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No change in position vector, yet a non-zero magnitude for change in magnitude for the vector?

Physics Asked on December 24, 2020

In the book "Fundamental Laws of Mechanics" by I.E Irodov, on page 15, he writes as a footnote,

note the in general case $ |mathrm{d}textbf{r}| neq mathrm{d}r$ where $r$ is the modulus of the radius vector $textbf{r}$, and $ v neq frac{mathrm{d}r}{mathrm{d}t}$. For example, when $textbf{r}$ changes only in direction, that is the point moves in a circle, then $ r= textit{const}$; $mathrm{d}r=0$ but $ |mathrm{d} textbf{r}| neq 0$.

The reason this doesn’t make sense to me, is suppose we have a position vector modelling circular motion, suppose,

$$vec{ r(t)} = vec{r_o} + ( cos theta vec{i} + sin theta vec{j})$$

Where, $ theta$ is a function of time (with $ frac{d theta}{dt} =omega$ and $r_o$ is a constant vector, then,

$ mathrm{d(vec{r})} = omega ( -sin theta vec{i} + cos theta vec{j}) dt $

now,

$ |mathrm{d(vec{r})}| = omega dt$

Now, neither of these are zero, so I’m looking for a simpler explanation of what Irodov is saying

2 Answers

In your example $mathbf{r}$ does not change in direction only. There is no reason to believe that (for $theta=0$) $vec r_0+hat i$ has the same magnitude as (for $theta=pi/2$) $vec r_0+hat j$.

In the statement “the point moves in a circle”, the center of the circle is assumed to be at the origin.

Correct answer by ZeroTheHero on December 24, 2020

I think the author is not refering to the same situation you are thinking on: if the origin does not coincide with the center of the circle, $|dr|$ is never zero, because $|vec{r}|$ varies between $r_0+R$ and $r_0-R$. The affirmation the author makes is only valid if $r_0=0$. In that case, $vec{r}$ changes its direction, but not its modulus.

Answered by Pablo Lemos on December 24, 2020

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