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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?

enter image description here

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  • $\begingroup$ You just need to find the projection of the 3-d gradient into the tangent plane at whatever point you are considering. $\endgroup$ – Paul Feb 26 '16 at 19:01
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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 $$\nabla \phi_t -proj_n = \hat{i}+\hat{k}-\hat{k}=t \hat{i}$$

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  • $\begingroup$ Thank you for your help. Can you comment on how $\vec{v}$ can be found? $\endgroup$ – AFP Feb 26 '16 at 19:35
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    $\begingroup$ @Ahmad, It's a typo. I'll change. $\endgroup$ – Paul Feb 26 '16 at 19:36
  • $\begingroup$ How is this constrained to the shape of the surface $S$? $\endgroup$ – AFP Feb 26 '16 at 19:41
  • $\begingroup$ Taking the projection onto the tangent plane takes care of that $\endgroup$ – Paul Feb 26 '16 at 19:54
  • $\begingroup$ @ Paul: can you also please tell me how $\nabla\phi$ can be calculated on a 3D discrete surface (e.g., when it is triangulated)? I have found a post here but it is not very clear: math.stackexchange.com/questions/386770/…. $\endgroup$ – AFP Feb 27 '16 at 0:50

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