Is Symmetry Present in the Derivatives of Normal Vectors on Surfaces?

[rych]

Are there any facts about the derivative of the normalised normal vector n to a surface embedded in n-dimensional Euclid space? Is it true, for instance, that
\frac{\partial n_j}{\partial x^i} = \frac{\partial n_i}{\partial x^j}
The context is as follows. The surface is defined implicitly by a constraint function; there's a Hamiltonian in reduntant coordinates and the canonical Hamiltonian equations of motion for (q,p) ensuring that trajectories lie in the constraint surface. I need to find acceleration \ddot{q}; there the time derivative of n appears. By the way, how could I reformulate this task in the language of differential geometry?

 

[Perturbation]

If the surface is given by \phi (\vec{r})=const. then the normal to the surface is n_i=(\vec{\nabla}\phi)_i, it's not necessarilly unit normal, but this is easily reconciled. So, in Cartesian coordinates

\frac{\partial n_j}{\partial x^i} = \frac{\partial n_i}{\partial x^j}

is correct.

 

[rych]

\frac{\partial n_j}{\partial x^i} = \partial_i \frac{\partial_j \phi}{(\nabla \phi \cdot \nabla \phi)^\frac{1}{2}} = \frac{\partial_{ij}\phi}{(\nabla \phi \cdot \nabla \phi)^\frac{1}{2}} - \frac{\partial_j \phi \sum_k \partial_k \phi \partial_{ik}\phi}{(\nabla \phi \cdot \nabla \phi)^{\frac{3}{2}}}

Thanks for the response! But I can't easily reconcile the second term: it doesn't seem to be symmetric.

 

[Perturbation]

\frac{\partial n_j}{\partial x^i} = \partial_i \frac{\partial_j \phi}{(\nabla \phi \cdot \nabla \phi)^\frac{1}{2}} = \frac{\partial_{ij}\phi}{(\nabla \phi \cdot \nabla \phi)^\frac{1}{2}} - \frac{\partial_j \phi \sum_k \partial_k \phi \partial_{ik}\phi}{(\nabla \phi \cdot \nabla \phi)^{\frac{3}{2}}}

Thanks for the response! But I can't easily reconcile the second term: it doesn't seem to be symmetric.

It's not, ignore me when I put the bit about "reconcilitation", I didn't mean to put that.