# Minimal surfaces and gaussian and normal curvaturess

If $M$ is the surface $$x(u^1,u^2) = (u^2\cos(u^1),u^2\sin(u^1), p\,u^1)$$ then I am trying to show that $M$ is minimal. $M$ is referred to as a helicoid.

Also I am confused on how $p$ affects the problem

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What have you tried so far? Where did you get stuck? –  Jesse Madnick Nov 3 '12 at 6:47
I tried getting the normal vector, the x1, x2, x11, x22, and Christoffel symbols, but am lost then –  Buddy Holly Nov 3 '12 at 21:03

Note that $$x_{1}=\frac{\partial x}{\partial u^1}=(-u^2\sin(u^1),u^2\cos(u^1),p)$$ and $$x_{2}=\frac{\partial x}{\partial u^2}=(\cos(u^1),\sin(u^1),0).$$ This implies that the unit normal is given by $$x_1\times x_2=(-p\sin(u^1),p\cos(u^1),-u^2)$$ which implies that
$$n=\frac{x_1\times x_2}{\|x_1\times x_2\|}=\frac{1}{\sqrt{p^2+(u^2)^2}}(-p\sin(u^1),p\cos(u^1),-u^2).$$ Also, $$E=x_1\cdot x_1=(u^2)^2+p^2, F=x_1\cdot x_2=0, G=x_2\cdot x_2=1$$

On the other hand, we have $$x_{11}=(-u^2\cos(u^1),-u^2\sin(u^1),0),$$ $$x_{12}=x_{21}=(-\sin(u^1),\cos(u^1),0).$$ $$x_{22}=(0,0,0).$$ Therefore, we have $$e=x_{11}\cdot n=0, f=x_{12}\cdot n=p, g=x_{22}\cdot n=0.$$

Therefore, we can calculate the mean curvature using the formula (see equation (6) here): $$H=\frac{eG+gE-2fF}{2(EG-F^2)}=0$$ which shows that $M$ is minimal.

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Thanks for your help and the link was very helpful –  Buddy Holly Nov 4 '12 at 5:01
This is not correct. $x_{12} \cdot n$ is not $0$. Result is still correct since $F=0$, however. –  Tyler Brabham Nov 5 '12 at 21:44
@TylerBrabham: Thanks for pointing it out. I corrected it. –  Paul Nov 6 '12 at 1:51

There is a good reason that the value of $p$ does not matter, as long as $p \neq 0.$

If you begin with a sphere of radius $R$ and blow it up to a sphere of radius $SR,$ the result is to multiply the mean curvature by $\frac{1}{S}.$ This is a general phenomenon. A map, which is also linear, given by moving every point $(x,y,z)$ to $(\lambda x, \lambda y, \lambda z)$ for a positive constant $\lambda,$ is called a homothety. A homothety takes any surface and divides the mean curvature (at matching points, of course)) by $\frac{1}{\lambda}.$ This can be done in any $\mathbb R^n,$ I guess we are sticking with $\mathbb R^3.$

So, what I need to do is show you that your helicoid with parameter $p,$ expanded or shrunk by a homothety, is the helicoid with a different parameter, call it $q.$ I'm going to use $u = u^1, v = u^2.$ And that is just $$\frac{q}{p} x(u, \frac{pv}{q}) = \frac{q}{p} \left(\frac{pv}{q} \cos u , \frac{pv}{q} \sin u, p u \right) = (v \cos u , v \sin u, q u).$$

Well, the mean curvature of the original helicoid is $0$ everywhere. So the new helicoid is still minimal.

There is a bit of work showing that a homothety changes the mean curvature in the way I described, no easier than your original problem. True, though.

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