1

An axisymmetric vortex, for which the azimuthal velocity $u_\theta$ is proportional to $r^{-\beta}$. What are the values for $\beta$ so that the circulation ($\Gamma(r))$ is finite

Abigail
  • 73

1 Answers1

1

Around a circular contour of radius $r$:

$$\Gamma(r) = \int_R \nabla \times \mathbb{u}\cdot d\mathbf {A} = \oint_C\mathbf{u} \cdot d\mathbf{l} = \int_0^{2\pi}u_\theta r \, d\theta = 2\pi r u_\theta.$$

With $u_\theta$ proportional to $r^{-\beta}$, say $u_\theta = Cr^{-\beta}$, we have $\Gamma(r) = 2\pi Cr^{1-\beta}$ and the circulation is finite as $r \to \infty$ if $\beta \geqslant 1$.

RRL
  • 90,707
  • I'm sorry i should have put that I knew that part already, if $u_\theta$ is $r^-\beta$ then the circulation as a function of r is $2\pi r u_\theta$ ? – Abigail Oct 22 '16 at 19:04
  • sorry $2\pi r^{-2\beta}$ – Abigail Oct 22 '16 at 19:05
  • Do I just take the limit as beta approaches infinity? – Abigail Oct 22 '16 at 23:03
  • 1
    @Abigail. See above. Just note that the circulation is proportional to $r^{1-\beta}$, not $r^{-2\beta}$. – RRL Oct 23 '16 at 02:39
  • why does the exponent have to be less than or equal to zero? I'm having a hard time conceptually understanding what is happening in this problem – Abigail Oct 23 '16 at 02:50
  • As a radial coordinate $r$ is nonnegative. Let $\alpha = 1 - \beta$. If $\alpha = 0 ( \beta = 1)$, then $r^\alpha = 1$ for all $r$ and the circulation is bounded. If $\alpha > 0 (\beta < 1)$ then $\lim_{r \to \infty}r^\alpha = +\infty$. This follows from the limit definition -- for any $M >0$ if $r > M^{1/\alpha}$ then $r^\alpha > M$. Can you show that $\lim_{r \to \infty} r^\alpha = 0$ if $\alpha < 0$.? In that case $r^\alpha$ must be bounded for large $r$. – RRL Oct 23 '16 at 03:29