure [8] and path $K$ author says: "Clearly $\overline{H(z)} = z$ has no circulation along either of the arcs". Is that because the vector field points radially outwards in a direction that is normal to these arc therefore there is curl round $z$ instead the field only moves away from $z$? Next, I have no idea how in the equation below: (fluid out) - (fluid in) = $2[\frac{1}{2}a^2\phi - \frac{1}{2}b^2\phi]$. Why $\frac{1}{2}$ and why the square braces is said to be the shaded area inside $K$? Mondo Yes, this vector field has no component tangential to the arcs. We can see on page 484 that $\nabla\cdot\overline{H}=2$, and remember that the area of the sector of a circle with radius $r$ and angle $\theta$ is $\frac{1}{2}r^2\theta$. The shaded area is the difference between the two sectors with radii $a$ and $b$ in figure 8 and angle $\phi$. Vasco |
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Post by mondo on Apr 16, 2022 5:13:51 GMT
Ok so it was arranged to fit in the formula for a section of circle. I think at least short comment should be given.
Thank you Vasco.
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