I Why is the Gradient of Displacement Vector Parallel to \( \vec{r} - \vec{r'} \)?

[TheCanadian]

I have a quantity defined as $r = \left|\vec{r} - \vec{r'} \right|$ and am trying to take the gradient of this quantity. Now the gradient is with respect to the ordinary vector, $\vec{r}$, and not $\vec{r'}$. But after looking at a solution, it says the direction of the gradient is in the direction parallel to $\vec{r} - \vec{r'}$ and not $\vec{r}$. My apologies for being a bit vague here, but shouldn't the direction of the gradient be pointing in the direction in which the derivative is being taken (assuming only a radial component exists)? So if I am taking the derivative with respect to $r$, then it points in a direction parallel to $\vec{r}$, while a derivative wrt to $r - r'$ would point in $\vec{r} - \vec{r'}$?

 

[Orodruin]

No, you are wrong. I suggest actually doing the math rather than trying to reason your way forward.

If you definitely want to reason, you should use a translation argument from the gradient of the distance from the origin.

 

[ytht100]

you should write out the equation in Cartesian coordinate

 

[Guneykan Ozgul]

If you want to see why direction of the gradient does not have to be in the direction in which derivative is taken, here is an intuitive argument:
Gradient of a scalar function means that the direction in which the function changes most. So when you change r infinitesimally the change in the magnitude of the displacement vector cannot be independent of the direction of the r' because it affects the displacement vector.
Also, if you write it mathematically you will encounter a term like rr'cosa where a is the angle between r and r'. So you cannot assume there is no angular dependency.