Confusion regarding spherical coordinates

Question

I have a few dots in my brain that I need to connect, one of them is:

1. Is $\dot{\vec{r}}=\vec{v}$ true always? But $\dot{r}=v$ isn't always true? If so, in which cases both are true or not?

2. Regarding one of my old questions something still not clicked and I didn't know what it was until today: Why does $\vec{r}\cdot\dot{\vec{r}}=r\dot{r}$? How are we using in my reasoning there that $\dot{r}=v$ at any moment? I understand that $\dot{r}\neq v$ in general but I try to see my reasoning and I don't use so in any moment, what is going on?

3. What is the difference between $v_\theta$ and $\dot{\theta}$? I know both are different mathematically because by spherical coordinates I know $v_\theta=r\dot{\theta}$ but what do both mean? I read both as how theta varies with time?

Answer 1

By definition, $$\tag1\vec v=\frac{\mathrm d\vec r}{\mathrm dt}=\dot{\vec r}$$ But also by definition of the magnitude and direction unit vector of position, $$\tag2\vec r=r\hat{\vec r}$$ so that $$\tag3\vec v=\dot{\vec r}=\dot r\hat{\vec r}+r\dot{\hat{\vec r}}=\dot r\hat{\vec r}+r\dot\vartheta\hat{\vec\vartheta}$$ you should now be able to see that the velocity vector $\vec v$ has not just the radial component that is proportional in magnitude to $\dot r$, but rather also has a ``tangential" component that is proportional to $r\dot\vartheta$.

Why does $\vec r\cdot\dot{\vec r}=r\dot r$? How are we using in my reasoning there that $\dot r=v$ at any moment? I understand that $\dot r\neq v$ in general

While the velocity has two components, by definition of the position vector in spherical coördinates, in Equation (2), there is only one component. The scalar dot product between the velocity and the position vectors will thus ignore the perpendicular component and pick out only the parallel component, leaving us with the equality that you wanted.

What is the difference between $v_\vartheta$ and $\dot\vartheta$? I know both are different mathematically because by spherical coördinates I know $v_\vartheta=r\dot\vartheta$

But, if you already know the formula linking the two things, what is more to it to be asked? What is the actual confusion that you want sorted out?

Answer 2

Okay, this is a post for myself, I have usually got confused because my brain kept thinking that $|\dot{\vec{r}}|=\dot{r}$ while it ISN'T true (at least for this choice of variables), because by definition $\vec r= r\vec{u_r}$ and $\vec{v}=\dot{\vec r}=\dot{r}\vec{u_r}+r\dot{\theta}\vec{u_{\theta}}$ so the modulus of the velocity is $$|\dot{\vec r}|=v=\sqrt{(\dot r)^2+(r\dot{\theta})^2}\neq \dot r$$ and the only moment where $v=\dot r$ is true or written in other words, $\dot r=|\dot{\vec r}|$ is when $\dot\theta=0$, that means, when it is a linear (straight path) motion.

But then as well, how does that relate with the question? Because it is a general truth that $$\dot{\vec r}\cdot \vec r=\vec v \cdot \vec r=\dot r r$$ and for a specific case where it is a circular motion (only circunference, not for elipses or any sort), where in that case, we know it applies that $\vec v\perp \vec r$ and hence $$\vec v\cdot \vec r=rv\cos 90=0$$ by applying that perpendicular notion of the velocity and position in a circular motion. And as well by knowing that in a circular motion, the $\dot r=0$ since the radius is constant throughout the circunference, $\dot r=0$ in a circular motion, so you also get $$\dot{\vec r}\vec r=r\dot r=0$$ which both then make sense. ($\dot{\vec{r}}=\vec{v}$ can equal $\dot{r}$ if you take only the radius component, for examole if you wanted to use the centripetal force on the radial axis)