Vector Calculus: Gradient of separation distance

[WWCY]

Homework Statement

Could someone explain how the property,
$$\nabla (\frac{1}{R}) = -\frac{\hat{R}}{R^2}$$
where ##R## is the separation distance ##|\vec{r} - \vec{r'}|##, comes about?

What does the expression ##\nabla (\frac{1}{R}) ## even mean?

Homework Equations

The Attempt at a Solution

I know this attempt misses the mark completely, but I'd like to know what I'm getting wrong:

The separation distance ##R## is a function of ##x,y,z## since ##r## is too. Thus
$$ \nabla ( \frac{1}{R} )= \frac{-1}{R^2} \frac{dR}{dx} \hat{x} + \frac{-1}{R^2} \frac{dR}{dy} \hat{y} + \frac{-1}{R^2} \frac{dR}{dz} \hat{z} $$
which doesn't resemble what I wrote at the beginning.

Thanks in advance!

 

[PeroK]

##\frac{1}{R}## is a scalar function of position. Try expressing this in terms of Cartesian coordinates ##(x, y, z)##.

##\nabla## is an operator:

##\nabla f(x, y, z) = (f_x, f_y, f_z)##

where I've used ##f_x = \frac{\partial f}{\partial x}## etc. But, you seem to be using this correctly in any case.

Now, just keep going.

 

[WWCY]

Thank you for your response.

##\frac{1}{R}## is a scalar function of position. Try expressing this in terms of Cartesian coordinates ##(x, y, z)##.

So I let ##\frac{1}{R} = f(x,y,z)##. So,

## \nabla f(x, y, z) = (f_x, f_y, f_z) = \frac{-1}{R^2}(dR/dx, dR/dy, dR/dz) = \frac{-1}{R^2} \nabla R ##

Is this what you meant?

 

[PeroK]

Thank you for your response.
So I let ##\frac{1}{R} = f(x,y,z)##. So,

## \nabla f(x, y, z) = (f_x, f_y, f_z) = \frac{-1}{R^2}(dR/dx, dR/dy, dR/dz) = \frac{-1}{R^2} \nabla R ##

Is this what you meant?

Why don't you express ##R## as a function of ##(x, y, x)##? Then you can differentiate it. And all will be revealed!

 

[Ray Vickson]

Thank you for your response.
So I let ##\frac{1}{R} = f(x,y,z)##. So,

## \nabla f(x, y, z) = (f_x, f_y, f_z) = \frac{-1}{R^2}(dR/dx, dR/dy, dR/dz) = \frac{-1}{R^2} \nabla R ##

Is this what you meant?

Yes, that is exactly what the notation ##\nabla (1/R)## means. Now just finish the job by computing ##\partial R/\partial x,## etc.

 

[PeroK]

## \nabla f(x, y, z) = (f_x, f_y, f_z) = \frac{-1}{R^2}(dR/dx, dR/dy, dR/dz) = \frac{-1}{R^2} \nabla R ##

Note that, as in all your posts, these should be partial derivatives.

 

[WWCY]

Thank you both for your input, here's what I came up with.
$$R = \sqrt{ (x - x')^2 + (y - y')^2 + (z - z')^2 }$$
$$\partial _x R = \frac{x - x'}{\sqrt{ (x - x')^2 + (y - y')^2 + (z - z')^2 }} = \frac{R_x}{R}$$
Doing this for all components of the gradient vector, I get
$$\frac{1}{R} (R_x , R_y, R_z)$$
which is the unit vector pointing in the direction of separation, ##\hat{R}##

Is this right?

 

[PeroK]

Thank you both for your input, here's what I came up with.
$$R = \sqrt{ (x - x')^2 + (y - y')^2 + (z - z')^2 }$$
$$\partial _x R = \frac{x - x'}{\sqrt{ (x - x')^2 + (y - y')^2 + (z - z')^2 }} = \frac{R_x}{R}$$
Doing this for all components of the gradient vector, I get
$$\frac{1}{R} (R_x , R_y, R_z)$$
which is the unit vector pointing in the direction of separation, ##\hat{R}##

Is this right?

Yes. In other words ##\nabla R = \hat{R}##.

 

[WWCY]

Yes. In other words ##\nabla R = \hat{R}##.

Thank you and @Ray Vickson for the assistance!