Consider a vector function $r: \mathbb{R} \to \mathbb{R}^n$ defined by $r(t)$. We use $\hat{r}$ to denote its normalized vector, and $\dot{r}$ to denote $\frac{d}{dt}r(t)$. We know that the derivative of a normalized vector is orthogonal to itself. It would be suggestive to write\begin{equation} \label{eq_ddtrt} \frac{d}{dt} \hat{r}(t) = a(t) N(\hat{r}(t)), \tag{1} \end{equation}where $a(t)$ is a scalar function and $N(\hat{r}(t))$ is a vector orthogonal to $\hat{r}(t)$ and it is a function of $\hat{r}$ explicitly. Consider the 2D case; that is, $n=2$. Then we can find out that\begin{equation} \label{eq_ddtrt2} \frac{d}{dt} \hat{r}(t) = \underbrace{\left(-\frac{1}{\Vert r\Vert} \hat{r}^T E \dot{r} \right)}_{a(t)} \underbrace{E \hat{r}}_{N(\hat{r}(t))}, \tag{2} \end{equation}where $E=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ is a rotational matrix. So $N(t)=E \hat{r}$ is indeed orthogonal to $\hat{r}(t)$ and it is explicitly a function of $\hat{r}$. The advantage is that I could take $N(\hat{r}(t))$ out, and combine different coefficients, say, $a_i(t)$, to simplify some other computations.
The detailed calculation can be seen in the appendix. The problem is that this seems to work for $n=2$, but it is difficult to obtain similar result for $n>2$.
Question: How to obtain a similar equation like \eqref{eq_ddtrt} for the higher-dimensional case where $n>2$? By similarity, I mean it can be written as $a(t)N(\hat{r}(t))$, where $N(\hat{r}(t))$ is explicitly a function of $\hat{r}$. Equation \eqref{eq19c} is not in this form.
Question: Is it possible to write $N(\hat{r}(t))$ as a cross product of $\hat{r}(t)$ and some function?
Appendix: Calculation of \eqref{eq_ddtrt2}\begin{equation} \frac{d}{dt}{\frac{r(t)}{\Vert r(t)\Vert}} = \frac{d}{dt}{r(t)} \cdot \frac{1}{\Vert r(t)\Vert} + r(t) \cdot \frac{d}{dt}{\frac{1}{\Vert r(t)\Vert}} = \left(\frac{I}{\Vert r\Vert} - \frac{r r^T}{\Vert r\Vert^3}\right) \dot{r} \label{eq19c} \tag{3} \end{equation} If we define an operator $\hat{\cdot}: \mathbb{R}^n \rightarrow \mathbb{R}^n$ such that $\hat{r}=\frac{r}{\Vert r\Vert}$, then \eqref{eq19c} can be re-written as follows:\begin{equation} \frac{d}{dt}{\hat{r}(t)} = \frac{1}{\Vert r\Vert}(I - \hat{r} \hat{r}^T) \dot{r} = -\frac{1}{\Vert r\Vert} E \hat{r} \hat{r}^T E \dot{r} = \left( -\frac{1}{\Vert r\Vert} \hat{r}^T E \dot{r} \right) E \hat{r}, \end{equation}where note that$$ I - \hat{r} \hat{r}^T = (E \hat{r}) (E \hat{r})^T = - E \hat{r} \hat{r}^T E. $$
$\endgroup$ 72 Answers
$\begingroup$Here is my interpretation of your question: For $n\ge 3$ you want to have a function $E: S^{n-1}\to {\mathbb R}^n$ (where $S^{n-1}$ is the unit sphere centered at the origin in ${\mathbb R}^n$) such that for every smooth nowhere vanishing function $$ r: {\mathbb R}\to {\mathbb R}^n, $$ for every $t\in {\mathbb R}$, the time derivative $r'(t)$ is a scalar multiple of $E(\hat{r}(t))$.
You also want to have an explicit function $E$, but this does not matter: Such a function simply does not exist (as soon as $n\ge 3$).
