Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

If $\vec r$ and is the position vector of a point in motion and $r$ is its length/modulus/magnitude/size, then:

Can it be true that:

$$\|\mbox{d} \vec r\| \neq \mbox{d}r? $$ I think that this is true in circular motion, but if this question is generic, i.e., for almost all cases, then do you think this is correct to say?

EDIT: Thanks for the very prompt answers and comments, but please tell me, if this question is given in an exam, which option would you mark? (As a student, I am very interested in understanding physics, but at the end of the day, I also have to think of the exam point of view...)

share|improve this question
    
In circular motion $|d \vec r| \not = 0$ while $dr = 0$. In fact, $|d \vec r| = dr$ is true only for radial motion. –  Johannes Jul 6 '13 at 12:12
    
@Johannes what do you mean by radial? –  Saurabh Raje Jul 6 '13 at 15:34
    
any motion along a straight line through the origin. –  Johannes Jul 6 '13 at 15:56
    
@Johannes How important is through the origin? Is that a compulsary condition? –  Saurabh Raje Jul 6 '13 at 16:01

2 Answers 2

For any generic motion, $|d\vec{r}|$ represents the magnitude of the infinitesimal change in the vector $\vec r$. That means that both change in magnitude and in direction will be reflected in $|d\vec r|$.

Whereas $dr$ represents the infinitesimal change in magnitude(look how those two words reverse) of $\vec r$. This means that it will only reflect change in magnitude, and not the change in direction of the vector.

From this we can say that $|d\vec r| = dr$ if and only if the direction of $\vec r$ is not changing, i.e. the particle is executing linear motion. I think in all other cases the equality won't hold true.

share|improve this answer

Generally,

$$\|\mbox{d} \vec r\| \not = \mbox{d}r$$

A simple counter example:

$$\vec r =\begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix}$$

$$r=1$$

$$\frac{\mbox{d}\vec r}{\mbox{d}\theta} =\begin{bmatrix}-\sin\theta\\\cos\theta\end{bmatrix}$$

$$\frac{\mbox{d}r}{\mbox{d}\theta} =0 $$

but please tell me, if this question is given in an exam, which option would you mark?

Two things.

  1. Nobody is here to help you in exams. The site is for helping people learn physics, and discuss physics. Not for exams.

  2. If you're fed up with the false things you are taught in school, then, I would say that write the correct thing, without thinking whether you'll be marked wrongly, and if you are marked wrongly, then, go and argue!.

but at the end of the day, I also have to think of the exam point of view...)

Not if you are honestly interested in Physics.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.