# Total derivative involving rigid body motion of a surface

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1. Jul 31, 2015

### user11443

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?

2. Aug 2, 2015

### Orodruin

Staff Emeritus
It is the total derivative of $\vec n$ that is given by $\vec w \times \vec n$, not the partial derivative wrt $t$.

3. Aug 2, 2015

### user11443

Thanks Orodruin. Is there a physical interpretation to the partial derivative of the surface normal with respect to time? Or in other words, it seems from applying the chain rule for $\displaystyle\frac{d{\mathbf{n}}}{dt}$ that

\frac{\partial {\mathbf{n}}}{\partial {\mathbf{x}}} \frac{d{\mathbf{x}}}{dt} + \frac{\partial {\mathbf{n}}}{\partial t} = {\mathbf{w}} \times {\mathbf{n}}.

But I am having a hard time seeing how the two things above turn out to be equal.

4. Aug 2, 2015

### Orodruin

Staff Emeritus
The partial derivative of n wrt time is the change in the normal at a given point in space. Since the normal is defined only on a surface, this derivative may not even make sense (you could extend the definition of n to the full volume by considering normal unit vectors to a family of non-intersecting surfaces, such as the level surfaces of some function). When it does make sense, you will find that this changes depending on the surface as well as the rotation.

5. Aug 2, 2015

### user11443

Thanks, Orodruin.