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user11443
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This stems from considering rigid body transformations, but is a general question about total derivatives. Something is probably missing in my understanding here. I had posted this to math.stackexchange, but did not receive any answers and someone suggested this forum might be more suitable.
A rigid body motion consisting of ##3 \times 3## rotation ##{\mathbf{R}}## and ##3 \times 1## translation ##{\mathbf{p}}## transforms a point ##{\mathbf{x}} \in {\mathbb{R}}^3## to ##{\mathbf{R}} {\mathbf{x}} + {\mathbf{p}}##. For a small motion, the displacement of a point ##{\mathbf{x}}## is ##\Delta {\mathbf{x}} = ({\mathbf{R}} {\mathbf{x}} + {\mathbf{p}}) - {\mathbf{x}}##, which is rewritten as ##\Delta {\mathbf{x}} = {\mathbf{w}} \times {\mathbf{x}} + {\mathbf{p}}##, where ##{\mathbf{w}}## is the angular velocity using ##{\mathbf{R}} \approx {\mathbf{I}} + [{\mathbf{w}}]_\times##, with ##[{\mathbf{w}}]_\times## the infinitesimal skew-symmetric matrix.
For a surface with unit surface normal ##{\mathbf{n}}##, suppose we have a unit direction vector ##{\mathbf{s}} \in {\mathbb{S}}^2## that depends only on time ##t## and is independent of ##{\mathbf{x}}##. Consider the function ##f({\mathbf{x}}, t) = {\mathbf{n}}({\mathbf{x}}, t)^\top {\mathbf{s}}(t)##. Suppose the same rigid body motion is applied to both the surface and the direction vector ##{\mathbf{s}}##, then it transforms the surface normal to ##{\mathbf{R}} {\mathbf{n}}## and ##{\mathbf{s}}## to ##{\mathbf{R}} {\mathbf{s}}##, so their changes are given by ##{\mathbf{w}} \times {\mathbf{n}}## and ##{\mathbf{w}} \times {\mathbf{s}}##, respectively. If a point ##{\mathbf{x}}_1## on the surface at time ##t_1## moves to ##{\mathbf{x}}_2## at time ##t_2##, we must have ##f({\mathbf{x}}_1, t_1) = f({\mathbf{x}}_2, t_2)##, since ##({\mathbf{R}} {\mathbf{n}})^\top {\mathbf{R}} {\mathbf{s}} = {\mathbf{n}}^\top {\mathbf{s}}##.
The above means the total derivative of ##f({\mathbf{x}}, t)## with respect to ##t## is ##0##. However,
\begin{align}
\frac{df}{dt} &= \frac{d}{dt} {\mathbf{n}}({\mathbf{x}}, t)^\top {\mathbf{s}}(t) = {\mathbf{s}}(t)^\top (\frac{\partial {\mathbf{n}}}{\partial {\mathbf{x}}} \frac{d{\mathbf{x}}}{dt} + \frac{\partial {\mathbf{n}}}{\partial t}) + {\mathbf{n}}({\mathbf{x}},t)^\top \frac{d{\mathbf{s}}}{dt} \\
&= {\mathbf{s}}(t)^\top \frac{\partial {\mathbf{n}}}{\partial {\mathbf{x}}} ({\mathbf{w}} \times {\mathbf{x}} + {\mathbf{p}}) + {\mathbf{s}}(t)^\top ({\mathbf{w}} \times {\mathbf{n}}) + {\mathbf{n}}({\mathbf{x}},t)^\top ({\mathbf{w}} \times {\mathbf{s}}) \\
&= {\mathbf{s}}(t)^\top \frac{\partial {\mathbf{n}}}{\partial {\mathbf{x}}} ({\mathbf{w}} \times {\mathbf{x}} + {\mathbf{p}}).
\end{align}
So, we have that ##{\mathbf{s}}(t)^\top \displaystyle\frac{\partial {\mathbf{n}}}{\partial {\mathbf{x}}} ({\mathbf{w}} \times {\mathbf{x}} + {\mathbf{p}})## must be ##0##, for any choice of ##{\mathbf{s}}##, ##{\mathbf{w}}## and ##{\mathbf{p}}##. Something seems to be wrong here, since ##\displaystyle\frac{\partial {\mathbf{n}}}{\partial {\mathbf{x}}}## is a property of the surface, independent of ##{\mathbf{s}}##, ##{\mathbf{w}}## and ##{\mathbf{p}}##. Can someone see what is incorrect in the above?
