# What causes angular deformation in an inviscid free vortex?

We can describe a two-dimensional (i.e. planar), inviscid, irrotational, free line vortex in cylindrical coordinates with the stream function $\psi = -K\ln{r}$, velocity potential $\phi= K\theta$, tangential velocity component $v_{\theta} = \frac{1}{r}\frac{\partial \phi}{\partial \theta} = K/r$, and radial velocity component $v_r = \frac{\partial \phi}{\partial r} = 0$, where $K$ is a constant. The motion of mutually perpendicular lines in a fluid element is given by

$$\dot{\gamma} = \frac{1}{r} \frac{\partial v_r}{\partial \theta} + \frac{\partial v_{\theta}}{\partial r} - \frac{v_{\theta}}{r}$$

where $\dot{\gamma}$ is the rate of angular deformation of the angle between the lines. In addition, because the flow is irrotational, $$\frac{\partial (r v_{\theta})}{\partial r} = \frac{\partial v_r}{\partial \theta}\,,$$ such that $\dot{\gamma}\neq 0$. However, to construct the stream function and velocity potential, we must assume that the flow is inviscid and the only forces acting on the fluid element are the normal stresses (i.e. the pressure) and any body forces. My understanding is that normal stresses and body forces cannot cause angular deformation, and the shear stresses are zero due to the neglect of viscous terms. Thus, what force is causing the angular deformation of the fluid elements in this flow?

This Phys.SE question is related, but does not answer my question: When is a flow vortex free?

$$\frac{d \gamma}{dt} = \frac{1}{r} \frac{\partial v_r}{\partial \theta} + \frac{\partial v_{\theta}}{\partial r} - \frac{v_{\theta}}{r} = -2\frac{K}{r^{2}}$$ Now integrate over one complete revolution, which is the complete circle $2 \pi r$ divided by velocity $K/r$.
$$\gamma(per \ cycle) = -\int_0^{2\pi r^{2}/K}2 K/r^{2}dt = -4 \pi$$
Which is the total amount of deformation of the axes in one cycle,i.e., it is the vorticity $\omega$ of the particle, since the system is in steady state. But we know that $\omega$ is defined as twice the rotation of the particle around its axis $\omega = 2\Omega$, there fore the total rotation is $\Omega = -2\pi$ which counteracts the revolution of one cycle of $\Omega_{rev}=2\pi$. Remember that deformation is seen from the frame of the particle but for the frame of the whole flow, it is a irrotational flow, since there is also a revolution around the center axis.
• I get $-2 K/r^2$ if I do out the derivatives – darthbith Dec 6 '17 at 21:08