Rigorous proof needed for: If $\vec{r}(t)\times d{\vec{r}(t)}=0$ then $\vec{r}(t)$ is a constant vector.

Question

If $\vec{r}(t)\times d{\vec{r}(t)}=0$, prove that $\hat{r}$ is a constant vector.

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My Attempt:

Let $\vec{r}(t)=x \hat{i} + y \hat{j} +z \hat{k}$.

Thus, $d\vec{r}(t)=dx \hat{i} + dy \hat{j} +dz \hat{k}$

$\therefore \vec{r}(t)\times d\vec{r}(t) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ dx & dy & dz \end{vmatrix}=0 $

From here, we get the following:

$ydz-zdy=0$

$zdy-xdz=0$

$xdy-ydx=0$

After this, we can write $\frac{dx}{x}=\frac{dy}{y}=\frac{dz}{z}$.

From there we can conclude $x=Ay$ and $z=By$, where $A$ and $B$ are constants.

So, $\hat{r}=\frac{\vec{r}}{|\vec{r}|}=\frac{Ay\hat{i}+y\hat{j}+By\hat{k}}{\sqrt{A^2y^2+y^2+B^2y^2}}$

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I feel my solution is not rigorous enough. I think it is not always correct that

$\frac{dx}{x}=\frac{dy}{y}=\frac{dz}{z}$

If $x$ or $y$ or $z$ were $0$ for some $t$, then that would not hold true. Is there any more rigorous proof for the quoted claim?

Answer 1

Here is a completely different approach. I always encourage working with vector derivatives abstractly, making use of vector identities. So,

$$ d\hat r = d\left(\frac{\vec r}{|r|}\right) = \frac{d\vec r}{|r|} - \frac{\vec r d|r|}{|r|^2} $$

using product rule. But $|r|d|r| = (1/2)d(|r|^2) = (1/2)d(\vec r \cdot \vec r) = \vec r \cdot d\vec r$, so $ d|r| = (\vec r \cdot d\vec r)/|r|$ and the above becomes

$$ d\hat r = \frac{d\vec r}{|r|} - \frac{\vec r (\vec r \cdot d\vec r)}{|r|^3} = \frac{(\vec r \cdot \vec r)d\vec r}{|r|^3} - \frac{\vec r (\vec r \cdot d\vec r)}{|r|^3}$$

which by another vector identity, $\bf a \times (b \times c) = b(a\cdot c) - c(a \cdot b)$, becomes

$$ d\hat r = \frac{-\vec r \times (\vec r \times d \vec r)}{|r|^3} = 0 $$ so $\hat r = const.$

Answer 2

A counterexample is $${\bf r}(t):=\left\{\eqalign{t^2{\bf e}_1\quad(t\leq 0)\cr t^2{\bf e}_2\quad(t\geq0)\cr}\right.\ .$$ Therefore assume $${\bf r}(t)=\rho(t)\,{\bf u}(t),\qquad \rho(t)>0,\quad {\bf u}(t)\in S^2\qquad(t>0)\ .$$ Then $$\dot{\bf r}(t)=\dot\rho(t){\bf u}(t)+\rho(t)\dot{\bf u}(t)$$ and therefore $${\bf 0}={\bf r}(t)\times\dot{\bf r}(t)={\bf 0}+\rho^2(t)\>{\bf u}(t)\times\dot {\bf u}(t)\ .$$ As $\dot{\bf u}\perp{\bf u}$ we conclude that $$0=|{\bf u}(t)\times\dot {\bf u}(t)|=|{\bf u}(t)|\ |\dot {\bf u}(t)|=|\dot {\bf u}(t)|\ ,$$ which proves $\dot {\bf u}(t)={\bf 0}$ for all $t>0$.

Answer 3

I dont' know if you look for this, but

$$\vec{A} \times \vec{B} = 0 \implies \vec{A} \parallel \vec{B} \implies \vec{B} = k\vec{A}$$

Use this and triangle law of vectors gives: $$\vec{r}_f = \vec{r}_i+d\vec{r}$$

$$\vec{r}_f =(1+k)\vec{r}_i$$

So $\hat{r}$ is constant.