I was reading today, and I came upon this identity:
$\frac{\partial \mathbf e_p}{\partial p} + \frac{\partial \mathbf e_q}{\partial q} = -(\nabla \cdot \mathbf n) \mathbf n$,
where $\mathbf e_p$ and $\mathbf e_q$ are unit vectors in the directions p and q of a general surface $x(p,q,t)$ and $\mathbf n$ is the unit vector normal to the surface, such that $\mathbf n = \mathbf e_p \times \mathbf e_q$. The unit vector $\mathbf n$ points in the direction that the surface is propagating.
I'd like to figure out how to prove this mathematically, but also understand it intuitively. I see the curvature, in the form of $\nabla \cdot \mathbf n$, but I haven't gotten any further.
Edit: Here is my idea for a possible solution, but I think it is still a bit hand-wavy, so hopefully someone will be able to sharpen up the details.
First of all, I could align $\mathbf e_p$ and $\mathbf e_q$ with the principle directions of the surface. Then, the two principle curvatures would each have a magnitude given by $||\frac{d\mathbf T}{ds}||$, where $\mathbf T$ is the tangent vector to the surface in the principle direction. In the p direction this magnitude would be $||\frac{\partial \mathbf e_p}{\partial p}||$ and in the q direction it would be $||\frac{\partial \mathbf e_q}{\partial q}||$. I know that the mean curvature is given by $H=\kappa_1 + \kappa_2$, where 1 and 2 are the principle directions, so, with my alignment of p and q, $H=\kappa_p + \kappa_q$. Since $H=\nabla \cdot \mathbf n$ (up to a sign (?), and a factor that depends on convention), this suggests that I'm nearly there, but I need to get the sign and direction right.
The Direction
The change in a tangent vector should point in either the n or -n direction. This is well illustrated by the answer to What is the intuition behind the unit normal vector being the derivative of the unit tangent vector?
So at this point I'm happy with $\frac{\partial \mathbf e_p}{\partial p} + \frac{\partial \mathbf e_q}{\partial q} = (\nabla \cdot \mathbf n) \mathbf n$ up to a sign. The question is, what should the sign really be?
This is where I start to get somewhat confused.
The Sign
In the set up described here, there is a surface with the normal vector pointing in the direction of propagation. Imagine that the surface is shaped like a hill. Then $\nabla \cdot \mathbf n$ will be positive, because moving the normal in any coordinate direction (x,y,z), assuming that z is upwards, will make the component of the normal in that direction larger. Likewise, if I imagine a valley, $\nabla \cdot \mathbf n$ is negative. (So at this point I think that the choice of the direction of n determines the sign of $\nabla \cdot \mathbf n$. Is that really true?).
But if I look at $\frac{\partial \mathbf e_p}{\partial p}$ and $\frac{\partial \mathbf e_q}{\partial q}$ for the hill case, it seems like they should point in the -n direction because the change in the tangent vector points into the hill. This would imply that there should be a negative sign because $\nabla \cdot \mathbf n$ is positive, but I need the final result to be in the -n direction. So:
$\frac{\partial \mathbf e_p}{\partial p} + \frac{\partial \mathbf e_q}{\partial q} = -(\nabla \cdot \mathbf n) \mathbf n$
At this point, though, I'm pretty suspicious of this negative sign. Should it really be there?
A final note
This argument depended on aligning the p and q directions with the principle directions. This isn't actually necessary as Euler's theorem can be used to show that the sum of the curvatures associated with any two orthogonal directions is equal to the sum of the curvatures of the principle directions.