I have a surface, $S$, defined in three-dimensional space. For the sake of this question, let's assume it is a sphere with unit radius although the surface in my problem can be any shape. I have a distribution of a scalar function, $\phi$, plotted over the surface $S$. $\phi$ is a function of time and space, i.e., $\phi=\left(x, y, z, t\right)$. From time $t_1$ to $t_2$, the distribution of $\phi$ changes.
So, given $S$ in ${\rm I\!R}^3$ and $\phi=\left(x, y, z, t\right)$, I am trying to find the vector field $\vec{v}$ over the surface $S$ that represents the direction of the change of $\phi$ from $t_1$ to $t_2$. Can the community give me some ideas how this can be done?
As a simplified example, consider the following distributions of $\phi$ on the surface at time $t_1$ and $t_2$. The solid black line in the figure is for visualization purposes only. By comparing $\phi$ from $t_1$ to $t_2$, it is clear that $\phi$ is moving upward and my question would be how the velocity vector field, which is always tangential to the surface, on the sphere can be found?
The direction of change is going to be given by $$\frac{d}{dt}\nabla \phi$$
Once you have this, you just need to find the projection of the 3-d gradient into the tangent plane at whatever point you are considering.
For example, let $\phi=xt+z$. Then $\nabla \phi = t \hat{i}+\hat{k}$. At the north pole, the unit normal for the surface is $\hat{k}$. and the projection of the gradient on the normal vector is $\hat{k}$, so the projection onto the tangent plane is $$\nabla \phi_t -proj_n = \hat{i}+\hat{k}-\hat{k}=t \hat{i}$$