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I'll admit I'm not really an expert on coordinate transformations, so my error/misunderstanding may be a simple fix; in fact, that would be ideal.

I'm trying to determine the eigenvalues of the Lamb-Oseen vortex velocity gradient field. The Lamb-Oseen velocity field in polar coordinates is given as $$ \vec{u}(r,t) = \begin{bmatrix} u_r \\ u_\theta \end{bmatrix} = \begin{bmatrix} 0 \\ \frac{\Gamma}{2\pi r} (1 - e^{-r^2/(4 \nu t)}) \end{bmatrix} $$ When I compute the gradient in polar coordinates, I get the following form: $$ \nabla \vec{u} = \begin{bmatrix} \frac{\partial u_r}{\partial r} & \frac{\partial u_\theta}{\partial r} \\ \frac{1}{r} \frac{\partial u_r}{\partial \theta} & \frac{1}{r}\frac{\partial u_\theta}{\partial \theta} \end{bmatrix} $$ (Maybe it should be transposed, but that doesn't end up mattering here.) Clearly, $\frac{\partial u_\theta}{\partial r}$ is the only non-zero term, so the eigenvalues will both be zero as well.

The physical problem with this is that the eigenvalues should have at least some non-zero imaginary part because the swirl criterion for vortex identification should identify pretty much the entire field as a vortex, and it is defined as the magnitude of the imaginary part of the velocity gradient eigenvalues. Indeed, when I compute the eigenvalues in Cartesian coordinates using the following form, I do get non-zero imaginary eigenvalues. $$ \vec{u}(x,y,t) = \begin{bmatrix} u \\ v \end{bmatrix} = \frac{\Gamma}{2\pi (x^2 + y^2)} \bigg(1 - e^{-(x^2 + y^2)/(4 \nu t)}\bigg)\begin{bmatrix} -y \\ x\end{bmatrix} $$ Edit: I originally (mistakenly) put $\frac{\Gamma}{2\pi \sqrt{x^2 + y^2}}$, but that was simply an error writing it. I'm assuming one of two problems:

  1. I'm just not doing the transformation/polar gradient correctly, so the terms should look different.
  2. The eigenvalues are unique to the coordinate system. I could see how the values themselves might change between coordinate systems, but all zeros to imaginary non-zero numbers seems wrong.

I would appreciate any explanation of what I might be misunderstanding or someone pointing out a mistake I made in my process.

Thanks!

Ryan Kelly
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  • I don't understand why $u_r$ is zero, since that would mean that the field would be zero everywhere. But the $u(x,y,t)$ vectors are not zero everywhere. OTOH, possibly I am mistaken, I admit that I have no familiarity with the Lamb-Oseen vortex velocity gradient field. – Ron Kaminsky Jul 06 '23 at 17:35
  • The gradient of a vector field in polar coordinates isn't what's shown. See these notes for more info on the more general cylindrical coordinates case. – A rural reader Jul 06 '23 at 17:58
  • OK, from the answer I understand that what is meant is that $u_r$ is the derivative of the $r$ component of the vector, not the $r$ component itself. – Ron Kaminsky Jul 06 '23 at 18:05
  • @RonKaminsky, no $u_r$ is the radial component of velocity. It is a vortex with Gaussian vorticity and no radial component. – Ryan Kelly Jul 06 '23 at 20:21
  • @RyanKelly Thanks, now I remember why I hated curvilinear coordinates so much. :-) – Ron Kaminsky Jul 08 '23 at 08:12

2 Answers2

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  • Coordinate transformations are always just based on the chain rule.

  • When we transform vector fields we cannot transform their components as if they were scalars. That was the (forgivable) sin you committed :). It happened to me as well.

In polar coordinates your vector field is of the form $$ \textstyle\boldsymbol{u}=u_r\,\partial_r+u_\theta\,\frac{1}{r}\partial_\theta $$ where \begin{align} u_r&=0\,,&u_\theta=\frac{\Gamma}{2\pi r} (1 - e^{-r^2/(4 \nu t)}) \,. \end{align}

According to first principles (chain rule), \begin{align} u_x&=u_r\cos\theta-u_\theta\sin\theta=\frac{\Gamma}{2\pi\sqrt{x^2+y^2}} (1 - e^{-(x^2+y^2)/(4 \nu t)})\,\color{red}{\frac{-y}{\sqrt{x^2+y^2}}}\,,\\ u_y&=u_r\sin\theta+u_\theta\cos\theta=\frac{\Gamma}{2\pi\sqrt{x^2+y^2}} (1 - e^{-(x^2+y^2)/(4 \nu t)})\,\color{red}{\frac{x}{\sqrt{x^2+y^2}}}\,. \end{align} These expressions can be simplified by which your $\sqrt{x^2+y^2}$ becomes $x^2+y^2\,.$ Now I am curious to see that Jacobian $\nabla\boldsymbol{u}$ again.

Kurt G.
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  • Dang, in reality I just wrote the equation down wrong. That square root on the Cartesian form was not supposed to be there, I was just incorrectly remembering it (incidentally, I should remember to double-check that when I'm doing calculations). Good catch, but it still doesn't quite get to the answer I'm looking for which is why the polar form is zero, but I think someone's comment answered that by pointing out my gradient is a little off. – Ryan Kelly Jul 06 '23 at 20:30
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Thanks to @A rural reader for pointing out the issue. For completeness, I'll spell out the answer here:

The velocity gradient has a couple more terms I've neglected by accident, and it should look like this:

$$ \nabla \vec{u} = \begin{bmatrix} \frac{\partial u_r}{\partial r} & \frac{1}{r} \frac{\partial u_r}{\partial \theta} - \frac{u_\theta}{r} \\ \frac{\partial u_\theta}{\partial r} & \frac{1}{r} \frac{\partial u_\theta}{\partial \theta} + \frac{u_r}{r} \end{bmatrix} $$

Now with $u_\theta$ and $\frac{\partial u_\theta}{\partial r}$ on the off-diagonal elements the eigenvalues are non-zero.

Ryan Kelly
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