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I just asked this question on ask.metafilter, and it was suggested that I ask here. Though I'm talking about coding something up, this question is about the math behind it, not the implementation.

We have done analysis in the past where we've computed an approximation for Gaussian curvature of a surface in Cartesian coordinates.

What we've been doing for Cartesian (in MATLAB) is

[fu,fv] = gradient(Z)

[fuu, fuv] = gradient(fu)

[fvu,fvv] = gradient(fv)

GC = (fuu*fvv - fuv*fuv)/(1 + fu^2 + fv^2)

So now I have a surface that I'm modeling in cylindrical coordinates, and I can do the same thing as above for $r$ as a function of $\theta$ and $z$. The problem is that it's only taking into account the change in $r$, not the fact that there is curvature inherent in it being a cylinder.

Looking on Wolfram (equations 27, 32 and 37 and thereabouts), it seems like there's a centripetal component that I don't know how to apply. Dividing by the (constant?) radius doesn't seem like it would work, so I think I'm missing something.

Any help would be appreciated, either explaining how to modify these equations to work correctly, or some other approximation that has worked for you.

Thank you.

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Thank you for your answer. It was very helpful. As you can see, I'm pretty much lost. I do have a followup question, however, and that is about units. Correct me if I'm wrong, but K has the units of m^-2? In your equation, am I correct in thinking that dr/dtheta is in meters and dr/dz is dimensionless? If I'm wrong, I apologize, but it appears that units for the four terms in the numerator of your equation are m^3, m^4, m^3, and m^2, all over m^4. I hope you or someone else is able to clear up my confusion. I don't mean to question, but this is for my work, I have to. Thank you. –  ABC Dec 29 '10 at 20:26
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It would have been preferable to comment under @J.M.'s answer, as then he would be notified of your comment. But I can answer your query: $\partial r/\partial z$ is indeed dimensionless, but $\partial^2 r/\partial z^2$ has dimensions of $\text{length}^{-1}$, and $\partial^2 r/\partial\theta\partial z$ is dimensionless (the easy way to work this out is to just ignore all the $\partial$'s). So all terms in the numerator have dimensions of $\text{length}^2$. –  Rahul Dec 29 '10 at 23:01
    
That makes sense. Thank you. –  ABC Dec 30 '10 at 0:10
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1 Answer

up vote 6 down vote accepted

Your cylindrical coordinate surface $r(\theta,z)$ in Cartesian coordinates is

$$\begin{align*}x&=r(\theta,z)\cos\;\theta\\y&=r(\theta,z)\sin\;\theta\\z&=z\end{align*}$$

which now allows you to apply the usual Gaussian curvature formula. In particular, you should get the expression

$$K=-\frac{r^3\frac{\partial^2 r}{\partial z^2}+r^2\left(\left(\frac{\partial^2 r}{\partial \theta\partial z}\right)^2-\frac{\partial^2 r}{\partial z^2}\frac{\partial^2 r}{\partial \theta^2}\right)+2r\frac{\partial r}{\partial \theta}\left(\frac{\partial^2 r}{\partial z^2}\frac{\partial r}{\partial \theta}-\frac{\partial r}{\partial z}\frac{\partial^2 r}{\partial \theta\partial z}\right)+\left(\frac{\partial r}{\partial \theta}\frac{\partial r}{\partial z}\right)^2}{\left(r^2+\left(r\frac{\partial r}{\partial z}\right)^2+\left(\frac{\partial r}{\partial \theta}\right)^2\right)^2}$$


For completeness, if you have $z$ as a function of $r$ and $\theta$, your Cartesian parametrization is

$$\begin{align*}x&=r\cos\;\theta\\y&=r\sin\;\theta\\z&=z(r,\theta)\end{align*}$$

and the corresponding Gaussian curvature expression is

$$K=\frac{r^2\frac{\partial^2 z}{\partial r^2}\left(\frac{\partial^2 z}{\partial \theta^2}+r\frac{\partial z}{\partial r}\right)-\left(\frac{\partial z}{\partial \theta}-r\frac{\partial^2 z}{\partial r\partial \theta}\right)^2}{\left(r^2\left(\left(\frac{\partial z}{\partial r}\right)^2+1\right)+\left(\frac{\partial z}{\partial \theta}\right)^2\right)^2}$$

I will leave the derivation of the Gaussian curvature expression for

$$\begin{align*}r&=f(u,v)\\\theta&=g(u,v)\\z&=h(u,v)\end{align*}$$

to the interested reader.

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