Why does $\vec{r}\cdot\dot{\vec{r}}=r\dot{r}$?

Question

Why is $$\vec{r}\cdot\dot{\vec{r}}=r\dot {r}$$ true? Before saying anything, I have seen the proofs using spherical coordinates for $$\dot{\vec {r}}= \dot{r}\vec{u_r}+r\dot{\theta}\vec{u_\theta}+r\sin\theta \vec{u_\phi}$$ and $\vec{r}=r\vec{u_r}$ but I was thinking, why is that true if $$\vec{r}\cdot\dot{\vec{r}}=\vec{r}\cdot\vec{v}=rv\cos\alpha$$ and since in spherical coordinates or any circular motion, the velocity is perpendicular to the position vector? So then it should be $$\vec{r}\cdot\vec{v}=rv\cos \frac{\pi}{2}=0~?$$ Isn't it? What from my logic is wrong?

Answer 1

What's wrong with your argument is that $\dot r \neq v$. In your example $\dot r$ is indeed zero (because velocity is perpendicular to $r$, therefore, the magnitude of $r$ doesn't change) and so is $\vec r \cdot\dot{\vec r}$.

$\dot r$ is the rate of change of $r$ coordinate while $v$ is the norm if $\dot{\vec r}$, which you can see from your own formula is not the same as $\dot r$.

Answer 2

To sum up the previous answers: it is indeed true that $\vec{r}\cdot\dot{\vec{r}}=r\dot{r}$, in fact: \begin{equation} 2r\dot{r}=\frac{\mathrm{d}}{\mathrm{d}t}(r^2)=\frac{\mathrm{d}}{\mathrm{d}t}(\vec{r}\cdot\vec{r})=2\vec{r}\cdot\dot{\vec{r}}, \end{equation} as it was pointed out by @Quillo. Regarding the fault in your logic, in general it is not true that $\dot{r}$ is equal to $v$, namely the norm of the velocity $\vec{v}=\dot{\vec{r}}$. By the way, in a circular motion, the radial velocity $\dot{r}$ is zero by definition and it is in fact true that $\vec{r}$ and $\vec{v}=\dot{\vec{r}}$ are always perpendicular. As suggested by @Miguel Garcia, you can employ spherical coordinates to describe any kind of motion, whereas circular motion is a special kind of motion with $\dot{r}=0$.

Answer 3

The reasoning using spherical coordinates is correct as it applies to all kind of motions, is just a description in polar coordinates but nothing is assumed about the particular kind of motion. In your second argument you are assuming a particular kind of motion (i.e. circular motion), of course the description of circular motion is particularly easy in polar coordinates but in principle you can use polar coordinates to describe any kind of motion. If in your first argument you add the additional supposition of circular motion then you also get zero as a result so there is no contradiction.

Answer 4

the position vector $~\vec r~$ is

$$\vec r=r\,\vec e_r\quad, \text{with}\quad \vec e_r\cdot \vec e_r=1 $$

thus

$$\vec v=\dot r\,\vec e_r+r\,\vec{\dot{e}}_r$$

and $$\vec r\cdot \vec v=r\,\vec e_r\,(\dot r\,\vec e_r+r\,\vec{\dot{e}}_r)= r\,\dot r\quad,\text{with}\quad \vec e_r\perp \vec{\dot{e}}_r $$

Answer 5

The thing that is missing in the statement is the angle between two vectors. So the complete equation should be:

$\vec{r}\cdot \dot{\vec{r}} =r\dot{r} \cos\alpha$.

$\alpha$ is the angle between the vectors $\vec{r}$ and $\dot{\vec{r}}$. In the Cartesian coordinate, $\alpha = 0$, while in Spherical coordinate $\alpha = 90^0$.