A rigid body motion consisting of ##3 \times 3## rotation ##{\mathbf{R}}## and ##3 \times 1## translation ##{\mathbf{p}}## transforms a point ##{\mathbf{x}} \in {\mathbb{R}}^3## to ##{\mathbf{R}} {\mathbf{x}} + {\mathbf{p}}##. For a small motion, the displacement of a point ##{\mathbf{x}}## is ##\Delta {\mathbf{x}} = ({\mathbf{R}} {\mathbf{x}} + {\mathbf{p}}) - {\mathbf{x}}##, which is rewritten as ##\Delta {\mathbf{x}} = {\mathbf{w}} \times {\mathbf{x}} + {\mathbf{p}}##, where ##{\mathbf{w}}## is the angular velocity using ##{\mathbf{R}} \approx {\mathbf{I}} + [{\mathbf{w}}]_\times##, with ##[{\mathbf{w}}]_\times## the infinitesimal skew-symmetric matrix.
For a surface with unit surface normal ##{\mathbf{n}}##, suppose we have a unit direction vector ##{\mathbf{s}} \in {\mathbb{S}}^2## that depends only on time ##t## and is independent of ##{\mathbf{x}}##. Consider the function ##f({\mathbf{x}}, t) = {\mathbf{n}}({\mathbf{x}}, t)^\top {\mathbf{s}}(t)##. Suppose the same rigid body motion is applied to both the surface and the direction vector ##{\mathbf{s}}##, then it transforms the surface normal to ##{\mathbf{R}} {\mathbf{n}}## and ##{\mathbf{s}}## to ##{\mathbf{R}} {\mathbf{s}}##, so their changes are given by ##{\mathbf{w}} \times {\mathbf{n}}## and ##{\mathbf{w}} \times {\mathbf{s}}##, respectively. If a point ##{\mathbf{x}}_1## on the surface at time ##t_1## moves to ##{\mathbf{x}}_2## at time ##t_2##, we must have ##f({\mathbf{x}}_1, t_1) = f({\mathbf{x}}_2, t_2)##, since ##({\mathbf{R}} {\mathbf{n}})^\top {\mathbf{R}} {\mathbf{s}} = {\mathbf{n}}^\top {\mathbf{s}}##.
The above means the total derivative of ##f({\mathbf{x}}, t)## with respect to ##t## is ##0##. However,
\begin{align}
\frac{df}{dt} &= \frac{d}{dt} {\mathbf{n}}({\mathbf{x}}, t)^\top {\mathbf{s}}(t) = {\mathbf{s}}(t)^\top (\frac{\partial {\mathbf{n}}}{\partial {\mathbf{x}}} \frac{d{\mathbf{x}}}{dt} + \frac{\partial {\mathbf{n}}}{\partial t}) + {\mathbf{n}}({\mathbf{x}},t)^\top \frac{d{\mathbf{s}}}{dt} \\
&= {\mathbf{s}}(t)^\top \frac{\partial {\mathbf{n}}}{\partial {\mathbf{x}}} ({\mathbf{w}} \times {\mathbf{x}} + {\mathbf{p}}) + {\mathbf{s}}(t)^\top ({\mathbf{w}} \times {\mathbf{n}}) + {\mathbf{n}}({\mathbf{x}},t)^\top ({\mathbf{w}} \times {\mathbf{s}}) \\
&= {\mathbf{s}}(t)^\top \frac{\partial {\mathbf{n}}}{\partial {\mathbf{x}}} ({\mathbf{w}} \times {\mathbf{x}} + {\mathbf{p}}).
\end{align}
So, we have that ##{\mathbf{s}}(t)^\top \displaystyle\frac{\partial {\mathbf{n}}}{\partial {\mathbf{x}}} ({\mathbf{w}} \times {\mathbf{x}} + {\mathbf{p}})## must be ##0##, for any choice of ##{\mathbf{s}}##, ##{\mathbf{w}}## and ##{\mathbf{p}}##. Something seems to be wrong here, since ##\displaystyle\frac{\partial {\mathbf{n}}}{\partial {\mathbf{x}}}## is a property of the surface, independent of ##{\mathbf{s}}##, ##{\mathbf{w}}## and ##{\mathbf{p}}##. Can someone see what is incorrect in the above?