So. Take a circle on a plane. Make it a complex plane, just for kicks. The unit tangent to the circle is i *z/Length[z]. That's because if our coordinate on the circle is z, then to get a unit vector we can divide by the length of z, and then we turn it by Pi/2 by multiplying by i, and now we've got a unit tangent. Now, we're interested in how the curvature of a given shape changes under an analytic mapping. It's possible to show that if we pick the above-mentioned circle as our 'shape', then the image curvature under an analytic mapping f(z) is given by

k_image = (1 + Re[z*f''/f'])/(z*f')

Now, the question is, without actually cranking out a calculation, what *should* the image curvature be if our mapping is given by f(z) = z^m? And then check your prediction using the formula.

So. z^m is certainly analytic. Hmm. Well. z^m basically dilates the plane, more or less, right? I mean, it takes each point, and pushes it out z^m. Which suggests that the curvature, which starts out as 1/Length[z] is going to be something like 1/Length[z^m]. Ok.

Doing the calculation, though, leads to a rather different conclusion. The image curvature is instead:

image_curvature = 1/Length[m*z^m] + m*(m-1)/Length[m*z^m]

WTF? The scary thing is that this bizarre result matches intuition if m = 1. Because then it's just 1/Length[z], which is the same as the initial curvature, which damn well better be the case, because m=1 we're just doing a linear mapping, that is, we're not changing jack shit.

I've got a sneaking suspicion I'm misunderstanding how z^m works geometrically, which would be embarassing...

## 0 Comments:

Post a Comment

<< Home