Calculating Gradient of 1/|r-r'|: Tips & Results

[TheGreatDeadOne]

Homework Statement

calculate the function gradient with respect to r

Relevant Equations

gradient in spherical coordinates

Doing R=|r-r'|, i get the expected result: \nabla \frac{1}{|r-r'|} = -\frac{1}{R^2}\hat r=-\frac{(r-r')}{|r-r'|^3}

But doing it this way seems extremely wrong, as I seem to be disregarding the module. So I tried to do it by the chain rule, and I got:

\nabla \frac{1}{|r-r'|}=-\frac{1}{2}(r-r')^{-\frac{3}{2}}

but that doing so looks much more correct using Cartesian coordinates. So, does anyone have any tips?

 

[anuttarasammyak]

Let us see physical dimension of the quantity. It should be $L^{-2}$ but your calculation says $L^{-3/2}$.

 

[PeroK]

... not to mention that $\nabla$ is a vector operator. I.e. $\nabla f$ is a vector.

 

[Delta2]

Homework Statement:: calculate the function gradient with respect to r
Relevant Equations:: gradient in spherical coordinates

So I tried to do it by the chain rule, and I got:

\nabla \frac{1}{|r-r'|}=-\frac{1}{2}(r-r')^{-\frac{3}{2}}

but that doing so looks much more correct using Cartesian coordinates. So, does anyone have any tips?

Something must have gone terribly wrong in your calculation cause the end result doesn't even make sense: r and r' are vectors and so is their difference r-r', But how do you raise a vector to the power of (-3/2)?

Please show us your calculation in cartesian coordinates to allow us to see where you go wrong.

 

[TheGreatDeadOne]

Something must have gone terribly wrong in your calculation cause the end result doesn't even make sense: r and r' are vectors and so is their difference r-r', But how do you raise a vector to the power of (-3/2)?

Please show us your calculation in cartesian coordinates to allow us to see where you go wrong.

Spherical coordinates. Using Cartesian coordinates I solve correctly, i am trying for spherical coordinates.

 

[Delta2]

Spherical coordinates. Using Cartesian coordinates I solve correctly, i am trying for spherical coordinates.

Ok, well then please show us your calculation in spherical coordinates how you got $$\nabla \frac{1}{|r-r'|}=-\frac{1}{2}(r-r')^{-\frac{3}{2}}$$ with as much detail as possible.