# Surface gradient and curvature

I am trying to get familiar with the surface gradient operator, i.e. $\nabla_S = (I-\mathbf{n}\mathbf{n})\cdot \nabla$, where $\mathbf{n}$ is the unit normal to the surface and $I$ the identity tensor.

I have seen in a paper that, for an axysimmetric surface, the mean curvature is $\kappa = \nabla \cdot\mathbf{n} = \nabla_S \cdot \mathbf{n}$, however I cannot get the latter. We can define the surface as $f = r - a(z)$, thus $\mathbf{n} = \nabla \, f / ||\nabla \,f|| = 1/\sqrt{1+(a')^2} \, \mathbf{e_r} -a'/\sqrt{1+(a')^2} \, \mathbf{e_z} = (1/\sqrt{1+(a')^2},0,-a'/\sqrt{1+(a')^2})$, and:

$\kappa = \nabla \cdot \mathbf{n} = \dfrac{1}{a\sqrt{1+(a')^2}} - \dfrac{a''}{[1+(a')^2]^{3/2}}$

However I do not obtain the same result with the surface operator,

$\nabla_S = (I-\mathbf{n}\mathbf{n})\cdot \nabla = \left(\frac{(a')^2}{1+(a')^2}\frac{\partial}{\partial r} + \frac{a'}{1+(a')^2}\frac{\partial }{\partial z},0,\frac{a'}{1+(a')^2} \frac{\partial}{\partial r} + \frac{1}{1+(a')^2} \frac{\partial}{\partial z}\right)$,

$\nabla_S \cdot \mathbf{n} = - \dfrac{a''}{[1+(a')^2]^{3/2}}$,

which is just the axial contirbution of the mean curvature. Is this correct and the paper is wrong? Or am I making a mistake in the procedure?

• The mean curvature is in fact always $\nabla \cdot \mathbf{n}$ (up to a constant depending on the definition). Commented Apr 7, 2017 at 15:49
• Yes I know, but what I am trying to find is if $\nabla \cdot \mathbf{n}$ = $\nabla_S \cdot \mathbf{n}$, being $\nabla_S$ a gradient operator which is tangent to the surface (i.e. the projection of the conventional gradient onto the surface). Commented Apr 7, 2017 at 15:55
• If you're using polar coordinates, I think you need $\partial_{\theta}$s. Certainly $\mathbf{n}$ varies with $\theta$. Commented Apr 7, 2017 at 16:10
• Think about a cylinder. Or a cone. The normal points at least partially in the direction $\mathbf{e}_r$, and that definitely varies with $\theta$. Commented Apr 7, 2017 at 16:23
• The unit normal has no $\mathbf e_\theta$ term, but, as @Chappers has stated, $\mathbf e_r$ depends explicitly on $\theta$. As a point rotates about the $z$-axis, $\mathbf n$ varies as well by rotating about the $z$-axis. You are correct, however, that the curvature is rotationally symmetric, i.e., independent of $\theta$. Commented Apr 8, 2017 at 7:51