The reason is that if $E$ existed then for every unit vector $u\in S^{n-1}$ we had a plane $P(u)\subset {\mathbb R}^n$ (maybe a line in some cases) spanned by the vectors $u$ and $E(u)$. For simplicity, I will work with functions $r: {\mathbb R}\to S^{n-1}$ so there is no need to normalize. For such functions you would require that $r'(t)\in P(r(t))$ for all $t$. But, given any unit vector $u\in S^{n-1}$, it is easy to find a function $r$ as above, such that $r(0)=u$ but $r'(0)\notin P(u)$. For instance, take a function parameterizing a unit circle passing through $u$ and orthogonal to $P(u)$. (This, of course, does not work if $n=2$.) The precise example will depend on what the map $E$ is, but, assuming for concreteness that $u=e_1$ and $E(u)=e_n$ (which one can always achieve by choosing suitable Cartesian coordinates once $E$ is given), then I would take
$$
r(t)= (\cos(t), \sin(t), 0,...,0).
$$
Of course, if you want $E$ to depend on more data, namely be a function on the tangent bundle of ${\mathbb R}^n$, then such $E$ does exist, namely, it is the function given by your equation (3).
$\endgroup$ 4 $\begingroup$Assume that (your vector is)$$ C:\overline{r}=\overline{r}(t)=\{r_1(t),r_2(t),\ldots,r_n(t)\}. $$This is a curve in Euclidean space $E^n\equiv\textbf{R}^n$. We know that $\frac{d}{dt}\overline{r}(t)$ is a vector tangent to curve $C$. If we assumre that $\overline{r}_0(t)=\frac{\overline{r}(t)}{\left\|r(t)\right\|}$, then $\overline{r}_0(t)$ is the normalized of $\overline{r}(t)$ and$$ \left\|\overline{r}_0(t)\right\|^2=1\Leftrightarrow\left\langle \overline{r}_0(t),\overline{r}_0(t)\right\rangle=1 \Rightarrow\frac{d}{dt}\left\langle\overline{r}_0(t),\overline{r}_0(t)\right\rangle=0\Rightarrow $$$$ \left\langle\frac{d}{dt}\overline{r}_0(t),\overline{r}_0(t)\right\rangle+\left\langle\overline{r}_0(t),\frac{d}{dt}\overline{r}_0(t)\right\rangle=0\Leftrightarrow \left\langle\overline{r}_0(t),\frac{d}{dt}\overline{r}_0(t)\right\rangle=0\tag 1 $$But$$ \frac{d}{dt}\overline{r}_0(t)=\frac{d}{dt}\left(\frac{1}{\left\|\overline{r}(t)\right\|}\right)\overline{r}(t)+\frac{1}{\left\|\overline{r}(t)\right\|}\frac{d}{dt}\overline{r}(t)= $$$$ =\frac{d}{dt}\left(\frac{1}{\left\|\overline{r}(t)\right\|}\right)\left\|\overline{r}(t)\right\|\overline{r}_0(t)+\frac{1}{\left\|\overline{r}(t)\right\|}\frac{d}{dt}\overline{r}(t). $$But $\frac{d}{dt}\overline{r}(t)=k\overline{e}_0(t)$, where $\overline{e}_0(t)$ is the tangent unit vector of the curve $C$ and $k:=\left\|\frac{d}{dt}\overline{r}(t)\right\|$. Hence$$ \frac{d}{dt}\overline{r}_0(t) =\frac{d}{dt}\left(\frac{1}{\left\|\overline{r}(t)\right\|}\right)\left\|\overline{r}(t)\right\|\overline{r}_0(t)+\frac{k}{\left\|\overline{r}(t)\right\|}\overline{e}_0(t).