B Lightwatch/Lightclock (Lichtuhr)

[weirdoguy]

velocity is the only vector quantity for which English-speaking physicists and textbooks (American ones at least) use a separate word for the magnitude (speed)

Same in polish. But some authors try to avoid it. It really gets messy in basic schooling in the context of mean velocity and mean spead, where we have 'magnitude of mean velocity' and 'mean magnitude of velocity'. It's hard for children to grasp the difference. I prefer to use word speed, it makes life easier.

 

[vanhees71]

Ah, I just remembered another such pair: [vector] displacement vs. [scalar] distance, whose time-derivatives give velocity and speed.

That seems not to be true since on the one hand speed is
$$|\vec{v}|=|\dot{\vec{x}}|,$$
i.e., indeed the magnitude of the velocity, but
[edit: corrected typo in view of #56]
$$\dot{r} = \mathrm{d}_t |\vec{r}|=\mathrm{d}_t \sqrt{\vec{r}^2}=\vec{r} \cdot \dot{\vec{r}}/r,$$
which in general is not $|\vec{v}|$.

 

[Martin68]

dt√˙→r2=→r⋅˙→r/r

Not necessarily I have to understand it ... ... but, why is the derivation of r squared and drawn root the same as the amount of r? And the next step I do not understand at all
(But, no problem)

 

[DrGreg]

Not necessarily I have to understand it ... ... but, why is the derivation of r squared and drawn root the same as the amount of r? And the next step I do not understand at all
(But, no problem)

$$
r = |\mathbf{r}| = \left( \mathbf{r} \cdot \mathbf{r} \right) ^{\frac{1}{2}}
$$
Therefore
$$
\begin{align*}
\dot{r} = \frac{\mathrm{d}r}{\mathrm{d}t}
&= \tfrac{1}{2} \left( \mathbf{r} \cdot \mathbf{r} \right) ^{-\frac{1}{2}} \frac{\mathrm{d}}{\mathrm{d}t} \left( \mathbf{r} \cdot \mathbf{r} \right) \qquad \: \textrm{(chain rule)}\\
&= \frac{1}{2r} \left( \frac{\mathrm{d}\mathbf{r}} {\mathrm{d}t} \cdot \mathbf{r} + \mathbf{r} \cdot \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} \right) \quad \textrm{(product rule)}\\
&= \frac{\mathbf{\dot{r}} \cdot \mathbf{r}} {r}
\end{align*}
$$

 

[PeroK]

$$
r = |\mathbf{r}| = \left( \mathbf{r} \cdot \mathbf{r} \right) ^{\frac{1}{2}}
$$
Therefore
$$
\begin{align*}
\dot{r} = \frac{\mathrm{d}r}{\mathrm{d}t}
&= \tfrac{1}{2} \left( \mathbf{r} \cdot \mathbf{r} \right) ^{-\frac{1}{2}} \frac{\mathrm{d}}{\mathrm{d}t} \left( \mathbf{r} \cdot \mathbf{r} \right) \qquad \: \textrm{(chain rule)}\\
&= \frac{1}{2r} \left( \frac{\mathrm{d}\mathbf{r}} {\mathrm{d}t} \cdot \mathbf{r} + \mathbf{r} \cdot \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t} \right) \quad \textrm{(product rule)}\\
&= \frac{\mathbf{\dot{r}} \cdot \mathbf{r}} {r}
\end{align*}
$$

Hence, rather logically:
$$ \dot r = \mathbf{v} \cdot \mathbf{\hat r} $$

 

[Ibix]

If you are trying to reconcile this, @Martin68, note that there's a typo in @vanhees71's version - there should not be a dot over the $\vec r$ inside the square root.