We have $$\nabla \cdot \mathbf{n} - \nabla_S \cdot \mathbf{n} = \mathbf{n} \cdot ( (\mathbf{n} \cdot \nabla)\mathbf{n} ) = ( (\mathbf{n} \cdot \nabla)\mathbf{n} ) \cdot \mathbf{n}.$$ But this is equal to $\frac{1}{2} (\mathbf{n} \cdot \nabla)(\mathbf{n} \cdot \mathbf{n}) = \frac{1}{2}(\mathbf{n} \cdot \nabla)1 = 0$. (To see this more clearly, use summation convention: $n_i n_j \partial_j n_i = \frac{1}{2} n_j \partial_j (n_i n_i)$.)
How does this translate to the example? Borrowing the formula for material derivative from Wikipedia, $$(\mathbf{A} \cdot \nabla) \mathbf{B} = \left(A_\rho \frac{\partial B_\rho}{\partial \rho}+\frac{A_\varphi}{\rho}\frac{\partial B_\rho}{\partial \varphi}+A_z\frac{\partial B_\rho}{\partial z}-\frac{A_\varphi B_\varphi}{\rho}\right) \hat{\boldsymbol \rho} \\ + \left(A_\rho \frac{\partial B_\varphi}{\partial \rho} + \frac{A_\varphi}{\rho}\frac{\partial B_\varphi}{\partial \varphi} + A_z\frac{\partial B_\varphi}{\partial z} + \frac{A_\varphi B_\rho}{\rho}\right) \hat{\boldsymbol \varphi}\\ + \left(A_\rho \frac{\partial B_z}{\partial \rho}+\frac{A_\varphi}{\rho}\frac{\partial B_z}{\partial \varphi}+A_z\frac{\partial B_z}{\partial z}\right) \hat{\mathbf z},$$ most of which disappears (all the components are independent of $\theta$, $n_{\theta}=0$, $n_r,n_z$ depend only on $z$). The terms that remain are $$(\mathbf{n} \cdot \nabla) \mathbf{n} = \left(n_z\frac{\partial n_r}{\partial z}\right) \mathbf{e}_r + \left(n_z\frac{\partial n_z}{\partial z}\right) \mathbf{e}_z = n_z \left( \frac{\partial n_r}{\partial z} \mathbf{e}_r + \frac{\partial n_z}{\partial z} \mathbf{e}_z \right).$$ Dotting with $\mathbf{n}$, we find $n_z ( n_r \partial_z n_r + n_z \partial_z n_z ) = \frac{1}{2} n_z \partial_z (n_rn_r+n_zn_z) = 0$ as before. Going deeper, $$\partial_z n_r = -\frac{a'a''}{(1+a'^2)^{3/2}}, \qquad \partial_z n_z = -\frac{a''}{(1+a'^2)^{3/2}},$$ and then the result is obvious.
Going the other way, $$\mathbf{n} \cdot (\mathbf{n} \cdot \nabla) \mathbf{B} = \left(n_r \frac{\partial B_r}{\partial r}+n_z\frac{\partial B_r}{\partial z}\right) n_r \\ + \left(n_r \frac{\partial B_z}{\partial r}+n_z\frac{\partial B_z}{\partial z}\right) n_z = \frac{\partial_r B_r-a' \partial_z B_r - a' \partial_r B_z + a'^2 \partial_z B_z}{1+a'^2},$$ Of course if $\mathbf{B}=\mathbf{n}$, only two of these survive, and they cancel out as before.
• Nice! Thank you so much. Now it is pretty clear that the curvature is both $\nabla \cdot \mathbf{n}$ and $\nabla_S \cdot \mathbf{n}$. However I still do not know why I do not get the same result with, $\nabla_S = (I-\mathbf{n}\mathbf{n})\cdot \nabla = \left(\frac{(a')^2}{1+(a')^2}\frac{\partial}{\partial r} + \frac{a'}{1+(a')^2}\frac{\partial }{\partial z},0,\frac{a'}{1+(a')^2} \frac{\partial}{\partial r} + \frac{1}{1+(a')^2} \frac{\partial}{\partial z}\right)$. What am I doing wrong? Commented Apr 8, 2017 at 14:19
• I have a feeling it's unfortunately rather simple: the other term comes from the radial part of the divergence, $$\frac{1}{r} \frac{\partial}{\partial r}(r n_r) = \frac{1}{r\sqrt{1+a'^2}}.$$ (And of course, $r=a$, which gives the expression you expect.) Commented Apr 8, 2017 at 16:40
• The problem is that when you do $\nabla_S \cdot \mathbf{n} = [(I-\mathbf{n}\mathbf{n})\cdot \nabla ] \cdot \mathbf{n}$ you have some terms multiplying $\frac{1}{r} \frac{\partial}{\partial r} (r n_r)$, as you can see in my comment above. I guess I am missing something about the scale factors. Commented Apr 9, 2017 at 14:29