\tag 2 $$Hence the vector $\overline{e}_0(t)$ belongs to the space spaned by $\overline{r}_0(t)$ and $\frac{d}{dt}\overline{r}_0(t)$, which (by (1)) are orthogonal to each other. Hence if the angle between $\overline{r}_0(t)$ and $\overline{e}_0(t)$ (chord and tangent) is $\theta$, then$$ \cos(\theta)=\left\langle\overline{e}_0(t),\overline{r}_0(t)\right\rangle=\left\langle \frac{\frac{d}{dt}\overline{r}(t)}{k},\frac{\overline{r}(t)}{\left\|\overline{r}(t)\right\|}\right\rangle=\frac{\frac{d}{dt}\left\langle \overline{r}(t),\overline{r}(t)\right\rangle}{2k\left\|\overline{r}(t)\right\|}. $$Hence$$ \cos(\theta)=\frac{\frac{d}{dt} \left(\left\|\overline{r}(t)\right\|^2\right)}{2k\left\|\overline{r}(t)\right\|}=\frac{\frac{d}{dt} \left(\left\|\overline{r}(t)\right\|\right)}{k}\tag 3 $$and in a better way we can say$$ \overline{e}_0(t)=\frac{1}{k}\frac{d}{dt}\overline{r}(t)=\cos(\theta)\overline{r}_0(t)+\frac{\sin(\theta)}{k_S}\frac{d}{dt}\overline{r}_0(t),\tag{4.1} $$where $k_S=\left\|\frac{d}{dt}\left(\frac{\overline{r}(t)}{\left\|\overline{r}(t)\right\|}\right)\right\|$ and $\theta$ is the angle between chord $\overline{r}(t)$ and tangent $\frac{d}{dt}\overline{r}(t)$ given by (3). Hence$$ \frac{d}{dt}\overline{r}_0(t)=-k_S\cot(\theta)\overline{r}_0(t)+\frac{k_S}{\sin(\theta)}\overline{e}_0(t),\tag{4.2} $$
Given two constant vectors $\overline{a}$, $\overline{b}$, we define the space $P$ of vectors $\overline{x}$, such that$$
P=P(\overline{a},\overline{b})=\{\overline{x}\in E^n:\overline{x}=\kappa\overline{a}+\lambda\overline{b},\kappa,\lambda\in\textbf{R}\}.
$$Every $\overline{x}$ of $P$ can be obtained by a rotating $\overline{a}$ with angle $\theta$ and scale the vector $\overline{a}$by $\frac{\left\|\overline{x}\right\|}{\left\|\overline{a}\right\|}$. That is,$$
x=T(P(\overline{a},\overline{b});\theta).
$$This happens in such a way that during the rotation process the vector $\overline{a}$ never gets out of $P(\overline{a},\overline{b})$ and the final vector becomes of lenght $\left\|\overline{x}\right\|$. Then relation (4.2) can be written as$$
\frac{d}{dt}\overline{r}_0(t)=-k_S\cot(\theta)\overline{r}_0(t)+\frac{k_S}{\sin(\theta)}T\left(P\left(\overline{r}_0(t),\frac{d}{dt}\overline{r}_0(t)\right);\theta(t)\right),\tag{5.1}
$$
or equivalently$$
\frac{d}{dt}\overline{r}_0(t)=-\frac{k_S\cot(\theta)}{\left\|\overline{r}(t)\right\|}\overline{r}(t)+\frac{ k_S}{\sin(\theta)\left\|\overline{r}(t)\right\|}T\left(P\left(\overline{r}(t),\frac{d}{dt}\overline{r}(t)\right);\theta(t)\right),\tag{5.2}
$$where$$
\cos(\theta)=\frac{\frac{d}{dt} \left(\left\|\overline{r}(t)\right\|\right)}{k}, \, k_S=\left\|\frac{d}{dt}\left(\frac{\overline{r}(t)}{\left\|\overline{r}(t)\right\|}\right)\right\|, \, k=\left\|\frac{d}{dt}\overline{r}(t)\right\|.\tag 6
$$By this way $\frac{d}{dt}\overline{r}_0(t)$, is expresed by $\overline{r}(t)$, $\frac{d}{dt}\overline{r}(t)$, which define $P$ and an angle $\theta=\theta(t)$ which is known $\cos(\theta)=\frac{\frac{d}{dt} \left(\left\|\overline{r}(t)\right\|\right)}{k}$.
NOTES
- The vector $\overline{e}_0(t)$ belongs to the space spaned by $\overline{r}_0(t)$ and $\frac{d}{dt}\overline{r}_0(t)$ (this is by (2)). But $\overline{r}_0(t)$ and $\frac{d}{dt}\overline{r}_0(t)$ are orthogonal to each other. Hence we get easily equation (4.1). Solving with respect to $\frac{d}{dt}\overline{r}_0(t)$, we get (4.2).
