Surface gradient (${\nabla }_{s}f$)

For an embedded surface with unit normal vector $\mathbf{n}$, the surface gradient is ${\nabla }_{s}f=(\mathbf{I}-\mathbf{n}{\mathbf{n}}^T)\nabla f$. It projects the ambient gradient onto the tangent space and is independent of how $f$ is extended normally off the surface.

Let ${\nabla }_{\tan }F=\nabla F-{\mathbf{e}}_r{\partial }_{r}F$ be the tangential surface gradient on a sphere of radius $r>0$. For smooth scalar fields, set $\mathbf{f}=R{\mathbf{e}}_r+{\nabla }_{\tan }G+{\mathbf{e}}_r\times {\nabla }_{\tan }F$. Its radial curl detects only $F$:To prove this, use the radial curl formula $(r\sin \theta {)}^{-1}[{\partial }_{\theta }(\sin \theta f_{\varphi })-{\partial }_{\varphi }{f}_{\theta }]$. The radial component $R$ never enters it. The $G$ terms cancel as $G_{\varphi \theta }-G_{\theta \varphi }=0$ by symmetry of second derivatives, while the $F$ terms give the unit-sphere Laplace-Beltrami operator. Taking the dot product with the position vector $\mathbf{r}=r{\mathbf{e}}_r$ gives the displayed $r^{-1}$ version. This tangential gradient decomposition does not assert that the whole vector field is solenoidal.

On the graph of a function $z=f(x,y)$, assume $f$ is differentiable, its planar gradient $p$ is nonzero, and speed $U>0$ is measured by the three-dimensional Euclidean norm. The upward unit normal vector is $n=(-p,1)/\sqrt{1+|p{|}^2}$. Projecting the downward vertical vector $-e_z$ onto the tangent plane gives $-(p,|p{|}^2)/(1+|p{|}^2)$. By the Cauchy-Schwarz inequality, normalizing this projected vector produces the tangent direction with greatest height decrease at the prescribed speed, giving the displayed velocity. Its vertical component is $-U|p|/\sqrt{1+|p{|}^2}$, and its horizontal direction is parallel to $-\nabla f$. Consequently its planar paths agree with negative gradient flow up to a change of time, although the speeds differ. Where $p=0$, every tangent direction has zero first-order height change and this normalized formula is undefined.