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One question is bugging me for a long time but I couldn't make out anything nor could my friends. Here it goes:

We know, $\vec r$ is regarded as the position vector. So we can say, $$\vec r \cdot\vec r = r^2$$

Differentiating both sides with respect to time $t$, we get $$\vec r \cdot\frac{d \vec r}{dt} + \frac{d \vec r}{dt}\cdot\vec r = 2r \frac{dr}{dt}$$ or,$$2 \vec r \cdot\frac{d \vec r}{dt} = 2r \frac{dr}{dt}$$ or,$$\vec r \cdot\frac{d \vec r}{dt} = r \,\frac{dr}{dt}$$ or,$$\vec r \cdot\ d \vec r = r \,\ dr$$ or,$$r \,\ dr \cos \theta = r \,\ dr$$ where $\theta$ is angle between $\vec r$ and $d \vec r$

or,$$\cos \theta = 1$$ or,$$\theta = 0^\circ $$

Question no.1:So can I conclude that $\vec r$ and $d \vec r$ have the same direction? The above calculation suggests so but the diagram below does not. Why? enter image description here

Also if $\vec r$ and $d \vec r$ have the same direction, then $$\vec r \times \frac{d \vec r}{dt} = 0$$ Now,$$\vec r = r \hat r $$ where$$\hat r = \hat i \cos \theta + \hat j \sin \theta$$ So,$$\frac{d \vec r}{dt} = \frac{dr}{dt}\hat r + r \hat \theta \frac{d \theta}{dt}$$ and$$\vec r \times \frac{d \vec r}{dt} = r^2 \frac{d \theta}{dt} (\hat r \times \hat \theta) \not = 0 \,\ \text{(in general)}$$

Why does this contradiction arise?

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    $\begingroup$ Question for the student: must $\vec{r}$ and $\vec{v}$ have the same direction? $\endgroup$ – dmckee Oct 7 '15 at 16:47
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    $\begingroup$ You are almost right, except for supposing that $\vec{r} \cdot d\vec{r} = r \, dr \cos \theta$. The true equality is $\vec{r} \cdot d\vec{r} = r \, |d\vec{r}| \cos \theta$, where in general $|d\vec{r}| \neq dr$ (in fact $|d\vec{r}| = \frac{dr}{\cos \theta}$, as you derive) ;) $\endgroup$ – Newbie Oct 7 '15 at 18:37
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Despite what some of the other answers are mentioning, the following equation you have is correct $$ \vec{r} \cdot d \vec{r} = r dr $$ You can check this by noting $$ \vec{r} = x {\hat i} + y {\hat j} + z {\hat k} \implies d \vec{r} = d x {\hat i} + d y {\hat j} + d z {\hat k} $$ Then $$ \vec{r} \cdot d \vec{r} = x dx + y dy + z dz $$ Further note $$ r = \sqrt{ x^2 + y^2 + z^2 } \implies dr = \frac{1}{r} \left( x dx + y dy + z dz \right) $$ implying that it is true that $\vec{r} \cdot d \vec{r} = r dr$.

Where you go wrong is the next step. You say $$ \vec{r} \cdot d \vec{r} = r dr \cos\theta $$ In doing this you are assuming $$ | d \vec{r} | = d r $$ Is this really true? Let us check. We have already computed $d r$. We have $$ | d \vec{r} | = | d x {\hat i} + d y {\hat j} + d z {\hat k} | = \sqrt{ dx^2 + dy^2 + dz^2 } $$ This is clearly not equal to $dr$, thereby making any conclusions based on this incorrect.

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    $\begingroup$ True. This failure is typical and mostly comes from using the same letter r for position vector and for radial coordinate. $\endgroup$ – Nicolas Oct 7 '15 at 19:17
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Instead of $r\, dr \cos \theta = r\, dr$, that line should read $r \, ||{d\vec{r}}|| \cos \theta = r\, dr$. Since $||d\vec{r}|| \neq dr$, the argument does not follow.

If you are not sure why $||d\vec{r}|| \neq dr$, ask yourself whether $||\frac{d\vec{r}}{dt}|| = \frac{dr}{dt}$.

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The mistake is in the step where you go from $$\vec{r}.\frac{d\vec{r}}{dt} = r\frac{dr}{dt}$$to$$\vec{r}.d\vec{r} = r.dr$$ The integrand is time dependant (and involves a dot product as well) and hence the result is non-trivial. You simply cannot "cancel" off the times as you have done.

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  • $\begingroup$ Leave the final step then.... just tell me if $\vec r\cdot\vec v = 0$ is true or not? For I believe this is not true. $\endgroup$ – SchrodingersCat Oct 7 '15 at 17:12
  • $\begingroup$ Mathematicians say we can't do stuff like this all the time. We do it anyway, and it works. Unless you're breaking new theoretical ground, I say go ahead and cancel your differentials, commute your partials, and let your boundary conditions go to zero even without writing a limit sign!! $\endgroup$ – user1717828 Oct 7 '15 at 19:19
  • $\begingroup$ That depends. Position and velocity are independent in general (you might have any velocity in a given position) so they may be perpendicular (as in circular motion) or not. $\endgroup$ – Nicolas Oct 7 '15 at 19:19
  • $\begingroup$ $$\vec{r}.\vec{v} = 0$$ may or may not be valid depending upon the situation and the co-ordinate system you are using. For one co-ordinate system, you might have $$\vec{r}.\vec{v} = 0$$ and for another one you might not when you are analysing the same situation. $\endgroup$ – Aritra Oct 8 '15 at 0:29
  • $\begingroup$ This answer is wrong. See the accepted answer as for why. $\endgroup$ – thermomagnetic condensed boson May 14 '18 at 19:05
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$$\vec r \cdot\frac{d \vec r}{dt} = r \,\frac{dr}{dt}$$ Is correct. But $\vec r \cdot\ d \vec r = r \,\ dr$ could be wrong or just misleading, because I can't tell if you are meaning the correct things by the symbols. $\frac{d \vec r}{dt}=\vec v$ so the first equation says

$$\vec r \cdot \vec v= r \,\frac{dr}{dt}.$$

And I know people like to talk about a $dr$ but really you have to fully specify your parameter space and have a clear change in parameters.

You could try writing $r \,\ |d\vec r| \cos \theta = r \,\ dr \;\text{where}\; \theta \;\text{is angle between} \;\vec r\; \&\; \vec v$ If that helps you avoid getting confused.

When you write math, it should always mean something to you and the things you do should mean something. You can't just manipulate expressions based on what they look like.

can I conclude that $\vec r$ and $d \vec r$ have the same direction?

No. One points in the direction from the origin to the location of the particle, it depends for instance on your origin. The other points in the direction of the velocity of the particle, and does not for instance depend on your choice of origin. They are not the same direction.

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