That is of course if we define for $\overline{a}=\{a_1,a_2,\ldots,a_n\}$ and $\overline{b}=\{b_1,b_2,\ldots,b_n\}$, the inner product of $\overline{a}$ and $\overline{b}$ as$$ \left\langle \overline{a},\overline{b}\right\rangle=a_1b_1+a_2b_2+\ldots+a_nb_n\tag 7 $$and$$ \cos(\theta)=\frac{a_1b_1+a_2b_2+\ldots+a_nb_n }{\sqrt{a_1^2+a_2^2+\ldots+a_n^2}\sqrt{b_1^2+b_2^2+\ldots+b_n^2}}.\tag 8 $$
- The rotation is in the sense of the above definitions (7),(8) of angle $\theta$. The space $P=P(\overline{a},\overline{b})$ is spaned by the two vectors $\overline{a}$, $\overline{b}$, which is a hyperplane of $E^n$. Hence if $\overline{x}\in P$, then $\overline{x}$ must be defined by $\left\|\overline{x}\right\|$ and its angle $\theta$ (of $\overline{x}$ with $\overline{a}$). This is true because (assume with no loss of the generality $\left\langle\overline{a},\overline{b}\right\rangle=0$):$$ \overline{x}=\kappa\overline{a}+\lambda\overline{b}\Rightarrow \left\langle \overline{x},\overline{a}\right\rangle=\kappa\left\langle\overline{a},\overline{a}\right\rangle+\lambda 0\Rightarrow \left\|\overline{a}\right\|\cdot\left\|\overline{x}\right\|\cos(\theta)=\kappa\left\|\overline{a}\right\|^2\Rightarrow $$$$ \kappa=\frac{\left\|\overline{x}\right\|\cos(\theta)}{\left\|\overline{a}\right\|}. $$Also$$ \left\langle \overline{x},\overline{b}\right\rangle=\kappa 0+\left\langle\overline{b},\overline{b}\right\rangle\Rightarrow \left\|\overline{b}\right\|\cdot\left\|\overline{x}\right\|\cos(\psi)=\lambda\left\|\overline{b}\right\|^2\Rightarrow $$$$ \lambda=\frac{\left\|\overline{x}\right\|\cos(\psi)}{\left\|\overline{b}\right\|}. $$But$$ \left\langle\overline{x},\overline{x}\right\rangle=\left\|\overline{x}\right\|^2=\kappa^2\left\langle\overline{a},\overline{a}\right\rangle+\lambda^2\left\langle\overline{a},\overline{b}\right\rangle+2\kappa\lambda\left\langle\overline{a},\overline{b}\right\rangle=\kappa^2 \left\|\overline{a}\right\|^2+\lambda^2\left\|\overline{b}\right\|^2\Rightarrow $$$$ 1=\cos^2(\theta)+\cos^2(\psi)\Leftrightarrow \psi=\pm\left(\frac{\pi}{2}-\theta\right). $$Hence$$ \lambda=\frac{\left\|\overline{x}\right\|\sin(\theta)}{\left\|\overline{b}\right\|} $$and finaly$$ \overline{x}=\frac{\left\|\overline{x}\right\|\cos(\theta)}{\left\|\overline{a}\right\|}\overline{a}+\frac{\left\|\overline{x}\right\|\sin(\theta)}{\left\|\overline{b}\right\|}\overline{b}.\tag 9 $$Hence for given two vectors $\overline{a}$, $\overline{b}$, every $\overline{x}\in P\left(\overline{a},\overline{b}\right)$ is defined only by $\left\|\overline{x}\right\|$ and $\theta$.
Hence for a $t=t_0$,$$ T\left(\underbrace{\overline{r}_0(t_0),\left(\frac{d}{dt}\overline{r}_0(t)\right)_{t=t_0}}_{plane-P};\theta(t_0)\right)=\overline{e}_0(t)= $$$$ =\frac{1}{\left\|\overline{r}(t_0)\right\|}T\left(\underbrace{\overline{r}(t_0),\left(\frac{d}{dt}\overline{r}(t)\right)_{t=t_0}}_{plane-P};\theta(t_0)\right) $$is a rotation of $\overline{r}_0(t_0)$ in angle $\theta(t_0)$ in thedirectional (hyper-plane) of $\overline{r}_0(t)$ and $\left(\frac{d}{dt}\overline{r}_0(t)\right)_{t=t_0}$. This directional hyper-plane is needed and that's make the diference for $n=2$ and $n>2$. For $n=2$ the hyper-plane does not change with $t$ and when $n>2$ it depends on $t$.
- In the case of definition (8), the number $\theta$ denotes always the angle in the sense that is measures how large or how small is the opening between $\overline{a}$ and $\overline{b}$. This (the concept of angle) is an invariant of any space Euclidean or not (Riemannian). However you must have a plane, defined by the two vectors.
